Will a solid object lose or gain atoms on standing over time (without being acted upon by external influences)? Suppose I'm in my room with my keyboard, my desktop and my desk. Are these objects losing or gaining atoms every second? Or are they forever stuck with the same number of particles until you scratch, break or pulverize them?

Or suppose, for simplicity's sake, I have a pure gold sphere in my room. Neglecting the presence of dust or oxide layers...would it be losing or gaining atoms over time on account of simply standing untouched in one place?

Short answer: Yes they do.

Firstly, you need to arrive at a proper definition for an "object". Why you ask? Because it would otherwise present philosophical problems which become apparent when we begin to work with those objects.

  ... what is an object? Philosophers are always saying, “Well, just take a chair for example.” The moment they say that, you know that they do not know what they are talking about any more. What is a chair? Well, a chair is a certain thing over there … certain?, how certain? The atoms are evaporating from it from time to time—not many atoms, but a few—dirt falls on it and gets dissolved in the paint; so to define a chair precisely, to say exactly which atoms are chair, and which atoms are air, or which atoms are dirt, or which atoms are paint that belongs to the chair is impossible. So the mass of a chair can be defined only approximately. In the same way, to define the mass of a single object is impossible, because there are not any single, left-alone objects in the world—every object is a mixture of a lot of things, so we can deal with it only as a series of approximations and idealizations.

If you can't define an object (more specifically: If you can't define its location in space) exactly...you cannot know its mass exactly. If you do not have its exact mass, you cannot make an exact/complete description (be it physical or chemical) of the body. [In fact, our ability to provide accurate descriptions of bodies is because we make good approximations]

I get what you are trying to ask in your question, and it is per se, a reasonable question; but the example (your room) is (unfortunately) difficult to deal with...the single glaring issue being the inability to define an object in an absolute sense (I'm not blaming you, in fact, no one can define any real-life object exactly :P)

Just picking out some other holes in your "room" example: Is you room sealed (can it be considered a closed system)? Is your room evacuated (I mean a "Vacuum")? Are the different "objects" present in the room in contact with each other? Are there foreign particles present on the walls, ceiling and floor? You can see where this is going...

You go on to mention using a "pure glass/gold sphere" in order to circumvent some of the difficulties of defining an "object". Unfortunately, 100% purity is inconceivable (the main issue being the dust/oxides that are present at the surface). To neglect the presence of these layers would be to forgo reality...and the question (using the "pure" spheres) would be reduced to a purely theoretical proposition.

Trying to address you question, I'll do this in two ways: first I will deal with an idealized case and secondly, I shall deal with the abomination you call your "room" (introducing a few constraints of course, to keep things simpler) :D

First: The (highly) idealized version of your example:

You have a really large ("infinitely" large, if you will, since this is a thought-experiment) sealed region of space, and this "chamber" (as we shall now call it) shall be totally isolated from its surroundings: No energy transfer, no mass transfer, no (external) gravitational field, no (external) electric or magnetic fields...no nothing. The walls of the chamber have no substance and are infinitesimally thin. The inside of the chamber is maintained at an arbitrary temperature of 25°C.

Suppose your chamber has a small amount of a mono-atomic gas (take neon for example) in it. Now you introduce an "object" in the chamber instantaneously. The "object" is a 1000 atom x 1000 atom x 1000 atom cube of absolutely pure, solid gold (and we shall assume there's no oxide layer on the surface).

Now what happens?

Simply put; the neon atoms will impinge on the gold cube and some may end up "sticking" to it. The gold atoms will slowly (like reeaaally slowly) "diffuse" out of the original block and co-exist with the neon atoms as a gas (you can think of this as the "evaporation" of the solid gold cube). You'll observe that even though neon diffuses into the gold cube, and the gold atoms diffuse into the neon gas...over a period of time there is a net diffusion of gold atoms out of the block. Keep watching, and you'll find that you have a gaseous mixture of neon and gold, with no solid gold left to speak of. Handwaving the kinetics involved here, this is an example of the Second law of Thermodynamics at play. The entropy (or the "randomness") of the system goes on increasing over time, and will continue to do so until an equilibrium condition has been achieved. {You might also want to read about the Heat Death of the Universe  <--- Interesting thought}.

Even if you didn't have the neon in the chamber (I put it in there to "simulate" the air in your room), the cube would still "evaporate" into a gas, albeit, at a slightly faster rate.

Second: Your room (with a few restrictions):

I shall proceed to define your room (albeit with a few constraints). You room is, also sealed so that no mass can be exchanged between the inside of your room and the surroundings. Everything inside your room is still subject to external gravitational, electric and magnetic fields. The temperature inside your room is 25°C. You room is not dust-free either. Your room is now set.

Start the clock...and what happens?

The results are similar (albeit, a scaled down version) to the idealized case. Every "object" in your room loses atoms/molecules and gains some too. Like before, there is a net diffusion of particles out of the objects in your room...entropy increases yet again. Leave it there for a few hundred million years, and you'll find that the objects in your room will have changed significantly (remember, the time frame was in hundreds of millions  of years).

As @Ruslan pointed out in the comments: Unlike the ideal case where the gold atoms can evaporate from the cube and continue to spread ad infinitum...your furniture and computer don't have that kind of luxury (their ability to spread out in space is limited by the walls of your cramped room) and hence won't evaporate completely (but there is evaporation nonetheless).

To understand that last bit ("incomplete" evaporation in your room, but "complete" evaporation in the idealized chamber), it will help to consider an analogy with the evaporation of water:

Here's an excellent piece (The Feynman Lectures on Physics, Volume 1, Chapter 1) you must read through before we continue:

  What happens at the surface of the water? We shall now make the picture more complicated—and more realistic—by imagining that the surface is in air. The figure shows the surface of water in air.

  We see the water molecules as before, forming a body of liquid water, but now we also see the surface of the water. Above the surface we find a number of things: First of all there are water molecules, as in steam. This is water vapor, which is always found above liquid water. (There is an equilibrium between the steam vapor and the water which will be described later.) In addition we find some other molecules—here two oxygen atoms stuck together by themselves, forming an oxygen molecule, there two nitrogen atoms also stuck together to make a nitrogen molecule. Air consists almost entirely of nitrogen, oxygen, some water vapor, and lesser amounts of carbon dioxide, argon, and other things. So above the water surface is the air, a gas, containing some water vapor.

  Now what is happening in this picture? The molecules in the water are always jiggling around. From time to time, one on the surface happens to be hit a little harder than usual, and gets knocked away. It is hard to see that happening in the picture because it is a still picture. But we can imagine that one molecule near the surface has just been hit and is flying out, or perhaps another one has been hit and is flying out. Thus, molecule by molecule, the water disappears—it evaporates. But if we close the vessel above, after a while we shall find a large number of molecules of water amongst the air molecules. From time to time, one of these vapor molecules comes flying down to the water and gets stuck again.

  So we see that what looks like a dead, uninteresting thing—a glass of water with a cover, that has been sitting there for perhaps twenty years—really contains a dynamic and interesting phenomenon which is going on all the time. To our eyes, our crude eyes, nothing is changing, but if we could see it a billion times magnified, we would see that from its own point of view it is always changing: molecules are leaving the surface, molecules are coming back.
  Why do we see no change? Because just as many molecules are leaving as are coming back! In the long run “nothing happens.” If we then take the top of the vessel off and blow the moist air away, replacing it with dry air, then the number of molecules leaving is just the same as it was before, because this depends on the jiggling of the water, but the number coming back is greatly reduced because there are so many fewer water molecules above the water. Therefore there are more going out than coming in, and the water evaporates.

Now back to where I left off; why do we see partial evaporation in your room, but total evaporation in the "infinite" chamber?

Based on the excerpt, I'll try to develop an analogy for what's going on in 1) Your room and 2) The "infinite" chamber I mentioned first.

1) Analogy for what's going on in your room:

You fill a perfectly dry bottle halfway up with water and screw on the cap. The water molecules constantly evaporate from the surface and molecules of oxygen, nitrogen, etc from the air inside the bottle constantly impinge on the water surface and may get dissolved into the water as a result. However, their ability of the water molecules to spread out in space (increasing entropy) is limited by the bottle's boundaries. Though we have an initial net evaporation of water...we're soon faced with an equilibrium state (described in the excerpt) where there is no net evaporation of water. You're still left with water that is mostly in its original (liquid) state.

It's pretty much for reasons similar to this, that the object in your room won't completely evaporate. There simply isn't enough space for the particles to "comfortably" move out in to.

2) Analogy for what's going on in the idealized case:

You pour a glass of water on the side-walk. Unlike the case of the bottle, water molecules can evaporate and expand till infinity (well, it's "infinite"  relative to the the space you had in the bottle).

It's pretty much for reasons similar to this that the gold cube in the idealized chamber will completely evaporate.There is enough space for the particles to "comfortably" occupy outside of the solid that they were initially constrained to.

So yes... over time, all the "objects" in your room lose and gain particles over time. Ultimately however, the "object" you once knew will no longer be the same >:)


Why opposite of electrophilic is not Protophilic In organic chemistry , considering definitions of  electrophilic and nucleophilic:

Electrophilic:having an affinity for electrons : being an electron acceptor.
nucleophilic : having an affinity for atomic neuclei (also can we say proton?) : being an electron donor.

So both being opposite to each, why dont we say nucleophilic as protophilic?

Because not all "nucleophilic" agents react equally well with all types of relatively exposed nuclei.  Phosphines are very unwilling to extract a proton but quite nucleophilic if the available electronic vacancy is in carbon or a transition metal.


Is there any book or text where ionic radius of elements are given? So my research has me looking for ionic radius characterization. If anyone could post any link or give the name of any book where ionic radius of all elements are given that would be really great?

There is "classic" work by Shannon and colleagues [1, 2], also available online as Database of Ionic Radii.

Extended ionic radii for lanthanides with higher coordination numbers were published by Jia [3].

The values from the references above are also compiled in recent editions of CRC handbook of chemistry and physics, e.g. [4, p. 12-12].

More recently, an updated series of calculated ionic radii has been published by Rahm et al. [4].

References

Shannon, R. D.; Prewitt, C. T. Effective Ionic Radii in Oxides and Fluorides. Acta Cryst B 1969, 25 (5), 925–946. https://doi.org/10.1107/S0567740869003220.
Shannon, R. D. Revised Effective Ionic Radii and Systematic Studies of Interatomic Distances in Halides and Chalcogenides. Acta Cryst A 1976, 32 (5), 751–767. https://doi.org/10.1107/S0567739476001551.
Jia, Y. Q. Crystal Radii and Effective Ionic Radii of the Rare Earth Ions. Journal of Solid State Chemistry 1991, 95 (1), 184–187. https://doi.org/10.1016/0022-4596(91)90388-X.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97.
Rahm, M.; Hoffmann, R.; Ashcroft, N. W. Atomic and Ionic Radii of Elements 1–96. Chem. Eur. J. 2016, 22 (41), 14625–14632. https://doi.org/10.1002/chem.201602949.


Will the temperature of a faucet actually be lower than the countertop or will it only feel cooler due to specific heat? Due to specific heat capacity of metals, they feel cooler because they don't retain  as well as other materials. But is their temperature actually lower if it could be measured?

The short answer:

A metal object as the faucet drains heat from fingers due its high thermal conductivity faster than the countertop.

The fingers get therefore colder than if touching the countertop.

The faucet perceived temperature is therefore colder, even if the faucet and countertop temperature are the same.

If it is really colder due cold water, the perceived coldness is even stronger.

The longer answer:

Metals have high thermal conductivity.
That causes higher thermal flow at the same temperature difference than for most non metal materials of the same temperature. ( Rare exception are diamonds, as their thermal conductivity is superior to all metals.).

That leads to perception like if metals were colder ( if being colder then body temperature ) or warmer ( if being warmer ) than they really are.

OTOH, metal (mass-)specific heat capacity  is mostly lower than for non metals, due heavy metal atoms, that in fact somewhat decreases the effect of high thermal conductivity.

For your specific case, it depends also on the real metal temperature, e.g if it is partially sharing temperature with cold or warm tap water.

Similarly, water feels like colder than air of the same temperature, as thermal flow from human body is higher. But at this time, the higher volume specific thermal capacity plays major role.

2 examples how our senses can be deceiving:

If you had not too much hot herbal tea in an aluminium cup, you could still be sipping the tea itself without burning your tongue, but the cup would feel like burning your lips. If you did the same with the stainless steel cup, it would not be feel like burning as much, but still hot. If we used thermometer sensors, we would realize the metal surface not in direct contact with tea is very slightly colder than tea. But in both cases you would have impression the metal is warmer than the tea itself.

This time our senses monitor real thing ( temperature ), but rather their own temperature, instead of the temperature  of the sensed object.

Let have 3 basins with cold water, lukewarm water and warm water. You measure their temperature with thermometers, so it is a real thing. Now you put your left hand to cold water and your right hand to warm water for a minute. Then you put both hands to lukewarm water. The left hand will give you perception it is warm water, while the right hand will give you perception it is cold water.

This time our senses process not the temperature itself, but rather the temperature spacial and temporal gradient.


Computational program not using full cpu power I started some computations on Gamess.

For instance, when I compute the frequencies of an optimized structure, I see in the terminal that the job only uses 40% of the CPU core most of time (sometimes it goes to 100%). When I decide to provide 2 cores for the job, it only uses something like 20-25% of each core most of the time (and sometimes 100% of each core too). Why would not the task take maximum power to achieve the computations ?

Most applications in theoretical chemistry are memory and or memory bandwidth bound. Especially due to shared memory bandwidth it can be bad for your performance if you use too many processes.

In addition you benefit from additional processes only if your algorithm is more or less parallel. If there are too many serial parts/communication overhead you will have idle cores, if your process number is too high. (Amdahl's law)

Take home message:

Time the execution of your code with a non trivial example.
More is not always better.


Why are some materials more flammable than others? Wood and oil are flammable materials, but water isn't. Moreover, flames spread a lot faster in oil than in a piece of paper. What is the reason that it is so?

I also don't know why a small flame can spread out to a very big flame; what is the physical reason for it? I would like to know the molecular reasons in detail what makes fire so powerful.

Firstly, for a material to be flammable it needs to contain something that is oxidisable by air oxygen, typically carbon. This explain why hydrocarbons and other carbon bearing materials typically burn well but minerals, rock, glass, water etc do not. In the latter materials the elements are already in a high Oxidation State and cannot be further oxidised.

For a generic hydrocarbon the oxidation reaction by air oxygen is given by the following reaction equation:

$$\ce{C_nH_{n+2}(g,s,l) +\frac{3n +1}{2}O_2(g)-> nCO_2(g) +\frac{n +2}{2}H_2O(l)}$$

The second major factor that affects flammability is volatility. The more volatile a material is, the more flammable it will be. The hydrocarbons illustrate that well. Methane is the most volatile and flammable of all of them and by the time we get to the solid ones like paraffin wax, flammability has been greatly reduced because these aren't volatile at all.


Why boiling can't take place in closed containers I am being told that boiling can't take place in closed containers. Can someone please explain why this is so.

You can understand this well by thinking about why a liquid boils in the first place.  Imagine you have a pot of water, and it's just sitting there in your kitchen.  Intermolecular forces (specifically hydrogen bonding for H20 molecules) keep the water in liquid state.  The air molecules zipping around the pot and above the water's surface represent a certain pressure.  Let's say your kitchen has standard conditions, so 1 atm.  Then, you put the pot on a stove, and heat the water up.  This heat breaks the intermolecular bonds between the liquid molecules.  Now, the only reason the molecules rise up and become water vapor is because the kinetic energy of the water molecules matches that of the surrounding air.  Pressure moves from high to low, and as the water boils, certain molecules acquire the necessary speed (and therefore pressure) to enter the gaseous state.  Keep in mind that heating the pot negligibly changed the kitchen pressure.  The pressure in your kitchen should still be about 1 atm, since the vast air space should distribute the heat produced by the stove.

In a closed container, however, the story is different.  Let's say you put a lid over your pot.  Now, there is a relatively small amount of air trapped between the lid and the water's surface.  Heating the pot heats the air as well, and pressure does not remain at 1 atm.  That is because that small amount of air has to handle all that heat, whereas in the situation without the lid, a bigger volume (the kitchen) allowed for more heat dissipation.  So now the small amount of air molecules are moving really fast.  The heat is still breaking the intermolecular forces of the water, but this time, there is no (entropic) reason to move up into the gaseous state.  The air molecules are moving just as fast, and therefore have just as much pressure, as water molecules that have broken bonds.

The key to your question is rooted in why the water molecule (without a lid) moves up out of the surface of the water and joins the air molecules, effectively becoming water vapor.  Simply put, it is because the water molecule (because of the heat) has enough speed to zip out of the water.  Pressure is intrinsically based on speed of molecules, and so you can think about this situation as: the water molecule(s) has a higher pressure than the air molecules, and thus they move up to join the air molecules, in order to balance out the pressure.  A lid prevents this from happening because the water molecule(s) pressure will never be higher than the air molecules, since the air takes the same amount of heat as the water.  (The preceding statement is true only in usual cases.  There are caveats to this - see comments.)

Therefore, water does not (usually) boil in a closed container.  A couple of things: There are many more explanations/videos online about this topic (boiling water, pressure cookers, more altitude = lower pressure, higher boiling point etc..)in case you want to strengthen your understanding further and explore the core of this question from different angles.  Also, the intuition in this Khan academy video https://www.youtube.com/watch?v=hA5jddDYcyg is pretty good.


Which quantities change when choosing a different standard state? Chemistry and biochemistry use a different standard state. For example, the standard hydrogen ion concentration is the one at pH = 0 in one case, and the one at neutral pH in the other case.

Which quantities change when choosing a different standard state, and which quantities have special values at standard state?

Quantities that are dependent on the standard state

When calculating the equilibrium constant K or the reaction quotient Q from a set of observed concentrations, these concentrations are given relative to the standard concentration of the species. For example, in aqueous solutions the solvent water has a different standard concentration (that of pure water) than solute (roughly 1 M). Therefore, the value of K and Q depend on choice of standard state.

Whenever a thermodynamic quantity has a Plimsoll symbol appended, it refers to a system at standard state, e.g. $\Delta_r G^\circ, \Delta_r H^\circ, S_f^\circ, \Delta H_f^\circ$ or to the substance at standard state. The choice of standard state changes the value of these quantities. For the biochemical standard state, the quantity is appended with a prime, e.g.  $\Delta_r G^\circ {'}$ to indicate that difference.

Quantities that have special values at standard state

At standard state, activities are equal to one. Consequently, the reaction quotient Q is also equal to one.

$$Q = 1 \ \ \ \ \text{(at standard state)}$$Also, and this is a bit of a tautology but useful in deriving some relationships, when concentrations are at standard state (or more generally if $Q = 1$), quantities are equal to the standard quantities. Take for example the Gibbs energy of reaction:

$$\Delta_r G =  \Delta_r G^\circ + R T \ln Q$$

When concentrations are at standard state, this reduces to:

$$\Delta_r G =  \Delta_r G^\circ \ \ \ \ \text{(at standard state)}$$

Of course, that is the idea in calling $\Delta_r G^\circ$ the standard Gibbs energy of reaction.


Formal charge and net charge What is the basic difference between formal charge and net charge . Does formal charge take part in resonance .

I searched it on wikipedia , but it is mostly discussing how to find formal charge.

https://en.m.wikipedia.org/wiki/Formal_charge

Net charge is the sum of all formal charges of the atoms in a molecule. Net charge is the charge of the molecule.

Formal charge is the charge of an atom in a molecule. Formal charge varies when you look at resonance structure. See this post of the nitrate resonance structures. It can be obtained through:

\begin{equation}
Formal\ charge\ =\ Valence\ electrons\ -\ no\ bonding\ electrons\ -\ \frac{bonding\ electrons}{2}\quad\quad\quad\quad\quad\text{(Eq. 1)}
\end{equation}

Example: I take the example given previously with $\ce{NO3^{-}}$, in order to give a complete answer to the present post.

So, Having in mind the rules to draw the Lewis Structure:

Operations

Sum the valence electrons of all atoms. When the system is an anion, an electron is added. When the system is a cation, an electron is subtracted.
Sum the 'octet-electron duet' of all atoms.
Subtract step 2 of 1. Results corresponds to shared or bonding electrons.
Subtract step 3 of 1. Result corresponds to the nonbonding electrons.

Indications

Assigns two electrons to each bond.
If electrons remain from step 3. Add them in pairs creating double or triple bonds. Having in mind the duplet/octet rule.
The obtained electrons from step 4. Add them in pairs to the more electronegative atoms.
Calculate the formal charge of each atom.

Then:

Operations

$$valence\ electrons\ \cdot\ number\ atoms\ of\ \ce{N}\ +\ valence\ electrons\ \cdot\ number\ of\ \ce{O}\ +\ anion\ condition $$ Then $$5\cdot 1+6\cdot 3+1=24$$
$$octet\ of\ \ce{N}\ \cdot\ number\ atoms\ of\ \ce{N}\ +\ octet\ of\ \ce{O}\ \ \cdot\ number\ of\ \ce{O} $$ Then $$8\cdot 1+8\cdot 3=32$$
$$32-24=8$$
$$24-8=16$$

Indications

After doing the first three steps, you will get:

 (Source: Stackexchange)

Now, taken equation 1 to nitrogen:

$$Formal\ charge\ of\ \ce{N} =\ 5\ -\ 0\ -\ \frac{8}{2}\ =\ +1$$

Now, taken equation 1 to oxygen bonded to nitrogen through single bond:

$$Formal\ charge\ of\ \ce{O} =\ 6\ -\ 6\ -\ \frac{2}{2}\ =\ -1$$

Now, taken equation 1 to oxygen bonded to nitrogen through double bond:

$$Formal\ charge\ of\ \ce{O} =\ 6\ -\ 4\ -\ \frac{4}{2}\ =\ 0$$

Therefore

Net charge of $\ce{NO3^-}$ is the sum of all formal charge of the atoms in the molecule equals to -1


Chalky residue from melted ice cubes I bought a new refrigerator that makes ice cubes. The instructions said to discard the first 3 full bins of ice in order to ensure the new water line is flushed clean. I dumped the full bins in my stainless steel sink each time. After the ice melted each time a chalky residue was left in the sink. I emptied a couple of extra full bins and still was left with the chalky residue. I cleaned the sink after each dumping to be certain there wasn't any left over residue from the previous dump. What is the chalky substance? Is it something to be concerned about or not?

It is possible that the white material is calcium carbonate or magnesium carbonate from your water source precipitating.

See if vinegar dissolves the white residue.

If the white residue dissolves in vinegar, it is probable calcium carbonate and/or magnesium carbonate.


Why is the attached hydrogen of 1,3-diethyl-5-(1-phenoxyethyl)benzene homotopic? The circled hydrogen is considered homotopic with reference to the hydrogen identified by the arrow. I predicted this as they are on identical substituents but am now confused because there orientations with respect to the molecule with the benzene are different. Am I misinterpreting something about the molecular structure?

Which hydrogen is homotopic to which is rather hard to see in this compound without actually sitting down and drawing the relevant structures. Before I do this, I am going to get a few obvious points out of the way:

The compound is chiral (see the $\ce{PhO}$ group on the third side chain). This not only greatly reduces the number of enantiotopic hydrogens but also the symmetry of the molecule.
Further analysis shows that there is, in fact, no element of symmetry, i.e. the molecule’s point group is $C_1$. This means that we cannot resort to any ‘cheap hacks’ to determine homotopic hydrogens. We have to use the old-fashioned methods.
Homotopic hydrogens are those that can be transformed onto each other by rotations only (of the entire molecule or rotatable bonds) in such a way that the final depiction is identical to the initial depiction.

We have a few rotatable bonds at our disposal; I will be using the $\ce{Ph-CHXMe}$ bonds, where $\ce{X} = \ce{H, OPh}$ — i.e. I will be rotating around each carbon-carbon bond that connects a residue to the central aromatic ring. Following are the two rotation steps necessary to transform the arrow-hydrogen onto the circled hydrogen:

Scheme 1: The required rotations. The rotated bonds are shown by curly arrows; non-moving parts of the molecule are grey. The circled hydrogen (red) and the other hydrogen on the same carbon (blue) have been highlighted.

I chose to rotate both ethyl groups first and to rotate the entire benzene ring second. However, the order is not important. As we can see, the first rotation transforms the circled hydrogen behind the plane. (I need to rotate the other ethyl group, too, so that the final rotation will give me the initial product again.) The second rotation of the entire benzene ring with the two ethyl groups again puts the circled hydrogen in front of the plane.

Thus, your analysis was correct; the two hydrogens in front of the paper plane are homotopic as are the two behind it. The two hydrogens on a single carbon (red and blue in the picture) are diastereotopic as they cannot be transformed onto each other preserving the symmetry of the molecule.


Thermal decomposition of calcium propionate What are the decomposition products?

For example:

Calcium formate decomposes into formaldehyde and calcium carbonate.

$$\ce{Ca(HCOO)2 -> CaCO3 + HCHO}$$

Calcium acetate decomposes into acetone and calcium carbonate.

$$\ce{Ca(CH3COO)2 -> CaCO3 + (CH3)2CO}$$

How does calcium propionate decompose?

There is a paper (1) where TG-GLC-MS with subsequent XRD analysis of solid products has been utilized for the analysis of evolved gases during the decomposition of calcium propionate at various heating rates. Detected products were $\ce{H2O}$, $\ce{CO2}$, $\ce{C2H5CHO}$, $\ce{C2H5COCH3}$, $\ce{C2H5COC2H5}$, $\ce{CaCO3}$.

The Discussion section of the paper suggests the following mechanism of formation:

  Following the free radical mechanism of Hites and Biemann (2), proposed for
  the decomposition of the calcium salts of monocarboxylic acids we suggest that the initial stage in the decomposition of calcium propanoate can be represented as:

  $$\ce{Ca(C2H5COO)2 -> CaCO3 + C2H^.5 + C2H5C^.=O,}$$

  Combination of the two free radicals results in the formation of the major product, 3-pentanone:

  $$\ce{C2H5C=O + C2H^.5 -> C2H5 - CO - C2H5}$$

  The other trace products observed, 2-butanone and propanal, can be accounted for by disproportionation reactions involving the ethyl radical:

  $$\ce{C2H^.5 + C2H^.5 -> C3H^.7 + CH^.3}$$

  and

  $$\ce{C2H^.5 -> C2H5 + H^.}$$

  The methyl and hydrogen radicals so produced may react with the large excess of the original propanoyl radicals to give 2-butanone and propanal respectively.

  $$\ce{C2H^.5=O + CH^.3 -> C2H5COCH3}$$

  and

  $$\ce{C2H5C=O + H -> C2H5CHO}$$

  The $\ce{C3H7}$ may combine with any of the other radicals or may disproportionate. In any event there was no evidence for compounds larger than 3-propanone. However on close examination of the GC curves there were indications of a very small peak before that due to $\ce{CO2}$ which could be attributed to low molecular weight alkanes and alkenes.

(1) Barnes, P. A.; Stephenson, G.; Warrington, S. B. Journal of Thermal Analysis 1982, 25 (2), 299–311. DOI: 10.1007/BF01912955

(2) Hites, R. A.; Biemann, K. J. Am. Chem. Soc. 1972, 94 (16), 5772–5777. DOI: 10.1021/ja00771a039


Rearrangment during alkyne synthesis What I have got as (G) is this.

$\hskip1in$

My questions

How the rearrangement from (G) to $non-1-yne$ can happen i.e. its mechanism?
Why Sodium Amide in particular yield terminal alkyne, unlike the fused Potassium hydroxides why almost always yield internal alkyne?
Is there any limitation by how much this alkyne formation site can shift?

Please note that this is not my homework, I have done this question and I am interested in the mechanism.

Your questions:

Q1) How the rearrangement from (G) to non-1-yne can happen (i.e., its mechanism)?

For example, see the following mechanism suggested in Ref.1 as the Acetylene Zipper reaction:

Q2) Why Sodium Amide in particular yields terminal alkyne, unlike the fused Potassium hydroxides why almost always yield internal alkyne?

To answer your question, I have to dig down literature to the 19th Century. To my knowledge, sodium amide ($\ce{NaNH2}$) is not the best base to use in this purposes. The best base, in particular, is the monopotassium salt of 1,3-diamminopropane or simply, KAPA (see mechanism), a difunctional “superbase.” According to the Ref.1:

  KAPA produces exceptionally rapid migrations of triple bonds from the interior to the terminus of the carbon chain in seconds at $0^{\circ}$.

On the other hands, from a synthetic standpoint, $\ce{KOH}$ as a dehydrohalogenating agent, tends to promote the shift of the triple bond away from the end of a chain, which is sometimes considered as a disadvantage (Ref.2).

Q3) Is there any limitation by how much this alkyne formation site can shift?

There are no limitations. Yet, if one substitution of disubstituted acetylene is branched, the triple bond migrate through the straight chain to the end (Ref.2). Mean time, the reaction has some interesting facts:

An equilibrium between alkynes and allenes can be established using basic catalysts (see the mechanism). For example, when very strong bases such as sodium metal ($\ce{Na}$) or sodium amide ($\ce{NaNH2}$) are used, the rearrangement of a triple bond of a disubstituted acetylene tends to migrate to a terminal position. This isomerization equilibrium shifts toward forward direction in favor of 1-alkyne formation due to the fact that formation of the terminal metal acetylide salt, which is stable, precipitated out from the solution, thus automatically removing itself from the reaction mixture.
Favorskiî was the first to discover that 2-alkyne could be irreversibly removed from solution by isomerization to the 1-alkyne and resultant precipitation of the sodium salt in 1887 (Ref.3). This procedure used metal $\ce{Na}$ as the strong base, and thus, has the disadvantage that some olefin would produce  as a result from hydrogenation of the alkyne by the hydrogen liberated upon acetylide formation. On the other hand, $\ce{NaNH2}$ when used in inert solvents, was shown to accomplished the same rearrangement but avoids the reduction problem (Ref.3,4). Temperatures greater than $\pu{100 ^{\circ}C}$ are employed with $\ce{Na}$ metal while the best results with a $\ce{NaNH2}$ catalyst are obtained at $\pu{150-160 ^{\circ}C}$. The data are summarized in the Table (vide infra). The rates of the rearrangement differ with the structures of the compounds isomerized. In general, 2-alkynes rearrange faster than 3-alkynes, and 1,2-alkadienes (allenes) rearrange faster than 2-alkynes.  The amount of 1-alkyne formed is proportional to the reaction time, and equilibrium is obtained earlier at higher reaction temperatures (Ref.3).
Later studies have found that $\ce{NaNH2}$ brings about the propargylic rearrangement in liquid ammonia at room temperature with essentially quantitative hydrocarbon recovery. (Ref.3). The ratio of 1-alkyne, 1,2-alkadiene, and 2-alkyne varied with the amount of $\ce{NaNH2}$ used. High $\ce{NaNH2}$ concentrations favored the formation of 1-alkynes. Equivalent or excess amount of $\ce{KNH2}$ in liquid ammonia, sometimes in admixture with HMPA, have also been shown to convert 2-alkynes to 1-alkynes (Ref.3,5).
The advent of newer strong bases has made the isomerization to terminal acetylenes a very facile reaction. The addition of dialkylacetylenes to a slightly excess of KAPA (the monopotassium salt of 1,3-diamminopropane) results in extremely rapid, quantitative isomerization of the triple bond to the terminal position where it is trapped as the acetylide ion (Ref.1).

Table:
$$
\begin{array}{ccc}
\hline
\text{Disubstituted alkyne} & \text{Base} & \text{Temperature} & \text{1-Alkyne} & \text{Disubstituted alkyne} & \text{Ref} \\ \hline
\ce{C2H4C#CCH3} & \ce{Na} & \pu{100 ^{\circ}C} &  \pu{4\! -5 h} &  n\text{-}\ce{C3H7C#CH} & 3\\
n\text{-}\ce{C3H7C#CCH3} & \ce{Na} & \pu{100 ^{\circ}C} &- &  n\text{-}\ce{C4H9C#CH} & 3\\
n\text{-}\ce{C13H27C#CCH3} & \ce{Na} & \pu{200\! -220 ^{\circ}C} &  \pu{15\! -20 h} &  n\text{-}\ce{C14H29C#CH} & 3\\
n\text{-}\ce{C4H9C#CCH3} & \ce{NaNH2} & \pu{160 ^{\circ}C} &  \pu{12 h} &  n\text{-}\ce{C5H11C#CH} & 3,4\\
n\text{-}\ce{C5H11C#CC5H11} & \ce{NaNH2} & \pu{210 ^{\circ}C} &  \pu{7 h} &  n\text{-}\ce{C10H21C#CH} & 3,5\\
n\text{-}\ce{C4H9C#CC2H5} & \ce{NaNH2} & \pu{170 ^{\circ}C} &  \pu{9 h} &  n\text{-}\ce{C6H13C#CH} & 2\\
n\text{-}\ce{C6H13C#CC6H13} & \ce{K(NH2C3H6NH)} & \pu{20 ^{\circ}C} &  \pu{1 min} &  n\text{-}\ce{C12H25C#CH} & 1\\ \hline
\end{array}
$$
Entry 4: The catalyst used was finely divided $\ce{NaNH2}$ suspended in mineral oil.

References:

C. A. Brown, A. Yamashita, “Saline hydrides and superbases in organic reactions. IX. Acetylene zipper. Exceptionally facile contrathermodynamic multipositional isomeriazation of alkynes with potassium 3-aminopropylamide,” J. Am. Chem. Soc. 1975, 97(4), 891-892 (DOI: 10.1021/ja00837a034).
T. L. Jacobs, “The Synthesis of Acetylenes,” Organic Reactions 1949, 5, Chapter 1 (78 pages) (https://doi.org/10.1002/0471264180.or005.01).
Hermon Pines, Wayne M. Stalick, In Base-Catalyzed Reactions of Hydrocarbons and Related Compounds; Academic Press, Inc.: New York, NY, 1977, “Chapter 3: Isomerization of acetylenes and allenes,” pp. 124–204.
H. H. Guest, “Rearrangements of the Triple Bond,” J. Am. Chem. Soc. 1928, 50(6), 1744–1746 (DOI: 10.1021/ja01393a036).
L. Brandsma, In Preparative Acetylenic Chemistry (Studies in Organic Chemistry Series: Vol. 34); 2nd Ed.; Elsevier Science: New York, NY, 1988, “Chapter XI. Base-Promoted Interconversions of Acetylenes,” pp. 231–246.
T. H. Vaughn, “A New Reaction of 1-Iodoacetylenes and Some New Mercury Acetylides,” J. Am. Chem. Soc. 1933, 55(8), 3453–3458 (DOI: 10.1021/ja01335a073).


How to remove sugar from juice My son has autism and we noticed that blueberries make him feel much better. The problem is blueberries contain sugars which interfere with diet he has (low carbohydrate).

How do I filter out the sugars (fructose?) without interfering with other compounds present in the juice? We can't add yeast and wait. Perfect option would be to add something that interacts only with the sugars and produces something neutral to the human body or is easily removed.

A blueberry contains ~85% water, 5% glucose, and 5% fructose (and it is generally considered a low glycemic index food, see e.g. this site). The solubility of fructose in water is extremely high (wikipedia says ~4000 g/L (25 °C), pubchem says "freely soluble in water"), and that of glucose is pretty high too, so there is essentially zero chance of removing these sugars through crystallization or other method without significantly altering the fruit or its juice.

If you are still interested in generating an extract of the fruit, this article might be useful. It suggests two different approaches, juicing and drying. The first will generate juice and a press cake, and since the sugars are highly water-soluble the cake should be the reduced-sugar component of interest to you. You could also freeze dry the juice, assuming this contains the important components, then partially rehydrate the dry product and attempt to crystalize out sugars. Or you could dry the whole fruit and then attempt a separation with ethanol, with sugar extracted into the alcohol to obtain a reduced-sugar extract which can then be dried. Further separations with more apolar solvents might extract hydrophobic "phytonutrients" from the juice. But you are likely to remain somewhat in the dark with respect to what you've removed from the berry without further analytical tools to help you characterize the products.

There are alternatives to extraction of the sugar that involve biochemical conversion of the monosaccharides. One alternative is to treat crushed blueberry with xylose isomerase, which would convert the glucose to fructose. Fructose has a low GI ranking, which might be a benefit. Another alternative is to treat the blueberry with a mixture of enzymes (perhaps including sucrose-sucrose 1-fructosyltransferase (1-SST), see e.g. this example) that converts the glucose/fructose mixture into largely indigestible polysaccharides (soluble dietary fiber).

Otherwise the best solution might remain to eat the fruit raw or to seek other foods with similar composition and without sugar (perhaps you should consult a dietitian).


3rd analytical group of cations If a mixture of Mn$^{2+}$, Zn$^{2+}$, Co$^{2+}$, Ni$^{2+}$ and Al$^{3+}$ salts is dissolved in water, a white dim solution is obtained. I guess that the aluminium and zinc ions hydrolysed and the respective hydroxides are formed. Am I right, or maybe one of them won't hydrolyze so much that the solution becomes dim or maybe some other cation could have hydrolysed, too?

Aluminium and zinc will produce a white precipitate due to formation of hydroxide. Both these hydroxides dissolve in excess acid or alkali due to their amphoteric character.

Manganese, nickel, cobalt hydroxides are probably soluble as they are precipitated in the 4th group as sulphides.


Look for superabsorbent balls whose expansion is reversible I am glad to find very cheap superabsorbent balls (see image below). However, I found their expansion is not reversible. The wiki entry 'Superabsorbent polymer' also says that the material will 'retain' extremely large amounts of a liquid relative to their own mass.

Is there any other superabsorbent material whose expansion is reversible? If yes, is it available as tiny balls?

Yes. The ones I have are known commercially as Jelly Marbles or Orbeez. They're sold at gardening stores, home decor, and online. They're used as toys but also as gardening products that absorb excess moisture in wet conditions and supply water in dry conditions. Mine grow up to about 1.5cm from a starting diameter of ~0.1cm. I've been told that some grow up to ~7.5cm; and they're advertised as increasing by 100-300 times in volume at full saturation. The process is reversible (e.g. by putting them in a cup with a wad of paper towels). I use them as a drying agent where powdered $\ce{CaCl2}$ etc. aren't ideal.

Note the actual identity of the polymer in the beads is a trade-secret, so I don't think you'll be able to find that information.


Gay-Lussac law explanation I was wondering if someone could explain me what exactly is Gay-Lussacs law?

And my textbook says that Berzelius and Dalton concluded based on that law that all gases have same amount of particles. How?

Gay-Lussac's Law, in its most basic form, says that, assuming constant volume (V) and amount/moles (n), the pressure (P) of a gas is directly proportional to the absolute temperature (T) of a gas.  We can write this mathematically in a couple different ways:

$$ P = kT$$
 (assuming constant V and n), where k is some constant.  More often, we use it in the form below, where we know the pressure and temperature under one set of conditions, and then change those conditions (again, keeping V and n constant):

$$
\frac{P_1}{T_1} = \frac{P_2}{T_2}
$$

As for the next statement, "all gases have the same amount of particles", as stated this makes no sense.  Perhaps what you meant to say was that, under identical conditions, equal volumes of an ideal gas contain equal number of moles (and therefore equal numbers of gas particles).  For example, it can easily be shown using the Ideal Gas Law that one mole of an ideal gas occupies 22.4 L at STP.


Why does a carbocation undergo covalent bonding with, say, chloride, rather than ionic bonding, like what occurs between Na+ and Cl-? I've been trying to look for an answer for a while but I can't find one. Why does a carbocation accept an electron pair from a nucleophile and form a covalent bond rather than just accepting one electron like what happens between a sodium ion and a chloride ion in the formation of sodium chloride?

Some carbocations do form true ionic compounds.  Many are known in which the positive charge is incorporated into an aromatic ring (e.g., cyclopropenyl, tropyl) or conjugated with multiple aromatic rings (e.g., triphenylmethyl).  But in most carbocatioms you have a compact, relatively low-energy vacant orbital that will overlap to form a bond with an electron pair from, say, a chloride ion.  Alkali metals don't have such compact valence orbitals and thus do not form a strong covalent bond even with the positive charge all on one atom.


Bioplastics from eggshells There are many resources online which talk about projects making bioplastic from egg shells. However none of these actually show or describe the procedure needed to make it. Does anyone know a site/paper which one could use to figure out how to do this for a science fair project?

Eggshell is 95% calcium carbonate. It seems more likely to be used as an additive in a composite plastic.  Testing composite plastics produced from adding ground eggshell to common plastics may make for an interesting science project.

References:

Dissertation on meat and bone meal derived plastics

Patricio Toroa et. al. Eggshell, a new bio-filler for polypropylene composites


Why is a metabisulfite / starch / HCl solution unstable? In one variation of the iodine clock demonstration, you add two colorless solutions together and apparently nothing happens. After about 30 seconds to a minute, you get an "instantaneous" change to a dark blue color.

Solution A is $\pu{0.006 M}$ $\ce{KIO3}$ and is stable over at least a year or two.

Solution B is made with starch, $\pu{0.2 g}$ of sodium metabisulfite ($\ce{Na2S2O5}$), and $\pu{1 mL}$ of $\pu{1 M}$ $\ce{HCL}$ in $\pu{200 mL}$ of water. Starch can be from "reagent grade starch", or starch from a packing peanut, or from laundry starch bought at the grocery store. The concentration of the starch ends up at about 0.05% w/v.

Solution B must be made up no more than 24 hours in advance or the reaction doesn't work: the blue color doesn't form or the timing of the reaction is way off, depending on how old the reagent is.

What happens in solution B that makes it unstable, and is there an easy way to stabilize it so that I could have the chemicals needed for the demo ready on the shelf?

The reactions that happen in the demo are:
\begin{align}
\ce{S2O5^2- + H2O &-> 2 HSO3-} & \tag{1}\\
\ce{IO3- + 3 HSO3- &-> 3 SO4^2- + I- + 3 H+} &\text{(slow)} \tag{2}\\
\ce{IO3- + 8 I- + 6 H+ &-> 3 I3- + 3 H2O} &\text{(slow)} \tag{3}\\
\ce{I3- + HSO3- + H2O &-> 3 I- + SO4^2- + 3 H+} &\text{(fast)} \tag{4}\\
\ce{2 I3- + starch &-> blue complex + I-} &\text{(fast)} \tag{5}
\end{align}

I don't see anything in these that tell me why the metabisulfite / $\ce{HCl}$ / starch reagent is unstable.

I can partially answer the question. I know what is happening to solution B. I don't know of a more stable version.

One potential problem is the acid-catalyzed hydrolysis of the starch into glucose, which does
not form the blue complex.
$$\ce{nH2O + (C6H10O5)}_n \ce{->nC6H12O6}$$

For the timing to be off, the metabisulfite must also be decomposing, either by reducing glucose or by reacting with dissolved oxygen. I'm not sure what the specific process is, nor do I know how to stop it.


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

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

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

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

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

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

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


What can cause artesian well water to smell? Recently, our water from the artesian well started to smell (again). It only smells bad for only few seconds than the smell disappear.  We have sulfur-iron removal system, water softener, UV lamp. Only the water from the system I listed above smell, because we also have a Reverse Osmosis installed in the kitchen and the water filtered from it does NOT smell.

We did a hydrocarbon test that was negative. We also did a normal "drinkable water" analysis and it passed with nothing ususual. My question is what are the possibilities of that smell?

PS. I can confirm the smell is not sulfur (different from rotten egg)
See this post for our water results before installing the system if it may be useful

As mentioned in the comments, water is often treated by ozonation in order to sterilize it and ensure it can be drank with no ill effects. Ozone is a very reactive gaseous allotrope of oxygen, produced in situ by exposure of atmospheric oxygen to UV-B light (in fact this happens naturally in the stratosphere, generating a thin ozone blanket covering the world). Ozone can be produced artificially by using a mercury lamp, which emits large amounts of UV light. The water is then brought into contact with the ozone in order to be purified.

When water is collected after being ozonated, typically a small amount of ozone is still dissolved in the water. If the water is allowed to rest for a few minutes after being collected, the residual dissolved ozone diffuses out. Ozone has a very strong, characteristically sharp odour, which can be detected even at very low concentrations. This is likely why the asker's water had an odd smell. Ozone is toxic, though, so exposure should be avoided.


Finding major product after dehydration here,should I prefer hydride shift or a product with more $\alpha$ hydrogens?
my doubt is...as hydride shift is easier than methyl shift,the answer should be $a$ or $c$.if methyl shift takes place the alkene formed has more $\alpha$ hydrogens than $a$ and $c$.what should I prefer? The question is to find major product.also,irrespective of the answer I would like to know which is more stable $a$ or $c$?

Compound $a$ is stabler than compound $c$ because the alkene in $a$ has $7$ hyperconjugative structures due to $7 \alpha$-Hydrogen, whereas $c$ has only $5$ hyperconjugative structures.
Hyperconjugation or Nathan-Baker Effect is a stabilizing phenomenon due to resonance-like charge delocalisation. Also, the reason why here hydride shift takes place is because it leads to a stabler carbocation.
Hence the product should be $a$.


What are the units for m in the average kinetic energy formula? For the equation $E_k=\frac{1}{2}mu^2$ what does the $m$ variable represent?

I would say mass but the textbook does not say explicitly that it is anything.

$$\ce{Kinetic ~Energy ~= ~1/2 (mass~of ~the~object~ [kg]) * (velocity~of~the~object~[m/sec])^2}$$


In synthesizing PbS nanocrystals will there be any problems if I use lead IV acetate instead of lead II acetate? I am just about to start my senior project synthesis of PbS quantum dots and we ordered the wrong lead acetate. The method calls for lead (II) acetate and we got lead (IV) acetate. If we use the lead (IV) acetate will our synthesis be contaminated and not work or can we still use that chemical with two extra acetates attached?

You might run into problems. $\ce{Pb(OAc)4}$ is a pretty strong oxidant!

What will be your sulfide source? Are you sure that it isn't oxidized these conditions?

Edit 1

At least N,N'-diarylthioureas are efficiently desulfurized by lead tetraacetate and furnish ureas (and sulfur) via a carbodiimide intermediate, see Bull.Chem. Soc. Jpn., 1984, 57, 315-316.

Edit 2
If you don't mind a personal advice: I wouldn't want to add complexity to the protocol just to maneuver around problems (and create others) I wouldn't have if I were using the right starting materials in the first. (The German term for that is verschlimmbessern, which translates to something like disimprove).


Is temperature of crystallisation same as melting point temperature for polymers? Melting point tends to be a temperature range. Is it true that the temperature of crystallisation is the highest temperature at which the polymer melts and thus there is some distinction between temperature of crystallisation, Tc and temperature at melting point, Tm?

Is temperature of crystallization same as melting point temperature for polymers?

In a word: NO.

If you undrestand what the glass transition, crystallization and melting temperatures are it will make sense why they are not the same. This paper does an excellent job at explaining the three concepts but I will summarize.

The glass transition temperature is the temperature where polymer chains become sufficiently mobile that the material transforms from being brittle to being soft and plastic (neither exo- nor endothermic, but does the heat capacity does increase). The crystallization temperature is the temperature where the molecules are mobile enough to rearrange into ordered arrangements (crystallize not decrystallize; this is an exothermic process; see plot below). The melting temperature  is where the polymer chains can freely move and ordered arrangements are disrupted (this annihilates any prior crystallization; an endothermic process).

Source: Plastic Technology Laboratories


Where did all the elements come from? I heard somewhere (I think is was a show called "How The Universe Works") that the big bang was the result of a hydrogen atom and a helium atom. Is this true? If so, where did all the other elements come from?

This is a very broad question, so below I mention just the very key points.

Big Bang

According to the prevailing cosmological model for the early development of the Universe, known as the Big Bang theory, the Universe begun at about 14 billion years from a singularity that can be thought of a state of an infinite density and temperature, and since its creation the Universe began to expand and cool.
Right after the Big Bang the Universe was immensely hot, the energy density was incredibly high and no particles existed, but the Universe was extremely rapidly expanding and cooling, so that soon enough energy was allowed to be converted into various elementary particles: quarks, leptons, bosons.

Baryogenesis

It is generally assumed that initially there was an equal numbers of particles and corresponding antiparticles, but at some point an unknown physical process referred to as baryogenesis lead to a very small excess of quarks and leptons over antiquarks and antileptons.
As the Universe continued to decrease in density and fall in temperature quarks and gluons combined to form baryons such as protons and neutrons, and the small excess of quarks over antiquarks led to a small excess of baryons over antibaryons, which in turn after mass annihilation resulted in the predominance of matter over antimatter in the Universe.

Big Bang nucleosynthesis

After annihilation it was just the matter of time for the Universe to cool down to a point where remaining protons, neutrons and electrons could combine to form atoms.
Already a few minutes after the Big Bang neutrons combined with protons to form the lightest nuclei (H-1, H-2 or D, H-3, He-3, He-4, Li-6, Li-7, Be-7, and Be-8) in a process called the Big Bang nucleosynthesis, and roughly 378,000 years after the Big Bang the Universe cooled to the point when the formation of light neutral atoms (mostly H and He with trace amounts of Li) became energetically favored.
Eventually, most of the protons in the Universe were bound up in these neutral lightest atoms and matter was uniformly distributed in the Universe.

Stellar nucleosynthesis

But the gravitational attraction of the nearby matter by slightly denser regions and the subsequent growth of such regions resulted in formation of gas clouds, stars, galaxies, and other large structures.
The formation of stars meant that the extreme conditions (temperatures and pressures) for the nuclear fusion were recreated and light elements (H, He and Li) from which stars were formed started to turn into heavier one by a process referred to as the stellar nucleosynthesis. Stellar nucleosynthesis has its endpoint, namely, when nuclear fusion reaches nickel-56 which then undergoes radioactive decay into iron-56 the process stops. The reason is that iron-56 has one of the highest binding energies of all of the isotopes, and thus, no more energy can be extracted from an iron-56 nucleus, either by fusion or by fission. But it is not the end of the story.

Supernova nucleosynthesis

When the core of a star converts to iron-56, the nuclear reactions stop. And that is a big problem, since so far the star balanced the forces of gravity pulling its materials inward by nuclear reactions pushing them outward, but now all the reactions pushing outward stop. What happens then depends on how massive the star is. Supergiants, that have masses from 8 to 12 times the Sun, gravitationally collapse in supernova explosions accompanied by a large burst of neutrons. These neutrons are captured by heavy nuclei, such as iron-56, and the resulting neutron-saturated isotopes then undergo beta decays producing stable isotopes of elements heavier than iron.


Is it safe to heat fluorite (CaF2) to 140 °C? For the manufacture of the optical system, we need to heat the fluorite ($\ce{CaF2}$) lenses to $\pu{140 °C}$. Is it possible:

for some poisonous gas to be emitted? ($\ce{F2, HF}$ or other)
for defects to appear on the glass? (scratches, chips, cracks and the like)

If yes, please supply sources describing the corresponding scenarios.

The melting point of $\ce{CaF2}$ is over $\pu{1400°C}$. There is no way $\ce{F2}$ can be generated without electrolysis of the molten $\ce{CaF2}$. $\ce{HF}$ is produced by reaction of conc sulfuric acid on $\ce{CaF2}$. Concentrated $\ce{HCl}$ does react with fluorite to some extend, but the reaction is very slow so, unless your operators drool like the monster from Alien, there is no chance of $\ce{HF}$ formation.


Why other benzene derivatives do not undergo nitrosation Solomon and Fryhle $11 th$ ed :

Most other benzene derivatives except phenols and tertiary aromatic amines do not undergo C- nitrosation reaction.

I agree that if there is a primary or secondary aromatic amine, the reaction will follow other routes but in case of other activating substituents EAS is the only route left for the electrophile $\ce{-^+N=O}$ for example with anisole.

Also it seems to contradict  ron's answer

So which one is correct?

Anisole can be nitrosated in good yield reference here.  This, and other references, note that the PhOMe group is unstable to the reaction conditions being cleaved to give phenol.

m-Xylene can be nitrosated in good yield, o-xylene and toluene in modest yield in TFA or sulfuric/acetic acid mix under nitric oxide reference here

Nitrosonium tetrafluoroborate will also nitrosate a range of electron-rich aromatics, including anisole and xylenes reference here

This suggests to me that the authors of the textbook quoted are in error, as two of the references predate the publication date of the textbook.


How can I calculate the pH of the solution created by reacting sodium with water? $\pu{23mg}$ of sodium was added to $\pu{100cm3}$ of water.

  What is the pH of the solution produced?

$$\ce{2Na + 2H2O -> H2 + 2NaOH}$$

So firstly to find the moles, I've divided $\pu{0.023g}$ by $23$ to reach $\pu{0.001 mol}$.

Am I correct that the concentration of $\ce{NaOH}$ would be $\pu{0.01mol\,dm^{-3}}$?
$[\ce{OH-}] = 0.01$ which should be multiplied by $2$ due to the balancing numbers.
So $K_\mathrm{w}$ divided by $0.02$ gives $5\times10^{-3}$
$$
\mathrm{pH} = -\log[\ce{H+}]\\
\mathrm{pH} = 12
$$

Is this correct?

You answer is right, but your reasoning is not.

Amount of $\ce{Na} = \frac{0.023}{23} = 0.001$ mole

$[\ce{NaOH}]= \frac{0.001}{0.100} (\mathrm{moleL^-1}) = 0.01$ M

$\downarrow$

$p\ce{OH} = 2 \to p\ce{H} = 14 - 2 = 12$


Whats between molecules? Okay, this will be tricky to explain. But, this is a question that i have been wondering about in quite a long time, okay lets get started, Molecules moves faster in higher temperatures and there will be more space between the molecules than in cooler temperatures, right? i wonder what is between the molecules? in pure water to an example, there should be only oxygen and hydrogen. But, what is between the H2O molecules? it got to be something, right?

I hope you guys can understand my question, my english is not the best and i am no expert in chemistry. :)

Just because they move faster doesn't mean there's more space between them. Here's why:

We need to look at several important things in this case: how many molecules, what's the temperature and what's the space they are in. For this example it is most easy to understand if we say we got the same space and the same amount of molecules, we just want to change the temperature.

If we got 1000 molecules (to make the math easy) in a liter (=1000 mL) then in average each molecule has free space around it of 1 mL. It doesn't matter how fast the molecules move, there are always 1000 molecules in 1000 mL so on average each molecules does always have 1 mL of space.

And in addition: between molecules is just nothing.


Why are acetals stable to bases and nucleophiles? Why are acetals stable to bases and nucleophiles?

Could it be due to electronic effects? Lone pairs on the two oxygen atoms create an unfavorable environment for the approaching nucleophiles and bases. Or could it also be steric factors?

Keep in mind that in these reversible reactions of carbonyl compounds, the mechanism for the forward and reverse reactions must go through the same intermediates. The result is that if you can't make a particular functional group under particular conditions, that functional group can't be destroyed under the same conditions.

Hydrates (as shown in the second reaction in ron's answer) and hemiacetals can be formed/decomposed under either acidic or basic conditions. In the hemiacetal formation under basic conditions, an alkoxide (methoxide below) attacks the carbonyl. Protonation gives the hemiacetal. In order to proceed to the acetal, the alkoxide would have to displace the hydroxide in an SN2 reaction. Hydroxide is not a good enough leaving group for SN2 reaction, so it is impossible to form the acetal under basic/nucleophilic conditions.

Hemiacetals can be decomposed to a carbonyl and alcohol under basic conditions. Deprotonation of the hydroxyl starts the mechanism, and the resulting anionic oxygen can kick out an alkoxide as a leaving group to give the carbonyl compound. Note that in this carbonyl chemistry hydroxide and alkoxides are competent leaving groups in an anionic elimination mechanism, such as the one here.

In an acetal, without the free hydroxyl group, alkoxide/hydroxide has nothing to deprotonate and there are no suitable leaving groups for SN2 reaction. There are no reasonable pathways for a nucleophile or base to react with the acetal.


Purity of calcium carbonate in limestone $\mathrm{150~mL}$ of $\mathrm{0.1~N}$ $\ce{HCl}$ is required to react with $\mathrm{1~g}$ of a sample of limestone. Calculate percentage purity of calcium carbonate.

Reaction: $$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

Number of milli equivalence of $\ce{HCl}$ used initially is $\mathrm{15}$

Number of milli moles of $\ce{HCl}$ used initially is $\mathrm{\frac{15}{2}}$

$\mathrm{2}$ moles of $\ce{HCl}$ reacts with $\mathrm{1}$ mole of $\ce{CaCO3}$.

So amount of calcium carbonate is $\mathrm{\frac{15}{4}\times 10^{-3}~mol}=\mathrm{\frac{3}{8}~g}$

But answer given is $\mathrm{75\%}$

We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.

Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$

...implying the purity is 75%.


What do final zeros to the right of decimal point indicate? Final zeros to the right of the decimal point are considered significant. What do those zeros indicate and why are they significant? For example, in 2.000 there are four significant figures.

"$2.000$" does not mean $2.000 \pm 0.0005$.
"$2.000$" does not mean the interval [1.9995,2.0005].
"2.000" means that there is an unspecified amount of uncertainty in at least the last, and possibly the last two digits.
See the NIST Good Laboratory Practice for Rounding Expanded Uncertainties and Calibration Values:

Example 5
The correction for a weight is computed to be 285.41 mg and the uncertainty is 33.4875 mg.
First, round the uncertainty to two significant figures, that is 33 mg. Then, round the correction
to the same number of decimal places as the uncertainty statement, that is, 285 mg

So in other words, for people who follow the NIST standard of stating uncertainty to two significant digits:
2.000 can mean anything from 1.990-2.010 to 1.901-2.099
Others only express uncertainty using one significant digit, in which case:
2.000 can mean anything from 1.999-2.001 to 1.991-2.009


Disposing of hazardous waste indicated by numbers on chemicals I'm doing a book for the African market and the syllabus asks for discussions on proper disposal of chemicals and waste products, particularly in the classroom setting

The authors said something along the lines of:

  Chemicals numbered 1 to 4 can be rinsed down the drain with water. Chemicals labeled 5 to 8 must be disposed of with the help of an Environmental Agent or official.

What are the numbers they mention? Is there some numerical system (similar to scheduling on prescription meds) which guides us on how to dispose of certain chemicals?

The numbering clearly is not a part of the international GHS (Globally Harmonized System of Classification and Labelling of Chemicals) labeling scheme.

On a side note, GHS requires

hazard pictograms,
coded hazard statements (Hnnn and EUHnnn)
coded precautionary statements (Pnnn)

but typically does not give any information on the how/where to dispose chemicals.

For classroom settings, the questions are:

How many different waste containers are needed?
How to label them?
Which chemicals go where?

To my knowledge, a global or EC-wide (European Community) scheme does not exist for this and I'd be astonished to find that the African Union has agreed upon a common scheme.

Instead, this is typically regulated on a nation-wise basis only.

To give an example: In Germany, the classification scheme is provided by the Deutsche Gesetzliche Unfallversicherung (mandatory accident insurance) and you may find their suggestions for waste disposal in schools here (PDF in German, Stoffliste zur Regel
„Unterricht in Schulen
mit gefährlichen Stoffen“, 2017).

Consequently, without knowing the (national) origin of the source you are citing in your question, it is very difficult to say which number refers to which compound class.


Deliquescence/efflorescence: How can I convert between salt concentration and RH? I am looking at two phase diagrams, one has axes of $T$ and $\%\ce{MgClO4}$ in a $\ce{H2O-MgClO4}$ mix, and the other has axes of $T$ and relative humidity. The authors of the second phase diagram say "The stability diagram is modiﬁed from [the earlier phase diagram] by converting salt concentration to relative humidity using the Pitzer model and parameters." How does changing the salt concentration affect the relative humidity, and how can I reproduce this calculation?

(I am a geologist with minimal chemistry background).

Top figure from Chevrier et al., GRL, 2009; bottom figure from Gough et al., EPSL, 2011.

You can't. The two phase diagrams are for two different parts of the system. There isn't enough data to go between the two phase diagrams. You have to use the two diagrams together to estimate the missing data (Relative humidity as a function of temperature over wt% magnesium percolate).

The top figure (Chevrier et al., GRL, 2009) is for a salt in solution. Think of having a sealed bottle with air over solution. The phase diagram shows  the relative humidity of the gas as a function of temperature. So at about 240 degrees the equilibrium system has about 45 wt% magnesium percolate in solution.

The solid circles in the lower figure (Gough et al., EPSL, 2011.) show relative humidity as a function of temperature. So at 240 degrees the relative humidity is about 50% for the the equilibrium system.

So at a temperature of equilibrium 240 degrees a solution of magnesium percolate will have about wt 45% of salt and the relative humidity of the air over the solution will be about 50%. You have to combine data from the two phase diagrams to figure out the whole system.

The open circles in the bottom bottom figure (Gough et al., EPSL, 2011.) shows a solid salt in various hydration states (5 vs 6 waters). This sort of detector would be like a crystal oscillator with a coating. As the hydration in the coating changes, then mass of the oscillator changes which changes its fundamental frequency. The chemical reaction (between hydrates with 5 and 6 molecules of water) is not temperature dependent hence the near vertical line.


Does an axis of symmetry determine chiralty? What I've understood about chirality and elements of symmetry:

A molecule that is not superposable on its mirror image is said to be chiral.

Plane of symmetry:
An imaginary plane cutting a compound into two symmetric halves.
Any compound possessing it is achiral, so any compound without it is chiral?

Axis of symmetry:
I have never spent so long trying to understand a concept.
An axis of symmetry is an imaginary axis around which by rotating through a minimum angle of rotation, the original compound is obtained.

1. Any compound, be it chiral, achiral or dissymmetric, can have an AOS.

2. You cannot say a compound is achiral just because it has an AOS.

3. Unless, I just had a thought. If on rotation it gives the same compound, making it superposable with the original, thereby making it achiral.

However, wouldn't all compounds (chiral or achiral) on rotation through a particular degree give back the same compound?
Please tell me if everything I've mentioned is right.

Chirality is left hand—right hand. The left hand is a mirror image of your right hand. If you take this mirror image and then rotate or shift it and you are able to superimpose it onto the original image, it is achiral. This is not possible with the hands, the thumb will always be wrong.

A plane of symmetry is sufficient but not mandatory to give an achiral compound. But it should be clear that mirroring an image that contains an intrinsic mirroring plane will give the same.

Another sufficient but not mandatory symmetry operation to generate a chiral compound is inversion. If some compound is point symmetric, it is never chiral.

And there are other elements of symmetry that are rare if not accompanied by planes or centres of symmetry but which are sufficient for generating achiral compounds: Improper rotation. In a nutshell, this is a symmetry element that requires a rotation with a subsequent mirroring perpendicular to the rotation axis. Take a tetrahedral compound such as methane: If you select the correct axis, rotate by $90^\circ$ and then mirror it perpendicular to that axis, you will receive the same product.

Strictly speaking, planes and centres of symmetry are special cases of improper rotation with a rotation of $360^\circ$ and $180^\circ$, respectively. So one could say that improper rotation is sufficient and mandatory to give an achiral compound.


Why Does an Area in Phase Diagram Considered as Having Degree Of Freedom=1? Why did the area in the center are Considered as Having 1 Degree Of Freedom. Aren't we able to change the concentration and the temperature at the sametime without changing the Phase?

Here is the link to the document https://www.google.co.id/url?sa=t&source=web&rct=j&url=http://www.colby.edu/chemistry/PChem/notes/GibbsPhaseRule.pdf&ved=0ahUKEwjw7Laq7M3QAhXGQI8KHXwKB3sQFggcMAE&usg=AFQjCNHWRoz8CuB0BSOoLJm-K0iPayNvgQ&sig2=vYhsTSV_woeOd2_GPPLdEw

The Gibbs phase rule is $f = c - p + 2$, where $f$ is the degrees of freedom, $c$ is the number of components, and $p$ is the number of phases.  Your diagram says $P = \mathrm{cst}$, which I interpret to mean that pressure is constant.  That specification lowers the number of degrees of freedom by one, so $f = c - p + 1$.

Inside the "center" region, liquid and vapor phases coexist.  Thus there are two phases, and (as with every point on this particular diagram) two components, thus $f = 2 - 2 + 1 = 1$.

Your question is why.

  Aren't we able to change the concentration and the temperature at the same[ ]time without changing the [p]hase?

Essentially, no, you can't do that.  Inside the center area, liquid and vapor coexist at equilibrium.  And -- this is the key point -- that equilibrium is another constraint that lowers the degree of freedom.  The compositions of the both the liquid and the phase vapor are not independent of temperature, given the constraint that the liquid is in equilibrium with the vapor.

The composition of the liquid phase will be given by the lower curve.  And the composition of the vapor phase is given by the upper curve.

Temperature as the single degree of freedom

So if you decide to use temperature as your (first and only) degree of freedom, then the composition of both the liquid and the vapor is fixed.  Say you pick an arbitrary temperature $T_{arb}$, which must obviously be between $T_{bB}$ and $T_{bA}$.  This temperature could be represented by a horizontal line on your chart.

The composition of the liquid phase is given by where that horizontal line for $T_{arb}$ intersects the lower curve on your chart.  And the composition of the vapor phase is given by where that same line intersects the upper curve.

Thus now that we've specified temperature, the compositions of both phases is also specified, and (given that pressure is constant) thus the system is fully specified.  No more degrees of freedom.

(Liquid) composition as the single degree of freedom

Instead of picking a temperature then let's say you pick an arbitrary liquid composition $x_{Aarb}$.  Then, given that vapor and liquid are in equilibrium, then the temperature is specified.  Your specified $x_{Aarb}$ is a vertical line on your chart.  The temperature of the system is given by the value of the ordinate where your $x_{Aarb}$ line intersects the lower curve on the graph.

Now that you know temperature, you can find the corresponding $y_{Aarb}$ by drawing a horizontal line at the newly identified temperature and finding where this horizontal line intersects the upper curve.

Extensive vs. intensive properties

The goal of the Gibbs phase rule is to describe all the intensive degrees of freedom of a system.  It doesn't tell you about extensive degrees of freedom.  The extensive degrees of freedom tell you "how much" of the system you have.  In this system, the extensive degrees of freedom are the absolute amounts of each phase.  (The compositions of each phase are intensive properties.)  In general for $p$ phases there are $p$ extensive degrees of freedom.

As a simple example, consider a single component system (e.g. pure water) at a constant pressure and in equilibrium with its vapor.  The Gibbs phase rule says $f = c - p + 1 = 1 - 2 + 1 = 0$, i.e. that there are no degrees of freedom.  But there can be any amount of water, from one picomole to one examole, and the equation is still true.  Since there are two phases, there can be one picomole of steam and one examole of liquid water, or vice versa.  The Gibbs phase rule isn't going to tell us how much of each phase we have.

The same idea holds in the two-component case that you are asking about.  The total amounts of liquid and vapor can vary in the center region.  The relative abundance of each phase would be given by the "lever rule" as described in this nice electronic lecture note from the University of Omaha.


How many d-electrons are in the outer shell of the metals in following compounds: How many d-electrons are in the outer shell of the metals in following compounds:

ZnS

Do I just use Zn^2+ and as I know that in Ions the s orbitals are removed first I count 10 d electrons?

However ZnS makes a metal lattice. Can I use Zn^2+ then?

Help would be really appreciated

In zinc sulfide, two s-electrons have been removed from $\ce{Zn(0)}$ to make $\ce{Zn^{2+}}$. Therefore, yes, it does have 10 d-electrons.

In its lattice, zinc is bound to four sulfur atoms which fulfills the 18-electron rule (10 from $\ce{Zn^{2+}}$ and 2 each from four separate $\ce{S^{2-}}$; the 18-electron rule for transition metals is the equivalent of the octet rule for main-block elements). In this case, however, the metal (zinc) still only has 10 d-electrons. The additional electrons are in hybrid, non-d orbitals.


Spectrochemical series of TMs: why does the field-splitting parameter NOT steadily increase along a period? Why doesn't the field splitting parameter increase steadily along the period for 3rd-row transition metals, as one would expect due to a better energy match between ligands and the metal as $Z_\mathrm{eff}$ increases?

The series is (in the order of increasing $\Delta$: $\ce{Mn^2+}, \ce{Ni^2+}, \ce{Co^2+}, \ce{Fe^2+}, \ce{V^2+}$ etc. Source: Spectrochemical series – Wikipedia, confirmed by Shriver and Atkins.

I would guess that this is to do with $\mathrm{3d}$ orbitals getting too contracted on the right of the period and, and hence, having a poorer size match, but I couldn't find this explanation mentioned anywhere.

A greater octahedral field split would be associated with a shorter $\ce{L\bond{->}M}$ bond length. Bond length is a function of:

size of both partners
strength of the bond

In particular, as we are examining metals in $\mathrm{+II}$ oxidation state, the metal ion’s size should increase somewhat steadily along the period while the ligand size should stay the same. This means that we need to be looking at bond lengths to understand the discrepancy.

For reference, here is a typical octahedral $\ce{[ML6]}$ complex ignoring π effects.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

We should immediately be able to see how field split is generated: it is the difference between the $\mathrm{t_{2g}}$ orbitals and the $\mathrm{e_g^*}$ orbitals; the latter of which are antibonding. Thus, having any electrons in $\mathrm{e_g^*}$ will weaken the $\ce{L\bond{->}M}$ bonds. This corresponds nicely with vanadium(II), a $\mathrm{d^3}$ ion, having the largest field split, and manganese(II) ($\mathrm{d^5}$) having the lowest.

However, the final value is dependent on very subtle differences and not easily predicted a priori.


Stereochemistry of cycloalkanes having substituents My teacher said that from $3$ to $6$ member cycloalkanes having any number
  and type of substituents we can always draw the planar structure and
  decide the specifics of stereochemistry like:

  Are they optically active or not?
  Are they meso compunds or not?
  Do they have a plane of symmetry/centre of symmetry/axis of symmetry/alternating axis of symmetry or not?

But I feel that the molecules are hardly in planar form. It would be incorrect to assume they are planar and find the specifics of stereochemistry. Only for cyclopropane the points should be correct as it is always planar. Maybe my teacher said that mistakenly in a hurry.

So regarding points $(1),(2)$ and $(3)$ who is correct? Why? Or are we both partially correct? I need a slightly detailed answer.

For a compound to be optically active it must not be superimposable with its mirror image. It's as simple as that. Also note that stereogenic centres are not always required for chirality. There are other types of chirality such as axial and planar chirality where a compound possesses either an axis or plane of chirality. Sometimes it can be the case that just the sterics of a molecule does not allow for free rotation of bonds and the molecule is locked one configuration. I think in the simplest cases of open chain molecules, you could indeed use the structural formula, but with rings which adopt specific 3D configurations, you must refer to the 3D structure.
Again, as said in 1., to determine if a molecule is chiral, it's better to refer to its 3D structure, especially when with meso, when you are dealing with more than one stereogenic  centre. For a compound to be meso, it must not be chiral, but posses stereogenic centeres (look up the definition).
No, you cannot use the simple structure to determine symmetry elements present if that structure is not really planar. It's not necessary to build a model, but you must use wedged/dashed bonds to show stereochemistry.

If someone thinks differently, please comment and correct me.


What is the electronic configuration of Cr (+1) ion I believe it should be $\ce{d^5}$ , but i found out that it is $\ce{d^4s^1}$. Is it so?
I have not found any explanation for the following. So why the strange configuration?

I assume we are talking about the free ions in the gas phase.

I checked my notes and it seems that the electronic configuration of $\ce{Cr+}$ should be $[\ce{Ar}]\mathrm{3d^5}$, the reference being a rather old book: Phillips & Williams' Inorganic Chemistry (1966). I tried to look in slightly newer textbooks, but I couldn't find any information on the monovalent ions.

I would not claim to be 100% certain on this, but it can certainly be rationalised by the increase in $Z_\mathrm{eff}$ in the $\ce{Cr+}$ ion as compared to neutral $\ce{Cr}$ (one less electron means less shielding and greater effective nuclear charge). That would serve to stabilise the 3d orbital more than it stabilises the 4s orbital, and therefore favour the $\mathrm{3d^5}$ configuration over the $\mathrm{3d^44s^1}$ configuration.

In general, the electronic configurations of transition metals and their ions are a balance of delicate factors. It's not easy to rationalise the electronic configurations fully and there are several factors that will affect the configuration adopted, e.g. 1) nuclear charge and relative energy of 3d and 4s orbitals, 2) relative e-e repulsions in 3d and 4s orbitals, 3) exchange energy.

Some interesting discussion on the topic can be found here:

J. Chem. Educ., 1978, 55(1), 2
J. Chem. Educ., 1989, 66(6), 481
J. Chem. Educ., 1994, 71(6), 469


What do the different grades of chemicals mean? When purchasing chemicals from Sigma, Fisher, or wherever, there are often -grade's attached to their description like reagent-grade, technical-grade, analytical-grade, or more niche-sounding biotech-grade, HPLC-grade, DNA grade (DNase free perhaps?)

Is there some sort of standard for what these actually mean or are they arbitrary, differing from supplier to supplier or even chemical to chemical?  Beyond the 'niche' grades, is there a common order of purity between the others or do they depend on the types of impurities?

Sigma-Aldrich gives a very useful table outlining what the different purity levels are and suggested applications. I was less successful at finding equal documentation from some of the other suppliers, but the analysis for the purity of the chemicals they sell is in the catalog and on the bottle.

In general, technical grade or laboratory grade are the lowest purity. ACS Reagent grade means that the chemical conforms to specifications defined by the Committee on Analytical Reagents of the American Chemical Society (but Aldrich "ReagentPlus" means >95% pure). So, "ACS Reagent grade" chemicals should be comparable from different suppliers.  Analytical grade is generally the most pure.

For some of the other grades, such as HPLC grade solvents, the issue is less about overall purity than about being free of substances that would interfere with a particular application. I've reproduced a piece of the table below to give you an idea of the information available.


Is constant pressure heat capacity = entropy during adiabatic reversible expansion of perfect gas? For any process, ∆G = ∆H - ∆(TS).

Apply this on reversible adiabatic expansion of perfect gas, since the entropy of system does not change during this process, S is constant so

∆G = ∆H - S∆T

The process is reversible, so

∆G = 0

and hence

∆H = S∆T

For perfect gas, assume heat capacity independent of temperature,

∆H = Cp∆T

and so

Cp∆T = S∆T.

∆T is not zero during adiabatic reversible expansion- temperature drops. So divide both sides by ∆T gives

Cp = S

Is this legit? If not where is my mistake? I have a hunch that I did something terribly stupid. The final result seems to be so counter-intuitive because

The Cp and S are constant throughout the process, so at the end of the process the two will still be the same. By measuring Cp, even though your system is not doing adiabatic reversible expansion, you will get S. So we are able to measure S?

Inspired by Chet Miller.
The fallacy in the argument is that, ΔG = 0  because the process is reversible. This is not true and so leading to wrong final result.
ΔG is a signpost for spontaneity of a process only if the process occurs through constant temperature and pressure. Likewise, that ΔG = 0 means reversibility, is true only if the process happens through constant temperature and constant pressure.
Adiabatic expansion clearly involves change in temperature, so ΔG is not a signpost for spontaneity of adiabatic change, and hence ΔG ≠ 0 for reversible adiabatic expansion.


Trends in solubility of group 2 nitrates In my lab report, we are required to explain the trends in solubility of group 2 salts, going down the group. I had explained all of the trends except one, group 2 nitrates. The following is the data provided.

$\ce{Mg(NO3)2}$ – $\pu{0.49 mol}$ per $\pu{100 g}$ of water

$\ce{Ca(NO3)2}$ – $\pu{0.62 mol}$ per $\pu{100 g}$ of water

$\ce{Sr(NO3)2}$ – $\pu{0.16 mol}$ per $\pu{100 g}$ of water

$\ce{Ba(NO3)2}$ – $\pu{ 0.04 mol}$ per $\pu{100 g}$ of water

So as the data shown, the solubility first increases and then decreases. My speculation is that the general trend should decrease but $\ce{Ca(NO3)2}$ has some special properties which makes it more soluble or $\ce{Mg(NO3)2}$ has some special which makes it less soluble. I searched the internet and all I got is "All group 2 nitrates are soluble" with no explanation regarding the trend.

Please correct my speculations if they're wrong and also provide an explanation regarding this trend.

Solubility is one of the present scientific problems that nobody is able to explain  thoroughly. As you are speaking of Calcium compounds, look at the series of Calcium compounds made with halogens. $\ce{CaCl2, CaBr2, CaI2}$ are so soluble in water that they can be dissolved in less than their weight of water. $\ce{CaF2}$ should be similar. It is not. On the contrary, $\ce{CaF2}$ is one of the most insoluble compounds, because it forms the mineral called fluorite, which is one of the main sources of Fluorine atoms on Earth. If it would have been at least partly soluble, it would have been washed away by the rains a long time ago. Nobody is able to explain such a difference of solubility.

There are plenty of theories explaining the solubilities of some groups of substances using electronegativities, ionic or covalent radius, and other parameters, which work pretty well with a lot of substances. But there are always exceptions. The final explanation remains to be discovered.


Defining sensitivity of TOF Mass Spec - Does the detector play a role? Looking at TOF MS systems, my current understanding is that temporal resolution results in mass resolution - so by controlling the flight time (with distance) and sampling rate, you can start to distinguish smaller and smaller changes in mass. Hopefully, this is correct.

Where I'm confused is when people start talking about sensitivity.
I heard a talk where they mentioned femtogram and attogram sensitivity.

Does this refer to the amount of the sample they are using?
OR is this referring to the minimum amount of the analyte they can detect?
Does that translate to needing a certain number of ions to hit the detector in a single scan to have a signal significantly above the noise?
Finally, what types of detectors are capable of measuring so low and are there any issues with them?

Thanks for your help!
Cheers

The term sensitivity has been misused in analytical chemistry a lot (even in German scientific literature). Be careful if you read articles especially the older ones. Sensitivity, in a true sense, refers to the slope of the calibration curve. Thankfully IUPAC sorted this out and defined it as "The slope of the calibration curve. If the curve is in fact a 'curve', rather than a straight line, then of course sensitivity will be a function of analyte concentration or amount. If sensitivity is to be a unique performance characteristic, it must depend only on the chemical measurement process, not upon scale factors."

When we wish to refer to the minimum amount that can be detected, it is called the limit of detection (LOD). Hope this clarifies the terminology. Note that sensitivity and limit of detection are related. Higher the sensitivity (= slope of calibration curve), lower the detection limit, provided that noise does not increase.

Coming to the detectors: The limit of detection of all detectors which are based on counting (basically we count ions indirectly in mass spectrometry) is limited by statistics rather not by electronics! This is the fundamental limit put by mother Nature. It is called shot noise. Whenever you count discrete random events, in case of a mass spec. random impinging of ions, there is always an error associated with it which is proportional to square root of N, where N is the number of events.

You can read the article on shot noise (for educational purposes).The Ultimate Limit in Measurements by Instrumental Analysis: An Interesting Account of Schroteffekt and Shot Noise, https://pubs.acs.org/doi/abs/10.1021/acs.jchemed.8b00202


Why doesn't ionization energy decrease from O to F or F to Ne? I know that in general, the first ionization energy increases across a period due to increasing nuclear charge, reasonably constant shielding & decreasing atomic radius.
From N to O, however, the I.E anomalously decreases. The reason usually given is that the spin pair repulsion between the 2px2 electrons in oxygen outweigh the effect of the increased nuclear charge.
However, the additional electrons in F & Ne are also spin-paired. Why doesn't the extra spin pair repulsion outweigh the extra nuclear charge in this case? http://www.chemguide.co.uk/atoms/properties/ies.html#top says

Between nitrogen and oxygen, the pairing up is a new factor, and the
repulsion outweighs the effect of the extra proton. But between oxygen
and fluorine the pairing up isn't a new factor, and the only
difference in this case is the extra proton. So relative to oxygen,
the ionisation energy of fluorine is greater. And, similarly, the
ionisation energy of neon is greater still.

But I don't understand how being a new factor or not makes any difference on its effect. It's not like the atom adapts to the new factor in O, reducing its effect its effect in F & Ne

It's a good question, and I agree that it can be quite confusing, but Chemguide is (at least partly*) correct about this.
I'd phrase it this way, which I hope is clearer. First, think about what would happen if the "paired electron" factor was not a factor at all. We would then expect that, going from B to Ne, there would be a roughly constant increase in ionisation energy. It would look something like this. Please don't take the exact shape or numbers too seriously; they don't mean anything, only the trend does.

This is obviously not the reality, because in O, F, and Ne there is the additional factor to consider, which is that the electron being removed is already paired in an orbital, making it easier to remove. This means that the ionisation energies of O, F, and Ne are each decreased by a roughly constant amount (red arrow).

You can see that we recover the observed trend in ionisation energies, and we have indeed accounted for the pairing energy in all of O, F, and Ne.
There are two alternative ways of looking at it which may "click" better for you. I'll describe these in text.

Going from N to O, the nuclear charge increases by 1 unit, which on its own should cause the ionisation energy to increase. The pairing energy consideration, however, overrides this. So far so good.
Going from N to F, the nuclear charge increases by 2 units, which on its own should cause the ionisation energy to increase roughly twice as much. Therefore, to make IE(F) smaller than IE(N), you would need two counts of the pairing energy. But we can't do that: you are only ionising one electron each time, so you can only subtract one count of the pairing energy.
The argument can be easily extended to N vs Ne.

Going from O to F, the pairing energy does decrease the ionisation energy of F. However, the pairing energy also decreases the ionisation energy of oxygen, so this factor on its own cannot be used to say that IE(F) should be lower than IE(O), as it affects both. The same is true of F and Ne.

* Exchange energy, which Chemguide does not cover, is also a consideration. However, that's a smaller effect, and it's not necessary to invoke that to answer this question.


Multiplicity of orbitals in a magnetic field Solving past questions preparing for a test, I am stuck at number 11. Is multiplicity the ml value

$m_l$ is the magnetic quantum number, and multiplicity refers to the number of allowed values that $m_l$ can be.


What is a common cause of peak tailing? What is a common cause of peak tailing?
a. The distribution constant decreases as analyte concentration increases.
b. The distribution constant increases as analyte concentration increases.
c. The plate height increases as analyte concentration increases.
d. The plate height decreases as analyte concentration increases.

Simple answer to your question should be - None. Whoever designed this question for students does not understand the basic principles of analytical chromatography.
Under low concentration of analyte, which is the usual case, peak tailing has kinetic origins, or it could be from secondary interactions or due to the structure of the packed bed. All peaks in real chromatography tail to some extent i.e., none of them are 100% perfect Gaussians.
In rare cases, you can inject a lot of sample on an analytical column. There peak tailing  originates due to non-linear isotherms. This is what the test question is about. You have to read this section on isotherms Distribution Isotherms in Chromatography to choose the right answer.
There is another form of chromatography, called preparative mode, where you really inject a lot and this is where isotherms come in.


What is the meaning of "no Hessian Eigenvalue "? When running GAMESS for finding Hessian Eigenvalues and frequencies ( using Optimization with hessian calculation at the end ) , I have seen there are times that it comes out "without any frequencies".

I think even a non-optimized system should have some vibrational properties and I can not understand what such a " no-frequency state " means physically.

I suspect the problem is pretty simple: hessian is not evaluated when geometry optimisation does not converge to a stationary point. It is actually mentioned in GAMESS manual (emphasis mine)

  HSSEND = a flag to control automatic hessian evaluation at the end of a successful geometry search. (default=.FALSE.)

And such behaviour is quite standard. The thing is that the conventional vibrational analysis (as descibed in most texts and implemented in most programs) is valid only when the first derivatives of the energy with respect to displacement of the atoms are zero. Or, in other words, the geometry which is used for vibrational analysis must result from a previously converged geometry optimisation run (at the same level of theory, of course).