What metals have been used in photography? I'm interested in knowing about the various processes that have been used in chemical photography. As I understand it, most are based on silver. Are there any major ones which have not used silver? What other metals have been used? Are there any processes which don't use metals at all?

I wanted to contribute what I know so far:

Most photography processes are based on silver halides, which are light sensitive, decomposing into metallic silver and halogens. Here is a claim that silver salts are the most light-sensitive, but it is not well-supported. Silver processes include black and white, as well as more complex color processes like E-6 and C-41 which use dye couplers and color-sensitive layers. I don't know what makes the layers color sensitive, but the reaction is like $$\ce{AgCl + photon -> Ag(s) + {1\over 2}Cl_2(g)}$$ Apparently, after a photon is received, freeing a silver atom, this atom must combine with another free silver atom through diffusion over the lattice of the silver halide crystal before it can become part of the image. There may be a better reference, but this fact is mentioned in the photographic developer Wikipedia article.
Platinum and palladium photography exists. These are also based on metal halides, but they usually include ferric oxalate as a sensitizer. Also, they appear to be based on alkali salts like sodium tetrachloropalladate. Palladium is above platinum on the periodic table, and was used as a cheaper alternative. I saw some old black-and-white photographs in a museum recently, which were labelled as platinum or platinum-palladium prints.
Other metals with light-sensitive salts include mercury and gold, but I am not aware of these being used in photographic paper (although mercury saw early use as a developer in the Daguerreotype process). These undergo a different reaction than silver halides, at least in the case of mercury - undergoing disproportionation rather than releasing free halogen.
The Cyanotype process is based on iron, using both ferric ammonium citrate and potassium ferricyanide. It is the source of the name "blueprint". The light-sensitive compound is ferric ammonium citrate; the citrate reduces the ferric ion under exposure to UV light. The resulting ferrous ion then reacts with potassium ferricyanide to form insoluble ferric ferrocyanide (Prussian blue).

Light-sensitive compounds outside of chemical photography:

In biological systems, chlorophyll is mostly organic, but contains magnesium in a key role. This contrasts with the opsins such as rhodopsin which are responsible for vision. They are proteins bound to cofactors, I think the cofactor is always retinal which is organic - so no metals in vision as far as I know.
Light-sensitive synthetic organic molecules exist, such as aryl phthalate esters. However, I am aware of no photographic process based on organic molecules.
Glow-in-the-dark toys are usually based on metallic phosphors such as zinc sulfide and strontium aluminate.
Wikipedia has a category of light-sensitive chemicals which seems under-populated, but includes many inorganic compounds, as well as hydrogen peroxide and LSD.
Digital photography of course uses metals, with CCDs containing capacitors formed from metal-oxide-semiconductors. Whether this can be usefully related to the chemistry of chemical photography I do not know.

I'm not sure what the overall conclusion is. Why metals are important but not necessary; why silver seems to be uniquely sensitive to light, and how to classify and understand this uniqueness...


Why is H2O2 named hydrogen peroxide? Why is $\ce{H2O2}$ named hydrogen peroxide and not dihydrate dioxide?

Oxide is a term reserved for the oxide anion $\ce{O^2-}$. This anion is typically characterised as rather stable since the very electronegative element oxygen is in its maximum negative oxidation state. Dioxide means, that there are two oxides present in whichever structure we are looking at.[1] A typical example for a dioxide would be manganese dioxide, also called manganese(IV) oxide $\ce{MnO2}$. As shown in the image below, it crystallises in the rutile structure, like titanium dioxide $\ce{TiO2}$.

Figure 1: Crystal structure of rutile, $\ce{TiO2}$ which is equivalent to β-$\ce{MnO2}$. Red spheres represent oxides, grey ones the metal in question. Image taken from Wikipedia, where a full list of authors is available.

Hydrogen peroxide, on the other hand, as is evident from the structure in figure 2 below, contains not two $\ce{O^2-}$ units but a single $\ce{O2^2-}$ unit, composed of two oxygens bound to each other. This means that oxygen’s oxidation state is $\mathrm{-I}$ rather than $\mathrm{-II}$ and that hydrogen peroxide is readily reduced. (It can also easily be oxidised back into $\ce{O2}$. The functional group $\ce{O-O}$ is labelled peroxide and it would not make sense calling it dioxide since it does not feature oxides.

Figure 2: Structure of hydrogen peroxide. Red spheres are oxygen, white ones hydrogen. Image taken from Wikipedia, where a full list of authors is available.

Aside from that, IUPAC also allows for nomenclature to entirely ignore the molecular structure and name entirely by molecular composition. In that compositional nomenclature, dihydrogen dioxide is an acceptable name like dihydrogen monooxide would be for water. Note, however, that this only tells us the number and type of atoms involved while not giving us any clue to how they are connected. It is thus typically better to refer to a peroxide as a peroxide to emphasise the actual functional group present.


How do I separate graphite powder from light oil? I have a sample of very fine graphite powder suspended in light oil. I've attempted to separate the two by placing the mix in a 2 micron filter sock. The oil will slowly drip through. If I apply pressure the graphite and oil will force their way through. No pressure, I believe the graphite will collect on the filter and eventually blind the filter preventing oil from coming through.

Any ideas on how I can separate and re-use both items?

"Any ideas on how I can seperate and re-use both items?"
The devil is in the details:

what is particle size and concentration?
what is the light oil?
what do you mean by reusing them?

Depending on particle size (pumped) filtration (paperfilter, colloid filter, celite ) should work as well as centrifugation as was recommended by @Jon Custer. Is it that urgent? Also, cannot you just try it instead of asking it?

What is the oil?

Whatever you do, your graphite will be oily. It does matter if this oil has high-boiling point/waxy components... On the other hand, you may able to use something with low boiling point and just remove it under vacuum / heating, maybe can skipp the whole filtration part.

What do you mean by reusing them?

As I said, there will be remains of oil on the graphite. Also, most probably, you will have fine powder of graphite remains in the oil whatever you do. How do you clean those up it depends on what do you need it for. Do you need to wash the graphite and re-filter the oil? You should know/test that. Also, cost-wise does all this fuss make any sense? Neither graphite nor the oil sound something expensive/special. Also, the more unique, expensive they are, most probably their cleanup will be more demanding (as you need to shoot for higher purity).


A mixture of 4 compounds is extracted with HCl. Which one would go to the aqueous layer? So this has to do something with the H+ ion bonding with the compound. I dont know how to approach it. The answer given is 2

This is a classic example of an acid-base extraction. If a mixture of organic compounds in some organic solvent is treated with an aqueous acid, any bases in the original mixture react to form their conjugate acids. These conjugate acids tend to have ionic functional groups that increase their solubility in water.
The key skills required to solve this problem are 1) identifying the principal functional group in each structure and 2) recognizing the characteristic acid-base properties of each function group.
These compounds are (in order), a sulfide, an amine, an ether, and a ketone. While you could envision perhaps three of them being polar and suitable hydrogen bond acceptors (sulfuer is approximately as electronegative as carbon, and sulfur does not participate in hydrogen bonds), only one is a base - the amine. The characteristic acid-base behavior of functional groups could be constructed each time you need it, but may best be memorized. There are only a few common functional groups that typically behave as acids or bases in aqueous solutions.

Acids: carboxylic acids, phenols, thiols, sulfonic acids, 1,3-dicarbonyl compounds, imides
Bases: amines, imines, amidines, guanidines


Where the equibrium will shift when we add one mole to both reactants and products in PCl5 giving PCl3 and Cl2 I know that if we add 1 mole to reactant side then the reaction will shift forward and if on product side the it will shift backward according to Le Chatelier's principle.Then what will  happen if I add 1 mole on both sides?Please explain.

In a reaction of the form $\ce{A <=> B + C }$ if $a_0$ is the initial amount of A (and B and C are initially zero) then if an amount x reacts the equilibrium constant $K_e = x_e^2/(a_0-x_e) $ where $x_e$ is the equilibrium amount reacted so equilibrium A is  $a_e = a_0-x_e$. Solving gives $x_e = (-K_e+\sqrt{K_e^2+4K_ea_0})/2$.

Now adding 1 mol to each side gives $K_e= (x_e'+1)^2/(a_0+1-x_e')$. Solving again gives $x_e' = (-K_e-2+\sqrt{K_e^2+4K_ea_0+8K_e})/2$. You can work out the ratio  $x_e'/x_e$ for different values of $k_e$ for simplicity, say, with $a_0=1$.

You can repeat the calculation for different initial amounts of B and C and the equations are far more messy but the final result similar.


What is CIF (a PCB developer) and what does it do? I made a PCB (printed circuit board) for the first time, and the process included covering the board with the shape we want, exposing the board to UV light and then putting it in a developer which removes the part of the resin that was exposed to the UV radiation.

My question is, what is this developer? I think it has the commercial name of CIF (http://uk.farnell.com/cif/ar46/developer-ready-for-use-1l/dp/1783557), but what is it really? What does it do to the board and why only to the UV-exposed part?

Thanks!

Developers are usually solutions of $\ce{NaOH}$ or $\ce{KOH}$ (perhaps with some additives). Negative photoresists uses organic developers/solvents. From your description, I assume, that you use a positive photoresist. Positive photoresists are more soluble after exposition to the UV light - polymer bonds are disturbed and therefore more soluble (in areas where exposed). Please see illustration. For more detailed information look for Photoresists and Photolithography.


FMO analysis of the [3,3] Sigmatropic reaction regarding geometry. Is it accurate? While studying the reasons behind the stereospecificity of [3,3] Sigmatropic reactions, I ran into an image portraying both possible transition states for the reaction: The Chair TS and Boat TS. The complete FMO orbitals were only drawn for the Chair TS and the final geometries of the products was shown for both transition states:

I find it a little bit hard sometimes to visualize the three dimensional transitions that result in a specific configuration, so I tried to complete the FMO analysis with the orbitals for the Boat TS and to draw the sigma-bond rotations necessary to create constructive overlap between the lobes. I came up with the following:

On the Chair TS we see the σ* shaded lobe is perpendicular to the shaded lobe of the π HOMO from the double bond next to it (having a 3D model would make it easier to see because in the drawing they seem aligned, but they are actually perpendicular). The sigma-bond would then have to rotate 90º inward for effective constructive interaction, "throwing" the R-Group outwards and resulting in the E-configuration seem in the product. The other σ* lobe also rotates 90º in the same direction (I forgot to draw the arrow there, but it would be the same as the one beside the shaded lobe), but for some reason this rotation result in the regular "cyclic-like" geometry for that double bond (Z-configuration?).

If it wasn't hard enough to visualize it on the Chair TS, it gets worse on the Boat TS. I came up with the orbital configuration as seem in the image. First, I got a little confused about the σ* shading, because the whole premise for the shading had to do with constructive overlap where the double bonds would occur. But there I had to draw opposite phases on the lower part. That, I believe, is because initially they were aligned with the same phases (as in the Chair TS), but the conformation twisting threw the constructive phases in opposite directions. Anyways, to reach constructive overlap the sigma-bond next to the substituted carbon would have to rotate close to 90º counter-clockwise, while the other sigma-bond would have to rotate almost 180º. That explains why this transition state is so unfavorable, but I'm still struggling a bit with the geometry. The 180º rotation sounds like it would result in the unsubstituted double bond pointing out of the "cyclic-like" shape as seem in the product. But I can't clearly see how the 90º rotation throws the R-Group inward in a Z-configuration geometry. Instead, I came up with another orbital configuration possible that might be the correct one:

In that orbital configuration the sigma-bond of the substituted carbon wouldn't have to rotate, and we would have the R-Group pointing inward like in the product.

Which orbital configuration is correct, if any? Is my reasoning about the sigma-bond rotations and the resulting products geometry correct? Do you have any tips when it comes to visualizing geometrical issues on more complicated molecules?

For chair-like transition states the angle at which you draw your chair can make a significant difference to how easy it is to visualise stereoelectronic interactions. Sadly, there is no real way to know which is the best way to draw it, except by either trial and error, or prior experience (or leveraging on other people's experience, i.e. copying it from somebody else!).

If you rotate the chair a bit so that the forming bond is in front (as opposed to at the side), then you can have something like this:

You don't need to overthink how the sigma bond rotates; this is a sigmatropic rearrangement, not a electrocyclic reaction where there are conrotatory/disrotatory modes. The geometry at the substituted carbon definitely changes (from tetrahedral to trigonal planar), so there is some kind of "flattening" as the reaction proceeds, but all you really need to pay attention to is the relative disposition of the R substituent and the rest of the carbon chain: here they are clearly trans to one another, which is an (E)-alkene. This is stereospecific, so if you start with the conformation shown in the left, you will end up with a trans alkene. There isn't a different "rotation mode" that leads to the cis alkene.

The main way of getting a cis, or (Z)-alkene, is not actually via a boat TS, but rather via a chair TS with the R substituent axial instead of equatorial. Note that this is a different conformation from the previous image, so again it's not about which direction the sigma bond rotates in, but rather the original location of the substituent (axial/equatorial). The orbital diagram is exactly the same, so there's no need to redraw it, but it should be easy enough to see that this leads to the reverse stereochemistry.

The boat TS is trickier to draw. This is my preferred representation:

but I am not entirely sure which TS is higher in energy (i.e. whether the substituent prefers to point up or point down), even after consulting several references. I don't think it's a matter of great significance; after all, the boat TS usually only plays a very minor role, as far as I know.


Why is the combination of hydrated aluminium oxide and magnesium hydroxide used to treat heartburn? Why is the combination of hydrated aluminium oxide and magnesium hydroxide used to treat heartburn?

I get why $\ce{Mg(OH)2}$ would be used since its a strong base and is badly soluble in water. Thus it won't do harm to the oesophagus in comparison to something like $\ce{NaOH}$. Aluminium oxide is also poorly soluble, however it is a weak base so what is the point of combining this substance with magnesium hydroxide instead of just increasing the dose of magnesium hydroxide?

Hydrated $\ce{Mg(OH)2}$, aka milk of magnesia, is an osmotic (saline) laxative, as well as antacid.

$\ce{Al(OH)3}$ is also an antacid, but has the opposite effect on the bowels: it's costive, i.e. a binding agent. Aluminum has also been linked (with much uncertainty!) with Alzheimer's disease, so it has become less popular as an antacid.

The combined effects of $\ce{Mg(OH)2}$ as laxative and $\ce{Al(OH)3}$ as a costive might tend to counteract each other, so as to avoid interfering with colonic function.

However, it would be far simpler to use $\ce{CaCO3}$ as an antacid, unless the release of $\ce{CO2}$ bubbles or excess calcium in the diet is an issue.


Replacement for cyclohexane I am attempting to isolate pure iodine from kelp using the method described here
but I do not have any way of possibly obtaining cyclohexane.  I was wondering if any easier to obtain solvents would work as well to isolate the iodine from the rest of the mixture.  I was hoping to use a solvent that is easy to obtain in a grocery or hardware store.

Cyclohexane is used for this extraction because it is nonpolar, as is iodine. A more polar solvent like acetone or ethanol will also extract a whole bunch of other stuff like essential oils, proteins, and other polar compounds.

You want Mineral Spirits which you should be able to purchase at any quality hardware store or Walmart. Mineral spirits, mineral turpenine, or whatever you want to call it, is a mixture of saturated hydrocarbons distilled from petroleum. It is a general purpose paint thinner for oil-based paints. Since it is mainly C7-C12 alkanes, it will have a higher boiling point than cyclohexane $(140\ ^\circ\mathrm{C}$ vs $81\ ^\circ\mathrm{C})$., but it will have the same general polarity as cyclohexane.

If you can find products labeled as petroluem ether, light naphtha, or ligroin, all are alkane mixtures with lower boiling points than mineral spirits.

Another common low polarity paint thinner is toluene, which has a lower boiling point at $110\ ^\circ\mathrm{C}$.


What does negative pressure mean in the context of this manometer? I have a basic question on a closed manometer:

I see that in this case the pressure of gas is lower than that of the liquid, which is why the liquid is higher on the left side. Now, what I don't understand is why do we say that the gas is at a negative pressure?  Wouldn't that mean there's no gas at all? Also, how do we know that it's negative? Usually, I only take the difference between the highest point and the lowest.

The manometer shown doesn't measure absolute pressure, just relative pressure. In this car the pressure of the gas in the closed tube versus the pressure of the gas in the bulb.

As you have drawn it, the pressure of the fas in the tube is higher. If you measure pressure in the bulb relative to the pressure in the tube the answer is negative. If you measured absolute pressure (which this manometer can't do because of the lack of any external reference in the closed system) both would be positive but one would be higher than the other.


How is taste of chemical compounds determined? How do we know the taste of compounds, and their smell too, when some of them are poisonous. My textbook had that Lead is sweet in taste, but it's poisonous. So, how exactly do we determine their taste and odour of such compunds?

When a compound is synthesised or isolated, it needs to be characterised. This characterisation serves to determine whether something has been synthesised/isolated before and as a template for later generations to compare their synthesised/isolated compounds to the one you have.

Nowadays, one reports NMR and mass spectrometry data; for chiral compounds specific rotation is measured. Sometimes, elemental analysis is included. Depending on the journal’s requirement you might also need to measure IR and/or melting points. Furthermore, you typically describe your compound (is it a yellow oil or white needle-like crystals?) although the description is considered secondary and the NMR/mass data is what you look at for matches.

In the old days, these analytical methods were not available. Instead, analyses would rely heavily on what a compound looks like and how it behaves. Thus, it was attempted to generate various crystalline forms and reactions with various standard reagents were investigated. Additionally, melting points were measured and chemists used all available senses to characterise. Thus, when the paper says that boron phosphate is a white, odourless and tasteless compound, the chemist in question not only sniffed the flask but also put a bit on his tongue to assess taste.

The problem with this approach is that you have no idea whether something is toxic/harmful or not. As there are many mechanisms by which a compound can interfere with the tightly controlled reactions in our body and cells or the macromolecular structures therein, many (if not most) compounds will be harmful in at least one way. While the common approach was to taste as little as possible, the risk remains as sometimes the $\mathrm{LD_{50}}$ is in the $\pu{\mu g/kg}$ range so even the smallest sample could kill you (although compounds that toxic are thankfully rare). Furthermore, a lot of compounds are able to induce DNA mutations which likely led to increased cancer rates and an overall lower life expectancy for chemists.

This is, quite obviously, unacceptable from a workplace safety point of view. Thus, tasting chemicals was prohibited practically everywhere. Furthermore, to prevent accidental contamination or confusion of glasses, not even the consumption of food and drink is permitted in most laboratories. Therefore, for anything that has been synthesised or isolated for the first time after these measures were in effect, we typically do not know taste any more. In fact, unless we are certain that we have evidence a compound is harmless, it is to be treated as highly toxic in the lab with all according security measures.

A compound’s odour is a slightly different story. It is said that the chemist working ideally will never get a wiff of their compounds because they will make sure that none access the common lab air. In practice, that is practically unattainable especially since lab scales are often in the middle of the room and most solids are in glass bottles that need to be opened fully to take out a sample. Thus, it is unavoidable that chemists pick up the smell of a number of (typically particularly volatile or particularly odourous) compounds. Systematic odour examinations are no longer performed but where the odour is characteristic, it is sometimes included in a compound’s characterisation (e.g. ‘sulphurous odour’). Some chemists take this a step further: a former colleague once sniffed at the flask of another former colleague and exclaimed ‘it smells like your product’, explaining that tertiary alcohols apparantly have a distinctive odour.


What is the oxidation state of carbon atoms in Ethanoic acid? The oxidation states of $sp^2$ and $sp^3$ carbon in ethanoic acid  are $-3$ if we do not consider the electronegativity difference of these two carbons due to their different  hybridisation  but this is not the case with the thiosulphuric acid ($\ce{H2S2O3}$); here the oxidation states of sulphur here are +6 and -2. So why are we considering the electronegativity difference of two $sp^3$ and $sp^2$ sulphurs here?

Typically when figuring oxidation state of an atom, electronegativity of different atoms of same element is considered equal. This means that in thiosulphate oxidation state of central sulfur would be somewhere around 4.

However!
Let us condider a very tricky case: ozone. Technically it consists of atoms of same element and the oxidation state of oxygen atoms should be zero. But ouch, auch, the central atom bears full positive chanrge in all feasible structures.... So, should ozone, disregarding electron shifts in covalent bonds, have oxidation state of atoms +1 (for central atom) and -1/2  (for two other atoms)? But this doesn't make much sense. This is a place where oxidation state abstraction breaks.

So, there are situations when oxidation state abstraction doesn't make much sence.

Thiosulphate is somewhate similar. The central sulfur behaves very muc like sulfur in sulfate, so by analogy it should be in +6 oxidation state. However, formally it is not the case. Here a clear conflict between formal abstraction and real life occurs, so people cope with it variously. One possibility is to arbitrarily say that the central sulfur of thiosulfate is in +6 oxidation state (and it makes sence). Another is to go formally (and it also makes sense). I saw both options in the wild.

TL;DR : formally the central sulfur in thiosulphate is +4; but people sometemes disregard it as the ion is very similar to sulphate with central sulfphur atom being +6.


Why does carbon dioxide not form addition compounds? Although $\ce{CO_2}$ has four lone pairs on the two oxygen atoms, but it still does not form addition compounds by donating lone pairs. Why?

$\ce{CO}$ on the other hand, does form addition compounds by donating the lone pair from C atom.

A fairly simplistic argument is that in CO, the lone pair on oxygen is of a much lower energy than the lone pair on carbon (because oxygen has a greater effective nuclear charge and therefore pulls electrons closer to itself). Therefore it is quite rare (very rare indeed) to find any compound that has the oxygen acting as a donor atom; CO only really reacts via its lone pair on C.

Carbon dioxide only has low-energy lone pairs on O, and is consequently a lousy Lewis base.

One could easily refine this by bringing in MO schemes, but at 3 a.m. in the morning, this is good enough to convince me.


Why do isotopes have different chemical reaction rates? Isotopes of a given element differ in the rates at which they undergo chemical reaction.

Why is this so? The only explanation I could think of is that the gravitational force on the electrons increases and hence the electrons are better bound to he nucleus. But this force is very negligible.
Is there perhaps another contributing factor?

Isotopes differ in mass.
Otherwise the same molecules but with the different mass have the different mean speed.
The different speed means the different molecular collision rate.
The different collision rate means the different kinetic constants.
There are some indirect effects of different isotop mass on the kinetics, but these are much smaller than the direct kinematic effect.


Biological contamination from ground I took samples of tapwater in order to examine concentrations of Escherichia coli and enterococcus in a rural area. I was told, from an analytical lab, to take samples in four sterilized vessels bought from pharmacy.

The problem is that at the completion of the last vessel, its cap fell frοm my hand down into the soil and some grains of sand were stuck on the inner surface of the cap. As fast as I could I picked it up and washed it off totally with the water of the sampling tap. After that I closed my last sampling vessel with it and this is the completion of the sampling.

How large is the possibility my sampling has a serious contamination of these twο biological pollutants after this incident?

The point is not how large the possibility is.
There is a possibility that the flask has been contaminated, so you should have taken a new flask.
I suppose you did not have a spare one with you.
Then, the best should have been to sterilize again that flask, but that is not easy in the field.
I mean, it is much easier to take more sterile flasks with you than to sterilize them in the field.

Now, if this sample shows that there are no bacteria, you can conclude that there are no bacteria in the water.
But if there is some concentration of bacteria, you can not know if they come from the water or from the soil or how much come from each part.


Why does naphthalene have a higher melting point than both water and diphenylamine? We learn that polar compounds have higher melting and boiling points than nonpolar compounds due to strong intermolecular forces such as hydrogen bonds and dipole-dipole interactions.

Naphthalene is an extremely simple and nonpolar compound, yet it has a significantly higher melting point than diphenylamine (a similar enough structure that is not only heavier, but has a dipole-inducing amine) and water (famously super polar).

What's going on here?

In comparisons such as this water needs to be excluded as it is exceptional in almost every respect due to its highly directional and strong hydrogen bonds.

As a broad generalisation the melting of solids is more dependant on the nature of the repulsive forces between molecules than on their attractive ones. Experimentally the enthalpy of melting ( Latent Heat or just Heat of Fusion) is approximately ten times less than the corresponding heat of vaporisation which indicates that repulsive forces play a greater role in melting than in vaporisation: in a solid molecules tend to be slightly closer together than in a liquid, thus may experience larger repulsive forces. Van-der-Waals forces tend to have weak orientational dependence, but repulsive forces do not as they are short range and  depend on the asymmetric shape of the molecules and can have a large effect.

Thus melting points are rather dependent on the geometry of the molecules, if they can pack well into a lattice then a high melting point is expected. If the molecule has a dipole this can cause both attractive and repulsive interactions depending on the relative positioning of molecules; head to tail can be attractive if molecules are face to face, but repulsive if in line one behind the other and vice versa if their relative orientation is changed.

In the case of naphthalene and the amine, clearly the naphthalene can stack nicely and so increase its attractive interaction relative to repulsive ones, whereas the amine is twisted and cannot so easily pack meaning that repulsion can be stronger and attractive interactions weaker (even with dipole) and melting points lower. As an example consider the  $\ce{C6H14}$ isomers, n-hexane mp -95 C, 2-methyl-pentane mp -154 C but boiling points are 69 & 60 C respectively and $\ce{C8H18}$ isomers, n-octane mp -57, 2-me-heptane mp -120 and bp 126 & 119 C respectively.

( Attractive forces largely determine the enthalpy of vaporisation, and by Trouton's rule their boiling points. The enthalpy of vaporisation is closely related to the (attractive) cohesive energy in a liquid.)


Why in thermodynamics, some reactions are taken at constant temperature and pressure? What is its significance? To be more specific while deriving relation between cell emf and free energy, constant temperature and pressure  assumptions were taken. Why?

If you read my answer to this question: Why does the Gibbs free energy only correspond to non-expansion work?, you will see that

$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + đW_\text{add}$$

Of these three terms, only $đW_\text{add}$ is related to the electrochemical work. Therefore, in order to equate $\mathrm{d}G$ to $đW_\text{add}$, you have to assume constant $p$ and $T$ such that both $\mathrm{d}p$ and $\mathrm{d}T$ are equal to 0.

In general, this applies to chemical equilibria as well. Because of the above expression, we can say that $G$ is the "natural quantity" to use in systems with fixed pressure and temperature. If instead you had a system with fixed volume and temperature, the Helmholtz free energy would become much more relevant.


Does solid fat float in liquid fat? Ice floats in its liquid state, water, because it is less dense, and I know fat also floats in water BUT will a piece of fat dropped into a pot of melted fat also float? or will it sink and what is it about its molecular structure that allows this?

I've searched google for about three hours and all I can find are articles on fat floating in water, so any help is greatly appreciated!

I thought this question was a good idea for a home experiment to do with my 8th grade son.

Into a block of dry ice, we carved a conical mold hole with the pointed end of scissors and poured canola oil into the hole:

We covered this with a second block of dry ice for about a minute and extracted the frozen oil with forceps:

Then we dropped the frozen oil into a jar of liquid canola oil:

And it sank.


What is the purpose of balancing a redox half equation using H+ and H2O and electrons? Is it possible to say that:

H+ is used due to the provision of an acidic medium?

H2O used as it is an aqueous medium and products have oxygen, so logically oxygen must come from H2O?

Electrons added to conserve charge?

One reason to track solvated hydrogen (or hydroxide) ions and water in a redox reaction is that the hydrogen ions often get transferred too, and you need to know how that takes place to render the products.  We should expect hydrogen-ion transfers as part of the process in a protocol solvent such as water or a concentrated protic acid, as the electrons are charged and some compensating charge-transfer process must take place to keep the charges in balance.
Copper in concentrated nitric acid, where nitrogen is reduced (primarily) to the +4 oxidation state, is a well-worn example.  The oxidation half-reaction is, of course, copper being oxidized to $\ce{Cu(II)}$.  So we look at the reduction half-reaction.  In its "raw" form we would have
$\ce{HNO_3 -> NO2}$
We balance the elements other than hydrogen and oxygen, which is already done, and then compare the combined oxidation states of those elements in the reactants and products to find that the nitric acid must have taken one elecron:
$\ce{HNO_3 + e^- -> NO2}$
Now the charges are not balanced, and so we invoke that compensating hydrogen ion transfer.  In acid this is set up by adding (solvated) hydrogen ions to balance the charges:
$\ce{HNO_3 + \color{blue}{H^+} + e^- -> NO2}$
Water is used to balance the hydrogen which, if we did things correctly, balances the oxygen as well giving our familiar result
$\ce{HNO_3 + H^+ + e^- -> NO2 +H2O}$
and upon combining with the copper oxidation:
$\ce{Cu + 2 HNO_3 + 2 H^+ -> Cu^{2+} + 2 NO2 + 2 H2O}$
in which we see that in the final reaction, hydrogen ions are indeed transferred.  So we needed to account for them.


Why are tertiary alkyl halides obtained in higher yield than secondary or primary during free radical halogenation of alkanes? From my textbook:

  Halogens ($\ce{Cl_2}$ and $\ce{Br_2}$ ) react with alkanes in the presence of ultraviolet light to form haloalkanes.This reaction is free radical substitution and gives a mixture of mono, di, or polysubstituted haloalkanes which are difficult to separate into pure components. Moreover, the yield of any one compound is low because of the formation of other compounds. For example, in case of butane, a mixture of 1-chlorobutane (28%) and 2-chlorobutane (72%) is obtained, where the secondary alkyl halide is obtained as the major product rather than the primary one.
       $$\ce{CH_3CH_2CH_2CH_3->[\ce{Cl_2,~UV~light}]CH_3CH_2CH_2CH_2Cl + CH3CH_2CHClCH_3}$$
    In general, the ease of substitution of various hydrogen follows the sequence:
      $${\rm tertiary > secondary >primary}$$

After reading through all the above statements in my book, I now have the following questions:

Why are tertiary alkyl halides obtained as major products instead of secondary or primary ones? It seems that a hydrogen from a tertiary carbon can be more easily replaced by a halogen than that of a secondary or primary carbon. If it is so, why is hydrogen from a tertiary carbon more easily replaced than a hydrogen from a secondary or primary carbon?
Why are fluorine and iodine not considered in case of free radical halogenation (the first two lines of the above passage extracted from the book mentions only chlorine and bromine)?
Why is a tertiary alkyl halide not produced in the above example? Even if we think it might be produced in a lesser amount, there should have been more of it than the primary or secondary alkyl halides. It seems the reason for tertiary alkyl halide not being in the equation is because it might not have formed or it might have formed in a lesser amount than primary or secondary alkyl halides. If it is either any case, it would violate the order mentioned in the book.

The reason for the ordering is that tertiary radicals have a lower energy (and are thus easier to form) than secondary radicals, which are in turn easier to form than primary radicals. This is because of a phenomenon called sigma conjugation - the weak overlap between the half filled orbital and the adjacent side group orbitals gives a slight lowering of energy.
This means that it is easier to remove a hydrogen to give a tertiary radical intermediate than to give a secondary radical intermediate and so on.
This is similar to Markovnikov's rule.

The reason a tertiary product is not formed is that for butane there is no tertiary centre to begin with - nowhere that a Cl radical can react to give a tertiary butyl radical. Forming a tertiary product would require a rearrangement of the carbon chain.


Relation between Gibbs energy and net reaction rate I have come across some papers in biochemistry that seem to claim that a reaction can be "near equilibrium" even though the flux through the reaction (net reaction rate) is large. For example, this paper measures absolute metabolite concentrations in E.Coli cells grown on two carbon sources, performs some thermodynamic calculations, and concludes that "lower glycolysis is near equilibrium on both carbon sources, with $\Delta G$ approximately 0". Yet this pathway carries net flux in these conditions according to the authors, and usually glycolytic flux is rather large.

How can this be? In my understanding, the net rate of a reaction depends on the Gibbs energy $\Delta G$ so that, near equilibrium where $\Delta G \approx 0$, there is no net flux through the reaction. And I think this dependence should be smooth, so that $\Delta G$ must be large when the net rate is large?

The authors suggest that reaction rate depends steeply on $\Delta G$, so that "Small changes in $Q$ are accordingly adequate to tip the thermodynamically favored flux direction." (From the supplementary it looks like the reaction quotient $Q$ varies between 0.5 and 6.) I have also encountered this idea in works on thermodynamic flux analysis, where only the sign of $\Delta G$ is considered important, its magnitude $|\Delta G|$ is ignored. But if this is true, the reaction direction seems unpredictable and the system should be difficult to control?

Is this all just a question of what "small" or "large" $\Delta G$ means? Can the dependence on $\Delta G$ be sharp, so that the sign of $\Delta G$ acts like a "switch"? Can a reaction really be considered "near equilibrium" if the net reaction rate is large?

Thank you for putting together an excellent, well-specified, thoroughly documented question.  The answer, as you suspected, largely comes down to what "small" and "large" mean.

  I have come across some papers in biochemistry that seem to claim that a reaction can be "near equilibrium" even though the flux through the reaction (net reaction rate) is large[....] How can this be?

Consider the equilibrium approximation of Michaelis-Menten kinetics:

$$  \ce{E + S <=>[k_f][k_r] E\cdot S ->[k_3] E + P}$$

When $k_r \gg k_3$, the enzyme-substrate complex $\ce{E \cdot S}$ is assumed to be in equilibrium with $\ce{E + S}$, i.e. free enzyme and free substrate.  Making this equilibrium assumption eventually provides the rate law for overall flux through the reaction:

$$\frac{dP}{dt} = k_3 E_0\frac{S}{S+\frac{k_r}{k_f}}$$

Note that the flux is not always zero even though we have assumed that $\ce{E\cdot S}$ is in equilibrium with $\ce{E + S}$!  This is one hint -- hopefully a familiar example if you have studied biochemistry -- that assuming (quasi-) equilibrium doesn't mean zero fluxes.

  In my understanding, the net rate of a reaction depends on the Gibbs energy $\Delta G$ so that, near equilibrium where $\Delta G \approx 0$, there is no net flux through the reaction.

This isn't quite right.  First, you are right that the net rate of a reaction depends on the Gibbs energy $\Delta G$.  But near equilibrium where $\Delta G \approx 0$, it would be better to say "there is little net flux through the reaction".  Then the question is little relative to what?  The answer is that "little" is relative to the exchange flux.  So mathematically,

$$\nu_{net} = \nu_f - \nu_r = k_f(E)( S) - k_r(E\cdot S) $$
$$\nu_{exchange} = \nu_r = k_r (E \cdot S) $$

Reactions that are "approximately" at equilibrium have $\nu_{exchange} \gg \nu_{net}$.  Thus, "little" is relative to the exchange flux, and not necessarily the rates of other reactions in a metabolic network.

  ... "Small changes in $Q$ are accordingly adequate to tip the thermodynamically favored flux direction."

Remember that $\Delta G = \Delta G^\circ + RT \ln Q$.  The question of whether a reaction is reversible is asking about the sign of $\Delta G$.  The authors' statement that small changes in $Q$ can change the sign of $\Delta G$ only make sense if $\Delta G^\circ \approx 0$.  From the context of their remark, I think they are aware of this.

  But if this is true, the reaction direction seems unpredictable and the system should be difficult to control?

I wouldn't say "unpredictable" but I think you're right to say "difficult to control".  Another way of saying it would be that those reactions in the metabolic network that are approximately at equilbrium are not likely to be control points.  This 2006 paper examines that topic in more detail.

  Is this all just a question of what "small" or "large" $\Delta G$ means?

Yes!

  Can a reaction really be considered "near equilibrium" if the net reaction rate is large?

It depends what you mean by "large".

If you mean "large" relative to some other flux in the network, then yes, a reaction can be near equilibrium even though its flux is "large".
If you mean "large" relative to the exchange/equilibration flux, then no, a reaction near equilibrium has a net flux that is very small compared to the exchange flux.


Can a solution of FeCl2 go bad over time? My lab has been using solution of FeCl2 for an assay that was mixed about three years ago. Recently the assay has failed twice. I was wondering if it is possible that the iron ions in the solutions have changed oxidation over time?

Water evaporates. Water dissolves the fixed gases ($\ce{N2}$, $\ce{O2}$, $\ce{CO2}$). A well-sealed bottle of water will last for years - that's assuming that we understand that 'bottles' are made out of heavy glass*, and seals are permanent (in the sense that opening the bottle is irreversible). That's not a particularly useful way to store stuff you need access to. (In other words, most bottles allow only limited storage life.) $\ce{O2}$ can oxidize $\ce{Fe(II)}$. Unless working under (and stored under) inert atmospheres, the titer will inevitably change over time. Even when working in inert atmosphere, there will be evaporative losses. All that said and assuming that there wasn't any accident involving the reagents, nor changes to their storage conditions (temperature, location, etc.) then the decline in $\ce{Fe(II)}$ should be roughly proportional to the number of times the bottles are accessed. Assuming that that is regular, with no recent upsurge, then the change in titer should be gradual over time; no sudden change should have occurred.(*please don't tell me that they're storing reagents in plastic! yet still any changes should be gradual, not sudden.)


Are there any crystals that are theoretically possible, but haven't yet been made? Are there any crystals (that exist at standard temperature and pressure) that are theoretically possible, but haven't yet been made?

For example, the predicted requirements for temperature, pressure, materials, etc. for crystallisation to occur are too difficult to achieve.

When we are talking about the prediction of materials, it is generally about the structure and corresponding free energy, not the crystallization process. That being said there are a huge number of inorganic materials that predicted to be stable yet no one has synthesized them.

There are computational databases like Materials Project that contain automatically generated (theoretical) structures and their estimated free energy, solubility, magnetic properties, bandgap, stability compared to competing phases, etc.


How Many mL of 1.0M HCl Would be Required to Completely Decompose Sucrose into Glucose and Fructose? I am currently conducting an experiment into the glucose content of certain sports drinks. However, I have hit a roadblock. A majority of the drinks that I intend to test, contain sucrose. I am thinking that by adding $\pu{1.0M~ HCl}$, to say, $\pu{100mL}$ of solution (sports drink), it will decompose the sucrose into its constituents. I'm planning to use Benedict's solution and a spectroscope in order to determine the amount of glucose present in solution (I'm going to add activated charcoal to the solution to remove the coloring from it before placing in the spectroscope). As a high school student I only have limited resources.

So my question is,

Assuming there is 10 grams of sucrose per $\pu{100mL}$ of solution, what volume of $\pu{1.0M~ HCl}$ would be required to completely break apart sucrose into glucose and fructose?

Would I have to use stoichiometry to find the amount?

So far I know I have $\ce{C12H22O11 + HCl}$, What happens to the $\ce{Cl}$ from $\ce{HCl}$ when fructose and glucose is formed?

The hydrolysis of sucrose is acid-catalysed, so HCl will not be in the chemical equation (it will be listed above the arrow). In theory, this means that you could have little HCl, as it will never be used up and will keep reacting to decompose all of your sucrose. So it isn't so important how much you use, but you should still use enough that it won't take ages to decompose all the molecules. If you're interested you could look up mechanisms for acid catalysis, but you wouldn't be expected to know this at high school level.

To find the chemical equation for the decomposition, you can add up the number of elements of fructose + glucose and compare to sucrose. You will find that two hydrogens and one oxygen is needed, so it is simply a hydrolysis reaction (adding a water molecule).


Why does the outside of a plastic cooking oil bottle get sticky over time? Cooking oil (I've observed it with canola oil, if that matters) is not ordinarily sticky. When I buy a plastic bottle of it, the outside is not sticky.

But when use some of it, then return after a few weeks, I find that the outside of the bottle is sticky. Why is this?

It does not appear to happen with olive oil in glass bottles.

Some drying oils, such as linseed oil, react with oxygen in air to polymerize, eventually forming a hard film. When the oil is in the midst of change, i.e. short-chain polymers, it's a sticky mess (don't touch wet paint).

Most cooking oils are unsaturated. Having double bonds makes the oil easier to digest and less likely to deposit in the circulatory plumbing. It also makes the oil more chemically reactive. When in a closed bottle, it reacts fairly slowly with residual air, eventually becoming rancid. Spread over the surface of a bottle, the oil is exposed to more air and oxidizes and polymerizes more rapidly.

An experiment to test this would be to coat two pieces of the plastic used for bottle, usually PETE, one with cooking oil (unsaturated) and one with mineral oil (mostly saturated) and check if the cooking oil becomes sticky, while the mineral oil should not.

One other factor, though, is that oils will cause dust in the air to adhere, so that might also cause some stickiness. Place the test swatches in a container covered with paper towels to keep off dust.


Continuous chemical reaction network versus chemical reaction network - difference? I'm reading a paper entitled Reachability Problems for Continuous Chemical Reaction Networks and as I was reading it I realized: I have no idea what the difference is between a continuous chemical reaction network (CCRN) and a chemical reaction network (CRN). The paper does distinguish between the two, so they aren't two terms for the same thing. The wikipedia article doesn't distinguish/give a definition for a CCRN. Most of the results when I search google are papers that are way over my head - the one I'm reading right now is to be quite honest over my head.

Any help would be appreciated. Thanks!

I suspect that the distinction is that a CRN considers discrete species concentrations whereas a CCRN considers continuous species concentrations: "The CCRN model is continuous, dealing with real-valued concentrations of species...". More precisely, whereas a CRN deals with concentrations $x(t), y(t) \in \mathbb{Q}$, a CCRN deals with concentrations $x(t), y(t) \in \mathbb{R}$.

On a microscopic level, it's easy to see why $x(t), y(t)$ should be elements of $\mathbb{Q}$: a microscopic concentration is simply a ratio of the number of one molecule to the total number of molecules in the system, and the numerator and denominator must thus be integral.

For macroscopic systems, however, since the denominator---the total number of molecules in the system---is so large, it becomes a very good approximation to take $x(t), y(t) \in \mathbb{R}$. Moving to $\mathbb{R}$ is convenient, since then by physical considerations $x(t)$ and $y(t)$ are continuous, and we can, under mild assumptions, take derivatives and return to a rate-law formalism involving differential equations rather than difference equations.

Remark. I'm not familiar with this field, and I don't know how common the CRN/CCRN distinction is. Indeed, the Wikipedia article assumes differentiability of $x(t), y(t)$ and hence that $x(t), y(t) \in \mathbb{R}$.


Why does mercury decolourise a gold ring? I saw that a gold ring decolourised as it got in contact with mercury .
Why does this happen ? Is there any way to reverse this?

The grey colour is an amalgam of mercury and gold.

Mercury forms amalgams with many other metals. Some are used as chemical reagents in laboratory chemistry as they have different properties than the original metals involved. Gold amalgam is much greyer than gold. Silver amalgam has been used in dentistry.

Mercury has been used in the extraction of gold in mining as it can extract gold from low grade ore. But the recovery process involves distilling off the mercury which is dangerous and polluting.

Chemists used to work with mercury a lot and, when they did, they were recommended to remove their wedding rings or other gold jewellery as the mercury would damage the gold.

Once the amalgam is formed there is no easy way to reverse the process short of resmelting the ring. If the contact with mercury has been brief, you might be able to polish it to remove the amalgam layer, but this won't be easy.


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

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

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

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

Examples that invoke hydrogen bonding make this more apparent.

F-C...(tertiary) + H2O

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

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


Molality of solution to be used when finding water separated from solution as ice Question

$\pu{1 kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$?
$K_\mathrm{f} \ \text{(water)} = 1.86\ \mathrm{K\ kg/mol}$

Solution given:

$\Delta T_\mathrm{f} = 0 - (-4)=4$
$i = 1$ for sucrose.
Therefore, since $\Delta T_\mathrm{f}$ is defined as
$$\Delta T_\mathrm{f}= K_\mathrm{f}\cdot i \cdot \frac {\text{amount of sucrose in } \pu{mol}}{\text{weight of water left in }\pu{kg}}$$
[...]

My doubts
$\Delta T_\mathrm{f}$  is the absolute value of the difference of freezing point of pure solvent and the solution with the non volatile solute. Here, the freezing point of the solution is not $\pu{-4 ^\circ C}$. So how can we say $\Delta T_\mathrm{f}$ is equal to $4$?
Why does the molality term contain amount of remaining water in the denominator? In the original equation shouldn't it be the original molality here?

The question: $\pu{1kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$?
Given: $K_\mathrm{f} \text{ (Water)}= \pu{1.86 K kg/mol}$

The solution is given assuming you understand the concept. It seems you did not completely understand the colligative properties of solutions. Thus I'm giving some clues you how to approach this problem.
The textbook equation you are going to use is:
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times \text{molality of the solution} = K_\mathrm{f}\times i \times \frac {\text{amount of sucrose in }\pu{mol}}{\text{weight of solvent in }\pu{kg}} \tag 1$$
where $i = 1$ for sucrose (sucrose does not ionize in water).
The solution is kept at $\pu{-4 ^\circ C}$. The freezing point of pure water is $\pu{0 ^\circ C}$ ($\therefore \ \Delta T = 0 - (-4)=4$). First what you should do is find the freezing point of given solution using equation $(1)$. If $\Delta T_\mathrm{f} \gt 4$, the game is over. That means, this solution does not freeze at $\pu{-4 ^\circ C}$. If $\Delta T_\mathrm{f} \lt 4$, the game is on. That means, some water will freeze out until the final molality reached the molality of solution, which freeze at $\pu{-4 ^\circ C}$.
For given solution (no units was used):
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m = 1.86 \times 1 \times 0.75 = \pu{1.395} $$
Thus, this solution freeze at $\pu{-1.395 ^\circ C}$.
Now, you have to find out at what molality of a solution would be just about to freeze at $\pu{-4 ^\circ C}$. That means, $\Delta T_\mathrm{f}= 4$ (still, $i = 1$):
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m \ \Rightarrow \ m= \frac{\Delta T_\mathrm{f}}{K_\mathrm{f}} = \frac{4}{1.86} = 2.15$$
This means, water in original mixture freeze out until the molarity of solution becomes $\pu{2.15 mol/kg}$.
Put it this way:

The original solution has $\pu{0.75 mol}$ of sucrose in $\pu{1 kg}$ of water. The amount of sucrose would not freeze.
Thus, how much water would remain with $\pu{0.75 mol}$ of sucrose to keep molality at 2.15?

I'm sure it is given in your solution manual. Regardless, I'll show you the calculations. Suppose the weight of water is $W \ \pu{kg}$ (ignore the units in equation}:
$$ \frac{0.75}{W} = 2.15 \ \Rightarrow \ W = \frac{0.75}{2.15} = 0.349$$
Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.349 kg}$ (the amount of sucrose is still $\pu{0.75 mol}$).
However, this calculation is in significant deviation from actual amount since the given amount of original solution is $\pu{1.0 kg}$. That means it does not contain $\pu{0.75 mol}$ of sucrose. Now, molecular weight of sucrose is $\pu{342.3 g/mol}$. If amount of sucrose in $\pu{1.0 kg}$ of original solution is $x$, then:
$$\frac{\frac{x}{342.3}}{(1000-x)\times 10^{-3}} =0.75  \ \Rightarrow \ 1.257x = 0.75 \times 342.3   \ \Rightarrow \ x = \frac{256.725}{1.257} = \pu{204.2 g}$$
Therefore, the amount of water in that solution is $\pu{1000-204.2) g} = \pu{795.8 g}$ or $\pu{0.796 kg}$. Thus, amount of sucrose in the of original solution is: $\frac{\pu{204.2 g}}{\pu{342.3 g/mol}} = \pu{0.597 mol}$.
Using these values, we can recalculate the water amount in final solution. Suppose the weight of water in final solution is $W_f \ \pu{kg}$:
$$ \frac{0.597}{W_f} = 2.15 \ \Rightarrow \ W_f = \frac{0.597}{2.15} = 0.278$$
Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.278 kg}$ (the amount of sucrose is $\pu{0.597 mol}$ making it $\pu{2.15 molal}$).
Thus, the amount of water freeze out from the original $\pu{1.0 kg}$ of solution is:
$$\pu{(0.796-0.278) kg} =\pu{0.518 kg}$$


Concentration of product at a time t in a reaction A reaction was given $\ce{A<=>B}$ The rate constants for forward and  backward reactions were given as $k_\mathrm{f}$ and $k_\mathrm{b}$. The initial $\ce{[A]}$ was given as $a_{0}$ and the $\ce{[B]}$ at a time t (before equilibrium is attained) was to be calculated.

I tried solving it as follows -

For A I wrote -

$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k_\mathrm{b}\ce{[B]}-k_\mathrm{f}\ce{[A]}$$

and similarly for B -

$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k_\mathrm{f}\ce{[A]}-k_\mathrm{b}\ce{[B]}$$

Combining these two I got a 2nd order differential equation and wasn't able to proceed further. Is there something wrong with my approach or my equations?

Your method should work as you know the boundary conditions for your scheme, but it is not the easiest method. Instead, it is easier in this case, as you are given the initial amount of A, to let $x$ be the amount of B produced at time $t$, then $a_0 - a = b = x$ where to save writing brackets, $a,\;b$ are the amounts of A and B at time $t$. The the rate equation is

$$\frac{\mathrm{d}x}{\mathrm{d}t} =k_\mathrm{A}a -k_\mathrm{B}b$$

and substituting gives

$$ \frac{\mathrm{d}x}{\mathrm{d}t}= k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x$$

This can now be integrated with limits $(0,x),\; (0,t)$ to give

$$\ln \left( \frac{k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x}{k_\mathrm{A}a_0}   \right)= -(k_\mathrm{A}+k_\mathrm{B})t$$

At equilibrium, the rate of reaction is zero, then $k_\mathrm{A}a_e=k_\mathrm{B}b_e=k_\mathrm{B}x_e$ where subscript $e$ means that it is the equilibrium value. Also, we know that $\displaystyle K_\mathrm{e}=\frac{k_\mathrm{A}}{k_\mathrm{B}}=\frac{x_e}{a_0-x_e}$, where $a_0-x_e$ is the amount of A remaining at equilibrium.

Substituting and rearranging gives the amount of $x$, which is B as a function of time:

$$x=x_e -x_ee^{-(k_\mathrm{A}+k_\mathrm{B})t}$$

Both this and the previous equation show that the equilibrium is approached with a rate constant that is the sum of the two rate constants.


How to identify ions by their quantum number? If the quantum numbers of an ion are given, how can we identify this ion by its quantum numbers?

For example :

  n = 3; $l$ = 2; $m_l$ = 0; m$_s$ = + ½

What I have so far: $n$ is 3 and $l$ is 2 which corresponds to $d$, so the last shell is $3d1$, therefore we have 21 $e^-$ and the answer will be $\ce{Sc}$ ion (from the periodic table) and the answer I got is $\ce{Mn^{2+}}$.   Is this the correct approach?

A pattern-based approach: Remember that the values for ${m_l}$ are centered about 0. This means for a d-orbital, where ${l=2}$, the values of ${m_l}$ = ${(-2, -1, 0, +1, +2)}$.

So, for ${l=2}$ and ${m_l=0}$ and ${m_s=+1/2}$, you should arrive at ${d^3}$. The orbitals each get an electron for all the positive spins first, then are filled completely with the negative spin ones.

That means you "visit" ${m_l}$ twice (in rising order: -2, -1, 0, +1, +2, repeat).

This means the first d-electron (↑ _ _ _ _) such as ${Sc}$ has the configuration ${m_l=-2, m_s=+1/2.}$

The 3rd electron (↑ ↑ ↑ _ _) would have ${m_l=0, m_s=+1/2}$.

A 5-electron configuration (↑ ↑ ↑ ↑ ↑) would have ${m_l=+2, m_s=+1/2}$.

Now on the right-hand side of the d-group, the 6th electron (↑↓ ↑ ↑ ↑ ↑) would have ${m_l=-2, m_s=-1/2}$, and then the magnetic quantum numbers rise to +2 again at ${d10}$.

Anyways, ${Mn^{+2}}$ is ${d^{3}}$ (5 to start, then subtract 2 because of its +2 charge) which matches the quantum numbers you specified.

It's not an 8 electron configuration because of the spin specified. That'd be (↑↓ ↑↓ ↑↓ ↑ ↑).


Melting ice off of trucks fast When a truck comes into the shop with extreme ice that is rock hard, is there a way to remove it fast? Any tricks of the trade?

Water has a high heat capacity, so I agree with JoeHobbit that a spray of hot water via hose would probably work best. A power washer that is capable of heating the water might be worth getting. A secondary advantage of doing this is that the water spray would remove accumulated road salt, which accelerates corrosion. Short of that, a space heater that blows hot air on the truck would be a less effective but also less messy solution. A moving stream of hot water or air should be far more effective than static immersion.

Small amounts of ice deposits can be removed with an aerosol deicer spray.

See, there is actually a lot of chemistry going on here.


What are the criteria for a molecule to be chiral? What are the criteria for a molecule to be chiral?
(We frequently get questions of this type, so this is an attempt to construct a suitable general answer for duping.)

Mirror images
The very first definition of a chiral molecule is one where it is not superimposable on its mirror image. Therefore, one of the most straightforward ways to determine chirality is to construct the molecule and its mirror image (perhaps via a model), and then to see whether they are superimposable.

Symmetry elements
A fully equivalent definition, rooted in molecular symmetry, is that a species is chiral if it does not have any improper axes of rotation.
An improper rotation is a rotation of the species (say 90° clockwise in the plane of the page), followed by a reflection perpendicular to the axis of rotation (in this case, the page). The improper rotation is generally denoted by $S_{n}$ where $n$ is the number of times you need to rotate to get back to the start, e.g., $n=4$ for a quarter rotation (90°).
So, molecules where you can rotate and reflect them to make them look like the original are not chiral. In other words, it's easier to say when a molecule is achiral. Let's look at some examples of improper rotation that make molecules achiral.
The simplest case to look at is $S_{1}$, but no one calls it that. This is a trivial rotation (a rotation that does nothing) followed by a reflection, or just a reflection, usually denoted by σ. Molecules with planes of symmetry are achiral. This molecule, (2R,4S)-2,4-dichloropentane, is achiral because there is a plane of symmetry that reflects the left side onto the right side.

Another example is $S_{2}$, rotation by 180° then reflection. Generally, we also don't call it that. This is the same as an inversion through the center, which is frequently denoted $i$. Molecules with centers of symmetry are achiral. The following molecule has a center of symmetry. You can also see that rotating it 180° and then reflecting through the page gives the same thing back. For clarity, the mirror image is shown on the right. You can rotate it to see that it's the same.

The above two are the most common symmetry elements that make molecules achiral. Occasionally, you'll run into a higher-order improper rotation axis, like $S_{4}$ in the following example. Rotation by 90° and then reflection through the page gives back the original molecule. The mirror image is shown on the right, and you can rotate it to see that it's the same, thus making this molecule achiral. There are more examples here.

So, looking for the above features will let you know if something is achiral.

Stereogenic elements
To figure out whether or not you even need to consider chirality, you will to find a stereogenic element. Keep in mind that having a stereogenic element doesn't mean that molecule is chiral, because it could have one of the symmetries discussed above and thus, be achiral.
Molecules with stereogenic elements should be considered for chirality, but are not guaranteed to be chiral. In other words, having a stereogenic element is necessary but not sufficient to make a compound chiral.
The main source of chirality that introductory students should worry about is the stereogenic center, usually a carbon with four different substituents:

More generally speaking, a stereogenic center such as this falls into the category of point chirality. Other examples of point chirality include inorganic complexes such as the octahedral $\ce{[Co(en)3]^3+}$ (link, also illustrated below), as well as substituted adamantanes.

Apart from point chirality, here are also two other ways to generate chirality, which are typically seen in more advanced courses.
The first is axial chirality (Wikipedia, IUPAC). This is a twist that can be described as either left-handed or right-handed. These include unsymmetrically substituted allenes, unsymmetrically substituted biphenyls, rings with exocyclic double bonds, and helicenes (image adapted from Wikimedia):

The second is planar chirality (Wikipedia, IUPAC). This is when a plane, which has high symmetry, has its planes of symmetry successively broken by substitution. An example is this borabenzene metal complex. A borabenzene has two planes of symmetry, but these can broken by putting a methyl group on the ring and then binding a metal to the face of the ring.


Preparation of phenylamine/aniline This is the procedure I carried out in class:

Reduction of nitrobenzene by using tin and $\ce{HCl}$ acid
After the completion of the reaction, $\ce{NaOH}$ added to neutralize excess acid and dissolve any $\ce{Sn(OH)2}$
Steam distillation is carried out to obtain suspension of phenylamine
Salting carried out where $\ce{NaCl}$ salt is added. Aqueous layer discarded
Solvent extraction: dry ether added and mixture shaken vigorously. Aqueous layer discarded
Distillation under reduced pressure carried out and solid phenylamine left in flask

My questions:

Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?
Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?

Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?

You could have performed any procedure to remove the solvent:

Simple distillation  - This would be slower than what you did and could overheat the undistilled aniline causing decomposition.
Vacuum distillation - This is what you did, perhaps even on a special apparatus called a rotary evaporator. This method is still a distillation, even if your product was not distilled. Something was distilled.
Evaporation on a hot plate (or steam bath)-

  Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?

Steam distillation is suitable because the aniline is already in an aqueous suspension. If do vacuum distillation, you would need to

Separate the aniline from the aqueous mixture (extract with ether)
Remove the ether (simple distillation or reduced pressure distillation)
Vacuum distil the aniline

If you look at what you did, it is the same steps in a different order:

Distil the aniline
Separate the aniline from the aqueous mixture (extract with ether)
Remove the ether (simple distillation or reduced pressure
distillation)

Either is probably suitable. In the end, the steam distillation is easier and less hazardous than the full-blown vacuum distillation you would need.


Equilibrium constant for the neutralization of weak acid by strong base Acetic acid has a $K_\mathrm{a}$ of $\pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $\ce{NaOH}$?

Given acetic acid

$$\ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} \qquad K_\mathrm{a} = \pu{1.8e-5}$$

$$\ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$

So, if we do $K_\mathrm{w} = K_\mathrm{a}K_\mathrm{b}$, then we get $K_\mathrm{b} = \pu{5.55e-10}$. How do I use this to find $K_\mathrm{eq}$?

I know how to find $K_\mathrm{eq}$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?

Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:

$$\ce{HOAc + OH- <=> OAc- + H2O}$$

$$K' = \frac{[\ce{OAc-}][\ce{H2O}]}{[\ce{HOAc}][\ce{OH-}]}$$

Since $[\ce{H2O}] = \text{const}$ (reaction medium), $K'[\ce{H2O}] = K = \text{const}$:

$$K = \frac{[\ce{OAc-}]}{[\ce{HOAc}][\ce{OH-}]}$$

By multiplying both numerator and denominator by $[\ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_\mathrm{a}$ and ionic product of water $K_\mathrm{w}$:

$$K = \frac{\color{red}{[\ce{OAc-}][\ce{H+}]}}{\color{red}{[\ce{HOAc}]}[\ce{OH-}][\ce{H+}]} = \frac{\color{red}{K_\mathrm{a}}}{K_\mathrm{w}}$$

For acetic acid:

$$K = \frac{\pu{1.8e-5}}{\pu{1e-14}} = \pu{1.8e9}$$


How does the solvent determine whether mono- or tribromination of phenol occurs? In phenol, the $-\ce{OH}$ group activates the benzene ring towards electrophilic substitution. But why is it that the electrophilic bromination of phenol gives trisubstitution (forming 2,4,6-tribromophenol) when aqueous bromine is used, but monosubstitution (either 2- or 4-bromophenol) when a $\ce{CCl4}$ solution of bromine is used? How does the solvent control the extent of bromination?

Related but does not explain completely: How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?

In aqueous medium, phenol is deprotonated to a certain extent, forming the phenoxide ion in which the ortho and para positions are even more activated than in phenol itself. Hence, trisubstitution occurs here.

$$\ce{PhOH + H2O <<=> PhO- + H3O+}$$

In an organic solvent, phenol is not deprotonated and monosubstitution tends to occur.

ron also pointed out in a comment that:

  The concentration of $\ce{Br^+}$ is higher in water than in $\ce{CCl4}$ making the reaction faster in water.
  The dielectric constant of water is much greater than that of $\ce{CCl4}$, so the former solvent will better stabilize the charged intermediate, again accelerating the reaction.


several coefficient differences in UV-vis Spectroscopy Does anyone know what is scattering coefficient, absorption coefficient and extinction coefficient, and how to separate them experimentally?

In most applications, these terms are used interchangeably.

Most absorbance spectrophotometers work by measuring the intensity of the light beam that is transmitted through the sample.  A strong light beam goes in, and a weaker one comes out the other side.  The transmission is the fraction of the light that made it through.

There are two main ways in which the light that goes in gets diminished in intensity before it goes out.  The first is absorption.  Molecules that absorb light of a certain frequency make photons of that frequency disappear.  The energy of those photons excite electrons into higher energy states.  These states may eventually decay, releasing heat or lower-energy photons, but not photons of the same energy that were absorbed.  The propensity of molecules to absorb can be called the absorption coefficient.

Scattering just deflects the incident light to a new angle.  Now, when it exits the sample, it won't come straight out on the side opposite where it went in, and it won't be detected by the detector.  Molecules can cause scattering if they are very large.  The "scattering coefficient" describes a molecule's propensity to scatter light of a certain frequency.

The "extinction coefficient" is -- in most applications -- just equal to the scattering coefficient plus the absorption coefficient, i.e., it's proportional to a molecule's ability to disrupt incident light in some way or other, regardless of the mechanism.

In the usual, transmission-based spectrophotometer, only extinction coefficients can truly be measured.  To resolve these measurements into scattering contributions and absorption contributions would require measuring the intensity of the scattered light, i.e. detectors would need to be present at multiple angles.


Could muscarine be an antidote to atropine poisoning? According to Wikipedia,

  "Muscarine poisoning is characterized by miosis, blurred vision, increased salivation, excessive sweating, lacrimation, bronchial secretions, bronchoconstriction, bradycardia, abdominal cramping, increased gastric acid secretion, diarrhea, and polyuria. If muscarine reaches the brain it can cause tremor, convulsions, and hypothermia. Cardiac ventricles contain muscarinic receptors that mediate a decrease in the force of contractions leading to a lower blood pressure. If muscarine is administered intravenously, muscarine can trigger acute circulatory failure with cardiac arrest... The specific antidote is atropine."

However, the wikipedia article on atropine also states,

  "In overdoses, atropine is poisonous."

Could muscarine be used to treat atropine poisonings, just like atropine is used to treat muscarine poisonings?

In short: no.

It is not that simple. According to Wikipedia, atropine works as a muscarinic antagonist, which blocks the action of acetylcholine at the muscarinic acetylcholine receptor. So, the mode of action as an antidote is to prevent muscarine from binding to the receptor, thus inhibiting its effects on the body.

The poisonous effects of atropine cannot be reversed by muscarine, though. That may be due to the fact, that it is able to displace muscarine and not the other way around.


Should chemical compounds and elements be capitalized? In the case of a name of a person, the first letter is written in capital letters.

Should the first letter of name of a chemical compound or element be written in capital letters?

The names of chemical compounds and elements should be capitalized if they appear at the beginning of a sentence or in a title - that is, they are treated just like any other common noun.
For example, a title:

Why I Don't Like Zinc

or a sentence:

Boron is my favorite element.

Within a sentence:

We used boron and zinc in the experiment.
Vinegar contains acetic acid.

The symbols for chemical elements are always capitalized, no matter what:

We combined $\ce{As}$ and $\ce{W}$ to make a new alloy.


Silver identification of the group I of cations I was wondering if this actually happens in a chemical reaction
$$\ce{AgCl_{(s)} + 2NH_4OH_{(aq)} \rightleftharpoons  Ag(NH_3)_{2(aq)}^+ + Cl^{-}_{(aq)} + 2H_2O_{(l)}}$$
My teacher wrote something like this but on Internet I saw that it yields to $\ce{Ag(NH_3)_2Cl}$ and water, so can someone explain it to me.

Both formulas for diamminesilver(I) chloride essentially mean the same. Writing it as

$$\ce{Ag(NH_3)_{2(aq)}^+ + Cl^{-}_{(aq)}}$$ emphasizes that the compound is dissolved in water as discrete ions (the cationic silver complex and chloride anions). One can also write $\ce{Ag(NH_3)_2Cl}$ instead because it is known that the compound exists in the aforementioned way in aqueous solution. In the latter case, it would still be more precise to use brackets to indicate the complex, i.e. $[\ce{Ag(NH3)2}]\ce{Cl}$, so that it becomes clear that the two $\ce{NH3}$ are the ligands of silver.


How many boiling points does a mixture of liquids have? Let's say I have a mixture of two miscible liquids. I want to get the liquids by fractional/pure distillation. My textbook says a mixture boils over a range of temperatures. On the contrary, another book I read says a mixture has only one boiling point and that the liquid with a lower boiling point adds more of itself to the mixture. I am wondering which of those is correct and will work with distillation.

I do use my comment above because the meaning of the books is, with little interpretation, clear and the answer is simple and don't require digging in theory of fractional distillation. So it would be pity to let the question unanswered.
A mixture as those discussed boils over a range of T in the sense that its boiling point BP depend on its composition.
The actual mixture, i. e. at a given molar ratio, has one well defined BP (at given P).
If you do not look at a given composition, but you actually conduct a distillation, as the boiling change composition the BP changes all along the process/column/plates, actually spanning a range of T.


Different techniques of distillations I know that fractional distillation is used to separate two liquids that are miscible in one another and have close boiling points.

However, I am really confused about vacuum and steam distillation - in which cases do we use these separation techniques? Can someone please give examples.

Vacuum distillation is employed when the boiling point of the substance is too high (400 Celsius for example). Under reduced pressure, the boiling point will be reduced to the range that is easier to handle.
Steam distillation take advantage of azeotropic effect, water steam is bubbled into your liquid (aromatic compound for example) and carry out the mixture of water vapor and your liquid vapor. This technique can also avoid handling the high boiling point distillation because the azeotropic boiling point is usually lower than pure liquid.


A question about ligand binding I have done part a and part b but I keep getting a negative concentration for part c.

For part a I got $\pu{2 \mu M}$
For part b I got $\pu{5.051 \mu M}$

For part c, I get  $2=3.2 + \text{[bound ligand]}$, but obviously that is wrong as you cannot have negative concentration for the bound ligand.

You have calculated correctly for parts $\bf{a}$ and $\bf{b}$. However, to answer part $\bf{c}$, you need to use the given plot, the interception and slope of which can be calculated as follows:

The slope is $\tan \alpha = - \frac{1}{K_d}$:
$$\tan \alpha = \frac{1-0.2}{0.1-0.8} = -1.14 \ \Rightarrow \ \therefore \ \frac{1}{K_d} = 1.14$$
According to the given Scatchard equation (where  each receptor has a single ligand binding site), interception of the plot is also equal to $\frac{1}{K_d} = 1.14$. Therefore, the Scatchard equation can be rewritten as:
$$\frac{r}{[A]} = 1.14(1-r) \tag1$$
where $[A]$ is actually free ligand concentration (Wikipedia). Therefore, when $[A]= \pu{3.2 \mu M}$, find the relevant $r$ using the equation $(1)$, which is the average amount of $A$ per $\pu{mol}$ of $P$ $(r = \pu{0.785 \mu M})$.
This means: Total number of binding site per receptor is $1$ and only $0.789$ is bound by $A$. Protein concentration in the dialysis tube $= \pu{2.0 \mu M}$
Therefore, $A$ bound concentration of $P$ $ = \pu{2.0 \mu M} \times 0.785 = \pu{1.57 \mu M}$


Why the radius of noble gases is more than the halogens or the previous groups? Noble gases have a larger radius than the halogens.

Sometimes it is greater than the radius of group I elements.

Why it is like that? When we talk about radius of noble gases what type of radius is referred? Is it Van der Walls or atomic?

Unlike a ball, an atom doesn't have a fixed radius. The radius of atom
  can only be found by measuring the distance between the nuclei of two
  touching atoms and then halving that distance.

According to my findings, I think the reasoning lies in the way in which the measurements are done:

i. Metallic/Covalent radius (left)

ii. Van der Waal radius (right)

It is believed that Noble gases don't form (or rather don't easily form bonds). This way only Van der Waals radius is measured.

The rationale here is based on the fact that noble gases have very low chemical reactivity and their atomic radii are non bonded.
Another explanation lies on electronic configuration of noble gases. They have completely filled outer orbitals with increased repulsion.

In addition, following Hund's rule, in a set of degenerate orbitals, spin-paired electrons exhibit more repulsion compared to singly occupying orbitals which have parallel spins (and consenquently less repulsion).

This also implies that increased electron repulsion between the completely-filled orbitals (as seen in noble gases which all have spin paired electronic configuration) has an effect of increasing "Van der Waals radius" as electrons tend to move as further away as possible to minimise repulsion.

However there is some disparity when comparing Noble gases radii with Group 1 elements (metallic radii) or Halogens (covalent radii) because you are comparing two different unrelated properties,
nevertheless it makes sense to compare radii between different noble gases.

References

Atomic and Ionic radius
Conceptual Chemistry


Why does lactose-free milk last so much longer than regular milk? I buy lactose-free milk. The sell-by date is usually as much as eight weeks from the date of purchase, which is in the neighborhood of twice that of regular milk. This seems puzzling; lactose-free milk, which still has the same mass of sugar, has twice as many sugar molecules as regular milk, since all the lactose molecules have been split in to galactose and glucose. With twice as many sugar molecules per unit volume, one would think lactose-free milk would last half as long as regular milk, not twice as long.

Apparently the process used by manufacturers (ultra-pasteurization) to produce lactose-free milk is responsible for the longer shelf life.

One source reports:

  Lactose-free milk is pasteurized at a higher temperature than regular milk. The process, known as ultra-pasteurization, is designed to remove the bacteria content entirely, giving lactose-free milk a refrigerated shelf-life of 60-90 days, compared with regular pasteurized milk, which retains some bacteria. It has a shelf life of 1-3 weeks.

A producer (in the United States) of organic dairy products (Organic Valley) describes ultra-pasteurization and the extended shelf-life here as:

  Ultra Pasteurization (UP), or Ultra High Temperature (UHT) pasteurization, is the process of heating milk to approximately  280 °F for just 2 seconds and then chilling it back down rapidly.  The result is milk that's 99.9% free from bacteria.  This pasteurization process creates an extended shelf-life for milk products of up to three times the length as HTST pasteurization; while providing the same wholesome, quality dairy product.  This allows us to distribute UP milk products regionally as well as to other areas of the country that might not have access to our dairy products. This is referred to as 'ultra pasteurized' on the milk package.  Organic Valley offers its quarts and some of its half gallons of milk with Ultra Pasteurization (UP).


What do typical results from inductively coupled plasma mass spectrometry (ICP-MS) look like, and how do I interpret them? I have a degree in chemical engineering and remember having seen a inductively coupled plasma mass spectrometer (ICP-MS) during the laboratory part of a chemistry course I took while at university, but it is many years ago. Now I would like to know what a typical output from a ICP-MS looks like and how I interpret the results.

There is no picture at the wikipedia page but some frequency domain like plots can be found with a Google search and I understand the Fourier transform and how it is used in number theory.

Are the ICP-MS results anything like the spectrum from the Fourier transform of some signal?

I am not an expert in ICP-MS, but an ICP-MS should be considered an ICP ion source coupled with a mass spectrometer. There are some key issues raised by the coupling between both instruments due to the high temperature of the ICP source as well as due to the fact that one wants to analyse atomic ions and remove any interfering molecular ions (mostly oxides, but other molecules and clusters are possible).

Therefore, to attempt an answer to your question, the standard output of an ICP-MS will be presented as a mass spectrum (intensity versus m/z or intensity versus amu). There are a couple of sample spectra in the brochures published by instrument makers, see for instance the Bruker brochure or a Thermo application note. How this mass spectrum is obtained from the detector will lead to the type of data that is acquired:

For a quadrupole mass filter, as well as for magnetic sector instruments, two modes of operation are possible. Either a single mass is selected, and therefore ont will only get a trace of the abundance of a given mass as a function of time (chromatography or on-line sample analysis). Otherwise the mass filter can be scanned, leading to a mass spectrum.
For time of flight detectors, the mass spectrum is built by the measurement of a flight time for an ion to reach the detector. Therefore the mass spectrum is converted from a time dependent signal to a spectrum.
Finally there have been some attempts to couple an ICP ion source with a Fourier transform mass spectrometer. As far as I know, these are not commercial instruments. In such instruments, the ions oscillate at m/z dependent frequencies, and the recorded image current is converted to a spectrum through the use of Fourier transform.


Difference between reversible and irreversible thermodynamic process I have been reading up (on this website) about the difference between reversible and irreversible thermodynamic process and it defines them saying that reversible processes are ideal processes while irreversible processes are the natural processes. Are these definitions correct?

If not how are they defined?

The simplest answer in thermodynamic terms is "a reversible process is one in which the entropy generation term is zero."

  Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible.

For any process, all of these would be the same whether it was reversible or not. The only one that would differ would be change in entropy.


How do we know a reaction is in equilibrium? In teaching secondary science equilibrium, how can we actually show there is a backwards reaction occurring during equilibrium rather than no reaction. A lot of equilibrium experiments involve external changes such as temperature or change in pressure. This to me shows that a reaction can occur backwards, i.e reversible but it doesn't show that there is a forward and backwards reaction occurring at the same time at equilibrium.  What's a simple science experiment or everyday life experience that equilibrium can explain where the idea that an reaction just stops taking place doesn't?

Every video just claims that there is a forward AND backwards reaction but doesn't really explain why and is also something I've taken for granted

Thanks

By definition, the macroscopic observables do not change once equilibrium is reached, and experiments that probe the microscopic domain are usually expensive. I use an analogy to show the microscopic changes:

The model

Take two beakers A and B, and fill one or both with some water (and add some food color for better visibility), and use two straws F and R to continuously transfer from beaker A to beaker B (with straw F) and from beaker B to beaker A (with straw R). The transfer is by having the straw touch the bottom, and then covering the top with your index finger or thumb.

At that point I ask students what they expect will happen if I continue for the rest of the class period. At one point, the level of liquid does not change anymore even though I am still transferring liquid. This is the state of dynamic equilibrium.

To make it more interesting, take straws of unequal diameter (or two vs one straw, like in the looped video).

The analogy explained

The liquid level in the two beakers represent the concentration of reactant A and product B. The diameter of the straws represent the rate constant of the forward reaction (straw F) and the reverse reaction (straw R).

$$\ce{A <=>[k_f][k_r] B}$$

Limitations

There is only one reactant and one product. The reaction will always be first order.

Avoiding misconceptions

If you take a single straw to transfer back and forth, students might think that equilibrium is like a pendulum. If you take straws of equal diameter, they might think equilibrium means that concentrations are equal (rather than correctly, rates are equal).

  Every video just claims that there is a forward AND backwards reaction but doesn't really explain why and is also something I've taken for granted

Starting with just A, you learn that the A molecules have the capability of turning into B. Starting with just B, you learn that the B molecules have the capability of turning into A. If equilibrium were static, someone would "have to tell" the molecules to stop reacting. With the dynamic equilibrium model (which has support from expensive molecular-level experiments), molecules keep on doing what they always did - react every now and then. There is no requirement for some kind of switch in behavior, it just is an emergent property of having both forward and reverse reactions going on and influencing concentrations of species.


Radial pair distribution function (VMD). How to define it for a water NaCl system with multiple Na and Cl? I'm struggling to understand a paper in which someone has presented radial distribution functions (RDFs) for a system that contains multiple Na, Cl, and water atoms/molecules.  I don't know exactly how the RDFs have been calculated--there must be some standard method.

I'm a newbie to molecular dynamics and have run my own simulations.  I would like to compare my RDFs to the ones in the paper.  However, I am not sure that I understand RDFs.  One reason I'm worried is that I keep receiving an error when I try to plot my own RDFs using the software, VMD.  The error is something like "cannot measure RDF."  It is either a bug in VMD or a conceptual misunderstanding on my part.  I don't know exactly how VMD calculates an RDF.

Let's say we have an RDF from Na to Cl in my system that contains multiple Na and Cl. Is it correct that an RDF is based on an average of what ALL Cl atoms are doing in relation to ALL Na atoms?  Or do I have to somehow define pairs of Na and Cl (I don't have these--they're all far apart in my system) and create RDFs for each pair?

I'm not sure whether "Chemistry" is the right StackExchange section for this question.  If it isn't and anyone has any advice on a more appropriate section, I'd be very grateful.  Thank you.

I haven't done it in vmd, but a rdf is really kind of like a histogram. For a particle of interest, you count how many particles are between r and $\Delta r$ from it and record that number in a bin, then move outwards, recording numbers in each shell into a bin for that shell. That tells you the rdf of that particle interacting with everything else. If there are three other types of particles in the simulation, you will get 3 rdfs from that one rdf calculation centered at the particle of interest.Repeat for other particles once you are done so you can get a nice average.

So all you need to do is count particles in $\Delta r$'s (and from this you get the volume too!, that is trivial though. I would do this for all Na's and Cl's and then average each of their bins to get one rdf for Cl-Cl, one for Cl-Na, one for Cl-O, one for Cl-H.

So that is how it should work, but I haven't used vmd to do it myself :S

RDF's don't have to be atom-atom, they could be molecule-molecule, or molecule-atom... anything really.


Why does a glass tube show white bands after being heated with a torch? When I use a torch to split a glass tube, there's always a hazy white band which appears. Does anyone know what the cause of this is?

It was a hydrogen-oxygen torch. The glass is fused quartz glass, which should be pure silica.

It is sublimed silica according to the following reference at page 8.

http://www.public.asu.edu/~aomdw/GLASS/book.pdf


Effect of dilution on pH for HCl Is it necessary to consider common ion effect when suppose a strong acid like HCl is added to water?

For example in this question: 10 mL of $10^{-6 } \ce{mol/L \ HCl}$ solution is mixed with 90 mL of $\ce{H2O}$. pH will change approximately?
Normally by using $SV=constant$ I get pH will change by 1 unit (without taking common ion effect).

How will common ion effect affect the answer? Help.

Yes, you will definitely need to account for common ion effect.

However, use Kw to compute the pH. Assume that there is a complete dissociation by $\ce{HCl}$. However, you would need to solve the according quadratic for Kw.

as such, $\ce{H+= concentration~~ H+ of~~ HCl + x ~~and ~~OH- = x}$.

You should get a result around 6.79=pH.


Why must the movement of electrons through the wire be balanced by a movement of ions in the solution? I have read through Barron's SAT chemistry prep book, in the topic of electrochemistry, the book says

  The movement of electrons through the wire must be balanced by a movement of ions in the solution.

It explains that

  Anions move toward the anode to replace the negatively charged electrons that are moving away.

Can someone help explain these two statements? I don't quite get this concept.

Karl Ratzch is basically correct, but I will try to provide a little more detail.

Electric charges, when separated, exhibit a very large tendency to try to get back together again.  This is represented in basic Physics by Coulombs Law and the large value of the constant in that relationship.

I presume we are talking about a battery here. If the flow of electrons out of the wire is different from the rate of ions arriving at the plate, the net charge on the plate will change.  Depending on which flow rate is greater, the voltage of the plate will instantly go up or down.  This will then change the rate of both charge flows such that the two almost instantly get back to the same value.

More detail: Assume that the flow of electrons out of the plate through the wire is larger than the flow of anions into the plate from solution.  This reduces the net negative charge on the plate.  This reduces the "force" that is pushing electrons down the wire.  The actual name of that "force" is voltage.  The voltage will change such that it doesn't push electrons out the wire as fast and will repel anions in solution less aggressively allowing more of them to reach the plate (per second).  The opposite is also true.  As a result, the flow in and out of the plate will always remain almost exactly the same, with small differences existing for very brief periods of time when the voltage changes.


What is an intuitive explanation for Electron Configurations I'm sort of self studying Chemistry. I'm now getting stuck wrapping my head around a concept (and I realize it was the same concept I had difficulty grasping years ago) and that is Electron Configurations.

I know that perhaps due to the nature of chemistry it is non-intuitive. However, I want an explanation that is more than "This is a bunch of numbers and letters that follow a pattern. Live with it" - which seems to just be the explanations I'm getting when I went to school.

So, my question is, what are Electron Configurations, and how would you explain them intuitively?

Electron configurations are a bunch of numbers and letters, and they also follow a pattern. But there is also a logic behind the whole thing.

Basically, chemists use electron configurations to see in which atomic orbitals the electrons are found. From this several things can be derived, most of all the reactivity of said electrons (which is where the chemistry happens).

The intuitive answer is this: Electron configurations are a way of telling in which atomic orbitals the electrons are put in, according to the Aufbau principle and the Pauli exclusion principle. Decoding it is pretty easy, as it always follows this pattern:
$n$$(s|p|d|f)^x$ where $n$ is the main quantum number (or shell number), $(s|p|d|f)$ denotes the atomic orbitals with the quantum number and $x$ is the number of electrons found in the orbitals.

The orbital "groups" ($p$, $d$, $f$) contain several "orbitals" (side note: an orbital is strictly seen a one-electron function) where two electrons with opposite spin fit in. $p$ has 3 ($p_x,\; p_y,\; p_z$), $d$ has 5 ($d_{xy},\; d_{xz},\; d_{yz},\; d_{x^2-y^2},\; d_{z^2}$) and $f$ even more. This means that the maximal number of electrons in the $p$ group is $6$, and in the $d$ group $10$. So when you fill in the electrons according to the aufbau principle, you have to go to the next orbital group when you reach this number.

Bottom line: Every electron in an atom is defined by the four quantum numbers $n$ (main or principal quantum number), $l$ (azimuthal), $m_l$ (magnetic) and $m_s$ (spin). Using the electron configuration we can derive the first three of these four numbers.