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Why does the water become positively charged?
I'm a little confused studying reaction mechanics and I saw this, where does the water electrons go when the bond is formed
I mean, the carbocation needs two more electrons to get its octect filled and the oxygen in water has 2 lone pairs, when the bond is formed, I think one oxygen's lone pair is used. And now the molecule has one lone pair.

Everything you say about octets and lone pairs is true, so what's the question? The whole thing has got to remain positively charged, because charge can't just go nowhere.

I wonder if by the water providing the necesary electrons, doesn't the molecule become neutral?

Ever heard about the charge conservation law? The whole thing (I mean, the carbocation and the water molecule) was positively charged; it totally can't just up and become neutral.

count all electrons and protons on the left, than count all protons and electrons on the right.

Oh Didn't know water was positively charged, I thought it was neutral, now I get it, thanks mate.

For most of organic chemistry reactions, octet rule is followed (notable exceptions are carbenes and nitrenes). The octet rule states that atoms are stable when they are surrounded by exactly 8 electrons (for hydrogen, it's two).
Oxygen has 6 valence electrons in atomic state. The net charge on the atom is zero.
Oxygen shares its electrons with other atoms to obtain an octet configuration. For example in a water molecule, water shares one of its electron with hydrogen. By doing so, oxygen can get the hydrogen atom's electron in return. There are 4 bonding electrons (2 $\ce{-OH}$ bonds) and 4 lone pairs on the oxygen. This gives oxygen an octet configuration around the atom. The oxygen owns 4 of its lone pairs and half of the bonding pairs (2 in this case). It has 6 electrons. A neutral oxygen atom owns 6 electrons.
Consider the case you have shown. There are 3 bond pairs (6 bonding electrons: two $\ce{-OH}$ bonds and one $\ce{-OC}$ bond) and one lone pair.
The oxygen has a total of 8 electrons surrounding it. The octet rule is satisfied. However, it does not own 6 electrons. It owns the lone pair (2 electrons) and half of the bonding electrons, i.e: 3. You get a total of 5 electrons. A neutral oxygen molecule owns 6 electrons. As the atom is one short of an electron, it is unit positively charged.


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Acid-Base buffer question

Calculate the change in pH of a buffer made by combining $\pu{500 ml}$ of $\pu{0.1 M}$ $\ce{CH_3CH_2COOH}$ and $\pu{500 ml}$ of $\ce{CH_3CH_2COONa}$, when $0.04$ moles of $\ce{NaOH}$ is added to the solution. Assume the addition of sodium hydroxide has no effect on volume.
  ($\pu{\mathit{K_a} = 1.3 \times 10^{-4}}$)

My solution:
$$\pu{0.10 M \times ~ 500 ml = 0.05 mol} \ \ce{CH3CH2COOH}$$
$$\pu{0.12 M \times ~ 500 ml = 0.06 mol} \ \ce{CH3CH2COO-}$$
$$\pu{pH = p\mathit{K_a} + \log \left(\frac{0.06}{0.05}\right) = 3.965}$$
Using the equation again where the salt is $\pu{0.1M}$ and the acid is $\pu{0.01M}$ due to the addition of $\ce{NaOH}$ yields a wrong result equal to $1$.
I honestly find the wrong answer to be logical, however I can't think of a way to find the right one.

The reaction is $\ce{NaOH + CH3CH2COOH -> Na+ + H2O + CH3CH2COO-},$ so adding $\pu{0.04}$ moles $\ce{NaOH}$ will result in the solution having $\pu{0.06 +0.04 = 0.1 mol CH_{3}CH_{2}COO^-}$ and $0.05 - 0.04 = 0.01\, \pu{mol} \, \ce{CH3CH2COOH}.$ Then, you can use the Henderson-Hesselbach equation to find the final pH is equal to $-\log(1.3*10^{-4})+\log\left(\frac{0.1}{0.01}\right) \approx 4.886.$

Huh?!? You start out with 0.05 moles of acetate anion and make 0.04 more. That is 0.09 moles total.

@MaxW I believe the question has a typo because in the OP's attempted solution, he states that the acetate anion has a concentration of 0.12 M.

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Having trouble understanding activity series of metals & electronegativity
I want to know how do metals replace hydrogen in a substance such as water. Isn't water happy with its configuration? It has shared enough electrons so that both hydrogen and oxygen complete their orbitals.
I know that alkali metals are very active and they have very low electronegativity. And elements such as gold have much higher electronegativity while being considered very inactive. Maybe I have a wrong understanding of what electronegativity is but isn't the element with higher electronegativity supposed to be more active and be able to replace those with lower ones?

What do you mean by Aurum? Shouldn't it be gold?

Yes that's what i meant. But in my language we refer it to as Aurum instead of gold

Alright! Nevertheless I'll have to edit your question so that it confirms to IUPAC standards. For your question, do you know what electropositivity is?

I'm just trying to understand it now, from what i've gathered it's how much element wants to attract electrons and i think it's only relevant in covalent bonds, but i've never got a clear explanation from my teacher.

2 Na + 2 H2O  ->  2 NaOH  + H2 < for instance in this example, Does the sodium want to give away its electrons more than hydrogen does and that causes replacement of hydrogen or are there other reasons?

https://chemistry.stackexchange.com/questions/53907/can-someone-explain-the-unintuitive-ordering-in-the-metallic-activity-series

https://chemistry.stackexchange.com/questions/20211/how-can-i-relate-the-reactivity-series-to-electronegativity-and-ionization-energ

There's electronegativity and then there's electropositivity. Electronegativity is the tendency of a substance to gain electrons and electropositivity on similar grounds is the tendency to donate electrons. What actually is a redox reaction? Exchange of electrons?
A substance with high electronegativity would like to snatch any electron it finds, while a substance with high electropositivity would like to force-fully hand over its electrons, however unwilling the substance on the receiving end may be. Metals are electropositive.

how do metals replace hydrogen in a substance such as water. Isn't water happy with its configuration?

There was water, it was living happily. Now, water had two components. Oxygen and hydrogen. Let's pull up their electronegativities. Oxygen is at $3.44$ and hydrogen is at $2.2$. This means that the bond that oxygen and hydrogen shared in water was a polar bond. Oxygen being powerful (higher electronegativity) kept the bond in it's favor. The electrons of the bond were more towards oxygen. This gave oxygen a partial negative charge ($\delta -$) and hydrogen a partial positive charge ($\delta+$).
In came a metal which was highly electropositive (think alkali metals, as an example you can take up sodium whose electronegativity is at $0.93$). He gave oxygen a tempting offer:
"Oxygen, you seem to be caught up in a tussle, look at hydrogen, he doesn't want a positive charge because he's quite electronegative. You're constantly fighting with him to keep the electrons of the bond pair pulled towards yourself. Why don't you pair up with me? I'm low at electronegativity. This means I'm electropositive. As much as electropositive species like bearing a negative charge, we electropositive species like bearing a positive charge!"
And oxygen was greedy. A greater difference in electronegativities of elements forming the bond meant a bond which would be richer in ionic character, which appealed to oxygen.

Wow, that's such a good explanation, thank you very much. It's much more clear now, but with that i have few more questions :
1)Do the 'partial charges' only occur during hydrogen bonds? or is it just a bond with smaller charges?
2)My teacher said that metals always have the positive charge when interacting with hydrogen, but what about the ones that have higher electronegativity than hydrogen? for example Gold has 2.4. I now understand from your comment that they woudn't be the best to pair up, but do they not interact at all? and is it possible to somehow make them react?

@Ebrin The pleasure's all mine. Let me say a bit more about partial charges. Say if you had the same elements forming a bond, i.e. $\ce{A-A}$ , you wouldn't have partial charges at all, when a difference of electronegativities comes into play, there's this thing called bond polarization. It can be due to various other factors, you might want to look up dipoles, induced dipoles etc. It occurs basically in every bond, as long as there's a electronegativity difference between the constituents.

@Ebrin In a bond, the element which has a higher electronegativity would bear a negative charge, that's true. http://pubs.rsc.org/en/content/articlelanding/cc/2016/c5cc06118k#!divAbstract this article hints at the formation of a $\ce{Au-H}$ bond, win!

Okay that makes sense, So what about Ionic bonds? Are ionic bonds just the same as covalent bonds, but the difference between electronegativites is much higher? My teacher said the ionic bonds occurs only between metals and nonmetals, she wasn't too clear about why or how it happens. From what I've gathered, there are partial charges between covalent bonds, and actual charges between ionic bonds? Do the covalent bonds give us cations and anions ?

The thing that confuses me the most, is that in class we would treat Ionic and covalent bonds the same exact way, only difference being that ionic was between metal and nonmetals. And when i was studying on the web, people would say that covalent bonds don't really have a charge and ionics do. So are they pretty much the same? what are the key differences? Sorry for asking so many questions, i just have to understand this once and for all.

@Ebrin The key point here is that every bond has some percentage of ionic character, I'm sorry for giving a short reply but if you can wait for a day, I'll reach home and revert back with more information.

That would be much appreciated, I have no problem waiting a day.

Interesting answer.

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Litmus paper reactions for mixed solutions
I am wondering what could cause my litmus paper to first react to the alkaline. turning blue, and then changing to an acidic red color?

Was the paper being held in solution during the color change, or was it whetted then turned red while sitting in the air?

Litmus paper contains a pH indicator, which is a compound that changes color in the presence of acid or base. Natural litmus contains numerous pH-sensitive dyes, but most of them are based on the 7-hydroxyphenoxazone core. In the presence of acid, the imine nitrogen is protonated; in base, the phenolic oxygen is deprotonated. These reactions are reversible, so the color will change when the pH is changed.

While you successfully explained how an pH indicator works, the question at hand remains unclear because it is missing details. I do, however, expect that the OP knows that the pH has changed. I guess the OP is concerned with why.

@Martin-マーチン I'm uncertain of what you would expect to be done in this situation. The two options, as I see it, are to try to answer the question the best I can given the inherent vagueness of the question, or to downvote the question without attempting to provide any answer in the hopes of the OP editing the original question for clarity. The OP refers specifically to the colour change of litmus as opposed to why the pH has changed, and I would argue that this therefore answers the question.

I would expect the question to be closed as unclear, additionally I think it warrants a down-vote. I don't think your answer is bad, but it won't be helpful in the long term of this site. Questions as vague as this clutter the site and prevent other questions to be effectively found. Your answer prevents the question from being deleted, which means it might have to be cleaned up manually in the future.


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Plastic enclosure - what is that smell?
I have a Raspeberry Pi in a small plastic enclosure. The whole thing smells intensely of some chemical I normally associate with burned electronics. It is quite choking. Any ideas what it might be? And is it dangerous? [I do not know the plastics involved -  Straight of the shelf, not powered up]

Is your setup generating lots of heat? If its a low weight plastic polymer you might be getting decomposition products of the plastic breaking down.

@J.Ari No. Straight of the shelf, not powered up.

I upvoted the question because I don't have any leads if this is off the shelf packaging so I am also curious if others could shed some light on this. I would edit your question to clarify with the info you gave to my comment so that the questions is more accurate to what your situation is.

'New in box' generally has a smell from outgassing plasticizer and molding compounds from the enclosure. These are lower molecular weight compounds in the plastic. Heating said plastic up both (1) releases more plasticizer more quickly, and (2) starts breaking down the long chain polymers and releasing the short chain remains. Add in a little trace elements (Na, Cl, ...) and you get a similar smell between the two situations.


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Adding fragrance oil to paraffin wax
I am trying to make a highly scented jar candle. I'm having trouble wicking the candle when I put in larger amounts of fragrance oil. It seems to be either too small and drowns or is too large and smokes. Is there a way to add more scented oil to the wax blend. Or are there things I can add to my candle to help this.


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Why is there no electron swapping after excitation?
If we have 2 electrons in the 1st energy level named X & Y, and one electron in the 2nd energy level named Z.

If the electron X was excited and transferred to the higher energy level, why didn't Z replace it in the 1st energy level?
Why don't we say that there is a type of nuclear gravity, and it increases when the atom becomes unstable and that gravitates Z electron towards the nucleus instead of the excited one?

Z might well replace it. But, you put energy in to the system to get X in to another level, so the system will have to release energy to drop Z. I wouldn't quite call this an Auger process, but...

On the first part: it will, or won't, in a subsequent, separate process.
If you excite the system and promote an electron from the lowest orbital to the next orbital, the immediate result is that you have a lowest orbital populated by one electron and a higher orbital populated by two electrons. It is an excited state, and will tend to relax back to the fundamental state if there is an available mechanism to do so - but that will be a second distinct event after the excitation. In fact, the excited state can, depending on the circumstances, last long.
The second part makes no sense. The nature of the interaction between nucleus and electron is not gravitational (it's electromagnetic), and the lowest energy level isn't "below" in any spatial sense. The gravitational analogy is misleading.

Regarding the gravity part, the electron doesn't fall into the nucleus because of its kinetic energy.

If we look at this figure: https://en.wikipedia.org/wiki/Atomic_orbital#/media/File:HAtomOrbitals.png we can see that with same probability the electron in 1S orbital occupies smaller volume than the electron in 2S orbital. We can can say that 2S is above 1S. And electrostatic and gravitational forces shares the very same Faraday's and Newton's equation with appropriate constants and measures. The gravitational analogy is quite good in educational terms.

its most probably a typo but the upper orbital will have one electron not two as you write in line 3.

@porphyrin If I'm not mistaken, OP asks about a system where the second orbital is already half-occupied (he calls that electron Z).

@Crowley All hydrogen-like orbitals occupy exactly the same volume: infinity. That's the danger in oversimplifying analogies, we can end up forgetting about the actual properties of the complex system because the simpler image is easier to think about.
Yes, the density of an electron occupying a 1s orbital is, on average, closer to the nucleus than that of an electron occupying a 2s orbital. To say that 2s is "above" 1s opens a can of worms of misconceptions.

@VicLineal If you look for probability that the electron is in given volume equal to one you are correct. If we look for 0.90 probability or 0.60 probability, the volumes will change. I think the key point here is that 2s orbital has higher energy than 1s orbital. The analogy is perfect when used during introducing a newbie to quantum mechanics but dangerous oversimplifying when used at conference on quantum mechanics.

@Crowley And expected distance to the nucleus vives you a different ordering. And most likely distance to the nucleus gives you a different ordering. And orbital penetration, which is the most chemically relevant measure, gives you yet another ordering - actually the one we learn by heart as newbies to quantum mechanics.

@Vic Lineal , yes you are correct, I misread .  And as you state the excited state will last for some time until the energy is released and the ground state is recovered. As electrons are indistinguishable we cannot know which one returns.


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Comparison of ways to provide energy to a reaction
As we all know, you can provide energy to a reaction to overcome the activation energy of a reaction by heating the reaction.
I have, however, recently sat through a talk where someone 'powered' a reaction through ultrasonic waves. Apologies, but it was a rather small symposium, so I don't have the details readily available to me to illustrate it.
This was very interesting to me. I'd love to be able to just put my mixture in a ultrasonic bath and not have to fool with heating (Or at least have an educated option of trying it).
There's a fairly large field of photochemistry that I believe is based around this same principal, and I have a general understanding of the concept.
However, how would one go about exploring alternate pathways to replace heating a reaction? I think I grasp the concept theoretically, I just do not understand the origins of it or how to apply it to a new situation.

Start by having a look at Wikipedia pages on sonochemistry. (photochemistry by definition does not use acoustic energy to excite reactions)

@porphyrin Thanks Ill take a look. And yes I know, I was providing a similar phenomenon to illustrate what i was looking for

Note that interesting as sonochemistry may be, it is fundamentally just the study of a different way of heating a solution.

@airhuff Yes. I know that all of the mentioned material is fundamentally the same thing: Provide energy to the reaction. My question is how are these different ways analyzed to gauge their effective and appropriate usage


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How were weights of elements and compounds determined?
I have browsed through many questions on this site but I still don't understand this:
How chemists in the past used to determine the weight of substances (which I believe they randomly called elements or compounds) without using modern techniques?
For example ... we know the molecular formula of water is $\ce{H2O}$
But how did chemists determine the composition of water?
How did they know hydrogen and oxygen, and not some other substance, made water and how could they determine the weight according to which combined?
What was the complete procedure behind it?
EDIT: Just like I said.... That answer still doesn't show us the complete picture... Moreover, the answerer has used phrases like (I don't know) and (Maybe) many times....

also https://chemistry.stackexchange.com/questions/15386/how-were-the-first-molecular-formulas-determined

@Mithoron...How did Chemists know they were burning Hydrogen and Oxygen and not something else? Moreover the answer in the duplicate question doesn't tell us How was the amount of Oxygen determined? That answer ,literally, doesn't tell us anything

@Mithoron Could you please remove the duplicate tag? I think you don't completely understand my point...

@Mithoron Imagine yourself in that era , Could you possibly explain how gases being burned made something liquid? Moreover can you even tell there were two gases? How would you isolate them?

@Mithoron At least please provide a reply or a reason why my question is not appropriate?

https://chemistry.stackexchange.com/help/duplicates

If you check out the Wikipedia page on Properties of Water, it has a very small history section which says Cavendish (not a chemist, arguably) determined the composition of water in 1781. Then following that lead to Cavendish's article, you'll learn quite a lot, unless you already knew it. As far as us putting "the complete picture" in front of you on a plate, yeah, that's not going to happen. We'd have to know what your background was, and even then what exactly would satisfy your concept of what the "complete" picture was. The up-side is that although satisfying yourself is up to you, the web has a lot of resources you can get help from. When it comes to a lot of scientific, mathematical, and technical subjects, though, you'll have to put in the study (and practice) time to learn and understand it, course work and textbooks will be difficult to replace with unorganized web research. OK, enough philosophy. I encourage you to put together a time-line of alchemy, natural philosophy, and science at around the start of the 19th Century. You probably know that the idea that matter is composed of small indivisible particles (atoms) was first seriously considered by the ancient Greeks, right? And that various technologists have used recipes (based on weights or volumes or both) to repeatedly create (or destroy) various materials of interest, from time immemorial. Alchemists were dedicated transmutation, so what they knew and when they knew it is important here. Proust didn't demonstrate his law of definite proportions until almost 20 years after Cavendish discovered the constituents. So first we knew what it was made of (oxygen and hydrogen, although the concept of elements was not yet firmly established.) And see also https://en.wikipedia.org/wiki/Oxygen#Early_experiments as well as the following discussion on the Phlogiston theory. Probably the reason you haven't gotten a clear answer here is because this isn't a technical question, it is a historical question. To get a clear complete "picture" you need to literally waste your time learning about all sorts of discredited theories (Phlogiston, and the earlier even more unfortunate Earth, Air, Fire and water "theory") and then learn how the various people overcame the blunders made before them to find the "right" explanations for the world around them. Needless to say, that is an ongoing effort but to recapitulate all of the dead-ends isn't something we're going to do in a short answer, frankly. Anyway, put together a time-line and you'll begin to see how the various strands of the story wove together. HTH.

Thanks for your answer....I will try to find whatever I can ...Thanks once again...


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Whats the difference between 2O and O2
I just saw something in a chemistry lesson what got me confused.
What is the difference between $\ce{2O}$ and $\ce{O2}$?
Thanks for the help!

The 2 O represents two separate  oxygen atoms, ( not connected to one another so free to move around and react independently  ), the $\ce{O2}$ is molecular oxygen.

@Porphyrin Why dont you post it as an answer? I think its the simplest solution I've heard yet. My teacher was just talking weird stuff!

@BerryHolmes No actually not I saw that question but its not that what I was looking for.

$\ce{2O}$ is basically two atoms of oxygen, unbonded and separate.
On the other hand, $\ce{O2}$ is oxygen molecule, which is more commonly called oxygen gas.


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Why add acid to rid of soluble carbonate or hydroxide impurities? Will it not also react with halide I'm testing for
When identifying metal halides with silver ions, the positive ion $\ce{Ag+}$ will form $\ce{AgX}$ with halide in a solution. However, $\ce{H+}$ is added (typically in the form of an acid like $\ce{HNO3}$) to solution to get rid of $\ce{OH-}$ and $\ce{CO3^{2-}}$ ions first. Will the $\ce{H+}$ not also react with the halide $\ce{X-}$?

Hydrogen halides are strong acids, meaning that their conjugate bases $\ce{X-}$ are extremely weak. Thus, adding $\ce{H+}$ will neutralize stronger bases such as $\ce{OH-}$ first before $\ce{X-}$.


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Color fading in contact with air over period of time
I am a product designer and I am wondering if there is a pigment or ink that is able to fade over time by staying in contact with air? I am designing a label that  should change color after approximately 6 months, in order to detect that the product is expired. What type of support is needed? I found these type of examples:
http://pubs.acs.org/subscribe/archive/ci/31/i02/html/02haas.html
Thank you!

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center.

As a side note, if you're a product developer, and hope to eventually come up with a product that you or your company would be interested in patenting, be care what you post here, as this is a public forum, and including too many details may constitute public disclosure.

Maybe look for something that reacts water vapour slowly? This wouldn't work if your product will be in a low humidity, cold storage situation for a significant period of time.

In its current form, th question may be voted as too broad for ChemSE.  If you were still interested in a solution based on slowly advancing chemical reactions, perhaps a platform like innocentive (https://www.innocentive.com/) represents an alternative.

Whether the color change is caused by chemical or physical phenomena, the rate of change almost certainly would be influenced by external factors, such as temperature or humidity during storage.


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The eigenvector matrix C

I was wondering what the difference between $c_n$ and $c_{jn}$ is? In the context of molecular orbital construction, for example, if $c_{jn}$ is the coefficient of the basis functions $\phi_j$ (contribution of the constituting atomic orbitals), what would $c_n$ be?

$c_n$ is an eigenvector, $c_{jn}$ is a value of an eigenvector

Ah, the commonest of sins in mathematical chemistry, elimination of indices without changing the parent symbols...

Isn't Math.SE a proper place for this question?

@pentavalentcarbon yeah wherever that's from, it's really bad....

@Ruslan Under the strictest rules, this does belong on Math.SE. However, I argue that this is a prime example of fundamental quantum chemistry (with the example given) and one where notation is important. IMO, chemists tend to be sloppy here.

$\mathbf{c}_n$ is a vector with $N$ entries (I switched to boldface to indicate - hopefully slightly more clearly - that it is a vector):
$$\mathbf{c}_n = \begin{pmatrix}c_{1n} \\ c_{2n} \\ \vdots \\ c_{Nn}\end{pmatrix}$$
This effectively means the same thing as writing
$$\psi_n = c_{1n}\phi_1 + c_{2n}\phi_2 + \cdots + c_{Nn}\phi_N = \sum_j c_{jn}\phi_j$$
I'll try to contextualise it. The columns of the eigenvector matrix are simply the eigenvectors themselves, so if (for example) the eigenvectors of an operator $\mathbf{A}$ are
$$\mathbf{c}_1 = \begin{pmatrix}p \\ q\end{pmatrix}; \qquad \mathbf{c}_2 = \begin{pmatrix}r \\ s\end{pmatrix}$$
then the corresponding eigenvector matrix $\mathbf{C}$ is
$$\mathbf{C} = \begin{pmatrix} \mathbf{c}_1 & \mathbf{c}_2 \end{pmatrix} = \begin{pmatrix} p & r \\ q & s \end{pmatrix}$$
[As an addendum: If the eigenvectors $\mathbf{c}_i$ are normalised, then you will find that
$$\mathbf{C}^{-1}\mathbf{A}\mathbf{C} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
where $\lambda_i$ is the eigenvalue of $\mathbf{c}_i$, i.e. $\mathbf{Ac}_i = \lambda_i\mathbf{c}_i$.]


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What are the boiling points of DMSO-water concentrations?
What are the boiling points of DMSO-water concentrations with respect to wt% DMSO? I have found extensive information on the freezing point change of wt% DMSO but nothing regarding boiling.

Figure 14 from http://www.gaylordchemical.com/products/literature/dmso-physical-properties/ seems relevant.

DMSO was FDA approved to medically store human tissue (organs, samples, etc.) down to much lower temps than the freezing point of water, when tissue is immersed in a 50/50 mixture of DMSO and water. Method of storing tissue was Proposed to FDA in the 1960’s by Dr. Stanley Jacob, a surgeon working at an Oregon hospital.

The binary mixture of water and DMSO do not form an azeotrope (a constant boiling mixture, where the vapour produced has fixed percentages of each component). Thus, water will boil from the solution first at a bit above 100 ºC, followed by the boiling of DMSO at around 189 ºC.
The reason for the drastic change in freezing point is a little more complicated. Freezing point depression is a common phenomenon; for example, it's the reason sea water freezes at a lower temperature than pure water. In these cases, the reason for the freezing point depression are typically entropic.
However, in the case of DMSO and water, the depression is of particular note—mixtures can remain liquid down to -140 ºC, compared to the normal freezing points of 0 ºC and 19 ºC for water and DMSO, respectively. Quantum mechanical studies have suggested that the formation of DMSO-water clusters is at least partially responsible for the massive freezing point depression. These clusters are strongly hydrogen-bonded, but interact far less strongly with other clusters, making it more difficult to form a solid lattice (and thus a solid).

Kirchner, B.; Reiher, M. J. Am. Chem. Soc., 2002, 124 (21), pp 6206–6215

Thank you. Why does the freezing point change so drastically then?
http://chemistry-chemists.com/N3_2011/U/DMSO-technical_bulletin.pdf

I've edited the solution to discuss the reasons behind the freezing point depression;it may be good to edit your question and title to reflect this.

The freezing part is fine, but your description of boiling is misleading. Water will not boil at around 100 ºC (unless your solution is mostly water), nor will it boil alone.


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Finding the molar concentration of reactant used
We were given this problem at my university, and I can't seem to figure out how it was solved. The question says we have $\ce{NaHCO3}$ and 100 ml of $\ce{HCl}$ react and one of their products is $\ce{CO2}$ gas whose volume is 11.2 L, find the molar concentration of $\ce{HCl}$.
I first found the moles of $\ce{CO2}$ by dividing 11.2 L / 22.4 L which gave me 0.5 mol. Then I know concentration is number of moles/volume, so should I assume the moles $\ce{HCl}$ are also 0.5 by ratio and divide 0.5 by 0.1 L?

Try to solve it yourself first. Don't give up so easily. Edit your post showing your efforts.

Note: This is a homework question, so I will not provide a complete answer.
You have found the amount of $\ce{CO2}$, where did it come from? If you guessed sodium bicarbonate, congratulations, you win a cookie! Now you should be knowing the amount of $\ce{NaHCO3}$ that underwent the reaction.
Write the balanced equation for the reaction of $\ce{HCl}$ and $\ce{NaHCO3}$. Assuming that $\ce{NaHCO3}$ is present in excess, how much $\ce{HCl}$ was present in that $\pu{100mL}$ sample. What is the concentration now?

Ok, if NaHCO3 was in excess the balanced equation would be 1NaHCO3 + 0.5HCL = 0.5CO2 because the reaction ratio is still 1:1 so even if sodium bicarbonate was in excess , the HCL would be limiting how much HCL is produced


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Balanced equation for the hydration of cyclohexene to synthesise cyclohexanol
I'm trying to determine the full balanced equation for a hydration of cyclohexene to synthesise cyclohexanol using concentrated sulfuric acid as catalyst and distillation as the extraction method. I'm not quite sure if I have done the balancing right, and if the notation is correct.
Here is my attempt:
$$\ce{C6H10 ->[][{\ce{H2SO4}}] C6H12O4S ->[][{\ce{H2O}\, + heat}] C6H13O^+ + HSO4^- <->[\ce{H2O}] C6H12O + H3O+}$$

Why is the sulfate bonded with the cyclohexyl ring? This should just be a protonation. Generally, in organic chemistry, plain old balanced reactions are of limited use. You almost always want to draw out an arrow pushing mechanism instead.

@Zhe I know cyclohexyl hydrogen sulfate is an intermediate compound of the reaction which is why I included it here.

Except it's not. This is a classic $S_{\mathrm{N}}1$ reaction. The hydrogen sulfate might be coupled to the cation as an ion pair, but you're really not going to bond them.


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VSEPR Theory and Predicting Shapes of N2F2
I had asked a question on a IA Past Paper:

Predict the structure of $\ce{N2F2}$ using VSEPR Theory.

Now I'm not that strong at VSEPR theory. For the most part, I understand it when it involves a central atom and a ligand, but how does it work for other molecules, for example:
$$\ce{N2F2}$$
Any clearer understanding of this or any help learning about the specifics of VSEPR theory, as in how it is generally use regarding or irregular molecules, is greatly appreciated.

Point your finger at one of the nitrogens and pretend this is the central atom.

Have you drawn the Lewis structure?

As far as you understand VSEPR theory (which is around a central atom, and actually yes! That's all you need to know to answer this) , you know that first of all you have to draw the Lewis structure of the compound, which you can draw like this and now you have got the preliminary linear structure of the molecule $\ce{N2F2}$.
Here, FYI, there might be a mistake. It can be done by considering $\ce{N2F2}$ to be analogues to $\ce{NF3}$, or anything pyramidal-like that which is totally wrong!
Now you just have to apply the VSEPR theory on the $\ce{F-N=N}$ moiety in the left side and then on the $\ce{N=N-F}$ moiety in the right side(which is now your cup of tea!). And after applying the VSEPR theory you will get these two structures and both these isomers do exist!

Please comply with the homework policy. Don't give complete answers

@Pritt Balagopal Yes, see that's why I just have given the outline of how will he proceed, I haven't explained in details how will he apply the VSEPR theory on the mentioned two moieties. And if that goes against the homework policy then I will keep this in mind next time. Thank you.

Based on your answer, he doesn't have to see all that, you already mentioned the cis and trans isomers to him. He wouldn't need more.

Okay! Now I got the point. I'll keep it mind.

One way to solve this problem is to treat both of the nitrogen atoms as central atoms. Each nitrogen has 4 electron pairs. However, because two of these pairs are in a double bond each nitrogen atom could be treated as having 3 electron pairs. This leaves two possible configurations: trigonal pyramidal and trigonal planar. Trigonal planar is the preferred structure because the repulsions between the other nitrogen and the adjacent fluorine are too high in the trigonal pyramidal structure.

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Is it safe to combine these?
It was suggested to me by a co-worker that I could create a very powerful and highly effective weed killer by combining One gallon of chlorine bleach,One gallon of vinegar and One pound of salt. Would these react to each other? would it be dangerous to combine them?

it's dangerous, could make chlorine gas

https://chemistry.stackexchange.com/questions/34006/does-bleach-without-rinsing-leave-residues-and-can-vinegar-baking-soda-neutra

I'm voting to close this question as off-topic because it's simpley "layoing" with  chemicals.

Read the bleach label. Does it or does it not answer your question? The bottle I have in front of me states:"DO NOT use or mix with other household chemicals, such as toilet bowl cleaners, rust removers, acid or ammonia containing products." In case you don't know: vinegar contains acetic acid. Life skill: before you take random advice, have a pretty good reason to trust the person giving it to you. The person giving you this advice may not be an idiot; it's possible s/he's just an idiot when it comes to chemistry...or perhaps s/he is out to get your job. Mixing vinegar with chlorine bleach generates chlorine gas. Chlorine gas is a poison gas. Poison gas kills people. Interestingly, the chlorine poison gas used in the First World War killed very few, but injured many - mostly blindness and (often permanent) lung damage. Salt and bleach might kill weeds, but it will kill most other plants, too. And could hurt any animal that touches it. If you want a "very powerful and highly effective" weed killer, why not just pour gasoline onto the area and light a match? At least then you won't leave behind an area where little or nothing will grow (salt makes soil infertile). Once it's burned off, it's gone (well, any left might cause cancer, but don't sweat the small stuff...). [I'm hoping I don't have to tell you not to do this...] In all fairness to your co-worker, I just mixed some of bleach with vinegar and while it did generate chlorine gas, there wasn't much generated. You might be able to do this without killing yourself, especially if you did it in a well ventilated area. Although I only tried it with a couple of tablespoon fulls, I suspect using a gallon would still not be lethal. I sure wouldn't try it. Nor do I think it would make a particularly effective herbicide. If you really really need to check this out, forget the vinegar, buy some thickened chlorine bleach (will better cling to leaves) and mix with salt and use that. I have no idea what it will do. Its possible that any animal seeking salt will be (severely) hurt, and it will damage lots of surfaces it comes in contact with. But I'd expect (I don't know, I've no experience nor direct knowledge on this) that it would kill the leaves it was on (I would expect the plant to survive). But its safe enough (unless children are around) for you to try that. Keep away from adding cleaners containing acid, including vinegar, and keep away from adding ammonia.


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Column pressure control in distillation
In the industry , it's important to maintain the column pressure control in the distillation column .
" Pressure alterations will alter column vapour loads and temperature profiles. .
So when using Temperature control as a substitute for composition control, pressure compensation is essential to achieve a desired composition . "
"Pressure variations change relative volatiles and affect fractionation process.  "
From this , I understand that if pressure acting upwards on the vapour is lesser than the pressure acting downwards , then it will affect the fractionation process.
But I do not understand the top part -
"
" Pressure alterations will alter column vapour loads and temperature profiles. .
So when using Temperature control as a substitute for composition control, pressure compensation is essential to achieve a desired composition . "
Why does it link to temperature ? And in conclusion , I'm confused about how pressure is maintained in the distillation column and for what for .

things heat up when you apply pressure to them and cool down when you reduce pressure; I suggest that you take a look at the ideal gas law

The most common way to vary pressure in distillation is with a vacuum pump.


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Mass of produced carbon dioxide
I am studying for a big test and I ran across something that I couldn't solve no matter how hard I tried. (reading through old tests)
Burning ethanol produces carbon dioxide ($\ce{CO2}$) and water ($\ce{H2O}$).
a) Write down a balanced chemical equation that illustrates the complete burning of ethanol.
b) What is the mass of produced carbon dioxide ($\ce{CO2}$), when completely burning 575 liters of ethanol?
NTP conditions apply. Ethanols density is 0.8 g/cm³.
MC = 12 g/mol, MH = 1 g/mol, MO = 16 g/mol
For the first part I came up with $\ce{C_2H_6O (l) + 3O2(g) --> 2CO_2 (g) + 3H2O (g)}$
Then I calculated that $M_\ce{CO_2}$ = 44 g/mol and $M_\ce{H_2O}$ = 18 g/mol
I also calculated that 575 liters of ethanol weighs 460 kg
I know that m = nM
However I am not sure what to do next.
I may also be using some weird terms here as this is my translation and I've never dealt with math/chemistry in English.

Ethanol's molecular mass $M(\ce{C2H6O})=46\ \mathrm{g/mol}$
$$n=\frac mM,$$
so
$$\frac{460\,000\ \mathrm g}{46\ \mathrm{g/mol}} = 10\,000\ \mathrm{mol}$$
So having 10 000 moles we can multiply that by the molecular mass of carbon dioxide and multiply again by two because there are two carbon dioxides.
$$10\,000\ \mathrm{mol}\times44\ \mathrm{g\ mol^{-1}}\times2=880\,000\ \mathrm g=880\ \mathrm{kg}$$

1.u can easily analyse that 1mole of etanol produces 2 moles co2.
2.so first calculate M for ethanol=46g/mol.
3.m calculated by u for C2H5OH is 460000gm
4.so no of moles=m/M=460000/46=10000 moles.
5.now 1mole C2H5OH produces 2mole co2 so
using unitary method 10000 moles of C2H5OH will produce=10000×2 Co2 moles.
6.molecular mass of co2=44g/mol,20000 moles=20000×44=880000gm=880kg of co2 is ur ans.

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Guidelines for when oxidation is mild and when it is strong in organic chemistry?
What are the rules, in organic chemistry, that determine whether an oxidation is mild or strong?
I know that potassium dichromate is a strong oxidizing agent and it can remove two hydrogen atoms from a primary alcohol to form an aldehyde.
So is strong oxidation (in organic chemistry) when more than one hydrogen atom is removed i.e. when oxidation is greater than the removal of just one electron?

No, these are totally unrelated things.

The ease of an organic redox reaction to occur correlates with the difference of electrode potentials between the half reaction of oxidation, and the one of reduction.  Hence, for example, $\ce{KMnO4}$ a more powerful oxidising agent under acidic, than under neutral or basic conditions (related: https://chemistry.stackexchange.com/questions/73350/is-there-any-difference-between-balancing-redox-reactions-in-acidic-and-basic-me/73357#73357).

If this is a serious post, which is not easy to believe, then:
Mild = sufficient or inadequate, depending on context.
Strong = adequate or excessive, depending on context.
There is no "rule", feel free to make your own. It needs to be consistent, however, and that is a problem since what is mild in one reaction context will be too strong in another. Also note that many organic reactions can be done electrochemically, which means the requisite potential can be determined. Unfortunately, oxidizing agents aren't one-dimensional. That is, their reactivity isn't based simply on their ability to accept electrons. Instead specific (element or group) interactions (additions, abstractions, substitutions, rearrangements) also matter. For example a reagent which had only a moderate tendency to oxidize but had a high tendency to add a particular group might be considered a "strong" oxidant.


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What is the point group of hexachloridotitanate(III)?
I found the answer to this question to be $D_\mathrm{3h}$ but I am not convinced. Can someone explain me with a diagram if possible and explain why it doesn't belong to the octahedral point group $O_\mathrm{h}$?

Six Cl can be arranged octahedrically all right, and the charges do not need to be arranged at all, so there must be some other reason... Jahn-Teller distortion, by any chance?

A figure would help here but if its regular then point group $O_h$ and if two opposite ligands extend or contract as @Ivan Neretin suggests then $D_{4h}$. $D_{3h}$ does not seem to fit at all but would for $\ce{TiCl5}$

Yeah, that too. If you distort an octahedron so as to keep the threefold axis, you get $D_{3d}$ at best.

Trigonal prismatic geometry would be D3h, but that would imply a very large distortion.

Just so we are explicit and clear, $\ce{TiCl_6^{-3}}$ implies $\ce{Ti^{3+}}$, which is $d^1$, and the Jahn-Teller effect should be weak.

Probably relevant: Jahn-Teller effect of titanium(3+) in octahedral coordination: a spectroscopic study of hexachlorotitanate (TiCl63-) complexes. Inorg. Chem. 1985, 24 (18), 2751–2754

I was able to pull a small snippet from the abstract of @orthocresol link and they seem to suggest a $D_{4h}$ distortion. Looking at "Compounds $\ce{A_2,A’TiCl6}$ (A, A’ = alkali-metal ions)" they write "The phase transitions at 210 and 155 K for A, A’ = Cs, K and Rb, Na, respectively, are caused by geometric packing forces and induce a tiny tetragonal distortion of the $\ce{TiCl6}$- octahedra($D_{4h}$ compression),which superimposes on the otherwise dominating dynamic Jahn-Teller coupling."

Getting straight to the point, I believe whichever reference you have that claims $D_\mathrm{3h}$ is incorrect.
Titanium(III) complexes were the first example presented in my coordination chemistry class to examine the energy difference between the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ orbitals, explain why it exists and where the colour of coordination compounds derives from including carefully examining the UV spectrum of titanium(III). Our lecturer, Professor Klüfers of the LMU Munich, explained that the hexaaquacomplex is practically octahedral in shape. However, there is not just a clean absorption band but rather a band with a shoulder. This is because the excited state of $\ce{[Ti(H2O)6]^3+}$ is Jahn–Teller distorted, having an uneven $\mathrm{e_g}$ population, meaning that two slightly different excitation energies exist.
One might go a step further and argue that even in the ground state a certain distortion should exist. If one were to pull (not push) the ligands in $z$-direction slightly closer to the central metal, this would destabilise any orbitals with $z$ contribution meaning that $\mathrm d_{xy}$ suddenly becomes the single most stable d orbital. This slight distortion which one might call anti-Jahn–Teller (since it is opposite to the classic Jahn–Teller distortion) could explain a reduction of symmetry from $O_\mathrm{h}$ to $D_{\mathrm{4h}}$.
$D_\mathrm{3h}$ symmetry is not consistent with octahedraloid coordination spheres at all. It is very common for pentacoordinated centres such as $\ce{PF5}$. The only way to have six ligands — as in hexachloridotitanate(III) — around a central metal in this point group would be a trigonal prism. That would be a very uncommon coordination sphere and I am sure I would have heard of it in the context of titanium(III) in said course above should it exist for this species. Indeed, the source orthocresol found and linked in the comments makes no mention of any distorted geometry at all:

The low-temperature structure results from rotations of the $\ce{[TiCl6]^3-}$ octahedra around a 4-fold axis. (Emphasis and square brackets mine)

Source: Amels, R.; Kremer, S.; Reinen, D. Jahn-Teller effect of titanium(3+) in octahedral coordination: a spectroscopic study of hexachlorotitanate (TiCl63-) complexes. Inorg. Chem. 1985, 24, 2751–2754. DOI: 10.1021/ic00212a009.

Hexachloridotitanate(|||) has D4h point group because structure distorted by Jahn -Teller distortion, two axial bonds get elongated.
Point group can be determined as following..

Aside from the illustration, this doesn't seem to add anything to the existing answer, which actually explains why there is JT distortion in the first place and cites literature sources, whereas your answer looks like a postulate. Also, please note that | and I are different symbols which are not interchangeable. In the absence of the proper Unicode symbol for Roman numeral three Ⅲ, typically three English letters I are allowed to be used: III.


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The MO and unpaired electrons
I am learning molecular orbital in general chemistry. I have come across this question that asks me to show that the Lewis structure of $\ce{O2}$ is actually an excited state when viewed in terms of the molecular orbital energy.(Zumdahl 9th edition, chapter 9, Excercise number 50) The solution book tells me that the Lewis structure of $\ce{O2}$ is an excited state because the two electrons in the $\pi ^*2p$ does not follow Hund's rule(the two electrons there exist in the same $\pi ^*2p$ orbital while leaving the other $\pi ^*2p$ orbital empty.) And it further states that the reason why the two electrons in $\ce{O2}$ Lewis structure are paired is because the Lewis structure of $\ce{O2}$ shows every electron to be paired. But my question here is, why does being paired in the Lewis structure equate to being paired in MO? I mean, when MO is formed, aren't all orbitals reconfigured, and therefore all electrons are also reconfigured?
While I was thinking about this, another question popped out of my mind. Why are the name of MOs like $\sigma$ and $\pi$? Doesn't $\sigma$ and $\pi$ get used when talking about $\sigma$ and $\pi$ bonds? Are $\sigma$ orbitals and $\sigma$ bonds correlated in anyway?

First things first, learn the difference between an apostrophe " ' ", and a backtick " ` ". The apostrophe is found above the question mark on the keyboard, while the backtick is found to the left of the "1" key.

Unfortunately you didn't keep it mind. You typed "I`ll keep that..." Backtick again. You shouldve type "I'll keep that...".

The reason why I tell you not to use backticks is because it's used to enclose syntaxes that you don't want to be rendered.

Oxygen can exist in two states, the singlet and triplet state. The triplet state is the ground state whereas the singlet state is the higher energy state. In the triplet state the highest energy electrons are unpaired whereas in the singlet state the highest energy electrons are paired.
Now, take a look at the MO diagrams of both states in Wikipedia. You'll be able to tell that the singlet state does not obey Hund's first rule.
Now lets look at the Lewis diagram for a oxygen molecule:

As you can see, all of the electrons in this diagram are paired. The oxygen represented in the diagram above is an oxygen that has two bonds and two lone pairs. The only MO state that has lone pairs is the singlet state.
Now on to "why does being paired in the Lewis structure equate to being paired in MO?" The reason is that paired electrons in Lewis diagrams represent paired electrons. The concept of paired electrons does not depend whether you are using MO theory or Lewis diagrams. Paired electrons are electrons that are in the same orbital.
Now, the next question. "Why are the name of MOs like σ and π? Doesn't σ and π get used when talking about σ and π bonds? Are σ orbitals and σ bonds correlated in anyway?" Sigma bonds and sigma orbitals refer to the same thing. Molecular orbitals are a way to represent the bonds formed by two atoms.
Sidenote: Mulliken first proposed the sigma, pi, delta notation according to this journal article..


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Baking soda + oil solubility
I have a question regarding the solubility of baking soda (bicarb soda) in oil. I am pretty sure that it has poor solubility, but I have a chemistry dilemma and cannot seem to solve it.
I make all-natural deodorant, and the main ingredients are coconut oil, shea butter and baking soda (amongst a few other ingredients). We have been making it for years, and it was always soft and smooth, but this time the baking soda is not dissolving and leaves a gritty texture (even after sifting baking soda to remove lumps).
Does anyone have any idea why this might be? We heat the ingredients to melting point on the stove, then let cool to room temperature and re-solidify in the fridge for a day, before using.
The only thing I can think of is that we have recently changed to a new brand of baking soda; however, with adequate mixing, surely that would not change the final texture so much.
Thank you!

Dissolve the bicarbonate in water first before mixing with your concoction. Add in emulsifier so that you get a nice single phase lotion.

The baking soda grain size might be bigger in this new brand. When you sift, you break up individual grains that have clumped together due to moisture or being pressed together tightly but there is a fundamental particle size from how the baking soda was manufactured that would require further processing to reduce (e.g. milling). The larger grain size would adversely affect the solubility. It could also be that the baking soda you buy now is either missing some additives or has some additives that are detrimental to your final product.

@EssElle Compare the ingredients list between the two brands to see if there is any difference, that may tell you the story from the additives side. The grain size issue would require a powerful microscope seeing as how baking soda is so fine. You could try running the baking soda you have now in a high power blender and see if that helps.

@EssElle If my answer is satisfactory, please consider accepting it.


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Should there be a difference in the absolute energy gaps between bonding and antibonding MOs from the energy level of the AOs in the H2 molecule?
I have been reading Molecular Orbitals and Organic Chemical Reactions (Student Edition) by Ian Fleming. On Pg 5, Fleming mentions that the diatomic helium molecule cannot exists because there is an overall energy increase from populating both the bonding and antibonding molecular orbitals equally (i.e. putting 2 electrons into each). However, why should there be a larger absolute energy difference between the antibonding MO and the atomic orbitals compared to the energy difference between the bonding MO and atomic oritals? Fleming explains this by saying that this is due to the presence of inter-electronic repulsion in the MOs as there are two electrons present in each MO. This may seem to make sense.
However, in most MO diagrams, the energy levels of the bonding and antibonding MOs are drawn equidistant away from the energy level of the atomic orbitals. Does there seem to be a contradiction?
It is not a duplicate as I am proposing a new explanation of "inter-electronic repulsion upon population of the MOs", which has never been discussed before.

To my understanding, the discrepancy is caused by most MO diagrams being drawn qualitatively, leaving off a spectroscopically determine energy scale. When you are just drawing an MO diagram from scratch, it makes sense to draw the bonding and anti bonding as equidistant from the atomic orbitals, because the difference is subtle unless you have experimentally determined orbital energy values to scale the difference.

Related (though there is some dispute on the accepted answer)https://chemistry.stackexchange.com/questions/4517/how-can-antibonding-orbitals-be-more-antibonding-than-bonding-orbitals-are-bondi?rq=1

May I take this to be the correct understanding: Initially, the energy levels of the bonding and antibonding MOs should be equidistant from the energy level of the uncombined atomic orbitals. However, after population by electrons shifts the energy levels of both types of MOs up by the same absolute energy, thus causing the energy levels of the MOs to be no longer equidistant from the atomic orbitals. Is my refined understanding now accurate?

@TanYongBoon No, the population of the orbitals is not the reason for the energy levels of the MOs not being equidistant from the AOs. It is rather the fact that the AOs overlap and are thus not orthogonal. If they were, the levels would be equidistant. The process of forming bonding and antibonding MOs, which are required to be orthogonal (due to the Pauli exclusion principle), leads to an overlap-dependent energy-raising term in the expressions for the energy levels of both the bonding and antibonding MOs. See this answer for the equations.

@Tyberius Just for the record, as far as I know the "Intuitive Explanation" in the answer you provided a link to is incorrect.

@Philipp Thanks for the clarification. However, I would like to know why is it an incorrect explanation. Inter-electronic repulsion should theoretically raise the energy of the orbitals, even if ever so slightly. Although I don't really understand, the explanation you provided is probably the more accurate one but I would like to also contend that this inter-electronic repulsion explanation is also accurate but just not so significant. Would you agree?

@Philipp Also, I was wondering if you could suggest to me some literature on this topic. I am really interested in MO theory but I don't really have any competency in QM and also, I don't have the luxury of time to dwell on these topics as I have school. I don't require a high-level understanding but I just want to learn it for practical use (i.e. predicting nucleophilicity and electrophilicity). Essentially, I just want to learn about the interaction of frontier MOs (HOMOs and LUMOs), if that simplifies it.

@TanYongBoon The problem with inter-electronic repulsion is that it is not the whole story when it comes to "occupation phenomena". You have for example also exchange energy contributions, so I'm not sure if the energy of an orbital necessarily rises when you fill it with electrons. This is rather complex. But more importantly it is simply not the reason why  the energy levels of the MOs are not being equidistant from the AOs. This is a result you get from the perturbational treatment of the problem of interacting orbitals which does not deal with different occupations.

@TanYongBoon If you only need MO theory for understanding reactivities  on a "pictorial level" especially in the context of organic chemistry then I'd recommend Molecular Orbitals and Organic Chemical Reactions by Fleming. This should give you most of what you will need to get a good understanding for orbitals in organic reactions.

@TanYongBoon Another good and short read which gives you a glimpse of orbitals in the context of solids - and is easy on the math while doing so -  is Solids and surfaces: A chemist's view of bonding in extended structures by Roald Hoffman. It is really remarkable how much it explains in so little pages.

@Philipp Thanks for the explanation and also for your recommendations.


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How do I determine which is the reduction reaction in a hydrogen peroxide and potassium permanganate solution?
How was it determined that $\ce{MnO4-}$ was reduction, and $\ce{H2O2}$ was oxidation?
Also, why wasn’t the equation $$\ce{MnO4- + 2H2O + 3e- -> MnO2 +4OH-}$$ used? Or the $$\ce{H2O2 + 2H+ +2e- -> 2H2O}$$? (These are from the half equation tables)

$\ce{KMnO4}$ is quite a powerful oxidizing agent, unless you add something really crazy like dioxygen difluoride to it, it's going to undergo reduction.

In the case of permanganate, it is in a very oxidised state, +7, and all the other stable oxidation states of manganese are lower (+6, +4, +2...) so it will typically reduce in redox reactions (i.e. it will be an oxidising agent). You can also know this from practical experience if you know that permanganate is used as a general oxidising reagent.
More generally, if you don't know whether a species is reducing or oxidising, try to build a picture of the entire reaction:
1) If you know what species it ends up as, it's easy: you only have to count electrons using standard charge rules. For instance, for MnO4- you know that oxygen atoms have a formal charge of 2-, so the manganese atom is formally Mn(7+); if you know that the product is Mn2+, it is reducing.
2) If you don't know that, look at what it is reacting with. Sometimes, one or two of the species will only oxidise or reduce (if it's the highest or lowest oxidation state you find in the table). For instance, MnO4- is in the highest oxidation state you'll find for manganese at the table (Mn7+), so you know it is reducing.
3) The trickiest case is when you have two species reacting that, depending on the circumstances, could both act as oxidising or reducing agents (i.e. species that appear at your tables both as reagents and as products). In that case, you have to use the standard reduction potentials to figure out.
About whether to use acidic or basic potentials (the tables with H+ or the tables with OH-), these are all equivalent and you are OK using either (as long as you are consistent!). Their difference in value comes from the water dissociation constant Kw, but if you write the semireactions and adjust the water, you end up with exactly the same results in both cases. If you know that your medium is acidic or basic it's generally a good idea to use the corresponding values (it saves you having to adjust the water), but it will still work as long as you don't mix them up and adjust the reaction (I can't stress that enough).


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What happens when utensils are washed in detergent or soap.what type of reaction it is?
What happens when plates after dinner are washed with soap or detergent ? What colour change is observed? What type of reaction it is?

The only color change I can think of is the dishes becoming cleaner

Broadly speaking, what you have is not a chemical reaction (no bonds are created and destroyed) but supramolecular processes (so molecules rearrange and interact differently with each other, without changing their formula). The main process when cleaning with soap is the formation of emulsions. Soap molecules typically have a polar head (which has affinity for water) and an apolar tail (which loves the fatty leftovers in your plate). The fatty residues won't dissolve in water, but the soap molecules will coat small droplets of fat with the apolar tails touching the fat (which is good) and the polar heads touching the water (which is also good). These droplets covered in soap are called "micelles", and they can be very stable (if you had the same fat without soap they wouln't be, and the fat and the water would separate, as happens with water and oil). This actually gets very complicated and there are many many details to it, but that's the basics.
Surfactants (the technical term for soap-like molecules) are also involved in the foams, aerosols, dispersions, wetting and many other supramolecular phenomena involving more than one phase (such as water/air, water/oil...).

There's no such thing like color change that I have observed but as far as the mechanism I can say it has to do with hydrophilic and hydrophobic parts of long chain carboxylic acid(soap)and long chain sulphonates of carbon (detergents) you can take reference from web for further research.


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Comparison of COSY and TOCSY for detection of small couplings between protons
I routinely run COSY spectra to help with assignment of my proton NMR.
The experiment is fast to run and generally useful, however some of the intermediates I've recently been making have had sufficiently small coupling constants that the COSY hasn't been terribly useful in certain regions(a 'standard' COSY isn't giving the resolution needed).
Typically, I get around this by running a set of HMBC and edited-HSQC, which allows me to trace the protons along the molecule, in a similar way as I'd be able to do with the COSY. The issue with this is that with the quantity of material I have, the HMBC takes significant amounts of time to run, making this method unhelpful for day-to-day work.
A colleague suggested using 2D-TOCSY to see which protons are in the same spin systems (not the same information as in the COSY, but would give me what I need).
Looking through the ever useful Claridge, it seems that the mixing pulse from the COSY gets replaced with a 'spin lock' sequence. What is less clear to me from this though, is whether TOCSY is also dependant on the magnitude of the coupling (i.e. if the J values are minimal, will the TOCSY suffer from the same issues as the COSY), or whether the spin systems are detected indirectly/through some other property.

Standard 2D COSY spectrum

Standard 2D TOCSY spectrum

Deleting my previous comment because on second thought, this is actually quite interesting. I don't know the answer, but I believe you are asking if a coupling that doesn't show up well in COSY will show up via an intermediate coupling that's part of the same system...

That sums it up pretty well.

http://chem.ch.huji.ac.il/nmr/techniques/2d/tocsy/tocsy.html suggests that there is a drop-off for coupling at longer range. I guess we're hoping that (a) the coupling exists and (b) that the falloff is less than the lower coupling observed in COSY.

One other experiment you could try is a delayed COSY, which should have better sensitivity for small coupling constants, the Bruker pulse sequence for that is cosylrqf.

Magnetisation transfer in a TOCSY experiment occurs via a different mechanism than for COSY. The spinlock generates an oscillatory exchange of coherence between spins that is propagated thoughout the coupling network. The efficiency of the spinlock transfer depends on a number of parameters; molecular conformation and topology which effects the coupling constants, and relaxation rates across the spin network being the most important. So, yes, signal does depend on the size of the coupling, but occurs via a different mechanism.
Once the spins are brought into the transverse plane through a standard excitation 90 and allowed to evolve during t1 period (as per a COSY), they are 'locked' into position using low cw power, or usually by a composite pulse train. Those signals that are in-phase following the evolution period are aligned with, and locked to, the transverse axis, effectively removing the influence of the primary magnetic field, Bo. The spin frequencies are now proportional to this new spinlock field, which is very small compared to Bo (so the chemical shifts are effectively all the same, because chemical shifts/transition energies are proportional to the strength of magnetic field), and so all of the spins are strongly coupled and this allows efficient mixing of magnetisation fur the duration of the mixing period. This is called isotropic mixing.
Typical values for spinlocks range from about 30ms-120ms. There is no hard and fast rule for predicting the optimum mixing time. Magnetisation transfer ocurs faster between spins with large coupling constants, so only 2,3J couplings will be seen with short mixing times. There are various rules of thumb used for selecting the appropriate mixing time. You can use 1/(10JHH) where JHH is the coupling across these transfer steps. A typical JHH is about 7Hz, so to see a 5 step transfer you might choose a mixing time of:
5 x 1/(10x7) ~ 70ms
Another guestimate is that around 15ms will give you results similar to a COSY, and you can add 15ms for each transfer step, which will give you something like:

A->B                15ms
A->B->C             30ms
A->B->C->D          45ms
A->B->C->D->E       60ms
A->B->C->D->E->     75ms

I would usually suggest values of 40-70ms as a good starting point. For larger molecules, you will lose signal through relaxation with longer mixing times. Also, longer mixing times give wider distribution of magnetisation, which means an overall reduction in total signal-to-noise
It is not uncommon for research samples to record 2 TOCSY experiments using different mixing times; one short and one longer.
A TOCSY is not necessarily a solution to a system that contains small or vanishing couplings (not weak, as you suggest). The TOCSY experiment still requires coupling to exist along the spin system, and it is the break in the spin-coupled pathway that halts the distribution of magnetisation that allows individual spin systems to be identified.

The TOCSY transfer also depends on the size of the J coupling, you don't get around that.

Generally speaking it is found that cross peaks due to direct coupling build up to their maximum intensity in a time which is of the order of $1/(2J)$. Peaks which arise from transfer through two successive couplings take longer to build up, and if three successive couplings are involved, even longer mixing times are needed
Understanding NMR spectroscopy, James Keeler, page 230

The TOCSY has the advantage of creating in-phase signals instead of the anti-phase signals of the COSY, which increases sensitivity. But that doesn't mean that it is generally more sensitive.

Whether or not the overall intensity of the COSY or TOCSY cross peaks is greater depends intimately on the choice of the experimental parameters and on the molecule being studied. However, the general experience is that for small to medium-sized molecules, COSY is the preferred experiment for detecting small couplings.
Understanding NMR spectroscopy, James Keeler, page 229

I'm not sure if you can get a general answer for your case, you might have to directly compare the experiments for the kind of molecule you want to look at. You could also try longer mixing times for the TOCSY and see if they help with your small couplings (which they should, as long as relaxation doesn't eat up that advantage).


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How does water freeze and crystallize on a cold superhydrophobic surface?
Superhydrophobic surfaces are nano-structured to "repel" liquid water and there has been a lot of work done to understand the microscopics of wetting of these nanostructures.
What I currently am interested in understanding is wetting and freezing of water on a cold, below-freezing superhydrophobic surface.
I would expect that the ice would still have mechanical adhesion to the surface, but I don't know to what extent.
So my question is the following: How does water freeze and crystallize on a superhydrophobic surface?
I know this is a general question that would depend on the nature of how superhydrophobicity is achieved, so feel free to specifically treat a certain branch of superhydrophobic surfaces.

As acknowledged in your question, there are a large number of factors at play here. The most basic of these is that there are three fundamental mechanisms by which freezing of a droplet can initiate when in contact with a superhydrophobic surface: 1) heterogeneous nucleation where an ice freezing nucleus (IFN) is present in the water droplet, 2) heterogeneous nucleation where the superhydrophobic surface serves as the IFN, and 3) homogenous freezing of the water droplet.
An excellent demonstration of the first or third scenario above (it is not clear from the article which is actually the case) is given in video format from an article published in Nature$^1$. In this experiment, the water droplet is evaporatively cooled until freezing initiates at the water/air interface, and continues without adhering to the aluminum-based superhydrophobic surface.
Whether or not freezing will initiate at a superhydrophobic surface depends upon the degree of supercooling, the presence or absence of potential IFN within the water droplet, and the potential for the specific material (the superhydrophobic surface) to serve as IFN. It would seem that the latter of these would be unlikely, except that a surface being hydrophobic doesn't mean that it is equally icephobic. Basically, the more the surface resembles the crystalline structure of ice, the more likely it is to act as an IFN.
Please don't hesitate to ask for any clarifications in the comments below.
1) Thomas M. Schutzius, Stefan Jung, Tanmoy Maitra, Gustav Graeber, Moritz Köhme & Dimos Poulikakos, Spontaneous droplet trampolining on rigid superhydrophobic surfaces, Nature 527, 82–85 (05 November 2015) doi:10.1038/nature15738

Thanks for the answer airhuff. If you don't mind, I'd like to ask a few more questions:

Are there any general features of how well ice adheres to hydrophobic surfaces after forming? Naively I would expect them to adhere quite weakly, but it would depend case-by-case. 2. Take the case of standard room humidity and introduce a cold superhydrophobic surface, how would ice form in this case? This seems to be different than if water was pre-existing on the surface

I agree with your expectations that ice would tend to adhere pretty weakly to a superhydrophobic surface. Your example of forming ice from the vapor is interesting. I once did a set of experiments where I cooled a glass rod in ambient air so that it would grow ice crystals. I had a terrible time with supercooled water depositing instead of ice because ice likes to form on crystaline-like materials, which glass is not. If I had been using a superhydrophobic material instead of glass, I suspect 2 things: 1) I would have had to cool to much lower temperatures to induce condensation (continued...)

(meaning higher levels of supersaturation of water vapor would be required to form condensation) and 2) since the focus of superhydrophobic materials is to be hydrophobic, not necessarily ice-phobic, I suspect that at low enough temperatures ice would have formed rather than supercooled water, and that the ice would have adhered to the surface more readily than if it had formed from the liquid phase. This is admittedly somewhat speculative, as I have little experience with superhydrophobic materials, though I do have a lot of experience growing ice ;)


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Will the solubility of a solute increase over the time a solute dissolve?
would like to ask about a question that I came across in one of my assessment books. Appreciate the help that I can get.
Qns: In an investigation, two substances X and Y of the same mass are dissolved in the same volume of water. It is found that X dissolves faster than Y. Which of the following graphs correctly shows the solubility of the substances against time in this investigation?

The answer given is A.
I would like to ask why is that so. Intuitively, I would think that the solubility would be a constant since solubility is defined as "the maximum amount of solute that will dissolve in a given volume of solvent at a given temperature and pressure", therefore it will not change with time as the solute is dissolving.
Even if we're talking about rate of dissolving here, shouldn't the rate be decreasing, as when a higher concentration of particles are dissociated, there would be more of the dissociated particles in the solvent which will shift the equilibrium to the left?
Thanks for all your help!

Yes, we are talking about the rate of dissolving. Yes, the graphs are named wrongly, because solubility is a constant; it should have been something like "Amount dissolved by the time t". Yes, in any realistic setup the rate should be decreasing. This is a dumb question intended to sort out even more dumb people; you're overthinking it.

The y axis is concentration. Downvote to whom wrote the exercise and chose the graph label.

Ivan's comment really nailed the problem, but I'll throw in a bit more.
In the question the term solubility would be ok in an English class, but the term is incorrectly used within chemistry. The more correct tense would be solvation. Likewise the diagrams should have solvation rather than solubility.

In an investigation, two substances X and Y of the same mass are dissolved in the same volume of water. It is found that X dissolves faster than Y. Which of the following graphs correctly shows the solubility solvation of the substances against time in this investigation?

According to question the rate of dissolving of X is higher than that of Y so the slope of the solubility-time curve will be higher for X that's why A should be the answer!


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Do processes involving synthesis need heat?
I came across a textbook that stated a combination process that required heat. There was no explanation whether it was a prerequisite for a combination process to have heat or it was just an example just that it was grouped under "heat process chemical changes".
Was doing a test paper that asked about 2 differences between thermal decomposition and combination processes and one of the difference was that "combination processes do not need heat but thermal decomposition processes do".
Would like to ask which interpretation is correct. From preliminary searches on Google it seems that the consensus is that any reaction that simply involves two separate substances combining in a reaction to become a single product is considered to be a combination reaction regardless of whether there is heat involved or not.
Would like to know what is the correct interpretation.

By combination process, you mean like: $$\ce{A + B -> C}$$

Hi! Thanks for the reply. Yes, I do mean: A+B⟶C. However, I would like to know if there needs to be heat involved in the reaction by definition.

Definitely. Although combination processes may be exothermic, they do have activational energy. As an example, consider the reaction:
$$\ce{H2 + I2 -> 2HI}$$
While the formation of two $\ce{H-I}$ bonds is quite energy releasing, you still have to go through breaking a $\ce{H-H}$ and a $\ce{I-I}$ bond. You need to supply heat to give thermal agitation to allow the molecules with sufficient energy to break the bonds. A typical reaction like this one would have this energy-reaction coordinate graph:

The reaction proceeds into a high-energy transition state, and then comes down to the low energy products. To reach the transition state however, you must supply heat.
Edit: As @porphyrin mentioned, there are some reactions that have essentially zero activational energy. Combination of radicals may be an example:
$$\ce{Cl. + Cl. -> Cl2}$$
This one has no activational energy because there is no unstable transition state involved. Another popular example is the basic neutralization reaction:
$$\ce{H+ + OH- <=> H2O}$$
Here again, there isn't any highly unstable transition state.

Thank you for the answer! Appreciate the detailed explanation!

@truetoall If this answer completely satisfies you, do click the ✅ button to accept it. This lets others know this question is solved.

As it happens the example you chose is a bit misleading as this reaction is not a simple bimolecular one but involves H and I radicals.  You could point out also that endothermic reactions have an activation energy and some reactions effectively have zero activation energy.


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Aside from carbon, what other substances can be made "superhard"?
So diamonds are an allotrope of carbon, that is formed when carbon is put under immense pressure, right? What other elements or compounds can you make into "diamonds", as in more durable, harder versions of themselves? What happens if salt or iron for example, is put under immense pressure, do they become harder too?

Well, taking it to the absolute extreme, any kind of atomic matter under sufficient pressure will convert to neutronium (neutron star matter), which is unbelievably hard. The speed of sound in such a material (a very rough measure of rigidity) can reach half the speed of light! Of course, if you release the pressure it will immediately explode with the force of a thermonuclear bomb...

Interestingly, nobody addressed the reason why diamonds are hard in the first place. The pressure (and temperature) are not the reason why they're hard, only the reason why they are formed. The diamonds are hard because the carbon atoms are bonded together by sigma ($sp^3$) bonds, which are the strongest chemical bonds. Other materials exhibiting the same kind of bonding are potentially also very hard. But the small atomic number of carbon (Z=6) means that the bonds will be stronger than in higher atomic number atoms, such as silicon.
However, amorphous (non-crystalline, like glass) materials can also be extremely hard. There is even an example that is even stronger than diamond (although not naturally occurring). And this material is made of... carbon! Exactly like diamond. This material, dubbed Q-carbon, was discovered in 2015, and is even created at atmospheric pressure, and around 4000 K with the help of a laser pulse. And it is estimated to be 10-20 % harder than diamond.
Bottomline: gigantic pressures are not required to create extremely hard materials, and diamonds is no longer the strongest known material.

Here are some compounds that have other structures, followed by their hardest structure (based on Moh's Scale).

Titanium dioxide: Rutile structure or Cotunnite structure
Aluminum oxide: Corundum
Silicon Oxide: Stishovite
Boron Nitride: Wurtzite Boron Nitride

There are many, many more.

Minor nit: it's not just the pressure, you also need the right temperature, and that's just scratching the surface. In general, most solids have multiple different structures with different properties, and adding other elements to the mix gives you even more possibilities. How and when exactly these structures form depends on a lot more than just pressure.
For example, pure iron can range all the way from tough and soft to hard and brittle. Pure iron grown in a single crystal (you'll need a vacuum or a reducing atmosphere for this, since iron is quite reactive) is exceedingly soft (softer than lead) and has a very low tensile strength (less than a human hair). Not very useful. As solid iron cools down, it changes crystal structure - each of these has different properties. The way (pure) iron metallurgy works is by exploiting the best structures and freezing them in place - for example, when you cool a piece of hot iron abruptly, it will form a dense and hard structure. The cooling is too fast to allow the whole piece to rearrange, so you get many grains of crystals placed one next to another. Add a bit of carbon, and you get the much stronger steel - the (slight) carbon impurity makes the whole thing tougher and stronger. Even smaller impurities of other elements (like nickel) add the whole thing some very interesting properties.
Sodium chloride (rock salt) doesn't really have any allotropes - it forms a very regular lattice with every sodium atom surrounded by six chloride atoms and vice versa. This is quite typical of giant ionic structures - the optimal structure is very stable. Just like diamond, salt is relatively hard and brittle.
The obvious comparison to diamonds are other gemstones. Most are compounds - like corundum (aluminium oxide) which forms rubies and sapphires, or beryl (beryllium, aluminium and silicon oxide) which forms aquamarines and emeralds. They tend to be hard, which is what makes them useful for jewellery - they resist scratching and keep a nice sheen. But gemstones are just the most obvious - the same minerals are present everywhere around you in the rocks and soils, and the only difference is the size and purity of the crystals. For example, most of sand is formed of silicon dioxide (silica, quartz) - it's basically the hardest thing that remains after all the softer minerals have been eroded (and formed various clays etc.). Silica itself has many allotropes, each forming under different conditions and with different properties. For your question, the most relevant is probably stishovite, a very hard form that forms under high pressures, but even less stable under standard conditions. It's quite similar to rutile, which is titanium dioxide. Silicon carbides are very interesting industrially nowadays - they're rather rare in nature (on Earth), but easily formed at high temperatures and stable at room temperature. Again, silicon carbide itself has many forms (about 250 - perhaps not too surprising given that it's formed from carbon and silicon, both very flexible elements). One especially relevant form is moissanite, which is used as a gemstone, and quite similar to diamond.
Some pure elements form multiple allotropes, though carbon is probably the most impressive element; there's a reason why carbon has its own branch of chemistry. Sulfur is pretty close to carbon, but not quite as useful for humans. You might have also heard about white phosphorous and red phosphorous - both are one of many allotropes of pure phosphorous, each with very different properties. Oxygen is an example of an allotropic element that isn't even solid - molecular oxygen and ozone coëxist with little trouble in both liquid and gaseous forms.
Among metalloids, the most notable is probably silicon - you see it everywhere around you in many of the minerals the Earth's crust is made of, but even pure silicon has interesting allotropes - amorphous (non-crystalline) silicon is used in solar panels, mono-crystalline silicon is used in integrated circuits, and it has forms analogous to carbon's diamond and graphene, with similar properties. Boron is quite interesting in having many different allotropes, including a super-conducting high-pressure phase.
About half the metals that occur naturally in significant quantities are allotropic at room temperature and pressure. The most notable is of course iron, and the flexibility allowed by this is one of the (many) reasons why iron is so important industrially. But I'll take tin for a closer inspection. Alpha-tin has a crystal structure like diamond. It is hard, brittle and non-metallic. Beta-tin is metallic, soft, malleable and ductile, and in cold conditions, tends to spontaneously transform to alpha-tin.
Some allotropes are stable at room temperature, others aren't. Diamond isn't stable - it spontaneously transforms to graphene, though very slowly (though consider that natural diamonds usually form over billions of years; in that scope, the transformation isn't slow at all). Beta-tin is stable close to 0 °C (up to about 13 °C, depending on the impurities), while alpha-tin is stable at room temperature.

Don't see how this "lecture" answers question.

Great answers here, just wanted to add a bit to this answer: the spontaneous transformation of tin allotropes is known as "tin pest" (see e.g. https://en.wikipedia.org/wiki/Tin_pest) and the failure of Robert Scott's expedition to Antarctica is in part credited to this effect (their liquid containers soldered with tin fell apart in that environment).

"just scratching the surface" I see what you did there.

@Mith True, it doesn't answer the question. However, I suppose it would be O.K to see this as a supplement to the other answers O:)

Don't know if this is exactly what you want, but Titanium metal has two allotropes ($\alpha$ and $\beta$).  $\beta$ phase is stable at high temperatures, but you can quench $\beta$ phase titanium so that it exists (metastable) at ordinary temperatures.  $\alpha$ phase is stronger and less ductile.  Small amounts of alloying elements are often added to stabilize either $\alpha$ or $\beta$ titanium.  (You get a similar polymorphism with zirconium metal and hafnium).  $\alpha$ has a hexagonal crystal structure and $\beta$ is body-centered cubic.

I'd like to add that technically, diamond is also a metastable phase of carbon at room temperature and atmospheric pressure. And steel (an alloy of iron, carbon and possibly other elements) has many many allotropes (pearlite, austenite, martensite, cementite, etc etc), which formations depend not only on temperature and pressure, but also on carbon content in the alloy. And obviously the hardness, strength, brittleness strongly depend on the considered allotrope


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Is it possible to "dope" glass to change it's refractive index?
For many purposes, like in the making of achromatic lens combinations, we need transparent media of different refractive indices. Let's assume I want a transparent medium of refractive index of 1.4, how would I go about getting such a material? Can I "dope" it by dissolve some impurities into molten glass mixture and then solidify it later? How would I choose my impurity to add? How do I calculate how much to add?

Refractive index depends strongly on wavelength but assuming you are in the visible then for a glass/quartz 1.4 is very low indeed and such a material may not exist. If you are in the UV then glass is no good as it absorbs and you need silica or calcite etc. If you want a liquid then there are plenty. Nowadays commercial lens design is hugely sophisticated and so buy the lens you need, don't even think of trying to make anything.

Well, yes. That is how they make lenses. If you don't know what impurity to add you can fiddle with the density, which - usually - changes the index of refraction. So you can take some $\ce{PbO}$ or $\ce{B2O3}$ and add to your $\ce{SiO2}$ to get different average densities. Higher density = (probably) higher index. But this is not a 1:1 relation, it is a very noisy correlation for glasses and minerals. The correlation is very good for liquids or mixtures of miscible solvents - and then it is called the Gladstone-Dale Relation
This used to be a "known" relation and referred to in science but it has fallen out of favor due to other, more explaining relations between more sensible parameters. It is a bit 1700-ish to promote a direct relationship between physical density and optical parameters, especially when you have to disregard the wave nature of photons. The "modern" ones relate to polarizability and permittance - which makes more sense.

Glass doesn't crystallize at all, that's the point. Consequently, unlike crystalline materials, it doesn't have fixed composition. In a way, it can be said that glass consists of impurities. And yes, of course, you may change its composition this way or that, so as to tune the refractive index to your purposes.
Further reading: crown glass, flint glass.

This quite doesn't answer my question. While I understood that glass doesn't crystallize, are there any calculations involved in finding the amount of impurity to be added? How do I choose the impurity?

You choose the impurity and the amount of it by looking up the industry reference books. Calculations are kinda possible, but insanely complicated and still not completely reliable.


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Why doesn't the bulb get brighter if I add another cell in parallel?

Suppose we add another cell and connect it to the same bulb shown(Two cells connected to the same bulb).
From the second cell perspective, it sees a resistor and the same chemical reaction should take place with in the cell, and it should try to push the same amount of charge per second ? If so, the current through the bulb should double ? This should violate the ohm's law (I=E/R) ?
I know that ohm's law cannot be violated. So bulb doesn't get any brighter. But how does the second cell know about the presence of the first cell in order to slow down its reaction rate ?

The second cell notices that pushing electrons through the wire suddenly became twice as hard.

May I know how it notices, as it sees the same resistor ?

True, but there is also a bunch of extra electrons trying to get through. Or let's put it this way: the resistor is the same, so is the voltage, hence so is the current, but somebody (the second cell) is stealing half of the current, so it doesn't reach us (the first cell).

It doesn't push charge. You are visualizing it wrong.
It sets up an electric field with a certain voltage. If you add another cell in parallell, it does nothing for the bulb, it already was under the influence of an electric field of similar strength. The original cell will "slow down", and your new cell will "speed up" since there are miniscule differences in the cell-internal resistance which will balance out when they are sharing the load (literally, and figuratively. I love when sentences turn out like that)
And Ohms law can most certainly be violated, it is violated every single day by any equipment using alternating current (AC).

it is violated every single day by any equipment using alternating current I have to disagree. Ohms law, in it's purest state is: $$j=\sigma E$$ where $j$ is current density and $E$ is electric field. $\sigma$ is conductivity. In a more simplified way, it would be: $$V=iR$$ In an alternating current, if you consider rms values, it seems to be violated, but it's followed at every single instant where ohmic resistors are present. Rms values are just averages, you cannot use them to generalise the law. Ohm's law is only violated when non-ohmic devices are used.

Yes, and everywhere AC is being conducted, you have non ohmic behaviour... Because no devices are purely ohmic.

Because no devices are purely ohmic are you attempting to conclude that Ohm's law is not at all valid? Your argument is not just for AC, but also for DC.

Yes, that is basically what is the conclusion of that. You can of course, construct a complex version of Ohm's law with impedance (and a phase relationship variable for AC) which would be much more correct. It is fairly important to realize the limitations of Ohm's law, and that is why I state that it is broken every single day. You can improve by translating to phasors and whatnot, but the point remains, Ohm's law is a simplification and should not be taken as any sort of truth about the natural world. It

I don't believe it is correct to state that another cell "does nothing for the bulb". Before I explain, I need to admit that this (simple DC circuits, batteries) is not an area I'm particularly competent in - I'm relying on my common sense and my decades and decades and decades old University studies. If it were true, then connecting 1000 identical cells would "do nothing", but I believe the filament would burn out in an instant. (I could be wrong here!) The place where you are clearly mistaken is your assumption that the resistance is constant. For a metal conductor, resistance is an increasing function of temperature - higher temperature, more resistance. This means that doubling the current flow will heat up the filament and increase resistance. So, one question you need to ask yourself is: is light intensity (brightness) proportional to work done? And are your eyes able to discern a increase in intensity? I don't know, but I suggest that since our eyes are can sense a single photon (under optimum conditions) and can also function under lighting conditions far more intense than the noon desert Sun, that its more of a log(I) measurement device than an I measurement device (I=intensity). My experience is that as you increase the current to a filament (starting from zero) at a fixed voltage the bulb will brighten until it reaches a "plateau" and little increase in brightness is noted (some of this is almost certainly my eyes adjusting to the brightness increase). I note some increase but not as dramatic as its ramp up to "full" brightness. If I continue to increase current, not much happens until pop! the filament fails (possibly catastrophically). So, evaluate this answer and see what others say about it. If it gets a lot of down votes, well I warned you. You know what they say about touching a live electrical wire, right? It's not the voltage that kills you, it's the current. Same idea with the filament. Power = IE. Work is power times duration. (I should also note that the relationship between resistance and temperature is more complicated that I care to discuss here).

Ideal case (internal resistance of the cell is zero):
The E.M.F across the cell does not change according to the external circuit. If you assume that the cell has no internal resistance, the potential drop across the cell is equal to its E.M.F in a closed circuit.
The potential drop across the bulb in the first case (one bulb connected to the cell) is equal to the potential drops across the bulb in the second case (two cells in parallel). This is because the bulbs are connected to the same ends of the cell and the potential difference across the cell does not change.
The power consumption of any bulb is given by:
$$P = VI$$
If the device obeys Ohm's law, the above equation can be written in terms of $V$ and $R$ as follows:
$$P = \frac{V^2}{R}$$
As the resistance of the bulb is a property of the bulb and does not change in the two cases, the power consumed by each of the two bulbs is the same.
Therefore, you shouldn't expect any appreciable increase in brightness.

Non-ideal case:
If internal resistance of the cell was taken into account, the potential drop acros the two bulbs would decrease (net resistance in the circuit decreased as the two internal resistances are now in parallel, current increased, net potential drop across the internal resitance increased and therefore, the potential drop across the cell decreased) and hence, they would glow less brightly than the bulb in the first case.

Unless I'm missing something, the question is about multiple batteries with a single bulb, but your answer addresses the case of a single battery with multiple bulbs.

I edited the answer. Thanks for pointing out.


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Electrolysis involving K+ and Li+ ions

Need clarification on Q31a) in the image.
Possibly a dumb question but the book's answer might be wrong. The book says lithium ions will be reduced instead of potassium ions. How is this true when K+ ions are stronger oxidants?
Their answer for part d even reiterates that's Li+ will react in preference to K+ so I think I may be wrong...not sure how though
Thanks

Who says K+ ions are stronger oxidants? It might seem so according to standard electrode potentials, but those only apply to water solutions.

How does a molten electrolyte change this?

everything is relative

$\ce{Li+}$ 's low standard reduction potential value applies to aqueous solutions only.
$\ce{Li+}$ has very high polarising power due to its small size (and in turn a high hydration energy). This is why $\ce{Li+}$ has a very low reduction potential compared to other alkali metals in aqueous solutions.
When you deal with a solution of molten metals, the idea of hydration energy makes no sense anymore. Therefore, the electrochemical series order for aqueous solutions isn't applicable to the molten solution of alkali metals.
As you go down the alkali metal group, the ionisation energy decreases. The tendency of an atom to give its electron (or get oxidised) increases. Therefore, potassium oxidises more easily than lithium or in other words, potassium has a lesser tendency to get reduced compared to lithium.
As lithium has a higher tendency to get reduced than potassium, lithium gets deposited instead of potassium.


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2-Deoxy-D-arabinohexose, 2-Deoxyglucose
Aldrich says that these two names are synonyms of one another.  Does anyone know the etymology of arabinohexose used in this context?

Actually arabinohexose got its name from the family of six carbon membered mono saccharide(i.e. aldohexose) and the first name arabinose is taken from the another popular name i.e. gum arabica because it was first extracted from it!

But if you look at the nomenclature of aldoses https://en.wikipedia.org/wiki/Monosaccharide_nomenclature arabinose has five carbons https://en.wikipedia.org/wiki/Arabinose and glucose has six.  Do you have a reference for your statement "[a]ctually arabinohexose got its name from ..."

In the statement you've replied you wrote about arabinose and actually you were asking for arabinohexose

OK.  Can you give me the link for arabinohexose please?

Sir, actually I have not copied it from somewhere rather i'd combined contents from internet and my knowledge of etymology. That's why I can't give you a exact link. I hope you understand!

D-glucose (a hexose) is a C6, normal (straight) chain aldehyde (in its non-cyclized form) bearing a hydroxyl group at each carbon (C2-C6) in a unique spatial arrangement at C2-C5. 2-Deoxyglucose is missing the hydroxyl group at C2. D-Arabinose is a pentose (C5) that has the same absolute configuration at C2-C4 as 2-deoxyglucose does at C3-C5. The prefix arabino refers to the configurations in 2-deoxyglucose. "Hex" means six carbons. "Ose" is the suffix for a sugar.