text:
Why does the distance between molecules increase when the temperature is raised?
I have learnt that when we heat ice-like substances it changes to water and when I asked my teacher she said  that the distance between molecules increases.
When I thought about it a bit more a question arose: Why does the distance between the molecule increase as we raise the temperature?

With ice and water, in fact, it is quite the opposite.

No, real world physical constants are mostly read-only.

@IvanNeretin tho' the actual  expansion occurs in water  as it cools just before freezing into a solid, not at the phase transition IIRC

I was expecting the energy vs interatomic distance graph.

Firstly, the generalisation isn't universally true: water below 4°C is denser than ice despite the extra temperature. Secondly, at least for liquids, the density doesn't usually decrease much as temperature is raised. So the generalisation is a poor one. Except for gases where the density at constant pressure changes a lot with temperature as a direct result of the higher kinetic energy of the molecules and the interrelationships of pressure, temperature and volume. Your teacher was fobbing you off with an over simplistic explanation.

The answer "the distance between molecules increase" is incomplete if not plain wrong.
Temperature is an effect of energy present. Basically, it's an effect of little movements and vibrations of molecules and atoms due to their energy.
In an crystal, the energy of the molecules is so low, that they don't vibrate and move enough to break the structure. The more energy you put into the system, the more the molecules move. At one point, the movement is too much to keep the molecules in place, the crystal structure breaks apart, the ice melts.
A liquid (NOT WATER, IT IS A SPECIAL CASE) has lower density than the crystal because the molecules are moving around a lot and "need more space".

It might be worth adding that it isn't true of water anyway - the solid state is less dense than the liquid.

OK, but for a solid, such as an iron rod in a 1atm environment. Its dimension will change with temperature, say from 0C to 200C; it will remain solid, but will expand with increasing temperature. Same for the alcohol in your average household thermometer over everyday temperature ranges (which is how it works to tell the temperature). What is going on at the atomic/subatomic level to explain this?

@AnthonyX vibrations, little movements.

I think it's also relevant to note that for water and for ice, density still decreases with temperature (for water above 4ºC, where it reaches a density maximum) - it's just that the pretty unique structural effects that take place between 0ºC and 4ºC override the effect of temperature.

In general, for any substance, molecules in the liquid state are far more free to move as the intermolecular forces of attraction are lesser here (Less-densely packed). But considering water(as pointed out by Ivan), it actually shows the reverse trend ie., density of ice < denisty of water. For more details on this, refer Physics.SE and/or Quora.
Also in general, for any substance, when temperature increases, the total energy (includes kinetic and vibrational) increases, as temperature is a measure of energy of a system. When considering the solid state, since they are tightly packed, the particles can only vibrate about their mean positions and when the vibrations becomes more, it is tougher for the system to hold onto, and hence the phase change occurs. (From solid to liquid state). But you’ve also got to note that some substances don’t exist in liquid state at a particular pressure.

Ever seen ice floating in water?

Yes, I get that ice is less dense. I referred the second statement to any substance in general. Will edit it to make it more understandable. Thanks.

To followup on other answers,  there is no guarantee that particle (molecule or atom) spacing will increase.  If you maintain sufficient pressure you can heat things well above their normal phase transition temperatures.
It is true that, in the absence of external pressure, liquids tend to have lower densities than their corresponding solids.  The particles no longer have shared orbital bonds as they do in a solid structure, and thus are held together via weaker forces such as ionic.   When they are heated to gas-phase temperature, even these forces are overcome and the particles move as far apart as possible -- the iconic "fill the container" behavior of a gas.  Due to statistical behavior, the density of the gas tends very quickly to be isotropic within that container.

I'm going to argue that the distance between molecules does generally increase with temperature for very fundamental reasons, although normal ice (Ih) and liquid water are an exception. At low temperatures, substances minimize their energy, which usually results in ordered crystalline arrangements. At high temperatures, substances maximize their entropy. That is, at high temperatures everything is as random and mixed up as possible.
Assuming that you have some empty space, there are more ways to put two molecules far apart than there are to put two moleucules close together. You can see this in the diagram below. There are more ways to place a second molecule in the blue spherical shell, at a long distance from the central water molecule, than there are to place water molecules in the orange spherical shell, at a short distance from the central water molecule. In fact, the number of ways to place the second water molecule increases with the volume of the spherical shell, that is like 4πR2, where R is the distance between the molecules. So, the state where two water molecules have a long distance between them has a larger entropy than the state where they have a short distance between them.

Simply due to geometry, at high temperatures, molecules will tend to be far apart. Water boils at high temperature simply because there are more ways to scatter water molecules throughout the whole kitchen than to place all of the water molecules in the pot. Most substances become gases at high enough temperatures given sufficient empty space. Most substances expand as their temperature rises.
Adding distance between molecules is not the only way to increase entropy, which accounts for the exception of melting ice. Entropy can also increase by making the arrangement of molecules and their orientations more disordered. Liquid water has more entropy than ice not because the molecules are farther apart, but because they arranged and oriented more haphazardly.

First you have to understand that molecules are randomly moving in matter.When you heat them,you essentially give them kinetic energy which increases their speed and hence the distance between molecules increases implying that interatomic forces decreases,leading to change in state from solid to liquid.The basic difference between solid,low,and gases are that molecules in gases have more kinetic energy than liquid which have more than solid.

How heat energy is converted to kinetic energy?

First you need to understand what is heat? Supplying heat basically means increasing temperature(provided it is not being utilized for  doing work).Now this means we need to understand what is temperature?Temperature is related to average kinetic energy of particles.Supplying heat increases temperature which means AVG kinetic energy increases


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Meaning of the word 'croton'

This is but-2-yne, which is also known as crotonylene.

This is but-2-enal which is also known as crotonaldehyde.
Coming to my query, what does the word croton indicate? Does it refer to $\ce{CH3-CH=CH-R}$? Can this prefix be used anywhere or only for pre-specified compounds? If it can be used anywhere, then can but-2-en-1-ol be called as crotonalcohol or hydroxycroton?

Note that the structure of but-2-yne shall be linear.

As with many chemical compounds (see the large class of terpenes, for instance) "croton" derivatives take their name from the natural source of their "father" compound. They are a family descending from crotonic acid, which was found in the oil from the seeds of the Croton plant.
"Croton" compounds are derivatives of the father compound "crotonic acid" (2-butenoic acid), so you can guess the names of the derived compounds by the logical chemical modifications of such compounds. For instance, as you might guess, crotyl alcohol is But-2-enol, from the reduction of crotonic acid; crotonylene's structure can be imagined with an analogy between acetic acid and acetylene, and so on.
I don't think that nowadays new names for the derivatives could be accepted, but for historical reasons many derivatives have these funny names.
And, knowing the stucture of the father compound, it's easy to imagine the structure of the derived ones.

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text:
Is it possible for the iodine atom in periodate ester to be octavalent?
Recently, I was reading Clayden's Organic Chemistry (2nd ed.) and I chanced upon this diagram on p. 443 (as shown below) showing an octavalent iodine in the periodate ester. I found this rather puzzling as I thought that the maximum possible valency of iodine was seven. I am well aware that bonding schemes such as the 3-centre 4-electron bonding scheme allow central atoms to make more than the number we would expect the atom to make conventionally. However, this is the first time I am seeing octavalent iodine and thus, I would just like to confirm if it is in fact, possible for iodine to be octavalent.

Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

These earlier Chem SE posts may prove helpful, link_1, link_2, and links therein.

It's just drawn wrong - it's a anion and there should be -O(-) instead of one =O - nothing interesting :/

@Mithoron I had the same thought too

I would suggest that a reactive intermediate/transition state was formed, and not an actual compound. I'd say both $\ce{C-O}$ bonds are partially formed in the intermediate

The diagram in the book does not show where is the $\ce{Na+}$ during the reaction (see above picture), probably for the convenience (but therefore, somewhat misleading). Remember, a periodate ester is an ester of periodic acid.  The most common periodate esters are cyclic periodate esters, form as the intermediate in oxidation of vicinal diols such as sugars and glycols. For example: The oxidative cleavage with periodic acid has been successfully applied in the constitutional analysis of sugars (see: Chemgapedia article).
Experimental evidence for presence of cyclic periodate esters (rightfully with hepta-valence iodine) can be found in: Buist, G. J.; Bunton, C. A.; Hipperson, W. C. P. The mechanism of oxidation of α-glycols by periodic acid. Part X. The oxidation of pinacol, and a general discussion of the stability of periodate esters and their role in the mechanism of oxidation. J. Chem. Soc. B 1971, 2128–2142. DOI: 10.1039/J29710002128.
Also, two forms of periodic acids can be found in the corresponding Wikipedia article.

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Color change of nitroglycerin. Potential hazard?
I recently nitrated around (~7mL) of glycerol through a process I have performed many times before. The nitroglycerin has been stored for ~3 weeks, and I noticed a drastic change in color from clear to an opaque dark green/blue.
I'm not sure exactly what occurred, but I am assuming some impurity in the final product was oxidized over the course of its storage. Most likely due to the fact that the container storing the fuming nitric acid I used for the nitration broke and the remaining acid was salvaged for this reaction.
Is there any chance the nitroglycerin could be unstable in its current state? I have not disturbed it and would like to dispose of it safely.

Well, nitro is a  hazard itself, and if something strange is happening with it then it's even less reliable. Hard to tell if it's, say, more prone to explosion, but lack of explosion when it's needed is also a problem.

You can easily absorb nitroglycerin in solid absorbent such as diatomite (like in dinamite) for safety.

I don't think you are understanding the situation correctly Mathew Mahindaratne.


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Reaction molecularity and order
Question:

A reaction involving two different reactants can never be a

unimolecular reaction
bimolecular reaction
second order reaction
first order reaction

The answer as per my book is 1. unimolecular.
But the nucleophilic substitution SN1 is unimolecular but it involves two different reactants. Let us take an example,
$$\ce{(CH3)3C-Cl + Br- -> (CH3)3C-Br + Cl-}$$ This follows SN1 mechanism. Yet, it involves two different reactants even though the R.D.S depends only on the formation of carbocation and the rate law expression being
$$\text{rate} = k[\ce{(CH3)3C-Cl}]$$ where $k$ is the rate constant.
So, is the answer that unimolecular reactions can never involve two different reactants wrong? Or am I missing something trivial?
Also it would be great if other options are discussed.

I think the confusion is primarily caused due to the lack in distinguishing between overall reaction and elementary reaction.

First of all,
Molecularity isn't defined for a complex reaction.
From Physical Chemistry by Ira. N Levine: (Page 531, Rate Laws And Equilibrium Constants For
Elementary Reactions  ):

The number of molecules that react in an elementary step is the molecularity of
the elementary reaction. Molecularity is defined only for elementary reactions and should not be used to describe overall reactions that consist of more than one elementary step. The elementary reaction $\ce{ A → products}$ is unimolecular. The elementary reactions $\ce{A + B → products}$ and $\ce{ 2A → products}$ are bimolecular. The elementary reactions $\ce{A + B + C → products}$, $\ce{2A + B → products}$, and $\ce{3A → products}$ are trimolecular
(or termolecular).

Atkins' mentions: (page 810, Elementary Reactions):

The molecularity of an
elementary reaction is the number of molecules coming together to react in an elementary
reaction.It is most important
to distinguish molecularity from order: reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism.

Now, from Wikipedia:

This difference can be illustrated on the reaction between nitric
oxide and hydrogen:
${\displaystyle {\ce {2NO + 2H2 -> N2 + 2H2O}}}$
The observed rate law is ${\displaystyle v=k{\ce {[NO]^2[H2]}}}$, so
that the reaction is third order. Since the order does not equal the
sum of reactant stoichiometric coefficients, the reaction must involve
more than one step. The proposed two-step mechanism is
$${\displaystyle {\ce {2 NO + H2 -> N2 + H2O2}}}$$
$${\displaystyle {\ce {H2O2 + H2 -> 2H2O}}}$$
On the other hand, the molecularity of this reaction is undefined,
because it involves a mechanism of more than one step. However, we can
consider the molecularity of the individual elementary reactions that
make up this mechanism:
The first step is termolecular because it involves three reactant
molecules, while the second step is bimolecular because it involves
two reactant molecules.

As far as your example of SN1 reactions is concerned, Peter Skyes'  acknowledges while discussing the kinetics of SN1:

The molecularity refers to the number of species that are undergoing bond-breaking or making in one step of the reaction, usually, in the rate limiting step.

Thus, when we say SN1 (S= substituition, N= nucleophilic, 1= unimolecular) we are essentially talking with respect to the RDS in which the carbocation is formed. Note that though most books identify 1 for unimolecular, Clayden's book refers to it as 1st order, which, in my honest opinion, is much more clearer.

The slope of the first graph is simply the first-order rate constant because rate = $k_1[t\text{-}\ce{BuBr]}$. But the slope of the second graph is zero. The rate determining step does not involve $\ce{NaOH}$ so adding more of it does not speed up the reaction. The reaction shows first-order kinetics (the rate is proportional to one concentration only) and the mechanism is called SN1, that is, Substitution, Nucleophilic, 1st order.

Conclusion:
The question is poorly framed. However, since molecularity is defined only for an elementary reaction, maybe after looking at the options, the question setters expect you to assume that they are talking about an elementary reaction. If this is the case, only then option 1 (but along with option 4) can be marked correct. However, because they haven't mentioned anything, the candidate attempting the question is bound to get confused. If they are talking about the RDS step, then your argument is valid. An SN1 reaction is indeed unimolecular wrt the RDS step (according to all standard books like Solomon Frhyle, Peter Skyes', etc.) which would deem all of the options incorrect.

A reaction involving two reactants can be:

a first order reaction, for instance when the rate solely depends on the concentration of one of the reactants. An example is the acid hydrolysis of sucrose, water being used in large excess (which makes the variation of its concentration negligible)
a second order reaction, for instance in basic hydrolisis of an ester (both the concentrations of base and ester affect the reaction rate)
a bimolecular reaction, for instance the classic SN2 reaction: both the reactants are involved in the transition state

It cannot be a unimolecular reaction: there is no way that two reactants could be involved in a transition state in which only one of them participates, after which you can expect a reaction between both of them (it would be bimolecular, if that happened).
There are indeed unimolecular reactions, for instance the thermal rearrangement of azulene to naphthalene, but they require a single reactant.

Why is my example/explanation not correct? Also can you explain “transition state”? Isn’t hydrolysis of ester-pseudo first order?

The example of the Sn1 is not correct because you are not asked about the molecularity of the rate determining step, but of the whole reaction!
Yes, my example is pseudo-first order.
Radioactive decay is an example of a pure first order reaction

Oh! So they are called unimolecular but actually molecularity of entire step is not 1?

This may be a good read: !https://en.wikipedia.org/wiki/Molecularity

Yep: in a Sn1 reaction you have two reactants and get two products, right? Bimolecular in the whole reaction, unimolecular at the intermediate step.
After all, since only one of the reactants partecipates in the formation of the carbocation, you see that until that step the reaction is unimolecular. But the whole Sn1 proceeds beyond that step.

Yes, I get why my example is incorrect now. Thanks. Is it just because a unimolecular reaction has to involve only changes within one molecule that we don’t have unimolecular reactions involving two reactants?

A unimolecular reaction is an elementary reaction of a single molecule, which produces one or more molecules of product.
e.g., thermal decomposition, cis-trans isomerization (by light or heat), radioactive chemical decomposition (or simply, decay), etc.
For great example, see: Unimolecular Reactions in the $\ce {CF3CH2Cl ↔ CF2ClCH2F}$ System: Isomerization by Interchange of $\ce {Cl}$ and $\ce {F}$ Atoms: J. Phys. Chem. A, 2011, 115(6), 1054–1062 (DOI: 10.1021/jp108955m).

Reaction between t-butyl chloride and a bromide ion is not unimolecular. There are two reactants, $\ce{Br-}$ and $\ce{(CH3)3Cl}$. Also, unimolecular and first-order are not synonymous. Molecularity refers to the amount of distinct molecules involved in a reaction while order refers to the exponent to which the concentration terms are raised to obtain the rate of reaction.
A unimolecular reaction isn't exactly a "reaction" if you look through common sense, as a single reactant cant actually "react" with anything. However unimolecular reactions are usually decomposition reactions, where a molecule simply dissociates into other molecules upon heat or certain other factors. Decomposition of calcium carbonate is a famous example for a unimolecular reaction.

Why is it not unimolecular? Molecularity = 1 and Being unimolecular are equivalent right? Also, I never involved order here.

No, $\pu{S_{N}}1$ reactions (while the name suggests) are not exactly unimolecular. They are pseudo-unimolecular. They appear to follow unimolecular rate statistics due to the rate determining step being unimolecular.

Can you elaborate on that please?

See my edit to my earlier comment. And what @The_Vinz mentioned in the comment to his answer is correct. Molecularity of a single step (rate determining step) and of the whole reaction are two different things.

Yup, just saw, so are there no reactions that involve different reactants but also unimolecular? I get what a unimolecular reaction is and why my example is incorrect, but not exactly how to explain it taking a general case. Any help? Thanks.

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When an electron returns to it's ground state does it release the same wavelength of light that promoted it in the first place?
If this is true, I fail to understand why we see colour at all -- wouldn't any wavelengths absorbed by electron promotion be inconsequential to colour because there would an equal amount, and identical wavelength of light emitted when the electrons returned to their ground state?
The only way this would make sense to me is if the energy of the photon absorbed by the electron is given off as heat or something besides an identical photon when the electron returns to its ground state.

See also https://physics.stackexchange.com/q/70478/162193   this very question was posed more than one time here or in Phys SE. The answer of Anna V is one of the reasons.  But this leads already to absorption and emission to happen (and their spectra to be recorded.  Consider that the most straightforward set ups include a detector along the light direction and perpendicular to it, in the two cases respectively)

See also https://physics.stackexchange.com/q/70478/162193

Not necessarily... it can return to the ground state in multiple steps, releasing different wavelengths each time.
Also, the excitation could be caused by something other than light in the first place (heat, for example).
Also, relaxation isn't instantaneous, so the emitted photon wouldn't really be "cancelling out" the color of the absorbed one as some time has passed.

Add that emission generally happens at random direction, a fact that itself guarantees some absorption along the path. As such you cover all possibilities

Also the direction of emission

For your understanding of how many ways an excited electron loose its energy see following Figure, which displays Jablonski diagram describing the electronic levels of common organic molecules and possible transitions between different singlet and triplet states (M. Sauer, J. Hofkens, and J. Enderlein. Handbook of Fluorescence Spectroscopy and Imaging).
![Figure 1: Jablonski diagram]!https://www.hindawi.com/journals/isrn/2013/230858/fig1/
If interested, please also read the article: !https://www.hindawi.com/journals/isrn/2013/230858/

Press help (button with ? sign) while editing and read relevant parts.

You are right about the heat, when electrons return to there ground state( directly or indirectly) a part or all of the energy can can be transformed to heat, this is called non-radiative relaxation.


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Preparation of an aromatic Grignard reagent
I was tasked with preparing 4-fluorophenylmagnesium chloride from the corresponding aryl chloride. But I've heard that aryl Grignard reagents are very difficult to prepare. I think that the optimal solvent for the reaction would be THF. Any advice would be greatly appreciated!
Also what are the downsides of using magnesium powder instead of magnesium turnings?

I have prepared many aryl Grignards in my career. Generally the aryl bromide or iodide is easier than the chloride, but it can be done. THF is certainly fine for this, run under inert atmosphere, of course. Magnesium powder may prove to have more of an oxide film than turnings and hence be less reactive. Most Grignard preps use turnings. It might be helpful to give your $\ce{Mg}$ turnings a brief wash with $\pu{2N}$ $\ce{HCl}$, then rinse with $\ce{EtOH}$ and $\ce{Et2O}$ just before you use them.
To initiate the reaction - do not stir the reaction but either hotspot it by using a heatgun to heat the region immediately around where the turnings sit at the bottom of the flask or put the whole flask in an ultrasound bath. A crystal of iodine in the reaction mixture will often serve to initiate the reaction, when it decolourises you know the reaction is proceeding. Depending on scale the heat of the reaction may bring the solution to reflux so you'll need a condenser on it.
Finally if all else fails a couple of mls of Red-$\ce{Al}$-$\ce{H}$ solution should start it.
If all fails with $\ce{Mg}$ metal, consider running a Knochel exchange with $\ce{iPrMgCl}$.

Sometimes these compounds need to be refluxed to get the reaction going. Other ways involve making the magnesium more reactive by including, for example, ground glass with the powder or turnings to abrade the surface oxide layer. Sometimes even using the intermediate formed when anthracene reacts with magnesium in THF can help (if you can get rid of the anthracene later without too much trouble.)

Besides iodine for initiation, 1,2-dibromoethane can be used along with the aryl halide to keep the magnesium clean. Reductive elimination occurs.


text:
slow down ice melting in water
Having to separate freshwater crabs for commercial fisherman while on the boat, we dump the crabs in ice water, as crabs aren't as aggressive and bite less when cold.  There's only so much room on the boat for carrying ice, and the ice melts too quickly in the wooden box filled with water and ice in order to "slush" the crabs.  We've been unable to work full days, running all of our crab traps, as we run out of ice before a full-day's work.  Any ideas of how to slow the melting process of the ice while in the water used for slushing the crabs?  Thanks for your time and knowledge.
"

Why do you need ice in the box? Is it the ice melting in and of itself that is the problem, or is the loss of a heat sink that is the problem? That is, if you were able to keep the ice water at the same temperature without any ice, would that work just as well?

It's a wooden box on a workboat that the captain built to throw crates of crabs in in order to slush the crabs. The box cannot have a top on it, as you need to be able to dump the crabs in it either from the traps or from the crates; so, therefore, we are unable to lesson the size of the box-top opening.  Thanks again for your help.

Insulating the container is a start. You might look into using "dry ice" (solid CO2) to keep your water-ice cold or generate more. It is possible a sample of dry ice could provide more "cooling power" than a sample of water ice of the same size and weight.

@MicheleCelina I am baffled as to what your comment is in response to.

Unfortunately there isn't much you can do to slow down the melting of the ice-water slush besides insulating your container. If possible, you can enclose the wooden box in styrofoam to prevent heat from entering the system.
Another note you should keep in mind is that the water in the slush speeds up the melting of the ice. A better scenario would be to keep a large (minimized surface area) block of ice in one container, and the slush in another. Use an ice pick to break up the block and add pieces to the slush as needed.


text:
Why is it only electrons that take part in chemical reactions?
I learnt that only electrons take part in chemical reactions and the protons do nothing (neutrons are electrically neutral, so they do nothing). Why is this so if the protons are positively charged and without them, electrons in the other atom won’t be attracted (and since electrons are like charges I do not think they would interact with each other)?

It is wrong to say that protons do nothing. Take just a bunch of electrons without any protons, and you won't have any chemistry.

Well, it's often better idea to describe acid base reaction as transfer of proton (H+ cation).

All those protons in a nucleus have a little influence over what electrons want to do around them...

The nature of an electron is more complex than a single charged particle. Once you get down to the quantum level, our intuition about how matter interacts with other matter fails.
A simple answer to your question is that electrons occupy a much larger volume than nuclei. When two atoms get in close proximity to one another, their electrons will "bump into one another" and interact far before their nuclei will.
You are also correct that the nucleus of a neighboring atom is essential for the electrons to interact with one another. If there was no positive charge to attract another electron, a bond would not be formed. The point your instructor was trying to get across is that in most (excluding nuclear reactions) chemical reactions, electrons are the species moving around and forming/breaking bonds.
If you would like a more in-depth discussion on the true quantum nature of electrons and their interactions, you should check out Mulliken's Nobel Lecture from 1966. It is a great introduction.

link is dead @Vice Chemistr

Because chemistry is all about bonds and bonds are made from electrons
A reasonable definition of chemistry is the science of how substances change and react. Most of those changes and interactions are caused by changes in bonds (rearrangements,breaking and forming new bonds). So, since most chemistry is about what happens to bonds, most chemistry is about electrons.
But the number of electrons and the way they organise themselves in space is determined by the nuclear charges in the atoms involved in the bond(s). It isn't that protons are not relevant to the bonds, but they don't get close or involved when the electrons are rearranging themselves in chemical reactions.
And electrons do interact with each other not least because of the Pauli interactions that prevent more than two electrons occupying an orbital (orbitals are a way of describing the cloudy electron distribution in space and bonds are, to oversimplify a little, made from orbitals that span more then one atom).
In an atom, electrons also occupy orbitals and fill them up as more are added to a "naked" atom in order of the energy levels. Higher energy orbitals tend to be further from the nucleus than the lowest energy ones but electron interactions of a different sort matter here too as the "lower" orbitals shield the higher ones from some of the electric charge in the nucleus. Molecules are a bit more complicated, but the shielding still matters. Most chemistry happens with the outer orbitals/electrons and the inner ones in one sense "protect" the nucleus from any of the chemistry that is happening as electrons skip about to break or create bonds.
Also, in spatial terms, most of the "action" is happening a long way away from the nucleii, so they mind their own business and don't get involved.

"since most chemistry is about what happens to bonds" is it actually the premise or the conclusion? I mean to say, one could always ask, why do only electrons form bonds in the first place?


text:
Calculation of composition of drain cleaner
Important note: I'm an eighth-grader. I taught myself chemistry to organic chem using an arsenal of textbooks*, but I have gotten things wrong, namely resonance.

Please read above (also this if you want more background on the problems).

You have been given a sample of Crystal Drano®. There are two components in the Drano – some small shiny
  metallic pieces, and some pale green beads. (The green color is a dessicating substance.) The metallic pieces are
  either zinc, magnesium, or aluminum. The beads are either $\ce{NaOH}$, $\ce{Ca(OH)2}$, or $\ce{Al(OH)3}$.

I recognized this as a titration problem immediately. I dissolved $\pu{1.6 grams}$ of the green beads in $\pu{50 mL}$ of distilled water. I filtered out the metal and added a total of $\pu{6 mL}$ of $\pu{3M }\ce{HCl}$ into the solution, causing the indicator to change color. I added $\pu{1.75 mL}$ of $\pu{1 M NaOH}$ using a more precise pipette (it doesn't say that you can use a buret) to titrate it completely.
Then I calculated these values (molar mass):
$\ce{NaOH} = \pu{40 g/mol}$
$\ce{Ca(OH)2} = \pu{74 g/mol}$
$\ce{Al(OH)3} = \pu{78 g/mol}$
$\pu{Mystery} = \pu{(\frac{3M}{L} * \frac{1L}{1000mL} * 6mL) - (\frac{1M}{L} * \frac{1L}{1000mL} * 1.75mL) = 13/1000 M in 1.6 g \approx 48 g/mol}
$
The mystery material(still) seems to be off by a sizeable amount. Why? All I can think about is that maybe the base reacted with the metal in aqueous solution to create a metal salt.
*Not impressive, only about 7-10 textbooks relying mostly on Atkins

I don't understand very well what you did. Do the $\pu{1.6 g}$ are just for the beads? Or does that include the metal as well? Also, the molar mass of $\ce{NaOH}$ is $\pu{40 g mol-1}$.

If your titration and values are correct, then $\pu{16.25 mmol}$ of $\ce{HCl}$ reacted with the bead compound. This would mean that you initially had either $\pu{650 mg}$ of $\ce{NaOH}$, $\pu{1.20 g}$ of $\ce{Ca(OH)2}$, or $\pu{1.27 g}$ of $\ce{Al(OH)3}$.

Oh shoot. Have to edit...

This said aluminum does form complexes with hydroxide, and aluminum metal reacts with base to form aluminate, $\ce{AlO2^2-}$ (and the hydroxide complexes).

The titration is not necessary to determine which hydroxide you have though! What other thing can you think of regarding these compounds (you already did the experiment for this part, actually).

The salts would be different... probably not what you're shooting for though. Also, the experiment has to take place within 3 hours, so evaporation is not an option.

Not really, do you have access to the three compounds separately? Or calcium hydroxide ?

Unfortunately no, because the test conditions state I only get $\ce{NaOH}$

I understand that, but I meant to try to get you see what you may be missing, I could tell you or you can see it happening and learn it on your own :) (I'm assuming you're preparing for the exam and not actually taking the 1999USNCO test right now)

Let us continue this discussion in chat.

Your assumption of "the base reacted with the metal in aqueous solution" is not correct since you have filtered out metal from aqueous solution before you add $\ce {3M HCl}$. Therefore, I assume the error is from your calculations. Your answer, $\ce {13/1000 mol in 1.6 g}$ of beads, is off the mark. The correct answer should be $\ce {16.25 mmol of OH^- ions in 1.6 g}$. It means, either $\ce {\frac{16.25}{1} mmol of NaOH}$ or $\ce {\frac{16.25}{2} mmol of Ca(OH)_2}$ or $\ce {\frac{16.25}{3} mmol of Al(OH)_3}$ in $\ce {1.6 g}$ of beads. Therefore, your $\ce{\% NaOH}$ contained in beads is $\ce {\frac{16.25 mmol NaOH}{1} \cdot \frac{40 mg}{mmol of NaOH}\cdot \frac{1 g}{1000 mg}\cdot \frac{100 g of beads}{1.6 g of beads}}=40.6\%$ ($w/w$).
Similarly, your $\ce{\% Ca(OH)_2}$ contained in beads is $\ce {\frac{16.25 mmol Ca(OH)_2}{2} \cdot \frac{74 mg}{mmol of Ca(OH)_2}\cdot \frac{1 g}{1000 mg}\cdot \frac{100 g of beads}{1.6 g of beads}}=37.6\%$ ($w/w$), or your $\ce{\% Al(OH)_3}$ contained in beads is $\ce {\frac{16.25 mmol Al(OH)_3}{3} \cdot \frac{78 mg}{mmol of Al(OH)_3}\cdot \frac{1 g}{1000 mg}\cdot \frac{100 g of beads}{1.6 g of beads}}=26.4\%$ ($w/w$).
PS: Please also use unit $\ce{mol}$ to represent moles. The unit $\ce{M}$ is for $\ce{mol\cdot L^{-1}}$.

Ok, thank you! I have recalculated the values.


text:
Why does this elimination reaction occur?
I was confused by why the following elimination reaction hits only one of the hydrogens instead of both isomers that are possible. Please help me out!

Why do you think that might be?

@ralk shouldn't the bulky base form the less stable product? (i.e. in this case, the non conjugated diene isomer instead)

@GaurangTandon you are misunderstanding the rule. Using a bulky base makes it harder to reach a more hindered proton. There's no much steric hindrance in this molecule that would make that happen

Conjugated dienes are significantly favorable to non-conjugated dienes (because of delocalization and hybridization, etc.), so this will be the major product. It is not to say that the other reaction cannot happen, but that a combination of factors leads to that product being major. All of chemistry occurs in an equilibrium of some degree, and the reaction above will become most stable in conjugated diene form.


text:
Usage of lowest locant rule

How should we name this compound by IUPAC system?
Is it
2,3-dimethylcyclohexene
 (or)
 1,6-dimethylcyclohexene?
In the first case locant position adds to 2+3 = 5, while in second it is 1+6 = 7. However, in second case, first substituent gets lower position, i.e., 1 , while in first case it gets 2.

According to latest IUPAC recommendations, lowest locant rule is the valid rule followed for organic nomenclature and presently, the lowest sum rule is no longer followed.
So, as you have discussed in the question, the name that follows the lowest locant rule is the proper name of the compound.
So, it is 1,6-dimethylcyclohexene.

The "lowest sum rule" has never existed as far as I know.


text:
How can we separate a mixture of two sulphide ores using froth floatation method?
As we know froth floatation method makes use of the wetting properties of mineral particles.

Mineral particles (sulphide) become wet with oils, while gangue particles get wet by water

According to my textbook (NCERT Class 12), we can use the froth-flotation method to separate a mixture of sulphide ores also using depressants.
How do these depressants actually work in this process?

In the froth floatation process, the role of the depressants is to separate two sulphide ores. They do so by  preventing one ore from forming froth.
In this process, we want some substances to settle down and some to float. Now ideally, we want the sulphide of the substance that we wish to extract tonfloat and the gangue to sink. However, sometimes the ore we are using also has other sulphides present in it as impurities. For example, lead sulphide generally has zinc sulphide also present as impurity which will also float along with the lead sulphide. To make sure that lead sulphide floats but zinc sulphide do not float, we add depressants.
Commonly used depressant is $\ce{NaCN}$.
To separate two sulphide ores ($\ce{ZnS and Pbs}$), $\ce{NaCN}$ selectively allows $\ce{PbS}$ to come with froth, but prevents ZnS. This happens because of the reaction of $\ce{NaCN}$ with $\ce{ZnS}$ to form $\ce{Na2[Zn(CN)4]}$.
The complete reaction is,
$$\ce{4NaCN + ZnS  →  Na2[Zn(CN)4] + Na2S}$$


text:
How to justify this contradiction in acid-base equilibrium of methanol and water?
Let's talk about the following compounds - water and methanol.

We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding.
The $\mathrm{p}K_\mathrm{a}$ values are approximately $15.5$ and $15.7$, for methanol and water respectively.
Now consider their conjugate bases, i.e. methoxide ion and hydroxide ion. It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

How is this possible? Isn't it contradictory?
It is a well known fact that a strong acid has a weak conjugate base, and a weak acid has a strong conjugate base.
This is a consequence of  $\mathrm{p}K_\mathrm{a}$ + $\mathrm{p}K_\mathrm{b}$ = $\mathrm{p}K_\mathrm{w}$, where Kw is the ionic product of water, a constant at a given temperature.

It's not a duplicate, hence the downvote is completely not justified.

In "methanol is slightly more acidic than water, because [...]" and "methoxide is a stronger base compared to hydroxide, due to [...]", "should be" would be more appropriate than "is" -- the "should be" only becomes an "is" once you account for the superposition of the various relevant effects in an actual system, including, in particular, the influence of the solvent.

It's not a duplicate! None of the links posted in the comments have mentioned the comparison between methoxide and hydroxide basic strength comparison - so why are you spamming this post?

(1) Note that me and Avnish Kabaj have linked to two different questions -- I believe the second one addresses more directly why the contradiction you point out is only apparent. (2) If the contradiction is only apparent, there is no need to make a separate mention of basic strength -- that is why the linked Q&As don't do that.

We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. [...]

[...] It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

You are pointing out competing factors that, in abstract, should lead to either methanol or water being more acidic than the other (that is. lead to either hydroxide or methoxide to be more basic than the other, respectively). Once we move from considering properties in abstract to an actual system in which an acid-base equilibrium is present, the competing factors are superimposed, with a non-contradictory result: either methanol will be more acidic (and hydroxide more basic) or water will be more acidic (and methoxide more basic), depending on which factors dominate. (In particular, as pointed out in Why is methanol more acidic than water? and Is methanol really more acidic than water?, the relative acidities depend on the solvent.)


text:
Finding direction of the reaction using Affinity of the reaction
I encountered following reaction:
$$\ce{CO + 1/2 O2 -> CO2}$$
I am provided with the composition of the various gases at $\pu{1000K}$ and am asked the direction of the reaction by finding the affinity of the reaction. I know the way to find affinity is:
$$A= \mu_\ce{CO2}-\mu_\ce{CO} -\frac{1}{2}\mu_\ce{O2}$$
I know the change in chemical potential considering ideal mixing. But how do I go about finding the absolute value of chemical potential, rather than the change, because the expression requires the chemical potential itself rather than the change? I thought of writing $A$ as $-\Delta G$, but for that also I need some reference value of $G$ and from there I can integrate $\Delta H-T\Delta S$, but again I can't figure out how reference state is chosen?
Please suggest something.

Potentials are standard not "absolute"...

I meant the "actual value of chemical potential". Suggest something.

Firstly, since $\overline{\mu_\alpha} = \overline{g_\alpha}$, the two approaches that you propose are basically the same: $\overline{\mu_\ce{CO2}}-\overline{\mu_\ce{CO}} -\frac{1}{2}\overline{\mu_\ce{O2}}$ is the same as the change in $G$ per mol of forward reactions.
Let's suppose that we can calculate the true, absolute value of the molar Gibbs energy at some reference physical and chemical state, labelled $^*$. The molar Gibbs energy at any arbitrary state is then
$$ \overline{g} = \overline{g}^* + \Delta \overline{g}^\text{chem} + \Delta \overline{g}^\text{phys} $$
Where the to $\Delta g$ terms account for the fact that the state of interest may have a different chemical and/or physical state than the reference state. If we substitute this back into your expression, we find
$$\begin{aligned} A &=\overline{\mu}_\ce{CO2}-\overline{\mu}_\ce{CO} -\frac{1}{2}\overline{\mu}_\ce{O2}
\\
&=\overline{g}_\ce{CO2}-\overline{g}_\ce{CO} -\frac{1}{2}\overline{g}_\ce{O2}
\\
&=(\overline{g}^*_\ce{CO2}+\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys})
-(\overline{g}^*_\ce{CO}+\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys})
-\frac{1}{2}(\overline{g}^*_\ce{O2}+\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys})
\end{aligned}
$$
If we collect the $g^*$ components, their sum is
$$ \sum \overline{g}^* = \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^* $$
But then, by definition of the reference chemical state,

$\overline{g}^*_\ce{CO2} = \overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO2}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)
$\overline{g}^*_\ce{CO} = \overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)

Substituting this back:
$$\begin{aligned} \sum \overline{g}^* &= \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^*
\\ &= (\overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}) - (\overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}) - \frac{1}{2}\overline{g}_\ce{O2}^*
\\ &= 0
\end{aligned}$$
(The term cancels to zero because the reaction is balanced)
Thus $A$ has no dependence at all on the absolute value of $g$. We can cancel the sum of the $g^*$ components and write
$$ A = (\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys})
-(\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys})
-\frac{1}{2}(\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys}) $$
In practice, we typically define state $^*$ by:

Physical State: $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$
Chemical State: pure elements in their most stable form at $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$

and then re-label/re-name the chemical component of $\overline{g}$:

$\Delta \overline{g}^\text{chem} \rightarrow \overline{g}^\circ_f$ the "Gibbs energy of formation" (change in Gibbs energy associated with forming the substance from pure elements at the reference physical state)

To actually evaluate this, you'd

Look up $\sum \overline{g}^\circ_f$ for this particular reaction (which might be labelled as $\Delta G$) in a table, if one is available. If one isn't, you could look up the individual species' $\overline{g}^\circ_f$ (if you have tables which present them), or calculate them from the species' $\overline{h}^\circ_f$ and $\overline{s}^\circ$ (again, from tables).
For each species, evaluate $\Delta \overline{g}^\text{phys}$ (the difference in $\overline{g}$ associated with changing the species' physical state from $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$ to the $T$ and $P$ of interest) using some property model (equation or table) for each substance in turn

My Notation: For some extensive quantity $X$

$x$ represents $X$ per mass
$\overline{x}$ represents $X$ per amount


text:
Why are DOM lithiations sometimes conducted at −78 °C and othertimes conducted at elevated temperatures?
I'm looking specifically at n-butyllithium. I've noticed a number of lithiation reactions are conducted at −78 °C, whereas I've encountered papers conducting these reactions at elevated temperatures, up to 60 °C or 70 °C in some cases, under an inert atmosphere of course.
Is there a particular reason for lowering the temperature in some cases and not in others? Does it relate to minimizing side reactions of other groups in the molecule, or does it primarily have to do with stability of the ending lithium reagent at various temperatures? If the latter, does having multiple directing groups assist in stability?

Solvent is also a factor. nBuLi reacts with THF at rt

there is no uniform answer, it is a case-by-case situation.  But generally you want to perform any reaction at the lowest temperature it proceeds at, because the lower temperature, the less side reactions occur. Buthyllitium is a very active compound, so many reactions could proceed at liquid nitrogen temperature (!) but liquid nitrogen is costlier than dry ice, so people usually settle at -78 as a practical temperature. However, when substrate needs a little push, or if you want thermodynamic control, elevated temperature might be used.

The reason a given set of conditions used in any given reaction is likely fairly arbitrary unless you're following a paper where the conditions have been extensively optimised (for instance in an OPR&D paper where a process group has rigorously screened) - as has been discussed here before, −78 °C  is a convenient temperature to run reactions, so if it works at −78 °C, its unlikely anyone will ever bother to warm it up.
That said, in Clayden's Organolithiums: Selectivity for Synthesis (which is probably the most comprehensive reference of R–Li chemistry I can think of), a table of directing groups and representative conditions is presented. I've included the table below, but the actual text has a discussion of the various groups and a discussion of the representative conditions used. As a general guide, it seems that temperature is largely dictated by:

the activating/deactivating nature of the directing group
the sterics of the directing group
the stability of the directing group, with low temperatures required in the case of electrophilic groups like the carbamate to prevent nucleophilic addition that would occur at higher temperature

As a general point, as temperatures get higher, the stabilisation of the anion by the directing group becomes less important to the overall energy and as such at higher temperatures, one would (generally) expect selectivity to drop.

Summary of directing groups used in DoM reactions with typical reaction temperatures. Taken from Clayden, J. Organolithiums: Selectivity for Synthesis; Pergamon: Oxford, U.K., 2002; p 34.


text:
How does conjugation affect reactivity?
From this topic on MOs of butadiene vs ethene, it is clear that when two ethene molecules are combined to extend the conjugate chain, the HOMO is raised in energy and the LUMO is lowered in energy. This implies that the resultant compound is more reactive towards nucleophiles(Lower LUMO more easily matched in energy with HOMO of Nu), and also towards electrophiles (Higher HOMO more easily matched in energy with LUMO of E).
In "Why chemical reactions happens" by Keeler and Wothers, "The conjugative effect" is mentioned as something that decreases reactivity (see p. 163-165):
"The delocalization of a lone pair from X into the carbonyl π system decreases the reactivity of the carbonyl towards nucleophiles by raising the energy of the π LUMO" .
The second argument seems to contradict my original argument, so what am I missing?

Perhaps extending conjugate system by pi bonds, empty p-orbitals or filled single orbital (LP) etc. have different effects on the reactivity?

I think an extensive answer explaining the effects of conjugation on reactivity(towards nucleophiles and electrophiles), perhaps also touching on the common statement that "presence of conjugation stabilizes a compound", would be very helpful for the site.

EDIT: Seeing as my original question is very broad, i would appreciate if someone could simply explain the apparant contradiction between the previous post and the Keeler&Wothers quote.

My first thoughts are that this is very broad. The effect of conjugation on each compound is slightly different because the MOs are not the same (both the forms and the energies). Therefore one cannot really make generalisations about whether the energies of the HOMO/LUMO will increase/decrease. That’s just reactivity, but if you want to talk about stability as well then you need another nontrivial section. Maybe useful reading for you is Ian Fleming’s “Molecular Orbitals and Organic Chemical Reactions”

This dicotomy seems to stem from the fact that in your first argument you mention alkenes and in the second you talk about carbonyl compounds, which have very different reactivities and molecular orbitals. A computational treatent of the molecules in question, using quantum mechanics to calculate the orbitals, might help you understand this question.

The effect of conjugation on electrophilicity and nucleophilicity of chemical species varies depending on the type of conjugation. There is no contradiction between the two cases you have presented in your question because in each case, the conjugation at work is different.
The conjugation which occurs in conjugated polyenes has been well elaborated upon by Philipp (and thus, I shall not elaborate further), in the answer you have linked to. This type of conjugation occurs as a result of interactions between parallel p orbitals in a chain of carbon atoms.
The other sort of conjugation you mentioned reading in the book by Keeler and Wothers is an interaction involving an interaction between the nonbonding molecular orbital, which holds the lone pair, localised on ${X}$ (e.g. $\ce {NH2,Cl, Br}$) and the antibonding $\pi$ molecular obital of the $\ce {C=O}$ bond, which is more localised on $\ce {C}$. This is a $\ce {HOMO-LUMO}$ interaction between the two orbitals and electron density is donated into the antibonding $\pi$ MO from the nonbonding MO on $X$. This interaction produces yet another set of bonding and antibonding molecular orbitals. Since the bonding MO produced is of a lower energy than the nonbonding MO on $X$, the lone pair is stabilised and its reactivity is decreased. Also, since the antibonding MO produced is of a higher energy than the previous antibonding MO of the $\ce {C=O}$ bond, the LUMO is raised in energy as a result, decreasing reactivity of the carbonyl group towards nucleophiles. One of the consequences of this conjugation is the lower reactivity of amides towards nucleophilic addition (Clayden et al., 2012).
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.


text:
Why does copper gain electrons in redox reactions, when it instead has a tendency to lose them?
Cu has tendency to lose electrons. But in some redox reactions, it gains 2 electrons and become neutral. Why is it so? How can it gain them?

Pretty much all chemistry is like that: sometimes you lose electrons, sometimes you gain.

Isn't this the intuitive one? A positively charged ion and negatively charged electrons ... I think the other way around, the reason why copper would want to give up electrons is worth asking questions about. But that's just me (don't bother answering my question btw, I'm aware ...)

As you already said it, it only gains electrons when it's oxidized, not like a Cu(-II) would form or so. Therefore it's just regular redox-chemistry. Copper is somewhat noble, or at least it has a mediocre redox-potential. When dissolved in acids it only reacts with oxidizing ones like nitric acid. On the other hand if you have Cu(II) and add an iron nail, the redox-potential of iron is much lower making it a good reducing agent. But that's basically the same for every element. Once you have an oxidizing agent it will be oxidized, if you add a reducing agent it will be reduced. Gold would do the same, sodium as well if you used lithium to reduce although gold doesn't readily give off it's electrons and sodium will do it quite easily. It really depends on the other redox-system you add.
You may also take a look at the redox-series:

Here the stronger reducing ones are on top (but depending on the artist it may also be the other way). Here it means the upper one will go from left to right if the lower one goes from right to left. To take an example:
Zn will form Zn(II) and Cu(II) will form Cu(0) in a system where both are mixed. But Cu will form Cu(II) and Pt(II) will form Pt(0) if those are mixed.


text:
Form ice from water in a non-deformable, completely sealed box
Suppose that I put some water in a non-deformable, completely sealed box, in a way that all the space is occupied by water, and reduce the temperature to transform water to ice.
Since ice takes more space than water, what would happen to the water and to the box?

Your non-deformable box stands a decent chance of being deformed. Other than that, water may postpone the transformation.


text:
What would happen to a diamond in the presence of oxygen and elevated temperature and how fast?
Let's say I have a $\pu{1cm^3}$  diamond and I put it into an oven where the temperature is about $\pu{ 400°C}$. We all know that diamond oxidizes. Would it evaporate or turn into graphite? If so, what would be the rate of reaction? Is there some way to calculate it?
From Wikipedia:

Above $\pu{1700^\circ C}$ ($\pu{1973 K}$/$\pu{3583^\circ F}$) in vacuum or oxygen-free atmosphere, diamond converts into graphite; in air, transformation starts at $\pu{\approx 700^\circ C}$. Diamond's ignition point is $\pu{720–800^\circ C}$ in oxygen and $\pu{850–1000 ^\circ C}$ in air.
[....]
But diamonds (sp3 carbon) are unstable against high temperature (above about $\pu{400 ^\circ C}$ ($\pu{752 ^\circ F}$)) under atmospheric pressure. The structure gradually changes into sp2 carbon above this temperature. Thus, diamonds should be reduced under this temperature.

So even if the diamond doesn't reach the temperature of $\pu{700^\circ C}$  it still oxidizes. That's why I'm asking.


text:
Compare rate of hydrohalogenation between 1,2-diphenylethene and 1-phenylpropene
Consider the compounds, (I) - 1,2-diphenylethene and (II) - 1-phenylpropene. Compare their rate of hydrohalogenation (HX addition)

I feel that the answer should be (II) > (I), but to my surprise it's given to be (I) > (II).
My thoughts:

In (I), phenyl group consists of sp2 carbons, operates -I effect and destabilises the carbocation intermediate.
In (II), protonation should occur at the carbon directly attached to the phenyl group. The methyl group (instead of phenyl, as in (I)) operates +I effect and stabilises the carbocation.
In both cases, the carbocation formed enjoys resonance stabilisation from a phenyl group - so it's not a factor we need to compare.

So, what is the real order of rates of $\ce{HX}$ addition? Is my reasoning correct, or have I missed something?

Assumung fruitfull collissions ,for structure (II) , intermediates 1 and 2 are formed.  Intermediate  2  is relatively more stable then intermediate  1 .
The intermediate I has 5 resonance structures ( scheme 1) .Further stablised by inductive effects of ethyl group.
For structure I, intermediates 3 and 4 are formed (Scheme 2).And 3 & 4 are similar. Intermediate 3 has 5 resonance structures. Similarly Intermediate 4 also has 5 resonance structures.
Totally 10 resonance structures are possible for structure I.(Since delocalization is more, energy of such intermediate should be less)
Stable carbocations for both I & II are formed but , they are dissimilar.
Assuming fruitfull collisions lead to intermediates 1,2,3 & 4 ,statistically since number of resonance structures are more for intermediate formed from structure I , it should lead to formation of major product.

Hence hydrohaogenation for (I) - 1,2-diphenylethene ais greater then (II)  1-phenylpropene

Why does the nucleophilicity of alkene not in question here? That in consideration, yields the reverse order.

I would suggest that there is a potential kinetic effect here.  While the most stable carbocation formed in either case is similar, the number of collisions that result in the formation of the more stable carbocation is higher in I than in II.
If we take a very simplified view of the possible collisions in II, we can see that in a reaction with $\ce{HCl}$ only "one" of the orientations of the collisions will result in the formation of the desired carbocation.

In the case of I, "both" of the possible collisions will lead to the formation of the same stable carbocation.

If the number of possible effective collisions is higher, the reaction will be faster.


text:
Are these alternate names for long chain hydrocarbons with complex substituents accepted?
Let's consider the following compound:

$\ce{CH3-CH2-CH(CH3)-CH2-CH2-CH(CH(CH3)-CH(CH3)-CH2-CH(CH3)3)}$

Its IUPAC name is 2,2,7-trimethyl-4-(1-methyl-propyl)-nonane

$\ce{CH3-CH2-CH(C2H5)-C(CH3)(C(CH3)3)-CH2-CH2-CH(CH3)-CH2-CH2-CH3}$

Its IUPAC name is 3-ethyl-4,7-dimethyl-4-(1,1-dimethyl-ethyl)-decane
I wish to ask if the following names are acceptable (respectively):

4-(1-methyl-propyl)-2,2,7-trimethyl-nonane
4,7-dimethyl-4-(1,1-dimethyl-ethyl)-3-ethyl-decane

Do complex substances follow a rule regarding where they should be placed in the IUPAC name or can they be placed anywhere?
Do prefixes like bis, tris, tetrakis, etc. need to be considered while naming?

The suggested names “2,2,7-trimethyl-4-(1-methyl-propyl)-nonane” and “3-ethyl-4,7-dimethyl-4-(1,1-dimethyl-ethyl)-decane” seem to be based on obsolete IUPAC recommendations from 1979. The preferred names according to current (2013) IUPAC nomenclature are 4-(butan-2-yl)-2,2,7-trimethylnonane and 4-tert-butyl-3-ethyl-4,7-dimethyldecane, respectively.

Okay so here is the situation: You have two subsitutions, methyl and 1-methylpropyl as the two substituents here. You wish to know in which order the substituents must be written in the name. You knew that substituents are arranged alphabetically, and both methyl and 1-methylpropyl start with the same letter m.
Consider a dictionary. It has over thousands of words that start with the letter C. But we can still arrange the words in order by looking at the second, third, fourth, fifth, etc. letters of the names. Similarly, if the first letters are the same, look at the second, third and fourth letters.
Here's where we arrive at the conclusion. methylpropyl will be ranked below methyl as per dictionary order. Thus the name of the compound would, in fact, be 2,2,7-trimethyl-4-(1-methyl-propyl)-nonane and the other name is not accepted.
I believe you can continue from here to name the second compound.
Important notes:

The inclusion of di, tri in the name must only be applied for complex subsituents such as in the case of 5-(dimethylethyl)decane. In cases such as 4-ethyl-5,5-dimethyldecane, do not use d as the starting letter. This is becase d is not the part of the name of the subsitituent (methyl) and only indicates how many times this subsitituent occurs in the compound.
Do not add hyphens between methyl and decane such as "methyl-decane". Use "methyldecane" Hyphens must only be used between numbers and subsituents for which the number represents its position in the name.

So prefixes like di, tri, tetra, etc need to be considered in complex substituents?

Exactly. It must be considered where it is a part of the name. (dimethylethyl) is the complete name of the substituent and hence d is considered as the starting letter.

This is strictly for complex substituents right?(Just clarifying)

Substituents are always written in alphabetical order (without considering the numerical prefixes such as di, tri, etc). Also, in case of trimethyl and methyl propyl, having same first alphabet (i.e. m), generally the substituent with shotter name is written first. But as per my knowledge, there is no such official rule. So the name that you have suggested for the first case is acceptable. But your second name is incorrect as alphabetical order is not considered.

2,2,7-trimethyl-4-(1-methyl-propyl)-nonane
3-ethyl-4,7-dimethyl-4-(1,1-dimethyl-ethyl)-decane                                     Is this right?

Yes. That's correct.


text:
Is osmosis due to osmotic pressure?
Okay, I am getting real confused on how water flows from a hypotonic solution to a hypertonic solution through a semipermeable membrane. It is known that hypertonic solutions have higher osmotic pressure than hypotonic ones, so shouldn't it be the opposite?
Is osmosis even related to osmotic pressure or is it just because of difference in concentration?

Osmotic pressure is the additional external pressure that needs to be applied to a solution to compensate its osmotic flow to a second solution, so that there is not net flow between them.
Let's imagine a tube with a piston in each extreme, so we can control the pressure we apply to both sides, with a semipermeable membrane inside it. If the concentration of solutes on both sides is different, there will be an osmotic flow from the hypotonic side to the hypertonic side. To compensate for this osmotic flow, we will need to apply more pressure to the hypertonic side than to the hypotonic side - and in mechanical equilibrium, this additional pressure is equal to the difference in osmotic pressure between the solutions.
Alternatively, if we want to get rid of external pressures, osmotic pressure can be understood not as a pressure exerted on a solution, but by the solution with a certain osmotic pressure on neighbouring solutions with which it can exchange solvent. So, the hypertonic solution exerts more pressure on the hypotonic solution than viceversa, and the result is that solvent moves from the hypotonic solution (which receives more osmotic pressure) to the hypertonic solution (which receives less osmotic pressure) until they become isotonic, just as would happen if it exerted more hydrostatic pressure. This is mathematically equivalent to the original definition.


text:
Determining the stereochemistry of the product(s) when an alkyl shift in a carbocation generates a chiral carbon
Consider this carbocation:

We know that this carbocation would immediately rearrange - by a methyl shift - to form the more stable carbocation.
Now, notice that the migration will produce a chiral center at carbon 1. Assuming that the leaving group was a chloride, will the configuration at 1 - after the methyl shift - be retained, inverted, or racemized?
Interestingly, assuming the nucleophile is $\ce{EtOH}$, in the final product - 2-d-3-ethoxy-3-methylpentane, carbon 2 is also a chiral center. I assume that it will be racemized, as in usual SN1 reactions, but I am not sure. Am I correct?
At a very detailed level, I was also taught that in SN1 reactions, 100% racemization does not occur. The inverted product is slightly more than the retained product. This is because - while the leaving group is still leaving - backside attack is preferred. So, does this phenomenon also have an effect on the molecule above?
To conclude, I wish to know the stereochemistry at both chiral carbon 1 and 2. Are they retained, inverted, or racemized? If racemized, then is the racemization 100%, or is the inverted product formed more than the retained one?

Isn't it a primary carbonium ion to begin with (in addition, in here, $\ce {C-D}$ is smaller than $\ce {C-H}$)? In that account, I'd say $100%$ inversion because reaction is  $\ce {S_N2}$ rather than  $\ce {S_N1}$.

As a general rule, if you start with an optical inactive solution, you'll end up with an optically inactive solution. Since the carbocation is not optically active, then you'll get a racemate.

Stereochemical information is lost once you form a carbocation, since in general it presents a planar geometry and $sp^2$ hybridized geometry; barring some special cases involving bulky substituents etc. there is no preferred face of attack.

Since this problem is about the stereochemistry, I'd try to address it according to sophomore undergraduate Organic Chemistry text books used in the USA.
Briefly, any nucleophilic substitution reaction at primary carbon predominantly undergoes with SN2 mechanism while those reactions at tertiary carbon undergo SN1 mechanism exclusively. Meanwhile, reactions at secondary carbon would undergo either mechanism, based on reaction conditions, solvents, etc.
The "carbocation" given is a primary carbon even though it is a chiral carbon due to the presence of deuterium atom in place of hydrogen. Note that the stereo-effect caused by deuterium can be considered similar to that of hydrogen. Nonetheless, the bond energy difference (the $\ce {C–D}$ bond dissociation energy listed as $\pu{341.4 kJ/mol}$ while that of $\ce {C–H}$ as $\pu {338.4 kJ/mol}$) may cause the isotope-effects in kinetic of reaction mechanisms (that is a different avenue than what in discussion). Since $\ce{-CH2-Cl}$ and $\ce {-CHD-Cl}$ are similar and $\ce {C}$ atom in both cases is primary, the mechanism of the reaction should be SN2 exclusively. Therefore, the stereochemistry will be $\ce {100\%}$ inverted.
For reading for substitution reactions involving alkyl halides: article. Please also note that, one of example given on that website, the compound $\ce {Ph-CH(CH3)CH2-Cl}$ does not undergo methyl group migration to give benzylic carbocation during its nucleophilic substitution reaction.

Thanks for your answer. First of all, I am not sure why you're assuming that the reaction proceeds by $\ce{SN2}$ instead of $\ce{SN1}$. I have said a carbocation was formed, that means it is $\ce{SN1}$, so, please don't assume $\ce{SN2}$. Thanks

Thank you for your editing and clarification about the carbocation. I'll think about tomorrow. :-)

Did you happen to think about it again? :)

Let's take your question back a step to a single chiral diazonium salt (R)-1, prepared by treatment of the corresponding amine with nitrous acid. This reaction (1 --> 2) is typified by the Demjanov ring expansion. Although there are examples of direct SN2 displacement of N2 by water, they are not on such hindered primary carbons. A quick Chemical Abstracts search did not reveal any concerted alkyl migrations in this reaction type which would cause inversion at C1.
When the primary carbocation 2 is formed, it is renderd achiral with enantiotopic faces at C1. There is also free rotation about the C1 - C2 bond.  The migratory aptitude of alkyl groups is tertiary > secondary > primary > methyl. Thus, racemic 3 would lead to racemic ether (alcohol) 4. [E1 products are ignored.] While migration of ethyl is preferred over methyl, methyl does have a statistical advantage in that there are two of them. In this case, racemic carbocation 5 is formed, which captures solvent to form racemic 6 as a near equal mixture of diastereomers. There is always a chance that the solvolysis of the cationic site in 5 may not be totally racemized. A priori, it is difficult to predict that outcome.

(not my downvote) But I am finding it difficult to infer what your final conclusion is regarding the stereochemistry of the products. Are they racemic or diaseteromers? You've mentioned both in your answer.

I feel there are two options. Either 4 and/or  6. Structure 6 as a mixture of diastereomers. I don't have a precise answer, haven't done the experiment.


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Cyclohexene conformation in fused ring

The picture (a) is from a organic chemistry book by Clayden, Chapter 32. That means drawing (a) is right expression of the molecule.
What I wonder is that is it possible for (a) to ring-flip? If so, is (b) correct as a ring flip version of (a)? If it's not possible to ring flip, why?
Second question: Is drawing (c) also valid to draw this molecule? If it's so, is it same with (a)? or is it ring flip of (a)? If it's not, what's wrong with it?
Personally, I thought (c) was the right drawing for the molecule. Below (c), I drew the conformation of cyclohexene. The first conformation has a pseudoaxial hydrogen and the inner side of carbon faces downward. This is why I drew (c) with methyl at pseudoaxial, and inner carbon facing downward. But the book only has a drawing of (a).

You can also just tell me where I can find those kind of conformation, in book, website, anything because my four organic chemistry books are needless now.

I noticed that you had typed your question in proper paragraphs to insert a line break use double space or <br>.

I suggest you build a model (simple molymod is fine) - it will all become clear (and much easier than any 2D representation that I could draw for you)

@NotEvans thanks maybe I will buy one later. I don't have one now. Do you know any website that I can see the ringflip molecular form? Chemspider shows me a good 3d image but no function to change conformation.

You can ring-flip the enone ring but the methyl must be axial to both rings. If the methyl were equatorial, the double bond would be too twisted. In other words in this example, if a double bond is exocyclic to ring B (steroid lettering), there aren't enough carbons in ring A to attached axially to ring B. Models would be helpful.


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Residue from boiling bottled water
I boil natural spring water (Evian) in a stainless steel kettle and there is a white residue left behind that I can wipe off with a wet paper towel. What is this?
I also notice residue floating in water remaining on bottom of kettle. This does not happen with regular tap water.

Both spring water and tap water will contain dissolved minerals, e.g. calcium carbonate. Potable water just needs to be feed of poisons and reasonably sterile. Typically potable water isn't pure.

on the bottle there is usually a list of the mineral content. You've just isolated those minerals.


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Finding stationary solutions and verifying differential equations

Consider a trimolecular system where the following chemical reactions occur:
  $$\ce{A + B ->[k_1] A + C}\tag1$$
  $$\ce{A ->[k_2] B}\tag2$$
  $$\ce{C ->[k_3] C}\tag3$$
  Write the differential equations that model the evolution of the three component molar concentrations and determine the stationary solutions (there are two). Finally, if initially $[\ce{A}]=0.75$, $[\ce{B}]=0.25$ and $[\ce{C}]=0$, calculate the time when the mole fraction of $\ce{A}$ is half of its initial value.

I think that the differential equations, following the expression for the reaction velocity, are
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=-k_2[\ce{A}]$$
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]$$
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t}=k_1[\ce{A}][\ce{B}]$$
However, I don't know how to proceed with the stationary solutions. I guess that one of them would be obtained doing:
$$-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]=0$$
but I'm not sure.
Finally, I got the relations
$$[\ce{B}]=\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+\frac{k_2}{k_1}$$
$$[\ce{C}]=-\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+1-\frac{k_2}{k_1}-[\ce{A}]$$
but I have no idea how to obtain the time. Any help would be appreciated.

What is the meaning of $C-> C$ with rate constant $k_3$ ??

That's in the problem statement, and it doesn't give more information about it... I'm physicist and I got no idea about chemistry, so I cannot give you the answer

Shouldnt the rate equation for A be $\\frac{d[A]}{dt} = - k_2[A] + k_1[A][B]\$ ?

@SoumikDas No, because A is neither spent nor produced in the first reaction.

And, is the time required for the mole fraction to become half or the moles? Can you check again ?

I got $\frac{d[A]}{dt}=-k_1[A][B]$ (that's the left term of the first equation)$+k_1[A][B]$ (that's the right term of the first equation)$-k_2[A]$ (that's the left term of the first equation). Why do you get this equation?

The third equation still makes no sense, even if it is a part of the problem statement. Also, what is a stationary solution and why do you think you need one?

A stationary solution will verify the time derivative is null. And I don't know why I need them, it's part of the statement

Well, then you don't need many equations to do that. A goes to B and never goes back, B goes to C and never goes back, so in the end there will be only C (and maybe some B which can't transform due to the lack of A).

From $k_2$ simply calculate the half-life of reaction (2).

I don't understand what you're saying

Your differential equations are correct.
From $\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_1[\ce{A}][\ce{B}]+k_2[\ce{A}] = 0$ follow the stationary solutions $[\ce{A}] = 0$ or $[\ce{B}] = \frac{k_2}{k_1}$.
Reaction (2) obeys a first order decay law. Its half-life is $t_{\frac{1}{2}} = \mathrm{ln}(2) / k_2$.
The image below shows how $[\ce{B}]$ reaches the stationary state soon after the reaction has started ($k_1 = 0.3, k_2 = 0.2$).


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How to determine the phases in a complex phase diagram?
In my school notes, we have the following phase diagram that shows an example of how complex things can get when we have intermediate phases. I tried filling in phases, but I have no idea how to check if I am doing a good job or not. What is the mechanism to determine the phases starting with a diagram like this?
This is a question I have made for myself, NOT homework.

Pretty close, but you have a line compound near 60 wt.% that you haven’t identified.

That is where I had more trouble working. What is a line compound? It made sense for there to be another phase in between $\beta$ and $\epsilon$, but I never thought a whole compound could be represented by a line. Would you mind explaining further? Thank you!

A line compound is one that has a single composition. For example, Al2O3 does not have any range of composition.


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Synthesis Golf IX: Tezacaftor
A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

The target of this Easter special round of synthesis golf is tezacaptor, which has recently been approved by the FDA as a treatment for cystic fibrosis in combination with ivacaftor:

InChI=1S/C26H28F2N2O6/c1-24(2,14-32)22-10-15-9-17(4-5-19(15)30(22)12-18(33)13-31)29-23(34)25(7-8-25)16-3-6-20-21(11-16)36-26(27,28)35-20/h3-6,9-11,18,31-33H,7-8,12-14H2,1-2H3,(H,29,34)/t18-/m1/s1

Any commercial starting material is acceptable (as previously, commercial meaning you can buy it from Sigma-Aldrich) providing it has 10 or fewer carbons
The synthesis must include a cyclopropanation (i.e. the starting materials cannot contain a cyclopropane)
The synthesis should ideally provide a single stereoisomer of the hydroxyl stereo centre, but to keep things more accessible, a racemic synthesis won't be criticised

I am really glad that you are reviving these challenge questions NotEvans.
As for the synthesis, the amide disconnection leading to fragments A and B seems like the most simplifying first disconnection:

Synthesis of fragments A and B is outlined below:

Esterification of the readily available acid with isopropyl alcohol under Yamaguchi conditions;
Demethylation (this is why I've chosen to make the isopropyl ester, it shouldn't be destroyed with $\ce{BBr3}$);
Formation of a thiocarbonate crucial for the introduction of two fluorine atoms;
JACS 2010, 132(51), 18199;
Aldol condensation of the ester with formaldehyde;
Ester hydrolysis;
Conversion of the acid to acyl chloride.

p-Nitroaniline is brominated;
Sonogashira coupling;
Gold-catalyzed hydroamination;
Fmoc protection of the indole nitrogen turned out to be necessary at this point (so it doesn't interfere with the coupling with fragment A);
Reduction of the nitro group.

Now comes the endgame:

Amide formation;
Corey-Chaykovsky reaction of a stabilized sulfure ylide is selective for enone-like systems;
Piperidine removes Fmoc;
The indole nitrogen attacks glycidol thus opening the epoxide on the more accessible site;
Reduction of the ester with $\ce{LiBH4}$ should be selective. Several equivalents are needed because of the free hydroxyl groups. My only concern is the stability of the cyclopropane, but other hydride donors could be tried.

EDIT: Here is the synthesis of the alkynyl ester needed to make fragment B:

18. Methylacetoacetate is methylated twice;

Formation of enol triflate;
Base-induced elimination of the triflate gives the ester.

Total step count: 20
Longest linear sequence: 12

The alkyne you are using in the Sonogashira is really expensive - you'd probably choose to make it. Otherwise I'm a huge fan of the gold

Yeah, its preparation should be rather simple. I will add it later.

Now that I think about it, the stability of the gem-difluoride with LiBH4 is questionable.

I was trying to come up with something more elegant, but ended up using fairly classical chemistry, which (imo) doesn't need all that much explanation; 13 steps, 9 in longest linear sequence.
Fragment 1

(a) BBr3
(b) CDI
(c) PCl5 forms the geminal dichloride
(d) Et3N•3HF replaces Cl with F [ref J. Fluorine Chem. 2014, 160, 72]
(e) LiHMDS, 1,2-dibromoethane
(f) nitrile to amide, probably acidic conditions – not sure how sensitive this cyclopropane is.

Fragment 2

(g) (CH2O)n, TFA
(h) TBSCl, imidazole
(i) 4-Chlorophenylhydrazine
(j) K2CO3, allyl bromide

Union

(k) Pd2(dba)3, Me4t-BuXPhos, K3PO4 [ref J. Am. Chem. Soc. 2007, 129, 13001]
(l) AD-mix-β
(m) TBAF

I like it. I'm only worried that in step f) the acetal would hydrolyse before the amide in acidic conditions. Maybe the usage of Parkins catalyst could circumvent this issue, but still it's problematic: http://www.mdpi.com/2076-3417/5/3/380/htm

BF3 acetic acid complex might be a good choice for hydrolysing the nitrile to the amide

How to avoid  acid in step f. Alkaline peroxide in the presence of a nitrile is an epoxidizing reagent with the by-product an amide. What is the stability of the geminal fluorine moiety in base?

FIRST Fragment

a - acetone dimethyl acetal, zirconium tetrachloride
b - 2-Chloro-p-phenylenediamine, cobalt-based catalysis

c - paraformaldehyde, TFA
d - TBSCl, imidazole
The combined product:

Conditions: Pd(tBu3P)2, potassium phosphate, magnesium sulfate, AcOH, DMA
SECOND Fragment

e) Carbonyl fluoride, H+
f) paraformaldehyde, TFA
g) Me3S, NaH
Union to Product
h) Both fragments, DCC, catalyst
i) Deprotect
Overall: 10 steps
Longest linear sequence: 5 steps

The carbonyl fluoride step is stretching it. It will probably have to be done in a few (3) steps - similar to how orthocresol or Marko did their syntheses, making the total step count 12 and longest liner sequence 7.

In step b, you have two amino groups and you want one to react with the chloride, I don't think that you will be able to achieve the selectivity you desire.


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What is the purpose of the equation
In my chemistry class we solved the equation wavelength = h/mv for finding the mass of a baseball and were given all of the other variables. I know that this equation has to do with wavelength of objects that contain mass but since objects with mass such as a baseball don't move constantly up and down in a wave pattern as they move towards something, they just arc down then what is the purpose of this? My teacher said it had something to do with waves emitted by the baseball or something like that but then why would it need mass?

How do you know that a baseball doesn't move constantly up and down? Maybe the waves are just so small you can't see them? Maybe the said wavelength has something to do with this? Maybe you should have found it? Maybe you did?

I'm voting to close this question as off-topic because it's not about chemistry.

deBroglie wavelength of a particle: λ=h/mv
Yup - it seems to be for getting the wavelength of an electron or a proton.
https://www.thoughtco.com/definition-of-de-broglie-equation-604418
http://www.softschools.com/formulas/physics/de_broglie_wavelength_formula/150/
It's an entertaining equation to use with a baseball, though, even if it might not be a good application of the equation.
The baseball has 'huge' masses and velocities.  What sort of a wavelength will you get if you divide h by a 'huge' number?
Light can be either a wave or a [very very tiny] particle.
https://en.wikipedia.org/wiki/Wave–particle_duality - this wikipedia article talks about how, according to quantum mechanics, every particle also has a wave nature.
I hope that helps!


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How was water formed if oxygen and hydrogen exist as molecules?
I have learnt that both hydrogen and oxygen  exist as molecules. If so how was $\ce {H2O}$ formed?
In water there is two hydrogen molecule and one oxygen atoms but how is this possible? Oxygen only exists as a molecule in our earth then how does the molecule oxygen react with molecule hydrogen to form water.
Moreover, if water is formed by combining of hydrogen and oxygen molecules, then the chemical formula of water should be $\ce{H2O2}$, shouldn't it?

So what if oxygen exists as molecules? Molecules can be broken.

If you look at Standard Gibbs free energy of formation of $\ce {H2O}$ and $\ce {H2O2}$, you would find some reasoning (for reasonable values, see: !https://en.wikipedia.org/wiki/Standard_Gibbs_free_energy_of_formation).

I didn't understand Standard Gibbs free energy of formation of H2O and H2O2

Water does not have two oxygen atoms (has one) and one hydrogen (has two) hence the molecular formula of water is H2O. Water forms when there is enough energy to break the molecular bonds within the Hydrogen and Oxygen diatoms, they then combine and form water. As Mathew said, look at the Standard Gibbs free energy of formation. It will explain that H2O is much more favourable to be formed than H2O2.


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Why is a [Cu(SCN)2] complex black?
I've been creating various Copper(II) complexes using different ligands and predicting their relative hues using Crystal Field Theory by using the spectrochemical series to predict the extent of d orbital splitting on the Copper(II) ion and the resultant colour of the copper(II) complex.
When I added potassium thiocyanate ($\ce{KSCN}$) to a solution containing hexaaquacopper(II) the entire solution turned deep black. Doing some research online and knowing that $\ce{SCN-}$ tends to form complexes with planar geometry I'm fairly certain that the resultant copper(II) complex was $\ce{[Cu(SCN)4]}$.
Because of the relative strength of the $\ce{SCN-}$ ligand compared to the pale blue complex formed with $\ce{H2O}$ I assumed the solution would appear redder in hue as higher frequency wavelengths would be absorbed, however, I fail to see why the entire visible spectrum would suddenly be absorbed. Can this phenomenon be explained with Crystal Field Theory?
Also perhaps my solution was not dilute enough?

Redox, maybe?$!$

As the user above states, if you mix concentrated solutions of $\ce{Cu(II)}$ salts and $\ce{NCS}$ you get $\ce{Cu(NCS)_2}$ which is a black solid. I'd disagree with them that it's not longer a complex because it 100% is - and you are correct to assume that the local coordination is $\ce{Cu(NCS)_4}$. To be more precise, each copper is coordinated by $\ce{Cu(NCS)_2(SCN)_2}$, and there's actually a large Jahn Teller distortion so it could also be described as $\ce{Cu(NCS)_2(SCN)_4}$. They're right that the colour can't be easily explained by $d-d$ transitions (i.e. the kind you're thinking of w.r.t. ligand field splitting) though.
$\ce{Cu(NCS)_2}$ is black probably because of ligand to metal charge transfer (the same reason $\ce{Fe(III)NCS}$ is blood red) - i.e. on absorption you transiently form $\ce{Cu(I)}$ and $\ce{NCS}$ from $\ce{Cu(II)}$ and $\ce{NCS–}$. It's more complex that that for sure, but I think it's fair to say no one knows yet, because we only worked out the structure two years ago.
If you want a lot more information about $\ce{Cu(NCS)_2}$ we published on it here https://journals.aps.org/prb/abstract/10.1103/PhysRevB.97.144421 (on the arxiv at https://arxiv.org/abs/1710.04889).

When you add $\ce{SCN^-}$ ions to aqueous solution of $\ce{Cu^{2+}}$, the thiocyanate ligands will start to substitute the water molecules from the octahedral $\ce{[Cu(H_2O)_6]^{2+}}$ complex.
Though $\ce{SCN^-}$ is a weak field ligand, It can substitute relatively stronger field ligand $\ce{H_2O}$ from $\ce{Cu^{2+}}$ complexes, which can be explained by Pearson's HSAB Theory. $\ce{Cu^{2+}}$ is a very soft acid, as it has less positive charge on it, more no. of $\ce{d}$ electrons etc. which matches with the properties of the soft acid. Between, $\ce{H2O}$ and $\ce{SCN-}$, $\ce{H2O}$ has its ligand site as $\ce{O}$ which is a hard base centre, but in $\ce{SCN-}$, the ligand site $\ce{S}$ is a relatively softer base centre, as it has a larger size, lesser electronegativity, and lesser electron density than $\ce{O}$,and these properties make $\ce{SCN-}$ a soft base. Now, according to HSAB theory, soft acids prefer to bind with soft bases. Thus $\ce{Cu^2+}$ has more affinity towards $\ce{SCN-}$ rather than $\ce{H2O}$.
Thus when you have a aqueous solution of $\ce{Cu^2+}$, and add $\ce{SCN-}$ gradually by little amounts, the following complexes will start to form.
$$\ce{[Cu(H2O)6]^2 ->[SCN-] [Cu(H2O)4(SCN)2]}\text{( apple-green colour)}$$
$$\ce{[Cu(H2O)4(SCN)2]^2 ->[SCN-] [Cu(H2O)2(SCN)4]^2-}\text{ (pale -yellow colour)}$$
This shift in the colours of the new complexes can be explained by CFT. As more weak field ligands ($\ce{SCN-}$) is introduced in the complex, the octahedral crystal field splitting energy ($\Delta_\mathrm{o}$) of the complexes decreases and thus the wavelength absorbed shifts to higher wavelengths and complementary colours also become higher in wavelengths.
Now, if you add sufficiently higher amounts of $\ce{SCN-}$, all the complexes will be destroyed and you will get only a compound of $\ce{Cu^{2+}}$ and $\ce{SCN-}$, which is a normal ionic compound, and no complex is left thereafter.
$$\ce{[Cu(H2O)2(SCN)4]^2- ->[high SCN-] [Cu(SCN)6]^4- (unstable) -> Cu(SCN)2(black)}$$
Thus the final compound is not at all any complex and just a black coloured compound, which might have formed if you have added significantly higher amount of $\ce{SCN-}$, in aqueous solution of $\ce{Cu^2+}$. So, it is irrelevant to judge the colour of final product i.e. $\ce{Cu(SCN)2}$ through CFT as it is now no longer a complex at all.

If I recall that correctly the problem with copper is that the copper is right at the border between thiocyanate and isothiocyanate, and thiocyanate is a pseudohalogen which can also reduce the Cu(II) to Cu(I). So you have a mixture of ionic, coordinated, oxidized and reduced linkage isomers which all have different colors.

Yes, this answer is thoroughly wrong, even maybe not an answer at all.


text:
Can strong acids be amphoteric?
There are amphiprotic (or amphoteric) substances, like $\ce{H2O}$ and $\ce {HCO3-}$, which act both as an acid and base.
If you mix a strong acid like $\ce{HCl}$ ($\mathrm{p}K_{\mathrm{a}} = -6.3$) with an even stronger acid, such as perchloric acid ($\ce{HClO4},~\mathrm{p}K_{\mathrm{a}} \approx -10 $) can it act as a base?

I'm totally not sure what you mean by "If you mix a strong acid like HCl with an extremely strong acid, HCl, can it act as a base like them?". $\ce{HCl}$ is not an amphoteric acid.

That said, it will be protonated, if you push really hard.

@GaurangTandon yes, HCl is amphoteric: https://www.aanda.org/articles/aa/ps/2010/13/aa14959-10/aa14959-10.ps.gz

@DavePhD Oh! I didn't know that. Well, that paper seems to be using some pretty extreme conditions though, to forcibly protonate $\ce{HCl}$. It doesn't seem to be as "amphoteric" as the regular bicarbonate anion...

@GaurangTandon That paper is reporting what occurs naturally in outer space.  This table lists 12 common substances that are weaker bases than HCl.  https://en.wikipedia.org/wiki/Proton_affinity_(data_page)

Well an acid can act as a base when there is a stronger acid present, But if you look at the molecular level acidity depends on how well the negative charge is balanced. so out of the two molecules present the one which can disperse(stabilize) the negative charge to a greater extent will act as acid.

Yes.
For example, $\ce{H2SO4}$ can be protonated to $\ce{H3SO4+}$
See THE SULPHURIC ACID SOLVENT SYSTEM Canadian Journal of Chemistry (1960), vol. 38, pages 1363-1370.

Chemists see this happen all the time when they use a mixture of nitric and sulfuric acids for nitration.  The nitric acid acts as a Bronsted-Lowry base, but forms the powerful Lewis acid $\ce{NO_2^+}$, the nitryl ion, which then acts as the nitrating agent:
$\ce{NO_2(OH) + 2 H_2SO_4 -> NO_2^+ + H_3O^+  + 2 HSO_4^-}$


text:
How to know if a sample suffered descomposition during melting point?
If I take a sample in order to determine its melting point, how do I know if that sample suffered a descomposition?

How about color change?


text:
How can endothermic reactions be spontaneous?
If endothermic reactions absorb heat from the surroundings, how can they be spontaneous, if heat only flows from high temp to low temp objects?

Heat does not necessarily flow from high temp to low temp objects. Open your fridge, and you'll see. How did it get cold? When you bought it, it was room temp, wasn't it?

Another example would be the evaporation of water at room temperature. Evaporation is endothermic, yet water slowly evaporates into vapor in the atmosphere.

A very similar example (which I find a bit easier to visualise) to Pritt Balagopal's is the melting of ice at room temperature.

A spontaneous endothermic reaction can occur when the changes in enthalpy and entropy yield a negative Gibbs free energy. An endothermic reaction can be spontaneous if entropy increases by more than the change in enthalpy.
$\Delta G = \Delta H - T\Delta S$
For endothermic r̥eactions, $\Delta H$> 0 and thus if the value of $-T\Delta S$ overshadows the $\Delta H$ value (usually at very high temperatures) then the overall value of $\Delta G$ becomes negative and thus the reaction becomes spontaneous.

At the end of the day spontaneity boils down to the sign of "Gibbs free Energy"
so if the $\Delta G= \Delta H- T\Delta S$ gives you a negative value then the process would be spontaneous.


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What is the difference between air and evaporated water (steam)?
I was thinking about the difference between air and water, and how water evaporates as steam and becomes air. This led me to believe that air is just evaporated water. However when I read about air it said it consists of nitrogen and other gases, which water doesn't contain. So my question is what is the difference between air and evaporated water (steam) and what happens to steam when it mixes with air?

Look at pretty much everything around you. Wood is very different from concrete, which is different from metal, which is different from glassware. Why?

@IvanNeretin: I haven't studied any chemistry yet, but is it because of the different arrangement of atoms?

Well, yes, and sometimes different types of atoms, too. Same thing with air and water steam: they are about as different as it gets.

Evaporated water doesn't become air it becomes a gas, water vapour. Your confusion is between the terms gas and air: air is a specific mixture of gases; water vapour is just water in gas rather than liquid form.

Air is a mixture of many gases. It contains elemental gases such as $\ce {N2}$ (nitrogen; $\ce {78.1\%}$), $\ce {O2}$ (oxygen; $\ce {20.9\%}$), and $\ce {Ar}$ (argon; $\ce {0.9\%}$), as well as gaseous compounds such as $\ce {CO2}$ (carbon dioxide; $\ce {0.03\%}$). The percentages are given by volume for dry air. That means, the first two gases, $\ce {N2}$ and $\ce {O2}$, make up $\ce {99.0\%}$ of the atmosphere by volume.  It also has some low level Ozone ($\ce {O3}$) and under ambient conditions, has varying quantity of water vapor as well (as clouds in the sky).
That water vapor in air comes from different sources such as steam water, water evaporated from sea bed, etc. You always hear how much vapors in air during everyday weather forecast. That has specific name--humidity. And there is an another measure to give you an idea how much vapor in air in a given day. That called the dew point. The dew point is the temperature the air needs to be cooled to (at constant pressure) in order to achieve a relative humidity (RH) of $\ce{100\%}$. At this point the air cannot hold anymore water in the gas form. If the air were to be cooled even more, water vapor would have to come out of the atmosphere in the liquid form, usually as fog or precipitation (you'd see this more often in spring day mornings). Hope, this'd be enough to give you an idea what happens to vapor when they get into air.

Note that the percentages mentioned in the first paragraph are on a dry basis, i.e. without considering water vapour. The percentage of water in air saturated with water vapour, i.e. at 100%RH, is around 4%, for instance.

@Eashaan Godbole: Thank you for the clarification. I have edited the phrases, accordingly.

santimirandarp: If actual temperature falls behind the dew point (say, dew point is $\ce {65}$ degrees but during the night atmospheric temperature down to $\ce {55}$ degrees), then precipitation occurs. That's why we see water droplets on plant leaves (called dew!).


text:
What are the optimal conditions for the complete acid hydrolysis (using HCl) of potato starch?
I'm hoping someone can provide the specifics on the optimal conditions required for the complete acid hydrolysis (using $\ce{HCl}$) of potato starch?
I will be using an autoclave (~$\pu{15 psi}$ @ $\pu{120 degrees Celsius}$) to facilitate hydrolysation.
Specifically I'm after for the following information:

Concentration of dry starch required for the starch slurry.
Concentration of $\ce{HCl}$ required to adjust $\pu{pH}$ to required level.
What is the required $\pu{pH}$ for hydrolysation?
Time required to complete hydrolysation under above autoclave conditions.

Plus anything else that I may have missed....
Thanks!

I have suggested some MathJax to make your post look a little prettier.

If you are able to measure the amount of the starch you hydrolized, why not to make some experiements to find by yourself your process' variables? You can make a little design experiment or just two or three experiments with a fixed volumes of starch and different volumes of HCl and then compare and try to find something optimum.


text:
Can heat capacity at constant volume for a gas decrease as temperature increases?
I found a function for the $C_V$ of a gas which is being heated under constant volume,but the function shows that the $C_V$ is decreasing. I was wondering if this is possible for any gas since I know the $C_V$ is supposed to increase with temperature. Is there any reason why this would happen?

Handbook of Chemistry and Physics, Chemical Engineers' Handbook, NIST, Fundamentals of Engineering Thermodynamics (Moran et al)

As temperature increases, it activates the rotational and vibrational degrees of freedom, so $\ce{Cv}$ increases with increase in temperature. The graph $\ce{CO2}$ would be: