text:
Are Tellurium Fluoride or Antimony Fluoride Ionic?
I would like to know if any of the tellurium fluorides or antimony fluorides are ionic.  I am aware that the difference in electronegativity is greater than the 1.665 cutoff indicating ionic bonds.  Melting point and reactivity seem to somewhat support that tellurium(iv) fluoride is ionic but this is not proof. Being an unusual case of an ionic bond between nonmetallic atoms only though, I would like to know if anything empirical (e.g. spectroscopic) has been observed to support this.

Which ones? Electronegativity and more importantly degree of covalency depend on oxidation state

https://en.wikipedia.org/wiki/Tellurium_tetrafluoride - it melts at about 130C  and is coordination polymer.

TeF4 is polymeric and low-melting.  Not ionic.  Covalent.
In general, the elements that are metalloids (on the "border" of metal and non-metal) like Te and Sb, tend to behave chemically like nonmetals (looking at their compounds).   But physically as metals (shiny, malleable, conductive).  So that's another good clue that these elements would be form covalent compounds (molecules or polymers ["network solids"]) with fluorine.
Finally even for hard core transition or p group metals, you will find that the lower oxidation count species tend to be ionic.  But as you get to tetravalent species a fair amount are molecular or polymeric (i.e. covalent) and few are ionic.  No penta or higher flourides are ionic.  And the few tetravalent ionic compounds tend to be pretty far to the left of the table (e.g. Zr)...pretty hard/electropositive atoms, not soft (i.e. polarizable) or less electropositive like those in the p group.  (To which Te and Sb are even more marginal metals.)
The Wikipedia article on Compounds of Fluorine has some good details on the concepts and good references to review articles (and Greenwood/Earnshaw Chemistry of the Elements).
https://en.wikipedia.org/wiki/Compounds_of_fluorine#Nonmetal_fluorides
P.s. Note also that it is important to specify which species of fluoride you mean (for elements below the second row).  Te forms a tetra and hexa fluoride (and some more complicated variants)--you did specific tetra, but that's not the only choice.  Also for Sb, you didn't specify which compound and both the tri and penta fluoride are well known.


text:
S-N bond IR vibrations
I am working on organosulfur compounds where i have S-N bonding, according to literature it might be having an ir vibration of 1100cm-1, could anyone confirm it with reference. (the complete bonding will be like S-S-N-C)
Thanks

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Why weren't the moles of oxygen calculated the same way as C?

Combustion analysis of a $\rm1.500~g$ sample of ascorbic acid yields $4.023\rm~g$ of $\ce{CO2}$ and $\rm0.96~g$ of $\ce{H2O}$. What is the empirical formula of ascorbic acid?

Actually my question is that in many solutions I saw it's solved like that:
In $\ce{CO2}$ the moles of C is calculated as $$\frac{4.023}{44.02}=0.9139\rm~mass=1.0976~g$$
and the $\ce{H2O}$ is solved with getting out the H moles with same way that C was solved and the mass is
$$\rm m~H=0.10761~g$$
 the mass of O will be $$\rm 1.500-(0.10761~g-1.0976~g)=0.29489~g~moles=0.1842$$
My question is why didnt we use the same procedure we did with H and C to compute the moles with O? Why did we make it in the last step?

Actually guys, let me edit this. Since it's a homework question with some effort and solution.

@IͶΔ Well, it was in dire need of editing ;)

You have done combustion analysis, so you have introduced a large amount of oxygen to the ascorbic acid sample.  Because of this, you can't know how much of the oxygen in the $\ce{CO2}$ and $\ce{H2O}$ you produced actually came from the acid sample and how much came from the atmosphere.
But you do know that the mass of your ascorbic acid is the mass of all the carbon plus the mass of all the hydrogen, which you calculated, plus the mass of oxygen.  The difference in mass between what you have calculated and what you started with is the mass of oxygen that was in the ascorbic acid.
If there were any other elements in your sample, such as sulphur or nitrogen, those amounts would also have to be calculated before finally working out how much mass you haven't accounted for which must have been oxygen.

I really appreciate your helping and very thankful , but in fact I used the fact that like 0.25 mol of AgCl means 0.25 mol of Ag and 0.25 mol of Cl and it also doesn't work. are my calculations wrong you think or my principle is typically wrong?

Where has the AgCl come from? Seems like a completely different question.

Im so stupid sorry because of my confusion I just said without indicting that its a different question .yah its different,actually its the question posted here chemistry.stackexchange.com/questions/13751/… – I thought if I used the mass og AgCl to have its moles and the say Cl has 2.103/143.3=0.0147 mol ^is this right?doing the same way that C moles have in previous question?.Im really sorry to mix the thing up but I have a crucial exam tomorrow.

I think you are correct.  because the ratio of Ag:Cl is 1:1, 0.0147 moles of AgCl means 0.0147 moles of Ag+ and the same of Cl-, just as in this question 0.91 moles of CO2 means 0.91 moles of carbon atoms.

You reference to the AgCl question, is answered using a different method. The final result mass of Cl should be the same if you use the method in this question, but the intermediate steps will be completely different so you can't compare your progress.

I just would say that I dunno how to thank you , you really helped me and Im very very thankful and grateful.^ignore the bad English Im not native :(

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Is there any way to carry out voltammetry without a potentiostat?
I'm trying to do a voltammetry experiment, but my high school doesn't want to have to buy a potentiostat (understandably, I know they are very expensive). Is there any way to carry out a voltammetry experiment, either linear sweep or cyclic, without a potentiostat? Or am I doomed to have to go back to the drawing board...
I should mention that the physics lab has some equipment like variable power sources, but not much--my school kind of under-appreciates science--and I was planning on creating a cruddy makeshift three-electrode cell.

Are you good with op-amps?   Do you have access platinum, or some other type of highly inert electrodes?

No I don't really have access to those things, but thanks.

For good science projects keep it simple. A very prominent scientist once told me if an experiment cannot be explained in a simple manner, then it is not worth doing. I have seen this same scientist turn down countless requests for equipment, so making do with what you have is part of scientific research. Three electrode experiments do form the basis of many chemical sensing and corrosion resistance technologies, but these experiments are tough to get to work, especially on the budget of a high school lab. Best of luck on the project.

Thanks for the help, but I've already moved back to brainstorming. I'm not an experienced user here so I don't know how to close a question, but if a moderator could/should(?) do that, then double thanks.

This is a valid question. This project is entirely possible, but I would not advise it for a high school research project based on my experience building these. Much better projects are available.
A nonreactive electrode, for example gold or platinum, is required for good results.
Knowledge of op-amp design, or some other feedback control system, which is taught in advanced electronics courses is needed.  The control system fixes the potential between the the reference and the working electrode by drawing current from the third electrode.  The current drawn is measured to give some information about the system.
The Wikipedia article shows the circuit need, https://en.wikipedia.org/wiki/Potentiostat.
The open source cheapstat is available: http://web.chem.ucsb.edu/~kwp/cheapstat/
Integrated versions of the circuit are available. http://www.ti.com/product/LMP91000.  Evaluation kit ~$100 http://www.ti.com/tool/LMP91000EVM.
Information on making electrodes: https://publiclab.org/notes/JSummers/01-09-2014/potentiostat-notes-5-how-to-make-low-cost-electrodes
It is possible to build a potentiostat without advanced knowledge of electronics; however, this runs into costs. Getting experiments done with the equipment, resources, and funding available is something every scientists faces.

Maybe you can raise $100 for CheapStat, http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0023783
It does not solve the electrode problem, but at least you can try something

Edited answer based on comment from ssavec.

@ssavec, I can't find the article I saw a while ago, but I remember seeing a way of making a reference electrode with a sieve stuck to the bottom of a glass tube + a silver wire (silver is not too expensive). Counter electrode could be graphite, and working electrode could also be graphite.

@Sean_Doris Is this the article?  https://publiclab.org/notes/JSummers/01-09-2014/potentiostat-notes-5-how-to-make-low-cost-electrodes

@Agriculturist, that's the one!


text:
Semipinacol-type rearrangement leading to ring expansion

What will be the major product in the following reaction?

I tried to work out the mechanism, which starts with protonation of the alkene, and I got C to be the answer (which is the answer in the book as well). But, I've heard that three-membered rings generally don't expand due to their exceptional stability. Is this true? And between the alkene and alcohol functional groups, which is more likely to get protonated?

Try clicking on the link. I have tried to upload it from the SE app. Anyway here is the link http://postimg.org/image/izfeyg5uj/

How are 3-membered rings exceptionally stable? If anything, they are unstable because of torsional & angle strain... they would love to expand. This reaction is a semipinacol rearrangement, you protonate the alkene, and then push arrows from the OH with the alkyl migration to expand the ring

Btw, that non-linear allene (I directly copied it from your original image) triggers me...

@orthocresol The orbitals bend for better overlap.I was taught this. Although I may be confusing it with the stability of the cyclopropyl methyl cation. Correct me if I am wrong.

In general, as far as I know, the cyclopropyl methyl cation is very stable, and the final products of a reaction with a nucleophile have nearly equal amounts of the nucleophile attacking the cyclopropyl methyl cation and the cyclobutyl cation.However it may not be very easy to say what might happen in this case, as the final product itself involves an sp2 carbon in the ring.

can anybody share some insight on the alkene vs alcohol problem?

Three-membered rings are not exceptionally stable, on the contrary, they suffer
a lot of ring strain and readily take part in ring-opening reactions.
Exceptions include the Hückel-aromatic cyclopropenyl cation, and the
resonance-stabilized cyclopropylmethyl cation.
Your assumption that the alkene is protonated is correct, as this generates a
substituted cyclopropylethyl cation. The regioselectivity of the protonation occurs to give the more stable secondary carbocation instead of a a primary carbocation. A 1,2-alkyl shift then occurs, which is in this case assisted by the lone pair on oxygen. Deprotonation of the carbonyl then yields the four-membered ring ketone C (2-methylcyclobutanone). [The carbons are numbered in blue to make it easier to see what's going on: the C1-C3 bond is broken and C3-C4 bond is formed.]

This is similar to the semipinacol rearrangement; in this case, the carbocation is generated via protonation of an alkene instead of by loss of a leaving group.
The formation of products A or B would require protonation of the alcohol group, followed by loss of water in an E1 reaction. In either case, the final deprotonation yields a three-membered ring with an endocyclic or exocyclic double bond, which is more strained than a saturated three-membered ring. This reaction pathway is therefore less favorable than ring expansion. Protonation of the alcohol can certainly happen, but since it does not lead to the formation of any stable products, it is said to be unproductive.


text:
Rice Water for hair
I have been leaving fermented rice water on my hair after shampooing and conditioning without rinsing it out. I can't find any websites that say you can do this, they say to rinse out after applying.  I love the way my hair feels when it dries,  but I'm not sure if I'm doing any harm to my hair like weakening it or maybe the fermented rice water could lighten my color treated hair.   I just don't know what is in rice water.   I hope it's okay because as I said, I love how it feels but there may be something in it that could do a lot of damage if I don't rinse it out.   What say you?  Thanks for any help you can give.

I'm voting to close this question as off-topic because I'm not comfortable giving advice that involves personal grooming.

So, the short- and long-term effects of various substances on hair is a complicated area and one I'm not really equipped to advise on (and this site generally doesn't and shouldn't advise on health issues).
However, I did manage to find, among all the hip information-free websites, a break-down from a study on the effects of rice water on hair with the various components of unfermented rice water (other than water, obviously).

16% proteins
10% triglycerides (i.e. ordinary oils/fats)
10% other lipids (probably slightly less common oils/fats)
13% phytic acid (a chemical found in lots of everyday foods)
4% inositol (a sweet sugar alcohol and useful but not essential nutrient)
9% starch
17% other carbohydrates
21% other inorganic substances (esp. ash)

From: The effect of rinse water obtained from the washing of rice (YU-SU-RU) as a hair treatment by S. Inamasu, R.Ikuyama, Y. Fujisaki and K.-I. Sugimoto: Beauty Care Laboratory, Kracie Home Products, Ltd., 134, Goudo-cho, Hodogaya-ku, Yokohama 240-0005, Japan (Original in Japanese, translated abstract here.)
Fermentation will likely break down some of those carbohydrates into ethanol and methanol, but probably not all of them.


text:
Does humidity increase oxygen levels in the air we breathe?
I'm testing a new form of meditation and it requires certain breathing exercises in order to generate a specific effect on the nervous system; I have noted significant differences in the results being generated based on the setting in which i practice the exercises... here is some data, hopefully you can help me put all of these pieces together, id like to base this off of solid data so i can present my findings for further evaluation.
QUESTIONS:Why do the effects of the technique increase in different settings?
Am I breathing in more O2 based on the environment i do the exercise?
Does high humidity increase the levels of oxygen in the air?
Or are other factors such as temperature & pressure contributing to the effects?
SETTING:
Steam Room:
- 100% humidity, temperature
- HIGH heat
- Pressurized steam room
Normal Room Temp Setting:
- "Normal" Dry Colorado Air (Assumed approx 78% nitrogen, 21%o2, 1% misc)
- Normal home setting
-the "effects" are intensified when performed in the sauna as compared to normal, dry Colorado air, i am trying to pin point why this might be.
Hopefully this is clear, thanks.

I am not trying to say that meditation is a scam or whatnot, but I think that chemistry cannot explain how meditation works (which is likely to be psychological in nature) nor can it answer your question.

I'm voting to close this question as off-topic because ^

I can't vote for closing things, but the question is asking something related to chemistry 'Does humidity increase oxygen levels in the air we breathe?'. Yes, the asker did give a lot of context about meditation, but the basic premise of the question is about chemistry, not psychological or off topic. @orthocresol

As to the answer in the title, no, increased humidity does not increase oxygen levels in the air.  In fact it does the exact opposite, since the air in each breath will have some fraction displaced by water molecules instead of the normal nitrogen and oxygen mix we breathe.  It's a relatively small effect, with water making up at most a few percent of the air under high humidity/high temperature conditions that are safe for people to inhabit for reasonable periods of time.  That would mean that oxygen content would be decreased by about 1% in your steam room (again, this assumes you're not doing something dangerous with the temperatures and humidity.)
As to your other questions, you've got serious methodological issues to deal with.  Your first goal in finding the reason for the significant differences you've noticed should be verifying whether those significant differences actually exist, or whether they are based on things like confirmation bias or the simple effects of being in a boiling hot room vs outdoors in Colorado.  How you do that is really more of a psychology experiment, as has been suggested by the comments above, and would be better suited for the Cognitive Sciences Stack Exchange.

I was simply trying to identify a few variables that exist in my experiment; i needed to start with finding the answer to the question regarding oxygen content in humid conditions & now that i have it, i can proceed with the rest of the experiment. The comments on meditation were obviously not appointed to the correct party, moving forward, ill be sure not to mention anything outside of the table of elements to this forum!

@JamesJimenez Understood.  I was hoping that a straightforward, serious answer with a pointer toward the area of SE that would better suit the other points you mentioned would help keep your question open.


text:
Does Methyl Iodide (CH3I) react with water (H2O) and if it does what would be its byproducts?
Does Methyl Iodide ($\ce{CH3I}$) react with water ($\ce{H2O}$) and if it does what would be its byproducts?
A project I am designing requires a reaction with methyl iodide and a polar compound, so I was wondering if methyl iodide reacts with water.

Methyl iodide is hydrolyzed by water in an $\ce{S_{N}2}$ reaction ($\ce{S_{N}1}$ is disfavored by the instability of the primary methyl cation), which yields methanol and hydroiodic acid as products.
$$\ce{CH3I + H2O <=> CH3OH2^+ + I-}$$
$$\ce{CH3OH2^+ + H2O <=> H3O+ + CH3OH}$$
A possible side reaction is the formation of iodine and methane by the redox reaction
$$\ce{CH3I + H3O+ + I- <=> CH4 + I2 + H2O}$$
However, this reaction becomes only important when a large concentration of methanol pushes the hydrolysis equilibrium to the educt side. It is negligible in pure aqueous solution (reference).

In common condition, it reacts very slowly and will generate HI and CH3OH. It's soluble. If you don't heat in reaction, I think it's OK. It's very very toxic so take care..


text:
What could cause one brand of carbonated water to be superiorly carbonated compared to another?
I buy a lot of carbonated water, its pretty much all I drink. There is something I've noticed that just made me curious though.
The usual brand I buy is the cheapest one available, 14 cents per 1,5l. This one almost always tries to give me a surprise shower when I open it, no matter how still I keep it prior to opening. The bubbles are large and plentiful at first but it goes flat pretty quickly, usually within the day but thats fine since I'll drink it in a day anyways.
The other brand I buy when the first one isn't available, its the second cheapest one and is 35 cents per 1.5l. Opens up pretty calmly, the bubbles are much smaller but lots of them around. The interesting bit is that it doesn't go flat nearly as quickly and I can expect some carbonation to be left even a few days later. Another point to make is that the second one behaves much more like various other brands.
The question is, what could cause such difference in the bubbling behavior?
The bottles are the same kind of clear plastic, the ingredients aren't listed which should mean that its just water + $\ce{CO2}$. Over here, if they put any other chemical in it, they are required to name it on the bottle, many water bottles have chlorine in them for example.
Could it be the carbonation process? Or the natural water composition? Maybe the transportation or storage?

Are the dates of manufacture the same for the bottles you compared?

They don't have a manufacturing date but the "best before" date is similarly ~11-12 months from the date of purchase

Most of the carbon dioxide is present as dissolved gas. I'd guess that the cheaper bottle that fizzes easily has a much higher concentration of particulates which become nucleation sites.

Carbonated water is supersaturated in carbon dioxide, and the only thing preventing all that excess $\ce{CO2}$ from bursting out all at once is the lack of nucleation sites for $\ce{CO2}$ bubbles to form. These nucleation sites can be almost any solid (including the sides of the container to some degree) from undissolved salts to soil particles or tiny particles of most anything.
The most likely reason for the formation of such a copious amount of bubbles forming in the cheap water is the presence of a large amount of undissolved solids relative to the more expensive waters. Depending on the product, one step of the processing of carbonated water is filtering in order to make the product appear clearer and more sparkly (and maybe to prevent rapid formation of $\ce{CO2}$ bubbles). Of course this process costs money, and it is likely that the cheaper brand either skips the filtration process or uses a much less efficient (and cheaper) process.


text:
diH2O, dH2O, and DI H2O. What do they mean?
What do $\ce{diH_2O, dH_2O}$, and $\ce{DI H_2O}$ mean?
I'm not asking for a description of deionized water and distilled water. I'm asking what the three abbreviations/formulas mean.
In other words, I'm asking someone to fill in the blanks below.

$\ce{diH_2O = \_\_\_\_\_}$
$\ce{dH_2O = \_\_\_\_\_}$
$\ce{DI H_2O = \_\_\_\_\_}$

diH2O = deionized water**
dH2O = distilled water
DI H2O = deionized water**

Thanks @CurtF. I found a book that referred to diH2O as distilled, not deionized water. Is there no commonly accepted standard, or is this just a fluke reference I found? https://books.google.com/books?id=KZateS3byQgC&pg=PA8&lpg=PA8&dq=distilled+water+%22dih2o%22&source=bl&ots=bj1HcMCZY-&sig=p10fa2fkSHtIYrdpCJPJorc_XXM&hl=en&sa=X&ved=0ahUKEwja2azhgv7KAhVMxGMKHRp1AcYQ6AEIIjAB#v=onepage&q=distilled%20water%20%22dih2o%22&f=false

I don't think there is a commonly accepted standard.  If I had to guess, I'd say more often than not di$\ce{H2O}$ is used to mean deionized water, but I've nothing but my own experience to base that on.  Sorry for the lack of references.  Also, ddH2O is a common abbreviation that used to mean "double-distilled water" but now is used to refer to highly pure water produced by laboratory purification systems such as those from milliQ or similar.

As Curt F. indicates, what they stand for and what they are in practice are two different things. In practice, the four abbreviations "mean" the same thing: high purity water. If there's some practical difference between distilled, double-distilled and deionized water that means one is suitable and the other is not, you'll normally get a more extensive description of the minimal performance and purity characteristics necessary for the application.

Hello community. Can we not remove the "In other words" part of my question? I know it is redundant. I'd like to state it twice in slightly different ways. Sorry if that irks those out there who like conciseness.

OK. this has become a challenge. Rolling back for the fourth time. I'll eventually wear out.

I have rolled this back and locked it for the time being.  Rather than taking up people's time in this fashion, this would have better been addressed on Meta rather than as a "war."  The bottom line is that it's much better to keep questions more conceptual rather than about specific homework examples.

A meta discussion about the edits to this question can be found here: http://meta.chemistry.stackexchange.com/q/3467/7951

I don't think there is a commonly accepted standard. If I had to guess, I'd say more often than not $\mathrm{di}\ce{H2O}$ is used to mean deionized water, but I've nothing but my own experience to base that on. Sorry for the lack of references. Also, $\mathrm{dd}\ce{H2O}$ is a common abbreviation that used to mean "double-distilled water" but now is used to refer to highly pure water produced by laboratory purification systems such as those from Milli-Q or similar.

$\mathrm{DI}\ce{H2O}$ = deionized water
$\mathrm{di}\ce{H2O}$ = deionized water

Such water is often available from centralized water purification systems in buildings that host many chemical laboratories.  It isn't as pure as "$\mathrm{dd}\ce{H2O}$" (see below) but is subtanstially more pure than tap water.

$\mathrm{d}\ce{H2O}$ = distilled water
$\mathrm{dd}\ce{H2O}$ = doubly distilled water
$\mathrm{mq}\ce{H2O}$ = Milli-Q water

The brand name of a popular type of water purifier that gives purity similar to $\mathrm{dd}\ce{H2O}$


text:
Balancing disproportionation reactions
How to balance the following disproportionation reaction
 $$\ce {XeF2 + H2O → Xe + XeO3 + HF + O2} $$
 My attempt: I tried by making total number of reductions equal to total number of oxidation but I couldn't do because Xe is both oxidised and reduced also the solvent is water so what should I do to balance  protons and oxygen on both sides?.  Please help by solving it on the basis of same procedure that is making oxidation equal to reduction

since acid is in the product side balancing should be done as in acidic medium

It can't be balanced unambiguously, for this is not a single reaction, but a mechanical sum of two reactions, and these can be formally mixed in arbitrary proportion.
The first reaction is indeed a disproportionation. To balance it, use the common trick: consider these two xenons different elements for a while.
$$\ce{\underbrace{XeF2}_{for\;reduction} + \underbrace{XeF2}_{for\;oxidation} +H2O->Xe + XeO3 + HF}$$
The second is a typical redox:
$$\ce{XeF2 + H2O -> Xe + O2 + HF}$$

Nertin how do you know that it is a sum of two reactions? Could you please suggest any good textbook for it

This is a sum of two reactions because it has two independent oxidized products: $\ce{XeO3}$ and $\ce{O2}$. As for the book, I don't think there is one; the matter is too trivial to deserve a whole book on its own.

Nertin how did you figure out what reactants and products would be in each reaction?

Why, the OP specified them all in the original question. I just figured that the oxidized products are independent, so there would be only one of them in each reaction.

The original equation cannot be solved using mathematics because it gives the coefficient of $\ce{XeO3}$ zero. The reason for that is exact reason given by Ivan Neretin (see else where). Yet, I'd like to show a way of solving a balancing problem of true chemical equation by using mathematics. First, assume the compound is $\ce{XeF4}$, which undergoes disproportion in water. The chemical equation for that disproportion can be written as:
$$\ce{a XeF4 + b H2O  -> c Xe + d XeO3 + e HF + f O2  \tag 1}$$
So the atom balance of equation $(1)$ yields:
$$\text{Balancing } \ce{Xe}: \quad a = c + d     \tag 2$$
$$\text{Balancing } \ce{F}: \quad 4a = e     \tag 3$$
$$\text{Balancing } \ce{H}: \quad 2b = e     \tag 4$$
$$\text{Balancing } \ce{O}: \quad b = 3d + 2f     \tag 5$$
Now you have four equations relate to the six unknown coefficients $a$ through $f$, each coefficient is in general, an integer greater than zero. Yet, keep in mind that, a fraction of coefficients are allowed in chemistry if at least one coefficient is an integral. Since there are only 4 equations for 6 variables, we can give an arbitrary value for at least one variable. Since $\ce{H}$ and $\ce{F}$ are appeared in only one compound in either side, we can give our arbitrary value to either $b$ or $e$, respectively. I'd prefer to give $e = 4$ for the sake of argument. Hence, we can proceed as follows:
Assuming $e = 4$ gives $a = 1$ from the equation $(3)$ and $b = 2$ from the equation $(4)$. However this makes mathematics complicated because in right hand side of equation $(1)$, there are two $\ce{Xe}$ atoms and that makes $c$ and $d$ fractions (since $a = c + d$). Thus, we assume $e = 8$, doubling the original assumption. Then, from equation $(3)$, $a = \frac14e =\frac14 \times 8 = 2$. Similarly, from equation $(4)$, $b = \frac12e =\frac12 \times 8 = 4$.
Applying $a = 2$ and $b = 4$ in equations $(1)$ and $(5)$ respectively give:
$$2 = c + d     \tag 6$$
$$4 = 3d + 2f     \tag 7$$
By subtracting the equation $(7)$ from the equation $(6) \times 3$ gives you:
$$4 = 3c - 2f = 2    \tag 8$$
The smallest values to satisfy the equation $(8)$ are $c = 1$ and $f = \frac12$, which is allowed since other coefficients are integers. From the equation $(2)$, $d = a - c = 2 - 1 = 1$. From the equation $(4)$, $e = 2b = 8$. Hence, all numerical values of coefficients are: $a = 2$, $b = 4$, $c = 1$, $d = 1$, $e = 8$, and $f = \frac12$. Therefore, the equation would be:
$$\ce{2 XeF4 + 4 H2O  -> 1 Xe + 1 XeO3 + 8 HF + 1/2 O2  \tag 9}$$
If you multiply whole equation $(9)$ by 2, it becomes:
$$\ce{4 XeF4 + 8 H2O  -> 2 Xe + 2 XeO3  + 16 HF  + 1 O2}  \tag {10}$$
The equation $(10)$ is the correct balance equation, using mathematics.
The given equation in the question is not true redox equation. Yet, it is a combination of two redox reactions, one of which is a disproportionation reaction (of $\ce{XeF2}$) while the other is regular oxidation-reduction of two compounds $\ce{XeF2}$ and $\ce{H2O}$. This complex mixture can be solved by step-by-step analysis:
As Ivan Neretin correctly pointed out, the given reaction is sum of two reactions: The first reaction is redox reaction where one molecule of $\ce{XeF2}$ is oxidized while another $\ce{XeF2}$  molecule is reduced. Let's look at oxidation first:
$$\ce{XeF2 <=> XeO3 + 2 F-}$$
Balance oxygen by $\ce{H2O}$, hydrogen by $\ce{H+}$, and charges by $\ce{e-}$:
$$\ce{XeF2 + 3H2O <=> XeO3 + 2 F- + 6H+ + 4e-} \tag{11}$$
The reduction of $\ce{XeF2}$ is:
$$\ce{XeF2 <=> Xe + 2 F-}$$
Just balance charges by $\ce{e-}$:
$$\ce{XeF2 + 2 e- <=> Xe + 2 F- } \tag{12}$$
Now add the equations $(11)$ and $(12)$ in order to cancel $e^-$s ($(11) + (12) \times 2$):
$$\ce{3XeF2 + 3H2O -> XeO3 + 2 Xe + 6 F- + 6H+} \tag{13}$$
or
$$\ce{3XeF2 + 3H2O -> XeO3 + 2 Xe + 6 HF} \tag{14}$$
This is the first redox reaction, now balanced. The second redox reaction, which is correctly given by Ivan Neretin:
$$\ce{XeF2 + H2O -> Xe + O2 + HF} \tag{15}$$
To balance this equation, follow above methodology:
$$\text{[Ox]}: \quad \ce{2H2O <=> O2 + 4H+ + 4e-}\tag{16}$$
$$\text{[Red]}: \quad \ce{XeF2 + 2e- <=> Xe + 2F- }\tag{12}$$
Add the equations $(12)$ and $(16)$ in order to cancel $e^-$s ($(16) + (12) \times 2$):
$$\ce{2XeF2 + 2H2O -> 2Xe + O2 + 4H+ + 4F-} \tag{17}$$
or
$$\ce{2XeF2 + 2H2O -> 2Xe + O2 + 4HF} \tag{18}$$
Now, these two equations, $(14)$ and $(18)$ can be add together in any proportion to get the sought equation of OP's. For example, add one to one proportion gives:
$$\ce{5XeF2 + 5H2O -> XeO3 + 4 Xe + O2 + 10 HF} \tag{19}$$
This is one of balanced equation in the equation given in OP's question. There are many of them can be written by different combinations of  equations $(14)$ and $(18)$.

Excellent! The circle is now complete! ;)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Concentration of NaIO3 using hypo

$$\ce{2NaIO3 + 5NaHSO3 -> 3NaHSO4 + H2O + I2 + Na2SO4}$$
  $\mathrm{20ml}$ of $\ce{NaIO3}$ solution reacted with excess $\ce{NaHSO3}$ releasing $\ce{I2}$ that can oxidize $\mathrm{20ml}$ of $\mathrm{0.1M}$ hypo. What is the molarity of $\ce{NaIO3}$

The reaction of iodine with hypo is
$$\ce{I2 + 2Na2S2O3 -> 2NaI + Na2S4O6}$$
moles of hypo used is $\mathrm{20\times 10^{-3}\times 0.1=20\times 10^{-4}}$
Therefore moles of $\ce{I2}$ reacted is $\mathrm{10\times 10^{-4}}$
Hence, moles of $\ce{NaIO3}$ reacted is $\mathrm{20\times 10^{-4}}$
Therefore concentration of $\ce{NaIO3}$ is $\mathrm{\frac{20\times 10^{-4}}{20\times 10^{-3}}}=\mathrm{0.1M}$
But answer given is $\mathrm{0.02M}$


text:
Why and how do we get NH4NO3 when reacting very dilute HNO3 with Zn?
From Russian Wikipedia's article on nitric acid:

It's clear that as we decrease the concentration of nitric acid, we get more and more reduced nitrogen: in $\ce{NO2}$, the oxidation state of N is ($\mathrm{+4}$), then it goes to zero in $\ce{N2}$ (when we use $\mathrm{10\%}$ strong $\ce{HNO3}$).
But why all of a sudden we get $\ce{NH4NO3}$? In $\ce{NH4}$, the oxidation of $\ce{N}$ is ($\mathrm{-3}$), in $\ce{NO3}$, the oxidation state is ($\mathrm{+5}$). So the overall oxidation increases by $\mathrm{2}$.
I wonder how this ammonium nitrate is generated, i.e. where the ammonium ion evolves from.

P.S. Got an answer on Socratic.

Ammonium ion is generated from $\ce{HNO3}$ in full accordance with the previous trend which you correctly summarized as follows: as we decrease the concentration of nitric acid, we get more and more reduced nitrogen.

Very well you got answer for your question. Here, best thing is that you can answer your own question. So, feel free to write that here.


text:
Numbering and naming a disubstituted cyclohexene
I saw this cyclohexene in my textbook and wonder why it’s like that. I searched the internet and it’s correct but why?
As I know we try to keep the numbers low, so instead of 1,6-dichlorocyclohexene we can tell 2,3-dichlorocyclohexene and the numbers are lower. I appreciate your help.

@IvanNeretin is it the same for 3-brumo 1,4,6 cyclohepta tri en and 7-brumo 1,3,5 cyclohepta tri en? How can i determine such things in my exams actually ? Im a little bit confused


text:
Detection of chloride anion

In a sample solution, containing $\ce{Cl-}$, while conducting the test for detection of $\ce{Cl-}$ ion first it is acidified with conc. $\ce{HNO3}$ then cooled and then $\ce{AgNO3}$ solution is added. A curdy white precipitate appears which is soluble in $\ce{NH4OH}$ but reappears on adding $\ce{HNO3}$.

My question: is the $\ce{HNO3}$ added dilute or concentrated?

Could be either... In fact if the salt were say sodium chloride, then HNO3 (both dil and conc) would give HCl, which then reacts with silver nitrate to form silver chloride.

HCl + AgNO3 -> AgCl + HNO3? So the nitric acid is regenrated?

The question clearly states the HNO3 is conc.  The reason to add a few drops of conc. nitric acid is to remove any other anions contaminating the sample that might give a false positive to the silver nitrate test.

To best understand, let us consider the reactions occurring.
First, we have $\ce{Cl-}$ ions in solution. Acidifying the solution serves to remove other anions that might give a false positive for the chloride test. Upon addition of $\ce{AgNO3}$, the following reaction occurs:
$$\ce{Ag+(aq) +Cl-(aq) -> AgCl (s)}$$
When enough ammonia is added to neutralize the $\ce{H3O+}$ in solution:
$$\ce{NH3(aq) +H3O+(aq) -> NH4+(aq) + H2O(l)}$$
The ammonia can then begin to form a metal complex with the $\ce{AgCl}$ that is soluble in water:
$$\ce{AgCl(s) +2NH3(aq)->Ag(NH3)2+(aq) +Cl-(aq)}$$
When $\ce{HNO3}$ is reintroduced, the ammonia ligand is protonated because it is more basic than water, and the metal complex is destroyed, reforming $\ce{AgCl}$:
$$\ce{Ag(NH3)2+(aq) +2H3O+(aq) +Cl-(aq) ->AgCl(s) +2NH4+(aq) +2H2O(l)}$$
Since you will likely be working with relatively small amounts, it is likely that you will be using concentrated acid in stoichiometric amounts to produce these products.

I'd quibble with notion that the sentence "Acidifying the solution serves to remove other anions that might give a false positive for the chloride test." Since the acid is added to the same solution, no other anions are removed, with perhaps the exception of carbonate. The solution needs to be acidic so that silver hydroxide doesn't form when the silver nitrate is added.

I can't find any indication that this reaction would occur. Silver hydroxide appears to be too unstable a compound to even form.

You're right. It isn't silver hydroxide, but rather silver oxide that forms in alkali solutions. https://en.wikipedia.org/wiki/Silver_oxide#Preparation

@ringo What is the source of NH3 here? Is it NH4OH? But it gives NH4+.

$\ce{NH4OH}$ is just another way of saying $\ce{NH3(aq)}$.


text:
Does the volume of solution increase during a neutralisation reaction?
Say I mixed together 20 mL of HCl solution with 20 mL of NaOH solution.
$$\ce{HCl (aq) + NaOH (aq) -> NaCl + H2O}$$
Strictly speaking, would the water produced from the neutralisation process cause the volume of the overall mixture to slightly increase? So, would the volume of the final mixture (strictly speaking) be 40 mL + the volume of water produced from the reaction?

Yes but, a quick calculation shows that using 0.1M solutions, which is relatively strong for many applications, the amount of water produced would be 0.036g of water, which is less than 0.1% of the initial volume.  This is in the same order as the thermal expansion of about 10 degree temperature rise at roughly room temperature. Does that context help?

@spontification why not just write an answer? :)

The volume may change slightly, but not for that reason. After all, these water molecules did not appear out of thin air. Instead, they formed from some particles ($\ce{H+}$ and $\ce{OH-}$) that were in the the initial solutions and certainly did occupy some volume there.

a good question to me, I think I felt I would be making more out of the question than is actually there. Strictly speaking the answer to the OP is "yes" but my comment was "practically speaking this is almost always irrelevant".

Let consider an example:
According to one supplier, at 20 degrees Celsius, $1$ Molar $\ce{HCl}$ has a density of $1.010\ g / cm^3$
According to another supplier, at 20 degrees Celsius, $1$ Molar $\ce{NaOH}$ has a density of $1.04\ g / cm^3$
From Wikipedia (note the $1\ mg$ / $1$ part in 10000 error in mass conservation)
1 mole of $\ce{NaOH}$ weighs $39.997\ g$
1 mole of $\ce{HCl}$ weighs $36.46\ g$
1 mole of $\ce{NaCl}$ weighs $58.443\ g$
1 mole of $\ce{H2O}$ weighs $18.015\ g$
Mixing 1 litre of each solution to make 1 mole of $\ce{NaCl}$:
Going in:
Base:

$\ce{NaOH}$ : $39.997\ g$
$\ce{H20}$ : $1040 (\pm 10) - 39.997 = 1000 \pm 10g$

Acid:

$\ce{HCl}$ : $36.46\ g$
$\ce{H2O}$ : $1010 (\pm 1) - 36.46 = 974 \pm 1 g$

Result:

$\ce{NaCl}$ : $58.443\ g$
$\ce{H2O}$ : $1000 (\pm 10) + 974 (\pm 1)+ 18.105 = 1990 \pm 10 g$

The percentage by mass of $\ce{NaCl}$ is
$$
\frac{58.443}{58.443 + 1990 \pm 10} = 2.85 \pm 0.02 \%
$$
Feeding that percentage mass in to chemical engineering tables
the density is given as $1.0183 \pm 0.0002 \ g / cm^3$ at 20 degrees Celcius.
This is very similar to the $1.020\ g/cm^{-3}$ density given by a supplier of $0.5$ M $\ce{NaCl}$
This gives a final volume of
$$
\frac{2050 \pm 10}{1.0183 \pm 0.0002} = 2010 \pm 10 cm^3
$$
So this particular case is likely to be a very small amount larger in volume than the original two solutions, but this balance is going to depend on atmospheric pressure, temperature, and the specific acid and base used in the neutralization and their salt.

It is a pity you took a poor precision in the initial density of $\ce{NaOH 1 M}$. With your choice ($1.04 g/mL$}, the uncertainty is ±$1$% :  the final result has a large uncertainty : $2010 ± 10 cm^3$. So it looks as if there is no volume change. If you had opened the Handbook of Physics and Chemistry, you would have found the value $1.0433$ for the density of $\ce{NaOH 1 M}$. As a consequence, the final volume of mixture would have been $2010 ± 1 mL$. And indeed I have often made the mixture $\ce{HCl 1 M + NaOH 1M}$ in my classroom. I can state that the volume increase is quite visible.

Say I mixed together 20 mL of HCl solution with 20 mL of NaOH solution.
$$\ce{HCl (aq) + NaOH (aq) -> NaCl + H2O}$$

You are not telling us the concentrations of the two solutions, so the answer will be conceptual rather than quantitative.

Strictly speaking, would the water produced from the neutralisation process cause the volume of the overall mixture to slightly increase?

The water from the neutralisation would contribute to the total volume. As Ivan Neretin said in the comments, there would also be a loss of hydronium and hydroxide ions.

So, would the volume of the final mixture (strictly speaking) be 40 mL + the volume of water produced from the reaction?

Strictly speaking, any reaction in solution will change the volume of the solution (except when the reaction is "symmetrical" or by rare coincidence). The direction of change depends whether the volume-averaged density of the two components you are mixing is larger or smaller than the density of the final solution. If all components are dilute aqueous solutions, the volume change will be hard to detect because it is so small.


text:
Density of saturated KBr
Google search shows the density of saturated $\ce{KBr}$ is $\mathrm{3.119\ g/ml}$ at $\mathrm{25\ ^\circ C}$.
We have a solution of saturated $\ce{KBr}$ and density is about $\mathrm{1.5\ g/ml}$ when weighed on a calibrated analytical balance.
How do we account for this discrepancy?

This is a little off topic, but I just did a google search for 'saturated KBr density' and this question came up. So in a weird paradoxical way, a google search shows it as 3.119g/ml simply because you asked this question :D

According to “Aqueous Solubility of Inorganic Compounds at Various Temperatures”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the solubility (expressed as mass fraction) of potassium bromide $(\ce{KBr})$ in water is $39.4\ \%$ (at $T = 20\ \mathrm{^\circ C}$) and $40.4\ \%$ (at $T = 25\ \mathrm{^\circ C}$).
According to “Concentrative Properties of Aqueous Solutions: Density, Refractive Index, Freezing Point Depression, and Viscosity”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the properties of a potassium bromide solution of with a mass fraction of $w_{\ce{KBr}} = 40.0\ \%$ at $T = 20\ \mathrm{^\circ C}$ are as follows.

molality: $b_{\ce{KBr}} = 5.602\ \mathrm{mol\ kg^{-1}}$
concentration: $c_{\ce{KBr}} = 4.620\ \mathrm{mol\ l^{-1}}$
density: $\rho = 1.3746\ \mathrm{g\ cm^{-3}}$

It seems, that the discrepancy arises from the fact, that the mentioned density of $ 3.119\,\mathrm{g\,ml^{-1}} $ belongs to solid $ \ce{KBr} $ (Source - Which might be wrong as well, since wiki suggests a much lower value of $ 2.74\,\mathrm{g\,ml^{-1}}$, which is in good accordance with the information available on the GESTIS Substance database: $ 2.75\,\mathrm{g\,ml^{-1}}$ at $ 20\,\mathrm{°C} $) rather than to a solution.
This is why your measured density and the literature density are not consistent.
I don't know where you got your information from, since you did not name any sources, but it was either wrong there, or it was mistaken with a solid density (or both). In his very informative answer, Loong already named some citations for densities of aqueous solutions.
Addendum: The source stating the $ 3.119\,\mathrm{g\,ml^{-1}} $ does not seem to be reliable at all, since it states a $ \mathrm{mp} $ of $ 734\,\mathrm{°C} $ and a lower $ \mathrm{bp} $ of only $ 58.8\,\mathrm{°C} $, which does not sum up at all.

Amen.  The same source says that KBr has a vapor pressure of 175 mmHg, which is absurdly high.  The "synonyms" for KBr listed include  BROMINE TS; BROMINE WATER; BROMINE LIQUID; BROMIDE BROMATE and many other absurdities.  Potassium bromide is not the same as liquid bromine!  Nor is it a bromate!

My guess is that the density listed on the absurd site is the density for potassium bromate, not potassium bromide.  The properties on that page are an absurd scramble of the properties for bromine, potassium bromate, and potassium bromide.  None of them seem to apply to aqueous solutions.

@CurtF. Thanks for the supplementation. Yeah, this site is a colossal cheek - Calling itself "Chemical Book"


text:
When is an alkene formed in a Wurtz reaction?
I have read that the Wurtz reaction could take place via either a radical or a nucleophilic substitution reaction. Due to steric hindrance, an alkene would generally be formed on treating a tertiary alkyl halide with Sodium in dry ether. But, what about secondary halides???
For example, what would be the major product(s) of the Wurtz reaction of    2-chlorobutane?
Would we get a mixture of 2-butene and butane, or would we just get 3,4-dimethylhexane(a symmetric alkane)?


text:
the number of significant figures
An electronic balance was used to determine that a sample has a mass of 9.00060 g.  If the balance's precision is ± 1 mg, what is the correct number of significant figures for this measurement?
actually the answer is 4 but I cant figure out from where has it come from? Isn't it 5?

The significant digits you use are the values of which you are confident, plus one more value which may be uncertain.
In this case, the balance is accurate to $\pm0.001$ mg, so you're confident about the first 2 decimal places, but the 3rd place is uncertain.
The measurement given is $9.00060$, but the decimal part is smaller than our uncertainty, so we can't count on it.  The most accurate we could record is $9.001$, because the first decimal we're uncertain about is the 3rd place, which leaves us 4 significant figures.
If the measurement had been $19.00060$ instead, then the same process would result in 5 significant figures.


text:
Reactivity of KOH with PVC
I Would like to know if its possible to get a reaction between KOH at ~50% (w/w) and an usual PVC pipe? If yes what kind?

If you're just curious for practical purposes (i.e., "Can I put this KOH solution in this PVC container or will it ruin it?"), there are many material compatibility tables for this type of thing. This one suggests that your solution will be OK at low temperatures for a few days at least. It may react after more time or at higher temperatures.
If you want to know specifically about the chemistry, KOH is a strong base. Strong bases are known to crack polymer chains through solvolysis. PVC in particular loses some chlorine atoms to the solvent which weakens and ultimately shortens the polymer chains, as explained in more detail in the link.


text:
Where can I find the standard enthalpy of formation for Sodium Cyanamide (Na2NCN)?
I am trying to find the standard enthalpy of formation for sodium cyanamide. Ive looked on various databases that list data about specific compounds but I can't find anything on this particular one. Where should I be looking?

Cyanamide is a known compound, $\cf{CN2H2}$.  The NIST Chemistry WebBook contains information on its enthalpy of formation.
Sodium cyanide is a known compound, $\cf{NaCN}$.  The same source also has has information on sodium cyanide's standard enthalpy of formation.  It is much tougher to find information on monosodium or disodium cyanamide!


text:
Can 0-Carborane be stored at room temperature?
I am a physics post-doc at Oak Ridge National Lab (with a bit more training in chem than the average physicist). I ordered a small sample of o-Carborane from Sigma-Aldrich to test for doping liquid scintillators for neutron capture. Everything I have read about carborane says that it is crazy stable and should be fine at room temperature as long as it is kept dry.
However Sigma-Aldrich says to store refrigerated at 2-8C. I called them and found out that this is because they sell it without an expiration date "We cannot guarantee product quality and performance if the product is not stored at the recommended temperature." However, I know that they ship it at ambient temperatures. I am currently trying to get them to send me confirmation that it is not a safety concern to just keep it in a dessicator (which we have), merely a quality concern (and a mostly imaginary one at that).
I am posting this because I am not certain of all the places I could look to find information that would support this conclusion and I need the support to go to the safety people and show we do not need to buy a chemical fridge to store this stuff. I found found SDSs that simply say cool dry place, I have seen safety data sheets that say refrigerate, I even saw one that said to keep it below -18C.
Can anyone point me towards a good source for definitively saying "It is safe to store o-carborane at room temperature in a desiccator" (or similar)?


text:
R or S configuration of propane-1,2-diol
What is the specific configuration (R or S) of this compound?

My answer is R, because $\ce{H}$ is not in the back so I flipped the compound. Is my answer right?
My teacher is telling me it is S, though I'm pretty sure that it is R. That made me very confused.
What I did is that I got S at the beginning, but because $\ce{H}$ is not in the back, I flipped the compound and got the answer R. My teacher flipped the compound in a way that the answer was still S.
$\ce{OH}$ (1), $\ce{CH2OH}$ (2), $\ce{CH3}$ (3), $\ce{H}$ (4) from 1 to 3, anticlockwise which means it is S configuration. But $\ce{H}$ is not in the back, Therefore rotate the compound and the answer should be the opposite to S which means it is R.

Well clearly one of you is wrong. Perhaps you could add some diagrams to your question to show what you mean. I have some difficulty in visualising what you are trying to say.

@saf, here's a hint. The H isn't in the front either...

Words of caution about flipping structures: https://chemistry.stackexchange.com/questions/99721/twisting-stereoisomers-with-rings-to-determine-r-s/99750#99750

Your comment is correct where you assign the priority.
Rather than "rotating" or "flipping" the molecule, consider "looking" at it from the angle which makes the H atom be at the back.  In the graphic it is drawn to the left side, so place yourself on the right side and look down the bond.

The priority groups 1-2-3 run anticlockwise, hence it has the S configuration.

Thank you, I always thought that if number 4 was not in the back, I have to rotate the compound.

It depends. For example if you swap the H with the CH3 or CH2OH groups, the initial glance of the priorities still looks anticlockwise but when you look down the bond, you see the configuration is actually flipped!

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Inert cyanide groups in polyacrylonitrile(PAN)
Cyanide group undergoes hydrolysis and gets converted to Carboxylic acid. However, the cyanide groups in polyacrylonitrile(PAN) are inert. Why?

But PAN does hydrolyse to an extent. There are plenty of articles on it...

so PAN is not a good polymer?


text:
what is the mass of chromium in this sample?
If the total number of atoms in a sample of potassium dichromate is $2.13\cdot10^{24}$ , then the amount of chromium in this sample is?
the answer that I've come up with is $367.64\ \mathrm g$
but in fact the solution in my book is $33.4\ \mathrm g$.

Well how did you come up with your answer?

2.1310^24/6.02210^23=3.54 mol *2=7.07 since the molecular form of the compound is K2Cr2O7 , then I"v calculated the mass by multiplying moles by 52

I like to solve such assignments using Python as a scientific calculator.
import scipy.constants

# number of atoms in sample
N_ATOMS_TOTAL = 2.13E+24

# atomic mass of chromium
M_CR = 51.9961 # g/mol

# ratio of chromium atoms to total atoms in K2Cr2O7
CR_RATIO = 2/11

# moles of Cr in sample
n_cr = N_ATOMS_TOTAL * CR_RATIO / scipy.constants.Avogadro

mass_cr = n_cr * M_CR # mass of chromium in g

mass_cr

Let me explain:

You have an unknown mass (in gramm) of potassium dichromate, $\ce{K2Cr2O7}$.

The total number of atoms in this sample is $2.13\cdot10^{24}$.

How can you calculate the number of chromium atoms in the sample?

You need to determine the ratio of $\ce{Cr}$ to all atoms in $\ce{K2Cr2O7}$, and that's $\frac{2}{11}$ atoms.

I know the number of $\ce{Cr}$ atoms now, but how many moles is that?

You know that 1 mol of an element has $6.02214129\cdot10^{23}$ atoms of that element (Avogadro's number).
Divide the number of $\ce{Cr}$ atoms determined before by Avogadro's number to determine the number of moles of $\ce{Cr}$.

I know the number of moles of $\ce{Cr}$ now, but what is the weight?

You know that the standard atomic weight of $\ce{Cr}$ is 51.9961.
Multiply  the standard atomic weight with the number of moles to get the weight.

Im so confused :( , I know that when a substance like H20 has 2.5 mol it means H has number of moles=2*2.5 . but If I used the way u solved the question it will be 2 H atoms+1 O atom=3 atoms , to calculate H it will be (2/3)*2.5 and the result is different then :(

@119j7fly $\ce{K2Cr2O7}$ has  2 $\ce{K}$ +  2 $\ce{Cr}$ + 7 $\ce{O}$ atoms, that's 11 atoms in total. Two of these are $\ce{Cr}$ atoms. This means that the number of $\ce{Cr}$ atoms in the sample is 2/11 * the total number of atoms :)

its very clear right now thanx a lot mr Klaus.


text:
What should be the name of this substituted cyclopentane?

I think its name is 5-amino-3-bromo-4-chloro-2-hydroxycyclopentan-1-carbonitrile. Is that correct?

What should I do wait please sir ..

@_Pi5 what do YOU see in your browser? Try uploading again, after checking your picture locally. If this should be  a problem instead of simply a wrong click, try uploading a test image (labeled so, to avoid confusion and undeserved down-votes).

I think it is loading now previously it was also not loading in my browser

Yea, now I can see something!

Note your NH2 is the wrong way round, I've edited it

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred suffix for the principal characteristic group $\ce{{}-CN}$ is ‘carbonitrile’. Therefore, the name of the parent structure of the compound given in the question is ‘cyclopentane-1-carbonitrile’ (note that the locant ‘1’ would be omitted in monosubstituted homogeneous monocyclic rings, i.e. ‘cyclopentanecarbonitrile’).
The relevant rules for the numbering of locants for the four substituent prefixes (‘amino’, ‘bromo’, ‘chloro’, and ‘hydroxy’) are:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

and

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule (f) takes precedence over Rule (g).
In accordance with Rule (f), the compound given in the question could be named as 2-amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitrile as well as 5-amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile since both names correspond to the locant set ‘2,3,4,5’. However, according to Rule (g), this example is named as 2-amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitrile rather than 5-amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile since ‘amino’ is cited first as a prefix in the name.

Loong, do you lurk in the [tag:nomenclature] question list, poised to slam down Blue Book quotes? Wow.

@Brian Sure! Even if a question is not a duplicate, many nomenclature questions can be answered using parts copied from similar answers.

So can we say that we are following the priority order of functional groups just while making the backbone af5er that we follow alphabetical order ???

@Pi5 You use the priority order to identify the characteristic group that is expressed as a suffix (here: carbonitrile) and that receives the lowest locant (here: 1). Then, low locants are given to the prefixes, all considered together (here: 2,3,4,5). If a choice still remains, the lowest locant (here: 2) is given to the substitutent cited first in the name (here: amino), which corresponds to alphabetical order.

Thank you .Can you give some examples sir . ?

@Pi5 You can find examples for Rule (f) here and here, and for Rule (g) here, here, here, here, here, and here.


text:
Magnesium bicarbonate dissociation products
What is the equation for $\ce{Mg(HCO3)2}$ dissociation? Are there more steps because it's a complex substance? Could the possible products be $\ce{MgHCO3+ + HCO3-}$?

Typically dissociation refers to the ions that are formed when a substance dissolves. The most common solvent is water. So in aqueous solution:
$$ \ce{Mg(HCO3)2(s) -> Mg^{2+} + 2HCO3^{-} }$$
The bicarbonate will also form very minor amounts of other aqueous species:
$$\ce{HCO3^{-} <-> H^+ + CO3^{2-}}$$
$$\ce{HCO3^{-} + H^+ <-> H2CO3}$$
$$\ce{H2CO3 <-> CO2 + H2O}$$
The carbon dioxide formed in the last reaction can outgas from the solution as well.


text:
Hydrodehalogenation of aryl chlorides
Palladium is the "go to" heterogeneous catalyst for dehalogenating aryl chlorides using hydrogen. What are the chemical (not economic) reasons for other precious metal catalysts not being as common or as efficient in this reaction e.g. rhodium?


text:
Why is −78 °C a magic temperature for organic reactions?
In many organic reactions that I have seen, running the reaction at $\mathrm{-78\ ^\circ C}$ seems to be quite a popular choice, but I've never seen any explanation for this. What makes this temperature such a popular choice?

That's easy one - sublimation of CO2

A little like so many reactions being 'refluxed for 12-15 hours' => I put it on before I went home and shut it off in the morning.

Just like 'run for 72 hours' --> "I finally got it set up and running Friday afternoon, then didn't get around to shutting it off until after lunch on Monday."

Relevant answer to a different question.

And just like "the last reagent we tried worked!"

Dry ice (solid carbon dioxide) sublimes at −78 °C. Dry ice and acetone are a common cold bath for chemical reactions.  The melting point of acetone is -95 °C so the bath never gets cold enough to freeze the acetone. The bubbling of the carbon dioxide gas as the dry ice sublimes keeps the cold bath well stirred.
Typically, though, the temperature in the flask with an ongoing reaction is at least about 5 °C higher than the one in the cooling bath.  (Except if using a thermocouple, working at a scale of 10...25 mL and less often implies that the thermometer is in cooling bath, and not in the reaction mixture. Thus, this temperature difference isn't recorded.)

Could you elaborate a little?

As MaxW said, dry ice (solid carbon dioxide) sublimes at -78°C and it can be used in acetone, which is a cheap non-toxic solvent.
The biggest problem with cooling baths is that temperature must be monitored and corrected if necessary. It implies that the actual temperature is not constant but oscillates around an average value, which is highly undesirable. Also, constantly correcting the temperature makes it impossible to stray from the bench. Cryostats (cold generating devices) are useful sometimes but they usually are a real pain to use.
By contrast, the dry ice/acetone bath stays at -78°C, constantly and possibly for hours. When you put ice (a lot) in water, the water temperature will remain 0°C as long as there is still ice (roughly) and the same stands for the dry ice/acetone; as long as there is some dry ice left, the temperature will remain -78°C.
That is why the -78°C is so popular among chemists for low temperature experiments.

I don't think acetone is non-toxic.

@Joel Though it is non-toxic. It is considered "harmful if swallowed" (who would do that?). Of course precautions must be taken but it is one of the safest solvents in the lab from a health point of view.

I’ll partially agree with @Joel that nothing is non-toxic. But I’ll also have to say that acetone is a lot less toxic than many other things in the lab (e.g. methanol which one would use for $-95~\mathrm{^\circ C}$).


text:
What happens to potassium if potassium permanganate is reduced in a redox reaction? How can I identify spectator ions?
I'm trying to get a fool-proof way of finding the products of a redox reaction. I can do both the oxidation and reduction the half reactions. For example in the potassium permanganate with hydrogen peroxide I found the net ionic equation to be:
$$\ce{5H2O2 + 2MnO4- +6H+ -> 2Mn^2+ + 8H2O +5O2}$$
However, this troubles me. First, what happens to the $\ce{K+}$ that was attached to the permanganate ion? My guess is that it hangs around like a spectator ion to partner up with one of the products at the end, but it can't attach to the reduced manganese because that is also positively charged.  Would it do this with the anion that was bound to the $\ce{H+}$ (e.g., $\ce{SO4^2-}$)? Would it attach to the $\ce{Mn^2+}$. Then, what does the $\ce{Mn^2+}$ cation bind to?
I get that only including the reacting species in the net ionic equation is supposed to make things easier, but it always seems to confuse me more. How do I decide what is important and what's not?

You're right. There are "spectator" anions and cations.

Is this supposed to be a balanced equation? The charges don't balance.

@jerepierre you right - the charges should balance. The equation should be: $$\ce{5H2O2 + 2MnO4^{-} +6H+ -> 2Mn^{2}+ + 8H2O +5O2}$$

Spectator ions do indeed just ‘hang around’. They are unchanged during the reaction and after the reaction they just pair up with whatever they can find. Remember that redox (half) reactions are always charge-balanced (if you set it up correctly). Thus, including the spectator ions which also have a constant charge, the overall charge must be zero. Taking your equation $\text{(Redox1)}$ and its half reactions $\text{(Red1)}$ and $\text{(Ox1)}$:
$$\begin{align}\ce{MnO4- + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O}\tag{Red1}\\
\ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox1}\\
\ce{5 H2O2 + 2 MnO4- + 6H+ &-> 2Mn^2+ + 8 H2O + 5O2}\tag{Redox1}\end{align}$$
We added the permanganate as potassium permanganate so there should be a potassium ion. On the reactant sides of the equations $\text{(Red1)}$ and $\text{(Redox1)}$, it can happily hang around with permanganate but it seems a little out of place on the product side; see equations $\text{(Red2)}$ and $\text{(Redox2)}$:
$$\begin{align}\ce{KMnO4 + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O + K+}\tag{Red2}\\
\ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox2}\\
\ce{5 H2O2 + 2 KMnO4 + 6H+ &-> 2Mn^2+ + 8 H2O \\ &+ 5O2 + K+}\tag{Redox2}\end{align}$$
And obviously, that cannot be the end of the story. We need something for the product side — but we also need something to offset the protons. This is where the acid comes in. We generally write it simply as $\ce{H+}$, but there will always be a counterion. For example, we might have used sulfuric acid. In that case, instead of writing $\ce{H+}$ we can also include the counterion as $\ce{H2SO4}$, see the third set of equations:
$$\begin{align}\ce{2KMnO4 + 10 e- + 8 H2SO4 &-> 2 MnSO4 + 16 H2O\\ & + K2SO4 + 5 SO4^2-}\tag{Red3}\\
\ce{H2O2 + SO4^2- &-> O2 + 2e- + H2SO4}\tag{Ox3}\end{align}$$
As is, this seems kind of counter-intuitive since there are still naked sulfate ions. What balances those? It’s the electrons that we are transferring. Since the electrons cancel out when adding up the half-reactions, so do the sulfates, see $\text{(Redox3)}$:
$$\begin{multline}\ce{5 H2O2 + 2 KMnO4 + 3 H2SO4 -> \\ 2 MnSO4 + 8 H2O + 5 O2 + K2SO4}\quad\quad\tag{Redox3}\end{multline}$$
As you can see, only part of the sulfates pairs up with potassium; the remaining sulfates pair up with manganese. All in all, there are enough sulfates around for every species to be netural — as required by the law of conservation of charge.

Identifying spectator ions is rather easy: Simply, anything that doesn’t partake in the actual redox process can be determined a spectator ion. So if you realise that only the sulfite anion of sodium sulfite reacts, sodium is automatically a spectator ion and can be left out. Similarly, if you add nitric acid but no nitrogen oxides $\ce{NO_x}$ can be detected, the nitrate ion is a spectator ion.
I’ll admit, it gets a lot easier identifying spectator ions once you ‘get a feeling’. You can always imagine what a potential spectator ion could do in a redox reaction. For most, it’s either improbable (e.g. potassium being reduced) or would have a notable outcome (e.g. nitrate being reduced).


text:
Calculating equilibrium partial pressures of a gas given Kp

My work:
${K_p = 28.4 = \frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{[NOBr]^2}{[107]^2[160]}} \rightarrow$$\ce{[NOBr]=7212 torr}$.
This doesn't match up with the given answer. I don't see why 262 torr is the correct answer. I correctly set up the equilibrium constant expression and everything in the reaction equation is in the gas phase so they are all included and I used only equilibrium concentrations.
Another problem with the same issue:

Weird. You math is right. If the constant is based on mole fraction then I get 19.8 for the constant based on the answer of 262 torr.

It should be products over reactants, but inverted: $$ \dfrac{(107^2) (160)}{262^2}  = 26.7 $$ It seems problem/answer just messed up.

@MaxW this is a consistent issue. See my edit. I found that I get the correct answer after converting all the torr values to atm and keeping Kp as is. Why is it that I must convert torr to atm?

In general $k_{p}$ is not a dimensionless quantity. In the case of the reaction $$\ce{2NO + Br2 <=> 2NOBr}$$ the unit of measurement for $k_{p}$ is in fact $$\frac{pressure^2}{pressure^2 * pressure} = pressure^{-1}.$$ Thus the value 28.4 of the equilibrium constant given in the assignment is incomplete.
Given the correct answer, it turns out that $k_p$ is actually given in $\mathrm{atm^{-1}}$. Since $$\mathrm{28.4~atm^{-1}} = 28.4*(\mathrm{760~Torr})^{-1} = 0.0374~\mathrm{Torr}^{-1}$$ you should use this value for $k_p$ if you prefer to do your calculation with $\mathrm{Torr}$.

how do we know the units a priori

The unit must be given along with the numerical value.
The exponent can also be derived from the stoichiometric coefficients as shown above. In the special case that the exponent is zero the unit of measurement is 1, i.e. neither Pa, nor atm, nor Torr, or any other unit of pressure measurement.


text:
Polyatomic Reaction Charges
When a polyatomic ion reacts with another atom to create an ionic compound, often, we know the charge of the polyatomic ion. But how can we infer the charge of the atom it is reacting with?
For example, the question states, 'What is the correct formula for Calcium Phosphate?'
I know that the charge of the polyatomic ion named Phosphate is 3-. But does that mean the charge of Calcium, in this case, must be 3+ in order for it to react with Phosphate? If not, how can we figure out the charge of the Calcium ion in this ionic bond?

When a Calcium atom loses two electrons, it attains the stable electronic configuration of the noble gas Argon. Thus, Calcium cannot have a $\ce{+3}$ charge in any ionic compound. So, the ions forming your salt are $\ce{Ca^2+}$ and $\ce{PO4^3-}$. We need to balance the charges to obtain a neutral salt. We can see that the least common multiple of the charges (2 and 3) is 6. A total charge of $\ce{+6}$ can be obtained by taking three Calcium cations, and this can be balanced by taking 2 Phosphate anions.
Calcium Phosphate is $\ce{Ca3(PO4)2}$


text:
What are the physical properties of exotic ice?
This chart shows the phases of water that form at various combinations of pressures and temperatures.

It has multiple exotic forms of ice (labeled II through XV) listed. What are the physical property differences between these exotic forms of ice, regular ice, and each other?

What properties are you interested in?

I'm really curious as to what differentiates them all. Solid, liquid,and gas all have obvious differences. How can you tell the difference between the different forms of solids?

the gist is that the different forms of ice have different crystalline structures. So all sorts of properties would have (small) differences. Density, melting point, and heat of fusion for example.


text:
is it possible to make ethanol from ethane through a chlorination / hydrolysis two-step procedure?

$\ce{C2H6}$ is chlorinated in presence of sunlight to form $\ce{C2H5Cl}$ and $\ce{HCl}$.
$\ce{C2H5Cl}$ is then reacted with water to form $\ce{C2H5OH}$ + $\ce{HCl}$

Can someone please confirm this???

Fundamentally possible, but pretty uneconomic on both stages.

@IvanNeretin Especially the first stage. The yield there will be awful.

Yes, it is possible.
Although ,commercially(and ironically!) alkyl halides are produced from the corresponding alcohols.
The main reason why this is so because free-radical chlorination requires high temperatures and catalysts to promote the reaction.
Theoretically, the first reaction will give good yields(and not a mixture unlike most free radical halogenations because of the simple reactant involved)
Edit:As pointed out by bon, I forgot to take into account that because its a free radical reaction, the number of possible products is a lot.So no, the first reaction will not give you good yields.

The first reaction will give terrible yields. Multiple chlorinations are highly likely and there is also the possibility of a variety of other radical reactions.

:/....I don't know why I didn't take it into account.Thanks for pointing it out :)


text:
Organic reactions reactions memorising
What is the best possible way to learn organic chemistry reactions rather than memorising blindly?

undersanding reaction mechanisms will surely help you....but before that you need to have a thorough knowledge(not mandatory but useful) of acid-base theory,chemical kinetics,nature and type of reactants etc.

Recognize functional groups.
Be aware what nucleophiles and electrophiles are.
Apply 2 on 1.

Yes, there's more to it, and you still have to memorize name reactions, but the upper will definitely make your life a lot easier.


text:
How to check purity of mercury metal
Mercury is a very good solvent for many metals. We have mercury metal available, how do we check for its purity? Is there some kind of test available to verify its purity? And, more helpfully, which kind of metals are dissolved in it?

It depends on what is in the mercury, what you want to analyze for ( Zn may be ok, but not Ag for example) and what chemical instruments  or techniques that you have available. Most of all when working with a heavy metal like mercury you need a good plan to dispose of the waste.

The mercury I'd have would most probably be used earlier. I am looking to eliminate heavy metals like Cd, Pb, As etc .. so wanted to see if there would be any simple test available to see if they are present? I am afraid I wont have some high sophisticated equipment availabe.

In case your mercury didn't come with an assay of the supplier, or has been used before, you might want to remove oxidizable metals by repeatedly dripping the mercury though diluted nitric acid and distilled water.
Subsequent distillation of mercury is certainly possible (at 25 mm Hg or lower), and has been reported by E. H. Riesenfeld and W. Haase in Ber. Dtsch. Chem. Ges., 1925, 58, 2828-2834 (
DOI).
In their article Über die Destillation von gold-haltigem Quecksilber ( = On the distillation of gold-containing mercury) the authors examined the distillation of gold-saturated mercury and stated that after two distillations, the gold content in the purified mercury was lower than 1.5 ng $\ce{Au}$ in 1 g $\ce{Hg}$.

UPDATE
In Joel H. Hildebrand, PURIFICATION OF MERCURY, J. Am. Chem. Soc., 1909, 31, 933-935, (DOI), the author seems to discuss further different methods,
THE PURIFICATION OF MERCURY BY AN ELECTROLYTIC METHOD
Science, 1933,
78(2027), 414-415, (DOI) might be woth a look too.
However, please notice that these are rather old references. Please make sure, that the methods proposed are still in agreement with currents standards of lab safety and waste management!

But wouldn't treating mercury with dilute nitric acid also react with mercury? Would that be a substantial amount?
moreover which kind of metals should be expected from used mercury? and any guides for testing for their presence. I would prefer simple tests as having sophisticated laboratory scale equipment wont be available.

@Muzammil The reaction of mercury with cold diluted nitric acid is much slower than with dissolved traces of less noble metals. So, the amount of loss is not significant. I already assumed that sophisticated analytics might not be available. Therefore I suggested simple purification techniques, rather than expensive analytical instruments. I'm afraid that there might not be a simple bench thest for trace analytics. Metal traces detected in the aqueous solutions might also come from the nitric acid itself.

That explains a lot. Would this treatment remove Cd, As, Sb & Pb impurities ? If you have any documents or links to sophisticated techniques or purification methods, can you please share ? wouldn't hurt to take a look.


text:
Making N,N-dimethylaniline from aniline
Can I make N,N-dimethylaniline from aniline just by reacting it with $\ce{CH3X}$ ?
According to me, the lone pair of nitrogen in aniline is not nucleophilic enough (since it is in conjugation with the whole benzene ring) to be able to displace the $\ce{-X}$ group.

Certainly pKb of hydrochloric acid (poorest leaving group among halides ) is   21 but pKb of aniline is 9.1, so aniline being a better base (and so better nucleophile, considering no other factors affect nucleophilicity comparison in this case) and so yes.

Yes this reaction will definitely work. Use a nice leaving group like iodide and the reaction should go easily. Although the nitrogen lone pair is delocalised into the benzene ring it is still quite nucleophilic. Additionally, the delocalisation is restored when the nitrogen loses a proton to get the substituted amine. As such there probably won't be a big problem with forming a quaternary ammonium salt (which is what usually happens if you try to alkylate amines) because doesn't have any opportunity to maintain delocalisation in the product and so will be disfavoured compared to the tertiary amine.

Hey bon, the lone pair from the N atom resonates to the carbon atoms in the ring, imparting the ortho/para carbons a negative charge. The negatively charged carbon atoms would be a better nucleophile than the neutral ntirogen atom. So, my doubt is that wouldn't the attack on iodomethane take place from those carbon atoms instead? (I am talking about a mechanism similar to what happens in Riemer Tienman reaction of phenols) Thanks!

@GaurangTandon The nitrogen has a far larger negative charge than the o/p carbons so that's not going to affect the reaction. Resonance is only a minor consideration here and it actually probably helps the reaction as I pointed out in my answer.


text:
Do bonding and anti-bonding orbitals exist for ionic bonds?
Is there any concept of bonding and anti-bonding orbitals for ionic bonds, as there are in case of covalent bonds?
OR is it possible to break an ionic bond by pushing electrons into the antibonding orbitals?
I know that this question may sound stupid to some, but please excuse my ignorance.

Ionic bonds are not about orbitals at all. Think, if you like, of a covalent bond in which the bonding orbital is skewed to the point of being totally centered around one atom.

Bonding and antibonding orbitals in a homonuclear diatomic molecule  A-B should mix due to polarization to give rise to pair of atomic orbitals with delocalized tails  for an ionic heteronuclear diatomic molecule. Now suppose A more electronegative and is the rightful anion and B the cation. The tailed atomic orbital of A is the bonding orbital(ie. A:- B°+) and the one belonging to B is the antibonding (ie. A°+ B:-). See original paper on Coulson Fischer theory.


text:
Where can I find a collection of density functional parameterizations?
I want to implement a solution to the electronic Schroedinger equation using DFT. I know there are many, many functionals to choose from but don't know where to start. I'd like to start with a couple LDA and GGA functionals, is there a good place where I can look up some parameterization, or even the tabulated data itself to form my own parameterization?

Why don't you search the papers in which the functional that you want to implement is introduced?

For few years I am trying to find some time to put up wikipedia.org page on commonly used DFT functionals, just name, formula, citation. But I not a skilled wikipedian, so if someone provides template, I'd gladly fill few of them.

@ssavec We could talk about it in chat

Unless you are doing it just for fun and curiosity how the think works, you are better with some existing library. Most prominent is the libxc, offering the lowest level interface, i.e. given density (and possibly gradient) at series of points, it returns the energy. What is left for you is the construction of the grid and optimization of density, ie. you can go for the real Hohenberg-Kohn DFT, not just Kohn-Sham. For the usage examples (of the really nice and simple interface) see http://www.tddft.org/programs/octopus/wiki/index.php/Libxc:manual
And most importantly, you can find definitions of all the functionals in the list of implemented functionals

You could check out the DFT page from Truhlar who's done a lot of xc functional parametrization, and I thought there are code snippets available as well as the references.
For a more formal approach to parametrization within the density functional theory formalism, look up papers by John Perdew, e.g. his key publications here:
https://phys.cst.temple.edu/john-perdew.html


text:
What is the oxidation state of nitrogen in N2O5?
According to the formula it should be +5 but according to its structure shouldn't it be +4? How is it +5 then?

$5\cdot(-2) + 2\cdot(+5)  = 0$

But according to the structure it should be + 4

No. Why do you think so?

A coordinate bond gives a formal charge of +1 to N, implying that it effectively gets a raise in oxidation state by +2

It is +5 .Even by the structure.

Notice that it is a resonance structure of:

Now one of the bonds are coordinate bonds where nitrogen donates BOTH its electrons to oxygen (otherwise, oxygen cannot have a single bond without a negative charge and no peroxy type linkage).
So, while determining oxidation state from the structure :
O.N=+2+2+1=+5
References:
https://en.wikipedia.org/wiki/Dinitrogen_pentoxide

Your second image is better since it shows formal charges.

True, but that's not much of an improvement. Evidently, I picked it up for representation purpose.

Oh, I always advocate properly assigning formal charges to all atoms that need them, which is why I wouldn’t have upvoted your answer if only the first structure were shown ;)


text:
Laboratory preparation of dinitrogen
In my book the laboratory preparation of dinitrogen is as follows:

$$\ce{NH4Cl(aq) + NaNO2(aq) -> N2(g) + 2H2O(l) + NaCl(aq)}$$
Small amounts of $\ce{NO}$, $\ce{HNO3}$ are also formed in this reaction: these impurities can be removed by passing the gas through aqueous sulfuric acid containing potassium dichromate. $\ce{N2}$ is collected by downward displacement of water.

So my question is how does it actually happen that impurities can be removed by passing the gas through aqueous sulfuric acid containing potassium dichromate? What are the reactions involved? Any video in which the following process is being illustrated will be helpful to understand.
I have understood what does it mean when it is said that $\ce{N2}$ is collected by downward displacement of water.This video  https://www.youtube.com/watch?v=Ol-2JN01Cfo illustrates it nicely.

The collection of nitrogen $(\ce{N2})$ by downward displacement of water works since nitrogen is poorly soluble in water (at room temperature and standard pressure) and does not chemically react with water.
For the same reason, some impurities (aerosols, water-soluble gases and vapours – for example $\ce{HNO3}$) can be removed by passing the stream of nitrogen through a washing bottle that is filled with water.
Nitric oxide $(\ce{NO})$ is only poorly soluble in water and therefore cannot be removed by this method. However, $\ce{NO}$ can be oxidized. Strong oxidizing agents can convert $\ce{NO}$ into $\ce{NO3-}$:
$$\ce{NO + 2H2O <=> NO3- + 4H+ + 3e-}\quad E^\circ = +0.959\ \mathrm V$$
A suitable strong oxidizing agent is acidic dichromate $(\ce{Cr2O7^2-})$:
$$\ce{2Cr^3+ + 7H2O <=> Cr2O7^2- +14H+ + 6e-}\quad E^\circ = +1.38\ \mathrm V$$
The balanced overall reaction is:
$$\ce{2NO + Cr2O7^2- +6H+ <=> 2NO3- + 2Cr^3+ + 3H2O}$$


text:
Is lithium hydride a salt?
Hydrogen is generally considered a non-metal, and a non-metal and a metal often makes a salt. However, lithium hydride seems like a special case. The wikipedia article does not clearly state if the compound is a salt or not, only describing it as "salt-like".
Structurally, it makes sense to classify it as a salt, due to the network structure, but are there any clear definition to settle where the compound belongs?

Many metals form hydrides; what's so special about lithium?

@IvanNeretin Can hydrides be considered salts?

Salt doesn't mean ionic compound. And there's no purely ionic or covalent compound.

Ah! A commonly used classification may be the best way to handle this then

The Wikipedia article Salt (chemistry) starts off with "In chemistry, a salt is an ionic compound that results from the neutralization reaction of an acid and a base." This seems like a pretty good Chemistry 101 definition. So, in essence, this means having an anion and a cation in the salt, which further implies ionic bonding. This seems like a pretty reasonable place to start.
I'd think that the definition could be tweaked to say "a salt is a solid ionic compound..." That would indicate that one "molecule" isn't a salt. So one molecule of NaCl in the gas phase is a molecule, not a salt. I also have trouble thinking of molten NaCl as a salt. So I'd describe it as a "molten salt."

Kind of like way I think German works. I'd expect that German would have some compound word for "molten salt" which better denotes that there is one new idea. So in English if a "salt" is a solid that has a 3D crystal lattice how can a salt be "molten"?

In general, there are a lot more organic salts than inorganic, so dissolving in water isn't a good restriction.
These last paragraphs point out the sorts of problem in trying to make such a definition absolutely unambiguous. You get a one sentence definition, then two pages of qualifications which render the definition useless.
I can't think of a solvent in which you could dissolve lithium hydride and I don't see how you could get lithium hydride as a neutralization reaction, so I wouldn't think of it as a salt. I don't know a better word than "salt-like" to describe such a crystal structure of anions and cations.

According to your new definition, wouldn't sodium oxide also be a salt?

Wikipedia has edited the article to meet with IUPAC's latest definition. Now, "In chemistry, a salt is a chemical compound consisting of an ionic assembly of cations and anions."

Hydrides of metallic elements are ionic compounds. In the case of $\ce{LiH}$ there may be some covalent characters due to the small size of the $\ce{Li+}$ ion. This hydride on electrolysis in molten form gives $\ce{H2}$ on the anode, which is an indication of the presence of hydride ions and, further, this hydride reaction with water or protic acid evolves $\ce{H2}$ gas.

This answer is a bit vague. For instance, metal carbonyl hydrides often have particularly non-hydridic "hydride" ligands. When mixed with water, they may form $H_2$, but not before acting as a Lewis acid to form a more electron-rich metal complex (by displacing the carbonyl ligands with water/hydroxide ligands).

Can you give a reference for the hydrogen release? The WP article identifies these compounds as acids, and in fact hydrogen is considered as having oxidation state +1 instead of -1 in such compounds.


text:
Non-symmetric compounds with more than one Chiral Carbon
If I have a compound which has non-symmetrical ends, and it has more than one chiral carbon, with no other stereocentres.
Then can I say that all of these will be stereocentres, and so there will be 2^n isomers? Can you give an example where this is not true (if)?

If you have $n$ chiral carbons such that the overall symmetry of the molecule does not make any of them equivalent, then yes, it would be $2^n$.

Yes I mean if n is a big number and we can't check all isomers, then how can I tell? I mean, what kind of asymmetry should I look for in the molecule?

Why would you need to check all isomers to know there are $2^n$ of them (which you already knew before I said that)?

I mean, to check if any of them are equivalent. How do I do that?

Why, just find the point group of your molecule, and you'll see that right away.


text:
Why is dimethyloxonium cation stronger acid than hydronium?
$\ce{(CH_3)_2O^+ H}$ has $\rm pK_a \; -3.8$ while $\ce{H_3O^+}$ has $\rm pK_a \; 1.74\;.$ So, the former is stronger acid than latter hydronium ion.
But why is it so?
Is it due to the $\rm +I$-effect of $\ce{-(CH_3)}$? But it would rather decrease the attraction of $\ce O$ on the electron pair of $\ce H$ and that would make it indifferent to release $\ce {H^+}$ isn't it?
What is actually going that makes $\ce{(CH_3)_2O^+ H}$ stronger acid than $\ce{H_3O^+}$ ?

A powerful but often overlooked factor in determining a substance's acidity and basicity is the effect of the solvent (or lack thereof) in which it is dissolved.
You are not wrong; in the gas phase, $\ce{H_3O^{+}}$ is indeed a stronger acid than $\ce{(CH_3)_2OH^{+}}$, by a factor of about $5\times 10^{18}$. This can be seen by comparing the proton affinities of the conjugated bases, dimethyl ether ($\mathrm{804\ kJ\ mol^{-1}}$) and water ($\mathrm{697\ kJ\ mol^{-1}}$), respectively.
In other words, a mole of gaseous dimethyl ether reacting with a mole of gaseous protons releases $\mathrm{804\ kJ}$, while a mole of gaseous water releases only $\mathrm{697\ kJ}$. This means that in the gas phase, water is a weaker base than dimethyl ether, which implies that protonated water is a stronger acid than protonated dimethyl ether.
Why does the tendency invert in aqueous solution? Water can form hydrogen bonds, a very strong type of molecular interaction. $\ce{H_3O^{+}}$ has three protons available for hydrogen bond donation, whereas $\ce{(CH_3)_2OH^{+}}$ only has one. This means that the solvating water stabilises $\ce{H_3O^{+}}$ more than $\ce{(CH_3)_2OH^{+}}$. Higher stability means reduced acidity, and the difference happens to be enough that the relative order of acidity in water is the opposite of that in the gas phase.

So, it has no relation with inductive effect?

The inductive effect suggests that hydronium should be a stronger acid than dimethyloxonium, which indeed is observed in the gas phase. However, the solvent effect overpowers the inductive effect for determining the acidity in aqueous solutions.

Here (1, 2), are some older answers of mine which touch on the importance of solvation for acidity and basicity.

So, the solvation is more important?

In this case, yes. If you were to analyze the same substances in another solvent (e.g. DMSO, toluene, etc), induction could still prevail.

@NicolauSakerNeto would like to make a short comment to supplement your excellent answer, which the OP might find interesting (since the importance of solvents with regard to determination of acidity was mentioned). The upper limit to the strength of any acid is fixed by the solvent. In essence in a aqueous media, nothing is more acidic than the hydronium ion (final proton acceptor); thus, a weaker proton acceptor solvent is needed to discriminate between very strong acids.
read more here: https://en.wikipedia.org/wiki/Leveling_effect


text:
What is the Franck-Condon region
What precisely is the Franck-Condon region? My understanding is that the FC region is the "bonding" or "equilibrium" region in a potential energy surface, i.e. near the minimum. But when a transition takes place from S0 to S1, it is said that molecule is excited to the FC region of the S1 state - but there is no guarantee that going vertically from an S0 minimum will lead to a near S1 minimum. So something is probably incorrect with my current understanding.
Also, in theoretical minimum energy path calculations, we start from an optimised high energy state, often in a "Franck-Condon region" of an excited state.
So, what is the precise description of the Franck-Condon region?

The Franck–Condon region is a region of the final potential energy surface which is accessed by the vertical transition from the initial state. Usually, the initial state is the ground vibrational state of the ground electronic state and the final is some vibrational state of an electronically excited state. For instance, the Franck–Condon region is shaded below on the Figure from Wikipedia article on the Franck–Condon principle.

But when a transition takes place from S0 to S1, it is said that
  molecule is excited to the FC region of the S1 state - but there is no
  guarantee that going vertically from an S0 minimum will lead to a near
  S1 minimum. So something is probably incorrect with my current
  understanding.

Nope, what you are saying is correct. Indeed, according to the Franck–Condon principle vertical transitions are the most probable. The minimum of the upper potential energy surface might lie outside of the Franck–Condon region, so a vertical transition from the ground vibrational state of the ground electronic state to the excited electronic state might indeed result in one of the vibrationally excited states like it is shown in the picture above. Once a molecule is excited in such way it quickly relaxes to the lowest vibrational level of the electronically excited state from which it can decay then back to the electronic ground state. And again, on this way back to the ground state the most probable transition is the vertical one, so that photon emission can lead to a vibrationally excited state of the ground electronic state, not necessarily to the ground vibrational state of it.


text:
Remove aceton from denatured alcohol
I have bought a bottle of denatured alcohol. On the web I've found these solutions typically contain the following:

$85\%$ ethanol
$10\%$ water
$3\%$ methanol
$1\%$ aceton
$1\%$ pyridine

I would like to get rid of the acetone. How can I remove it? (and how do I know if it is in my solution?)

Without any hplc or other high end equipment (just in the kitchen)

You should be able to distill the acetone and methanol off.

@A.K. What height would you suggest for the column?

Depending which country you're in, this could be illegal

@Beerhunter I live in the Netherlands, 1 liter distillation devices are legal. I'm not concerned about the methanol, just the aceton since it's not good for my printed circuitboards.

@MehdiNellen Just distill until the vapors reach the boiling point of ethanol or rather a few degrees shy.

Ethanol and water are going to azeotrope.  Also word of caution, just in case anyone might decide to take a swig, pyridine is fairly toxic even at low doses particularly for males and their reproductive systems.

If you have access you can stir then distill from CaH2 and get as much water out. Then stir over clean Mg turnings, either under nitrogen or otherwise protected from the air. Add a crystal or two of iodine to activate, and reflux it for some time. Then distill off the ethanol and residual water. Distillation from Raney nickel might do it, as would shaking it with 4 angstom molecular sieves. Point is its going to be tough.  Not sure its worth it for 1%.


text:
Covalent and Ionic bonds with Semi-metals
why are some semimetals considered as a metal, and some, a nonmetal when paired with some atoms.
Like in BF, it is an ionic bond, but in SiCl, it is a covalent bond.
Do their electronegativities matter?

I would never call a $\ce{B-F}$ bond ionic.

Yes, in fact difference in electronegativity is the defining property.  Ionic character of a bond is given by:
$$ \mathrm{\% IC = 1-exp[-\frac{1}{4}(\chi_A - \chi_B)^2]}$$
Where $\chi_A$ and $\chi_B$ are the electronegativities of the anion and cation species.  When $|\chi_A - \chi_B|$ exceeds 1.665 the ionic character transitions from less than $50\%$ ionic to more than $50\%$ ionic, making the compound ionic.

oh so, if i subtract the electronegativity of a, for example, a semimetal and a nonmetal, and if it is greater than 1, will the semimetal become a metal?

No, not at all.  Electronegativity is only informative about the ionic nature of the bond.

It's only a suggestion about nature of bonding http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound


text:
(P1,T1) -> to (P2,T2) then what will be the change in enthalpy for ideal gas?
I need to prove that it is equal to the integral of $C_pdT$ from $T_1$ to $T_2$.
But should not be the case only when the pressure is constant?

The enthalpy of an ideal gas does not depend on pressure.  It is a function only of temperature.  So $C_pdT$ is always equal to dH for an ideal gas of constant composition.  The subscript p on $C_p$ merely indicates that $C_p$ can be measured directly by determining the heat Q absorbed by the gas in a constant pressure experiment.

Enthalpy is defined as $$H= E + PV$$
Its change is defined by the differential $$\mathrm dH= \mathrm d E +\mathrm d\, (PV) \;.$$
Using Thermodynamic Identity $$\mathrm d E=  - P\mathrm dV + T\mathrm dS  $$  for quasistatic process, we get $$\mathrm dE= \delta W + \delta Q$$
So, infinitesimal change in enthalpy becomes $$\begin{align}\mathrm dH &= \delta Q + \delta W +P\;\mathrm d V + V\;\mathrm d P\\ &= \delta Q + V\mathrm d P\;.\end{align}$$
Now, take again the thermodynamic identity
\begin{align}\mathrm d E& =   T\;\mathrm dS- P\;\mathrm dV\\ &=\mathrm d(TS)- S\;\mathrm dT -\mathrm d (PV)+ V\;\mathrm dP\\ \implies \mathrm d(E+ PV - TS)&=-S\;\mathrm dT + V\;\mathrm dP\\& = \mathrm dG  \;_;\end{align} this implied change in $G$ (which is Gibbs Free Energy) depends on the variation of $T$ and $P\;.$
Therefore we can write; $$\mathrm dG= \left(\frac{\partial G}{\partial T}\right)_P\, \mathrm dT +\left(\frac{\partial G}{\partial P}\right)_T\, \mathrm dP \,.$$ Comparing the above equations, we get $$\left(\frac{\partial G}{\partial T}\right)_P= -S\\\left(\frac{\partial G}{\partial P}\right)_T= V $$
Equating the cross derivatives, we get $$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P\;.$$
Also, $$\mathrm d S = \frac{\mathrm dE}{T} + \frac{P}{T}\, \mathrm d V\,$$ and this implies $S$ depends on $T, V\;.$
Now, the definition of $C_P$ is $$C_p = \left(\frac{\partial Q}{\partial T}\right)_P= T\,\left(\frac{\partial S}{\partial T}\right)_P\;.$$
Using all these, we can write the differential of $H$ as
\begin{align}\mathrm dH &= T\left[\left(\frac{\partial S}{\partial T}\right)_V \,\mathrm dT+ \left(\frac{\partial S}{\partial V}\right)_T \,\mathrm dV\right] + V\,\mathrm dP\\ &=C_P\,\mathrm dT - V\left[1 - \frac{T}{V}\,\left(\frac{\partial V}{\partial T}\right)_P\right]\,\mathrm dP\\ &=C_P\,\mathrm dT - V\left[1 - \frac{T}{V}\,\left(\frac{\partial \left(\frac{\nu R T}{P}\right)}{\partial T}\right)_P\right]\,\mathrm dP \\ &= C_P\,\mathrm dT - V\left[1 - \frac{\nu R T}{PV}\,\left(\frac{\mathrm d T}{\mathrm dT}\right)\right]\,\mathrm dP\\ &= C_P\,\mathrm dT\;.\end{align}
So, the total change in enthalpy $$\Delta H=  \int_{P_1,T_1}^{P_2,T_2}\; C_p\,\mathrm dT\;.$$

References:
$\bullet$ F Reif's Fundamentals of Statistical and Thermal Physics.
$\bullet$ E Fermi's Thermodynamics.

That equation for $\Delta H$ is incorrect.  It should be calculated from $$dH=C_pdT+\left(V-T\frac{\partial V}{\partial T}\right)dP$$For an ideal gas, this reduces to $\Delta H=C_pdT$.

The starting equation for dH should be dH=TdS+VdP.  One then writes $$dS=\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP$$.  Then, there is a Maxwell relation giving $$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$There is also the equation $$\left(\frac{\partial S}{\partial T}\right)_P=\frac{C_p}{T}$$If you combine all these equations, you obtain the equation that I wrote.

This is a well-known derivation that is found in most Thermodynamics textbooks.  Please just consider presenting the final result with a reference.


text:
Why is He2- unstable?
$\ce{He2-}$ according to MO theory should be possible with one electron in 2s sigma, but my professor said only at very low temperatures for some reason.
Why only at low temperatures? What is the relationship between the temperature and the stability of that bond? Why is this bond special in this way?

There's no such molecule. Even argon matrix wouldn't help, even hard vacuum as it would probably have negative binding energy

related http://chemistry.stackexchange.com/questions/32717/stable-natural-helium-hydride http://chemistry.stackexchange.com/questions/518/do-noble-gasses-besides-helium-form-diatomic-molecules-at-low-temperatures http://chemistry.stackexchange.com/questions/21820/why-do-noble-gases-bond-with-themselves-but-not-other-elements