What does equilibrium constant mean? I really don't understand equilibrium constants: they seem to change nonstop! Can someone clarify the definition of equilibrium constants and give me examples of them?

Also, I looked in my textbook and the explanation for heterogeneous equilibrium used an example of decomposing calcium carbonate.

$$\ce{CaCO3 (s) <=> CaO(s) + CO2(g)}$$

The textbook continued to explain that $K^{'}_{c} = \frac{[\text{CaO}][\text{CO}_{2}]}{[\text{CaCO}_3]}$

There is my first obstacle: What is the meaning of $K^{'}$ (I don't think they're talking about the reverse process equilibrium constant)? And the next part to me is even stranger...

I do get that the concentration of the two solids are basically constants, so we can treat those parts as constants.

But why would you do $K^{'} \times \frac{[\text{CaCO}_{3}]}{[\text{CaO}]}$, which equals $K$, according to the textbook, the actual equilibrium constant of this reaction?

I am sure the book, if it is any decent, should have said something like this already. $K'$ is not a measure of the forward reaction, nor is it a measure of the reverse reaction. It is a measure of the equilibrium position.

Equilibrium occurs when the rate of the forward reaction is equal to the rate of the backward reaction. By the laws of thermodynamics, this occurs at one special point where the instantaneous concentrations of all the compounds satisfy the equation

$$\frac{[\ce{CaO}][\ce{CO2}]}{[\ce{CaCO3}]} = K'$$

$K'$ for a reaction is a constant. Clearly, a large $K'$ indicates that when equilibrium is reached, there is a lot of $\ce{CaO}$ and $\ce{CO2}$, and only a little bit of $\ce{CaCO3}$. We say that the equilibrium position lies to the right. Conversely, a small $K'$ means that the equilibrium position lies to the left, i.e. reactants predominate.

The equilibrium constant is properly defined in terms of activities of the compounds, not concentrations. This means that for the reaction you gave,

$$K = \frac{(a_{\ce{CaO}})(a_{\ce{CO2}})}{a_{\ce{CaCO3}}}$$

The activity of a pure solid is defined to be equal to $1$, which means that you can simplify this by removing all the terms corresponding to pure solids:

$$K = a_{\ce{CO2}}$$

The "equilibrium constant" that you are learning that involves concentrations is a simplification.


Anti-aromatic or Non-aromatic Please tell if the 2 given conpound are Anti-aromatic or Non-aromatic,

I have confusions arising due to incosistent facts all around internet some of them are as follow :

1.) Arguments for Cyclopropenyl Anion

a.) Wikipedia which under the heading of "Effects on reactivity"
     clearly states that :

  "However, the cyclopropenyl anion has 4 π electrons in a cyclic system
  and in fact has a substantially higher pKa than 1-propene because it
  is antiaromatic and thus destabilized"

b.) This Answer on ChemSE which clearly states that its Non-Aromatic in
        Nature

c.) Also this NEPTL site which clearly states that:

  Cyclopropene is not aromatic because one of its ring atoms is sp3
  hybridized so it does not fulfill the criterion for aromaticity. But
  the cyclopropenyl cation is aromatic because it has an uninterrupted
  ring of p-orbital and (4n+2) π-system. The cyclopropenyl anion is
  antiaromatic as it has (4n) π-system.

2.) Arguments for cycloheptatrienyl anion

a.) This Question on ChemSE where the same question have 2 deflecting answer this answer is saying is that the anoin is antiaromatic and this states that the anion is nonaromatic

b.) Also my book (solomons and fryhle organic chemistry adapted for JEE) gives this as Antiaromatic

I don't know which one is correct, please help by providing a comprehensive answer with a reason that can be applied everywhere in such questions.

EDIT: Hello the one who closed the question can I ask the reason for closing as much I can do I abid by this guideline I asked my question straigthformward and then presented my research please tell me why my question is put on hold, I seek answers

Whenever you want a definition of a chemical concept, you should refer to the IUPAC's Gold Book which in this case states that antiaromatic compounds are

  compounds that contain  4 n(n ≠ 0) π-electrons in a cyclic planar, or nearly planar, system of alternating single and double bonds, e.g. cyclobuta-1,3-diene.

A corollary to this rule is that all the atoms must have an sp2-hybridization, so the orbitals can overlap. However, the carbon bearing the negative charge has an sp3-hybridization, i.e. they does not count as π-electrons but σ-electrons and that breaks the rule of alternating single and double bonds. So both anions are non-aromatic.

Edited after the OP opened my eyes on a big mistake I had made in answering his question. Sorry about that.


Entropy as Sum of Kinetic Energy? Is it possible to view entropy as the sum of the kinetic energy of all molecules in the system? Since a system with 0 entropy would have 0 motion, and as you increase the motion of the molecules, entropy increases as well. If not, could you provide a simple counterexample in which entropy does not make sense as the sum of kinetic energy?

In a crystal lattice, there are two contributions to the entropy: the vibrational entropy and the configurational entropy. The first is related to the kinetic energy; atoms with higher kinetic energies have tend to have larger displacements from equilibrium, and have a larger "footprint" in phase space, resulting in a higher entropy. The second contribution, configurational entropy, arises from disorder in the crystal lattice, namely the number of ways of arranging the atoms. In a perfect crystal, there is only one way, so the configurational entropy ($k\ln W$) will be 1.

An counterexample to the suggestion you make is any disordered solid, whether it's amorphous, a random alloy or some kind of crystal with geometric frustration. The most obvious example of this is water ice. Each water molecule occupies a lattice position and is tetrahedrally coordinated with its neighbours. In a defect-free ice crystal, the ice rules state that each molecule must accept exactly two hydrogen bonds and donate exactly two hydrogen bonds. This is quite a strong constraint, but allows orientational disorder --- the molecules can be oriented in one of six possible ways at each lattice position, resulting in an orientationally disordered structure. This will have a finite entropy even at absolute zero.


How to determine the hydroxide ion concentration in an acidic aqueous solution? When doing a titration with $\ce{HCl}$ and $\ce{NaOH}$, both being strong and dissociating completely, if you add a volume of $\ce{NaOH}$ that is less than needed to reach the equivalence point, you say all the $\ce{OH-}$ reacted with an equal amount of $\ce{H+}$ from $\ce{HCl}$.
It would be amount of substance $\ce{HCl}$ - amount of substance of $\ce{OH-}$ = $[\ce{H3O+}]$, and all of the $\ce{OH-}$ is gone, so $[\ce{OH-}] = 0$. But if you take the pH found using the $[\ce{H3O+}]$ value, use that to find pOH and then use that to find $[\ce{OH-}]$ you get a very small value, but not zero. So if I need to report a $[\ce{OH-}]$ value, do I just put 0 or the one gotten from pOH?

The problem that you are facing comes from the fact that technically, not all of the $\ce {OH-}$ reacts with the $\ce {H+}$.  It is only because the amount of OH- that doesn't react is so negligible compared to the amount that does, that the $\ce {[H_3O+]}$ can be taken to be $\ce{[HCl]-[$\ce{OH-}$]}$.

So, yes, to find the remaining (negligible) amount of $\ce{OH-}$, you must use the "very small value" given by either subtracting $\ce {pH}$ from $14 =\ce{pOH}$ or doing $\ce K_w/[\ce{H+}]=[\ce{OH-}]$ (which amounts to the same thing).


Iodoform reaction of alkyl halide? I read from a source that 2-iodopropane undergoes iodoform reaction. What reaction mechanism is undergone here, and why does this happen?

Yes, Aaron.

It will.

Because Iodide is very good leaving group. In iodoform test, we use base... which will convert 2-iodopropane into propan-2-ol. The latter will give haloform the same way as ethyl alcohol and any other methyl ketone does!


Le Chatelier's Principle: Temperature dependence of endothermic and exothermic reactions (chemical equilbrium) What effect does temperature have on endothermic and exothermic reversible reactions? Please also explain the reason why it responds like this.
And what do people mean when they say something like "increasing pressure or volume shifts the reaction to left/right"? What do people mean by "left/right reaction shift"?

In a reaction, there is an exothermic and a endothermic side. Increasing the temperature will favour the endothermic side.

So if your reaction is exothermic, that means it must release heat if it shifts to the left. Then, since the left is exothermic, it shifts to the right (endothermic) to minimise this disturbance.

If your reaction is endothermic, that means it must absorb heat if it shits to the left. Then, since the left is endothermic, if you increase the temperature it shifts to the left (endothermic) to minimise this disturbance.

The above is much easier to understand if you think of 'heat' as a product or reactant. In an exothermic reaction, heat is on the product side of the equation. So if you increase this component, to minimise this disturbance you must shift the equilibrium to the LEFT to decrease the 'heat' amount - and vice versa.


Do there exist metals that are composed of molecules? Most metals are composed of pure elements or are alloys (mixtures of pure elements.)

Are there metals that are composed of one type of molecule that is metallically bonded together?

The difference between a metal that is composed of one type of compound and an alloy is that in an alloy different elements are attached by metallic bonding but in a metal composed of one compound covalent or ionic bonding attaches multiple atoms together into compounds and then these compounds form metallic bonds with each other.

I have read that Gallium in its pure elemental state forms pairs which are then metallically bonded together which sort of is what I'm wondering about except that I'm wondering if metals can form out of more complicated compounds.

I would also expect there to be a lot of compounds which on their own do not form metals but do not degrade the metallicity of alloys they are put in very much.

I think a more specific and low level question might be: are there covalent compounds that can make metallic bonds with other things?

According to A Conducting Crystal Based on A Single-Component Paramagnetic Molecule,
[Cu(dmdt)2] (dmdt ) Dimethyltetrathiafulvalenedithiolate) J. Am. Chem. Soc., 2002, 124 (34), pp 10002–10003 :

  Single-component molecular metals should greatly extend the
  development of new types of molecular conductors. For example,
  while the first metallic molecule-based magnet has been reported
  recently, [reference 8] it may be possible to create such metallic molecular
  magnets from single-component molecules containing paramagnetic
  ions such as Cu2+ and Co2+.

Reference 8 is Coexistence of ferromagnetism and metallic conductivity in a molecule-based layered compound Nature 408, 447-449

  The temperature dependence of the resistance exhibits a metallic behaviour over the entire temperature range that we investigated (down to 2 K).


Coordinate Bond...(Dative Bond) Is NH4Cl Dative or Ionic or Covalent ?
I made the Lewis structure but I was not able to make it...
After many tries I posted this question
Some where I saw that they are made up of Ions but contain Covalent bond..

The bonding between $\ce{NH4+}$ and $\ce{Cl-}$ is ionic.

The bonding between $\ce{NH3}$ and $\ce{H+}$ in $\ce{NH4+}$ is a dative bond. (Since $\ce{H+}$ ion has no electrons, 2 electrons from $\ce{N}$ atom in $\ce{NH3}$ form the coordinate bond with $\ce{H+}$, thus forming $\ce{NH4+}$)

The bonding between $\ce{N}$ atom and $\ce{H}$ atom in $\ce{NH3}$ is covalent.


How do breaking chemical bonds turn into kinetic energy I’m trying to understand how breaking chemical bonds turn into kinetic energy. If AB is a chemical which breaks into A and B which zip away from each other, is it because when the bond breaks they suddenly repel each other? Another way I'm thinking of this is what happens if one single molecule of AB alone in a vacuum breaks apart.

I think the point I'm trying to make is the reaction is making the products move faster than when they started. It takes a force to make things move. So it has to be either gravity, electric, magnetic, or one of the nuclear. I'm pretty sure we can rule out nuclear and gravity.

Breaking chemical bonds does not turn their potential energy into kinetic energy. All bond dissociation enthalpies are positive (for stable bonds that actually exist), which means that breaking them is an endothermic or heat absorbing process, and therefore they absorb kinetic energy (heat is a transfer of molecular kinetic energy).

What you might be thinking of is exothermic reactions, where the enthalpy of the reactants is lower than the products, and so the reaction releases heat, or increases the kinetic energy of the surroundings.

The short answer to your question is "because energy is conserved, it has to go somewhere, and heat and light are the only places it can go."

I'll try to think about a better way to explain it that isn't too far away from reality, but at the same time isn't too complicated. If I can, I'll edit this answer later to include it.


What is the difference between an amorphous solid and glass? They are both formed when a liquid is supercooled rapidly, no free energy, and they both have irregular structures. What defines a glass other than how it is amorphous, transparent, and has a glass transition state?

While a glass is generally considered to be a supercooled, configurationally frozen liquid, not all amorphous solids are glasses. For example, amorphous silicon is a four-fold coordinated semiconducting solid, much like crystal silicon.  Liquid silicon is 8-12 fold coordinated with metallic bonding. Amorphous silicon has been shown to display a first-order phase transition to both the crystal and the liquid.

So, all glasses are amorphous solids, but not all amorphous solids are glasses.


Polymerisation of a UV-curing resin I have a photosensitive resin MSDS. It lists a mixture of 1,6-Hexanediol diacrylate ($\ce{C_{12}H_{18}O_4}$) with its proper CAS number and Polyurethane acrylate together with the CAS number of "Butyl acrylate/methyl methacrylate/methacrylic acid copolymer" with the formula (or rather, formulas) of $\ce{(C7H_{12}O2.C5H8O2.C4H6O2)_x}$. There's also the Photoinitiator 1-Hydroxycyclohexyl phenyl ketone listed. The additives of color and reaction inhibitor are most likely irrelevant for the actual question, so I leave them out.
My organic chemistry is somewhat rusty, but the first part of the compound I can identify as having the setup $\ce{H2C=CH-CO-O-C6H12-O-CO-CH=CH2}$. This appears to be a quite nice molecule that reacts evenly on both sides under polymerization. This material is also listed as photosensitive, which I interpret that it sheds something from both ends to activate the ends.
The other, of course is a mixture of three monomers: Butyl acrylate ($\ce{H2C=CH-CO-O-C4H9}$), methyl methacrylate ($\ce{H2C=C(CH3)-CO-O-CH3}$) and methacrylic acid ($\ce{H2C=C(CH3)-COOH}$). The first two have ester groups in them, the last has its precursor, the $\ce{-COOH}$ group (carboxylic acid). All these only seem to have one active side to me.
Assuming the four main chemicals are mixed somewhat evenly and in even numbers and what might be an ample supply of the starter Hydroxycyclohexyl-phenyl-ketone, they surely will react and polymerize under UV light of about 400 nm (+- 5), as this is how the mixture behaves in its native application, an SLA printer.
However I have a problem with these parts:

What kind of reaction does happen between the compounds? I remember vaguely that some polymerizations create waste in the shape of water (was that condensation?), but I don't really see that happening here for most of the combinations and, as a result, I fail to see what the parts of the polymer might look like in their bonded state.
What type of bonds does the resulting polymer have? This is pretty much the direct result of failing to see the reaction between the compounds. I thought I had seen a possibility to form an ester ($\ce{R_1-CO-O-R_2}$) between the methacrylic acid and the hexanediol diacrylate, but then realized the later misses an alcohol group ($\ce{R-OH}$) to attack and create water from to form the bond. So I am pretty sure that these don't bond as esters to one another, but what is the correct name for these bonds that elude me?!
What would this polymer be classed as? I remember that some polymers are classed based on their monomers, manufacturing process, the chain length, bond type, and the macro-structure, putting each into several groups. These would allowing to estimate some vague properties about them. It's most certainly not a Homogenous one like pure PolyLacticAcid, as it is basically made up of 4 different monomers. It also should not form an elastic bond due to what I know from the MSDS and that the processed resin is a somewhat brittle product. But what groups it can be put into from the polymerization process?

Polymerizations expelling water, equally known as polycondensation (PET is a typical product), are a subset of polymerizations. The radical polymerization here is more likely to proceed differently with three reaction types running concurrently: initiation, propagation, and termination.

During the initiation, light splits the phenylketone into two fragments with one unpaired electron each. This is symbolized by the dot next to the oxygen atoms, denoting that these chemicals are radicals:

These radicals may react with one electron of the double bond of a neutral acrylate in the propagation reaction. Symbolized by blue is the newly formed single bond between the two reagents (first line).  Simultaneously, the other electron of the former double bond represents a new site where again a neutral acrylate may react (second line).  This growth of the polymer chain may continue again and again, or propagate (third line).

As soon as there are radicals formed, there may be a number of reactions where these carbon-centered radicals are consumed.  This cancels the further growth of the polymer chains.  Among these termination reactions, hydrogen-transfer / redox reactions as well as dimerizations are the most prominent ones.  Some which seem likely to occur are depicted below:

Structure
Your starting material consists of a starter, and a blend of bifunctional acrylates.  Thus the curation will yield polymer chains that are cross-linked with each other.  The illustration below (for a different material) symbolized this by the blue zig-zag lines:

(Konuray et al., Polymers 2018, 10, 178 doi 10.3390/polym10020178, open access)
Because you started with multiple different monomers, it would be an example of branched copolymers.


Fireplace window: Can one influence how fast soot is building up? I'm sorry for such a stupid question, but after googling for half an hour I know two things

There are at least a zillion pages, blog-posts and articles about how to keep your fireplace window clean or how to clean it.
There is not a single page which explains what I'm looking for and which doesn't make claims without supporting them with scientific facts.

Therefore, here the short form of my question: Assuming I have a fireplace with a window, which looks like this after a while:

Further assume I'm able to keep all other factors constant (like type of wood, temperature of the fire etc.), is there something I can apply/do to the cleaned window to prevent the building of soot for a longer time?

The reason I'm asking is this article I stumbled over, which claims that normal corn-starch can be used to fill the pores of the glass. This left me with more questions

does each glass (especially my one) have pores with a size that starch particles can fit in? I read somewhere that starch has a diameter between 2-170 µm depending on the type.
I'm making a fire where I burn wood! Why wouldn't the starch simply burn and create itself the soot?
As we know for a while now making a surface really smooth doesn't mean it keeps clean. With a size of 10-20 µm of the lotus papillae, it seems to fall in the same range as starch.

And for the diligent, here two more sub-questions:

How does soot building work? I mean, why does the soot stick so hard at the window? After all it's just little particles flowing around and landing on the window. (I know it's called burning in, but how does the heat work on a already burned particle?)
Is there some connection to the pores (if any) in the glass in this process?

Update
After using my fireplace one winter, I'd like to make recommendation based on my experience so far. Important is that I can easily remove the window from my fireplace for cleaning. In case the window is fixed, cleaning might become a bit messy. Let me bring it down to 3 tips:

My current method is to use a cleaner for ceramic glass cooktops. This works like a charm. In case someone wants to know the specific product: it's a cleaning stone for class ceramic which is probably only available in Germany but I guess something similar can be found all around the world. This thing comes with a sponge and you just have to rub the wet sponge a few times over the cleaner to apply some. Cleaning the window with this is done in less than a minute, even if there is a lot of soot.

If you don't want to invest any money, the most simple and very effective cleaning method is to use the ash itself to scrub the window. Simply take water and a newspaper, if possible make the window itself wet or better rinse it with water. Take a sheet of newspaper, make it wet and apply ash directly from the fireplace itself. It's really surprising how easily the soot can be scrubbed away with this method.

Last tip: Heat! As already said in the accepted answer, soot is basically the deposition of incomplete combustion products from a flame. There is no doubt and I have tested this very often: a dirty window can partially be cleaned by make a hot fire with very dry wood and a lot of air. Therefore, try to fill your fire-room with the maximum possible amount of wood, open all air-valves and let it burn for a while.

I don't have a fireplace or wood stove, so I don't have a practical technique that I know works to prevent soot buildup on glass, but I can answer some of your questions about glass and soot. To start, I should add a caveat here: while many fireplaces use tempered glass windows, wood stoves generally have glass-ceramic windows to withstand the increased heat. As far as surface properties go for this, glass-ceramic isn't dramatically different from normal glass.

  does each glass (especially my one) have pores with a size that starch
  particles can fit in? I read somewhere that starch has a diameter
  between 2-170 µm depending on the type.

While special techniques exist for making porous glasses, normal glass is not porous and float glass (the standard technique for fabricating flat glass panes) produces quite smooth pieces even without polishing (hundreds of nm peak-to-peak), so probably whole starch particles are not getting stuck in the surface of the glass and the surface roughness probably has little impact on soot adhesion in this case.

  I'm making a fire where I burn wood! Why wouldn't the starch simply
  burn and create itself the soot?

Unless the flame is right up to the glass, it probably doesn't get hot enough to really burn starch (if any is even there). That site didn't compare it directly with not applying starch so it's not clear that it makes any difference.

  How does soot building work? I mean, why does the soot stick so hard
  at the window? After all it's just little particles flowing around and
  landing on the window. (I know it's called burning in, but how does
  the heat work on a already burned particle?)

Soot is basically the deposition of incomplete combustion products from a flame. A wood flame appears yellow or orange because tiny particles of mostly carbon are incandescing. There is insufficient oxygen to fully combust the carbon in the fuel to $\ce{CO2}$, so the combustion results in carbon molecules in various states of oxidation from small polymers to elemental carbon. These species are really only stable as gases within the heat of the flame itself and will thus readily deposit onto cold surfaces, like the window glass. Many modern wood stoves seem to have "air-wash" systems that flow air past the inside of the window to reduce how much soot can actually reach the surface.

What is actually deposited as soot is a complex mixture, but what is emitted from the flame is related to how much air is available and the temperature of the flame (the two are also connected with each other). If the oxygen supply is increased, more complete combustion is favoured so less carbon species that can deposit as soot are formed. The temperature that is attained influences the composition of soot particles: higher temperatures produce more aliphatic and less aromatic compounds. The practical significance is that reducing the flame intensity by limiting oxygen can greatly increase the production of soot. (and may have some effect on how well it adheres to the glass, though I wasn't able to find anything concrete to that effect)

From the non-scientific sources I glanced through, it seems to be well known that excessive soot buildup can be due to wet fuel or low airflow and that burning a very hot fire can help make soot easier to remove. As for the corn starch, it may indeed be leaving some deposit that hinders soot adhesion (there are commercial cleaners that claim to leave a film behind that makes soot easier to remove), the act of applying it may be removing existing soot that's too small to see but new soot adheres readily too, or it may do nothing. Perhaps you might want to try applying corn starch to half of a freshly-cleaned window to see if there is a noticeable difference.


When is a reaction reversible? When speaking in thermodynamic terms, when is a reaction reversible?

My attempt: A reaction is reversible when $\Delta\text{S} < 0$, because according to the second law of thermodynamics, "in every natural thermodynamic process the sum of the entropies of all participating bodies is increased" (Wikipedia.org), therefore if a reaction is caused and has negative change in entropy, a reversed reaction should be natural.

However, this is not the answer that is absolutely correct, nor a rigid solution or a rule. Can anyone help me out? (I have searched on the internet, and although thermodynamic reasons are preferred, I do not mind answers to be molar or solution chemistry, as long as a thermodynamic approach is included as well or a statement that there is no apparent thermodynamic rule that can be applied to this question.)

"Reversible" is not binary.  Both the forward and backward reactions always occur and the equilibrium system never has zero reactants or zero products.  Thus, irreversible reactions are called this not because they cannot be reversed - they absolutely can - but because reversal is impractical.  The equilibrium constant may be so skewed toward product that the reactant quantities at equilibrium would be undetectable.  The products may be gases that escape solid reactants, which ensures that the system never reaches equilibrium in the open anyway.  There may be a lot of side reactions on the products.

A "reversible" reaction just lacks factors like this so that we can have measurable quantities of reactant beside measurable quantities of product at equilibrium.  We add some reactant, watch the reaction go forward.  We add some product, it goes in reverse.  See?  It's reversible.  There is no exact cutoff in any measure.


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

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

My thoughts:

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

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

I would suggest that there is a potential kinetic effect here.  While the most stable carbocation formed in either case is similar, the number of collisions that result in the formation of the more stable carbocation is higher in I than in II.

If we take a very simplified view of the possible collisions in II, we can see that in a reaction with $\ce{HCl}$ only "one" of the orientations of the collisions will result in the formation of the desired carbocation.

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

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


Is the Aufbau principle violated in the diagram (b)? In which of the following diagrams is the Aufbau principle violated?

Ignoring the Hund principle, the Aufbau principle states that electrons have to first fill the lowest energy levels. Now the energy level of the s-orbital has to be the lowest, and only in the option (b) the s-orbital is not filled fully. So the answer is (b).
Is my reasoning correct?

Since the question specifically asks about the violation of Aufbau principle, your reasoning is correct.

Had it been based on the violation of Hund's rule then option (a) would be the answer.


Why does chlorination of toluene stop at two chlorine atoms? My instructor mentioned a reaction for preparation of benzaldehyde by chlorination of toluene in presence of sunlight to form benzal dichloride, which on hydrolysis gives benzaldehyde.

However, I'm not sure why the chlorination of side chain should stop after two chlorine atoms are added? At what step does the side chain chlorination of toluene stop?

It doesn't stop, it just gets a lot slower the more chlorine atoms you add. From the abstract of an article titled Photochemical side-chain chlorination of toluene[1]:

  Where $k_1$, $k_2$, and $k_3$ are the rate consts. for the successive reactions: $\ce{PhMe}$ → $\ce{PhCH2Cl}$ → $\ce{PhCHCl2}$ → $\ce{PhCCl3}$, resp.; $r = \frac{k_1}{k_2}$ and $s = \frac{k_2}{k_3}$, values of $r$ and $s$, resp., were 6.0 and 5.7 at 100° and 8.4 and 8.5 at 40°, for batch operation, with light varying from ultraviolet to yellow.

[1] Haring, I. H. G.; Knol, H. W. Chemical and Process Engineering (London) 1964, 45 (10), 560–567.


Inductive effects of halogens The Wikipedia page on inductive effects states that the order of inductive effects of halogens is as follows:

$$\ce{-F} > \ce{-Cl} > \ce{-Br} > \ce{-I}$$

From this, and from the fact that electronegativity of fluorine is higher than that of chlorine, I guessed that the acidity of p-fluorophenol is more than p-chlorophenol.

But the $\mathrm{p}K_\mathrm{a}$ values are:

p-fluorophenol: 9.8
p-chlorophenol: 9.4

Where did I go wrong? Why does this discrepancy occur? Detailed explanations are very much welcome.

This is due to the fact that we also consider +M effect . In fluorine it’s 2p orbital that participated in overlapping but in chlorine it’s 3 p orbital. 3p of chlorine does not overlap efficiently with 2p of benzene carbon so +m decreases in chlorine thus acidic nature increases


Can I split water without constant refill of solution? I've looked all over the place to find a way for me to separate water without needing to put in salt. What I've found is you can separate water into hydrogen and oxygen through electrolysis (which needs salt to make the water conductive), radiation (which I don't want to use - this is a house project that I want to get into), thermal (the water needs to be hotter than what I can provide), and photobiological (looks cool - but it looks like it won't produce much).

There's a lot of things out there for electrolysis water splitting, but every single one looks like it needs salt. Can I initially provide salt to a water splitting machine, then never refill it and still split water? Is there anything else, in which I could put in the water to make it split, and not have to ever refill said thing?

Really, I want to build the machine, and not have to keep providing resources that I'm going to have to get - just provide the machine water and electricity.

Salt is not needed. Salt merely allows the water to be somewhat more conductive.

The King of Random on YouTube has a great video on making one. All you need is electricity and water.
https://www.youtube.com/watch?v=cqjn3mup1So

He uses sodium hydroxide, but this not needed. It will just be less efficient.

If you want to be more efficient, you can add an electrolyte like salt. Whatever electrolyte you use, should not be used up. However, if the concentration of electrolyte gets to high (after you have used up a lot of water) it will split it too. This is bad with salt, as you can end up with chlorine gas and sodium hydroxide, which can be nasty to deal with.

So, as long as you replace the water fast enough and don't put in too much electrolyte, it should be fine and you will not need to replace it often.


Where is the center of symmetry in the PbO crystal? I'm very confused as to where the center of symmetry is for the PbO  (litharge form). I know that it is in the P4/nmm space group, and that it's centrosymmetric, but beyond that I'm clueless. Thank you.

The best thing to do is load up the CIF at the Crystallography Open Database (COD), e.g.
http://www.crystallography.net/1010977.html?cif=1010977

JSmol used for the visualization lets you visualize the symmetry and adjust the packing. In this case, look for the inversion center, e.g.:

The inversion center itself is the yellow ball. But it's probably easier to see it in the inverted Pb centers I've circled in yellow. You can see that there's an exact inversion center there.


why is NaOH used instead of KOH for preparation of K2Cr2O7? I have been told by my teacher that we use NaOH for the first step of preparation of potassium dichromate to form sodium chromate, which then reacts with conc H2SO4 to form sodium dichromate. Then it undergoes a double displacement reaction with KCl to form K2Cr2O7..My question is that why cant we directly use KOH in the first step so that we dont have to use the third step?

Manufacturers usually prefer sodium hydroxide over potassium hydroxide because it is cheaper. Sometimes, NaOH is preferred because a smaller quantity (as compared to KOH) serves the purpose. And also the product from by KOH is much more soluble than NaOH, make it harder to extract than using NaOH.


Why does vibrational cooling occur much faster than sponteneous emission in the condensed phase? Following the absorption of light by an arbitrary small molecule, the molecule is found to be excited both vibrationally and electronically. The excess vibrational energy of the molecule will usually be dissipated into the solvent in less than ~10 ps, and fluorescence emission will occur in about ~10 ns.

Since there are many transitions available to accept the excess energy form the vibrations in the surrounding solvent it makes sense that the relaxation occurs very rapidly. On the other hand, electrons can move much faster than nuclei can. What must happen for an electronically excited molecule to relax through the emission of a photon and why does it take so much longer than the vibrational relaxation?

As you point out, there are many places for this relatively small amount of vibrational energy to go. There are both radiative and non-radiative transitions which can take place to vibrationally quench the system in the electronic excited state. The explanation for both parts of your question (the vibrational and electronic parts) can be understood by Fermi's Golden Rule. I won't go into a lot of detail of deriving this or anything else because you can find anything you wanna know about it by searching online.

Fermi's golden rule says that the transition rate, $\Gamma_{i\to f}$, from some initial state, $|i\rangle$, to a set of final states, $|f\rangle$, is given by,
$$
\Gamma_{i\to f}=\frac{2\pi}{\hbar}|\langle f|H'|i\rangle|^2\rho
$$
where $\rho$ is the density of final states of the system and $H'$ is a perturbing Hamiltonian of some kind. For instance, the electric field of a photon. This is what it would usually be for absorption. In the case of decay, it may also be the field of a photon, but I am not as familiar with using Fermi's golden rule in this case.

Notice that this rule has dimensions of inverse time. It can be shown that the inverse of this rate is equal to the mean lifetime of the excited state (see this page).

Basically all of the physics you are asking about is contained within the fact that this transition rate is proportional to the density of possible states to which the system will decay. To put it simply, there are many many more intramolecular and intermolecular transitions which can take place to redistribute this energy. The vibrational density of states for this problem could probably be approximated quite well as being continuous because there will be many vibrational levels which are slightly perturbed due to distortions from equilibrium geometries. Think of the width of vibrational bands in IR.

On the other hand, a very large amount of energy must be dissipated for electronic relaxation to take place, and the density of these high-energy states is very low. Especially if one excites to the lowest electronic state. If, however, excitation occurs to a higher electronic energy level, decay should take place faster because the density of electronic states becomes much larger.

These processes have almost nothing to do with the speed of nuclear motion compared to electronic motion, as the condition which must be satisfied is really conservation of energy, and this is not always easy to do when the energy levels are quantized. From this perspective, the result is not surprising because vibrational energy levels are certainly closer to being continuous than electronic energy levels. Also, the spacing of vibrational state is generally smaller on an excited state than the ground state, so this adds to the quickness of the decay to the ground vibrational state.

As an aside, this paper is of some interest as it shows that vibrational relaxation times in the gas phase (for this one molecule) are slower than vibrational relaxation in solution. This can also be understood by the density of states argument discussed above.


Partial hydrolysis of aluminium chloride What is the meaning of partial hydrolysis of anhydrous aluminium chloride? What does the partial term signify?

Partial means that it is not fully hydrolyzed.

A partially hydrolyzed solution of aluminium chloride will be slightly acidic.

The chemistry behind the hydrolysis of AlCl3 is quite interesting due to the fact that many different complexes can be formed.


Difference between probability and probability density of finding a particle in space What is the difference between probability and probability density of finding a particle in space?

I am at basics of the wave function and this thing confused me.

Probability = $[\psi(x, y,z)]^2(\mathrm{d}x,\mathrm{d}y,\mathrm{d}z)$

Probability density at point $(x,y,z)$ in space = $[\psi(x, y,z)]^2$.

The latter one is the probability of finding particle in an infinitesimally small volume element $\mathrm{d}V=\mathrm{d}x, \mathrm{d}y, \mathrm{d}z$ situated at $(x,y,z)$. Then what's the former one?

The latter is the limit of probability per volume, as volume approaches zero, at a particular point in space.


Tape versus 1922 silver coin I have a 1922 silver half dollar with scotch tape on it.  The tape has been on it 50 years plus.  It’s dried out and totally adhered to the coin.  I’ve soaked it in lemon essential oil and googone but still a lot of tape left, adhered to coin.  Impossible to get off. What will melt tape but not damage silver coin?

Circulated Coin

I'll assume that the coin is circulated so a very gentle cleaning won't be a problem. You want to use some sort of organic solvent to loosen the glue, then very gently rub the residue off. You don't want to use any sort of silverware polish or anything abrasive when rubbing the coin.

Good solvents might be olive oil or nail polish remover. After the organic solvent wash the coin with dish washing liquid, rinse with plain water,  then dry by flipping coin over on a dry cloth or paper towel.

Uncirculated Coin

If the coin is uncirculated then you would want to use an ultrasonic bath with an organic solvent to prevent scratches on the coin.

Two tone coin after cleaning

Because the surface of the coin would have been sealed  where the tape was, it is possible that the coin will have a different color where the tape was. If that is the case then you might need to use a coin dip to get the surface looking even again. Unfortunately this will leave the surface shiny which screams that the coin has been cleaned. The coin will dull over time but never really develop a natural patina again since a truly natural patina also depends on wear.


Thickening and/or slowing evaporation of IPA-based ink The conundrum

I have a security printing ink that I make by adding an IR-absorbing dye to a 50/50 mix of Isopropyl Alcohol and distilled water. The ink solution is pale green but dries virtually invisible to the human eye.

The problem is that when the ink is applied to the stamp the ink evaporates within seconds but I need it to remain viable for ~10 minutes. Even if I shortcut the 10 minutes to just a few seconds, the ink evaporates enough to cause an uneven distribution (i.e., it is faded in some areas.)

Requirements

Provides an even distribution of ink (to me, this means viscous)
As little bleeding as possible (to me, thin coat on stamp, also viscous)
Stays viable on a stamp for ~10 minutes
Allows the final inked product to be virtually invisible after ~24 hours

A solution that works (almost)

I have solved this problem before with a different ink. It was a UV-reactive ink which uses an identical IPA/water solvent. I was able to meet all of my requirements by adding 14% Glycerin to the solution. The increased viscosity really helped with consistent and clean (no bleed) marks. The slower evaporation gave me plenty of time to stamp. There was much rejoicing.

However, the Glycerin has an unwanted side effect with the IR ink in that it remains visible on the substrate. I'm guessing this is because the Glycerin prevents the ink from fully drying. I tried heating final product to see if that dried out the Glycerin but it didn't help.

You may ask why the UV ink ended up being invisible (requirement #4). Simple - that ink is clear to begin with.

My questions

I think what I need is a viscous fluid (maybe not as viscous as Glycerin) that evaporates slowly but fully in ~24 hours. Does such a wonder exist?
Glycerine solves all of my problems and creates one (visible marks.) Is there a way to use Glycerin, while still getting an invisible mark? (I've tried varying the amount of glycerin, but it's a trade-off of usability and visibility.)
Is there another chemical (hopefully easily obtainable by an individual) that could work as well as Glycerine while also allowing invisible marks?
I would love to understand this problem better so I can troubleshoot it myself (I'm a software engineer, not a chemist.) Any random insights would be welcome.
It seems to me that viscosity is inversely proportional to evaporation rate. Is that true?

Background info

I am trying to apply this ink to long thin strips (~7mm each) of a porous paper substrate (wrapped around plastic.) This needs to be done hundreds of times. To simplify the alignment process, I have a large rubber stamp with a series of ~100 duplicate strips embossed into it. This stamp is placed face up into a rig with alignment grooves on either side. From here, I apply the ink to the stamp and then slide each strip, one by one, into a pair of adjacent grooves and press down to stamp. This is much faster than re-inking a single thin stamp for every strip. This process takes about 10-15 minutes.

This is more of comment than a solution.

It has often been noted here that chemistry is an experimental science so often you just need to try things.

There are a number of considerations here. It seems like it takes on the order of 10 minutes to stamp a batch of "strips" (whatever that is...) and that:

You need a stamp with multiple "stamp faces" to stay wet that long.
You also want the ink solvent to "dry" so that the ink solvent doesn't leave a noticeable "watermark" on the paper.
You want to use a minimal amount of solvent so that the dye stamp doesn't bleed into the paper.

(1) The dye must be fairly polar to dissolve in a solution of 50/50 mix of isopropyl Alcohol (2-propanol)  and distilled water. You can look at a table of solvents with their polarities listed at this webpage. You can try other solvents that you have, or mixtures that you can make, to get a rough idea of how polar the solvent (mixture?) must be.

Reasonable to assume about 1 gram/ml for any of the solvents listed. Estimate polarity of a mixture by assuming ideal solution, ie that total solvent polarity is equal to molarity of a solvent times its mole fraction. (ie 50 wt % is not the same as 50 mole %.)
You can check to see if the ink dissolved by seeing if the solution is "clear" - in other words shine a bright light through it. If you see scattering then the dye didn't dissolve.
You can probably estimate from your work with glycerine how concentrated the dye must be to be invisible when dried, but visible under fluorescence. (I'm guess that the dry dye itself is green. So you don't want to "paint" the surface with particles of the dye. Rather you want the dry to be carried into the fibers of the paper as the solvent carrying the dye is adsorbed.)

(2) Look at other solvents that have been used for inks, for example on this webpage. Choose some solvents to test which have a polarity that works from step (1). Now look at solvents vs vapor pressure. The lower the vapor pressure the slower the solvent will be to evaporate.

Note that a mixture won't evaporate homogeneously. So with an acetone/glycerin mixture the acetone evaporates preferentially to the glycerin.
There is also an effect here in that the ink solvent is adsorbed into the paper not just sitting on the surface of the paper like a drop of such solvent on glass. So the correlation between vapor pressures and drying time would be "good" but certainly not perfect.


Why do some salt ions in solution conduct electricity better than others? My experiment involves a potentiometric titration, and I have used $\ce{AgNO3}$ as a titrant and KCl/KI/deionised water only as an analyte.

In the image below of my data, the potential difference as recorded by a multimeter drops from 0.6V to 0.15V.

What causes the drop in potential difference when the silver nitrate and whatever analyte (all exhibited the same pattern of a sudden drop in PD) reaches its endpoint in the titration?

Why is there a higher potential difference when using KI as analyte than KCl?

As you add $\rm AgNO_3$ to $\rm KCl$, solid $\rm AgCl$ forms. The silver ion you add will at first be completely consumed by the chloride. Once all of the chloride is locked up in the precipitate, the silver ion concentration will rapidly increase. The Nernst equation shows how the potential is expected to change when this happens. If you're using a silver/silver chloride electrode with a reference electrode, the indicator potential is

$$\rm E = E^\circ_{\rm AgCl,Ag} - \frac{R T}{\cal F}\ln\frac{K_{sp}}{a_{\rm Ag^+}}$$

To see whether you'll get a drop or a rise in $E$, ask yourself what happens when silver concentration rises. If you've got the leads hooked up to your multimeter properly (negative lead to the silver cathode, positive lead to your anode), shouldn't you be seeing an increase in potential? What exactly are your $x$ and $y$ axes on that graph?

You'll get a bigger change when you form $\rm AgI$, because $\rm AgCl$ is about a thousand times more soluble than $\rm AgI$. The $\rm AgI$ will lock up relatively more ions when it forms (the $K_{sp}$'s of $\rm AgCl$ and $\rm AgI$ are $1.8\times 10^{-10}$ and $1.5\times 10^{-16}$, respectively).

Addendum 1: You say you're just using a silver electrode; the indicator potential is

$$\rm E = E^\circ_{\rm Ag^+,Ag} - \frac{R T}{\cal F}\ln\frac{1}{a_{\rm Ag^+}}$$

(but if you dip a silver wire into a chloride solution you'll get a very thin layer of AgCl and it'll begin to be a silver/silver chloride electrode (Source: G. D. Christian, Analytical Chemistry, 4th ed., Wiley 1986, p. 282)).

You also say you don't have a true reference electrode, just a graphite electrode... the fact that its potential can change as well complicates things. Your cell potential will depend on what the oxidation process is at the anode.

Addendum 2: You asked for a source for the $K_{sp}$'s. CRC Handbook 58th edition gives $K_{sp}$'s at $25^\circ$C for AgCl and AgI as $1.56\times 10^{-10}$ and $1.5\times 10^{-16}$; the 84th edition gives $1.77\times 10^{-10}$ and $8.52\times 10^{-17}$; Keller's Basic Tables in Chemistry gives them as $1.0\times 10^{-10}$ and $1.0\times 10^{-16}$ for "temperatures near 298 K". They're very temperature dependent and they depend on conditions like ionic strength as well, so it's not surprising that the values you're finding on the web are all over the place.  I took the values above from a genchem text that happened to be lying on my desk.


Entropy change of system in adiabatic irreversible process In an adiabatic irreversible process we know that q = 0. Now change in entropy is defined as q/T. So when q is zero, why is change in entropy not zero ?

The change in entropy is the reversible heat flow divided by $T$.  So if the process is irreversible, you can't use the irreversible process's heat flow to find the entropy change.


Molar conductivity of coordination compound The compound is $\ce{[Cr(H2O)6]Cl3}$. Which isomer has least molar conductivity ?
Correct answer : $\ce{[Cr(H2O)4Cl2]Cl.2H2O}$
My answer : $\ce{[Cr(H2O)3Cl3].3H2O}$
Molar conductivity is directly proportional to the number of ions on dissociation. If number of ions is same, the one with more net charge  has more molar conductivity. Therefore the compound I stated should have least molar conductivity as the only dissociation it undergoes is separation of coordination entity from water molecules. As none of them are charged ions, they wont increase molar conductivity. What is wrong with my concept ?

You are right. But for some reason, your compound has never been synthesized, although theoretically ti should exist. It may be a problem of entropy.


How to test for and remove lead(II) sulfate from a sample of 1 M sulfuric acid? If you had a clear, colourless sample of dilute sulfuric acid and suspected lead contamination, probably lead sulfate, how would you test for this? It's solubility is low in sulfuric acid solutions, and lower near freezing, so I thought cooling might produce a precipitate, but I'm not sure what level if any this would be able to detect. Raising the concentration by open boiling or distilling would, from what I can gather, produce a filterable precipitate, but that is not something I want to test.

Would there be any easy way to remove any detected contamination? By easy, I mainly mean not distilling or boiling and filtering conc. sulfuric. Ideally, I would be looking for a method that does not involve raising the concentration at all. Chemical conversion to something even less soluble? Electrolysis?

EDTA can be used to combat lead intoxication. Would there be a way of using it to remove lead compounds?

To answer your question on how to test for Lead sulfate, I refer you to this paper, to quote:

  In 1929 Fischer [1] introduced a sensitive reagent for small quantities of lead which has found interesting applications in widely diverse fields. Diphenylthiocarbazone, or dithizone as it is commonly called...

And further:

  We have made no study of this reagent, except to find means of using it in our problem of measuring the solubility of lead sulfate. Dissolved in chloroform, dithizone imparts a noticeable green tint to the solution, provided the solution is not too concentrated. The lead-dithizone complex, however, has a bright cherry-red color.

Also of import;

  Dithizone is known to react with metals other than lead. The precautions taken to avoid interference by other metals are discussed on page 58. These precautions, together with the spectroscopic analysis of the lead salt and the redistillation of water and acid, make it highly improbable that the results were affected by other metals. Confirmation of this is found in the analysis of solutions of known lead content.

If Pb is present, dilution appears to be the best path due to solubility.

[EDIT] Here is an theoretical path to the removal of heavy metals that does not require dilution. I am suggesting UV light therapy employing TiO2 photocatalyst to act on cold H2SO4 containing heavy metal impurities. Heavy metals are expected to precipitate.

With respect to the mechanics of the photolysis, I referred, for example, to this source. Per cited Equation [R1]:

$\ce{ hv -> e- + h+ }$

Per Equation [R3], expected reaction with an electron hole:

$\ce{H2O + h+ -> .OH + H+ }$

$\ce{H2SO4 = H+ + HSO4- }$

$\ce{H+ + e- = .H }$

where the hydrogen radical behaves as a e-/H+ pair, see Hydrometallurgy 2008: Proceedings of the Sixth International Symposium, p. 818, a commercial reductive leaching equation, to quote:

  $\ce{ PbS + 2 •H = Pb + H2S }$ (5)

In the current case:

$\ce{ .H + PbSO4 -> Pb (s) + HSO4- }$

However, the associated transient sulfate radical is also created:

$\ce{ .OH + HSO4- -> H2O + .SO4- }$

Leading to the persulfate ion:

$\ce{ .SO4- + .SO4- -> S2O8(2-) }$

So, the sulfate ions is not alone. Possibly more advantageous, further  pump air/oxygen into the solution during the UV light treatment. Then per Equation [R3], expect the superoxide radical anion formation:

$\ce{O2 + e- = .O2- }$

which becomes, in acidic conditions, the $\ce{.HO2}$ radical, which may similarly react with Lead sulfate as follows:

$\ce{ .HO2 + PbSO4 -> PbO2 (s) + HSO4- }$

or, just oxidize any liberated Pb.

I would even suggest you try this process out by creating a test solution (with at least added Lead) and heat in H2SO4 to form a heavy metal impurity and then apply the UV light/TiO2 (and also added oxygen) therapy.


Benzene and proton in 1H NMR The chemical shift in benzene is around $\pu{7.27 ppm}$, which is lower than, for example, aldehydes. However, the magnetic field due to the pi electrons are in line with the applied magnetic field $B_0$, resulting in $B_{eff}>B_0$. Whereas in aldehydes, the hydrogen is simply "less shielded". So shouldn't the $\ce H$ atoms in benzene (or alkynes etc.) have even greater chemical shifts than species that are deshielded "normally" through electron withdrawing substituents?

Also, somewhat of a side question: What kind of chemical shift would we see for a proton ($\ce{H+}$) in a proton NMR, since it is completely deshielded? Could this theoretically be measured?

(Following the previous logic, shouldn't benzene have a higher chemical shift than a naked proton?)

The aldehyde is both deshielded by reduction in electron density about the carbon it is bonded to through electron withdrawal and by the magnetic anisotropy of the double bond. Although a benzylic hydrogen is slightly more deshielded than the hydrogen attached to a two electron pi bond like a carbonyl or an alkene, it does not also have electron density withdrawn from its adjacent carbon. Generally electron withdrawal through resonance is more important than inductive electron withdrawal, so the difference in chemical shift between an aldehyde and an alkene will be greater than the difference in chemical shift between the proton proton geminal to a halogen in  a halogen-substituted alkene and the proton in its unsubstituted analogue. All that is to say the additive effects of resonance electron withdrawal and magnetic anisotropy make an aldehyde proton more shifted than a benzylic proton.


What kind of interacion is there between a nanoparticle and its functionalizing coating? The functionalization of nanoparticles (with coatings such as polymer ligands; see image) are very common (for example, DNA-functionalizied gold nanoparticles). What's the type of interaction between nanoparticles and their coatings? And why is it more favorable than forming aggregations of the the nanoparticles?

In small nanoclusters/nanocrystals/quantum clusters (my goodness, they have so many names nowadays) in the size range around 13, 55 and 144 metal atoms,1 the ligands enter a real orbital interaction (dative bond2). They are ligated to the metal core (Puls et. al., 2014).

In DNA-ligated gold clusters, the phosphate groups are able to bond to the metal core and substitute the phosphine ligands it had before (Yunping et. al., 2003).

In larger nanoparticles,3 the tendency of the particles to agglomerate is much smaller. Some of them can be measured with TEM (Transmission Electron Microscopy) without full agglomeration. This is not given for the small 13 metal atom clusters. The big particles enter interactions with their surfactants for which the main contribution is given either by van der Waals forces, or by electrostatic interactions.

1 Those are some of the most common numbers in gold cluster chemistry, like $\text{Au}_{13}(\text{PR}_3)_{12}$ for example. 55 and 144 can be found in thiolated gold clusters.

2 It has covalent portions as well as ionic. It is, however, better understood if you think of coordination complexes. Only in this case, there is not only a single metal atom in the centre

3 i.e. 5 nm, 12 nm. I stated this because I am not aware of any case in which a 13-metal-atom cluster was analysed by TEM (Transmission Electron Microscopy). The conditions are simply too harsh, the ligand shell is ripped off and the little clusters experience a very strong tendency to agglomerate. You need specialised TEM equipment to acquire meaningful data. Nanoparticles of a few nanometers are less prone to show such a fast agglomeration.


Why does shaking a match put the fire out? Move a match slowly and nothing happens but if you shake it violently the fire will extinguish. Oxygen makes fire grow so why does waving a flame through the oxygen rich air put the fire out? Does this primarily have to do with a decrease in temperature to the burning materials or is it something else?

Also what about forest fires, do high winds spread or kill the fire?

Combustion of small materials, such as a match or birthday candle, actually involve the release of volatile vapours, which themselves burn. It is not the solid material that burns. There needs to be a minimum amount of volatile material present in this combustion zone (just above the burning match) for the ignition to occur. As the combustion process continues, heat is given off, and more volatile materials are released, which in turn continues the combustion cycle. Now, if you shake a match or blow on a candle, you rapidly disperse these volatile fuels from the combustion zone, and there is no longer sufficient fuel to ignite. It is effectively removing the fuel component of the fire triangle, for just a brief moment.

Large fires can reignite because there is sufficient heat left in the fuel to further release volatile fuels, which can either self-ignite or ignite through the presence of embers.

Blowing gently on a small wood fire increase the oxygen content into the combustion zone, without dispersing the fuel. Similarly, if you experiment with a small enough wood fire, you will see that blowing on different parts of the fire will have a different outcome: blowing on the base of the fuels will increase oxygen content, and not affect volatile fuels. Blowing on the top of the fire (where the flame starts to burn) will probably put the fire out.

Will this put out a small paper fire? That will depend on the heat retained by the burning fuel. A single piece of A4 paper if shaken hard enough will extinguish. A ream of A4 paper that has burned half way down the page will be put out this way, but could easily reignite if the paper pages are pulled apart to allow oxygen into the released volatile fuels.

Generally, forest fires are accelerated by strong winds. Winds affect forest fires in a number of ways, and increase the rate of spread significantly. This is a topic for a whole other question.


Rock salt structure: chloride lattice or sodium lattice? Source

From this diagram of the rock salt structure ($\ce{NaCl}$) we see that both the chloride and sodium ions have the same environment. That is to say, they each have the same number of neighbours at every distance, except that the charges are opposite.

However, I've always been taught to consider the chloride ions as forming a face-centred cubic (fcc) [or cubic close packed (ccp)] lattice, with sodium ions filling octahedral holes.

Is there any reason why we can't think of it the other way round, with sodium ions forming the fcc lattice and chloride ions filling octahedral holes?

Is it perhaps because sodium ions are smaller than chloride ions, or something else?

Unit cells are just a way of describing a given crystal. You can certainly draw $\ce{NaCl}$ the other way, with $\ce{Na+}$ in the corners and it's still correct. There are a few ways of thinking about this structure:

as two interpenetrating FCC lattices
as a $\ce{Cl-}$ lattice with $\ce{Na+}$ in the octahedral holes
as a $\ce{Na+}$ lattice with $\ce{Cl-}$ in the octahedral holes

They're all equivalent. The figure above actually shows both types of octahedral holes. Some might prefer to visualize the structure as the second case since, as you say, the chloride ions are bigger, but going the other way isn't wrong, nor does it describe anything different.


How to calculate atmospheric composition Okay, so I'm a novice in chemistry and am trying to figure something out.

Say that you want to turn carbon dioxide into oxygen using photosynthesis, given an infinite supply of water. Say that the atmosphere's volume is 100km^3 and has a pressure of 5kPa. The current composition of the atmosphere is 90% Carbon Dioxide, 5% Argon, 5% Nitrogen. You want to get the Oxygen in the atmosphere to the point that it's at 20% of the atmospheric composition. How would you calculate the atmospheric composition after you do that?

A step by step guide with just formulas would be fine, you don't necessarily have to solve this exact problem; I just need to figure out how you go about solving it. I'm only asking because when I tried doing the math and putting it back into a percent via # mols or their mass, I thought that in the end it should still have 5% Argon and 5% Nitrogen in the atmospheric composition since the chemical equation is only using the Carbon Dioxide in the air to make Oxygen.

Keep in mind when explaining it to me that I've only just completed my first semester of college chemistry, so make sure you tell me what each variable is!

A really interesting problem, but I don't think you have enough information to solve it. At first glance, it seems like you would get one oxygen molecule in the atmosphere for every $\ce{CO2}$ molecule consumed. If we assume we are only making carbohydrates, then the process would be $\ce{CO2 + H2O -> [CH2O] + O2}$. That would give a $70:20:5:5$ ratio of $\ce{CO2:O2:N2:Ar}$. However, plants will do more than just produce carbohydrates into their structures. They also produce fats, proteins, etc. Some of the $\ce{N2}$ would become incorporated. In addition, $\ce{CO2}$ is appreciably soluble in water (more so at lower temperatures). I think both factors would reduce the $\ce{CO2}$ and $\ce{N2}$ levels in the final mix.


In solid-phase peptide synthesis, why are the formation of aspartimides a problem but not "glutamides"? I'm studying solid-phase peptide synthesis (SPPS) at the moment, and a common problem is the formation of aspartimides, which disrupt the synthetic process.
I understand how they form and why are they a problem. What I don't understand is why don't glutamic acid residues face the same problem? They have the same functional group, and a 6-membered ring is probably even more favoured than the 5-membered ring in an aspartimide.
Does anyone have any insight into this?

Since you understand how aspartimides (succinimides) form and why are they a problem in solid-phase peptide synthesis (SPPS), I'm not going to elaborate it further. Your main question is why don't glutamic acid residues face the same problem? Then, my question is who said they (glutamic acid residues) won't? Of course, they do. One example is discussed in Ref.1. You didn't recognized this cycle formation because the cyclic form is called Glutarimide Formation (contrast to aspartimides, which is actually a succinimide). The abstract states that:

[...] We found that the sequence of Glu(Gly)-OAll leads to glutarimide formation, which disrupts the formation of cyclic RGD peptides. However, our results show that glutarimide formation is sequence dependent and can be inhibited by incorporating an amino acid like Lys(Boc) with steric hindrance from the protecting group. To prevent glutarimide formation, Ser(tBu) was used to replace the glycine in the GGG spacer adjacent to the residue of Glu, and a tailed cyclic RGD peptide, c[RGDfE(SGGKK-NH2)] was successfully obtained.

Initially, $\ce{c[RGDfE(GGGKK-NH2)]}$ was designed with $\ce{c(RGDfE)}$ linked through $\ce{Glu}$ to a tail consisting of a spacer of three $\ce{Gly}$ residues $\ce{(GGG)}$ and a linker of two $\ce{Lys}$ residues $\ce{(KK)}$. We found that the sequence of $\ce{Glu(Gly)-OAll}$ leads to glutarimide formation, which disrupts the formation of cyclic RGD peptides (Note: $\ce{Glu(Gly)-OAll}$ is glutamic acid(glusine) $\alpha$-allyl ester):

The succinimide and glutarimide formations are not just common in SPPS. They can be formed under physiological conditions and can disrupt the biological processes (e.g., Ref.2,3).
References:

Junmin Zhu and Roger E. Marchant, "Solid-Phase Synthesis of Tailed Cyclic RGD Peptides Using Glutamic Acid: Unexpected Glutarimide Formation," J. Peptide Sci. 2008, 14(6), 690–696 (DOI: https://doi.org/10.1002/psc.975).
Josipa GrzeticJosipa Grzetic and Jos Oomens, “Spectroscopic Identification of Cyclic Imide b2-Ions from Peptides Containing Gln and Asn Residues,” J. Am. Soc. Mass Spectrom. 2013, 24(8), 1228–1241 (DOI: https://doi.org/10.1007/s13361-013-0661-6).
Josipa GrzeticJosipa Grzetic and Jos Oomens, “Effect of the Asn side chain on the dissociation of deprotonated peptides elucidated by IRMPD spectroscopy,” International Journal of Mass Spectrometry 2013, 354-355, 70-77 (DOI: https://doi.org/10.1016/j.ijms.2013.05.005).


Proof that Equilibrium Constant is a Ratio of Rate Constants I have a question about this proof that shows the relationship between the rate constant and the equilibrium constant. In this proof, the assumption is that the reaction is $(a+b)$th order in the forward direction and $(c+d)$th order in the reverse direction.

I had always thought reaction orders are empirically determined (and not necessarily whole numbers); can one simply add the stoichiometric coefficients like this? If not, how does one show that equilibrium constant is a ratio of the rate constants since the equilibrium constant requires the concentrations raised to the order of the species' stoichiometric coefficient?

For a reaction step, the assumption that the rate is proportional to the product of the concentration of reactants raised to their respective powers is valid.

However, generally, reactions are made up of many reaction steps (e.g. SN1 reaction). For such reactions, you cannot use the assumption above. You have to understand the reaction mechanisms in order to determine the rate equation.

(For reactions that involve decimal powers, usually the steady state approximation is applied, and some mathematics is done along with empirical findings in order to determine the rate.)

Addendum: Yes. A reaction step means an elementary reaction. From Wikipedia: "An elementary reaction is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.". Therefore, there is only one reaction step.

Likewise, for the equilibrium constant,$K$, in your ratio derivation $K_c = k_f/k_r$, the chemical species must also be from that particular reaction step, and not the global reaction step. It makes sense because the equilibrium constant, K, is related to the free energy change, $\Delta{G}$ of a reaction step. In a two step reaction (e.g. SN1), there are two $\Delta{G}$ and therefore two equilibrium constants.

However, in practice, people do use the overall equilibrium constant, $\beta{}$, which is the product of the equilibrium constants at each reaction step. $\beta{} = K_1K_2K_3...$ (E.g.If the first step gives you 70% yield, and the second step gives you 70% yield, the global step would be 49% yield).(http://en.wikipedia.org/wiki/Chemical_equilibrium#Multiple_equilibria)

In short, for your rate constant,$k$ and your equilibrium constant $K$, in $K_c = k_f/k_r$, they must be of that of a single reaction step.


Temperature in the Gibbs Free Energy equation My professor gave us 2 different ways to calculate this problem. I did it using the Clausis-Clapeyron equation.

However, I don't quite get the second method because I thought that if the change in enthalpy and entropy of formation are in standard states, the temperature (T) in the Gibbs Free Energy equation meant to be 298K? And then making the temperature in the second equation 1100K, to find the equilibrium constant?

In fact, Gibbs free energy is not defined at exact temperature. It is defined for standard states of all compounds in equation. So you really can use this solution of problem. However, it is a very simplified solution, because you assume that standard entropy and enthalpy do not change with temperature. This is not entirely correct assumption but if you took change of enthapy and entropy with temperature into account, it would be much more complicated.
You can find information about Gibbs free energy here
http://www.chem1.com/acad/webtext/thermeq/TE4.html#3A


Why and how are azeotropes formed? Why and how are azeotropes formed?

I could not find a detailed explanation of how azeotropic mixtures are formed anywhere on the internet. Now, that I figured it out myself, I thought I'd share it.

If we prepare a liquid-liquid solution of equimolar concentration, we know that the more volatile liquid component will have a greater composition in the vapour phase.

Let us keep increasing the concentration of the less volatile component in the solution. Initially, even though the solution is richer in the less volatile component, the vapour phase will be richer in the more volatile component but there will come a time when the vapour phase will be richer in the less volatile component. It may so happen that the composition of the vapour phase becomes equal to the composition in the solution phase. At this point, azeotrope is formed.

Note that if azeotrope is formed at all, both the solution and the vapour phase will be richer in the less volatile component.

Take the example of ethanol-water solution.
Ethanol boils at 78.4 °C, water boils at 100 °C. Azeotropic mixture is obtained at about 95% percent of ethanol. Note that the less volatile component is present in greater amounts when the azeotropic mixture is formed.

I will keep adding more to it as it comes to me.


What is the configuration of dimethyl[tris(trifluoromethyl)]-λ⁵-phosphane? I know that $\ce{(CH3)2P(CF3)3}$ is geometry of trigonal bipyramid and according to Bent's rule $\ce{CH3}$  goes to axial positions and $\ce{CF3}$s remains on equatorial positions. However, as per Bent's rule more electronegative ligand goes to axial position.

Now my question is how to determine which is more electronegative between $\ce{CH3}$ and $\ce{CF3}$?

You can note that $\ce{-CH3}$ shows electron releasing (+I) effect while $\ce{-CF3}$ shows strong electron withdrawing effect. Now, you can treat inductive effect as the organic analogue of electronegativity, giving axial positions to $\ce{-CF3}$  and equatorial ones to $-\ce{CH3}$ in accordance with Bent's rule.


3D printing of alumina I want to print a structure of alumina, and the 3D printer that I am using is an extrusion-based printer. So I need your help in knowing that what composition (binders and deflocculant) of alumina will be suitable to print the 3D structure in such a manner that it holds its structure (as it is printed)? And to what extent should rheological properties be maintained so that structure doesn't fail?

Use methylcellulose(MC) as a binder and Ammonium Polyacrylate (NHPA) as your dispersant or deflocculant. The concentration of methylcellulose can be set according to your requirement as its non-ionic (if you want liquid slurry use MC in large amount and if you don't use less amount). For NHPA you have to make different samples of fixed alumina concentration and NHPA of different concentrations. Check viscosity of each sample and the one which gives you least viscosity means the best-dispersed sample and hence your operating concentration of NHPA.


NaI in acetone v/s acetone and water When NaI is in acetone and we have R-X, then I replaces X by Finklestein mechanism (Sn2 => Inversion). But when we have acetone and water, why do we get a racemic mixture?

NaX (X = Cl, Br) is insoluble in acetone, but is soluble in water. So, if you have water/acetone system, the solubility of NaX is increased comparing to acetone. Solvation effect doesn't have a big role on iodide ions and does not greatly impact it's nucleophilicity. So, the mechanism is still going to follow SN2 path. So, after one halogen is displaced by I-, you have introduced a better leaving group in the substrate. Then, another iodide is going to attack the newly formed R-I, so the result is: 2 x inversion = configuration retention. As no NaX precipitate forms, it is better for I- to attack RI than RX. It turns out that after equilibrium is reached, you end up with racemic mixture, or something close to racemic mixture. Note that free iodide ions keep attacking R-I and that this is a dynamic process which constantly repeats in the solution.


Integrated rate laws My teacher gave us the equations for zero-order, first-order and second-order reactions as follows:

$$ [A]_t = -akt + [A]_0 $$
$$ \ln [A]_t=-akt+\ln [A]_0$$
$$\frac1{[A]_t}=akt+\frac1{[A]_0}$$

Where a is the coefficient of $A$ in the chemical reaction, but everywhere I've seen, the $a$ is not there. I have the Central Science 12th edition and there's no $a$ either, is my teacher right? Because that value does affect the answer.

From the reaction

$$\ce{2A -> B}$$

we have $a = 2$, which leads to the differential equation:

$$\frac{d[A]}{dt} = - k [A]^2$$

which then leads to the integrated rate law you've cited.  The $2$ is nowhere as a coefficient, it is related to the order of the equation, and determines which integral/form should be used.  Similar conventions also apply for 1st and zeroth order kinetics.  I disagree with your instructor.

It should be stressed that these rate laws apply to elementary reactions, not complex reactions.  Experimental measurement may differ from that predicted using an elementary approach, which then leads to the conclusion there are other steps involved.  In the case of OP's comment regarding oxidation of carbon monoxide with NO2, elementary reactions would predict a rate proportional to $$k [\ce{NO2}][\ce{CO}]$$ if the reaction is the result of a simple bimolecular collision. The fact that a different rate is observed indicates this process is more elaborate.


Why is neodymium the most paramagnetic lanthanide? My textbook says that paramagnetism rises to a maximum in neodymium.

I don't understand how gadolinium has 8 unpaired electrons, whereas neodymium has only 4. Shouldn't paramagnetism be higher  in gadolinium?

Has this got anything to do with the magnetic moment associated with orbital angular momentum?

This is from Shriver and Atkins' Inorganic Chemistry (p.586, 2009 ed) :

  The magnetic moment of many d-metal ions can be calculated by using the spin-only approximation because the strong ligand field quenches the orbital contribution. But, for the lanthanoids, where the spin orbital coupling is strong, the orbital angular momentum contributes to the magnetic moment, and the ions behave like almost free atoms. Therefore, the magnetic moment must be expressed in terms of the total angular momentum quantum number J:
  $$\mu = g_J{\{J(J+1)\}}^{1/2}\mu_B
$$
  where the Landé g-factor is
  $$g_J=1+\frac{S(S+1)-L(L+1)+J(J+1)}{2J(J+1)}
$$
  and $\mu_B$ is the Bohr Magneton.

Being a high school student, the only thing of matter for you is the key point at the top of the page:

  The magnetic moments of lanthanoid compounds arise from both spin and orbital
  contributions.


Can resonance take place in between biphenyls? I thought since both phenyl rings are not coplanar, the resonance between them is not possible. But my book says the following:

  Resonance effect is partially nullified

Partially nullified means resonance is relatively weak but it's taking place

I thought since both phenyl rings are not coplanar, the resonance between them is not possible.

This is not correct. If the rings are coplanar you get the best orbital overlap and the strongest resonance effect, while if you twist it the effect will be smaller but still present until you are at an dihedral angle of 90°, where no overlap can take place.

Let us look at phenyl-tetrazine. The phenyl substituent loweres the LUMO of the tetrazine through conjugation and raises the HOMO. The most favored confomer is the planar one, so the ideal system to study this. If it's correct that resonance only takes place at a planar conformation we should see a sudden jump in LUMO energy between planar conformation and any other and for all angles not equal to 0° the LUMO level should be the same.

In addition resonance is stabilizing, so the energy should drop rapidly between non-planar and planar.

Calculated energies (M06-2X/6-311+(d,p)) while twisting the bond looks like this:

So the energy goes up continuously if we twist the phenyl ring, there's no jump in energy. Since there's basically no steric effect, unlike in biphenyl, the energy is most likely only influenced by resonance.

Let us look at the HOMO level. I just picked some angles:

As expected the HOMO is raised in case of planar conformation in contrast to 90°, where we expect no overlap but we can clearly see that there's an influence on the HOMO at nearly every angle getting much stronger if we get near planarity.

Since the HOMO is completely localized at the tetrazine even in the planar form let's look at the HOMO-3 as an example. This orbital is part of the conjugated aromatic system and looks like this if the angle is 0°. Of course orbital energies will get shifted if we twist the rings and finding "the same" orbital is hard, but I'll show the orbital with the same symmetry at the tetrazine.

Here's the picture for 0°, a nice conjugated system:

Here's the picture for 90°, there's no conjugation with the pi-system, little bit with some, I assume, sigma* orbitals.

Now here for 15,30 and 60°:

We can see quite nicely that the orbital in case of 15 and 30° still looks quite similar to the one at 0°, while the one at 60° already looks quite like the one at 90°, with a little bit of orbital in para position at the phenyl ring. Showing that there's a gradual loss of conjugation.

In addition, if the conjugation would be possible only at exact 0° dihedral angle then two things could happen: Either we won't see the effect of conjugation ever, since the molecules move all the time and if rotation around that bond is possible then the chance that the molecule is at exact 0° is basically zero. But we know that conjugation exists. Or the planar conformation would be much more stable due to the conjugation and it would be exact 0° all the time, but from experiments we actually know that those bonds rotate quite well.

  coplanarity the topmost condition for resonance –

The most important condition is an overlap between the orbitals, this usually works best if planar, but it most certainly doesn't have to be planar.


Can I use a 1:1 mixture of isopentyl alcohol and water as mobile phase to carry out paper chromatography of a mixture of red and blue ink? Can I use a 1:1 mixture of isopentyl alcohol and water as mobile phase to carry out paper chromatography of a mixture of red and blue ink?

We generally use ethanol water mixture as mobile phase for separating mixture of blue and red ink using paper chromatography. Can I expect better results with isopentyl alcohol and water mixture?

The principle of chromatography is that you can always try (as per the question in the title). This can be beneficial: in my PhD thesis I had two diastereomers that were hard to separate on TLC using hexanes and ethyl acetate, but a hexanes/toluene mixture worked wonders.

However, in your case you should not expect better results. For one, as the commentators have pointed out the two liquids are not sufficiently soluble in each other. For two, your separation is already pretty good and it seems unlikely that it can be improved even more.


Does electrolysis of water alter the electrodes? Today I used a 12 volt power supply to observe electrolysis of water.
I mixed table salt into the water, nothing else.
The leads coming from the power supply had clamps on their ends.
When in water bubbles started evolving on the clamps.
Does this damage the clamps in any way ?
Is there something I should put in the clamps to improve this experiment ?

It depends on the substance you have used for your electrodes. If they are in platinum, they will not be altered. If they are in carbon (charcoal, graphite), they will not be chemically tansformed, but after a while, they will be slightly corroded, and some powder of charcoal will progressively appear under the electrode on the ground of the flask used for the electrolysis. If they are in copper, there will be oxidized and a chemical reaction will occur at the anode. The copper metal will be oxidized and transformed into $\ce{Cu^{2+}}$ ions, which will produce a precipitate if the solution is basic or contains an anion like phosphate or carbonate. If the electrodes are in iron or zinc, the anode will be quickly corroded and transformed into the corresponding ions. In the beginning, the cathode will not be altered, but in the long run, they will also be corroded, specially if the solution is acidic.
Anyway you have not described the composition of the electrolyte. Common metals like iron and zinc are altered by acids, with or without electrolysis.


Why are electrons shared equally when calculating formal charge, but unequally when calculating oxidation state? When calculating formal charge - electrons are shared equally between
the atoms in the bond.

When calculating oxidation state - electrons are both given to the
most electronegative atom.

Why is this different?

I am assuming it is something to do with the intended use of calculating formal charge and oxidation state, which I believe are:

Formal charge can be used to find the most stable structure.
Oxidation numbers can be used to determine what will become oxidised
in a redox reaction.

However, even based on their uses, I am unsure why they are calculated differently.

*Images from Wikipedia

The oxidation state is a form to distribute the charge of atoms in molecules considering to have only ionic bonds. In contrast, the formal charge is the distribution of the electrons considering to have only 100% electron-pair covalent bonds.

Actually, the formal charge is not very realistic and is more or less a bookkeeping tool. Now there exists theoretical and computational methods to calculate more reasonably charges on atoms from electronic structure calculations like Hartree-Fock, post-HF, semi-empirical and density-functional methods.

One example, taken from Anslyn's Modern Physical Organic Chemistry, is the case of the tetramethylammonium. One would place a positive formal charge on the nitrogen atom, but this is more electronegative than carbon, so there is a contradiction. The calculations mentioned before, indicate that the nitrogen is essentially neutral and the positive charge resides on the methyl groups. So the formal charge is more adequate for indicate the charge on the molecule and not on the atoms.


How is the rate of a reaction defined? I have learnt for many years that rate of reaction is the rate of change of concentration of a reactant or product with respect to time. I never questioned this before, but recently I have been learning about the rate equation and the order of the reaction with respect to a reactant. Seeing the graphs of how concentration changes with respect to time when you have a different order with respect to a reactant (as shown here)

It seems to me that choosing a different reactant to measure the rate of change of concentration of would give you a different rate of reaction? So how is the rate of a reaction defined?

Let's take a sample reaction $\ce{\it aA + bB-> pP + qQ}$ with stoichiometric coefficients $a$, $b$, $p$, and $q$ for reactants $A$ and $B$ and products $P$ and $Q$.

The rate of reaction is the rate of change of the extent of reaction.  Extent of reaction is usually given the symbol $\xi$, so reaction rate is $\frac{d\xi}{dt} = \dot{\xi}=r$.

If the reaction rate is $\dot{\xi}$, then the rate of change of concentrations is:

$$\frac{d[A]}{dt}=-a \dot{\xi}$$
$$\frac{d[B]}{dt}=-b \dot{\xi}$$
$$\frac{d[P]}{dt}= p \dot{\xi}$$
$$\frac{d[Q]}{dt}= q \dot{\xi}$$

That is the stoichiometric coefficients link the rate of formation/consumption of the products/reactants, sort of by the definition of stoiciometry.

Now, when talking about the order of a reaction, what we are doing in effect is giving formulas (or partial formulas) for $\dot{\xi}$.  If the reaction is "elementary", then it will be $a$th order with respect to $A$ and $b$th order with respect to $B$.  This doesn't have to be true but often is.  This means that if you change the concentration of A while holding everything else constant, the change in rate will be proportional to the concentration of $A$ to the $a$th power, etc.  In math form that looks like

$$\dot{\xi} \propto [A]^a$$
$$\dot{\xi} \propto [B]^b$$

where $\propto$ means "proportional to", which is saying that the things are equal except for some one unknown constant.  Thus those two $\propto$ "equations" are equivalent to

$$\dot{\xi} = \kappa_1 [A]^a$$
$$\dot{\xi} = \kappa_2 [B]^b$$

It's important to note that $\kappa_1$ could depend on everything except $[A]$, i.e. it could be a function of $[B]$ etc.  Similarly $\kappa_2$ could depend on $[A]$ (but not on $[B]$).  The only way for both equations above to be true is if

$$\dot{\xi} = k [A]^a [B]^b$$

This means that $\kappa_1 = (k [B]^b)$ and $\kappa_2 =  (k [A]^a)$.  Now that we know the reaction order with respect to each reactant, and have written it down in equation form, we can relate it back to the change in concentration of each of the reactants/products:

$$\frac{d[A]}{dt}=-a k [A]^a [B]^b$$
$$\frac{d[B]}{dt}=-b k [A]^a [B]^b$$
$$\frac{d[P]}{dt}= p k [A]^a [B]^b$$
$$\frac{d[Q]}{dt}= q k [A]^a [B]^b$$

These equations clearly show the relationship between the order of a reaction and the rate of change of reagent/product concentrations.

  I have learnt for many years that rate of reaction is the rate of change of concentration of a reactant or product with respect to time.

By looking at the four equations above, the rate of reaction will only be equal to the product or reactant concentration change when the stoiciometric coefficient of that product or reactant is 1.  And that's true regardless of what the "order" of the reaction is.  The reaction $\ce{\it aA + bB-> pP + qQ}$ doesn't have to be $a$th order with respect to $a$.  It could have been that $\dot{\xi} \propto [A]^{1.7}$ or even that $\dot{\xi} \propto \frac{[A]^{0.2}}{K_m+[A]}$ or whatever.  But in chemistry, it does often happen that reactions do have orders that depend straightforwardly on stoichiometry, which is why those types of reactions are studied so often.


Why is Omega-3 a polyunsaturated and not a monounsaturated fatty acid? If Omega-3 is a fatty acid with a single double bond at the third carbon atom from the omega end, then why is it regarded as a polyunsaturate and not a monounsaturate? Are there different Omega-3s with more than one double bond?

The term "omega-3" just denotes that the position of the first double bond in the fatty acid will be at position "3" (counting from the omega end of the chain).  It does not mean that there is only one double bond in the fatty acid.  The most common omega-3 fatty acids contain between 3 and 6 cis double bonds (therefore they are polyunsaturated).

Here are few examples, again note how they all have a double bond at the third carbon from the omega end of the chain.

α-Linolenic acid (ALA)

Eicosapentaenoic acid (EPA)

Docosahexaenoic acid (DHA)


Chemistry of Walter Mitty's negative developer/stop bath/fixer? I asked a question here How can Walter Mitty examine the negatives of photos in the bright sunny day?

Commonly known, there are three steps using chemical baths to bring out the images of negatives:

developer to bring out the image,
"stop bath" to halt the process,
"fixer" to make the image permanent.

So in details what are the Chemical procedure/compounds/reaction for the negative's developer/stop bath/fixer?

Wish the answer is in depth.

basics here: http://science.csustan.edu/nhuy/chem1002/photoexp.htm

In depth: there are many good books about the chemistry of photography, starting with this 120-year old classic:
https://archive.org/details/chemistryphotog02meldgoog

Here is another that's fully available online and recent enough to cover every major aspect of the subject:
https://openlibrary.org/books/OL5953553M/Photographic_chemistry_in_black-and-white_and_color_photography

You can buy more recent books, especially for developments in color.


How to identify which species will undergo Oxidation and which will go Reduction I have been given $\ce{B2}$, $\ce{N2}$, $\ce{(O2)-}$  and $\ce{O2}$.

I have to short out  which will undergo oxidation, reduction or both.

How can I tell it when reactants are not given?

You have to look at the possible oxidation states of the atoms involved.
For instance, if you were given the molecule $\ce{F2}$, you should consider that fluorine has two possible oxidation states: 0 (in the $\ce{F2}$ molecule) and -1.
Hence, $\ce{F2}$ can only undergo reduction.
For the other molecules, you should use the same strategy: try yourself!


Bond character and bond angles Somewhere along the line of researching why the $\ce{Cl-C-H}$ bond angle in methyl chloride is less than what is predicted ($\pu{109.5^\circ}$), I ran into a book which said that the electron withdrawing effects of the chlorine atom gave the $\ce{C-Cl}$ bond more "p-character".

What does this mean? Is he simply using this term to describe the elongated nature of the bond - i.e. s-orbitals are spherical and p-orbitals not spherical but rather elongated (at least as depicted by cartoon drawings in textbooks)?

Is the author saying that the electron-withdrawing effects of chlorine cause the electrons to spend more time further away from the nucleus; hence the "p-character"? I understand that in a spherical shell, electrons are basically equidistant from the nucleus. But in an elongated, "p-character" shell, electrons can go further away from the nucleus.

S orbitals are lower in energy than P orbitals.  Electrons prefer to be in as low an energy orbital as possible.  Therefor, when we mix S and P orbitals to make hybrid orbitals, the more S character the orbital has the lower the energy of electrons occupying that orbital. Let me use chloroform, $\ce{CHCl3}$, as an example to make my point (I have data for it and the $\ce{C-Cl}$ bonds are analogous to those in methyl chloride).  The chlorine atom is electronegative, so electrons in the $\ce{C-Cl}$ bond will spend more of their time closer to chlorine - away from the carbon, then they would in say a $\ce{C-H}$ bond.  So if the electron density in the hybrid orbital contributed by carbon to the $\ce{C-Cl}$ bond is going to be reduced, why put as much S character in it?  Instead save that S character for other hybrid orbitals emanating from the carbon that have higher electron density in them.  It turns out in the case of chloroform, knowing the various bond angles and the $\ce{C_{3V}}$ symmetry of the molecule, that the carbon portion of the $\ce{C-Cl}$ has a hybridization of about $\ce{sp^4}$ (more P character and less S character just as we predicted since it has a lower electron density) and the $\ce{C-H}$ bonds are approximately $\ce{sp^{2.75}}$ (using that S character we saved from the $\ce{C-Cl}$ bond to stabilize the electrons in these 3 orbitals, just as we predicted).  The same effects play out in methyl chloride, just not as dramatically (e.g. the hybridization index of the $\ce{C-Cl}$ bond will be greater than 3 but less than 4).


'Alcohol Free' toiletries Sometimes I see 'alcohol free' written in big letters on toiletry bottles, such as on deodorants and shaving foams.

You would think that the toiletries bearing these words contain 100% no alcohol, but looking at the ingredients suggests otherwise - I still see alcohols in these products.

So does 'alcohol free' only refer to one type of alcohol not being present in the toiletry, such as a type of alcohol which causes irritation etc?

Summarizing from the FDA website (some of this info could vary by country, but I think the general ideas should hold anywhere) commercial use of the term "alcohol" doesn't coincide with its use in chemistry. Cosmetic products use "alcohol" to mean ethanol, so you may find other alcohol compounds in a product that is ostensibly "$100\%$ alcohol free".


How to separate butane from air? I work in a manufacturing company specializing in polyethelene foam and I have been struggling to separate butane from air in our warehouse during curing period where all the butane gas used to make our product is just released in the air. I have been thinking of ways how to separate and recycle the butane? Can someone help me?

Butane has pretty high  boiling point (around $\ce{0 °C}$). This means, that just cooling air to $\ce{-30 °C}$ should condense a lot (if not all) of butane from it.

However, as I guess, most of butane used in foam production remains in the foam, otherwise the foam would collapse.  So, it is very possible that there is not much to recycle in air itself. The gas, of course, will be released eventually, but only when foam itself is destroyed.


Why are silver halides insoluble in nitric acid The silver halides $\ce{AgCl},~\ce{AgBr},~\ce{AgI}$ are insoluble in $\ce{HNO3}$. This is given in J.D.Lee's inorganic chemistry text. (The strength of nitric acid is not specified though).

But won't this reaction take place?
$$\ce{X- + HNO3 -> X2}$$

This example is taken from JD Lee's book itself:
$$\ce{6Br- + 8HNO3 -> 3Br2 + 2NO + 6NO3- + 4H2O}$$

because  concentrated Nitric acid is a powerful oxidizing agent.

Nitric acid (even concentrated) is essentially an aqeuous solution.

Effectively, silver halides will not dissolve in water (they  have incredibly low $K_\text{sp}$), so the reaction indicated is moot.... there is no appreciable concentration of bromide in solution to be oxidized.

$K_\text{sp}(\ce{AgBr}) = 5.0\times10^{-13} \Rightarrow [\ce{Br-}] \approx 700~\text{nM}$


Equivalence point of titration of Sodium carbonate The pH at the equivalence point of titration of $\ce{Na2CO3}$ solution with $\ce{HCl}$ is around 3.7, as shown in this titration curve:

At this point, the following reaction is completed:
$$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$
But considering the fact that $\ce{NaCl}$ is a neutral salt, shouldn't the pH be closer to 7? The $\ce{H2CO3}$ formed also decomposes to form $\ce{H2O + CO2}$, and shouldn't affect the pH either. Why then is the equivalence point at such a low pH value?

When the $\ce{Na2CO3}$ is completely neutralized by the $\ce{HCl}$, the solution will be saturated with $\ce{CO2}$, so the pH will be lower than 7 - think of carbonated beverages. Although the $\ce{CO2}$ bubbles out, not all of it bubbles out - not even if stirred. If you had stopped the titration at pH = 7, the pH reduction from 11 would have been partly from $\ce{HCl}$ and partly from $\ce{H2CO3}$.