Does the mass of sulfur really decrease when dissolved in water and increase when burnt? I was going through a bunch of interesting science 'facts' and one entry went this way:

  Name an element whose mass decreases when it is dissolved in water and increases if it is burnt.

I tried guessing, but that didn't help. So I had a look at the answer: it was Sulfur

There are two questions that struck my mind after I saw the answer:

What does mass decrease/increase as stated here mean?(Does it mean that mass of sulfur along with other elements combined with it changes/or anything ambiguous such as change in actual mass is taking place) )
Is sulfur really the correct answer to that question? If yes, then why does sulfur show this anomalous property? Are there elements that display a similar property?

Any suggestion is welcome.

Upon reading the answers on Quora (thanks S007 for pointing that out) I realized this trick question is a lousy play upon two somewhat peculiar features of sulfur:

When submerged in water (not "dissolved", mind you, for sulfur does not dissolve), its apparent weight becomes less, thanks to Archimedes and the buoyancy force. This is indeed true of any element and compound, but sulfur has density of about 2, so its apparent weight decreases about twofold.
When burnt, sulfur forms a compound with oxygen which (containing all that sulfur plus oxygen) naturally weighs more than sulfur alone. Again, this is true of any element or compound that can be burnt, which are numerous. If anything, sulfur is "special" in that its atomic weight is 32, and the molecular weight of $\ce{SO2}$ is 64, so the increase is exactly twofold.

The way it stands now, though, the question is quite meaningless. Mass of any element never changes (barring nuclear reactions), that's the law of conservation. If you count compounds, then the mass would increase in any reaction (presuming we started from the pure element) and stay unchanged otherwise. If you count all compounds except gases, then the mass would sometimes increase and sometimes decrease, so there might be an answer, but then sulfur fails both conditions: it does not dissolve in water at all, and it all transforms to gases (that is, "vanishes") when burnt.

Theoretically yes, if something loses energy it also loses mass (seriously no joke). Also if it gains energy it gains mass(By Einstein's mass energy equivalence)
Of course the energies involved in chemical reactions are in no way affecting mass of the molecule atom in any measurable way.

This applies to all forms of energy, potential, chemical, doesn't matter.

Example: the planet Saturn would be pulling harder at my toenail then any change might due to the potential energy of being in a gravity well (Earth).

What is the mass gain loss? We use Einstein's equation: $$\Delta M=\frac{\Delta E}{c^2}$$

So you see the  $c^{-2}=1.1126501 \times 10^{-17} $ part will slap the mass change to something almost nonexistent.
So yes,the mass will change,it depends on you how do you measure such a small change.


Method to synthesize KAu(CN)2 (or KCN)? Every reasonably effective method of electroplating $\ce{Au}$ onto another metal's surface (e.g. $\ce{Cu}$,$\ce{Ni}$,$\ce{Ag}$, etc.) involves use of $\ce{KAu(CN)2}$ or in some cases $\ce{KAu(CN)4}$ in acidic, neutral, or (in almost all cases less effective) basic solution. Ideally it should be $60$ to $80~\mathrm{^\circ C}$ with a certain current density applied (c. $0.5~\mathrm{A/dm^2}$). In the past I have tried a direct route of dissolving the gold in warm $\ce{HCl / HNO3}$ as the deposition solution to little effect (a brownish/blackish tint is left on the receiving surface with very little tint of gold). The short story is that I have found the use of cyanide unavoidable.

$\ce{KAu(CN)2}$ runs $157/g from Sigma-Aldrich so I would like to synthesize it, if possible.

I have:

1g solid $\ce{Au}$
Excess $\ce{KNO3}$, $\ce{KI}$
Various $\ce{C-N}$ organic compounds, e.g. urea, uric acid, etc.

I think my best route is indirect — to first synthesize $\ce{KCN}$ and then form the product:

$$\ce{4 Au + 8 KCN + O2 + 2 H2O → 4 K[Au(CN)2] + 4 KOH}$$

...and follow by vacuum filtration on the solid precipitate. Which begs the question of the best method for safe synthesis of $\ce{KCN}$.

Much of my information comes from this difficult-to-find and expensive book, "Properties of Electrodeposited Metals and Alloys: A Handbook" by Safranek. It's got quite a lot of proprietary data. Also Kohl's chapter on gold electrodeposition.

I read that you are a computational chemist at university, so I assume you went through chemical training. This is a good thing. But please remember that cyanides are potent poisons and that protonated cyanide ($\mathrm{p}K_\mathrm{a}(\ce{HCN}) \approx 8$) aka hydrogen cyanide is a poisonous gas.

If you are not a trained chemist (or if you are one but do not feel $100~\%$ safe about using cyanides), stick away from them!

With that in mind this leaves two methods of acquiring $\ce{K[Au(CN)2]}$ — either buying it directly from YourSuppliers™ or by synthesising it. My preferred route of choice would definitely be to buy it from YourSuppliers™, as that is the safest way possible — and always always ensure that it remains in an alkaline solution.

I have no idea about the experimental procedure to synthesise $\ce{K[Au(CN)2]}$ according to the reaction you propose, but it didn’t escape my notice that you are generating $\ce{4KOH}$. This means, I’ll have to assume that the reaction runs under acidic conditions. If that is the case, do not use that method unless you have access to a proper and tested fume hood! If there are other procedures out there that employ alkaline reaction conditions you may (as a trained chemist!) decide to use them at your own risk.

Assuming you did find a method that employs alkaline conditions, the question remains where to get the cyanide from. Well, SigmaAldrich gives me a price of 38 € for $100~\mathrm{g}$ which means that the chemical is very cheap. Note that acquisition of potassium cyanide may well be regulated in your jurisdiction — and rightly so!

Whatever you do, do not attempt to synthesise $\ce{KCN}$ (or any other cyanide for that matter) at home!

Remember that a certain percentage of the population can sense $\ce{HCN}$ which they perceive as an almond-like smell. Don’t attempt to test which part you belong to. If the concentration of $\ce{HCN}$ is such that you can smell it, you are in severe danger!

This bears repeating for future visitors:
Do not do this unless you are a trained chemist and have access to a proper fume hood! Never, ever try this at home! Never work with cyanide in the open air or in any other place where unsuspecting passers-by may be endangered!

Sodium cyanide can be very easily made by fusing sodium ferrocyanide with sodium carbonate. The guy in the video is still alive, but I would recommend wearing gloves at least.

Video including Au dissolution


What is the difference between polyethene and polyethane? The monomer in polyethene seems to drop the $\ce{C=C}$ bond and end up with a $\ce{C-C}$ bond. The polymer chain then looks like $\ce{[CH2-CH2]_{n}}$. If I were to look at that polymer chain I would probably call it polyethane not polythene (or I could even call it polymethane) because it doesn't have a double bond, so I wouldn't classify it as an alkene. Am I doing something wrong with my analysis here? Or is there some sort of rule about nomenclature that change when things become polymers and drop a functional bond (in this case the $\ce{C=C}$ bond)?

The reason it is called polyethene (or polyethylene) is because it was produced from the monomer of ethylene. To produce polyethylene, a radical initiator is added to ethylene, which attacks one carbon and opens up the double bond leaving a radical at the end. This radical attacks another ethylene molecule, etc. The chain stops forming when it encounters another radical. So, from ethylene, we get polyethylene (even though it looks like polymethylene).

There is not such correct term for either of the terms you've come up with. There is no polyethene. There is no polyethane, not in polymer or macromolecular chemistry and thinking.

Polyethylene (PET) a thermoset polymer (plastic) on the other hand, is ubiquitous.


Iodine formation reaction $$\ce{IO3-(aq) + 5I-(aq) + 6H+(aq) -> 3I2(aq) + 3H2O(l)}$$
Why does the reaction between iodide and iodate ions have to occur in an acidic medium ($\ce{H+}$) to form the iodine? what do the $\ce{H+}$ ions react with?

A bit more complex per a source 'Kinetics and mechanism of the iodate-iodide reaction and other related reactions', to quote from an article in the journal Physical Chemistry Chemical Physics (November 1999):

The reduction of iodate by iodide, known as the Dushman reaction, is one of the components of the oscillating Bray-Liebhafsky reaction. The large body of work devoted to the kinetics of the Dushman reaction gives seemingly contradictory results. This work analyses and summarizes them and shows that they are special cases of a complicated kinetics and that they can be explained by a common mechanism. This mechanism can be generalized to all the XO3- + Y- reactions (X, Y = I, Br or Cl). They all involve an asymmetric intermediate YXO2.

See reaction system depicted in Equations (8) to (13). Here is a partial discussion of the system from the available PDF to quote:

The essential hypothesis of this mechanism is the existence of as a definite intermediate species and not as I2O2 an activated complex. Its existence was already proposed in  1930 by Bray [17] in order to explain the classical rate law. We make no distinction between I2O2 and H2I2O3 (I2O2 + H2O) as these two forms are kinetically equivalent. In reaction (12), B~ represents the anion of the buffer....

As to the question on the role of H+, note that the starting Equations (8) and (9) of the system both involve H+, which results in the cited intermediate $\ce{I2O2}$:

$\ce{IO3- + H+ ⇌ IO3H}$  (8)
$\ce{IO3H + I- + H+ ⇌ I2O2 + H2O}$  (9)

Omitting the hydrogen ions would lead to
$$\ce{IO3-(aq) + 5I-(aq) -> 3I2(aq) + 3O^2-}$$
but $\ce{O^2-}$ would not exist in water. So an alternative could be to write
$$\ce{IO3-(aq) + 5I-(aq) + 3H2O -> 3I2(aq) + 6OH-(aq)}$$
Basically, the hydrogen ions are there to indicate that the reaction takes place in an acidic solution and hint that access hydrogen ions would improve the reaction kinetics.

The version with the hydroxide ions could be used to emphasize that the reaction would cause an increase of the pH value.


Why is copper a better conductor than iron? Metals are good conductors as the electrons in their outer shell are loose and can plunge out of the atom with the application of the slightest force (voltage).

Silver is the best metallic conductor, then comes copper.

Why is one metal better at conduction than the other one?
What determines the difference in conduction between 2 metals?
Why is silver a better conductor than copper, and why is copper a better conductor than iron if all of them have loose electrons?

This is, ultimately, a question on solid state physics rather than chemistry. Further, the OP indicates that they are in high school, which kind of limits the depth of the answer that might be useful to them. However, I will try to make a simple, yet detailed answer.

As atoms are brought closer and closer together, their electron clouds overlap and interact. When a crystalline solid is formed, one abandons consideration of the atomic energy levels. Instead, one seeks out the Bloch functions to describe the electonic states - the Bloch functions explicitly consider the symmetries of the crystal. Further, they represent extended electron states, those that extend across the entire crystal. They represent the allowed (energy and momentum) combinations of electrons.

For a metal, the highest Bloch states occur as a band of states which is only partially filled. Unoccupied states lie just above states with electrons. To move an electron about, it needs just a little bit of energy, and off it can go. (Fully filled bands do not allow conduction - for any electron going one way there is one going the other way. This needs a good grounding in solid state physics to fully grasp).

OK, in general then the conductivity should just be the number of electrons near the line between occupied and un-occupied. Kind of. This line is called the Fermi surface, and things get more complicated quickly. The shape and connectivity of the Fermi surface impact conduction dramatically. Lets take a deeper look at copper and iron.

Copper is a simple noble metal with a beautiful near-ideal Fermi surface (see Ashcroft and Mermin's Solid State Physics book, or C.Y. Fong et al., Comparison of band structures and charge distributions of copper and silver). It turns out that the Fermi surface of copper is almost exactly what one would expect from the free electron sphere. Electrons can move easily in any direction, the surface is fully connected, life is good.

Iron is much more complicated. Following along from J. Callaway and C.S. Wang, Energy bands in ferromagnetic iron one finds two major differences. First, because iron atoms indeed have individual magnetic moments, the overall band structure is split into two different spin states for the electrons. So, each spin state is separated from the others, reducing how each can move (relative to copper, were such splitting does not exist). Second, and more importantly, the Fermi surface looks nothing like a free electron sphere. Instead, for iron the Fermi surface is not fully connected, comprising instead of multiple 'pockets' of electrons that do not communicate directly with the other pockets (and each spin state has a different set of pockets). To move electrons around, they have to 'jump' from one pocket to another through scattering. This makes it much harder to move the 'free' electrons in Fe around to get net conductivity.

An analogy would be the difference of crossing a stream using either (1) a nice smooth bridge (copper) vs (2) jumping from rock to rock (iron). The band structure for copper just makes it so much easier, so the resistivity is much lower.

TL;DR.
To get the "real" answer, you have to have a good understanding of the underlying quantum mechanics. I'm pretty sure you're not ready for that, yet! (and note that this topic would require several semesters of quantum mechanics, it's in the sub-discipline of solid-state physics...) So, my answer will have to be simplified.
You're taught in school (perhaps) that there are 3 types of chemical bonds: ionic, covalent, and metallic. The world isn't that simple in reality, while all chemical bonding is electromagnetic, there is no "perfectly pure" ionic, covalent or metallic bond. If you accept the idea that metallic bonding involves all of the valence electrons of the atoms in the crystal (metals generally exist in small crystals called grains or crystallites) then we can simply say that Ag, Cu, and Au all have a single valence electron (per atom) and it turns out that this electron is fairly well shielded by each atoms' inner electrons from the nuclear charge which makes it much easier for these electrons to move, compared to, say iron. One way to look at it, I suppose, is that the road is less sticky (I wonder if calling it smoother would be a better analogy?) so that these puppies can move more easily than the valence electrons of iron. So, the answer is that there are degrees of "looseness". There are also grain defects and grain boundaries to cross. And also you should always keep in mind that electrical conductivity is temperature dependent. Stating categorically that silver is "the best" electrical conductor is more fan-boy enthusiasm than objective fact. The "best" electrical conduction is with superconductors - but of course then you need to consider the amount of current they can carry before they fail so "best" takes on a magnetic field dependence..."best" isn't a very scientific word is all I'm saying...there's almost always too many qualifications involved.
If you want to get a more correct answer, go to wikipedia's articles on Nearly free electron model, Metal, Fermi surface, Electrical resistivity and conductivity but I'm afraid that they're pretty hard to understand even for someone who has had some quantum mechanics.
I personally like my analogy that the road for silver's electrons isn't as sticky, but that's not to be taken too far... You actually have to consider not only the 'resistance' due to the nuclear (attractive) charges as the electron zooms by, but also the path the electron follows (the crystal lattice, how zig-zag the road is) and the defects and boundaries is must pass through (stop and pay a toll).


What are the resonance structures for chlorite anion? What are the possible resonance structures for $\ce{ClO_2^-}$? Assigning one double bond to the structure makes for formal charges of $\ce{O(-1)-Cl(0)=O(0) <-> O(-1)=Cl(0)-O(-1)}$. It appears that the same set of formal charges can be achieved with two double bonds (expanding the octet on $\ce{Cl}$ to 12 electron capacity, using the d orbital). That is, $\ce{O(0)=Cl(-1)=O(0)}$.

Is this feasible, and if not, why not?

Disclaimer
The here used methods are also only approximations and I do not claim that the provided numbers are accurate. However, they are used here to make the point, that bonding situations involving resonance are not easily explained.

Oxygen is slightly more electronegative than chlorine, therefore the bonds will always be slightly polarised towards the oxygen.

The following resonance scheme is based on an NBO analysis on the DF-BP86/def2-SVP level of theory. The percentage value gives an estimate how close the given conformation is to an idealised Lewis structure. From this you could also derive how much such a structure would contribute to a resonance situation (in terms of Valence Bond theory).
What we do see here is, that the electronic structure is better described with more formal charges. The contribution of a π bond is best described by an ionic donor-acceptor structure, rather than breaking the octet rule at chlorine.
Noteworthy is also that the doubly ionic donor-acceptor configuration is as equally good a lewis structure than any that breaks the octet rule.
The worst representation is the one with a four electron three centre bond (which is also not part of the Lewis bonding picture anyway).

Videos like the one mentioned in the Wikipedia article are easy to follow, but often they are too simplified to be correct.

That would be like calling it Chlorine dioxide

(http://en.m.wikipedia.org/wiki/Chlorine_dioxide)

While in chlorite ion, the ones you specified will only appear as stated here

  A chlorite (compound) is a compound that contains this group, with chlorine in oxidation state +3.

(http://en.m.wikipedia.org/wiki/Chlorite)


Is glass an amorphous solid or supercooled liquid? I have been informed that glass is a super-cooled liquid and is also considered to be an amorphous solid.

Can it be both and, if not, what category does it fall into?

You are correct: it's not both.

Why? What you've heard is nonsense. Very old window panes are thicker towards one side because of how they were made at the time. It became common practise to put the thicker edge at the bottom, because it was less likely to exceed the acceptable stress on the thinner area.

Contemporary windows have been found with the thicker edge to one side, or even the top. Gravity on Earth has been consistently towards the core since at least the Middle Ages, so the "flowing glass" hypothesis is false.

Glass is an amorphous solid by definition (per Karl's comment) and is not a super-cooled liquid (per Prabhdeep Singh's comment).

I think the thicker bottoms could be due to a phenomenon known creep (the internal flow of molecules under stress and over time). Amorphous materials below their glass transition temperatures show no molecular chain mobility, yet under a constant stress (gravity) and a very very long time (700,1000) years some very, very slow creep motion increases the volume at the bottom of the glass and thins it towards the rest of the body and in direction to the top.


Comparing nucleophilicity We have to compare the nucleophilicity of $\ce{OH-,NH2-,CH3CH2-}$ and $\ce{CH#C^{-}}$. I managed to find out that $\ce{CH3CH2-}>\ce{NH2-}>\ce{OH-}$ and that $\ce{CH3CH2-}>\ce{CH#C-}$. But I don't know how to compare $\ce{CH#C-}$ with either $\ce{OH-}$ or $\ce{NH2-}$.

My attempt:

I know that the acidity order of the 2nd period is $\ce{CH4}<\ce{NH3}<\ce{H2O,HF}$. Now the stronger an acid is, the weaker its conjugate base will be. Therefore, the conjugate base order will be:
$$\ce{CH3-}>\ce{NH2-}>\ce{HO-}>\ce{F-}$$

We know that nucleophilicity parallels basicity along the period, therefore the nucelophilicty order will be as the basicity order:
$$\ce{CH3-}>\ce{NH2-}>\ce{HO-}$$

Now we can compare $\ce{CH3CH2-}$ with $\ce{CH3-}$.

Since methyl is an electron donating group, it will increase the negative charge on the carbanion and therefore $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{CH3-}$.

Since $\ce{CH3-}$ is more nucleophilic than $\ce{NH2-}$ and $\ce{OH-}$. Also $\ce{CH3CH2-}$ is more nuceleophilic than $\ce{CH3-}$. As a result $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{NH2-}$ and $\ce{OH-}$ and $\ce{CH3CH2-}$.

We can also compare $\ce{CH3CH2-}$ and $\ce{CH#C-}$. Since $\ce{CH#C-}$ has sp hybridization, it will be less nuceleophilic than $\ce{CH3CH2-}$.

But how can I compare $\ce{CH#C-}$ to either $\ce{OH-}$ or $\ce{NH2-}$?

You also may want to consider Hard-Soft Acid-Base Theory. The carbanion is generally said to be a better nucleophile than hydroxide, presumably as measured in a highly controlled setting using a standard electrophile like methyl iodide. I'm under the impression nucleophilicity is very nearly defined as how fast an electron donor reacts with $\ce{CH3I}$. Note that there are no hydrogens available for elimination on this electrophile, which precludes the possibility of the nucleophile acting as a base instead.

In practical reaction conditions (protic solvents, beta-hydrogens on electrophile, etc), we may need to worry about our nucleophile acting as a base. This is where HSAB Theorycomes in. Hydroxide is considered a "hard base", which means it may be more inclined to react with "hard acids" like $\ce{H+}$ rather than "soft acids" like methyl iodide. As a "soft base", a carbanion will probably prefer to bond with methyl iodide (as a classic nucleophile) than with any available protons (as a Brønsted-Lowry base). So in practice, the difference between a carbanion's nucleophilicity and hydroxide's might be larger than expected due to competing "hard base" reactions (Brønsted-Lowry rather than Lewis). Just food for thought.

Qualitative nucleophilicty ratiomale

By removing H ions, negative charge on more electronegative more stable so acidic strength order that those EN is less easily lose its electron so order of nucleophilicity EN of NH3 is 3 o is 3.5 $\ce{C#C}$ is 3.25


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

It is +5 .Even by the structure.

Notice that it is a resonance structure of:

Now one of the bonds are coordinate bonds where nitrogen donates BOTH its electrons to oxygen (otherwise, oxygen cannot have a single bond without a negative charge and no peroxy type linkage).

So, while determining oxidation state from the structure :

O.N=+2+2+1=+5

References:
https://en.wikipedia.org/wiki/Dinitrogen_pentoxide

Nitrogen cannot exhibit a +5 oxidation state due to the absence of d-orbitals.

PROOF: Quantum Mechanics basics

As Nitrogen is in 2nd period in the modern periodic table, principal quantum number being n=2, the Azimuthual Quantum numbers = 0 to n-1
for Nitrogen it ranges from  0 to 1.

0 (zero) represents an s orbital and 1 represents p orbital hence no D orbitals are there.

... which clearly tells us that due to absence of d orbitals, nitrogen cannot expand its oxidation state greater than 4. Hence by the basic principle of quantum mechanics  we can say that in N2O5 nitrogen has +4 oxidation state.

NOTE: Formulae are made to simplify the things. But they have some limitations But in order to overcome those limitations one must use the basics of the subject.


What is the difference between a glassy, amorphous and a vitreous phase? I have been reading about some inorganic materials lately, and these three terms confuse me. If they all mean the same thing, would still one of them have preference over others?

After some reading I have decided to answer this question myself for anyone else with the same confusion. The info is from the downvoted answer of alphonse and the wiki-link, but with a better overview.

Both vitreous and glassy are amorphous. Amorhous is the broadest term to rule them all.

Some amorphous materials have a Glass Transition Temperature (GTT). Above this temperature the material becomes a viscous liquid. note that both below and above the GTT the material is amorphous.

An amorphous solid that has a GTT is called a glass.
The transition (heating) over the GTT results in a glassy phase.
The reverse transition (cooling) corresponding to the GTT is called vitrifcation. The result is a vitreous phase.

If I made mistakes please comment and I will update the answer.

Mostly a matter of preference. BUT. Consider the Glass Transition temperature. (see https://en.wikipedia.org/wiki/Glass_transition). It is important that you understand that while all glasses are, by definition, amorphous, that not all amorphous solids are glasses. Vitrification is the transition from liquid to a glass, such a glass could equally be called vitreous. I wouldn't call a solid, for example a rubber solid, below its Tg to be vitreous, no matter that it is amorphous. (Of course, if a solid is crystalline, then it's outside the context of this discussion.) There are contextual complexities, of course. Silica glass in it's molten state is still glass, while at the same time is not in the glassy state. So, there are different meanings for the term "glass". For instance, I'd not use it for polymers, even though polymer glasses is a legitimate area of molecular dynamics  study. Glass in the strictest sense is an solid, amorphous, brittle, rigid, inorganic, transparent material. As you remove those qualifiers the term tends to be used less often. (Some would even argue that glass has to be SiO containing, but I wouldn't go quite that far.) I wouldn't describe a solid as vitreous unless its appearance was "shiny" (and it was brittle), but that may be a personal preference rather than being representative of my technical background. My preference is that amorphous materials be called amorphous. If something can exist in both the liquid and the amorphous solid state, then I'd be comfortable with calling the solid state a glass. For polymers, the question is quite a bit trickier. Many polymers are semi-crystalline. And some (quite a few, actually) amorphous polymers do not exhibit a true liquid state (for example, due to physical cross-links). So, I'd be very cautious about the use of the term glass for them, although in certain contexts its appropriate. I'd avoid the use of "vitreous solid" completely, unless speaking about something after having vitrified.


Structures present in liquid water I know this may be a too generic question, but here it goes anyway.

It is already know that solid water (ice) organizes its molecules in a lattice (in a variety of 16 different crystals and one amorphous phase). See, for instance, http://en.wikipedia.org/wiki/Ice .

Now, what about liquid water? Is it the case that there are a (limited) number of local structures, like stable clusters or lattices configurations, combining and dissociating over relatively short periods of time, that dominate the liquid phase, loosely resembling the organization of a solid phase (the principal difference being that the hydrogen bonds would be weaker and would break more easily because of thermal agitation)? If so, what are those frequent structures?

Yes, there are various water clusters leading to local order on the scale of a few molecules holding stabilized orientation with one another for measurable timescales.  This article(1) is a pretty good introduction to the major varieties in liquid water.  This article(2) talks more about various structures available in any phase, including the ices, which gives some pretty good context.

(1) Keutsch, Frank N., Saykally, Richard J., Water clusters: Untangling the mysteries of the liquid, one molecule at a time, 2001, PNAS, 98, 19, 10533–10540, doi:10.1073/pnas.191266498

(2) Ludwig, Ralf, Water: From Clusters to the Bulk, 2001, Angewandte Chemie International Edition, 40, 10, 1808-1827, doi:10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1

There is no lattice or psuedo-lattice structure associated with water while it is the liquid state. Look at it as a medley of H2O molecules combining, breaking apart, and recombining with Hydrogen bonds as the main mediator of attraction.

Polar molecules (like the above water molecules) have a weak, partial negative charge at one region of the molecule (the O atm) and a partial positive charge elsewhere (The H atm) These bonds that form between water molecules account for some of the unique characteristics of water.


Is peroxoborate and perborate the same thing? Recently I came across the following question:-

  Preparation of Sodium Peroxoborate- Equations

The term peroxoborate came from the fact that it has the peroxo bond $\ce{-O-O}-$. So, it should be strictly called peroxoborate. But some books says  peroxoborate while other books say perborate. Is it true?

I googled sodium perborate.  Wikipedia recognizes the compound as sodium perborate. Then, I googled sodium peroxoborate, the following message appear:-

So, is peroxoborate and perborate the same thing? If so, why some book says  peroxoborate while othe books perborate? It should be strictly called peroxoborate since it contains the peroxo bond $\ce{-O-O}-$.

Update

This is what written in Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

  Sodium perborate is used as a bleach  in laundry powder, automatic dishwasher powders, and whitening toothpastes. Although the formula is given as $\ce{NaBO3.H2O}$ or $\ce{NaBO3.4H2O}$, the compound contains th peroxide anion, $\ce{O2^{2-}}$ and is more accurately given by $\ce{Na2[B2(O2)2(OH)4].6H2O}$.

But this is what written in Lee, J.D. Inorganic chemistry

  $$\ce{2NaBO2 + 2H2O2 + 6H2O -> Na2[B2(O2)2(OH)4].6H2O }$$
  Sodium peroxoborate is used as a brightener in washing powder. It is compatible with enzymes which are added to some biological powders.

So, what is the correct name:- sodium perborate or sodium peroxoborate?

This is necessarily  not  a chemistry answer but is a continuation of aventurin's answer.

Google ngram viewer is an online search engine h  that charts frequencies of any word or short sentence found in the sources printed between 1800 and 2012 in American English, British English etc.

So, here is the Google ngram for peroxoborate and perborate.

So, you can see that perborate is more often used than peroxoborate. Though, the correct name (IUPAC name) is tetrahydroxidodi-µ-peroxido-diborate, people often mention the compound sodium perborate. Sodium peroxoborate seems to be obsolete.

No, the names refer to different compounds. Attached below are their Lewis representations.

I should add that these structures were arrived at using Chemdraw's name to structure conversion. And it appears that this representation of perborate violates the octet rule. So this may be incorrect.


Weight of sodium acetate required to get to certain pH How many grams of $\ce{CH3COONa}$ we need to dissolve in $\pu{2.50 L}$ of water to get a solution with $\mathrm{pH} = 9.00$? Given $K_\mathrm{a}(\ce{CH3COOH}) = 1.75\cdot 10^{-5}$

Like solving a puzzle, I found from the asked $\mathrm{pH}$ the concentration of hydrogen cation and hydroxide anion, and wrote to myself that I need to find the final concentration of $\ce{CH3COONa}$ to seek from it the amount and then the weight.

But my problem was in all the middle steps. I understand the the concentration of hydroxide anions that I found is after the equilibrium, but I miss the point of how to use them to find the concentration on the acetic acid by given the $K_\mathrm{a}$. I'm guessing that I need to get the $K_\mathrm{b}$ from the $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$ equation. But why exactly and how then, I need to find the concentration of $\ce{CH3COONa}$?

Maybe I lack the knowledge and thus the imagination for it.

Support highly appreciated!

It's not really a matter of remembering a certain formula, this is not a conceptual thinking and should be avoided. When approaching these types of questions, always start with a balanced chemical equation. In this case we deal with a salt formed by a weak acid and a strong base, which hydrolyses according to the following net ionic equation:

$$\ce{CH3COO- + H2O <=>[$K_\mathrm{h}$] CH3COOH + OH-}$$

$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}\qquad\right|\cdot\frac{[\ce{H+}]}{[\ce{H+}]}\\
&= \frac{[\ce{CH3COOH}][\ce{OH-}][\ce{H+}]}{[\ce{CH3COO-}][\ce{H+}]}\\
&= \frac{K_\mathrm{w}}{K_\mathrm{a}}\label{eq:1}\tag{1}
\end{align}
$$

At the same time disregarding autoprotolysis of water, and taking into account that $[\ce{CH3COOH}]\approx [\ce{OH-}]$ and $[\ce{CH3COO-}]\approx c$ ($c$ is the concentration of sodium acetate), hydrolysis constant is

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{c}\label{eq:2}\tag{2}$$

Knowing that

$$[\ce{OH-}] = \frac{K_\mathrm{w}}{[\ce{H+}]} = \frac{K_\mathrm{w}}{10^{-\mathrm{pH}}}\label{eq:3}\tag{3}$$

one can equate \eqref{eq:1} and \eqref{eq:2}, substitute $[\ce{OH-}]$ with \eqref{eq:3}, and express concentration as

$$
\require{cancel}
c = \frac{K_\mathrm{a}[\ce{OH-}]^2}{K_\mathrm{w}} = \frac{K_\mathrm{a}K_\mathrm{w}^{\cancel{2}1}}{10^{-2\mathrm{pH}}\cancel{K_\mathrm{w}}} = \frac{K_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}}\tag{4}
$$

Finally, mass $m$ of sodium acetate can be found from its molar mass $M$ and volume of the solution $V$:

$$m = cVM = \frac{VMK_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}} = \frac{\pu{2.50 L}\cdot\pu{82.03 g mol-1}\cdot 1.75\cdot 10^{-5} \cdot 10^{-14}}{\pu{10^{-2\cdot 9} mol-1 L}} = \pu{35.89 g}\tag{5}$$

pH = 0.5(pKa + logC)
pKa is log of Ka

C is the concentration

pH is Given 9.

Hence,

9= 0.5[(log1.75) - 5 + logC]

18= 0.2430 - 5 + logC

22.75 = LogC

This equation will give you Molarity.

Use molecular mass of Ch3COONa to convert it to grams

I may be wrong but I think I've used the right formula


How was the diatomic nature of many common gaseous elements originally determined? How did scientists find out that  $\ce{Cl2, H2, O2}$ atoms have a two-atomic molecular structure ?

Here I'll only provide the reasoning for $\ce{O2}$ and $\ce{H2}$ diatomic structure, I think similar reasoning was to deduce that of $\ce{Cl2}$.
The divalency of oxygen, shown by Avogadro's hypothesis.
It is known that two volumes of hydrogen gas and one volume of oxygen gas combine to form water. From Avogadro's hypothesis (equal volumes of gas contain equal numbers of molecules) it can be concluded that water must have two atoms of hydrogen and one atom of oxygen (or conceivably some multiple). Hence, oxygen is divalent (valence of 2) while hydrogen has a valence of 1:
$$\ce{2H + O \to H2O}$$
This would indicate that $\frac12$ gram of hydrogen atoms reacted with 8 grams of oxygen atoms. Since the
atomic weight of hydrogen was defined as $H = 1$, then the respective atomic weights of hydrogen and oxygen would be $H = 1$ and $O = 16$ (the values accepted today).
The diatomic nature of hydrogen and oxygen, as shown by Avogadro's hypothesis.
It is observed that when two volumes of hydrogen gas and one volume of oxygen gas combine, two volumes of water gas are formed. If hydrogen and oxygen were monatomic (as shown in the equation above), then only one volume of water would be formed. The only explanation is that hydrogen and oxygen are diatomic:
$$\ce{2H2 + O2 \to 2H2O}.$$
(Copied almost verbatim from this page)

Hydrogen, oxygen, chlorine, etc, prefer to fill their valence electron shell by sharing electrons. Energy is released in the sharing a pair of electrons then individual atoms.

Sodium, potassium, etc, would rather give up an electron (to have a full valance electron shell), then to share electrons in an attempt to fill up the electron shell. In the case of bonding, a little energy is used to release an electron and a lot of energy is released when another atom pulls it into its valance shell.

Since the metal atom would use energy to take an extra electron, and the release uses energy, metal atoms will not take an extra electron.


Calculating reaction enthalpy given other reactions $$\begin{alignat}{2}\ce{X(s) + 1/2O2(g) &-> XO}&&\qquad\Delta H=-895.5\ \mathrm{kJ} \\
\ce{XCO3(s) &-> XO(s) + CO2(g)}&&\qquad\Delta H= +484.3\ \mathrm{kJ}\\
\ce{X(s) + 1/2O2(g) + CO2 &-> XCO3(s)}&&\qquad\Delta H = \;?\ \mathrm{kJ}
\end{alignat}$$

I wish to know how to do it. I have seen some questions where there's 3+ equations and at the end a final equation that I would need to determine the $\Delta H$ for the final equation. I have looked online on how to solve this but the tutorials are very confusing and I can't understand them. Can someone explain how to do this and other similar problems like this?

One of the tutorials talks about making sure they cancel out and flipping the equations? Can someone please explain this?

As enthalpy is a state function, so regardless the multiple stages of a reaction, the total enthalpy change for the reaction is the sum of all changes.This is Hess's law.
So, you can see clearly that the third reaction results from adding  the first reaction to the reverse  of the second reaction, and so is the enthalpy change.

Ok so for future reference if anyone needs help with this you look at the last equation and then at the other ones, if the compounds(ex: CO2) of the other equations don't appear to be on the same side of the final equation you flip that equation and as a result the sign. After you have done this add everything up and it will give you the answer


How to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture? A quantity of benzene, $\ce{C6H6}$, and toluene, $\ce{C6H5CH3}$, is placed in a $\pu{1 L}$ evacuated vessel at $\pu{25 ^\circ C}$. At equilibrium, a small volume of liquid is visible at the bottom of the container. A sample of the vapour phase is analysed and found to contain $\pu{53 mol-\%}$ benzene.
  What is the mole fraction of benzene in the liquid phase?

  Vapour pressures at $\pu{25 ^\circ C}$: $P_\mathrm{vap}(\text{benzene})= \pu{0.125 atm}$, $P_\mathrm{vap}(\text{toluene}) = \pu{0.037 atm}$

First of all, I am not even certain if this a Raoult's law problem. I just assumed so, since Henry's constant was not given in the problem.

First I'm getting the total liquid pressure using Raoult's law.
\begin{align}
P(\text{benzene}) &= X(\text{benzene}) \cdot P_\mathrm{vap}(\text{benzene}) &
                  &= 0.53 \times 0.125 &&= 0.06625~\mathrm{atm}\\
P(\text{toluene}) &= X(\text{toluene}) \cdot P_\mathrm{vap}(\text{toluene}) &
                  &= 0.47 \times 0.037 &&= 0.01739~\mathrm{atm}\\
P_\mathrm{total}  &&&= 0.06625 + 0.01739 &&= 0.08364~\mathrm{atm}\\
\end{align}

Then using Dalton's law, I am getting the mole fraction of Benzene in the liquid phase.
\begin{align}
P(\text{benzene}) &= Y(\text{benzene}) \cdot P_\mathrm{total}\\
Y(\text{benzene}) &= \frac{P(\text{benzene})}{P_\mathrm{total}}\\
                  &= \frac{0.06625}{0.08364} = 0.792.
\end{align}

However, my answer is incorrect according to the answer key that has $0.25$ as the answer. I have a feeling my whole approach may be wrong since my answer is off by a lot. What am I doing wrong?

The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.

It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)

First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):

$$p^*_\text{benzene} = 0.125\ \mathrm{atm}  \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm} $$

and we require $x_\text{benzene}$

We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$

Finally,

$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$

It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is:

Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.

According to Raoults law, it should be 0,25.
Lets say, you have the molar fraction of each component in vapor, and lets say you have the saturated pressure at some temperature, than according to Raoults law you can say:
0,53(molar fraction of benzene in vapor)Pi(which is equal to Xbpb(saturated, benzene)+Xtpt(saturated, toluene) = Xbpb
than:
Xb*0,125=0,53*(Xb*0,125+(1-Xb)*0,037)
Xb=0,25


Can heteroatoms with lone pairs be chiral centres? If a compound has a carbon atom with four different groups covalently bonded to it, it is called asymmetric and enantiomers of the compound can exist.

But imagine if one has a different central atom, such as a nitrogen or a sulfur where one of the four "groups" is an electron pair (see examples below).

Would such a thing still behave like a "normal" asymmetric carbon? If not, how else does it behave, i.e. are there enantiomers? Is my assumption that those will still have $\mathrm{sp^3}$ hybrid orbitals wrong? I would also enjoy pointers to literature, I was unable to find any.

Generally, amine nitrogens will not behave like a normal asymmetric carbon. Simple amines are roughly $\mathrm{sp^3}$ hybridiized and the molecules you use as examples do have 4 (we include the lone pair of electrons as a substituent) different substituents around the central nitrogen atom. So in principle me might consider it asymmetric or chiral.  But simple amines can undergo a process called nitrogen inversion, which essentially coverts one enantiomer into the other.

(image source)

However, if one can find a way to slow down or eliminate the nitrogen inversion process, then chiral amines can be isolated. One way to achieve this is to incorporate the amine nitrogen into a 3-membered ring (an aziridine). Now to achieve the planar state necessary for nitrogen inversion requires the bond angle in the 3-membered ring to open from 60° to 120°, an impossible task. Aziridines containing a chiral nitrogen atom have been isolated and characterized.

In the case of N, the compound is chiral but not isolable as it inverts. (compare quinuclidine). The diastereomors are not resolvable.

In the case of sulfoxides, these are isolable and chiral.

In the case of Carbon.Carbanion intermediates are nonresolvable but neutral ground state carbon does not invert spontaneously.


hydrated halide structure What is the structure of $\ce{LiCl.2H2O}$?
I think when $\ce{LiCl}$ is put in aqueous medium, it will break into $\ce{Li+}$ and $\ce{Cl-}$. Then $\ce{H2O}$ will give LP to Li and hydrate it. But why 2 $\ce{H2O}$ molecules? 2 means 2 coordinate bonds? Octet is not filled? Then why does this happen?

The radius ratio of $\ce{Li+}$ and $\ce{Cl-}$ is very small, this means the size of $\ce{Li+}$ is so small than $\ce{Cl-}$ that the crystal lattice of $\ce{LiCl}$ becomes unstable. So to make the size of $\ce{Li+}$ bigger $\ce{H2O}$ molecules surround the caption bigger and make the structure stable.


Which of the following possesses the highest energy electron? Which of the following possesses the highest energy electron?

  $\ce{Br-}$
  $\ce{Ca^2+}$
  $\ce{Cr+}$
  $\ce{As}$

Don't bromine and arsenic have equally high energy electrons (i.e. in the $\ce{4p}$ subshell)? My book says that arsenic is the correct answer. But I don't see why $\ce{Br^{–}}$ is wrong.

My best guess is that your book wants you to argue that, since the bromine nucleus has a higher charge than arsenic, electrons are more strongly attracted to it, lowering their energy. And indeed this would be true - if we only considered interactions between the electrons and the nucleus, and ignored electron-electron interactions. Unfortunately, however, the situation in real life is more complicated than this, because electrons repel one another, “shielding” each other from being attracted to the nucleus (which, of course, is the reason why subshells have different energies in the first place). And the bromide ion has three more electrons than the arsenic atom, which actually turns out to be quite significant.

In fact Webelements gives the electron affinity of $\ce{Br}$ (i.e., the first ionisation energy of $\ce{Br^-}$) as 324.8 kJ mol$^{-1}$ while the first ionisation energy of As is 947.0 kJ mol$^{-1}$. So in fact it seems that, experimentally, the bromide ion has the highest energy electron (which therefore requires the lowest energy to remove it).

This is a more difficult question than it looks - the problem of how to deal with interactions between lots of electrons is still an active research field! A fairer question to do in your head would be to compare, say, $\ce{Br^{2+}}$ with As: in this case the number of electrons of each species is the same, so that the argument I put forward at the beginning of this response is actually correct here.

$\ce{Br-}$ = $\ce{[Kr]}$. $\ce{Kr}$ is a noble gas and has a full  octet, meaning that the energy is SUPER low because of how stable it is.


Molecular formula of a hydrocarbon A is a non cyclic hydrocarbon chain with $\ce{C_{3x-1}H_{6x}}$ as a molecular formula which is mainly used in cars as fuel, its cracking leads to hydrocarbon (B) which belongs to the same family of hydrocarbon A, also two molecules of ethene form the second product in the above reaction.

My effort:
To determine if it is an alkane, we compare the given subscripts on $\ce{C}$ and $\ce{H}$ ($3x-1$ and $6x$, respectively) to those in the general formula for an alkane, $\ce{C_{n}H_{2n+2}}$. Let $n=3x-1$ where $n$ is the number of carbon atoms.

$$6x\buildrel{?}\over{=}2(3x-1)+2=6x$$

and therefore A is an alkane, with the molecular formula satisfying the form $\ce{C_{n}H_{2n+2}}$.

You're right in thinking that it's an alkane. Now I think I shouldn't blame you for getting confused by

  $\ldots$ as a molecular formula which is mainly used in cars as fuel $\ldots$

Because

First off, we don't know if $x$ in the molecular formula is an integer or a decimal number.
Second, we don't know what fuel is meant. It's just supposed to be a fuel used in cars. And what are the fuels used in motor vehicles again?

  Currently, the majority of motor vehicles worldwide are powered by gasoline or diesel. Other energy sources include ethanol, biodiesel, propane, compressed natural gas (CNG), electric batteries charged from an external source, and hydrogen.

What are these?

  The bulk of a typical gasoline consists of $\mathrm{\color{red}{hydrocarbons~with~between~4~and~12~carbon~atoms~per~molecule}}$ (commonly referred to as C4-C12).

  Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including n, iso, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\color{red}{\ce{C12H23}}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.

Ethanol is an $\color{red}{alcohol}$.

  Biodiesel refers to a vegetable oil - or animal fat-based diesel fuel $\mathrm{\color{red}{consisting~of~long-chain~alkyl~(methyl,~ethyl,~ or~ propyl)~ esters}}$.

Propane is an $\color{navy}{alkane}$.

  CNG is made by compressing natural gas (which is mainly composed of $\color{red}{methane}$, $\ce{CH4}$), to less than 1 percent of the volume it occupies at standard atmospheric pressure.

Electric batteries and hydrogen are $\ldots$ a bit irrelevant. Let's not make this answer longer.

Thus, the answer would be propane, since it's the only alkane among the common car fuels. But is it? Of course not; since propane cracking isn't in high school syllabuses anywhere in the world. There is a very very good chance the querent had gasoline mind.
Now that we trust our natural instincts (!) that they meant gasoline when they were referring to the fuel, we should remain perplexed and run in circles. Because gasoline isn't consisted of one or two distinctive and well-known substances, and it's a mixture of  It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes), where the usage of the terms paraffin and olefin is particular to the oil industry. To make matters worse, gasoline composition is obviously not the same around the world.

$\hspace{25ex}$
$\hspace{25ex}~\small{\mathrm{Some~of~the~compounds~commonly~found~in~gasoline}}$
$$\small{source}$$

So, let me rephrase your question into something actually meaningful:

  $\ldots$ as a molecular formula which is mainly known to be a really smooth-burning species in gasoline $\ldots$

Now we're talking. The best known gasoline would be one that consists of 100% isooctane, and that's why we have octane rating. It's not a real fuel, just the material in an imaginarily ideal mixture that would burn "real smooth". That's why your question is problematic and scientifically incorrectly asked.

Back to the question, 2,2,4-Trimethylpentane must have been the compound the querent had in mind. Now we have the answer for A.

$\hspace{25ex}$
$\hspace{28ex}$ Isooctane (PIN: 2,2,4-Trimethylpentane)
$\hspace{43ex}$ $\small{source}$

But what would the products of the cracking be?

According to the Wikipedia article on cracking the cracking process in entropy-driven and the radicals formed in certain conditions usually move the reaction forward. But the final products (and even which radicals form) highly depend on the methods of cracking (thermal, steam etc.) and/or the catalysts. There are numerous catalysts which demonstrate the diversity in the mechanism of cracking.

For example, take a look at this paper (It's just an abstract, but it gives you an idea)

  $\ldots$ The main reaction route on FAU is hydride transfer followed by β-scission leading to mainly C4 species, while on MFI protolytic scission is responsible for the formation of high amounts of C1−C3 species. $\ldots$

So, we're left with speculations. This isn't a real question. The only answer the examiners want is B's most probably butane, as it must be an alkane.

As this also demonstrates, a very simplistic view is that an octane can break into two ethene molecules and one butane molecule. So, the reaction they must have had in mind is
$$\ce{C8H18 -> C4H10 + 2C2H4}$$
Pfft.

TL;DR:

This isn't a good or correct question to ask at all; at the very least, it's very poorly worded. The answers are supposedly isooctane for A and butane for B, but the OP's confusion was innocent.

You have done the first step in the solution.

So we know now that A is an alkane and after cracking it leads to a molecule B and $\ce{2C2H4}$. But $\ce{2C2H4}$ contains $4$ carbon atoms so the number of carbon atoms in A must be $5$ or more (i.e. $3x-1\geq5 \Longrightarrow x\geq3$)

let us start trying:

$x=3$ will lead to $\ce{C8H18}$
$x=4$ will lead to $\ce{C11H24}$
.
.
.
.

And I suppose that you as a student only studied alkanes till octane ($\ce{C10H22}$)
so the answer will be $\ce{C8H18}$.

I know that is not a good reasoning but I am sure that it is the true answer.


Why does Ethylene Glycol have higher boiling point than Propylene Glycol? Looking at the two alcohols, both have hydrogen bonding and are singly bonded throughout the molecule. However, propylene glycol has a larger electron cloud which to my knowledge would increase its London dispersion forces and thus its boiling point. I already know both molecules exhibit hydrogen bonding since they're both alcohols.
However, it seems that ethylene glycol has a higher boiling point despite propylene glycol seemingly having stronger IMFs. I believe this has something to do with molecular shape but I can't seem to find the definitive answer explaining why ethylene glycol has a higher boiling point.
I've used https://pubchem.ncbi.nlm.nih.gov/compound/Propylene-glycol#section=RTECS-Number and https://pubchem.ncbi.nlm.nih.gov/compound/1_2-Ethanediol#section=Odor for comparison of the two alcohols.

This is all about linearity of the molecule and steric hindrance. Consider the linear structure of ethylene glycol versus the branched structure of propylene glycol. EG can "stack" more easily and therefore engage in more intermolecular bonding than PG. More attraction means more energy required to vaporize. On the other hand, 1,3 propane diol, which is the linear isomer of propylene glycol, has a higher boiling boiling point than EG at around 213C. To further illustrate - 1,4 butanediol (linear) has a higher boiling point than 1,3 butanediol (branched).


What determines how much water of crystallisation a salt has? Is there any reason behind the number of water of crystallisation?

For example, $\ce{LiCl.2H2O}$ has two waters of crystallisation, and $\ce{MgCl2.8H2O}$ has 8 waters of crystallisation. Why does $\ce{LiCl}$ have only 2 waters of crystallisation and not 8?

The water of crystallization in a crystal of any ionic salt can be associated with the size of the cation, magnitude of charge on it, tendency for co-ordination and hydrogen bonds, its lattice structure and other factors involving the anion. But we cant actually theoretically determine why a particular salt has these man water of crystallization. We determine this experimentally in the labs.Then after finding this data for a set of salts we try and link them with the above mentioned factors and try to give a suitable explanation which justifies real data to the greatest possible extent. A theoretical explanation would be very complex and definitely involve way more parameters than what I mentioned above to be accurate for max cases.Even there isn't any 100% full-proof theory or mathematical formula that would allow us to calculate the water of crystallization in any random salt.

In the solution then crystal, in the first example Li becomes Li+1 cation with one bonding positions, and Cl become Cl-1 ion, both having only one position so they can bond any way needed in 3D.  However, a direct bond between them will not create a crystal, as a crystal needs an continuous chain (one must be open to the next set) to continue.  You get Li-Cl single bond, and not crystallization.

Yet with H2O, the Cl-1 will bond width one H2O (hydrogen bond), and the another Hydrogen (in H2O) is available for crystal chain bonding.  That is why H2O are needed to create a crystal.  You get a +H-O-Cl.

The Li+ would bond to an outer electrons of the H2O, and again end the chain.  In that way, we only have the H2O Hydrogen bonds, and get a chain +H-O-H+ with Li attached in the other hemisphere.

With regular H2O in crystal as +H-0-H+ with two bonds. That means that you get a crystals with chains at the corners of the cube (8 positions).

The H2O in 3D has two Contributing Hydrogen bonds (proton outward), and six receiving outer electrons.  That leads to your conclusion that it would be 8H2O.

Yet, the Cl- ends the chain, Li+ does not.  That means you get a chainable structure with just 2 H2O where two Hydrogen are ready for other Hydrogen bonds:

                    H+
          +H-O-H+   O-H+Cl-
             Li+

There is two open H+ for hydrogen bonding along the crystal chain in a clean 2H2O:LiCl structure.  That makes this crystal possible with that low ratio (if build in that environment) as achievable.

However, Mg becomes Mg+2 with those bonding positions at endcaps.  That means the Mg+2 always creates bonds to two H2O atoms at the H+ positions.  The two contributing positions do not line up relative to the H2O H+ positions (or calc very well as attractive) as a double bond.

But, you cannot create a crystal without an H+ in H2O remaining open, so you get two H2O for every Mg, which then leads to 4 more H2O by hydrogen bonds.

                +H      H+
                  O+Mg+O
                +H      H+

The positions for Cl- still end the chain, so those can insert that pair anywhere; you can have a crystal ending one H+ in the Cl-1, and still build in the other three directions.

                           H+
                +H      H+ O-H+ Cl-
                  O+Mg+O
                +H      H+ O-H+ Cl-
                           H+

However, the +H bond to other H2O, so you get 2 direction from the Mg x 4 hydrogen H2O bonds = 8 H2O required for the second crystal.  It cannot work with just 2H2O in the 3D structure.

That is why the MgCl2 requires a ratio of at least 8 H2O, so MgCl2:8H2O, not :2H2O.

The main quality of water for creating crystals is its creation of Hydrogen bonds at two diagonal corners of the Scrunched (104.5 bonding from ~66 nucleomagnetics inclination angle) Cube in the same hemisphere.  In AVSC, the Hydrogen electron fills the Oxygen settling position.  That leaves the Hydrogen as settling Proton outward, which can create the location for Hydrogen bonds at specific angles.


Find the pH of a sodium chloride solution using the extended Debye-Huckel equation Question
  Use activities to calculate the pH of each of the following solutions, being sure to use $\mathrm{\alpha}$ values and the extended Debye-Huckel equation.

  The first one the concentration is $0.020 \mathrm{M}$ $\ce{NaCl}$
  with the answer being : $\mathrm{pH}$ = $6.99$

  The second one the concentration is $0.10 \mathrm{M}$ $\ce{NaCl}$
  with the answer being : $\mathrm{pH}$ = $6.98$

The thing is i have no idea how to find the activity coefficient of $[\ce{H}^+]$

I know ionic strength $u$ is calculated as
$\mathrm{\frac{1}{2}\left(\sum C_iZ_i^2\right)}$

but how does $[\ce{H+}]$ work into the charge balance?

I also know that the activity coefficient is calculated as

$$\log\gamma=\frac{-0.51(z^2)(\sqrt{u})}{1+(\alpha(\frac{\sqrt{u}}{305})}$$

The best I have is:
$$\ce{Na+ + Cl- + H+ + OH-}$$

and that raises the question of how do I know what $[\ce{H+}]$ is and $[\ce{OH-}]$ for calculating the ionic strength? Am I approaching this wrong? Where do I start?

Let's first look at ionic strength:

Ionic strength is a measure of the total ion concentration in solution, however, species with higher charge can also have elctrostatic interactions with other species.

$$\mu = \frac{1}{2} \sum_i C_iZ_i^2 $$

What is the strength of 0.02 M NaCl solution?

$$\mu = \frac{1}{2}([\ce{Na+}]z^2 + [\ce{Cl-}]z^2) = \frac{1}{2}([0.02](1)^2 + [0.02](-1)^2 ) = 0.02 M$$

With the ionic strength, we can now calculate the activity coefficients ($\gamma_i$) of ions in solution. The activity is merely the product of the concentration of each species and the activity coefficient.
$$\textrm{activity} = a_i = \gamma_i[i]$$

Let's look at a typical equilibrium expression:

$$\ce{A + B <=> C + D} $$

$$K_{eq} = \frac{[C]\gamma_c[D]\gamma_d}{[A]\gamma_a[B]\gamma_b}$$

The Extended Debye-Huckel equation is used to find the effect of ionic strength on the activity coefficent. It is given as above:

$$\log_{\gamma \ce{H+}}=\frac{-.51(z_i^2)(\sqrt{\mu})}{1+(\alpha(\frac{\sqrt{\mu}}{305})}$$

where $\alpha$ is the size of the hydrated ion in pm.

The Extended Debye-Huckel equation is valid from $\mu = 0 \ \text{to} \ 0.1 \ \text{M}.$

To find pH, you must first find $\mu$ of the hydrogen ion. You can then use the Extended Debye-Huckel equation to find the $log_{\gamma \ce{H+}}$. Take the antilog.

$$a_{\ce{H+}} = \gamma_{\ce{H+}}[\ce{H+}] $$

Take the negative log and you will get pH.

$$ pH = -\log( a_{\ce{H+}} ) = -\log( \gamma_{\ce{H+}}[\ce{H+}] ) $$

In context with the problem:

$$\log_{\gamma_{\ce{H+}}} = \frac{-.51(1^2)(\sqrt{0.02})}{1+(900 (\frac{\sqrt{0.02}}{305})}$$

$$\log_{\gamma_{\ce{H+}}} = -0.0508886 $$

$$\gamma_{\ce{H+}} = 1.1243166 $$

$$\textrm{pH} = -\log a_{\ce{H+}} = -\log (\gamma_{\ce{H+}} \cdot 1.0 \cdot 10^{-7}) = 6.95 $$

Your error is when you take the antilog of the $\log_{\gamma \ce{H+}}$. You raised it to the $10^{\log_{\gamma \ce{H+}}}$ which is $0.89429197$. It is actually supposed to be $10^{-(-0.0508886)}$ which comes out to be $1.124316$.

For the 0.10 M of $\ce{NaCl}$ I calculated the pH to be $6.92$.


Is there safe chemicals for making voice pitch or bass? I'm looking for safe experiment that can pitch and bass my voice.finally I find a video that do it but he said "not try in home,ok..?".Is there safe tool or chemicals for Manipulation of the larynx is used to generate a source sound with a particular fundamental frequency.

other my question is about below viedo link:How these chemicals make voice pitch or bass!(Helium and Sulfur Hexafluoride)

Pitch is a perceptual property of sounds that allows their ordering on a frequency-related scale,1 or more commonly, pitch is the quality that makes it possible to judge sounds as "higher" and "lower" in the sense associated with musical melodies.

video link

Assuming that helium and sulfur hexafluoride (as shown in the video) will give you the correct sounding "pitch" and "base", it would be fine if done correctly and safely.

Helium is incredibly inert and so will not cause any sort of toxicity. It is also a very light gas, so it will exit your lungs quickly. Just do not breath only helium, or for a long time or you could suffocate. Make sure to get lots of normal air.

Sulfur Hexafluoride is also rather unreactive, but more dangerous. This is because it is heavy and will tend to stay in your lungs. Only use it for a short period of time, or you can asphyxiate. While the same goes for helium, it tends to exit your lungs more easily.

The actual mechanism that causes them to work is based of how dense they are. It does not change the pitch, however. Instead, it changes something called timbre. This is basically the fact the helium allows sound to travel much faster through it, so it is a similar affect to playing a video at a faster speed.

Sulfur hexafluoride does the exact same thing, except it is much denser and does not allow sound to travel through it very quickly.

If you want to do this, be careful! You can suffocate, especially with sulfur hexafluoride.

A  possible safe way to inhale helium is to mix it with 20% oxygen.
Likewise, mix sulfur hexafluoride with 20% oxygen.


Cationic vacancies created when non-stoichiometric iron oxide is doped with silicon(IV) ions Problem

It is believed that non-stoichiometric compound $\ce{F_{0.93}O}$ forms by doping of $\ce{Fe^3+}$ ions in $\ce{FeO}$ crystal by replacement of $\ce{Fe^2+}$. Calculate the number of cationic vacancies if all of the $\ce{Fe^2+}$ ions are replaced by $\ce{Si^4+}$ ions in $\pu{0.1 mol}$ of $\ce{Fe_{0.93}O}$. Express your answer as a multiple of Avogadro's Number.

Answer stated by the book

 0.0465

My Solution
As the compound is electrically neutral I equated the net charges due to the cations and anions like this, assuming $x$ to be the fraction of $\ce{Fe^2+}$ in the compound.
$$[2x + 3(0.93-x)] - 2 = 0$$
This gave $x = 0.79$. Now, in $\pu{0.1 mol}$ of the compound there would be $\pu{0.079 mol}$ of $\ce{Fe^2+}$. As each silicon ion would replace two ions of $\ce{Fe^2+}$, one vacancy would be created for every replacement.
The number of replacements would be $\left(\frac{0.079}{2}\right)\times N_\mathrm{A} = 0.0395 N_\mathrm{A}$, and we would have as many vacancies. ($N_\mathrm{A}$ is Avogadro's Number)
Can anyone point out the error in my solution?

Your approach is correct, however, you forgot to account for the vacancies created when Fe3+ replaces Fe2+. First, calculate the cationic vacancies formed due to the replacement of Fe2+ by Fe3+.
Assuming  to be the fraction of Fe2+ in the compound, the fraction of Fe3+ in the compound is (0.93 - x).The total charge on x Fe2+ and (0.93 – x) Fe3+ should be equal to the charge due to O2- ions.
Hence, 2x + 3(0.93 – x) = 2
=> 2x + 2.79 – 3x = 2.00 or x = 0.79. So, fraction of Fe3+ ions in the compound is 0.14
Now, 2 Fe3+ ions replace 3 Fe2+ ions, so there's one vacancy created for every 2 Fe3+ ions. That means for 0.1 mole of F0.93O, the number of vacancies will be 0.007A due to Fe3+ replacing Fe2+ in FeO. This is the part you missed.
Now, as you mentioned, each Si4+ ion will replace two Fe2+ ions in the compound, so again, one vacancy per replacement. And, as you already solved, The number of replacements would be (0.079/2)A = 0.0395A. So, the total number of vacancies are 0.0395NA + 0.007NA = 0.0465NA.
Hope this helps!


Orthogonal Wavefunctions My current understanding of orthogonal wavefunctions is: two wavefunctions that  are perpendicular to each other and must satisfy the following equation: $$\int\psi_1\, \psi_2\, \mathrm{d}\tau =0$$

From this, it implies that orthogonality is a relationship between 2 wavefunctions and a single wavefunction itself can not be labelled as 'orthogonal'. They must be orthogonal with respect to some other wavefunction. However I have seen some textbooks refer to single wavefunctions as being orthogonal.

Am I missing some crucial thing in my definition of orthogonal wavefunctions? Could anyone provide me with a more complete definition of orthogonal wavefunctions from a chemistry point of view?

The notion of orthogonality in the context of the question referrers to the very well-known general concept of linear algebra, the branch of mathematics that studies vector spaces. Instead of going deep into the mathematics (that requires at least 50 textbook pages) let's just clear some OP's doubts.

First, a small (but important) correction: two wave functions $\psi_1$ and $\psi_2$ are called orthogonal to each other if
$$
  \int \overline{\psi}_1 \psi_2 \, \mathrm{d} \tau = 0 \, ,
$$
where the first function is complex-conjugated as indicated by a bar on top of it. Wave functions are complex-valued functions and complex-conjugation of the first argument is important.1

So, yes, orthogonality is a not a property of a single wave function. It either refers to a pair of them being orthogonal to each other as described above, or, in general, to a set of them, being all mutually orthogonal to each other, i.e. to a set $\{ \psi_i \}_{i=1}^{n}$ such that for any $i \neq j$
$$
  \int \overline{\psi}_i \psi_j \, \mathrm{d} \tau = 0 \, .
$$
In the last case it is said that the whole set $\{ \psi_i \}_{i=1}^{n}$ is orthogonal.

1) More precisely, one of the two functions has to be complex-conjugated in this expression, where which one is the matter of convention: in physical literature it is often the first function, while in mathematically oriented literature it is usually the second one.

The same wave functions can not be said to be orthogonal. the comparison must be between two different function.


Conversion of Hessian from internal coordinates to cartesian coordinates I have performed a CASPT2 frequency calculation using the MOLCAS8 program package. I need the full Cartesian Hessian matrix as input for a subsequent program package.

MOLCAS8 performs CASPT2 frequency calculations numerically. The program MCKINLEY is called for this purpose, however, it is now my opinion that all numerical second derivative calculations in MOLCAS are handled not by MCKINLEY but by the barely documented module supermac. Anyway, the Hessian is fully printed in internal coordinates once the calculation is finished. For a system of 10 atoms the internal Hessian is of size 24 x 24, which suggests non-redundant internal coordinates (MOLCAS calls them "primitive internal coordinates").

Does anyone know of a software capable of transforming a Hessian matrix from internal coordinates to Cartesian coordinates? I am aware that this will probably be complicated as internal coordinates are not as unambiguous and straight-forward as Cartesians. However, my initial search efforts did not yield a single software package capable of transforming a calculated Hessian in internal coordinates to Cartesians. Any help would be appreciated!

Alternatively, does someone know a quantum-chemical program package using which I might be able to obtain a full Cartesian hessian at CASPT2 level?

I don't know if this helps, but MOLCAS also seems to print out the force constants of each atom in Cartesian coordinates. Furthermore a .molden file is produced where the normal modes are included. The full Hessian does not seem to be written to this Molden file though.

BAGEL (http://nubakery.org/ and https://github.com/nubakery/bagel) has CASPT2 analytical gradients and can run frequency computations.

It's currently being prepared for 1.0 release, so it might still have bugs. On the other hand, it is freely available under the GPL.

As for converting the Hessian, it might be useful to look at the INTDER program, which is (among others?) part of the PSI3 suite of programs available here: https://sourceforge.net/projects/psicode/files/psi/3.4.0/psi-3.4.0.tar.gz/download - PSI3 is under the GPL as well.

The manual page for INTDER is here: https://manned.org/intder and it says:

"Force field transformations between Cartesian and general internal
coordinate spaces up to fourth order, including nonstationary
reference structures.  Both forward and reverse transformations
are possible; hence, by using intermediate Cartesian coordinates,
force fields can be transferred among different representations in
the internal space."

Whether or not it will be easy to coerce the MOLCAS Hessian into a format INTDER likes and then transform it as you want I cannot say as I don't have any direct experience with INTDER - it is typically run as a step in a chain of commands automatically generated by the PSI3 driver routines.

Maybe openbabel can do it.
I found this website which can convert z-matrix's to Cartesian coordinates:
http://www.shodor.org/chemviz/zmatrices/babel.html.
I know that the MolMod python library can make the conversion if you want to delve into that. And another python library that can do it chemcoord
I hope this helps.


Is a hydrogen bond considered to be a van der Waals force? Is a hydrogen bond considered to be a Van der Waals force?

According to the IUPAC gold book a van der Waals force is:

  The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces.

Hydrogen bonding is a type of dipole-dipole interaction, so it would fit the definition of a van der Waals force.

The way I think of it is: van der Waals forces are anything that make a gas non-ideal, since that's how they were originally discovered and defined.

I'm a physics grad student. From my understanding, H-bond is not what people typically mean by van der waals force.
H-bond is just basic dipole-dipole interaction with essentially fixed dipole moments (the dipoles are simple result of difference of electronegativity within each molecule).
Van der waals force (like interactions between H2 molecules) typically mean induced-dipole - induced-dipole interaction. (The dipole of an isolated molecule is zero)
One major consequence of the difference is that dipole-dipole force scale with $r^{-4}$.
I'm not totally sure of the above...my understanding is Efield_dipole scale as $r^{-3}$. F=-d/dr U = d/dr ( DipoleMoment $\cdot$ Efield_dipole ) = $r^{-4}$
Whereas van der waals have higher order distance dependency because each dipole moment themselves are induced by electric field which depends on distance. So, van der waals force scale with $r^{-6}$


Physical Meaning of the Difference between MO and VB Wave Function of H2? What would be the physical meaning of the difference between the MO wave function and VB wave function of $\ce{H2}$?

I'll give a somewhat simplistic answer. VB assumes that all bonding is between only two atoms. MO bonding considers bonding as a molecular phenomenon.

In $\ce{H2}$ there isn't really much difference. However consider benzene, $\ce{C6}\ce{H6}$. VB theory would have three double carbon bonds and three single carbon bonds which isn't really true. So VB bonding then hybridizes over the six carbon atom ring to predict the aromatic character of the molecule. However in MO theory the aromatic character is fundamental to the model.

From wikipedia

  "In particular, valence bond theory correctly predicts the dissociation of homonuclear diatomic molecules into separate atoms, while simple molecular orbital theory predicts dissociation into a mixture of atoms and ions. For example the molecular orbital function for dihydrogen is an equal mixture of the covalent and ionic valence bond structures and so predicts incorrectly that the molecule would dissociate into an equal mixture of hydrogen atoms and hydrogen positive and negative ions."


How to quench unreacted sodium azide from reaction mixture? The reaction involved synthesis of azide formation from alkyl bromide in DMF using sodium azide. After the reaction has taken place, how do I quench the unreacted sodium azide from the reaction mixture?

Excess of sodium azide can be quenched using nitrous acid.

$$\ce{2NaNO2 + H2SO4 → 2HNO2 + Na2SO4}$$

$$\ce{2NaN3 + 2HNO2 → 3N2 + 2NO + 2NaOH}$$

For further reading, Click here.


Is the phenoxide ion aromatic? Is the phenoxide ion aromatic? I read that technically it isn't aromatic because it violates some arcane rule, but I can't find any references in the literature.

I don't see why it couldn't be aromatic, since it's

flat/planar
has a system of conjugated p-orbitals.
has 4n+2 pi electrons: 2 from the oxygen lone pair that is delocalized and n = 0.
is cyclic.

I agree absolutely with Max, and while I was still running the calculations, he already provided the answer.
The goldbook provides a nice definition for aromatic compounds:

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Therefore not all aromatic compounds have to strictly full-fill the Hückel (4n + 2) rule, but all that obey that rule are aromatic. Phenols certainly behave like benzene and are therefore aromatic, please note that aromaticity is an experimentally observed feature.

You can see the aromaticity from the $\pi$ type orbitals, that are in both cases quite similar and extensively delocalised. The calculations were done at the DF-BP86/def2-SVP level of theory.

Benzene exhibits a very high symmetry $D_\mathrm{6h}$, therefore the last two of the shown orbitals are degenerate.

The restriction is not present in the phenolate, as the symmetry is only $C_\mathrm{2v}$.

Also due to the missing symmetry constraint the neutral phenolic radical, $\ce{H5C6O.}$, and the phenolic cation, $\ce{[H5C6O]^{\oplus}}$ would be aromatic.

As phenoxide ion has 8 $\pi$ electrons due to presence of lone Pair on oxygen, it counts in $π$ electrons so it is antiaromatic but not aromatic


Why does temperature affect the dissolved oxygen concentration? I know that an increase in temperature decreases the solubility of oxygen in water, but I don't know why. Could anybody explain?

The solubility of a many gases in a variety of liquids is observed to decrease with increase in temperature, for example oxygen, nitrogen and methane in water.  This follows the qualitative predictions of Le Chatelier’s principle, which in thermodynamics is supplanted by the Clapeyron equation. As one of the two phases involved is a gas the Clausius–Clapeyron equation can be used.

This assumes that the molar volume of any gas/vapour is far greater than that of the solution. Assuming that all values correspond to one mole the Clausius–Clapeyron equation is

$$\frac{\mathrm d\ln(p)}{\mathrm dT}=\frac{\Delta H}{RT^2}$$
which in its integrated form between pressure and temperature $p_1, T_1$ to $p_2, T_2$, and by assuming that $\Delta H$ is independent of temperature gives

$$\ln\left( \frac{p_2}{p_1}\right)=- \frac{\Delta H}{R}\left( \frac{1}{T_2} - \frac{1}{T_1}   \right)$$

Despite the approximations involved in its derivation this equation does describe well the measurements of the vapour pressure over many solids and liquids, for example depression of freezing point, elevation of boiling point and solubility of solids and gases in liquids, although the latter is not as frequently described as the other two.

In the elevation of the boiling point the solution at a temperature $T_1$ has a lower vapour pressure $p_1$ than the pure solvent. The solution only boils when the temperature is increased to $T_2$  where its vapour pressure has increased to  $p_2$ (normally 1 atm ), which is the pressure at which the pure solvent boils.  By Raoult’s law  $p_1=p_2x_{\mathrm s}$, where $x_{\mathrm s}$ is the mole fraction of solvent
$\ln(p_2/p_1)=\ln(1/x_s)$
As $x_\mathrm s=1-x_\mathrm t$ where $x_\mathrm t$
is the mole fraction of solute,
$$\ln\left( \frac{1}{1-x_t}\right)= -\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$
and as $x_\mathrm t$ is far smaller than unity
$\ln(1-x_t) \approx -x_t -x_t^2/2 \ldots$
from which the second term can be ignored as it is insignificant compared to the first then,
$$x_\mathrm t= \frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$

Thermodynamically the elevation of boiling point and solubility are very similar but we must swap solute and solvent over so what we formerly called solvent has now to be solute and vice versa.

Considering now solubility, the variation of the mole fraction dissolved $x_\mathrm t$ can be written as
$$x_\mathrm t=\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_\mathrm B}   \right)$$

where what was $T_2$ is now the normal boiling temperature of the liquefied gas. Also $\Delta H_\text{vap}$ (a positive quantity) is now the (molar) heat of vaporisation of the liquefied gas and we approximate this to be the heat of vaporisation from solution.

This last equation indicates that the solubility decreases with increase in temperature and should be the same in all solvents in which it forms ideal solutions.

In the solution a gas molecule has an attractive interaction energy between itself and the solvent. (Experiment shows that $\Delta H_\text{vap}$ is positive thus the heat to dissolve is negative). The solvent molecules also have an attractive interaction between themselves otherwise they would not form a liquid.  Thus for a gas-solvent interaction a solvent-solvent one has to be destroyed. Overall the difference in energy is probably small.  However, there is an entropic effect to consider. Clearly this will have increased when a gas dissolves as now there is another species in the solvent and thus the number of ways the molecules can be arranged has increased and this will have increased the entropy. Together these effect make the free energy negative as expected for a spontaneous process.

As the temperature is increased the molecules gain in average energy and so are more able to overcome the attractive potential that keeps them in solution and thus the vapour pressure of the solvent increases and the solubility of the gas decreases. It is also worth noting that the average kinetic energy $(3/2)kT$ is the same in the solution and in the gas phase, in solution it is merely restricted to a smaller range on motion. Thus the notion that it is the kinetic energy that is acquired that causes a molecule to enter the gas phase in incorrect.

For ideal gases (probably in solution the physics are a little different) but anyway as oxygen is an ideal gas, the equations from physics that you need to keep in mind are ...

kinetic energy = (1 / 2)(mass)(velocity)^2 = (3 / 2)(Boltzmann constant)(Temperature),

so therefore temperature is a measure of the velocity of your molecules, where as temperature increases, so does the velocity of the molecules increase. Thus, all molecules in solution, including gaseous oxygen move more rapidly gaining kinetic energy to escape solvation energy, and so they are more prone to bubble up the surface and escape acqueous solution because oxygen gas is less dense than liquid water.


Free refined value in Rietveld I have a question about Rietveld refinement

The author of this paper claimed that the free refined CN ligand is 1.081.
However, they does not show how they got that number. They stated it was based from this table however I am not sure how you can get 1.081 from this table. What is the calculation?

Thank you.

Source:

Reference

Wang, W.; Gang, Y.; Hu, Z.; Yan, Z.; Li, W.; Li, Y.; Gu, Q.-F.; Wang, Z.; Chou, S.-L.; Liu, H.-K.; Dou, S.-X. Reversible Structural Evolution of Sodium-Rich Rhombohedral Prussian Blue for Sodium-Ion Batteries. Nature Communications 2020, 11 (1), 980. DOI: 10.1038/s41467-020-14444-4

I guess that 1.081 refers to the distance length (in Angstrom) separating the C and N atoms. You obtain this value by applying by using the values listed in the table; this is quite tedious, takes some time and require quite good knowledge about crystallography. Otherwise (and much more easy), you can use the crystallographic data listed in the table to build the structure using a proper crystallographic software (such as Vesta); then the software will calculate bond distances for you.


Visualizing rotation about a double bond So I'm having trouble visualizing rotation about a double bond without using a model kit. The 2nd image has a red dot where I rotated it to make the molecule look exactly like the second image with my head and then my model kit. This is also consistent with my answer key that says these molecules are identical.

Now, the next two molecules:

I visualize rotating the left molecule at the red dot carbon and in my head, it matches up with the right molecule. However, when I pull out my model kit, it does not rotate like that and does not match up. This is also consistent with my answer key that says these molecules are diastereomers. So my question is, are there any rules that I could use or a way to visualize that will help me realize if a bond can be rotated to match a molecule without using a model kit.

Also, when people talk about it is not possible to rotate a double bond without breaking it, what exactly are they talking about that makes it different from this?

Adding one more element to help you long-term: I think the underlying reason you are having a problem with this is that the visualization you are using is a little misleading.

You're selecting an atom with your red dot.  You can't rotate around an atom. You can rotate around the bonds between the atoms, though, assuming that the constraints of that bond allow it.

So in the first example, there are two possible carbon-carbon bonds you can rotate around that would both correspond to that red dot. Your dot should be on the sigma bond, based on your diagram.

In the second, you have three carbon-carbon bonds connected to your red dot. The one you are rotating is the double bond, not either of the bonds in the ring.  So the dot should instead be on that bond.

If you change the way you're marking the connection in your head, it could help you think about the nature of that connection, and that could help you with the rest of the information everyone is providing.

It is not possible to rotate a double bond without breaking the double bond .
Below 300$\ce{^o}$ C, double bonds are configurationally stable; that is, cis stays cis and trans stays trans.$\ce{^1}$

In alkenes, there is no free rotation about the double bond, unlike in alkanes. Consequently, groups attached to a double bonded carbon cannot exchange places. Therefore, the cis and trans isomers are separable compounds because they cannot interconvert.

source :ORGANIC CHEMISTRY , PETER VOLLHARDT &NEIL SCHORE pg439

In the first molecule ,red dot is on carbon bonded to a double and a single bond .If rotated along an axis passing through the red dot same molecule is formed.

However,in the second molecule,rotation along the red dot is not possible since the bonds are locked.

Your molecular model kit is correct.


Is it possible to synthesise ethers from carboxylic acids? Is it possible to synthesise ethers from carboxylic acids, and if so, how? By ether I mean an ether in general ($\ce{R-O-R'}$), not just specific examples.

I theorised one could perform Fischer esterification, $\ce{ROH +R'COOH -> RCOOR'}$, and then perform Wolff-Kishner reduction on the carbonyl group. The problem with this theory is that it assumes that the single-bonded oxygen between the $\ce{R}$ and $\ce{R'}$ doesn't influence the reduction process.

This is an interesting question. I can tell you that Wolff-Kishner won't work - both the hydrazine and the hydroxide required will attack the ester in a substitution fashion (with a loss of -OR'). Other reductions might work.

For example a Google search for "reduce ester to ether" brought up the following:

The reaction appears to be complete within 3 hours with moderate to good (but not stellar) yields. The abstract and key figures are available for free at www.organic-chemistry.org. The paper is published in the Journal of Organic Chemistry.

It appears that $\ce{Et3SiH}$ is the terminal reductant, but most of the work is being done by indium. The combination of $\ce{Et3SiH}$ and $\ce{InBr3}$ produces the active reductant $\ce{HInBr3}$, which provides the equivalent of $\ce{H.}$. The second $\ce{H.}$ comes from $\ce{Et3SiH}$. The proposed mechanism from the paper is as follows:

Other mild sequential one-electron reductions with a source of $\ce{H.}$ would also likely work. The two-electron reductions, like Wolff-Kishner ($\ce{NH2NH2 +NaOH}$) and hydride-transfer (e.g. $\ce{LiAlH4}$), are all heavy-handed and nucleophilic with the tendency to take apart or over-reduce the ester.

Take a carboxylic acid and react with an alcohol. Distill off the ester to get the formula, RCOOH + R'OH <=> RCOOR' + H2O. Decarbonylate it with a decarbonylation catalyst like Palladium or its compound. Palladium(II) Acetate is also.

Sources:

Dalle-Donne, Isabella; Aldini, Giancarlo; Carini, Marina; Colombo, Roberto; Rossi, Ranieri; Milzani, Aldo (2006). "Protein carbonylation, cellular dysfunction, and disease progression". Journal of Cellular and Molecular Medicine 10 (2): 389–406. doi:10.1111/j.1582-4934.2006.tb00407.x. PMID 16796807. Grimsrud, P. A.;
Xie, H.; Griffin, T. J.; Bernlohr, D. A. (2008).
"Oxidative Stress and Covalent Modification of Protein with Bioactive Aldehydes". Journal of Biological Chemistry 283 (32): 21837–41. doi:10.1074/jbc.R700019200. PMC 2494933. PMID 18445586.

^ Jump up to: a b c Bertleff, Werner (2000).
"Ullmann's Encyclopedia of Industrial Chemistry". doi:10.1002/14356007.a05_217. ISBN 3527306730. |chapter= ignored (help)

Jump up ^ Arpe, .J.: Industrielle organische Chemie: Bedeutende vor- und Zwischenprodukte, 2007, Wiley-VCH-Verlag, ISBN 3-527-31540-3

Jump up ^ Hartwig, J. F. Organotransition Metal Chemistry, from Bonding to Catalysis; University Science Books: New York, 2010.

Jump up ^ Zoeller, J. R.; Agreda, V. H.; Cook, S. L.; Lafferty, N. L.; Polichnowski, S. W.; Pond, D. M. (1992). "Eastman Chemical Company Acetic Anhydride Process". Catalysis Today 13: 73–91. doi:10.1016/0920-5861(92)80188-S.

Jump up ^ Beller, Matthias; Cornils, B.; Frohning, C. D.; Kohlpaintner, C. W. (1995). "Progress in hydroformylation and carbonylation". Journal of Molecular Catalysis A 104: 17–85. doi:10.1016/1381-1169(95)00130-1.

Jump up ^ W. Bertleff; M. Roeper; X. Sava (2005), "Carbonylation", Ullmann's Encyclopedia of Industrial Chemistry, Weinheim: Wiley-VCH, doi:10.1002/14356007.a05_217

Jump up ^ El Ali, B.; Alper, H. "Hydrocarboxylation and hydroesterification reactions catalyzed by transition metal complexes" In Transition Metals for Organic Synthesis, 2nd ed.; Beller, M., Bolm, C., Eds.; Wiley-VCH:Weinheim, 2004. ISBN 978-3-527-30613-8

Jump up ^ Karlheinz Miltenberger, "Hydroxycarboxylic Acids, Aliphatic" in Ullmann’s Encyclopedia of Industrial Chemistry, Wiley-VCH: Weinheim, 2003.

Jump up ^ Riemenschneider, Wilhelm; and Bolt, Hermann (2000). "Esters, Organic". Ullmann's Encyclopedia of Industrial Chemistry: 10. doi:10.1002/14356007.a09_565. Retrieved 17 December 2013.

Jump up ^ Elschenbroich, C. ”Organometallics” (2006) Wiley-VCH: Weinheim. ISBN 978-3-527-29390-2


Is there a way to experimentally measure entropy? I've been doing thermodynamic chemistry, and recently focusing on Gibbs Free Energy. Whilst doing calculations using, $$\Delta G = \Delta H - T \Delta S$$ I have been given a table of values for $\Delta H$ and $\Delta S$, and a temperature to work with. I was wondering , as the unit for entropy is $\mathrm{J/K}$ or $\mathrm{J\ K^{-1}}$ how exactly one would experimentally measure entropy/change in entropy, or can you?

The most common way of measuring $\Delta S^\circ$ for a chemical reaction is probably by making a van't Hoff plot. You measure the equilibrium constant $K$ at different temperatures and plot $\ln K$ vs $T^{-1}$.  The $y$-intercept = $R\Delta S^\circ$ and the slope = $-R\Delta H^\circ$.

Another option is to measure $\Delta H^\circ$ by calorimetry and measure $K$ by some other means. Then compute $\Delta G^\circ$ from $K$ and solve for $\Delta S^\circ$

Entropy is a measure of randomness.
The definition itself is hard to comprehend at times. And when you go to think, about how one would measure randomness, you get no answer. The term Entropy was introduced as a State Function to check whether the reaction is possible or not.
Well there are formulae to determine the change in entropy. However as far as I have read, all books say that calculating the absolute value of entropy is quite difficult.
According to me if you know the values of Entropy at Standard States you can calculate the Change in entropy for a process from STP to the temperature at which you want its absolute value and then just subtract to get the absolute value. The Standard State entropies are mentioned in quite a few books.
Hope the answer helped.:-)


molecular orbital theory i read in my textbook that the energy diagram of moleculal orbits is different for atoms nitrogen and below them.
my question is which one of the energy diagrams to use in heteronuclear compounds like CO?

(In heteronuclear diatomic molecules, atomic orbitals only mix when the electronegativity values are similar.)
(While MOs for homonuclear diatomic molecules contain equal contributions from each interacting atomic orbital, MOs for heteronuclear diatomics contain different atomic orbital contributions.)
(Orbital interactions to produce bonding or antibonding orbitals in heteronuclear diatomics occur if there is sufficient overlap between atomic orbitals, as determined by their symmetries and similarity in orbital energies.)

It depends on how many electrons your compound has. In $\ce{O2}$ , there are 16 electrons.
Likewise all compounds with more than 16 electrons follow the $"\ce{O2}"$ diagram and those with less than 16 follow the other diagram.


How can elements have the same number of electrons, but different electron configurations? $\ce{Ti}$ and $\ce{Cr^2+}$ are isoelectronic, yet their electron configurations are different, with the former's being $\ce{[Ar] 4s^2 3d^2}$ and the latter's being $\ce{[Ar] 3d^4}$.
How can that be explained?

The reason is very complicated, but in short:
The energy gap between $\ce{3d}$ and $\ce{4s}$ is wider in $\ce{Cr^2+}$ than in $\ce{Ti}$ because $\ce{Cr^2+}$ has more protons.

When you promote an electron from $\ce{3d}$ to $\ce{4s}$, the electron gained the energy thanks to orbitals, but you also lost the energy thanks to repulsion (of electrons in the same orbital). Remember that electrons like to be in states of less energy, not more.

The energy gap between $\ce{3d}$ and $\ce{4s}$ is less in $\ce{Ti}$, so the electron, by promoting, lost more energy thanks to repulsion than it gained thanks to the orbital, so its configuration is $\ce{[Ar] 4s^2 3d^2}$.

In $\ce{Cr^2+}$, the enrgy gap widened, so the electron would gain more energy than its lost by promoting, making its configuration $\ce{[Ar] 3d^4}$.

See also chemguide.co.uk.

When you fill electrons in different orbitals, you keep in mind the $n+l$ rule, but when you remove electrons you have to consider the principal quantum number. Between 3d and 4s, 4s has a lower value of $n+l$, so electrons will be first filled in it, that's why titanium metal has the atomic configuration $\ce{[Ar] 4s^2 3d^2}$.
In case of $\ce{Cr^2+}$ you have to first write configuration of $\ce{Cr}$, which is $\ce{[Ar] 4s^2 3d^4}$ and then remove two electrons from the orbital having highest value of $n$ (principal quantum number) which is 4s. after removing 2 electrons from 4s you will get configuration of $\ce{Cr^2+}$ which is $\ce{[Ar] 3d^4}$.


Why does the acetyl group drop the pH? I have a molecule that is 8% of the R groups are replaced with Acetyl groups.  When I dissolve it in water it drops the pH.  I am trying to understand why.

Because the electrons in the double bond C=O help to stabilize the CH3 group after it has released an H+ (it is similar in this way to what happens with carboxylic acids), the Acetyl groups are acid. Hence the pH drop.


Is carbon dioxide organic or inorganic? Today in chemistry class we were discussing Organic Chemistry. We discussed what organic compounds basically are and then I asked the teacher whether $\ce{CO_2}$ is organic or not. She told that it is as it contains carbon and oxygen with a covalent bond. I told her it can't be as it is not found in animals (naturally). I am very confused about it.

I need some good reasons to agree with either explanation. (I have searched the internet already but found no great insights as of now).

It is entirely arbitrary whether you call it an organic compound or not, though most would not.

The distinction you make that organic compounds should be found in living things is not a useful criterion. Moreover you are wrong that carbon dioxide isn't: it is both made and used by living things. Animals make it when they metabolise sugars to release energy; plants consume it when they build more complex organic molecules through photosynthesis. In fact most organic molecules are, ultimately, derived from $\ce{CO2}$.

Even more importantly most molecules considered organic are neither made by nor are found in living things. Chemists make new carbon compounds all the time (tens of millions in the history of chemistry) and most have never been made by animals or plants.

The organic/inorganic terminology is mostly very simple: covalent compounds containing carbon are organic. The only fuzzy area is around very simple molecules like $\ce{CO2}$ where the distinction doesn't matter much. So we would not normally think of diamond or silicon carbide as organic. But we might (though many would not) call calcium carbide organic because it contains a $\ce{C2}$ unit with a carbon-carbon triple bond.

However since the terminology is mostly very obvious and also somewhat arbitrary, it isn't worth much argument to sort out those very simple but awkward edge cases.

organic usually refers to chemicals made from oil coal or gas, i.e the organic compounds are made from organic carbon sources and come from natural sources. This is why CO2 is not regarded as organic.


With which 2 of these 5 chemicals would Chlorine not react? N, NaOH, O2, NaBr, Fe A question in a Russian EGE test asks to choose 2 chemicals out of 5 with which Chlorine does not react:

Nitrogen - I checked and it reacts
$\ce{KOH}$ -- reacts
$\ce{O2}$ -- reacts
$\ce{NaBr}$ -- reacts
$\ce{Fe}$ -- reacts

I'm at a loss. Which two should I pick then? Maybe Fe and Nitrogen, because they would be the least likely to react?

Here's the question:

$\ce{NaOH}$

$$\ce{2NaOH + Cl2 → NaCl + NaClO + H2O}$$

  Reaction between chlorine and sodium hydroxide to form sodium
  chloride, sodium hypochlorite and water. The reaction takes place in a
  cold dilute solution. (Chemiday 1)

$$\ce{3Cl2 + 6NaOH → NaClO3 + 5NaCl + 3H2O}$$

  Reaction interaction chlorine and sodium hydroxide to form sodium
  chlorate, sodium chloride and water. The reaction proceeds in a
  concentrated solution with heating.(chemiday 2)

$\ce{NaBr}$

$$\ce{2NaBr + Cl2 → 2NaCl + Br2}$$

  Reaction between sodium bromide and chlorine to form bromine chloride
  and sodium. Hot - sodium bromide. The industrial method of producing
  bromine.(chemiday 3)

$\ce{Fe}$

$$\ce{2Fe + 3Cl2 ->[\Delta] 2FeCl3}$$

  The reaction of chlorine and ferric chloride reaction with the
  formation of iron (III). The reaction proceeds at a temperature above
  250 C °.(chemiday 4)

But I couldn't find any reaction of chlorine with neither nitrogen or oxygen. Guess what, chlorine doesn't have sufficient energy to break the triple bond of nitrogen or double bond of oxygen.

Its most likely $\ce{Fe}$ and $\ce{N}$ like you typed. $\ce{N}$ is most accurate because nitrogen is nonreactive so a spark will be needed to react it with chlorine. $\ce{Fe}$ will react with Chlorine but its reactivity is rather low compared to other transition metals such as zinc.