text:
What does non-integer bond order mean?
I am studying molecular bond theory, and I see that sometimes this happens.
Does this have some significance to the molecule?

It is one of the man-made tools to conceptualize the nature of a bond between two atoms, to know if there is a single / double / triple / quadruple *) bond between them, and to account for both electrons involved in bonding / antibonding molecular orbitals.  Because the nature of electrons is neither "true and only" particle, nor wave alone, hence sometimes escaping our day-to-day experience with macroscopic objects.
In addition, rarely you look at the bond order of two atoms in a molecule without keeping in mind the situation with other atoms present in the same molecule, too.
*)  Sorry, I forgot at least the three-centre-two electron bonds in diborane...

Bond order is non-integral when the difference in the number of electrons in molecular orbital and the number of electrons in anti-molecular orbital is odd.
I don't see anything "significant" in the difference being odd.

Bonding is more complex than a simple 2-electron picture suggests
In the teaching of chemistry and in the early development of the theory of chemical bonding, bonds are (or were) often thought of as consisting of two shared electrons. This is where the idea of bond order comes from: a bond consisting of two electrons is a single bond (as in a hydrogen molecule). Some elements have enough electrons arranged in the right way to give higher bond orders because more electrons are shared (in nitrogen molecules six electrons are shared between the two nitrogens giving a bond order of 3).
This picture, given how simple it is, does a surprisingly good job of explaining the structure of many chemicals and a great deal of organic chemistry. But it is an overly simplistic picture. Sure, ethene is easy to explain: there is bond between the carbons that has a bond order of two. That bond is shorter than the C-C single bond in ethane but longer than the C-C triple bond in ethyne (acetylene for traditionalists). But this simple picture falls apart for other molecules. In benzene, the archetypal example, we could write the structure as cyclohexatriene with alternating double and single bonds around the six membered ring. That certainly uses up all the electron pairs. But this neither explains the reactivity nor the measured structure where all the bonds are the same.
In reality six electrons in benzene are shared equally among all the bonds ("delocalised" is one way to describe this). Each bond is shorter than a single bond but not as short as a double bond. There is no point in trying to explain this using the simple picture where all bonds share two electrons and have an integral bond order. So we say those bonds have a bond order of 1.5 which is compatible with the actual physical structure.
A full explanation of how those bonds works requires some molecular orbital theory and more complex structures where bond orders can be other fractions are not uncommon. But without getting in the details of more sophisticated theories of bonding, the basic explanation of non-integer bond orders is that electrons are sometimes not very monogamous and share themselves between several atoms in which case we get bond orders that are not integers depending on the number shared and the numbers of atoms involved in the sharing.

Bond order not being a whole number usually happens when the molecule has different resonant structures....basically what one should understand first and foremost is that a chemical bond is not the stick like thingy we draw...the stick/line is just a representation of the bond...hence a non whole number bond order means a non whole number bond order...which you obtain from 1/2(Nb-Na)..where Nb and Na mean the no of electrons in the bonding and antibonding orbitals respectively
Im not really too sure it means anything more than that


text:
Aromatic nomenclature, when to use numbers or prefixes?
When you're naming aromatic nomenclature compounds, when do use numbers, such as 2,4-diiodonitrobenzene and when do you use prefixes such as p-xylenene? Thanks! I hope that makes some sense

1,4-dimethylbenzene and p-xylene are both equally correct nomenclatures. You could also say 4-methyl toluene if you wanted. I personally prefer the first example because it allows me to visualize the molecule immediately.
As for 2,4-diiodonitrobenzene, this is indicating three substituents. You have the nitro group, and two iodo groups - one para to the nitro, and one ortho to the nitro.

Recurring to ortho, meta, para constrains the description to two subsitutents, and more often seen for simple "tradtional"compounds; like your example of p-xylene, or something like o-cresol.
Using numbers allows more flexibility, not only in regard of the number of subsitutents and their relative position to each other, but in reference to a hetero-aromatic ring system, too.
Addendum:
Related / almost duplicate to this.  Somehow I thought "this is not new" ...

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text:
Identifying functional groups with labile hydrogens from IR hydrogen bonding region
Here is the $\pu{2000-4000 cm^{-1}}$ spectral region (known for showing hydrogen bonding interactions, $\ce{-H\bond{<-}X-}$) for three different compounds, each showing a different functional group:

where

A is a carboxylic acid,
B is an amine, and
C is an alcohol.

Is it possible to uniquely identify these three compounds just from peak positions, shapes, and intensities within this window?

Not quite sure anyone is really going to understand what your exact problem is? Do you have any specific reason to doubt your choices? If so, what was your line of thought in solving these? Give us specifics - we are not here to do your homework.

@Bob I just looked at pictures of graphs of similar groups and assigned them based on shape. We weren't told if we should do anything more analytic than that.

I would suggest reading your textbook to figure out why specific patterns appear in a spectrum.

@Bob I remember Pretsch et al., Structure Determination of Organic Compounds (http://link.springer.com/book/10.1007%2F978-3-540-93810-1) includes even "hypothetical IR spectra".  To clarify this:  how the spectrum should look like if only this functional group (amine, aldehyde, ester, ...) were present; presenting the whole range from about 4000 to down 600 (or so) 1/cm; followed by examples. Still today I find this presentation more instructive and helpful than "the simplier tables" alone.

@Buttonwood Then they're based off the max's.

@Buttonwood  A is a carboxylic acid because of a max at 3000. B is an amine because max is below 3000. C s an alcohol because max is at 2900.

Perhaps you do not know that the title of your question already contains an important key to interpret IR-spectra: not only to inspect rather small, isolated regions of the spectrum, but to see the spectrum as a whole.  By doing so, you have to watch out for additional absorption bands that were equally providing evidence for one, or the other structure; by scrutinizing both the position ($\tilde\nu$), shape, and relative intensity of the absorption bands in question.
As an example: is there evidence for a carbonyl group, indicative for the acid?
Why?  Because -- depending on the mode of data collection (e.g., transmission in a matrix of a KBr pellet, or neat film by ATR) the shape of the spectra recorded may differ.  In contrast to 1H NMR, IR offers you a view on structural elements .and. the whole molecule.  Then, indeed, IR spectroscopy may offer much insight.

P.S.:  From the baseline of the spectra shown it is likely the data were recorded on a ATR-FT-IR spectrometer.  If you look into the data recorded, you will find each point is typically about 0.43 wavenumbers a part from the next one.  Hence it may be more sound to perform the "peak pic" / stating the barycentre of the absorption bands only by whole wavenumbers, without two digits behind the decimal point (significant figures).

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text:
Is it always the case that when a reaction's products are higher in energy than its reactant, then the reaction is not favored?
If this is the case, can you give a brief explanation why is this always the case?

If you push a boulder up a hill, does the potential energy increase or decrease?

That actually depends on what you mean by "energy." If you are talking about free energy, then that would always be the case.

The reaction can still occur to some extent but the equilibrium condition favours the reactants, whereas if the reaction is exothermic the equilibrium lies on the product side.

This question brings up the broader point of what it means for a reaction to be favored. Your question describes thermodynamic favorability and that is just dependent on the relative energy of the products and the reactants. An analogy to this is to think of energy levels as literal heights subjected to gravity. If you have a ball on the ground (reactants) and you want to put it on a shelf (products), you have to put in energy. But, once it is in the product state, gravity will want to drive the ball back to the reactant state because it is lower in energy.
Another type of favorability is kinetic favorability; this relates to the rate at which the forward and reverse reactions can occur. There are cases where a thermodynamically favorable reaction doesn't occur (or rather occurs at an incredibly slow rate) due it being kinetically unfavorable. An example of this is diamond and graphite; it would be thermodynamically favorable for diamond to convert to graphite, however the reaction to do so is so kinetically unfavorable that you are unlikely to see it occur.


text:
How to understand reactivity rates
I have tried to rationalize the questions with resonance in their intermediate forms, but have not been able to see a difference. Does it have to do with steric interactions based on ortho, meta, and para positions? How do I approach this question?

Do you know about Hammett plots and linear free energy relationships (LFER)s?

RBr + MeOH → ROMe + H2O ?? I. don't. think. so. There is one reaction which the proximity of the ortho MeO on the ring might cause steric hindrance. The two other effects you should be considering are: i) the electron withdrawing/donating nature of the substituents on the ring and ii) their location (ortho, meta, or para) because their location has an electronic (not steric) effect.
Since these considerations are (obviously) the reason you are being asked this question, I leave it to you to digest what those effects will be. You should understand that their effects are not only on the reactant but on the transition states as well.

I totally did not think about activating groups, thank you!

Think about which direction the electrons will flow based on each group.
In A, the ester puts a positive charge on the carbonyl carbon, thereby drawing electrons away from the benzene ring, and increasing its electron withdrawing effect on the reaction site. With electrons pulled away from the reaction site, bromine will have a stronger bond to the reactive-carbon.
In B, the ethers are donating electrons. Thereby decreasing the strength of the bromine bond.
In C, the benzene ring has electron withdrawing effects, similar to in A.
In D, the ether is donating electrons, like in B (but to a lesser extent).
I trust you can figure it out from here!


text:
Is this a homogenous reaction?
$$\ce{HCl(aq) + NaOH(aq) ->  NaCl(aq) + H2O(l)}$$
Would this reaction be considered homogenous even though there is a liquid in the products?

Hello and welcome to Chemistry.SE! You can get a good overview of how this side works by taking the short [tour]. For any questions you can visit the [help]. Also, as this appears to be a homework-type of question, I suggest reading through this post regarding how to ask homework questions.  Best of luck!

What other phase is present besides liquid in the reaction? Is there one?

Yes, it would be considered homogeneous, since all of the components are in water. We don't write water as being aqueous because it would be somewhat redundant to say the water formed in water.


text:
Difference between change in energy level and hyperfine energy level?

The second is the duration of 9 192 631 770 periods of the radiation
  corresponding to the transition between the two hyperfine levels of the
  ground state of the cesium 133 atom.
--> This definition refers to a cesium atom at rest at a temperature of 0 K.

-From what I learnt, changing in energy level of atoms means change from the ground state to the "excited" state. Does "transition between the two hyperfine levels" as quoted above has the same meaning?

There is a nice description of the Cs transitions at http://hyperphysics.phy-astr.gsu.edu/hbase/acloc.html#c4. The definition of 'excited state' is somewhat subjective, since the atom is placed in an upper hyperfine energy by a microwave photon   it is certainly 'excited' as it can loose energy again. More common is to used 'excited state' to mean that the transition occurs between levels with different principle quantum numbers, 1s-2p for example. Others may have different views on this, however.


text:
How Can an Amino Acid Be Non-Polar?
Looking at the structure of an amino acid (as a zwitterion), it seems that the part of the molecule with the amino group and the carboxylic acid group are quite polar (the amino group has a positive hydrogen and the carboxylic acid group has a negative oxygen). However, my textbook says that the character of the R group completely determines the polarity of the molecule (i.e. if R is non-polar, the molecule is non-polar). Why is this?

Compared to other amino acids, that is.

What @IvanNeretin said, but taking it a little further, when amino acids are called "polar" or "nonpolar," I think it usually means the amino acid residues within a protein, i.e. whether the side chain makes that part of the protein hydrophilic (or so-called "polar") or hydrophobic (or "non-polar").

There are two things at play here and you need to mentally separate them.
When seen as molecules, all amino acids are polar to at least some degree. Most of them are very polar, especially when compared to other organic molecules. This is (obviously) due to the carboxylic acid and amino groups both of which are very polar functional groups and both of which are ionised in neutral aquaeous solution.
However, it is desirable to categorise the different amino acids into various groups, to better explain their individual contributions to an overall peptide or protein structure. As all amino acids have a carboxylic and amino group and as these groups are connected in peptide bonds to form the peptide or protein backbone, these two groups alone are not able to distinguish between the different amino acids.
For this reason, the amino acids have been classified according to their residue R. If the residue contains only carbon and hydrogen atoms (e.g. Gly, Ala, Val, Leu, Ile), it is a non-polar residue and thus these amino acids are classified as ‘non-polar amino acids’ in simplistic terms. In a peptide or protein, these amino acids’ non-polar side chains will usually act as non-polar domains within the tertiary structure
A similar case can be made for the ‘charged’ amino acids (Glu, Asp, Lys, Arg): obviously and trivially, all amino acids have charged groups when dissolved in water but only those four have a charged side chain. So only these four are able to build up salt bridges to cross-stabilise tertiary structures and are thus classed as charged.

All amino acids have the capability to form the zwitter ion...but that doesnt necessarily make all of them polar....consider the example of a saponification reaction...we have a polar end and a non polar alkyl group attacked to it making that whole part of the chain hydrophobic...and non polar
Indeed since the general structure of an amino acid being NH2-CH(R)-COOH
The R group has a significant part to play in determining if an amino acid is polar or not...which makes amino acids like  alanine, cysteine, glycine, isoleucine, leucine, methionine, phenylalanine, proline, tryptophan, tyrosine and valine show more of a non polar character than polar

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text:
Neutralization Reaction problem

$\frac x4$ equivalent of $\ce{NaAl(OH)2CO3}$ is neutralized by $x$ equivalent of $\ce{HCl}$
Milliequivalent of $$\ce{HCl}=\mathrm{{160}\times{0.25}\times{1}\times{40}}/100=16$$
$1$ is for one $\ce{H+}$.
Hence milliequivalent of $\ce{NaAl(OH)2CO3}$ required is $4$
$\mathrm{4 = {0.1}\times{V}\times{2}}$
$2$ is for two $\ce{OH-}$
So,
$V = \pu{20ml}$
But given answer is \pu{40ml}, I think its basicity is taken $1$, but why?

First, the reaction given is not balanced. 3 moles of water should be formed, not 2. However, that will not affect the calculation. Suppose the volume is x ml. The reaction will consume 160 (ml) * 0.25 mole/liter = 40 mmole of HCl. We have x (ml) * 0,1 (mole/liter) = 0,1 * x mmole of salt reacting with HCl. According to the reaction, each mmole of the salt will require 4 mmole of HCl. In order to set up a 1:4 relationship between the salt and the HCl, we have to divide the consumed amount of HCl by 4, i.e. x * 0,1 = 160 * 0,25/4 => x = 100 (ml)
We check:
Salt added: 0,1 (M) * 100 ml = 10 mmole.
HCl added: 0,25 (M) * 160 (ml) = 40 mmole.
Thus, we have a 1:4 relation between the salt and the HCl as the reaction states.


text:
Exothermic reaction

Please someone explain me the solution of this question.

Hello and welcome to Chemistry.SE! At this point I suggest you do a couple things. First, take the short [tour] to familiarize yourself with how this site works. Secondly, this is not a "homework" site. You will need to show some work and effort to start off with before others will put their efforts into helping you. How far can you get? What specifically don't you understand? This post give some good guidance for asking homework questions. Best of luck with it!

The question tells you that the reaction is exothermic, so option 4 is out as the products are at a higher energy than reactants.
Now we see that $\ce{A + B -> X}$ is the slow step or rate determining step of the reaction, so it needs a higher activation energy than the step $\ce{X -> AB}$.
This eliminates the option 1. Option 2 is nowhere close to a right answer as the intermediate is nowhere to be seen. Thus we see that option 3 satisfies all our conditions and hence comes out to be the answer.


text:
Enthalpy of reaction

Please someone help me with this question
How is d possible?

It isn't. The answer options are 1, 2, 3 and 4;  there is no option for d. Also, this is a simple question to write out in legible fashion; no need for difficult-to-read sideways photograph.

Ea = (delta)H + RT
i guess that ends the confusion.
Now you can adjust the last term and have all 3 cases as possible
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=13676


text:
Gibbs free energy=potential free energy? (Transition state theory)
Online I found the plot Gibbs free energy vs. reaction coordinate:
https://en.wikipedia.org/wiki/Transition_state_theory
and on many books I found the plot potential energy vs. reaction coordinate (e.g. page 237 Atkins, Physical Chemistry for the Life Sciences, 2nd Ed).
In the first case activation energy is defined as $\Delta G$, whereas in the second case is defined as difference of potential energy.
Thus is the Gibbs free energy a potential energy?
Thank you very much.

Gibbs free energy is defined for system/ensemble of particles. Potential energy surface is generally calculated to a single molecular complex. They are not in the same realm, but can estimated.

Yes both mean exactly the same thing.
By definition gibbs free energy is the total amount of non pv work that can be extracted from the system. Non Pressure-Volume work refers to the work which does not allow change in both volume and pressure of the system. An example of non pv work is electrical work which is done by cells and batteries in creating a potential difference about its ends.
Now pv work is related to kinetic energy of the molecules of the system. We can see this by two ways. First is that pressure and volume both are dependent on movement of molecules. Hence increasing pressure at constant volume is the same thing as increasing kinetic energy of the gas due to which there are more collisions of higher magnitude. Similarly increase in volume at constant pressure also means increase in kinetic energy of system. Second is that pressure and volume are both related and dependent on Temperature by gas equation where temperature of a system is directly related to kinetic energy of the gaseous system.
So whenever pv work is done by or on system changes take place only in kinetic energy of the molecules of system and not in the potential energy of system.
Hence whenever non pv work is done by or on system, changes in only potential energy of the system takes place. So Gibbs free energy is same as potential energy of system.


text:
Why tosyl group is better leaving group than benzenesulphonyl group?
I read a book on organic chemistry that stated that tosyl is better leaving group than benzenesulphonyl  group, but it gave no reason for comparison.
I think that benzenesulphonyl should be better leaving group since methyl group in tosyl should be destabilizing for the anion due to its +I and +R effect.
I tried searching on the internet but it was of no help.
Please give a reason why tosyl is better leaving group.

First of all CH3 group dosent show +R effect in benzene system .
Secondly Inductive effect wears out over distance and has maximum range of 4 carbons.The charge is delocalised on carbons which are unaffected by this effect so ultimately it has no role in destabilising the anion

Doesn't methyl group show +R by hyper conjugation.

Hyperconjugation and resonnance are two different things    http://www.askiitians.com/forums/Organic-Chemistry/21/39702/hyperconjugation.htm

moreover if the methyl group wants to hyperconjugate,the carbon on the benzene system bonded to it would become pentavalent

Well, I was trying to say that methyl groups reduces the electron deficiency on benzene ring by hyper conjugation thereby reducing the -I effect of benzene ring.

please do explain how it can hyperconjugate @nikola

As i said in my earlier comment,if the methyl group hyperconjugates,the carbon it is attached to would have to be pentavalent..Hopefully you understood,pls do accept the answer if you did

You can search the internet to see the hyper conjugating structures. I will post the link in a minute.

Let us continue this discussion in chat.

This has toluene's hyper conjugating structures http://www.chemvista.org/hyperconjugation.html


text:
If polar non-polar intermolecular forces are stronger, why do polar compounds elute first in GC?
I remembering reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules.
However, in analytical chemistry classes, we are told that polar molecules elute first in a Gas Chromatograph that has a non-polar column. According to http://www.chem.ucla.edu/~bacher/General/30BL/gc/theory.html

If the polarity of the stationary phase and compound are similar, the retention time increases because the compound interacts stronger with the stationary phase. As a result, polar compounds have long retention times on polar stationary phases and shorter retention times on non-polar columns using the same temperature.

These two concepts seem to conflict. Is there something else that is going on? Shouldn't polar compounds elute last even if the stationary phase is non-polar.
(Also sidenote: most mobile phases in Gas Chromatography are non-polar: helium, argon, or nitrogen)

I think you're incorrectly remembering something you've been told. People speak of the adage "like-dissolves-like" which really means that polar things interact best with polar things and nonpolar interact best with nonpolar. There are always exceptions and fringe cases, but that adage works pretty well here. Nonpolar molecules interact with other nonpolar molecules pretty much exclusively through dispersion forces. These are the so-called "instantaneous dipoles" you have probably heard of. Polar molecules usually have some way of interacting through dipole-dipole interactions, hydrogen bonding, or simple electrostatics.
Note that the mobile phase, i.e. the carrier gas, does not have a large effect on the retention time in GC, so it really is mostly due to the simplified explanation of intermolecular forces given above.
One might ask why the polar molecules don't also have large dispersion interactions with the stationary phase (they certainly have some), and this is due to two effects. First, polar molecules tend to interact fairly strongly with each other in the gas-phase, and so there is less of an energetic gain for interacting with the stationary phase. This is a problem because interacting with the stationary phase is entropically unfavorable. Second, many polar molecules are smaller than a lot of the nonpolar systems that are studied using GC, and dispersion is correlated with surface area, so the polar molecules tend to have weaker dispersive interactions even ignoring the first point.

I am surprised by the statement that "polar molecules tend to interact fairly strongly with each other in the gas-phase". Most of the time, molecules in the gas phase don't interact with other molecules at all.

Looking at a gas chromatogram region, it might seem that polar molecules are smaller than nonpolar molecules with similar retention times because if you compare molecules of similar molar mass, the polar ones bind more tightly to the stationary phase and will have much longer retention times. See discussion here

Your statement about intermolecular forces are correct.
Your statement about polar molecules eluting first from a non-polar gas chromatography column is debatable. It really hinges on comparing apples with apples instead of with oranges.
If we group molecules according to shape and molar mass, we ensure that molecules within a group will have similar strength of dispersion forces. Then, we can compare molecules that are non-polar with ones in the same group that are polar, for example nonane vs 1-octanol or dibutyl ether. It turns out that 1-octanol has a higher retention time (elutes later, i.e. interacts stronger) than nonane. Dibutyl ether is similar in retention time (and boiling point) to nonane.
There is no need to classify the mobile phase as polar or non-polar. The interactions in the gas phase are minimal (otherwise it would condense and form a liquid).
So there is a conflict between your two statements (strength of IMF, behavior on non-polar GC column) which resolves once you realize that one of the statements is incorrect.


text:
When was the symbol for potassium changed from Ka to K?
Wikipedia says that Ka was a symbol for potassium once.

Current symbol is K.
Name changed due to a standardization of, modernization of, or update to older formerly-used symbol.

When was the symbol for potassium changed from Ka to K? What was the rationale behind that?

Some findings. The INTERNET Database of Periodic Tables's first image with a symbol for potassium is of Charles Daubeny's system using K in 1831. Surprisingly I could not find any use of Ka. But there are many German-speaking books using Ka in the late 19th century, as well as some English ones. More recent findings of Ka: 1894, 1943, 2013

Probably better suited at [hsm.se]

We already had a very competent answer to a similar question here. http://chemistry.stackexchange.com/questions/42561/why-was-lawrenciums-symbol-changed

Here is an image that shows Charles Daubeny used K in 1831.

There are many German-speaking books using Ka in the later 19th century.  English KaCl findings (because that makes a better search term): 1894,  1943,  2013!

If guessing is OK I assume it is to avoid confusion with Potassium K and subscript a, like in pKa. There was no "K" so then the change is simple and easy.

@StianYttervik That was my guess also. But I need a definite answer (background for my question).

This is the original publication in which Jacob Berzelius first proposed the name "kalium" with the symbol "K" for potassium in 1814. Emphasis on proposed, I cannot yet find when this was accepted...we're getting there though!

I would speculate that Ka would be easily confused with K(alpha) and the "a" was therefore dropped .

The symbol for potassium was mostly K in the English speaking world and even some old German book use K (as early as 1829 search Kalium in this book). So there was no official change from Ka to K, because both symbols were concurrently used in various parts of the world, just one symbol, Ka, wiped out of fashion. Perhaps the German logic was to retain (Ka)lium and (Na)trium. It is even interesting to see Ka and K being used in the same paper by no other than Bunsen and Kirchoff (the discoverer of rubidium and cesium). For example,

And their beautiful spectra, note Ka on the left. Available for free from Taylor and Francis or use Google Scholar. G*. Kirchhoff & R. Bunsen (1861) XLII. Chemical analysis by spectrum observations.—
Second memoir , The London, Edinburgh, and Dublin Philosophical Magazine and
Journal of Science, 22:148, 329-349, DOI: 10.1080/14786446108643164*

Humphrey Davy did not propose a symbol for sodium or potassium (from soda ash and potash, in 1807) because he was not 100% sure if they were elements. He implied that even if someone later shows that they are elements, the names can be retained as elements obtained from those metals.

"Available for free from RSC"? Where?

Google Scholar the DOI, and the first result is that paper. Sorry, it is from Taylor & Francis.

Neither https://doi.org/10.1080/14786446108643164 nor https://doi.org/10.1039/QJ8611300270 are freely accessible for me.

If you really want to read this article, request a friend from a university. For some reasons, I could see pdf option in Google Scholar.

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text:
Why hydrogen peroxide sizzles in the bathroom sink?
Extremely short question. I pour a bit of hydrogen peroxide in my bathroom sink and it starts to sizzle, almost the same way it does on the open wound. Why? What elements in there it reacts with?

Decomposition of hydrogen peroxide is catalyzed by a great many compounds. It could be anything.

Hydrogen peroxide contains an peroxide linkage and it decompses rapidly to give oxygen.
But when substances like detergent come in its contact in a bathroom (as in the question) it not only  catalyses the reaction but if it was a foam making detergent it would make the decomposition process more evident and visible as oxygen bubble would produce effervescence in foam.

So, do I understand correctly that even common bathroom soap stains will make the H2O2 make visible and audible reaction there?


text:
Bubble formation when aluminium reacts with copper(II) sulfate solution
I recently did an experiment putting pieces of $\ce{Al}$ and $\ce{Fe}$ into copper(II) sulfate solution and I noticed when the piece of aluminium was reacting, tiny bubbles were coming out whereas with the piece of iron this didn't happen.
I think it was part of the bare aluminium that was reacting with water producing hydrogen gas but I'm not sure. Can anyone please explain this to me?

Your explanation is correct. Al is much more active than Fe, so once the protective oxide layer is breached, it starts reacting with pretty much everything around, including water.

Did the $CuSO_4 $ solution changed its color in both case ? I beleive it should have cause Al and Fe can displace Cu

As Ivan Neretin noted in a comment, your supposition is absolutely correct.  The aluminium metal is reactive enough with water that it forms hydrogen gas and aluminum hydroxide (or other aluminum oxyhydroxide species) where it's exposed to the solution:
$$
\begin{align}
\ce{6H2O + 6e-} & \ce{->~ 6OH- + 3H2\uparrow} \\
\ce{2Al(s) + 6H2O } & \ce{->~2 Al(OH)3 + 6H+ + 6e-}
\end{align}
$$
Overall reaction:
$$
\ce{2Al(s) + 6H2O -> 2Al(OH)3 + 3H2\uparrow}
$$
If the $\ce{CuSO4}$ solution is acidic enough, the $\ce{Al(OH)3}$ rapidly dissolves, exposing a fresh aluminium surface, ready for further reaction.

On the other hand, the redox reaction of $\ce{Fe(s)}$ and water is not favorable enough for hydrogen evolution to occur, which explains the lack of bubbles you observed.  The single-displacement reaction between $\ce{Cu^2+}$ and $\ce{Fe(s)}$ is quite favorable, however, leading to paired reactions like the following:
$$
\begin{align}
\ce{Fe(s)} & \ce{->~ Fe^2+ + 2e-} \\
\ce{Cu^2+ + 2e-} & \ce{->~ Cu(s)}
\end{align}
$$
Overall reaction:
$$
\ce{Fe(s) + Cu^2+ -> Cu(s) + Fe^2+}
$$
This results in the rapid exchange of metallic Cu for metallic Fe on the iron surface, leading to a film of plated copper on the surface.


text:
Suggest the product from two-substituted benzene ring
Which are the possible products from this reaction?
The answer in the mark scheme shows that it should be as demonstrated as pic A.
However, I think it could be either A. or B. since both substituted groups of the benzene ring, NH2 and Cl, direct the incoming electrophiles to attack the 2,4 and or 6 positions. Do I understand correctly?
I have never come across two substituted groups before. Should I consider both NH2 and Cl as substituted groups or only one of them, but which one? Also, how to count the position of carbon on the benzene ring in this case?

Please help! Any help would be much appreciated.:)

Aniline directs o/p, chlorobenzene is weaker but directs m.  The effects are additive

I thought halogens dieected o/p (except $\cw{F}$, maybe), bbut more weakly than an amino group.

On substituting an already substituted benzene ring, you need to know which kind of substitution takes place; It is electrophillic substitution that is taking place.
Then can you now think about the correct choice?
$\ce{-NH2}$ has a greater positive mesomeric effect (+M effect)
in comparison to the Halogen groups.
Since the para position with respect to $\ce{-NH2}$ is already occupied by chlorine the ortho position would be attacked.
What we come to know is that since $\ce{-NH2}$ creates more electron availability in comparison to halogen it leads to "electrophilic substitution" on ortho position with respect to $\ce{-NH2}$.
And moreover $\ce{-NH2}$ group since a very very strong electron donor so once again another substitution of next bromine takes place at the other ortho position with respect to $\ce{-NH2}$.

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text:
Rules for benzene halogenation
Can anyone please show me the rule on how a halogen like bromine reacts with benzene, such as when bromine replaces hydrogen? I'm wondering what happens after the first hydrogen is replaced? Does it replace the hydrogen in the 2- or 4-position after that or something else?

SEAr

The most common reaction of aromatic compounds is electrophilic
aromatic substitution, in which an electrophile (E1) reacts with an aromatic
ring and substitutes for one of the hydrogens.
In an unsubstituted benzene ring, the halogen will likely take any position (essentially any position is the same - just depends on how you draw the structure) since the ring is only attached to hydrogen atoms with similar effect  in the ring system.
The real question is what would happen if we were to carry out a reaction on an aromatic
ring that already has a substituent?
A substituent already present on the ring
has two effects:

Substituents affect the reactivity of the aromatic ring. Some substituents
activate the ring, making it more reactive than benzene, and some
deactivate the ring, making it less reactive than benzene. In aromatic
nitration, for instance, an -OH substituent makes the ring 1000 times
more reactive than benzene, while an - NO2 substituent makes the ring
more than 10 million times less reactive.
Substituents affect the orientation of the reaction. The three possible
disubstituted products—ortho, meta, and para—are usually not formed
in equal amounts.

Instead, the nature of the substituent already present
on the benzene ring determines the position of the second substitution.
An -OH group directs substitution toward the ortho and para positions,
for instance, while a carbonyl group such as -CHO directs substitution
primarily toward the meta position
All meta-directing
groups are strongly deactivating, and most ortho- and para-directing groups are
activating. The halogens are unique in being ortho- and para-directing but
weakly deactivating.
See screenshot below

Taken from  Organic Chemistry (Mc Murry) 8th ED

Does this also apply alkyl benzene?

Yes, the positioning of the electrophile will depend on the substituent, the substituent can be an alkyl group, alkene, aryl. What matters is whether the substituent group exterts an electronic withdrawal or donating effect. From the illustration above you can notice the alkyl group has an electron donating effect, conversely, the common characteristic of deactivating groups is that they are electron withdrawing.

So Bromine is ortho and para directing but which one is more likely to happen?

Both -ortho and -para intermediates are stable, however a higher ratio of the para substituted product is formed to enhance the stability of the Π system (since this configuration results in the substituents bring furthest away)

Meant being furthest*


text:
What is a functional group?
I know that a functional group gives definitive property to an organic compund.
But my text book claims carboxylic acid is a functional group but isnt carboxyl the functional group with a formula -COOH , is the text book wrong or am i missing something?
And similarly isnt hydroxyl the functional group of alcohols rather than alcohol being the functional group.
Could you please clear this out ?

You are right, -COOH is a functional group and carboxylic acid is a compound. Likewise, -OH is a functional group and alcohol is a compound. If the book says otherwise, don't trust the book.

@IvanNeretin Thank you so much for the answer , if i attatch an image of the text , could you clarify whether or not the book is wrong ?

If you would like to clarify what your book writes, consider rewriting the passage as text and adding a full citation to the book for context and reference. You can then also tag the question [tag:erratum].

it says carboxylic acids are functional groups with formula -COOH but a google search showed that functional group is called carboxyl and group of compounds is called carboxylic acids, am i right ?

Your books statement that "carboxylic acid is a functional group" doesn't seem quite right to me. I think it's better to say that a carboxylic acid is an organic molecule containing a carboxyl functional group.
Similarly, I wouldn't say that an alcohol is a functional group, but that an organic molecule containing a hydroxyl group is an alcohol.
Of course you can have single large molecules with multiple functional groups that could be called, at once, for example, an alcohol, a ketone and a carboxylic acid, because it contains a hydroxyl functionality, a carbonyl functionality and a carboxyl functionality.
I hope this addresses what you were asking, don't hesitate to ask if I missed something.

Carboxyl, hydroxy, chloro, amino etc are prefixes which represent the functional group used. They are standard 'PREFIXES', note that, and not the functional groups themselves. They represent the functional group in the compound while naming.

as my question stated , is -COOH a carboxylic acid or is it a carboxyl group which makes the compound formed a carboxylic acid . Moreover my textbook claims that carboxylic acid is a functional group with formula -COOH , i dont think this is right because a -COOH group is a carboxyl functional group

Carboxylic acid - Functional group;
Carboxy - Prefix of FG

what is the difference between carboxyl and carboxy and does it mean carboxylic acid is not the name of the homologous series of compounds that  have the -COOH (carboxyl  group) attached to it ?

Carboxylic acid is something similar to the term 'alcohol', whereas carboxy stands a counterpart to 'hydroxy'. Carboxylic acid is something similar to the term 'haloalkane', whereas carboxy stands a counterpart to 'halo-'.


text:
Why are isotopically pure diamonds 50% more thermally conductive than other diamonds?
A laboratory grown type-IIa diamond (no nitrogen defects) has a lambda of $\pu{1800-2200 W/mK}$, but an isotopically pure diamond of $\ce{^{12}C}$ can have up to $\pu{3320 W/mK}$.
Why are $\ce{^{12}C}$ diamonds so much more thermally conductive than diamonds with $\approx \pu{1.1\%}$ $\ce{^{13}C}$ atoms?
Sourcing for my question:
https://www.sciencedirect.com/science/article/pii/092596359290197V?via%3Dihub
https://adsabs.harvard.edu/abs/1990PhRvB..42.1104A

Less phonon scattering from the 'impurity' C13 atoms. I'm not equipped right now to dig up the original paper, but the predicted results matched quite well with reality.

From THERMAL  CONDUCTIVITY  OF  NATURAL  AND  ISOTOPICALLY  ENRICHED  DIAMOND  - EFFECT OF NEUTRON IRRADIATION, D.P. White, Department of Physics, Merrimack College, N. Andover, MA 01845:

The thermal conductivity was calculated using the Callaway method.  This method takes into account  the  effects  of  intrinsic  three-phonon  normal  (N)  processes.    These  processes  do  not create thermal resistance directly but affect the thermal resistance by transferring phonons from frequencies where they are not scattered to frequencies where scattering is more efficient.

[...]

In order to  explain  the  large  increase  (originally  much larger  than  expected)  in  terms  of  isotope scattering it was necessary to include the effects of N processes.  Subsequent analyses of the thermal  conductivity  of  enriched  diamond  have  included  N  processes.    Wei et.al.  used  the Callaway  expression  to  fit  experimental  thermal  conductivity  data  on  diamond  with  the  natural isotope  concentration  and  isotopically  enriched  (0.1% $\ce{^{13}C}$)  diamond  over  a  wide  temperature range (100-1000K).

OK, nice reference. But what does it mean?

The TL;DR is that the thermal conductivity of diamond is dependent upon it's intrinsic scattering phonon relaxation times. This property is at a maximum (meaning maximum thermal conductivity) with maximal isotopical purity (of either 12C or 13C), and at a minimum when there is a 50% mixture. This is not linear in nature, as even a diamond with 1% of one isotope has markedly lower relaxation times than one with 0.1%. This is a difficult topic (for me) to summarize, though I should probably have given it some discussion. Honestly there's not much of a shortcut here vs reading the cited article.

Airhuff - thank you! I've only found data on conductivity of 12C diamonds and not 13C - is the lambda of a 13C diamond similar to that of a 12C diamond (~3300 W/mK)?

Also, would a boron doped 12C diamond be more conductive than a pure 12C  as it's electrically conductive? Here's another good explanation of phonon conductivity: http://chemistry.stackexchange.com/a/17250/42303

I found then lost a link to an article in which they made pure 13C diamonds, pure 12C diamonds and many mixes in between. I'm sure I'll find it and I have some time to search now. Great answer by Dryden that you linked to. Off the top of my head, I would guess that a boron doped diamond would have increased phonon scattering, but maybe also increased electronic conductivity???

Still can't find that article, but here's one you might find interesting. Their claim (one of them anyway) is that (if I understood it correctly) that the heat conductivity of diamonds is all about phonons, and the electronic element is irrelevant. Isotopically-pure carbon-12 or carbon-13 polycrystalline diamond possessing enhanced thermal conductivity.


text:
Is this a violation of Equipartition theorem?
My book has a question asking to calculate average kinetic energy of the molecules in 8g of methane.
By the Equipartition theorem, I calculated that energy:
1. per molecule per degree of freedom = $(1/2)kT$ (by definition)
2. per molecule = $3kT$ (since methane is polyatomic => 6 degrees of freedom*)
3. per mole = $3RT$
4. for 8g methane ($=0.5$ moles) = $1.5\cdot RT$
But the answer stated by my book is $0.75\cdot RT$  - just half of what mine is.
This is not the only question where values have mismatched. I have started losing faith from Equipartition theorem.
Question:

What did I get wrong?

SUMMARY+UPDATE: Final conclusion reached is that while total kinetic energy is indeed the one Equipartition theorem gives (and so it's correct! :D) but the textbook assumes the molecule to be at at very low temperatures (although this is NOT mentioned) implying the rotational and vibrational energies are NOT counted.

*=>my textbook says so. I understand that there's a lot of complex things in this topic, but let's please assume it is 6 as my book says.

I don't thjnk you did anything wrong.

Regardless of the numerical answer, the book is slightly lax on units. $3RT$ per mole is correct, but in the next step that should be multiplied by 0.5 mol, not by 0.5. The final answer should not have units of $\mathrm{mol^{-1}}$. As for your mistake, which degrees of freedom do you think "kinetic energy" refers to? I'm afraid you might have to learn about the "complex things" in this topic.

You are right in your thinking, kinetic energy is that of 'moving things' as you put it, the total kinetic energy a molecule can have is the sum of translational, rotational and vibrational parts.

For a non-linear polyatomic molecule like methane, there are a total of $3N$ degrees of freedom, where $N = 4$ is the number of atoms in the molecule. This is broken down as follows:
$$\begin{array}{c|c} \hline
\text{Component} & \text{# Degrees of freedom} \\
\hline
\text{Translational} & 3 \\
\text{Rotational} & 3 \\
\text{Vibrational} & 3N - 6 \\ \hline
\end{array}$$
(Note that for a linear molecule, or for an atom, this breakdown is different.)
Under "normal" temperatures (close to room temperature) the vibrational contribution to the internal energy can be neglected. (The vibrational gap $\Delta E = \hbar\omega \gg k_\mathrm{B}T$ for "normal" $T$, so higher vibrational states are not populated and the equipartition theorem does not hold.) This is probably the reason why your book states that there are six total degrees of freedom.
The key is to realise that the kinetic energy of a molecule arises only from the translational degrees of freedom. From my limited knowledge of physics, it seems to be a common thing to separate translational degrees from the "internal" degrees of freedom, i.e. rotational, vibrational, and electronic. This allows the translational motion of a body to be modelled as the movement of a point mass, located at the centre of mass of the body.
In this paradigm, rotations and vibrations preserve the centre of mass of a molecule, and therefore don't contribute to the kinetic energy of the molecule. Therefore, the correct answer is $(1.5~\mathrm{mol})\cdot RT$.
Another way of looking at it is to consider the definition of the kinetic energy
$$E_k = \frac{1}{2}m\langle v^2\rangle = \frac{1}{2}m(\langle v_x^2\rangle + \langle v_y^2\rangle + \langle v_z^2\rangle)$$
where $v$ is the velocity of the centre of mass, $v_\mathrm{cm}$. For each quadratic term the contribution to the energy is $kT/2$, for a total of $3kT/2$ per molecule. These three terms correspond to precisely the three translational degrees of freedom.

The equipartition theorem is a well established result of classical statistical mechanics. It states that the mean value of each independent quadratic term in the energy (per molecule) is equal to $\frac{1}{2}k_BT$ where $k_B$ is Boltzmann's constant and T the temperature. Being 'classical' means that if $\Delta E$ is the energy of a quantum then the thermal energy is far greater than this and effectively all levels are populated at the given temperature, i.e. $k_BT >> \Delta E$. (A derivation of equipartition can be found in this answer Derivation of mean kinetic energy )
The kinetic energy is split into three parts, translational , rotational and vibrational. As motion is possible in three directions (cartesian axes x, y, z thus three degrees of freedom) the average translational kinetic energy is $<E_{trans}> = \frac{3}{2}k_BT$.
The rotational kinetic energy also exists and occurs because a molecule can rotate as a whole body about each of the three axes (x, y, z, and assuming a non-linear molecule such as methane). ( The angular kinetic energy is $\frac{1}{2}I\omega^2$ where I is moment of inertia and $\omega$ the angular velocity). Thus the average rotational kinetic energy is $<E_{rot}> = \frac{3}{2}k_BT$.
If there are N atoms in a molecule then there are in total $3N$ degrees of freedom. We have accounted for $6$ of them in whole body translation and rotation, which leaves $3N-6$ internal degrees which are attributable to vibrations. As methane has 5 atoms the average vibrational kinetic energy is  then $<E_{vib}> = \frac{9}{2}k_BT$. This value will not be reached at room temperature as typical vibrational quanta ($500 \cdots 3000 \pu{  cm^{-1}}$) are  far greater than $k_BT \approx 208 \pu{ cm^{-1}}$ at room temperature. Thus the vibrational kinetic energy can often be set to zero with little error. However, the temperature is not specified in your question so in any answer it is necessary to justify removing the vibrational energy.
The rotational quantum for a polyatomic molecule is by contrast often only a few wavenumbers, or less, and  $ \approx 10 \pu{cm^{-1}}$ for methane and so the average rotational energy will satisfy the energy criterion $k_BT >> \Delta E$ as will translational kinetic energy.
The total average kinetic energy (in Joules) is $<E_{tot}> = \frac{15}{2}k_BT$ per molecule. Removing the vibrational energy gives $<E_{tot}> = \frac{6}{2}k_BT$, which is $ \frac{6}{2}RT ~ \pu{J mol^{-1}}$ or $ \frac{3}{2}RT ~ \pu{J }$ for $0.5$ mol. It seems that to obtain the answer you quote, the rotational kinetic energy would also have to be removed, which then supposes that the temperature is very low.


text:
1 mole of substance is equal to atomic/molecular mass in grams?
I have been reading the answer of this question here- Why is weight of 1 mole of substance equal to atomic/molecular mass in grams

I feel that the answer writer was wrong to say that -"the ratio of g/mol to amu/atom is exactly one"
I  feel so because he proved that the ratio of g/mol C12 to amu/atom C12 is exactly one.He never proved that the ratio of g/mol to amu/atom is exactly one.If I am wrong then please correct me!

The statement is more or less correct.
But there are two caveats here:

There is no proving of these concepts since there are definitions.
In general, it's a bit strange to use the ratio of amu to atom. An atom has a well defined mass based on it's isotope, but the number of grams per mole is defined based on natural abundance and weighted average mass. Therefore, there is some discrepancy here. Of course, here, you're referring specifically to carbon-12, so there is no confusion, but just watch out in the general case.

Why is it that 12 grams of C12  atoms represent 6.022* 10^23 atoms of carbon? Similarly why 1 g of H represents 6.022*10^23 atoms of H?

That comes from the definition of mole and Avogadro's number. https://en.wikipedia.org/wiki/Avogadro_constant

According to Wikipedia- Mole  is defined as the amount of a chemical substance that contains as many representative particles, e.g., atoms,molecules, ions, electrons, or photons, as there are atoms in 12 grams ofcarbon-12 (12C), the isotope of relative atomic mass 12 by definition. But this definition has nothing to do with 6.022*10^23.How should I establish the relation between the number of atoms in one mole of C12 and it's gram atomic mass (that is 12 g,numerically equal to its atomic mass)?

Guess how many atoms of carbon-12 are in 12 grams of carbon-12...

It is 6.02210^23 atoms.I found it written on wikipedia.however,I do not know ,how chemist got to the point that 12 g of C12 atom always contain 6.02210^23 atoms??have they assumed it?

Again, that is the definition of mole. We know that we have a constant number of units due to the law of definite proportions. Measurement is another issue. See: https://en.wikipedia.org/wiki/Avogadro_constant#Measurement

ok!but why did chemists decide to use the number 6.022 * 10^23 as a mole?was there any specific reason?

We didn't decide. We measured it...

I think i am in a big confusion!

Let me start from beginning.

Wikipedia gives this definition of a mole - A mole is defined as the amount of a chemical substance that contains as many representative particles, e.g., atoms,molecules, ions, electrons, or photons, as there are atoms in 12 grams ofcarbon-12 (12C), the isotope of carbonwith relative atomic mass 12 by definition. so,my first question is that how did chemists come to know that 12 grams of C12 atom would contain 6.022 * 10^23 atoms?

You can perform experiments, see comment above: en.wikipedia.org/wiki/Avogadro_constant#Measurement


text:
What is the correct organic nomenclature of this substituted toluene?
Are both 2-ethyltoluene and 1-ethyl-2-methylbenzene correct? If yes, do they belong to the same compound or belong to different compounds?
I'm confused when I have to put organic nomenclature for a compound if there is a methyl group. When should I use toluene and when should I use benzene?

The names 2-ethyltoluene and 1-ethyl-2-methylbenzene are unambiguous and describe the same compound.
1-Ethyl-2-methylbenzene is a correct systematic name according to current IUPAC recommendations.
Various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations. In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name toluene is still retained as the preferred IUPAC name, but no substitution is allowed.

P-22.1.3 Parent hydrocarbons having retained names
Toluene, xylene, and mesitylene are specific parent hydrides that are composed of two components, one cyclic and the other acyclic and saturated. These names are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not freely substitutable; toluene is substitutable under certain conditions, but only for general nomenclature (see P-15.1.8 for a general substitution rules for retained names).
  (…)

For the compound that is given in the question, the systematic name 1-ethyl-2-methylbenzene is the preferred IUPAC name.

In general nomenclature, toluene is substitutable; however, substitution is permitted only by groups listed in P-15.1.8.2.2.

P-15.1.8.2.2 Substitution rules for Type 2b retained names
(…)
  The following characteristic groups cited can be used to substitute parent structures of Type 2b (ring and the side chain if required): halides $\ce{-Br}$, $\ce{-Cl}$, $\ce{-F}$, $\ce{-I}$, pseudohalides $\ce{-N3}$, $\ce{-NCO}$ (and chalcogen analogues), $\ce{-NC}$, substituent groups derived from the halogen oxo acids $\ce{-ClO}$, $\ce{-ClO2}$, $\ce{-ClO3}$ (similarly for groups in which $\ce{Cl}$ is replaced by $\ce{Br}$ or $\ce{I}$), $\ce{-NO2}$ and $\ce{-NO}$, and $\ce{-OR}$ ($\ce{R}$ =alkyl groups), and chalcogen analogues, and-$\ce{SO-R}$ and $\ce{-SO2-R}$, and $\ce{Se}$ and $\ce{Te}$ analogues.

Therefore, the name 2-ethyltoluene is not in accordance with current IUPAC recommendations, neither for preferred IUPAC names nor for general nomenclature.

I like to use ChemDraw program to help me to confirm in converting IUPAC name to molecular structure and vice versa. With your proposed two names, based on CD program, they are the same molecule.

While this provides a good answer to the first part of the question, you have completely ignored the second part: "I'm confused when I have to put organic nomenclature for a compound if there is a methyl group. When should I use toluene and when should I use benzene?"

@airhuff My answer is clearly itself. They are the same molecule. Whatever name what you call this structure, as long as those name are right and can be refer to the same molecular structure, it's ok. On the other hand, if I provide the 30 students just its 2D structure of this chemical on their final exam and ask them to provide the name of this structure, would their answer be the same ?

All I'm saying is that your answer would be more complete if it explicitly and directly addressed the second part of the question. I feel like I can't give it a UV without that and I suspect others will think the same. Just my advice, take it or leave it.

@airhuff Thank you for your advice. I got your point.


text:
Why is a C–D bond stronger than a C–H bond?
I have heard that the carbon–deuterium bond is stronger than the carbon–hydrogen bond. What are the possible reasons for it? Is this also the reason that C–H bonds participate more in hyperconjugation than C–D bonds? Please explain.

I sincerely recommend to look at the matter from a slightly different angle. C-D bonds are not stronger; for most purposes they are identical. Only if you measure the energies with really high precision, you might notice some very small, very insignificant difference, so small that most people don't even care which way is it. That's the difference you want to discuss. No, I don't think it has anything to do with hyperconjugation. My bet is on lower zero vibrational level for C-D because of higher equivalent mass.

The lesser contribution of C-D to hyperconjugation is the cause of some secondary kinetic isotope effects; it is related to the very slightly different bond strengths. See for example p 12 of this MacMillan group meeting (The classic text, Anslyn/Dougherty, doesn't actually mention it. I haven't checked Lowry/Richardson yet.) Warning: organic chemists' explanations tend to be a bit handwavy. However, the KIE is indeed there for all to see, so it's not fake.

While it is true that the C-D strength will have only very minute effects on reactions, it is worth noting that many many systems if placed in D2O will switch any protium that leaves at a reasonable pace with a deuterium. And like the answer given below indicates, this is almost solely due to a lower zero point vibrational energy because electronically the systems are the same.

@IvanNeretin No need to bet on it. That is exactly correct. :) You don't need to measure the energy with the precision you suggest those. The lower ZPE results in a measurable difference in the rate of bond cleavage.

The question focuses on the difference in bond strength, and the answers explain this in terms of ZPE. But for proton transfer, quantum mechanical tunneling has to be considered. Especially at low temperatures, this makes it possible to have huge differences between reaction rates with H and with D -- many orders of magnitude, not just small effects. There is a nice open-acess review article on this sort of thing: Scheiner, "Calculation of isotope effects from first principles," https://doi.org/10.1016/S0005-2728(00)00058-X

Physics is better able to answer "how" questions than "why" questions, but here goes. The quantum mechanical description of the C–D system versus the C–H system gives the former a lower zero-point energy, which is the minimum energy the quantum system can attain. A good conceptual model of this is to consider the C–X system as two masses connected by a spring. When X = D, the system will vibrate more slowly than when X = H simply due to mass.
Since energy is proportional to the frequency of vibration, the energy of the C–D system is less (lower). This table lists the C–D bond dissociation energy as $\pu{341.4 kJ/mol}$ and C–H as $\pu{338 kJ/mol}$. Since this is the energy to break the bond, the C–D bond is stronger.

As far as your question about hyperconjugation; I'm surprised that C–H would have greater hyperconjugation than C–D, but it is not something that I know anything about. I guess I could speculate and wave my hands around and say that since the C–D bond is stronger, it keeps the electrons "closer" and so more confined to the sigma orbital, but that is just hot air.
Here's something which claims that the reason D does less hyperconjugation is because it vibrates less (see slide 15). So, it seems (if this is correct) that it's the smaller bond deformations that reduce the hyperconjugation for deuterated systems. Here is a pretty clear explanation of the kinetic isotope effect.

I thought I would try to do a quantitative check of this explanation. The first vibrational excitation of CH (methylidine) occurs at an inverse wavelength of 2840.2 cm^-1 ( https://webbook.nist.gov/cgi/inchi?ID=C3315375&Mask=1000 ). This means that for CH we have a phonon energy $E=\hbar\omega=0.35$ eV. The zero point energy will be almost entirely in the hydrogen or deuterium, not the carbon, due to the smaller mass, so we expect $E_H/E_D\approx\sqrt{2}$. Since the ZPE is half the phonon energy, the difference in bond energies should be $(1/2)(E_H-E_D)=(1/2)(1-1/\sqrt{2})E_H=0.05$ eV.[...]

[...] This seems to be in excellent agreement with the figure I've seen, which is that the C-D bond is stronger than the C-H bond by about 1.2 kcal/mol=.052 eV.


text:
Redox reactions of halides in hexane
When we have two test tubes and we add:

In the first test tube: hexane, aqueous solution of $\ce{NaCl}$ and aqueous solution of $\ce{I2}$
In the second test tube: hexane, aqueous solution of $\ce{NaI}$ and aqueous solution of $\ce{Cl2}$

We are asked to write the reactions that take place in these experiments. Do we write only this:
$$\ce{NaCl + I2 → no reaction}$$
$$\ce{2NaI + Cl2 → 2NaCl + I2 (displacement)}$$
Or should hexane appear in the reactions too?

I'd write the reactions the same way you did. What puzzles me is why they bothered to mention hexane at all.

Hexane is the solvent for the two halogens. It does not react with either so leave it out of your answer.


text:
Chromium +1 is unknown while Mn+2 is known?
Although both $\ce{Cr+}$ and $\ce{Mn^2+}$ both ions have $d^5$ configuration, still only $\ce{Mn^2+}$  is well-known while $\ce{Cr+}$ is not common. What is so special about Cr that it does not form $\ce{Cr+}$?

Strongly related (possibly a dupe?): Cr(II) and Mn(III) - their oxidizing and reducing properties?

@orthocresol Even if the answer is basically identical, the question is different. So: not a dupe.

@hBy2Py I know that and to be honest, I did not think the questions were very different at all. :/ To me, this is a subset of the other. I did opt to leave it open, though.

@orthocresol $\ce{Mn^{2+}/Cr^{+}}$ have odd numbers of electrons; $\ce{Mn^{3+}/Cr^{2+}}$ have even numbers. That fact alone I would think would make for appreciably different behavior.

@hBy2Py The way I see it, the crux of the linked question is not about Cr(II), it is about "why can't Cr(II) get reduced to Cr(I) to attain a stable half-filled  orbitals", or essentially "why can't I make Cr(I) even though it is d5". I guess you are seeing it more as "why can't I reduce Cr(II)", and fair enough, but I think we have to agree to disagree on this one.

Even with the stabilization of the $d^5$ electron configuration, chromium is quite electropositive and the (second) ionization energy for removing an additional electron from $\ce{Cr^+}$ is rather low.  It's outweighed by the extra electrostatic attraction or solvation energy available by having two charges on the chromium, therefore you are ordinarily going to go on to $\ce{Cr^{2+}}$.
Chromium can adopt a +1 oxidation state in certain complexes, such as $\ce{Cr(CN)_5NO^{3-}}$.  See https://en.wikipedia.org/wiki/Chromium.


text:
What is the relation between the amu and avogadro’s number?
I know that the relation between Avogadro’s number and amu is a reciprocal relationship but the relation is slightly unclear.
Could anyone give me more clarification?

$$N_{A} = \frac{m_{\mathrm{in\ amu}}}{m_{\mathrm{in\ grams}}}$$

$\pu{1amu}$ is defined as one twelfth of the mass of one carbon-12 atom.
Avogadro's number is defined as the total number of entities in $\pu{12g}$ of carbon-12.
$N_\mathrm A=6.022\times10^{23}$
It is a measuring criteria, just like a dozen, which is used to put a number to a certain item. A dozen means 12 items (items like bananas, atoms, molecules, etc) and 1 mole means $6.022\times10^{23}$ items (items like bananas, atoms, molecules, etc).
So, $\pu{12g}\ \ce{^12C}$ contains $N_\mathrm A$ atoms
Therefore, 1 atom of carbon-12 weighs $\frac{12}{N_\mathrm A}\ \mathrm{g}$.
As per the defintion of atomic mass unit,
$$\pu{1amu}=\frac{1}{12}\frac{12}{N_\mathrm A}\mathrm{\ g}=\frac{1}{6.022\times10^{23}}\ \mathrm{g}$$
Hence, $\pu{1amu}=\pu{1.6\times10^{-27}kg}$

I don’t know where your obscure unit gram-metres derive from, but I’m reasonably sure you were trying to talk about grams all across this answer so I corrected that.

It wasn't meant to be interpreted as $gram-meter$, I used to the notation 'gm' for gram.  However, thanks for the edit.

If you wanted to write grams, please use the standard SI-conform abbreviation $\mathrm g$ and not any made-up non-standard ones. Everything else will cause confusion.

Yeah, I got that.

The atomic mass unit (amu or simply u) is the $1/12$ of the mass of a $\ce{^12C}$ atom.
Avogadro's number ($N_\mathrm A=6.022\times10^{23}$) is the number of atoms contained in $\pu{12g}$ of $\ce{^12C}$.
The relation between them has to do with relative atomic mass ($A_\mathrm r$), which is basically the number of atomic mass units an atom is equal to. One can easily understand that carbon is equal to $\pu{12u}$, hydrogen to $\pu{1u}$, helium to $\pu{4u}$, etc. (I'm referring to the most common isotopes)
Now, $N_\mathrm A$ atoms (or $\pu{1 mol}$) weigh $A_\mathrm r\ \mathrm{g}$. For example, $N_\mathrm A$ atoms of hydrogen weigh $\pu{1g}$. This is because they are the $1/12$ of $N_\mathrm A$ atoms of carbon (which weigh $\pu{12g}$).

There is a lengthy discussion of what an amu is, as well as on Avogadro's constant in Wikipedia. see https://en.wikipedia.org/wiki/Unified_atomic_mass_unit and see https://en.wikipedia.org/wiki/Avogadro_constant. An amu is a unit of mass, like a pound, gram, microgram, or tonne. An amu is a very, very small unit of mass, defined, as you know by now, as being exactly the mass of 12 isolated 12C atoms in their ground state (which is just a way to eliminate kinetic or potential energy which would give the atoms more mass (remember E = mc²)). Avogadro's Number is just like a dozen, a pair, a score, a gross - it is a definite number of things, of particles. It is, like "dozen" and the rest, dimensionless. A very important note is that the term has been depreciated by the IUPAC (the consensus "Authority" for Chemistry (and some Physics)). You should not be taught about Avogadro's Number; it has been replaced by Avogadro's Constant. Avogadro's Constant is not dimensionless - which is the big difference between the two. The Constant has units of reciprocal quantity. If that quantity is reciprocal moles, then the numerical value of the Constant and the Number are identical, but the quantity could be something else, for instance reciprocal pound-moles. This shouldn't confuse you, because it is almost always reciprocal moles, the important point is that it has units while Avogadro's Number does not. Unfortunately, both are symbolized NA, and it appears by your question that your teacher has failed to (or is unable to ) distinguish between the two. {Consider the common terms Atomic Weight and Molecular Weight. They are not weights, but masses and we are taught that. IUPAC also depreciates them for their "Mass" alternative, yet they are widely used, while Molecular Mass is rarely used (I don't see MM ever, but I see MW all the time). This is a similar situation, where the historical term persists although it has been depreciated.}
One dalton, 1 Da, is one amu, is one u, but guess what? Da is preferred as being more clear. (Guess whether amu has been depreciated...) In a perfect world, one mole would be a specific integer number of particles. Unfortunately, our measuring instruments aren't accurate to 24 decimal places, so the actual number is only specified to 10 to 12 decimal digits and there is a proposal to define the Constant to be exactly  6.022140 or 6.022141 x 1023 - they are, last I heard, still arguing over it. My point here is that the number itself has no fundamental importance, just like a baker's dozen is 13 pieces, while a dozen is 12, as long as we all know how many we're talking about either is fine. So 12 Da of 12C has a mass in grams of 12 ÷ NA. You may find it interesting to show this answer to your teacher and listen to what s/he says about it.


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Is it correct to consider dissolving HCl gas in water as oxidation –reduction process?
References define oxidation –reduction process as change in oxidation number ,Accordingly it is not appropriate to describe the mentioned  process as oxidation –reduction because there is  no change in oxidation number of atoms and it is not a chemical reaction .
But the teacher consider it as oxidation –reduction reaction because HCl gas ionize in water .
Iam confused , so any idea will greatly help.

No. The oxidation state of neither H nor Cl changes so it is not a redox reaction. Dissolving NaCl isn't a redox reaction either.

Teacher consider H as an atom in HCl molecule but separate as an ion in water and Cl atom gained electron from H to form negative ion ,accordingly H loose to Cl

Do you mean  in your above  comment  Dissolving HCl  or  Nacl isn't a redox  reaction ?

:( neither are redox reactions.

You need a different teacher! The Chlorine in HCl is already in the -1 oxidation state, dissolving it in water does not change this.

Can I consider ionization of acetic acid in water as oxidation - reduction reaction ?

Given the dissolution of $\ce{HCl}$ gas in water:
$$\ce{HCl(g) + H2O(l) <-->H3O+(aq) +Cl-(aq)}$$
The oxidation state of all hydrogen atoms is +1 and that of all chlorine atoms is -1. Thus, nothing has been oxidized or reduced, and no redox reaction has taken place.

just to make sure ,Can I consider ionization of acetic acid in water as oxidation - reduction reaction ?

For the same reason, no, you would not. Neither acetic acid nor water are being reduced or oxidized in that process.


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Using the Nernst equation with a gaseous reactant/product
I have a question about using the Nernst equation.
$E = E^\varnothing + (0.059/n)(\log\frac{[Ox]}{[Red]})$
Specifically, I am interested in the reaction of $\ce{MnO4-}$ and $\ce{C2O4^2-}$ below:
$$\ce{2MnO4- + 5C2O4^2- + 16H+ = 2Mn^2+ + 10CO2 + 8H2O}$$
How can I write Nernst equation for $\ce{CO2/C2O4^2-}$ in order to determine the titration curve of this reaction?. I'm confused because $\ce{CO2}$ isn't in the solution, and therefore there isn't a concentration for it.
Is it right if I remove $\ce{CO2}$ from the equation? For example:
$$E = E^\varnothing + \frac{0.059}{2}\left(\log\frac{1}{[\ce{C2O4^{2-}}]}\right)$$

Hello and welcome to Chemistry.SE! A good way to familiarize yourself with this site is by taking the short [tour]. Then you can always visit the [help] if you have more questions. Regarding your question, this is not a "homework service" site, in that it is against policy to just give answers to homework-type of questions. This post gives a good discussion for how to ask a homework question. The main thing is that you need to show effort towards solving it yourself.

Thanks for reply my question.
This is'n my homework. I'm having a experiment report in analytical-chemistry and I really need to solve this question. I've read many book and still can't find the right answer. All the document I've read just mention about the case that both Oxidizer and reducing are in the solid.

@SardinMike have you ever heard about a reaction constant? How do you use pressure for it then?

HINT
If we consider a reaction such that :
$$\mathrm{n^-\alpha^+Ox^+ + n^+\beta^-Red^- = n^+\alpha^-Ox^- + n^-\beta^+Red^+}$$
Where $\mathrm{Ox^+/Red^+}$ is the redox couple at the positive side of the battery and $\mathrm{Ox^-/Red^-}$ correspond to the negative side redox couple, then can show easely,
$$e=E^+-E^-=e^\circ+\frac{RT}{n^+n^-F}\ln\left(\frac{a_{Ox^+}^{n^-\alpha^+}\cdot a_{Red^-}^{n^+\beta^-}}{a_{Ox^-}^{n^+\alpha^-}\cdot a_{Red^+}^{n^-\beta^+}}\right)$$
Where $e^\circ$ is the stardart electromotrice force.
Now you just need to know what is the activity of a constituant when it is a gas or something else. In this answer you'll find some good informations.
Sorry in advance if there are typo but I think no.

I looked at this several times. I don't understand the notation.

@MaxW which one(s) ?


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Distinction between whole molecules and functional groups in IR spectra
Are infrared spectra specific to compounds or to functional groups?
For example, which of the molecules below best fits the IR spectrum shown?

I'd have to say #2, which is a ketone based off of looking at a similar graph of a linear chain ketone. Is there a better way to approach IR graphs than learning them for functional groups?

IR spectra have both function group peaks and peaks which are due to "larger molecular" vibrations. For the spectra shown the bands just below 3000 are C-H functional group bands, and the peak at 1700 is a C=O stretch. The mess below 1500 is really hard to identify peak by peak. One Cl for example on the ring would shift a lot of the bands below 1500 around.

Better maybe. As I know it, to had some lessons about did and used it a lot in organic experiment, you don't have to remember a lot of values. But if you have only one spectrum without any other informations you'll need a table and a bit of common sense. If I would had been able to predict a full spectrum by head I would had been glad. Here you see a peak at $1715 \mathrm{cm^-1}$ so it must be a ketone. And so on.

@MaxW So identification is based on the peak for the functional groups bond?

Well you're not really trying to analyze the "whole" spectra, you've already been told that out of trillions of organic molecules the spectra is for one of four compounds. The 1700 peak is for a ketone. So it can't be #3. All bands around 3000 are below 3000 so molecule is aliphatic not aromatic so #1 is out. So that quickly gets you to either 2 or 4.

You should probably base identification on the carbonyl band (a) whether you expect it to be there or not, and (b) if it is there,   look up shifts to values between aldehydes, ketones, esters etc. Tables are given in many organic textbooks, particularly for carbonyl frequency shifts.

Your question contains several points to address here.   And following Martin's suggestion, it is time to wrap up the points already mentioned in comments into a conclusive answer, too.

"IR spectrum graphs"-- there is nothing such.  The graph (or, the figure) represents the IR spectrum.  (And, for future reference, should there be need to deploy the plural form, it is "spectra".)
"Are IR spectra specific to compounds or to functional groups?"  If you record IR spectra in the typical interval of 4000 to 600 (or, depending on the window materials built in, about 500) 1/cm, then your IR spectrum will contain both: absorption bands that are characteristic for functional groups, like the classic ca. 1600 1/cm for carbonyls, or the ca. 2270 1/cm for the nitrile; AND between about 1400 to 500 1/cm what is called "fingerprint region" which contains absorption pattern that are characterisitic for the very molecule.
Attribution which absorption band in the fingerprint region belongs to what structural feature may be complicated enough that looking up a reference cataloge (by Sadtler, Aldrich, or other corporations) does not help, yet thankfully, modern computer chemical methods often allow a prediction of what vibrations may occur and where these should be observable in the IR spectrum. To render this fingerprint region more legible, often the $\tilde\nu$-scale is depicted with changes the scaling, like at 2200 and 1100 1/cm.
Which of the four compounds is more likely to be the one for the spectrum depicted?
 Well at first sight, 3 (1,4-dioxane) differs by the absence of a carbonyl group, typically yielding an intense absorption band around 1600 1/cm.  However, such an absorption band is recorded -- so it is either acetophenone (1), cyclohexanone (2), or $\delta$-valerolactone (4).
 There is not so much absorption in the region of 500--600 1/cm that were typical for aromatic rings, rendering 1 unlikely.
 Now, facing the distinction between either cyclohexanon 2, or the ester 4, it is indeed helpful that these compounds are literature known, already characterized by IR spectroscopy, and -- thankfully -- their data are made public; for example by AIST.  And as it turns out, it is likely cyclohexanone 2 which's spectrum was depicted here.

As a comment to the spectra by AIST:  They are searchable by chemical name, formula, or CAS number.  In the case of IR spectra, a little table ($\tilde\nu$), observed remnant transmittance) and a small image of the molecular structure is provided, too.
The excluded acetophenone 1:

The likely cyclohexanone 2:

The early exluded 1,4-dioxane 3:

The eventually excluded cyclic ester (lactone) 4:

So in turn, deciphering IR spectra is both learning / training to spot features (here: absorption bands), and comparison with reference data (here: spectra cataloges).  And typically supported by other means of analysis or characterization, like melting points / refractive indices compared with tables, or further NMR spectroscopy, for example.


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Amine and Amide reactions with ketone
Amines react with ketones whereas amides do not. Can anyone please explain why?

Protonation of the carbonyl group of a ketone allow amines to react via nucleophilic addition. This is because the added proton pulls electrons further away from the reaction site, thereby allowing the lone-pairs from the amine to attack the carbonyl carbon.
In the case of an amide, the C=O beside the nitrogen causes electrons to be pulled away because of the positive charge of the carbon. This means that despite the possibility of having a protonated ketone, there will still not be a reaction because the nitrogen is not nucleophilic.


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Mechanism of a cyclization reaction - why does an enol ether form?
I was asked to predict compound A from the follwing organic scheme:

Looking at the formation of A, my prediction was that since TsOH is a strong acid, it would dehydrate the alcohols (like sulfuric acid does). However, looking at the next reaction of A, it seems to me that what happened was that the ketone formed its enol and did an aldol condensation with the aldehyde. But then, if this was the case, why do we have to add TsOH to get to A, then add $\ce{H3O+}$ to get to the final product when we could've gotten to the final product from the start with an aldol condensation?
I was told that the structure for A is actually:

I was not expecting this at all, and I'm not sure how this was formed. I can see that the enol ether in A is hydrolyzed in acid to get the final product, but the formation of A seems strange. What is the mechanism for the formation of A and why were my predictions wrong?

It's just an Aldol reaction. A is probably just the enol

@Carbenoid: A year has passed since you requested a mechanism from @Mithoron. The statutes of limitation have run out. Your comment about why not do an aldol condensation (in base) is well taken. But the question was what is the structure of compound A.  I have no problem with the dienol ether 5 (aka A) in spite of many people's aversion to forming 8-membered rings.
Full stereochemistry of structure 1 could have been informative but not essential. I will forgo the mechanism of the acid-catalyzed aldol condensation and defer to @ringo's contribution. The bold bonds in structure 2 form a rigid framework so that the degrees of freedom are minimized in the formation of the hemiketal 3. Effectively, the cyclization is that of a much smaller ring. This ring closure is mechanistically more sound than the SN2 displacement with an enol particularly one with an (E)-geometry when at least the (Z)-geometry would be required. The red arrows terminate with dienolether 5. Aqueous acid and microscopic reversibility via the blue arrows return dienolether 5 to structure 2. In fact, structures 2 and 3 may well be in equilibrium with each other.

The structure for A will most certainly not be that which you were shown. I would say that you were right in your original assumption, with the reaction proceeding via the mechanism:

For the A you were shown, the aldol product would have to be dehydrated (why is this even considered an intermediate then?), tautomerized via the $\alpha–\beta$ unsaturated bond, and then attack the protonated alcohol with the much less nucleophilic oxygen site of the enol to form a 8-membered ring. While this isn't technically impossible, each of these steps are unfavorable, and reversible, this species would likely never be observed... at all... ever. See the rather strange looking mechanism:

Well, I can see a different mechanism in which this would be indeed an intermediate, starting with creation of tosylate.

@Mithoron is right. That intermediate explains the need for aqueous $\ce{H+}$. Similarly to how the intermediate shown in the question wouldn't be my first guess (or, apparently, yours), I'm guessing the authors of whatever paper it's from didn't expect it, either. Luckily, the enol ether is easy enough to hydrolyze.

@Mithoron can you write out the mechanism you're thinking of? Thanks!


text:
Absorption coefficient for liquid water
Wikipedia has this plot of the absorption spectrum of liquid water:
The vertical axis seems weird to me. The Beer-Lambert Law says $T=10^{Ax}$ where $T$ is the fraction of light transmitted, $A$ is the absorption coefficient, and $x$ is the optical distance through the material. If we take the values listed on the vertical axis to be $A$ as I've described it here, the transmittance ranges from 80% transmittance though a 1 nm (!) thick sample at 80 nm wavelength ($10^{-10^8*10^{-9}}$) to 80% transmittance through a 10 m thick sample at 500 nm wavelength ($10^{-10^{-2}*10}$).
Alternatively phrased, the absorption coefficient at 2 um is 10^4. Therefore the transmittance would be $10^{-10^4*0.01}$ through a 1 cm sample. That would be one in $10^{-100}$ photons transmitted through the sample. Given that IR spectroscopy is possible, that sure doesn't seem right to me.
Does the transmission of light through water really vary by 10 orders of magnitude or am I misreading this?

You are not misreading the graph. About 10 orders of magnitude difference.

Thank you for the accept! I think water is really fascinating, but I've always found the math and especially the units associated with absorption a real challenge.

The plot is from Wikipedia, where important/helpful links to the source information can be found, and is worth a look.
A similar and more detailed plot can be found on this page of Martin Chaplin's famous site on all things water. I can't recommend that site highly enough! The plot is shown below, and note that it's using centimeters instead of meters.

What happened to Martin Chaplin's famous Water Structure and Science website?

This section in Wikipedia gives us a clue. It says

Absorption coefficients for 200 nm and 900 nm are almost equal at 6.9 m−1 (attenuation length of 14.5 cm).

Those numbers are reciprocal, and reading the links shows the terms refer to $1/e$ lengths, natural units.
Both this plot and Wikipedia's are intended to give a global electromagnetic perspective rather than detailed quantitative information. I believe there really is a peak near 2 microns where a 1 cm cuvette of pure water would be nearly opaque. Remember it's using natural units, not 10, so ya, if you sit on that peak, using:
$$T = exp\left(-h(cm) \ \times \ ~100 (1/cm)\right)$$
and plugged in a 1 cm optical path, you'd get a transmission of 4E-44, or basically zero. You many find that there are different cuvettes with substantially reduced optical paths that can be used in this case. A sub-1 millimeter path length cuvette might do better.
However, a little bit above or below 2 microns, the attenuation drops to only 10 per centimeter, which gives a transmission of about 5E-05 for a standard 1 centimeter path length cuvette, and a quality instrument would have no problem measuring that with a suitable detector and integration time.
If you're interested in going farther, I'd recommend you find a plot that is nod drawn with such thick lines, or find tabulated data and plot it in detail by yourself. There are some numbers just for example in this 1975 paper but the hard short-wavelength cutoff at 2 microns may obscure the peak there. However let's look at the properties near 3 microns.
At 2.959 microns, $n, k = 1.329, 0.292$. Putting that into here
$$\frac{E}{E_0}  \ = \ exp\left(j  \left(n + jk\right) \frac{2\pi}{\lambda} x \right) $$
puts the electric field attenuation of 1 centimeters at $exp(-6200)$. You need to square to get transmitted intensity, so doubling the argument gives 12,400/cm or 1.24/um, which very nicely matches the "3 micron peak" in the plot, especially the inset in red(see below). This is really a mixture of states in the circa 3400 cm${}^{-1}$ water vibrational spectrum. Also, when you read, make sure you keep track of the isotopic mixture being discussed. H2O and HDO differ profoundly in their infrared optical properties.
All is well, and water continues to be amazing, profound stuff!

What the plot shows is the reciprocal path length needed for a given absorbance to apply, for example slightly less than 100 m in the visible to $10^{-8}$ m in the uv, as you point out, but this is for $55.345$ molar water which is  very, very concentrated compared to most solutions from which an absorption spectrum is obtained, typically $<< 1 $ molar.
As you also point out the absorbance is $A=\log_{10}(I_0/I)=\epsilon[C]l$ with $\epsilon$ as the extinction coefficient usually in units $\pu{dm^3mol^{-1}cm^{-1}}$ and which depends upon wavelength, and $[C]=55.345 \pu{ mol}$ for pure water and l the path length in cm. What seems to be plotted is
$$\frac{1}{l}=\frac{\epsilon[C]}{A}$$
so is still proportional to the extinction coefficient and so still indicates regions where water absorbs greatly or otherwise. But it is not clear what value for A is used. It may be understood that $A=1$ always applies or that the value is taken at $1/e$ absorption. The values may also be based on $\log_e$ not $10$ so a factor of $2.303$ may be involved. As I'm not familiar with this much detail I can't help any more other than to suggest that  you will have to check by finding a true absorption spectrum or  extinction coefficient and work backwards and converting units as necessary.
In general the amount of absorption a molecule has can vary a huge amount. In water in the visible part of the spectrum absorption is low as there are only weak overtone bands from vibrational transitions and nothing else. At shorter wavelengths electronic absorption begins and these can have extinction coefficients of tens to hundreds of thousands in units of $\pu{dm^3mol^{-1}cm^{-1}}$.
A  concentrated solution of a dye  at, say, $1$ molar with a typical extinction coefficient of $10^4$ would have an optical density of $4$ in a $1$ cm cell or a reciprocal length of $10^6$ $\pu{m^{-1}}$ if $A=1$. Hope this helps.

may need a physical chemist: Does water really have strong EM absorption at 3 kHz in solid and 2 GHz in liquid? Why the huge shift?


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Altering Electronegativity of metals
Can the electronegativity of metals be changed by applying an electric potential?

Electronegativity is a property of the element. Can you transmute an element by applying an electric potential?

Electronegativity of an atom is due to the pull of electrons by the nucleus of atom.
Applying electric potential will not change the force the nucleus applies on the electron and hence it will not change the electronegativity.
By applying electric potential, the electron can be pulled out of the atom, not because it decreased the electronegativity but because it could apply enough force on the electron to overcome the electrostatic attraction by the nucleus and hence let it come out of the atom.
ps: If you had this doubt, then probably you are not clear with the actual concept of electronegativity. You might want to read up on that.


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What is an imaginary boundary?
We know that anything that separates the system with the surroundings is called a boundary.
It is said that a boundary can be either conducting or non-conducting, rigid or nonrigid, and real or imaginary.
What is meant by "imaginary boundary"?

Boundaries are defined with respect to the system in observation.
Imagine you're in a room with the door closed. Say, you are measuring the temperature of the room, i.e. the room is your system. In that case, it's easy to understand that the boundaries are real. In this case, the walls, the ceiling, the closed door comprises the boundary of the system.
Now, say somebody opens the door. Understand that this does not change the system; you are still interested in measuring the temperature of the room only. So, we can say that at the place of the closed door, there is an imaginary boundary. By specifying the boundary, you are now able to define your system more clearly.
This concept of imaginary boundary helps us to clearly define our system.

"System" refers to the area you're studying in regard to heat changes.
"Boundary" separates the system from the surrounding and, in the case of a system, say a glass of water, there is an imaginary boundary on top of the surface of water.
While you may also consider that the glass itself acts as a real boundary in the same example.

If you have a piece of equipment for which fluid is flowing in and out, you can define your control volume as the entire piece of equipment. This control volume has mostly rigid boundaries, but it also has boundaries through which fluid enters and exits.  The latter are called imaginary boundaries.


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Do the letters ol in a substance only indicate that the substance is an alcohol if they came at the end of the name of a substance?
Do the letters ol if they are together only indicate the substance is an alcohol if they come together at the end of the name of a substance? for example,is glycerol an alcohol because the letters o and l come together at the end of its name and olives and olive oil are not alcohols because the letters o and l do not come together at the end of any words in their names?

Yes, the ending ol in a name is a good hint that the compound contains an $\ce{OH}$ group.
However, beware of trivial names:

$\alpha$-tocopherol is not an aliphatic alcohol, but a substituted phenol
mannitol does not contain one $\ce{OH}$ group, but six of them

It's getting even more confusing if you consider other languages, such as German. Benzol is the German name of benzene, which has no $\ce{OH}$ group at all. The same is true for styrol, the German word for styrene ;-)

Gotcha but one thing I do want to get clarified-if the letters o and l are together at any spot in the name of a substance other than at the end of a word in that name,that does not mean that the substance is an alcohol,right (to use my previous examples,that means olives and olive oil are definitely not alcohols,right?)

Right :) Bisabolen is not an alcohol, geraniol and farnesol are.


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Why do we measure the initial rate of a reaction?
Why do scientists usually measure the initial rate of a reaction, instead of measuring the rate some time into the reaction?

Initially you know exactly how much of each compound you've got. Some time into the reaction, you don't know a thing.

Hi Ivan, why would the concentrations of each compound some time into the reaction be of significance to scientists?

Because we don't measure the rates just for fun. Instead, we want to derive the rate constants, and to do that, we need the rate and the concentrations.

Some may do this, but if you are serious you will work out a way to measure  the change in concentration (or something proportional to this) as a function of time across the whole time course of a reaction. As you might expect there are several ways of doing this depending on the speed of reaction.

Who says we can't determine the reaction rate some time after time zero?

As the reaction proceeds the concentration of reactants decreases and it may become difficult to measure the changes accurately.The reverse reaction can also produce errors in the rate measurement.Both these complications are absent at the initial stage of reaction.

First, note that one does not always measure initial rates. It is fairly common to fit an entire reaction curve (or nearly entire), but it requires a little bit more math. It is also common to measure an "initial" rate that is not quite initial. That is, it is the rate after a burst phase has occurred, and the system has settled into something close to a steady state.
That said, measuring initial rates is done for two practical reasons.
1) we can approximate the product concentration as 0. For a reaction that is at all reversible or product-inhibited, the math is much easier if [P]=0.
2) we can often approximate the concentration of one (or more) of the reactants as fixed. For example, if you measure the rate only for an amount of time required for a few percent of the reactant to be consumed, you can get a pretty good fit to the data by assuming that reactant is fixed at its initial concentration. That makes it easy to produce plots of the initial rate vs the concentration of that reactant.
Both of these factors make it much easier to determine basic rate laws and the approximate value of the associated rate constants.


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Oxidation state of Cu
If copper has 1 valence electron in the 4s subshell, how is it able to form Cu2+ with oxidation state of 2? I know the energy level difference between the 3d and 4s subshells has a part to play here, but I can't seem to put my finger on it.

What makes you think electron cannot leave from 3d orbitals?

lets just put it like this.....you wouldn't be too excited to be without your parents.....and wouldn't be really stabilised in their absence.....but put forward a condition..like you get 6-7 of your best friends to stay with you all the while they're away.....now that is good enough compensation for your parents being away for a period of time....similarly....water....surrounds the $\ce{Cu^2+}$ ion...and stabilises it through hydration...as $\ce{Cu+}$ has a much higher hydration enthalpy that $\ce{Cu+}$....it gets stabilised..and is ready to lose that last electron..though left configurationally unstable

What do you mean by "configurationally unstable"? Is that distinct from energetic stability?

what i referred to by "configurationally unstable" is that it lacks the fully filled d orbital configuration @JonathanSmith


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Why do the nitrosonium and nitronium ions act as electrophiles in organic reactions?
Why does nitrosonium and nitronium ion

act as electrophiles when each atom in them have a complete octet of electrons?
Even if we consider that they are electrophile just because of the fact that they have a formal positive charge on O or N then why don't $\ce{H3O+}$, $\ce{Li+}$, $\ce{Ca^2+}$, $\ce{Mg^2+}$, $\ce{NH4+}$, etc. also act as electrophiles they also have a positive centre and a complete octet of electrons?

TL;DR All the other species you mentioned also act as electrophiles, just not in the way you are thinking of.

Both $\ce{NO^+}$ and $\ce{NO2+}$ have empty $\pi^*$ orbitals, which are low-lying in energy. The reaction, in terms of frontier orbitals, then consists of the addition of the benzene $\pi$ system to the empty $\pi^*$ orbitals of the electrophiles.
The lowest energy unoccupied orbital in $\ce{H3O+}$ would be the $\sigma^*$ orbital. You might ask, why can't it react as an electrophile via the $\sigma^*$? Well, it does. The $\sigma^*$ orbital is largely centred on the hydrogen atom, so that's where the benzene $\pi$ system would attack.
Pushing electrons into the $\ce{O-H}$ $\sigma^*$ orbital would lead to cleavage of the $\ce{O-H}$ bond. How might one draw this with curly arrows?

Yes, it's an acid-base reaction. (Obviously, this only happens to a tiny extent, but the equilibrium position is not relevant to the discussion.) However, that doesn't stop it from also being a reaction between an electrophile and a nucleophile.
At the end of the day, what's the difference between $\ce{NO2+}$ and $\ce{NH4+}$? It is that the LUMO of $\ce{NO2+}$ is centred on nitrogen, and therefore attack of the benzene ring occurs at nitrogen to form a $\ce{C-N}$ bond. The LUMO of $\ce{NH4+}$ is centred on hydrogen, and therefore attack of the benzene ring would occur at hydrogen to form a $\ce{C-H}$ bond.
Good old valence bond theory is slightly inadequate in explaining this, since there is no notion of an antibonding orbital within VBT. It is still possible to rationalise it by, for example, saying that the $\ce{N-O}$ bond in $\ce{NO+}$ is polarised towards oxygen, and that nitrogen is therefore the electrophilic site within the molecule. [Note that the Lewis structure, which places a formal positive charge on oxygen, is not necessarily representative of the true charge distribution.] The same applies for the $\ce{N-H}$ bonds in $\ce{NH4+}$, which are polarised towards nitrogen, and therefore the electrophilic atoms are the hydrogens.
As for the metal cations you mentioned, why can't they be electrophiles? They jolly well can be electrophiles.

So what if N is electrophilic site in $NO_2$ and H in NH_4^+$

Nitrogen and oxygen are more electronegative atoms. Positive charge on them is much less stable than on Li, Mg and Ca.
Electrophiles must also have a vacant orbital to accept lone pair of electrons from the nucleophile. Oxygen in hydronium ion or nitrogen in ammonium ion has no vacant orbital. They can simply lose a proton and become stable.

Neither does n or o in nitrosonium or nitronium have a vacant orbital

@RaghavSingal That is completely not true, read at least about mesomeric structures, or better MOs.