What happens if you make a gas of magnetic particles I was reading todays SMBC which asked what happens if you make a gas of magnetic particles

So what does happen if you make a gas of magnetic particles?

If you heat up the magnet above Curie temperature (or more drastically just evaporates it), the magnetic spins will be randomly aligned and the magnetic field  is not produced anymore (at higher temperature).


Comparing the stabilities of allylic, benzylic, and tertiary radicals Which is the most stable free radical among the given species?
  (1) $\ce{CH2=CH-CH2^.}$ (allyl radical)
  (2) $\ce{C6H5-CH2^.}$ (benzyl radical)
  (3) $\ce{(H3C)3C^.}$ (t-butyl radical)
  (4) $\ce{C2H5^.}$ (ethyl radical)

The book claims the answer to be the allyl radical, option 1. But how can we definitely say that the allylic free radical is more stable, compared to the benzyl radical (option 2) which is also resonance-stabilised, or the tertiary radical (option 3) which has more hyperconjugative donation from α-hydrogen atoms?

This is a tough question.  I think it might even be unfair to ask such a question on a test in non-advanced classes.  In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the "real" answer is known.

What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the t-butyl or ethyl radicals which are not resonance stabilized.  You can see from the following diagram that you can draw two resonance structures for the allyl radical.

For the benzyl radical you can draw even more resonance structures.

Superficially, this might suggest that the benzylic radical is more stable than the allyl radical. This was the "answer" and reasoning provided here.

However, if you do some thermochemical calculations (as done here), you arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/mol more stable than the benzyl radical. The number of resonance structures alone is not a perfect indicator of stability.

For me, that is a pretty small difference in energy, and I think it would be fair to answer that the allyl and benzyl radicals have comparable stabilities.

The answer is benzylic radical.

It is the most stable because it resonates more compared to allylic and alkyl radicals.
Regardless of its radical class, be it primary, secondary or tertiary, benzenlic radicals will always be more stable than any other types of radical.


Why is boron trifluoride a lewis acid? $\ce{BF3}$ does back bonding with fluorine and still accepts a pair of electron and is considered lewis acid why?

A Lewis acid can accept a pair of electrons from a Lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.

A lewis acid is defined as an electron pair acceptor. The boron in BF3 is electron deficient & has an empty d orbital,so it can accept a pair of electrons,making it a lewis acid.Also it contains only 6 electrons in outermost shell making it able to accept an electron pair to complete its octet.Thus it's lewis acid


What is responsible for the odor of gasoline? Gasoline as we know it is a blend of combustible hydrocarbons and has a characteristic odor. The natural gas we use to heat our homes and (for some of us) cook our food has its own odor, except...

Natural gas as piped to our homes consists of methane, which is odorless, plus an odorant so that we can be aware of its presence for safety reasons. So the odor we think of as "gas" isn't the fuel at all, but an odorant.

So this leads to my question: is the familiar odor we attribute to gasoline that of one or more of the components which are the fuel itself (heptane, octane, etc.) or an additive for the purpose of giving it an odor, again for safety reasons?

Many of the components of gasoline (petrol) have notable odours

A simple starting point for understanding the odour of gasoline is the typical composition of the product. That composition varies a lot and always has done as different refineries use different processing techniques and different crude oil sources have different components to start with. And regulators have changed the rules over time (for example to reduce the amount of benzene, a carcinogen). All that matters, in practice, is the octane rating which is what engines care about and this is achievable in a variety of ways by varying the composition and the additives that modify the combustion characteristics.

This (old) paper gives a crude overview of a typical composition.

  The typical composition of gasoline hydrocarbons (% volume) is as follows: 4-8% alkanes; 2-5% alkenes; 25-40% isoalkanes; 3-7% cycloalkanes; l-4% cycloalkenes; and 20-50% total aromatics (0.5-2.5% benzene).

Modern blends will now have far less benzene (but will still have a lot of other aromatics). Without getting too specific on the individual chemicals we can already tell some of the sources of the distinctive odour of the mixture. It is pretty meaningless to try to pin it to a single component.

Aromatics as a class are actually named because of their distinctive smell. even if we substitute other aromatics like toluene or xylene for the benzene, there will be a distinctive "aromatic" component in the product. But plenty of the other hydrocarbons also have notable odours. Cycloakenes (maybe 7% of the mixture) often have sharp smells (cycopentene is even said to have "a petrol-like odour"). Even linear and branched alkanes have some smell (though it is far less intense than the unsaturated components of gasoline).

In short, trying to pin down the typical smell of gasoline to a single compound is a little silly when there are often scores of components with well-known and strong odours and the specific composition varies a lot. aromatics are a big part with other unsaturated compounds providing some other odiferous components with a smaller but notable contribution even from unsaturated hydrocarbons.

Also, gasoline is not "odourised" as it doesn't need to be. But it sometimes has specific ingredients added to make tracing the source easier. But this is done with complex instrumentation not the human nose.

Gasoline has it's distinctive odor because of benzene, which is found naturally in crude oil, which gasoline is derived from. Benzene is a carcinogenic, highly flammable liquid with a high octane number, but because of it's toxicity is not really used for non industrial uses.

Source: https://en.wikipedia.org/wiki/Benzene


How can I perform structure alignment on an inorganic species? I'm attempting to perform atomistic optimal structure alignment on an inorganic species. Can anyone point me in the right direction to software or a technique for performing this?

There exists a wide range of structural alignment software, but this really appears to be geared towards aligning protein backbones, whereas I'm interested in aligning unlabelled Cartesian coordinates of small molecules.

Inorganic structures – that could be proper solid or molecular metalloorgnic. For the later one you can use xp (program written G.M.Shedrick - comes with Bruker diffractometers). If you have running license of Cambridge Crystal Data Base there is program called IsoStar, and I think that it should do the job. I haven't used it my self but apparently Olex2 can make aligning of fragments (this program is free so that's the place were you can start). I'm not quite sure but I think that old GROMOS code could also make a fit of orthogonal coordinates.

OpenBabel using SMARTS. pymol also does it.


How can you qualitatively tell when the hydrolysis of aspirin is over? I'm doing an experiment on the effect of temperature on the time taken to produce acetic acid and salicylic acid. Are there any qualitative observations I can make about when the reaction has finished?
I am using a 0.01 M aspirin in pure water.

First I was thinking that maybe a chemical indicator could be used, however the pKa values of all three components are relatively close to each other (especially aspyrin and salicylic acid). It might still be a bit of a help to stick a pH electrode in the system and observe the change - however, this is not exactly something you can see easily.
Another thing that came to mind is that salicylic acid (as all phenols) will form a colorful complex with Fe(III). However, the appearance of the red-ish color will only tell you when the reaction has started, since it's borderline impossbile to observe that the solution is not getting any darker anymore. Again, if you have some machinery, you could stick your solution into a cuvette and put it in a photometer.
Apart from the two ideas above, I don't really see an easy way of observing anything.

Why be qualitative when with just a little more effort you can be quantitative?
Acetylsalicylic acid has a pK$_a$ of 2.97, i.e., it is fairly acidic (Ref 1). Salicylic acid, one of the hydrolysis products, is comparably acidic (pK$_a$ = 2.79 Ref 2) and its sodium salt is fairly neutral (pH = 6.0 - 8.5 for a commercial material Ref 3; calculations suggest ~7.4 at 0.01 M).
Acetic acid, the other hydrolysis product, is not so acidic: pK$_a$ = 4.75, and its sodium salt, in solution, has a pH of 8.375 (Ref 4). This is how the calculation plays out:
pH = 7 + 0.5(pK$_a$ + logconcentration)
= 7 + 0.5(4.75 - 2) = 8.375
Approximately. The other ions in solution will probably have some effect.
One more fact: phenolphthalein is colorless below pH = 8.5 and pink/red above pH = 9 (Ref 5).
So, your 0.01 M acetylsalicylic acid will dissolve (soluble in water 3g/L = 0.0166 M). As it hydrolyzes, it will produce salicylic acid with essentially the same strength (pK$_a$ ~same) and concentration (1:1). But at the same time, it will generate acetic acid, somewhat weaker, but still titratable.
If you take 10.0 mL of your experimental solution, add a few drops of phenolphthalein indicator solution and then titrate with 0.01 N NaOH (or KOH), at time = 0, it will require 10.01 mL to turn the phenolphthalein red. As the hydrolysis proceeds, the acetic acid produced requires additional alkali to get the pH high enough to get a red color. In fact, at the end of the hydrolysis, it will require 20.01 mL of 0.01 NaOH to turn red. (The decimal places are for instructive purposes, not to indicate accuracy - you know what I mean.)
In fact, you don't need to use exactly 0.01 M alkali - just realize that whatever you need to turn red at time zero will double when the hydrolysis is complete. And you can weigh the amount of alkali added rather than use a burette. And you can do a reaction rate study by checking every 15 minutes or so.
Ref 1. https://www.chem.purdue.edu/courses/chm333/Spring%202013/Lectures/Spring%202013%20Lecture%202%20-%204.pdf
Ref 2. https://foodb.ca/compounds/FDB000882
Ref 3.  https://labchem-wako.fujifilm.com/us/product/detail/W01W0119-0314.html
Ref 4.  https://tardigrade.in/question/pk-a-of-ch-3cooh-is-4-74-the-ph-of-0-01-m-ch-3coona-is-bus7pexw
Ref 5. https://www.britannica.com/science/phenolphthalein


Why is "Spontaneous" a part of 2nd law wording? The way I understand the law, any process is implausible if the total entropy of the universe does not increase. Why is "Spontaneous" even a part of the wording, what does it mean?

Is it perhaps a part of the probabilistic theory of thermodynamics, in which it is not "impossible" for heat to flow from cold to hot, simply very very unlikely?

In chemical thermodynamics we are interested in the amount of work that can be done by a reaction as it proceeds towards equilibrium and whether a reaction will spontaneously proceed towards products or, if not, how much work is needed to bring it about. Spontaneous is used because experimentally reactions and processes occur without us having to do anything to bring them about, and this needed an explanation.

The second law can be stated as$^1$  'Spontaneous changes are those which, if carried out under the proper conditions can be made to do work. If carried out reversibly they yield the maximum amount of work. In natural processes the maximum amount of work is never obtained.' This is also a general definition of equilibrium. A shorter definition of the second law is 'heat does not spontaneously flow from from a cold to a hot body'.

If there are differences in intensive properties then a spontaneous change is always that which eliminates the difference. Thus if there is a (perfect) piston with different pressures on either side, it will move to equalise the pressure. Similarly if there is a temperature difference between two bodies in contact with one another then 'heat flows' from the warmer to the colder. Similarly, a concentration difference in a solution will equalise. However, this cannot be used to predict the direction of a chemical reaction in which there may already be constant pressure, temperature or concentrations.

In a purely mechanical system, the direction of spontaneous change seems obvious and is one in which the potential energy is lowered to a minimum. It was thought at one time that the same would apply to chemical systems, i.e that the direction of spontaneous change was one that minimises the heat content of the system. But experiment shows that this is not the case, there are, for example, spontaneous endothermic reactions$^2$, so that another criteria is needed to predict the direction of a spontaneous reaction.  This means understanding why for example, a gas expands to fill a container even though no loss of energy occurs on doing this. The missing idea is the entropy$^3$.

It used to be thought, before the second law was properly understood, that the maxiumum work a chemical reaction could do was to be found by converting all the heat of reaction into work. This was not bourne out by experiment and it is the free energy $\Delta G$ that gives the maxmum amount of work. Gibbs was the first to understood that the maximum amount of work is given by the first law unless some heat is taken in or given out to the surroundings. Thus the work done may be greater than the heat of reaction.

The second law requires that there be an increase in entropy during any actual process. Thus $dS_{sys} + dS_{surr} \ge 0$ for the system and surroundings. The sum of the two entropy production processes is always a positive quantity, but which can be zero in the case of a reversible process.

(Your speculation about statistical mechanics is partly correct, it is possible for entropy to 'run backwards' if only a very small number of molecules is considered and then only for a very short time. In 'normal' lab chemistry the number of molecules involved is vast say, $10^{18}$ so deviations from the average are minute.)

$^1 $ E. B. Smith, 'Basic Chemical Thermodynamics'.

$^2$ e.g. $\ce{Ba(OH)2(8H2O)_{(s)} + NH4SCN_{(s)} +heat \rightarrow Ba(SCN)2 + 2NH3 + 10H2O}$

$^3 $ The second law is needed because the first law offers too much latitude compared to experimental observations; by the first law heat could spontaneously flow from a cold to a hot body or a broken glass could spontaneously reform itself.

Refer to Wikipedia. Please note that your understanding (If dS/dt isn't positive...) is tautological. That is (as far as we know), our Universe requires dS/dt to be positive (or perhaps I should say we've never seen otherwise). The 2nd Law is most useful for nearly perfectly isolated systems and systems near (or, arguably, at) equilibrium. "Spontaneous" refers to time-forward processes in systems which are not perturbed (by outside events/forces). Our laws of quantum mechanics do not distinguish between positive and negative time. Supposedly, any event which can happen, can "un-happen" (i.e. reverse itself). This can be contrasted with macroscopic processes which are never perfectly reversible. This is one way that "the arrow of time" is seen. This problem remains unsolved and is recognized as one of the major fundamental problems in physics. We can't go backwards in time. So, we're limited to what experimental evidence we have when we talk about the "direction" of time.
You should also understand that the 2nd Law has no one single written form. It is expressed in different ways for different purposes. By this I mean that the word "spontaneous" need not be part of its expression.
See
https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Various_statements_of_the_law
https://en.wikipedia.org/wiki/Arrow_of_time#Arrows
(also Entropy_(arrow_of_time) and Entropy#Definitions_and_descriptions)
and maybe https://en.wikipedia.org/wiki/Equipartition_theorem
and of course Statistical_mechanics


If a gas absorbs heat, then how can its temperature remain constant? My book has this line:

  In an isothermal expansion, the gas absorbs heat and does work

But, if a gas absorbs heat, then how can its temperature remain constant? (as the process is isothermal)

From the first law, $\Delta U=Q-W$.  If the process is isothermal, then $\Delta U=0$ (assuming an ideal gas), so Q = W.  The heat energy is consumed by your gas doing work, so there is no change in internal energy (temperature).

Hopefully I can provide you with some physical intuition on this phenomenon through a simple example.

Imagine a gas heated by a flame in a closed cylinder with a piston on the top balanced by the same pressure as the gas inside. As you heat the gas, the particles move faster (gain kinetic energy) and collide with the walls of the container harder, increasing the pressure of your system. This pressure increase is then used to drive the piston out. As the particles of your hot gas collide with the piston wall to move it, they necessarily dump some of their kinetic energy into it, slowing down (i.e. cooling) in the process. On the other hand, kinetic energy is not lost when the gas particles bump into the other rigid walls of the container, since those collisions are basically elastic.

So overall you have a heat source dumping kinetic energy into the gas particles from one side of the cylinder, and the gas particles dumping their kinetic energy into the surroundings while pushing a piston on the other side. Now run this process very carefully and slowly, and the temperature of the gas can be held more-or-less constant!


Is there any difference between getting dissolved and getting dissociated? In the context of electrolytes only, is there any difference between electrolyte getting dissolved and electrolyte getting dissociated?

According to my professor, they are different. He says a substance may dissolve completely but it may not dissociate completely and a substance may not dissolve completely but amount that dissolves can dissociate completely and, thus, substances which dissolve (either wholly or partially), amount that gets dissolved if dissociate completely are called strong electrolytes and amount of substances which gets dissolved if dissociates sparingly are called weak electrolytes (although it may dissolve completely).

How is it possible? I mean, how can solubility be different from dissociating?

If something dissociates it means it’s dissolving and in the context of electrolytes if something dissolves in water it has to dissociate into ions and even Arrhenius’ theory of electrolytic dissociation sys that substance which are water soluble, dissociates into ions and get dispersed in water.

And if someone says they are different to them, I ask the following two questions:

If solubility and dissociation are different then what would be the criteria for getting dissolved and what will be criteria for getting dissociated (for me the criteria for getting dissolved is dissociation).
how would you arrange the following in increasing order of 1.solubility
2.dissociation

NaCl, MgCl2, FeCl2, ZnCl2

This is a simple logical and semantic problem, which is not a problem at all. Your professor is right and wrong- both at the same time. He is creating a classification which does not exist and which is meaningless.

Look at the word origin of electrolyte:

  Etymology from OED: < electro- comb. form + ancient Greek λυτός that may be
  dissolved, soluble (see -lyte comb. form). Compare electrolysis n.,
  electrolyse v.

  and the portion -lyte "< ancient Greek -λυτος, combining form (in e.g.
  εὐδιάλυτος : see eudialyte n.) of λυτός that may be dissolved, soluble
  < λύειν to loose (see lysis n.) + -τός, suffix forming verbal
  adjectives. Compare French -lyte, German -lyt.

  A substance which gives rise to ions when dissolved (typically in
  water) or fused; a liquid or gel which contains ions and can be
  decomposed by electrolysis; spec. the conducting fluid used in a
  battery.

Does this solve the semantics problem?

By definition, an electrolyte is soluble in a given solvent. It is also understood among electrochemists that an electrolyte is added to a solvent to impart conductivity to that solvent. I will never call sugar an electrolyte because nobody would use sugar to increase conductivity of water. You should ask: Does a given substance produce ions in a solution? If yes, it can be called as an electrolyte.

Look at this logic: grapes is a fruit but not all fruits are grapes.

Electrolytes are soluble in the solvent, but not all soluble substances in that solvent behave as electrolytes. Anything which produces ions in that solvent, can be labelled as an electrolyte.

Q-1 If solubility and dissociation are different then what would be the criteria for getting dissolved and what will be criteria for getting dissociated (for me the criteria for getting dissolved is dissociation).

The measure of dissociation is conductivity. How much electrical resistance is offered by that dissolved (and dissociated) substance in that solvent

Q-2 how would you arrange the following in increasing order of 1.solubility 2.dissociation

NaCl, MgCl2, FeCl2, ZnCl2

There is no correlation between solubility and dissociation.  If I want to study their dissociation I will measure the conductivity of their solutions systematically.

[EDIT] I start with the following definition:

  Dissociation, in chemistry, the breaking up of a compound into simpler constituents that are usually capable of recombining under other conditions. In electrolytic, or ionic, dissociation, the addition of a solvent or of energy in the form of heat causes molecules or crystals of the substance to break up into ions (electrically charged particles). Most dissociating substances produce ions by chemical combination with the solvent. The idea of ionic dissociation is used to explain electrical conductivity and many other properties of electrolytic solutions.

I now present an example of two molecules, chlorine oxides ($\ce{Cl_xO_y}$) and also chlorine, all of which are gases at room temperature, to state my case that there can be a major difference between dissolve and dissociate creating distinct and important properties in a solvent of water.

For example, $\ce{ClO_2}$ is a gas that is highly soluble in water (some 8 g/L at 20 °C). However, per Wikipedia , to quote:

  It does not hydrolyze when it enters water, and is usually handled as a dissolved gas in solution in water.

Although technically, there are many (including important) 'produce ions by chemical combination of the molecule with' water (see my note below).

In contrast, we have $\ce{Cl_2O}$, which per Wikipedia is described as:

  At room temperature it exists as a brownish-yellow gas which is soluble in both water and organic solvents. Chemically, it is a member of the chlorine oxide family of compounds, as well as being the anhydride of hypochlorous acid.

As a result of the latter, we have:

$\ce{Cl2O + H2O <=> 2 HOCl}$

And, further aqueous hypochlorous acid can create ions (hence an electrolyte, albeit, very weak):

$\ce{HOCl <=> H+ + OCl-}$

In fact, dichlorine monoxide is very soluble due to hydrolysis in water (some 143 g of $\ce{Cl_2O}$ per 100 g water), but also can dissolve in organics (like CCl4 and CHCl3).

So in water, $\ce{ClO_2}$ mostly dissolves and $\ce{Cl_2O}$ largely dissociates, which seemingly presents significant differences. And, finally, in contrast to chlorine, it is but mildly soluble (1/10 of $\ce{ClO_2}$) and mildly dissociates, per below, which varies significantly with temperature:

$\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}$

Therefore, it appears that all three gases have distinct (and important) properties connected to solubility and propensity to undergo dissociation. If a chemistry student of mine were to conflate solubility and dissociation, in regard to these compounds, I would feel bad if I contemplated assigning a final grade above C.

A technical note, with respect to $\ce{ClO_2}$, I explain the dissociation of chlorine dioxide due to its self-reaction in water as chlorine dioxide is an example of a stable free radical:

$\ce{.ClO2 + .ClO2 <=>  Cl2O4}$

$\ce{Cl2O4 + H2O <=>  HClO2 + HClO3}$

where both created acids form ions in water. [EDIT] A supporting source makes the statement:

  In air, chlorine dioxide readily dissociates both thermally and photochemically and may form chlorine, oxygen, hydrogen chloride, HClO3, HClO4.ClO, chlorine peroxide, and/or chlorine trioxide, dependent on temperature and humidity. Chlorine dioxide dissociates in water into chlorite and chloride, and to a lesser extent into chlorate (Budavari et al. 1996).

I would also note that $\ce{Cl_2O}$ dissolves before undergoing dissociation, and the dichlorine monoxide reversibly, to an extent, will escape from solution (especially concentrated or on heating) and is responsible for the 'chlorine-smell' of bleach (per Watts, p. 16) and as also reported by Watts, the ability to concentrate HOCl by distilling off half of the solution (due to the partial release of the $\ce{Cl_2O}$).

[EDIT] Now, as to the answer to the 2nd question, the answer depends on a substance's ability to dissolve and undergo dissociation, to rank one must measure the electrical current produced by my adopted definition. However, I would argue that the results can vary in the presence of dissolved oxygen for acidic ferrous salt, for example. The explanation is that the ferrous can be converted into ferric per an electrochemical reaction in the presence of H+ (per hydrolysis of FeCl2) and O2 as dissolved or from air, upon exposure over time. As such, the FeCl2 with time and air exposure may seemingly present different electrical properties (due actually to dissociation). Similarly, per this work: The reaction of oxygen with magnesium chloride, again measurable changes in voltage may occur. Apparently, MgCl2/O2 is NOT exactly inert either, so again a possible departure in ranking with time in the presence of gaseous or dissolved oxygen.


How many valence electrons does the azide ion have? We had a test and and we had a question asking the number of valence electrons in an azide ion, $\ce{N3^{-}}$.  I find out that the answer is 16 but I could not understand how.
 I saw the structure. It says that the end nitrogen have a -1 charge and the middle nitrogen has a +1 charge.

From where does the one of the end nitrogen get the electron assuming the other end nitrogen gets it from the middle nitrogen?

Nitrogen (the element) has 7 electrons. But two of those electrons are in the $1s$ orbital, and so are not considered valence.  Therefore, a single elemental nitrogen has five valence electrons.

There are three nitrogens, and so $5 \times 3 = 15$ electrons for $\ce{N_3}$.

Finally, azide is $\ce{N_3^{-}}$, there is an additional electron beyond the neutral form.  So, the total number of valence electrons is $15 + 1 = 16$.

Nitrogen is a 5 group element hence has valency 5 and AZIDE ion has an negative charge it means it gained an electron hence as AZIDE HAS 3 NITROGENS and a negative charge it will be 5 multiplied by 3 = 15 + one electron gained = 16


Is there an equivalent unit cell that can describe a bcc unit cell where all points are the same element? Similar to how an NaCl-structure can be described as just a primitive cubic structure if all the lattice points were the same element or ion. I'm asking if there any such alternative for the CsCl-structure.

You certainly can construct what is called a primitive cell containing one atom for a bcc lattice.  There could be more than one way, but here is my method.

Start with the usual cubic representation.  Pick any face and connect the midpoints of its edges to make a smaller square.  Do the same with the opposite face.  This gives a pair of squares which are then the bases of an elongated square prism, whose lateral edges connect the two opposing faces you "contracted".  This cell, which contains just the central atom of the original bcc structure, is a primitive cell.  In this cell the lateral edges, which we did not "contract", are about 1.41 (actually, the square root of 2) times as long as the basal edges.

How does this replicate to make the lattice? If you just scoot the primitive cell over to "catch" a neighboring atom on either side of the lateral faces, you find that you must move the adjacent cell "up" or "down", in the direction perpendicular to the bases, to get the second atom centered properly.  The "up"/"down" move is half the height of the prism.  But once you see this, you can repeat this process and get all the atoms involved in the entire cubic representation.

Last question:  I said the primitive cell is an elongated square prism.  It no longer has cubic symmetry.  Where did the cubic symmetry go?  The answer is you can't make a primitive cell in the bcc lattice with the full cubic symmetry.  Instead, note that when you modified the cube you had the choice of three pairs of opposite faces to "contract".  Thus there is really a combination of three primitive cells with different orientations.  The cubic symmetry emerges from the combination rather than any one primitive cell.

Yes there is. In your particular case, it is also the primitive cubic lattice.

A hint is in the name - BCC = Body Centered Cubic Any arrangement that is BCC is also primitive cubic, for the isolated constituent building blocks.


How does ammonium nitrate explode on its own? I thought ammonium nitrate was an oxidizer that needed to be mixed with fuel to form a high explosive (e.g., ANFO). But apparently there have been accidental explosions involving just the "fertilizer". Are these explosions also detonations? What is the chemical formula for the process?
$$\ce{NH4NO3 -> ???}$$
Part of my motive for asking is the news today (Aug 4, 2020) of an explosion in Beirut. Initial reports say it was caused by "2750 tons of stored ammonium nitrate".

It is known that ammonium nitrate decompose exothermically when heated to form nitrous oxide and water. This paper1 notes that the irreversible decomposition of ammonium nitrate occurs at the temperature range of $\pu{230-260 ^\circ C}$.
$$\ce{NH4NO3 ->[t >230 ^\circ C] N2O + 2H2O}$$
They also further noted that beyond $\pu{280 ^\circ C}$, $\ce{NH4NO3}$ is  capable of rapid, self-accelerating decomposition (to the point of detonation).
But at the detonation temperature, $\mathrm{t_d}$ (the temperature at which compounds detonate), ammonium nitrate fully decomposes to nitrogen, oxygen and water releasing a tremendous amount of energy.
$$\ce{2NH4NO3 ->[t_d] 2N2 + O2 + 4H2O}$$
In the context of Beirut explosion, the question that raised was "when did ammonium nitrate reached detonation temperature, and why did it suddenly explode?". According to a news report from cnet.com:

When heated to above 170 degrees Fahrenheit, ammonium nitrate begins
to undergo decomposition. But with rapid heating or detonation, a
chemical reaction can occur that converts ammonium nitrate to nitrogen
and oxygen gas and water vapor. The products of the reaction are
harmless -- they're found in our atmosphere -- but the process
releases huge amounts of energy.[...]
Additionally, in the explosion, not all of the ammonium nitrate is
used up and exploded. Some of it decomposes slowly creating toxic
gases like nitrogen oxides. It's these gases that are responsible for
the red-brown plume of smoke seen in the aftermath of the Beirut
explosion, Rae said.

So, my theory is that ammonium nitrate started heating (from the fire) releasing all sorts of nitrogen oxides (the red fumes). This fire further accelerated the reaction, further heating the remaining ammonium nitrate to the point of detonation and that's when ammonium nitrate exploded instantaneously releasing tremendous amount of energy which send shockwaves around the site along with a white mushroom shaped cloud (from @DDuck's comment) which could probably be nitrogen and/or water vapours where the humid air (water vapor laden air) condensed due to the explosion(@StianYttervik) with release of nitrogen. It is a sad and quite devastating incident.
References

On the Thermal Decomposition of Ammonium Nitrate. Steady-state Reaction
Temperatures and Reaction Rate By George Feick and R. M. Hainer, 1954 (PDF)
Reaction Rates of Ammonium Nitrate in Detonation
Melvin A. Cook, Earle B. Mayfield, and William S. Partridge
The Journal of Physical Chemistry 1955 59 (8), 675-680
DOI: 10.1021/j150530a002 (PDF)
https://en.wikipedia.org/wiki/2020_Beirut_explosions

edit: my original assesment was wrong
It does decompose ENDOTHERMICALLY, so it's not self propagating, and is not a low explosive:
NH4NO3 - > 2H2O + N2O
However a passing shockwave can induce this decomposition in a large quantity of ammonium nitrate at once, which is a detonation. The starting shockwave could have come from another explosive or from a fuel fire that was intensified by the oxidizing properties of ammonium nitrate. Not sure about exactly how the latter happens, but this is not the first time it did
https://en.m.wikipedia.org/wiki/Ammonium_nitrate_disasters


What determines the atom or molecule that receives electrons during oxidation? I'm no chemist, so bear with me.

From my understanding, the reason metal oxidation occurs is because the metal is losing electrons to the oxygen ions in the air; and that an ion is an atom or molecule which either has more or less electrons than the "standard" of that atom/molecule; which is why it's "taking" electrons from something else.

What prompts my question has to do with soldering. I'm learning that one reason for the use of flux is to prevent this oxidation and protect the soldering tip. So according to what I said before, a soldering iron becomes oxidized because it has lost electrons to the air, and the high temperatures hasten this process.

Now to get to the detailed question: I understand the very basic oxidation process. But, once the metal has lost electrons, doesn't that now make it composed of ions? If so, Why does the metal continue to corrode? Why isn't the metal now taking electrons and virtually "restoring" itself? What is it about oxygen ions that make them the ones that take electrons from metals such as copper? Why isn't copper the one taking electrons from the air? Does it have to do with valence electrons or electron shells or something?

When a metal like iron is oxidized, each iron atom loses 3 electrons. These electrons are kept by oxygen atoms from the air. The iron atom becomes an ion charged 3+, and the oxygen atom becomes an ion charged 2-. The positive ions still attract negative ions, and they will never separate. Two iron ions and three oxygen ions are necessary to make a new compound which must be not charged. This assembly is a new substance called rust, which has a brown color. Once such a transformation has happened, the next iron atoms can repeat this operation : the amount of rust will slowly increase.

All metals can be oxidized by more or less the same way. But it could be quicker or longer.  For copper, this reaction is not so frequent. Copper makes a copper oxide which is black, but you may wait for months to see it.


Do the mechanism (SN1 or SN2 ) which reaction follow depends more upon intermediate stability than on substrate and nucleophilic? For eg : would the reaction of 3-methylbut-2-en-1-ol with HBr follow SN1 mechanism due to stability of your intermediate carbocation stability  or SN2 due to the compound being primary alcohol and good nucleophilic. Also please correct me if I got something wrong.

It will follow SN1 as the intermediate carbocation formed is resonance stabilised.


Does CN4 with the following structure exist? I am a mathematician, not a chemist, so my questions are rather combinatorial. I am intrigued by the possibility of the following binary compound:
$$\ce{\overset{-}{N}=\overset{+}{N}=C=\overset{+}{N}=\overset{-}{N}}$$
which ought to be isoelectronic with carbon suboxide, and thus possibly stable. Does this compound exist, even in interstellar space?

All kinds of weird molecules exist in space. The molecule is formally known as bis(diazo)methane according to chemdraw. By looking at the structure it contains a Diazomethane moiety. Resonance would also take place to introduce the charges at both nitrogen atoms.  The structure doesn't seem to exist on any chemical database nor any official papers have been released relating to the exact structure. It may exist considering the fact that all atoms have a total of 8 surrounding electrons.
I hope this provides some clarity


Isothermal compression at critical temperature Suppose I am compressing a gas isothermally at its critical temperature. Initially, it has a large volume. I start compressing it and reach the critical point. At this point, the gas will start to form liquid and both of these phases cannot be distinguished, right? Now if I further compress it isothermally at its critical temperature so that the pressure reaches beyond critical pressure, what will my gas look like? Will it be a liquid or a gas or something in between, I don't know. Is there a video where I can see something of this sort? Just curious

https://youtu.be/RmaJVxafesU?t=242

carbon dioxide has a critical point at 31 degrees celsius.


Is it possible to curve fit GC, Mass-spectrometry and NMR peaks? I am trying to measure the amount of compounds by the peak area and I want to increase accuracy to 0.1% by curve fitting. Can I do it and what functions should I use?

You should not assume any given peak shape as in practice this may not be what it should be from theory.

You have a set of data points so once you have determined the baseline a numerical integration should work well. There are many of these in the literature. When fitting use the chi squared statistic as a measure of fit, the Pearson R is just useless and should never be used in fitting. (This is because a poor fit has, say, an R = .99 and a good fit .999 so its not anywhere near discriminating enough). You should also look at the shape of the residuals, autocorrelation function, runs tests etc to make sure you have a good fit. A one parameter good/bad test is often not good enough.

If you have overlapping peaks then you will have to assume some shape simply to use non-linear least squares to estimate one peak from another. This will lead to some more uncertainty but depending on what you are trying to determine this may not matter.

In gc data the peak shape is not necessarily gaussian, often it is lopsided so a numerical method is better here.

It may not be necessary to fit your data to try to get such high precision, the data simply may not allow it due to instrumental factors. But what is important is results not data, i.e. its what the data tells you that is important; this may not need such high precision as you seek. I'm absolutely not being patronising, but I would go as far as to say that if you need 0.1% accuracy you should perhaps think of another way of trying to learn what you want to know.

Yes, you can curve fit all three of them. For NMR use lorentzian shape.For other two use Gaussian peak shape.

But remember: measurements in chemistry are typically 10% accurate. Increasing accuracy beyond that is difficult.

Mass-spectrometry is only quantitative if you study isotopic molecules. You cannot compare alcohol with an acid and amine using MS. Amine will dominate in positive mode and acid will dominate in negative mode.

GC is the most reliable but you have to find the right dose. Not to small, so that your peak is visible, and not to high to avoid asymmetry. Tecknically, you can use asymmetric fitting curve, but those fits are less stable (less reliable, even though the R is better).


Why does wood burn but not sugar? Fundamentally, they're both carbohydrates, although the cellulose in wood is essentially polymerized glucose, which combined with its isomer fructose forms sucrose.

So why does wood readily burn while table sugar chars?

Combustion is a gas phase reaction. The heat of the flame vapourises the substrate and it's the vapour that reacts with the air. That's why heat is needed to get combustion started.

Anyhow, wood contains lots of relatively volatile compounds so it's not too hard to get combustion started. Once combustion has started the heat of the flame keeps the reaction going.

However sugar dehydrates and emits water when you heat it. Water isn't flammable (obviously) so there's no way to get combustion started. Dehydration leaves behind pure carbon and that is non-volatile so again there's no way to get this to burn. Carbon will burn of course, but you need a high temperature to get it going.

The oxygen (chemically reduced oxygen; that is, essentially, water) content of sugar is very high. It's already highly oxidised.

$$
\begin{array}{llr}
\hline
\text{Compound} & \text{Formula} & \text{Oxygen}/\text{wt %} \\
\hline
\text{Glucose} & \ce{C6H12O6} & 53 \\
\text{Saccarose} & \ce{C12H22O11} & 51 \\
\text{Cellulose} & \ce{C12H20O10} & 49 \\
\text{Dry wood} & & \sim 35 \\
\hline
\end{array}
$$

The oligo- and polysaccharides have lost water after polymerisation, but not very much. Cotton is not very flammable. That's why labwear is required to be made of cotton (plus it doesn't melt and then stick to your skin).

Wood however is not just a carbohydrate, but has a high content of lignin, which contains much less oxygen and has a sum formula in the range of $\ce{C3H3O}$.

Very simply put, you have to boil a lot of water out of sugar before anything can actually burn. If you look at the sum formulas above, about two thirds of the mass and all the hydrogen are gone at that point. Pyrolysis of sugar gives water + carbon, and very little flameable volatiles. Which are what you see on burning wood. When all the volatiles are pyrolysed out of the wood, the remaining charcoal is hot enough that it keeps burning.

Sugar has the additional problem that it melts, which makes it burn even worse, because the grainy or powdery structure fuses into a closed surface with good internal heat conductivity. There is an old trick, mixing sugar with ash, which makes it burn a lot better. Cellulose doesn't melt, that's why paper burns much better than sugar. (And because it's thin, a complete book is not so easy to burn completely.)

Btw. the autoignition temperature of sugar (probably saccharose) is ~410°C, wood 280-340°C. (from de:wp, without reference unfortunately)


Is beryllium difluoride covalent or ionic? My textbook  says that despite the large electronegativity difference $\ce{BeF2}$ is covalent since the beryllium ion will have too much charge density and it will attract the fluorine electron cloud and therefore forms polar covalent bonds.

I looked it up on Wikipedia and found out that it has a crystal lattice,  is soluble in water, and has a quite high melting point.

What else does one need to call a compound ionic? Am I or the textbook right?

There is a very relevant article by Gillespie Covalent and Ionic Molecules: Why Are BeF2 and AlF3 High Melting Point Solids whereas BF3 and SiF4 Are Gases? J. Chem. Educ., 1998, 75 (7), p 923.

According to the article, the charge on Be is +1.81 and the charges on the Fs are -0.91. (citing to his earlier article Reinterpretation of the Lengths of Bonds to Fluorine in Terms of an Almost Ionic Model Inorg. Chem., 1997, 36 (14), pp 3022–3030)

So while nothing is completely covalent or ionic, BeF2 is extremely toward ionic.

According to my chemistry book (Pearson Baccalaureate IB) beryllium forms covalent bonds due to it being one of the expectations to the intramolecular bonding.

Furthermore, according to the book the molecule should be non-polar because all the lone pairs are bonded and thus not exerting a repulsion and altering bond angles.


What causes the SN1 reaction? The SN1 is a two-step mechanism which begins with heterolysis of the C-X bond. What triggers this? Are there external causes that prompt the leaving group to simply leave?

This question is more conceptual than anything, but I can't seem to find answers anywhere.

Recall that SN1 heavily favors protic solvents (polar/H+). In such cases, the nucleophile to-be-leaving-group (to be Nu-) is going to be attracted to the positively charged molecules of the protic solvent (...H+) and away from the molecule it is attached to. The power of the attraction comes from the dipole between C-Nu, where Nu has the negative bit of the dipole.

At the end of this step, when the nucleophile attached group leaves, you often see "Nu-" hanging around, but really it's with the solvent, e.g. H3O+Br- , and it's fairly common to omit the solvent here and in other similar ionic situations, if I'm not mistaken.

Examples that invoke hydrogen bonding make this more apparent.

F-C...(tertiary) + H2O

The dipole here looks like (δ-)F-C...(δ+) because F is more electronegative

The F is strongly attracted to the closer H: (δ+)H-(δ-)O-H(δ+)(δ-)F-C...(δ+) (pretend the HOH is bent :)

Greater nucleophilicity of the attacking group than the nucleophilicity of the leaving group causes SN1 reactions.
The cleavage of the C-X bond keeps occuring always, both the reactant and the product remaining in the solution. The equilibrium shifts towards the final product.


Electrolysis of aqueous copper (II) nitrate There are two copper blocks sitting in the $\ce{Cu(NO3)2 (aq)}$ solution, a battery is attached onto both of them, providing enough energy to start the reaction.

Since solid pieces of copper are involved, $\ce{Cu}$ must be considered in the reduction potential as well.

However, looking at the half reaction for copper:
\begin{align}
  \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}&
  (E^\circ &= \pu{+0.34 V})
\end{align}

Compared to nitrate:
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) + 3e- &-> NO (g) + 2H2O (l)}&
  (E^\circ &= \pu{+0.96 V})
\end{align}

Since it produces more energy, I believe the second half reaction will occur instead.

I begin by finding all the species/ions I have:

$\ce{Cu^2+}$
$\ce{NO3^-}$
$\ce{H2O}$

Taking a look at the Standard Reduction Potentials at 25°C:

For my SOA (strongest oxidizing agent) half-reaction ($\ce{NO3^-}$):
\begin{align}
  \ce{NO3^- + 4H+ + 3e- &-> NO + 2H2O}&
  (E^\circ = \pu{+0.96 V})
\end{align}

As for my SRA (strongest reducing agent) half-reaction ($\ce{H2O}$):
\begin{align}
  \ce{2H2O &-> O2 + 4H+ + 4e-}&
  (E^\circ &= \pu{-1.23 V})
\end{align}

For the balanced reaction I have:
\begin{align}
  \ce{4NO3- + 4H+ &-> 4NO + 2H2O + 3O2}&
  (E^\circ &= \pu{-0.27 V})
\end{align}

Some observations that can be made are that:

Nitric gas is forming at the anode
Oxygen gas is forming at the cathode
At least $\pu{0.27 V}$ must be put in

My question is:

Did I do all of this correctly? There's no answer key and I'm pretty sure that I might have made a mistake somewhere.
What would happen if you placed a necklace at the cathode?

There is a misunderstanding in the analysis of your first two half-reactions. Your first two half-reactions are fine. Remember that positive values of $E$ mean the reaction is spontaneous. Negative values of $E$ mean the reaction is nonspontaneous, since

$$\Delta_\mathrm{r}G = -nFE_\mathrm{cell}.$$

Your first half-reaction is an oxidation (as written). You have the wrong sign on that $E$ value.
\begin{align}
  \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}&
  E^\circ_\mathrm{ox} &= \pu{-0.34 V}
\end{align}

Your second half-reaction is a reduction (as written)
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) +3e- &-> NO (g) + 2H2O (l)}&
  E^\circ_\mathrm{red} &= \pu{+0.96 V}
\end{align}

The combination of these two half-reactions produces a positive $E^\circ_\mathrm{cell}$ which is a spontaneous reaction. You do not need to use the oxidation of water as your oxidation half-reaction. Copper is a much better reducing agent than water:

$$
\ce{3Cu (s) + 2NO3- (aq) +8H+ (aq) -> 3Cu^2+ (aq) + 3NO (g) +4H2O (l)}\\
E^\circ_\mathrm{cell} = \pu{+0.96 V} +(\pu{-0.34 V}) = \pu{+0.62 V}
$$

How does this change your analysis?

This seems to be a thought experiment, and many factors have been considered, but not enough to correspond to reality.
One item to consider is that copper is electrolytically purified by psssing current between a thin cathode of pure copper and a thick cathode of impure copper. An electrical power supply drives the copper (in solution) from the anode (+) to the cathode (-), and less reactive metals precipitate while more reactive metals remain in solution. Sulfate is the most usual anion.
But nitrate could be used as well. Oh, but nitrate is an oxidizer, so it complicates the issue. Perhaps. At high concentrations (say, >25%), nitric acid is such an oxidizer that it will passivate iron, and not dissolve it. Reduce the acid concentration to 10% or less, and iron will dissolve in it, because the acidity function outweighs the oxidizing function. So let's consider our copper nitrate to be at a moderate concentration (~10%), and without added acid ($H^+$), which is an important reactant in all the oxidation reactions.
With pH near neutral (pH = 7), the potential of the nitrate reduction half cell drops from +0.96 volts to +0.408, so then the combined cell voltage is only +0.068. What this means is that a copper block in a solution of copper sulfate will corrode slightly, consuming hydrogen ion. The reaction will not proceed as far as developing pH = 8 before stopping. The copper blocks would show a bit of staining or etching.
When an external potential is applied, one reaction is that $Cu^{+2}$ is forced off the anode into solution and some other $Cu^{+2}$ in the solution is driven onto the cathode to maintain neutrality everywhere. But nitrate ion is a a bit more active than a mere spectator ion and some will be reduced at the anode to $NO$. The pH will increase because $H^+$ is consumed.
At the cathode, copper will plate out, not oxygen. I can imagine copper hydroxide being precipitated at the anode and $NO$ gas being liberated there as well, but it could get more complicated, as $NO$ reacts with $OH^-$.


Regenerating a Cobalt Acetate Catalyst I have an eluent from an ion-exchange column containing Cobalt (II) ions and Chloride ions. In order for the cobalt catalyst to be recycled to the main process, it must be in its acetate form.
Is there a route from cobalt chloride to acetate that is feasible at an industrial scale?

Thanks

Quick Laboratory method:

$$\ce {CoCl2(aq) + 2NaOH(aq)->Co(OH)2(s) + 2NaCl(aq)}$$
$$\ce {Co(OH)2(s) + 2CH3CO2H(aq)->Co(CH3CO2)2(aq) + 2H2O(l)}$$
$$\ce {Co(CH3CO2)2(aq) + \Delta->Co(CH3CO2)2\cdot 4H2O (s)}$$

You should separate $\ce{Co(OH)2}$ by filteration and washed with water ($\pu {K_{sp} = 1.0 \times 10^{-15}}$) to remove all $\ce{NaCl}$ before subjected to react with $\ce{CH3CO2H}$.

This might work in industrial scale, if be careful. $\ce{Co(OH)2}$ is unstable to heat but no need to dry before react with $\ce{CH3CO2H}$.

add sodium sulphide solution -> precipitate  cobalt sulphide complex ( highly stable ) -> filter - dissolve in acetic acid


Find the concentration of the HCl solution when mixed with 1.265g of ACES-K+ $\pu{1.265 g}$ of N-(2-acetamido)-2-aminoethanesulfonic acid potassium salt ($\ce{ACES^-K+}$, $M=\pu{220.29 g/mol}$) is dissolved in $\pu{88.42 mL}$ of water. $\pu{27.59 mL}$ of $\ce{HCl}$ is added to the solution, resulting in $\mathrm{pH}$ of $6.54$. Calculate the concentration of the $\ce{HCl}$ solution. The $\mathrm pK_\mathrm a$ of $\ce{ACES}$ is $6.85$.

I tried setting up the equation as follows:
$$\ce{ACES + H3O+ <=> HACES +H2O}$$

The moles of ACES initial is $\pu{0.0057 mol}$ $(1.265/22.290)$ and we are trying to find $x$ the amount of moles of $\ce{HCl}$ $\ce{(H3O+)}$ we have. Therefore, using an ice table $\ce{ACES}=0.00574-x~\pu{mol}$ and $\ce{HCl}= x~\pu{mol}$.
When you plus into the equation
$$\mathrm{pH}=\mathrm pK_\mathrm a+\log(\text{base}/\text{acid})$$ I did $6.54=6.85+\log(0.00574-x/x)$ and got $\pu{0.0038 mol}$ of acid divided by total volume = $0.03321\ \pu M$. Is that correct?

Let us call the potassium salt "A-" and its corresponding acid for "HA".
The total volume of the final solution is 88.42 + 27.59 = 116.01 ml.
The final concentration of [HA] + [A-] = 1.265/(220.29 *0.11601) = 0.049499 (M).
[K+] = [HA] + [A-] = 0.049499 (M).

The equilibrium equation gives:
6.54 = 6.85 + log ([A-]/[HA]) => [A-]/[HA] = 0.48977 => [A-] = 0.48977 * [HA].
=> [HA] + 0.48977 * [HA] = 0.049499 => [HA] = 0.03322 (M) and [A-] = 0.016279 (M).

The charge balance is:
[H3O+] + [K+] = [OH-] + [Cl-] + [A-]
At pH = 6,54 [H3O+] = 10^-6.54 and [OH-] can be neglected in comparison with [A-].
=> [Cl-] = [H3O+] + [K+] - [A-] = 0.03322 (M).

The amount of chloride ions in the final solution will be 0.03322 * 0.11601 = 0.003853 mole. Thus, the concentration of the hydrochloric acid used will be:
(0.003853 * 1000) /27.59 = 0.1396 or about 0.140 M.


How to write equations for dissociation of ionic compounds in water? Please write a dissociation-in-water equation for the compounds $\ce{BeI2}$ and $\ce{LiI}$. Make sure to add the states of matter after each compound.

Currently, for $\ce{BeI2}$ I have the equation $$\ce{BeI2 (s) -> Be^2+ (s) + I2^2- (g)}.$$ I have yet to attempt the second one.

Your ionic charges are not correct for iodine. Looking at your attempt.

As $\ce{Be}$ is in group 2, the ionic charge for beryllium ion is fine, but iodine is in group 17, so its ion is $\ce{I-}$.  When the ions dissociate, they become aqueous or (aq) as the state of matter.

Then the ionic charges need to balance, thus:

$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$

To balance the ionic charges in this example, you need 2 $\ce{I-}$ (iodine ions) to balance the $\ce{Be^{2+}}$ (beryllium ion).

A similar example (and further explanations) are provided on the UC Davis ChemWiki page Unique Features of Aqueous Solutions (including an example of the dissolution of $\ce{MgCl2}$ - another compound with group 2 and 17 elements).

So,

determine the group, hence ionic charge of each dissociated ion
balance these charges
state that the dissociated ions are aqueous

Now, use the process to determine the dissociation of $\ce{LiI}$

By definition, a dissociation-in-water reaction results in aqueous, independent ions, not negatively charged diatomic molecules like the $\ce{I_2^2-}$ in the original question (not to mention that diatomic molecules are generally not negatively charged).

The answer should result in solely single-atom ions, as so:

$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$

And using this same logic for $\ce{LiI}$:

$$\ce{LiI (s) -> Li+ (aq) + I- (aq)}$$


Is there any method to synthesize salicylaldehyde from salicylic acid? I know salicyclic acid can be produced from salicaldehyde by treatment with $\ce{KMnO4}$.

Is there any way to reverse this, i.e- produce salicaldehyde from salicyclic acid?

Salicylic acid can be converted to salicylaldehyde using a triangular or square wave alternating voltage. The process of this conversion is explained in this paper$\ce{^{1}}$but it is behind paywall so here's the abstract:

  The effect of superimposing sinusoidal, triangular, and square‐wave
  alternating voltage (aV) on the potentiostatic reduction of salicylic
  acid to salicylaldehyde has been studied. Polarization curves with aV
  modulation were measured with an amalgamated copper rotating disk
  electrode. Enhanced dc current densities were observed with increasing
  magnitude of superimposed aV. However, aV frequency and type did not
  have significant effect on the resulting dc current densities. Batch
  cell electrolysis experiments were conducted with a H‐type cell. The
  conversion of salicylic acid and yield of salicylaldehyde were
  measured as functions of electrolysis time, aV magnitude, type, and
  frequency. The superimposed aV significantly enhanced the reaction
  rate, conversion of salicylic acid, and the yield of salicylaldehyde.
  The energy consumption of ac electrolysis was higher than that of dc
  electrolysis.

doi: 10.1149/1.2100342 J. Electrochem. Soc. 1987 volume 134,issue 12, 3074-3077

It's simple. Reduce with LAH then oxidise it with PCC or PDC.


Is there an easy way to find number of valence electrons" I want an high-school level answer. What I mean with "valence electrons" is the outermost electrons in that atoms' electronic arrangement? For example, 3 in an atom with an electronic arrangement of 2, 8, 3.)

Without actually writing the electron configuration, or orbitals, and knowing electron number, is there any way to find this number easily?

Also, is there an easy way to find if certain atom fits "atomic/orbital symmetry" given the numbers of electrons? (I don't know the english words for this and couldn't find them. So I translated it literally, but what I mean is, if orbital is half-full or full i.e  $\mathrm{s^1}$ or $\mathrm{2p^3}$ or $\mathrm{2p^6}$ and so on... hope you understand what I am trying to say)

I am asking this because sometimes questions give me electron numbers in a range of 40+ and just writing it out doesn't seem like a good way.

You can get the valence electrons in an atom's electronic arrangement by consulting the periodic table:

The Group 1 atoms have 1 valence electron.
The Group 2 atoms have 2 valence electrons.
The Group 3 atoms have 3 valence electrons.
The Group 4 atoms have 4 valence electrons.
The Group 5 atoms have 5 valence electrons.
The Group 6 atoms have 6 valence electrons.
The Group 7 atoms have 7 valence electrons.

The atom with electric configuration 2, 8, 3 is Aluminium and you can find it in Group 3 of the periodic table.

The periodic table is usually given in exams so...

The first two groups of the periodic table are the groups concerned with the filling of the $s$ orbital. So, all Group 1 atoms will have an outermost $s$ orbital filled with a single electron.

Similarly, Group 3 - Group 8 (Group 8 is also referred to as Group 0) are where the $p$ orbital is being filled up. All the Group 5 atoms thus all have an outermost $p$ orbital filled with 3 electrons.

The Periods in the periodic table can be used as an indication of the number of shells the atom has. Sulfur for instance is in Period 3 and thus has 3 shells.

Yes there is! For all elements in the following groups;

Group 1A= +1
Group 2A= +2
Group 3A= +3
Group 4A= +4 or -4
Group 5A= -3

From group 5, the valency descends in this order;

-3, -2, -1, 0.

Group 8A elements have a valency of Zero.

I hate complicated stuff too lol.


The real structure of difluoro(dioxo)xenon (XeO2F2) I have a problem regarding structure of difluoro(dioxo)xenon ($\ce{XeO2F2}$). I have posted a image above. my teacher is saying that the structure B is correct as there is very less repulsion in structure B that is what I too think but many of the sources and books have written the structure A to be correct and me as well as my teacher is in a doubt that why structure A is correct not structure B as there is an angle of 180° between oxygen atoms. If I am wrong please guide me and also please tell me what are factors should I look at for judging a correct shape for a compound.
Any hint will be very very helpful to me .

Xenon is a noble gas and thus does not need to form any bonds to complete its octet. When such a bond is formed, this is known as a hypercoordinate structure: it contains more bonding partners than the most simple of simple Lewis structure theories would allow. Thanks to Linus Pauling, for many years chemists have been trying to invoke d orbitals to explain this bonding situation when the actual case is far more trivial and can be explained with simple octet considerations.

The two oxygen atoms which are depicted as double-bonded in the d orbital model are actually connected by a single bond. If you want, this bond can be formed by oxygen keeping its six valence electrons in three orbitals and using the forth, empty orbital to bond with a single orbital of xenon. Both electrons can be described as originally donated by xenon and a good way to draw it in the Lewis formalism is to use charge separation:

$$\ce{\overset{-}{O}-\overset{2+}{Xe}-\overset{-}{O}}$$

This leaves the two fluorine atoms. They actually form a slightly more complex bond known as a 4-electron-3-centre bond (thus fluorine only bonds to xenon in neutral compounds in multiples of 2). For a deeper explanation of the 4-electron-3-centre bond, see this previous answer of mine. In a Lewis formalism, this can only described with resonance structures as shown below:

$$\ce{F-\overset{+}{Xe}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Xe}-F}$$

This bonding situation means that the fluoride atoms must occupy opposite positions with $\angle(\ce{F-Xe-F}) \approx 180°$. Thus, A is the correct depiction of  $\ce{XeO2F2}$ which was confirmed in a neutron diffraction experiment by Peterson, Willett and Huston.[1]

The paper determines the $\ce{O-Xe-O}$ angle as approximately $105°$. This is somewhat in-between a pure p orbital and an $\mathrm{sp^2}$ description of xenon, $15°$ from either. At first approximation it would, however, be fine to assume unhybridised xenon and derive the bonding situation from there with each oxygen bonded to one of the p orbitals and the two fluorines bonding to the third.

Reference:

[1]: S. W. Peterson, R. D. Willett, J. L. Huston, J. Chem. Phys. 1973, 59, 453–459. DOI: 10.1063/1.1679827.

Think for yourself. The sole objective of the VSEPR Theory is to determine the nearest shape of a molecule such that there is minimum repulsion. In this example as we can see, we have to take into account lp-lp and lp-bp repulsions. Since Fluorine is more electronegative than Oxygen, the electronic cloud density is more in the former. Hence, it is more favorable to put the F-atoms in the equatorial plane with the O-atoms occupying the axial plane. This is to reduce the bp repulsions with the lp : Option B) correct.


How to determine the pH of a mixture of two weak acids? We have two solutions:

Solution 1 is $\ce{HCOOH}$, its concentration is $c_1 = \pu{10^-2 mol/l}$, its volume is $V_1 = \pu{50 ml}$, and its $\mathrm{pH}_1 = 2.9$.
Solution 2 is $\ce{CH3COOH}$, its concentration is $c_2 = \pu{10^-2 mol/l}$, its volume is $V_2 = \pu{50 ml}$, and its $\mathrm{pH}_2 = 3.4$.

How would be the equation and the ICE table, and what is the $\mathrm{pH}$ of the mixture of these two solutions?

I used numbers to differentiate the solutions, and to understand how it works, but for no other reason.

I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work.

Below a more general approach.

Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:

Reactions

\begin{align}
  \ce{HA + H2O &<=> H3O+ + A-} &
  K_{\mathrm{a},(\ce{HA})}
    &= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
  \tag{1}\label{eq:KAcidHA}\\
  \ce{HB + H2O &<=> H3O+ + B-} &
  K_{\mathrm{a},(\ce{HB})}
    &=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
  \tag{2}\label{eq:KAcidHB}\\
  \ce{2 H2O &<=> H3O+ + OH-} &
  K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}

Mass balance

\begin{align}
  C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
  C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}

Charge balance

$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$

Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
  \frac{K_\mathrm{w}}{\ce{[H3O+]}}
  +\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HA})}}
  + \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$

or as polynomial

\begin{align}
\begin{split}
 &\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
  - K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}

This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$, $K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}$, and $C^0_\ce{HB}$ values, we can yet obtain a simpler version[1].

The \eqref{eq:GeneralPol} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll \ce{[H3O+]}^4$:

\begin{align}
\begin{split}
 &\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
  - K_\mathrm{w} \right]\\
-&\left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  \right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}

The \eqref{eq:GeneralPolSimp1} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll \ce{[H3O+]}^3$:

\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}

The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.

$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
      + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})}
}{2}$$

Or using the initial concentrations

$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4\left(%\displaystyle
      \frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
        }{V_\ce{HA} + V_\ce{HB}}
      + \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
        }{V_\ce{HA} + V_\ce{HB}}
    +K_\mathrm{w}\right)}
}{2}$$

Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid[3], we have
$$\ce{pH}=3.05.$$

Addendum

@MaxW made some warnings about the assumptions which simplificate ($\ref{eq:GeneralPol}$)(see comment) and ($\ref{eq:GeneralPolSimp1}$)(see comment) equations.
To completeness and with the aim of show the limitations of present approach, consider the following graphics.
The ($\ref{eq:GeneralPol}$), ($\ref{eq:GeneralPolSimp1}$), and ($\ref{eq:GeneralPolSimp2}$) equations were solved using $\text{p}K_{\mathrm{a},(\ce{HA})}=\text{p}K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}=C^0_\ce{HB}$, and $V_\ce{HA}=V_\ce{HB}=\pu{50.0 mL}$.

As we can see, the ($\ref{eq:GeneralPolSimp1}$) equation can be used on any $\text{p}K_\ce{a}$ values, inasmuch as the concentration be major of $\pu{1E-6 mol L-1}$.
The ($\ref{eq:GeneralPolSimp2}$) equation becomes problematic when dilute solutions or strong acids solutions are used.

Thus, when applied to book-examples, which $\text{p}K_\ce{a}$ values from 3 to 6, and the concentrations from $1.0$ to $\pu{1.0E-4 mol L-1}$, the assumptions are valid and the equations can be used, providing good approximations of $\ce{pH}$ values.

Acknowledgement

Thanks to @MaxW for draw attention to limitations of assumptions.

Reference

Edited based on first @MaxW's comment.
Edited based on second @MaxW's comment.
Haynes, W. M.;  Lide, D. R.; Bruno, T. J., CRC Handbook of Chemistry and Physics. 97 ed.; CRC Press: 2017; pp. 5–88.

$-\log[\ce{H}]=\text{p(H)}$ you can calculate concentration of $[\ce{H}]$ for each acid, then add the $[\ce{H}]$ for two acids and calculate $\text{p(H)}$ by $-\log[\ce{H}]$.


Is the name of this chemical 3-oxopropanoic acid? Is the name of this chemical 3-oxopropanoic acid?

I included "oxo"- for the aldehyde group, and "-oic acid" for the carboxylate group.

However, I'm unsure whether this is the correct name since I cannot find it on the internet, while I can find almost all other names I've been trying to figure out.

The name ‘3-oxopropanoic acid’ for the compound $\ce{OHC-CH2-COOH}$ given in the question is correct.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  15. aldehydes
  (…)

Thus, the carboxylic acid group has priority to be cited as a suffix.

Systematic names of carboxylic acids are formed as follows:

  P-65.1.2 Systematic names

  Carboxylic acids are named substitutively using the suffix ‘oic acid’ or ‘carboxylic acid’ and the prefix ‘carboxy’ to describe carboxy groups that cannot be included as the principal characteristic group or to express the carboxy group in the presence of a higher principal characteristic group.

  P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; (…). Except for (…), systematically formed names are preferred IUPAC names; (…)

Therefore, the preferred IUPAC name (PIN) of the unmodified carboxylic acid is ‘propanoic acid’. (Note that the retained name ‘propionic acid’ may be used in general nomenclature.)

Aldehydes can be systematically named in various ways. Monoaldehydes derived from alkanes are preferably named substitutively using the suffix ‘al’. In this case, however, this method cannot be used since the carboxylic acid group has priority to be cited as a suffix. Therefore, the aldehyde group has to be expressed by a prefix as follows:

  P-66.6.1.3 In the presence of a characteristic group having priority to be cited as a suffix or when present on a side chain, a $\ce{-CHO}$ group is expressed by the preferred prefix ‘oxo’ if located at an end of a carbon chain, or, otherwise, by the preferred prefix ‘formyl’.

Thus, the PIN for the complete structure is ‘3-oxopropanoic acid’.

The corresponding name that is formed by using the prefix ‘formyl’ is ‘formylacetic acid’. Such names are not acceptable as PINs but may be used in general nomenclature.

Furthermore, a special rule for aldehydic acids applies to the given compound:

  P-65.1.6.3 Aldehydic acids

  When a dicarboxylic acid has a retained name (…) and when one of its carboxy groups is replaced by a formyl group, $\ce{-CHO}$ (…), the resulting structure is called an aldehydic acid and, in general nomenclature may be named by replacing the ending ‘ic acid’ of the name of the dicarboxylic acid by the ending ‘aldehydic acid’. Preferred IUPAC names for aldehydic acids derived from all dicarboxylic acids are constructed systematically. The prefix ‘formyl’ is used in preferred IUPAC names, except for a $\ce{-CHO}$ group at the end of an acyclic chain, which is designated by the prefix ‘oxo’.

For the compound given in the question, the PIN of the corresponding dicarboxylic acid is ‘propanedioic acid’.

This dicarboxylic acid also has the retained name ‘malonic acid’, which may be used in general nomenclature. Hence, according to P-65.1.6.3, the corresponding aldehydic acid may be named ‘malonaldehydic acid’ in general nomenclature.

By way of comparison, for the analogous example $\ce{OHC-CH2-CH2-COOH}$, all three corresponding names are explicitly mentioned in the Blue Book:

4-oxobutanoic acid (PIN)
3-formylpropanoic acid
succinaldehydic acid

3-Oxopropanoic acid (if carbon of aldehyde is numbered, use "oxo" as prefix) is correct IUPAC-name for the given chemical compound.

Formylethanoic acid (if carbon of aldehyde is not numbered, use "formyl" as prefix) is also correct IUPAC -name.


About the use of pascal or atmosphere in chemistry I was wondering if there are specific formulae in chemistry for which it is "compulsory" or convenient to use atmosphere (atm) rather than pascal (or even mmHg) as a unit of measurement.

Of course, we can convert one into the other, but the choice of using one of the two also conditions other units, such as in the case of volume.

Can someone provide me with a good picture of the relationship between these units and the others that we may need to use in the same context? (e.g. we use L  with atm; $\mathrm{m^3}$ with Pa).

Generally, the use of quantity equations is preferred and is strongly recommended. They have the advantage of being independent of the choice of units.

For example, instead of “speed in kilometre per hour is 3.6 times the quotient of distance in metres by time in seconds”, which corresponds to the numerical value equation

$$v/(\mathrm{km/h})=3.6\frac {l/\mathrm m}{t/\mathrm s}$$ or $$\left\{v\right\}_{\mathrm{km/h}}=3.6\frac{\left\{l\right\}_{\mathrm m}}{\left\{t\right\}_{\mathrm s}}$$

you should write “speed $v$ is the quotient of distance $l$ by time $t$”, preferably in form of the quantity equation

$$\displaystyle v=\frac lt$$

Generally, you can work with any units you like; however, this requires the use of conversion factors. The use of conversion factors is never required when only coherent SI units are used; therefore, the use of coherent SI units (the SI base units and the SI coherent derived units, including those with special names) are preferred.

In order to avoid large or small numerical values, decimal multiples and submultiples of units may be formed with the SI prefixes. However, when prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor in the expression for the unit in terms of the base units.

Furthermore, various non-SI units are accepted for use.

In the International System of Quantities, pressure $p$ is defined as
$$p=\frac{\mathrm dF}{\mathrm dA}$$
Where $\mathrm dF$ is the force component perpendicular to the surface element of area $\mathrm dA$.

The coherent derived units in the SI for the pressure has the special name pascal and the symbol $\mathrm{Pa}$. According to the definition of pressure, the pascal can be expressed in terms of other SI units as $1\ \mathrm{Pa}=1\ \mathrm{N/m^2}$. The pascal can also be expressed in terms of SI base units as $1\ \mathrm{Pa}=1\ \mathrm{m^{-1}\ kg\ s^{-2}}$. The last form is especially useful to check whether units in equations and calculations are converted correctly.

The SI prefixes may be used with any of the names and symbols, but when this is done the resulting unit will no longer be coherent.

The bar (symbol $\mathrm{bar}$) is included in the list of non-SI units that are used by special interest groups for a variety of different reasons. Although the use of SI units is to be preferred for reasons already emphasized, authors who see a particular advantage in using these non-SI units should have the freedom to use the units that they consider to be best suited to their purpose. In particular, the bar is included because since 1982 one bar has been used as the standard pressure for tabulating all thermodynamic data. (Prior to 1982 the standard pressure used to be the standard atmosphere, equal to $1.01325\ \mathrm{bar}$, or $101\,325\ \mathrm{Pa}$.)

The following other non-SI units can be used, which requires the use of conversion factors. However, the use of these units is deprecated.

$$\begin{array}{lll}
\hline
\text{Unit} & \text{Unit} & \text{Conversion} \\
\text{name} & \text{symbol} & \text{factor} \\
\hline
\text{standard atmosphere} & \mathrm{atm} & 1\ \mathrm{atm} := 101\,325\ \mathrm{Pa}\\
\text{kilogram-force per square metre} & \mathrm{kgf/m^2} & 1\ \mathrm{kgf/m^2} = 9.80665\ \mathrm{Pa}\\
\text{technical atmosphere} & \mathrm{at} & 1\ \mathrm{at}=1\ \mathrm{kgf/cm^2} = 98\,066.5\ \mathrm{Pa}\\
\text{conventional millimetre of water} & \mathrm{mmH_2O} & 1\ \mathrm{mmH_2O}:=10^{-4}\ \mathrm{at} = 9.806 65\ \mathrm{Pa}\\
\text{conventional millimetre of mercury} & \mathrm{mmHg} & 1\ \mathrm{mmHg} \approx 13.5951\ \mathrm{mmH_2O} \approx 133.3224\ \mathrm{Pa} \\
\text{torr} & \text{Torr} & 1\ \mathrm{Torr}:=\frac1{760}\ \mathrm{atm} \approx 1\ \mathrm{mmHg} \approx 133.3224\ \mathrm{Pa} \\
\text{pound-force per square inch} & \text{psi}, \mathrm{lbf/in^2} & 1\ \mathrm{lbf/in^2}\approx6\,894.757\ \mathrm{Pa} \\
\hline
\end{array}$$

Note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted.

For example, write “the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”.

(The symbol $p_\mathrm e$ is recommended for gauge pressure, defined $p-p_\mathrm{amb}$, where $p_\mathrm{amb}$ is the ambient pressure. Thus the gauge pressure is positive or negative depending on whether $p$ is larger or smaller than $p_\mathrm{amb}$.)

In the formula Delta G° = - RT lnK, the numerical value of the equilibrium constant K must be in atmospheres.


At what temperature (in kelvin) are most of the elements on the periodic table liquids? This question is out of pure curiosity.

At what temperature are a majority of the elements on the periodic table in a liquid state/phase of matter? For the purpose of this question, assume the standard pressure (100,000 Pa). Is there some sort of formula or theorem that explains this? Please give a detailed explanation on your conclusion.

Now that's a great question indeed! Evidently, at 0K all elements except helium are solids, at 10000K they are all gases, so someplace in between the number of liquids must reach a maximum; what and where might that be?

Well, there is no formula or theorem that says liquid hydrogen must boil at 20K, nor is there such a thing for any other element, so this line of thinking hardly provides any insight. Let's try another way.

Let's take a look at all elements' melting and boiling points. (I used the great site referenced in a comment by ringo, where all data are conveniently stored on line 47 in the file http://www.ptable.com/Static/interactivity-2fcd37b.js). Let's simply make a joint list of all melting and boiling points, sort it (keeping track of which is which), and then walk through it from bottom to top, calculating the number of liquids on each interval. If you know some kind of programming, you can do it with a script.

It turns out that the maximum number of simultaneously liquid elements is 49, and it is reached several times between 2183K (melting of vanadium) and 2792K (boiling of aluminium). By dragging the slider you may actually see which elements are included in these 49. Now, the data are not terribly reliable (for example, in higher actinides they claim some melting points obtained by God-knows-what extrapolation, but no boiling points at all), yet still this is something to think about.

The boiling point of hydrogen is $-252.87~^{\circ}\rm C$ or $20.28\rm\,K$, which is the least boiling point so you can expect that all other elements would liquidify or solidify at or below this temperature.

But the concept of phases is much more complicated than that. Aside from three states (namely solid, liquid and gas) there are other states such as plasma, fermionic condensate, Bose–Einstein condensate and top-quark condensate.

At Bose–einstein condensate, at 170 nano-kelvin rubidium atoms coalesce to form one particle that acts like a wave forming Bose-gas.
So, from this example you can perceive that on super-cooling you can form gas. Hence, it is very difficult to specifically answer your question as technology improves nature tend to show its diverse beauty.


Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number? Why does the melting point get lower  going down the Alkali Metal Group with increase in atomic number?

You have to understand that melting point is related to the bonds in a metal between the metal cation and the 'sea of electrons'. With an increase in atomic number, you have an increase in electron shells.

As the radius of atoms get larger down the group, you could say that the force holding them together is spread over the greater area and hence, the metal cations are more weakly bonded. And coming back to my first point, when the bonding is weaker, the metal's melting point will decrease.

You have to know that elements with high lattice energy has high melting and boilng point.And since lattice energy is inversely related with radius of atoms it increases down a group and hence the melting point increases.
I hope this will help you!


Why IUPAC name this amine is written as 2-methylpropan-2-amine and not as 2,2-dimethylethanamine? Why IUPAC Nomenclature of following amine is written as 2-methylpropan-2-amine and not as as 2,2-dimethylethanamine?

The basic principle in nomenclature is to look for the longest carbon chain: here it is the chain with 3 carbon atoms so the suffix prop; since it is an amine connecting to the carbon 2 it gives the suffix propan-2-amine.
As you have a $ \ce {CH_3} $ group grafted in position 2 of the main carbon chain, you must add the prefix 2-methyl
So you have 2-methylpropan-2-amine.

It's so because while doing IUPAC naming we have to always choose longest branch where the number of functional group (here Amine) should be at least number. That's why we here call it 2-methylpropan-2-amine and not 2,2-dimethylethanamine. You can clearly see in the actual name it's propane which mean 3 carbon atom while as the name you want to give it has name ethan which means 2 carbon. Also if you will write 2,2dimethylethananime it means you are considering 2 methyl group which means only 1 carbon is left which will be called methane not ethane.
Hope it helps


Separating citric acid from sugar I'm trying to get as pure citric acid as possible from a mandarin. It's not because I desperately need citric acid, but because I want to achieve this.

I started by grating the mandarine and filtering the juice. After filtering overnight, I was left with slightly yellow liquid. I heated it while avoiding boiling and I reduced the volume to 50%. When volume was reduced, the solution started to become saturated and crystals appeared. At this point, I stopped.

Now, I'm looking for a way to remove sugar from the solution, as this is the most annoying part. It's sticky.

When crystals appeared while I was boiling the solution, I came to think, that perhaps that could allow to separate sugar. Citric acid has solubility in water 1477.6 g/L (20 °C), while sugar has 2000 g/L (25 °C). Wikipedia unfortunately doesn't provide values for the same temperature.

From the above it seems that I can have more sugar dissolved than citric acid. I think that means that citric acid will crystallise first. Does it work like this? Are the crystals I've produced the citric acid crystals?

Is there another way to separate the compounds? Does something react with sugar to produce precipitate that can be filtered?

Citric acid is relatively soluble in ethanol (62g/100g) whereas sucrose is relatively insoluble in ethanol (1g/170ml). There are many different kinds of sugars and sucrose is probably not the primary one in this extract; branched sugars... oligosaccharides (which have generally higher solubilities than sucrose) are likely, and I'm sure there are a few other organic acids involved with a crude, fruit extract, but the different solubility characteristics should be relatively helpful for extracting citric acid from sugars and some other organic acids.

So, I would try extracting the citric acid from the dried fruit or dried water extraction with 99%+ ethanol. I would then allow the alcohol to evaporate to help concentrate the acid.

Dilute citric acid (from a concentrated extract) can be precipitated with calcium hydroxide as practically insoluble calcium citrate, which is then reacted with sulfuric acid to form (food grade) citric acid and calcium sulfate (gypsum) as a recoverable by-product.

To really purify the stuff, an ion exchange resin or some other chromatographic method would be the last step.

Try micro-filtration with gelatin https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0ahUKEwj_gs_7hMHNAhUG6SYKHXfyAnwQFggsMAI&url=http%3A%2F%2Fwww.nytimes.com%2F2007%2F09%2F05%2Fdining%2F05curi.html&usg=AFQjCNHNgFO_oZ5o_AfUU9dazQgt7wrzWw&sig2=ham-sb2aV1PEYkMSe8qutw&bvm=bv.125221236,bs.1,d.dmo

search gelatin filtration on google


Confusion Regarding Nitrogen Inversion In nitrogen inversion, the lone pair changes its direction continuously.

Does this in any way delocalize the charge density on the nitrogen making it less nucleophilic and less basic than another nitrogen attached to, say, a bridgehead carbon?

Consider the specific case of
Quinuclidine and tri-methyl amine. Which will be more nucleophilic among them and why?

Yes, amine inversion does effect(reduce) basicity and nucleophilic character of amines

In amine inversion a nitrogen which is sp3(tetrahedral) converts into sp2(planar) and then again coverts into sp3(tetrahedral).

This happens because nitrogen has a lone pair and this lone pair moves through nucleus towards the opposite side. This is known as quantum tunneling. While the electron pair is moving through the nucleus nitrogen has sp2 hybridisation. So, if nitrogen inversion is effective availability of lone pair decreases.

Now let's come to the case of quanidine. In quanidine, nitrogen inversion doesn't happen occur because of the impossible geometry it would have to form to do so.
However, in tri-methyl amine there is no such hindrance and hence achieve planar state relatively easily.


Diluting a sodium chloride solution with specific glassware I have a $\pu{0.3 M}$ $\ce{NaCl}$ stock solution and need to create a $\pu{0.01 M}$ $\ce{NaCl}$ solution that is $\pu{50 mL}$. The only tools I am given are volumetric pipettes that are $10.0, 5.0, 1.0 \text{ and }\pu{0.5 mL}$ and $\pu{50.0 mL}$ volumetric flasks. I understand the physical process where I would measure out a specific amount and place it in the flask using the pipette(s) and then add the required amount of water.

I need to use the dilution equation $M_1\times V_1 = M_2\times V_2$ which I did and I got $\pu{1.666667 mL}$ of solution that needs to be taken from the $\pu{0.3 M}$ stock solution to create the $\pu{0.01 M}$ solution. The only problem is that I can't do that since the materials I have do not allow me to measure out that amount.

I tried to "convert" the $\pu{0.3 M}$ to various other concentrations that would hopefully get me down to $\pu{0.01 M}$ but have failed so far. I tried to find concentrations that would result in a whole number or a value with $.5$ at the end after dividing by the product from $M_1\times V_1$ but failed every time. In addition, I combined the two concentrations and found the midpoints between them and tried to use those values, but they did not work either. Where is a good place for me to start?

Assuming you can use a beaker (or just one of those 50 ml volumetric flask for mixing), you could first dilute some of the 0.3 M solution to 0.1 M by adding 5 ml of stock to 10 ml of water. Now you have a 0.1 M solution of NaCl which you can calculate how to dilute using the M1*V1=M2*V2 equation. Doing this, I calculate that to make 50 ml of 0.01 M solution, you need 5 ml of the 0.1 M solution you made, which you can do with a pipette.

This question is years old, but I went through the calculations for my own interest.
The stock solution needs to be diluted to 1/30th of its original strength.
Take 5 ml of it and add 5ml water, resulting in 10ml of 0.15M solution.
Add 20 ml water, resulting in 30ml of 0.05M solution.
Take 10ml of that, add 40 ml of water. The result is 50ml of 0.01M solution.
To verify: 10 x 0.05 = 50 x 0.01
That's one way to do it.


Why draw "up" arrows first when filling orbital diagrams? I am a 11th/12th grade student studying orbital configurations. I have gone over this lesson and it says to always draw spin up first. However, it never says why. Is drawing down spins first incorrect?

There is no real reason about that and drawing them "down" would be exactly the same as long as you draw them always in the same direction (it needs to be consistent).

The same applies to orbital lobes which can be hollow (white) or plain black depending on their relative sign. When two molecules' orbitals interact, you just consider they have the right sign for an interaction.


Where does the characteristic peak for phtalic acid diethyl ester at m/z 149 originate? Which fragmentation is possible at this m/z value for phthalic acid diethyl ester?

I tried McLafferty but it didn't solve my problem.

First of all, you should indicate if you consider the fragmentation in electron ionization or in other ionization conditions. Considering it is the electron ionization conditions and that you refer to this spectrum, one can postulate for a reasonable mechanism, which can be derived from what is known on the fragmentation of benzoic acid and esters. I am not aware of any complete study of phtalate ester fragmentation, but I did not carry a complete litterature search.

In the figure below is a proposal for the mechanism leading to the formation of two possible structures for the $m/z$ 149 ion.

If one consider that is fragments as do benzoate esters, through radical induced $\alpha$-bond cleavage leading to the loss of $\ce{CH3CH2O^.}$ followed by standard decarbonylation of the acylium ion. For phtalate esters, the question at this point is whether the structure will remain as I or might undergo isomerisation to more stable structures.

I have provided a few alternative structures: simple addition of the neighbouring carbonyl would lead to structure II which although it has some constraints, allows a good stabilization of the positive charge both in the ester and by the aromatic ring. Other structures can be postulated, such as III after addition of the ethanoyl oxygen to the aromatic cycle. This four membered ring can reopen to lead to the acylium structure IV which can in turn undergo decarbonylation to form the ion $m/z$ 121, also present in the spectrum. So there is some indirect evidence that these structures might be involved in the fragmentation process.

Finally, in order to explain the formation of ions at $m/z$ 105 and 77, one can also propose a last isomer for ion $m/z$ 149, structure V, formed after 1,5 hydride transfer. This one can quite easily lead to loss of acetaldehyde, eventually followed by decarbonylation to lead to ion $m/z$ 77.

If phtalates present specific reactivity, the other option is that ion A at $m/z$ 177 can cyclize to structure B which is a very good candidate for ethene loss leading also to an ion at $m/z$ 149.

The question then becomes which of these two $m/z$ 149 is the actual ion. These two isobaric ions could be resolved by high resolution experiments, but after looking in publicly available databases, I did not find any high resolution spectra. Deuterium labelling of the ethyl groups could also prove handy, but once again I did not find any spectra of diethylphtalate labelled on the ethyl groups. (The usual diethylphtalate used in quantitation experiments seems to be the $d_{4}$ isotopomer, with deuterium on the aromatic ring.) Finally one can also have a look at isomers and related ions to see how they fragment. And indeed if one takes the dipropylphtalate ester, the ion at $m/z$ 149 is present (produced after propanoyl loss followed by propene loss) and there is no strong ion at $m/z 163$ which would tend to be a good indication of ethene loss for diethylphtalate. But, on the other hand, both the meta and para isomers of diethylphtalate have a much weaker ion at $m/z$ 149, which cannot be accounted so easily by an equivalent to intermediate B. So I would probably favor the concurrent occurence of both pathways, with a ratio that still needs to be measured.

In addition to the answer above, it should have been quite immediate that a McLafferty rearrangement was not an option: since there are no nitrogen, the nitrogen rule allows to conclude directly that the odd mass of the ion at m/z 149 indicates that it is an even electron cation, as opposed to radical cations produced by McLafferty rearrangements.

the fact that dimethyl phthalate does not produce significant amounts of the 149 m/z fragment, but all other phthalates with a longer alkane chain do, would put decarbonylation in favour, leading to fragments I - V from above scheme.


Why is the ground state electronic configuration of Group 10 (Ni group) elements so anomalous? While studying about d block elements I came across this table showing outer shell electronic configuration of group 10 elements

$\ce{Ni}$ $ $ $ $ $ $ $ $ $ $  $3d^8$ $4s^2$
$\ce{Pd}$ $ $ $ $ $ $ $ $  $4d^{10}$ $5s^0$
$\ce{Pt}$ $ $ $ $ $ $ $ $ $ $ $5d^9$ $6s^1$

(Source Concise Inorganic Chemistry, JD Lee, adapted by S. Guha Pg 563)
Now I cannot understand the reason behind this. If we consider that $\ce{Pd}$ has changed its config. to attain stability then what's wrong with $\ce{Ni}$ and $\ce{Pt}$? And should the outer shell electronic configuration of the ground state not be same in elements down a group?
Edit
To those who have marked my question as a duplicate of the one mentioned, I humbly request to explain me which answer to the above mentioned question (of which mine is marked as a duplicate of) tells about the anomalous behavior of group 10 elements and why $\ce{Zn}$ assumes $d^{10}$ config and none of the others in the group do the same and again $\ce{Pt}$ assumes a $5d^9 6s^1$ config, while $\ce{Ni}$ does not. Also please do point out  where can I find an answer to why in this particular group there is no similarity in ground state electronic configuration b/w elements of the group. Thank You.

I think the relevant comparison should be with the earlier transition metal groups where $s^2$ configurations are more predominant.

In multielectron atoms, electrons interact in such a way that for a given shell quantum number $n$, the orbitals with higher total angular momentum $l$ are raised, thus for example $3d$ ($n=3, l=2$) becomes higher than $3p$ ($n=3, l=1$).  This angular momentum effect is most important when the orbitals are well shielded with a low effective nuclear charge, where the electron-electron interactions don't get swamped by electron-nucleus interactions.

In earlier transition groups the valence shells have this low effective nuclear charge and the angular momentum effect is so strong by comparison that the $d$ valence subshell is raised above the $s$ subshell even with one higher $n$ quantum number.  So, for instance, in the fourth period we see $3d$ orbitals filled only after $4s$ in neutral atoms.  Not so much in the ions, however, where ionization leaves the remaining electrons with more effective nuclear charge, hence we often see $d$ instead if $s$ subshell electrons remaining in the early transition ions.

But by the time we get to the later transition metals the effective nuclear charge has increased even in the neutral atoms, the lower-$n$ $d$ subshell drops relative to the higher-$n$ $s$ subshell, and we begin to see a greater preference for the $d$ subshell being filled first.  By the time we get to Group 12, this transition is complete and only the $s$ electrons remain valence electrons except, possibly, in extreme circumstances.

Here Platinum shows d9 s1 configuration due to the relativistic effect of 6s orbital. Which is inert towards oxidation.
But in case of Pd it is d10 configuration due to the full filled highly stable 4d orbital.  But Ni cannot show this
configuration.  When it comes to the case of Ni , it has 3d orbital in its valence shell so it is not so much large and   cannot supply spin pairing energy. So the configuration show d8 s2.That's why this group element show analomous behavior.