How to light a candle with small amounts of safe to use chemicals? Already saw some youtube video where they lit a camp fire with glycerol and potassium permanganate and tried those but the amount of the stuff you need to create enough heat is just too much.

So is there a way to light a candle with small amount of something and something, like having the candle wick dipped/powdered with something and then just letting it get into contact with something else?

Safety and ability to light a fire, as far as I understand, directly contradict each other. What starts a fire on your wish, would occasionally start a fire against your wish. The glycerol/sulfuric acid/permanganate recipe, like it or not, is probably the safest of all.

Others include:

White phosphorus, not only flammable but also severely toxic by itself.
Butyllithium and many other organometallic compounds that would ignite instantly on air; that's not quite what you want, anyway.
Hydrogen against platinum black (danger of explosion).

I definitely would not try any of these at home.


Determining solvent and solute On which conditions does determining solvent depend on?
1. the higher boiling indicates the solvent

the material with the higher concentration is the solvent
the solvent is kept first in a pot. Then the solute is added.

Here, pls identify the solvent and solute. I hope that it will clear all my doubts about them.

Chemists need terminologies. NO terminology is perfect and you will always find exceptions. In general, in an n-component mixture, the component which has the highest concentration is the solvent.*

Therefore your point no. 3 is invalid because pouring does not determine the solute-solvent label. Only point 2 is relevant. Point 1 is irrelevant.

*Corollaries & Exceptions:

If we mix 50 mL of hexane and 50 mL of heptane, which one is the solvent and which one is your solute? It is up to you to label one as solvent and solute. Both components were liquids.

If I take 100 gram of water, I can dissolve about 140 grams of potassium iodide! The mixture remains a liquid. Now which one is the solute and which one is the solvent? Traditionally, you may call the liquid component as the solvent and the solid phase as the solute, because here water molecules are helping to solvate potassium and iodide ions.

The same goes with sugar in water.

So how to overcome these paradoxical labeling issues?

Physical chemists and analytical chemists prefer to use the term mole fraction x when they want to extra precise to what they are saying about mixtures. It is defined as "Amount of a constituent divided by the total amount of all constituents in the mixture. It is also called mole fraction. Amount fraction is equal to the number fraction: the number of entities of one constituent divided by the total number of entities in the mixture."

Just as nobody cares about "physical" vs. "chemical" changes taught in school, no one will chastise you for identifying one component as solvent and the other component as solute in complicated cases.


Is Sweetwater, Missouri, named due to water contamination with lead compounds? The town of Sweetwater, Missouri is known for its mineral resources, particularly rich lead sulfide (galena) deposits. I am curious to find the origin of the name "Sweetwater" and if it has any connection with lead compounds that at one time or another leached into the local water resources (I believe lead acetate tastes sweet). In any event, I can't seem to find any history or such a connection through internet research.

According to The State Historical Society of Missouri:

  Sweetwater is the name of a little trading-point established on Sweetwater Branch just after the Civil War. It was named for the branch.

and

  Sweetwater Branch was named by the pioneers for the unusually good water in the stream.

The place (and branch) are both in Newton County, Missouri. The same reference linked above indicates that what was once called Sweetwater is now referred to as Boulder City (because of the boulders nearby).


How does overvoltage affect the products of electrolysis? When you electrolyse an aqueous solution of $\ce{NaCl},$ the product formed at cathode is $\ce{H2}$ gas (preferential discharging), but the product formed at anode is $\ce{Cl2}$ gas. According to standard electrode potential, reduction of $\ce{Cl2}$ is 1.36 V and $\ce{O2}$ is 1.23 V.

Keeping the values in mind, $\ce{O2}$ gas should be formed at anode because it has a smaller electrode potential. But it is given that chloride ions is preferred by anode because of a phenomena called overvoltage.

I don't understand how overvoltage can effect the products of electrolysis.

When you electrolyse an aqueous solution of NaCl, the product formed at cathode is H2 gas (preferential discharging), but the product formed at anode is Cl2 gas.

There is another complication. The book is missing the fact whether chlorine or oxygen is evolved depends on the concentration of the salt. At low salt concentrations, oxygen will be formed and at high NaCl concentrations $\ce{Cl2}$ will be produced at the anode.

Keep in mind the thermodynamic "competition" is between the oxidation of water and oxidation of chloride ions. Electrochemists have spent their lives and careers in understanding the oxidation reduction of water for hundreds of years. So if you are slightly confused and you have a lot of questions that is a sign of a healthy scientific mind. I am glad that you did not digest what the textbook was saying right away.
Wikipedia hints as how complex is the process of hydrogen and oxygen production at a electrode.

The numbers which you are quoting are like comparing apples and oranges.
This half cell
$$\ce{2H2O(l)} → \ce{O2(g) + 4 H+(aq) + 4e−}$$ $E^o$= +1.23 V is valid for pH of 0, i.e., 1 M H$^+$
ion. Your aqueous NaCl solution is not present in 1 M H$^+$.

You might need to use the Nernst equation to see what is the true potential at a neutral pH.


Why should fire extinguishers containing carbon dioxide not be used on magnesium fires? As the title states, why should $\ce{CO2}$ fire extinguishers containing carbon dioxide not be used on fires with magnesium metal?

I understand that $\ce{CO2}$ extinguishers work by displacing oxygen, a crucial component of a fire. Does this somehow not work when magnesium is involved, or is there another reason?

The reaction of magnesium metal (Mg) with carbon dioxide is extremely exothermic (heat evolving). In other words, carbon dioxide makes magnesium metal burn.

$$\ce{2Mg(s) + CO2(g) -> 2MgO(s) + C(s)}$$


Dimerisation of HOOC-COOH in non polar solvents In a textbook I read, the following question is my doubt:

Why is the answer B)? Why do the other molecules not dimerise? How does it work?

(P.S- Why dosen't the B) option only form hydrogen bonds with the same type B)
?)

Edit: Turns out I was looking at the wrong answer key and the answer is actually a) :p

I believe your answer key is wrong on this one. Carboxylic acids like acetic acid (a) are well known to form doubly hydrogen bonded dimers in which each the carbonyl group on one acid is hydrogen bonded to the OH of the other.

For more explanation and structures, see https://www.chemguide.co.uk/organicprops/acids/background.html.

Oxalic acid, being a carboxylic acid, can participate in this kind of intermolecular association, but the result would be a polymeric structure, rather than a simple dimer.

The alcohols can hydrogen bond to each other only singly - too weakly to form stable dimers.

Kudos to you for having the right instincts to challenge your answer key an probe more deeply!


How could I cause metal to rust rapidly? I'm writing a novel set in the mid-nineteenth century. One of my characters needs to cause an iron padlock to rust rapidly, ideally within a matter of hours.

The lock doesn't need to rust away completely - the goal is just to make it noisey and annoying so that it can't be opened in silence.

Is this a realistic goal? If so, what methods could someone use to cause rapid rusting? Bonus points for suggesting period-appropriate materials.

Update: Based on the answer and comments below, it sounds a period appropriate approach would be to mix fine copper filings with saltwater and apply it to the lock.

I've read that bleach is an oxidizing agent which was available in the 19th century. Would adding bleach to this mixture increase the rate of rusting without creating an overly dangerous mixture?

Update: I conducted an experiment in my garage. Iodine worked pretty well.
Experiment Video

A few months later the iodine had eaten through the lid of the jar I tested it in:

Here are factors that would speed up rusting:

Presence of water: make sure that the iron is wet.
Presence of oxygen: make sure the iron have access to air (the dissolved oxygen in water also works).
Presence of metal below iron in the metal reactivity series (see the picture below): usually copper is used. Tie copper to iron. Make sure that they are in contact. The larger the surface area of contact the better.
Temperature: make the iron as warm as possible.
Presence of salt: explained here. In short, salt increases the conductivity of water which makes the iron rusts faster, since rusting involves electron transfer.
Acidic environment: since rusting involves hydrogen ions ($\ce{H+}$), an acidic environment would increase the concentration of hydrogen ions, making the iron rust faster. However, this would dissolve the rust.
Presence of hydrogen peroxide: hydrogen peroxide ($\ce{H2O2}$) is an oxidizing agent which would not dissolve the rust:
$$\ce{3H2O2 + 2Fe -> 3H2O + Fe2O3}$$

Metal reactivity series:

(source: bbc.co.uk)
Equation of rusting:

$\ce{Fe -> Fe^3+ + 3e-}$
$\ce{O2 + 4H+ + 4e- -> 2H2O}$
$\ce{2Fe^3+ + 3H2O -> Fe2O3 + 6H+}$

More equations can be found here.
Overall equation:
$$\ce{4Fe + 3O2 ->[H+,H2O] 2Fe2O3}$$


Can phenol be detected with a ceric ammonium nitrate test? Today, I came across a question which asked name three  reagents can be used for identification of phenol. The answer given was:

  Neutral $\ce{FeCl3}$
  $\ce{ NaNO2 +HCl}$
  $\ce {(NH4)2[Ce(NO3)6]}$.

I understand (1) and (2), but I am doubtful about CAN test (3). As far as I have studied in various books, it is used for testing presence of alcohol, and phenol is not alcohol.

Can phenol be detected with a ceric ammonium nitrate test?

With unsubstituted phenol ($\ce{PhOH}$) dark-brown precipitation is formed:

$$\ce{(NH4)2[Ce(NO3)6] + 2 PhOH -> [Ce(NO3)4(PhOH)2] + 2 NH4NO3}$$

Other phenols should produce wine-red solution as expected [1, p. 478]. Also, this and other phenol tests are discussed in [2].

Pavia, D. L. A small-scale approach to organic laboratory techniques, 3rd Ed.; Brooks/Cole, Cengage Learning: Belmont, CA, 2011. ISBN 978-1-4390-4932-7.
Atul, S. The Pearson Guide to Objective Chemistry for the AIEEE; Pearson Education India, 2010. ISBN 978-81-317-3381-3.


Why can't Schiff's reagent distinguish between glucose and fructose? I have heard that Schiff's base only gives a positive test for aldehydes. Glucose is an aldehyde. However, I was told that glucose doesn't give a positive Schiff's base test. Is this correct? If so, how?

Schiff test is a colorimetric method for determination of aldehyde groups. Both glucose and fructose are monosaccharides and exhibit oxo-cyclo tautomerism. Glucose is an aldose and has aldehyde group in linear form. Whereas fructose is a ketose and doesn't give a positive Schiff test.


Is the 'n' in Delta G = -nFE, ever negative? Is it possible for $n$ in $\Delta G=-nFE$ to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.

No. This result occurs from the Nernst equation.

The "n" is the number of electrons transferred. If no electrochemical reaction occurred, then n = 0.

Otherwise "n" is positive.

Having a negative number of electrons transferred would be impossible.


Relationship between microplastics and wildfire smoke My understanding is that:

microplastics are short fragments of previously long polymer chains;
wildfire smoke largely consists of incompletely burned organic materials such as wood. Wood burns by heat first breaking long polymer chains into small fragments, which then ignite (or not).

So ... it sounds like there is quite some similarity between (some parts of) wildfire smoke and microplastics.
What are the key differences? Does it boil down to nature's ability to recycle one but not the other because their composition is different because they came from different types of polymers?
This is a noob question. But I'm still curious :-)

The term «polymer» only refers to using small molecules (called «monomer», needn't be of one type) multiple times to build a larger molecule.  E.g., if you would use the small molecule thrice to get a larger one, it were a trimer, if four times, a tetramer to be more specific; but these were all examples of «polymers».  The term does not tell if the material were of industrial / artificial (like for a PET bottle), or biological origin (e.g., lignin, chitin, cotton, or kreatin).
Given the small size of molecules, even tiny particles known as microplastic consist of many polymer molecules each.  Like if the band of a tagliatelle would just be cut into smaller and smaller pieces and still count as a noodle.  The fine mechanical division  / chopping into smaller and smaller fragments only increases the accessible area per unit of weight.  Similar to a combustion engine, this material then spread out may yield a combustion more efficient with less smoke and soot than, e.g., a torch, or a candle. Not all polymers burn equally well, even if finely divided.
The «problem» with wood, compared to natural gas - chemically speaking - is that it consists of many different compounds.  Beside the lignin, there are a number of minerals, sugars, fluids, resins, etc. and water present in a tree; each with its own characteristics.  Some will catch fire quickly no matter what, like human hair.  Some need to be heated first so that fire nibbles them away.  Meanwhile others may be heated and then burn as gases, rather than as solid or liquid, sometimes liberated like in an explosion.  Some materials require much air (wind) per unit of mass to burn, or otherwise would just smoulder; so now equally soil, weather and topography of the terrain enter the the set of parameters yielding a typical wild fire with incomplete combustion and smoke.


Is electron density a good measure of similarity between molecules? I am looking for a good similarity measure between molecules in order to use it with machine learning algorithms. I found a paper, Carbó, R., Leyda, L. and Arnau, M. (1980), How similar is a molecule to another? An electron density measure of similarity between two molecular structures. Int. J. Quantum Chem., 17: 1185-1189. doi:10.1002/qua.560170612 where they propose a similarity measure based on the electron density $\rho$:
$$\varepsilon_{AB} = \int_V|\rho_A-\rho_B|^2\,dV$$
This can be expressed as the sum
$$\varepsilon_{AB} = \int_V\rho^2_A\,dV+\int_V\rho^2_B\,dV-2\int_V\rho_A\rho_B\,dV$$
The authors then retain only the third term of this sum because only this term refers to the difference between the molecules A and B, and then they normalize it arriving to their final similarity measure
$$r_{AB} = \frac{\int_V\rho_A\rho_B\,dV}{\left(\int_V\rho^2_A\,dV \right)^{1/2} \left( \int_V\rho^2_B\,dV \right)^{1/2}}$$
which is zero for completely different molecules, and one if A = B.

My question is: why the electron density $\rho$ is appropriate here? A large portion of the electron density is due to the core electrons that don't participate in chemical bonding. They spend most of the time near the nuclei, and don't influence the rest of the molecule very much.

Instead of using the total electron density $\rho$, wouldn't it be better to use the density of only valence electrons: $\rho_{val} = \rho-\rho_{core}$?

Elements along the same group of the periodic table have similar chemical properties but have very different $\rho$ because they have different number of core electrons. However, their $\rho_{val}$ will be similar.

Is my thinking valid or I am missing something? What is the influence of the core electrons besides screening a part of the nuclear charge and preventing the valence electrons to come closer to the nuclei? Are there any papers where the density of the valence electrons is used as a descriptor of the molecular structure?

TLDR: Defining the similarity between two molecules is an open field of research, but is fundamentally a subjective question which depends on your context. If you are interested in the similarity of molecular structures, then a graph-based measure might be a logical choice. If you're interested in similarities of reactivity, then a measure incorporating the electron density or valence electron density might make sense. Ultimately, only you can know which measure makes the most sense for the problem you are trying to solve.

This is a good question and one which does not (and likely never will) have a definitive answer. That is, there will never be a physically correct way to answer how similar two different molecules should be, or even what type of mathematical object that answer should be (i.e. a number, vector, matrix, etc.). Nonetheless, some measure of the similarity between systems is clearly a useful thing to know about, and it is becoming more practical as machine learning becomes more common in chemistry.

From what I have seen recently, many of the most commonly used similarity measures are derived from a representation of a molecule as a graph. I think there are two main reasons for this. First, most people who are interested in molecular similarity are also interested in machine learning. Hence, they are working with a relatively large dataset. Computing a graph representation of a molecule and performing operations on this graph is almost always significantly cheaper than computing some kind of quantum mechanical property for every molecule in your dataset. Second, machine learning projects in chemistry tend to be very collaborative and there many applied mathematicians and computer scientists that know a lot about graph theory and are hence eager to apply this knowledge in a new context.

I have found a review from 2003 which discusses some more chemically motivated methods of defining molecular similarity and some methods based on a graph representation of the molecule[1]. They also discuss a similarity measure based on the overlap a field. The most natural choice for the field being the electron density, though one could choose something like an electrostatic potential or even some kind of magnetic vector potential. What you choose would just depend on what type of similarity you're thinking of.

As an example, you propose using the electron density due to valence electrons. I would be surprised if this hasn't been done before, but I couldn't find any references where this is done. This would be very natural if one is interested in similarities with regards to reactivity. On the other hand, if you were interested in predicting how similar the x-ray absorption spectra of various compounds would be, then it would be stupid to leave out the core electrons from the density as those are the electrons which are most relevant. This further illustrates that the similarity between two molecules is unanswerable because it is not a well-defined question. That is, this similarity will inevitably be connected to some way in which two molecules produce a similar outcome in some physical situation. If you're interested in predicting a molecule which will behave similarly to some other molecule in a reaction, then valence electron density could be a great descriptor, but this descriptor might be entirely useless if you're actually interested in finding molecules with similar x-ray absorption spectra.

Now, chemists have a serious habit of trying to make a connection between everything a molecule does and the structure of that molecule. Very often, this works out extraordinarily well, so chemists keep doing it. I think this is another reason why graph based measures of similarity have become so popular. Virtually all methods of determining molecular similarity based on a graph begin by computing the adjacency matrix of the molecule based on connectivity along covalent bonds (or maybe hydrogen bonds). Then, some operation is performed on a pair of graphs which results in a number or vector or somethings else which is interpreted as a measure of similarity. Or, an operation is performed on each graph individually, and then some measure of distance is used to define the similarity of these graphs. These operations are usually called kernels. A review of some of the kernels which have been used to define graph similarity can be found here[2].

There are all sorts of tricks which can be played with the adjacency matrix to give some more physical meaning to the measure of similarity. For instance, you might weight each of the edges by the atom-atom distance defining that connection. You might also weight each of the nodes with an atomic charge or polarizability and incorporate this into the adjacency matrix. You might also choose to do this and then work with the laplacian of the graph rather than the adjacency matrix. Anything at all can be done really. You just have to choose what makes sense for your context.

More recently, people have come up with measures that also incorporate similarities of the subgraphs of two graphs[3]. This seems like it has some potential advantages if functional groups, etc. are important to you.

References:

[1]: Nikolova, N., & Jaworska, J. (2003). Approaches to measure chemical similarity–a review. QSAR & Combinatorial Science, 22(9‐10), 1006-1026.

[2]: Rupp, M., & Schneider, G. (2010). Graph kernels for molecular similarity. Molecular Informatics, 29(4), 266-273.

[3]: Kondor, R., & Pan, H. (2016). The multiscale laplacian graph kernel. In Advances in Neural Information Processing Systems (pp. 2990-2998).


Explanation of Lemon Juice-Invisible Ink A cute homegrown spy-trick we all know: lemon juice invisible ink. But there is no scholary article available in the internet about this phenomenon.

A rough explanation I found in Scientific American:

  Lemon juice—and the juice of most fruits, for that matter—contains carbon compounds. These compounds are pretty much colorless at room temperature. But heat can break down these compounds, releasing the carbon. If the carbon comes in contact with the air, a process called oxidation occurs, and the substance turns light or dark brown.

But, this is not the type of explanation to be content with.

Question: What is the mechanism going on here? Please, support your answer with sufficient information and references, don't give a rough explanation.

Edit: I don't think my question is duplicate as I am asking for a well-referenced and informative answer which the other question doesn't have.

Lemon juice works as a sympathetic ink partially due to its caramelization, as @Mithoron profoundly noticed -- especially upon extreme heating.
Recent review of the degradation process of lemon juice [1] suggests that browning of lemon juice is associated with 3 processes:

ascorbic acid degradation;
Maillard reaction;
caramelization.

One of the first in-depth investigations revealed the major role of ascorbic acid degradation referring to the browning phenomena [2]:

Compared with other fruit products, lemon juice contains few natural pigments to mask.
...
Because of the high acidity (pH 2.5) of this product, it was unlikely that browning was due to sugar-amine condensation and the results showed that ascorbic acid was the main precursor. Browning of lemon juice and model systems was proportional to the level of ascorbic acid; the presence of amino-acids in model systems increased the intensity of browning.
The principal role of amino-acids in the non-enzymic browning of fruit products appears not to be in the initial reaction leading to the formation of reactive compounds, but to increase the browning potential after the oxidation of the ascorbic acid to reactive carbonyl compounds. Thus, the pathway for the build-up of carbonyls depends on the pH and nature of the product, but in all cases the formation of melanoidin complexes originates from the polymerisation of carbonyl and $\alpha$-amino-groups.

It was also confirmed that the $\alpha/\beta$-unsaturated carbonyls are potent browning agents and also that dicarbonyls of the glyoxal type make a contribution to browning in the early stages.
More recent investigation at elevated temperatures (up to $\pu{120^\circ C}$) also supports the major role of ascorbic acid degradation among other factors of browning [3]:

Browning of clariﬁed cashew apple juice was caused by the degradation of ascorbic acid. Changes in the absorbance at $\pu{420 nm}$ and the ascorbic acid levels during thermal treatment of clariﬁed cashew apple juice were
described by ﬁrst order kinetics, showing the correlation between ascorbic acid loss and colour formation (browning).

It is concluded in all mentioned sources that formation of intermediate undesirable carbonyl compounds (mostly furfural and 5-hydroxymethylfurfural (5-HMF)) and various products of their polymerization and combination with amino acids (furfural is an aldehyde) upon heating is one of the major factors of browning.

Maillard reaction is a chemical reaction between an amino acid and a reducing sugar, usually requiring the addition of heat. The reaction is initiated by a condensation between the free amino group of an amino acid, peptide, or protein and the carbonyl group of a reducing sugar, leading to the formation of Amadori compounds (various furoylmethyl derivatives in case of fruit juices)[4].

Caramelization generally occurs at high temperatures (above $\pu{140^\circ C}$), when the juice is dehydrated and has high sugar content. Basically it is a pyrolysis and a final step of decomposition.

To sum it up, these are the products of first 2 processes (ascorbic acid degradation and reaction with amino acids (Maillard process)) (from [3], referred to as browning markers):

Bibliography

Bharate, S. S.; Bharate, S. B. J Food Sci Technol 2014, 51 (10), 2271–2288. DOI 10.1007/s13197-012-0718-8.
Clegg, K. M. Journal of the Science of Food and Agriculture 1964, 15 (12), 878–885. DOI 10.1002/jsfa.2740151212.
Damasceno, L. F.; Fernandes, F. A. N.; Magalhães, M. M. A.; Brito, E. S. Food Chemistry 2008, 106 (1), 172–179. DOI 10.1016/j.foodchem.2007.05.063.
del Castillo, M. D.; Corzo, N.; Olano, A. Journal of Agricultural and Food Chemistry 1999, 47 (10), 4388–4390. DOI 10.1021/jf990150x.


How to measure and transfer suspensions? I need to measure and transfer 10 ml of a fairly dilute suspension at a time to another container.

But no matter how diligently I do it, sedimentation happens quickly after mixing and successive transfers have different strengths or concentrations of solute.

Is there a way to achieve this without increasing the viscosity of solvent?

Note: I have to start with a dilute suspension and then transfer measured quantities of it. I have no other option.

Stir the dilute suspension rapidly, e.g., using a magnetic stirrer. Let a  vortex be generated whose size you can use as an estimate for vigor of stirring. With the same mixing vigor, and using the same size vessel and same original volume for the suspension, and withdrawing only a small portion (say, ~10%) for the other container(s) should enable fairly repeatable measurements.

For obsessive accuracy, you could do three withdrawals: #1 for reference, #2 for experiment, and #3 for reference. If #1 and #3 show the same concentration of solid, #2 will certainly be the same, and the concentration will be verified while sedimentation will be proven to be absent.


What is the significance of Law of Multiple Proportions? I cannot understand how the Law of Multiple Proportions is significant, or how does it further improve over Law of Conservation of Mass or Law of Definite Proportions. From Wikipedia:

  If two elements form more than one compound between them, then the
  ratios of the masses of the second element which combine with a fixed
  mass of the first element will be ratios of small whole numbers.

Now I must say that I don't understand anything from the "small whole numbers" part, since it is not rigorously defined. For example, $31454315/43546546$ may be a ratio of small whole numbers for me but not for you.

Nevertheless the following is the algebraic interpretation of the law, with substituting the "ratios of small whole numbers" part with: "a real constant", since the ratios of two whole numbers is always a real constant.

Compound 1: Has $m$ mass of component $A$ and $x$ mass of component $B$.

Compound 2: Has $m$ mass of component $A$ and $y$ mass of component $B$.

By the Law of Definite Proportions, the following are correct:

$x/m = c_1$

$y/m = c_2$

where $c_1$ and $c_2$ are real constants. This is because Law of Definite Proportions states that "a chemical compound always contains exactly the same proportion of elements by mass".

Dividing $x/m$ by $y/m$ gives:

$x/y = c_1/c_2$

Since the ratio of two real constants is another real constant, we can express this as:

$x/y = c_3$

Since a real constant can always be expressed as the ratio of two whole numbers, this gave us the Law of Multiple Proportions.

As I have shown above, the Law of Multiple Proportions is a corollary of the Law of Definite Proportions.

So again: What is the significance of the Law of Multiple Proportions? Isn't it just an application of the Law of Definite Proportions? How does Law of Multiple Proportions improve over Law of Definite Proportions and Law of Conservation of Mass? How does it contribute to the first scientific atomic theory?

These laws all date back from the very early days of chemistry and are only taught at school to give pupils a way to grasp parts of the subject more easily. Except for the law of conservation of mass, which some professors/supervisors love stating frequently, they are hardly mentioned in universities outside of science history classes.

That said, as I understood them, they refer to slightly different cases. The law of definite proportions should be understood to read:

  No matter how much mass of two elements will react with each other, the product will contain the masses of its elements in a certain ratio and this ratio only.

Taking an easy example, if I reacted $46~\mathrm{g}$ of sodium and $16~\mathrm{g}$ of oxygen, I would get $62~\mathrm{g}$ of $\ce{Na2O}$ where the ratio of sodium to oxygen is $46:16$. If I had instead reacted $92~\mathrm{g}$ of sodium with an unlimited amount of oxygen, I would have gotten $124~\mathrm{g}$ of the same product which contains sodium and oxygen in a $46:16$ ratio. No matter how much sodium I let react with how much oxygen, the product will contain the definite proportion of sodium to oxygen $46:16$.*

The law of multiple proportions now states that some elements are special. For example, iron can react with chlorine to give two different chlorides: $\ce{FeCl2}$ and $\ce{FeCl3}$. For each of the two you can, as you realised set up a law of definite proportions which often gives you a rather weird rational ratio. However, when comparing those two with each other, you find that the mass of chlorine in $\ce{FeCl3}$ is 1.5 times as much as the mass of chlorine in $\ce{FeCl2}$ — corresponding to a $3:2$ ratio. And it is these very small (usually single digit) whole numbers that are actually meant.

Taking a different example, the relative atomic masses of sulphur and chlorine are $32.07$ and $35.45$, respectively. So a sulphur dichloride would have mass ratios of $32.07:70.9$ — far from any small number, but still a rational ratio. Sulphur tetrachloride exists, too, with another pretty unclean ratio. However, the ratio of the chloride contents is $1:2$ — isn’t that a nice simple ratio?

*: This example is not technically correct. Burning sodium gives $\ce{Na2O2}$, sodium peroxide. But I wanted nice and whole numbers and oxygen and sodium have nice and whole mass numbers.


Why does the helium hydride ion have such a large dipole moment? According to Wikipedia, the helium hydride ion $\ce {HeH+}$ has a dipole moment of 2.26 or 2.84. What contributes to this large dipole moment in this molecular ion?

The dipole moment of an array of charges is defined as
$$
\vec{\mu}=\sum_i \vec{r}_iq_i
$$
where $\vec{r}_i$ is the position of charge $q_i$ in the molecule with respect to the center of mass. In a neutral molecule the sum of the charges is zero and therefore the terms in the sum partially cancel. For ions this is of course not the case and the dipole moment is much larger than for the corresponding neutral molecule (provided that the ion is not completely symmetric such as H$_2^+$).


Best chair configuration for trimethylcyclohexane matching a specific Haworth projection? Which option best depicts trimethylcyclohexane shown in the Haworth projection below?

I thought B because

All the methyl groups are equatorial and equatorial groups are more stable.
All the methyl groups look as spread out as much as possible, reducing steric repulsion.

But the provided solution states that-

Although A,B and D could all be said to represent the chair form, the best depiction will show the methyl groups repelling one another as far as possible. The arrangement in A with two equatorial and one axial methyl groups makes A the best depiction.

Also, when converting Haworth to Chair, do we have to match the upward facing groups on the Haworth to upward facing positions on the chair?
i.e. the Haworth has 2 upward methyl groups and 1 downward facing methyl. But in A, the left equatorial one is facing upward (as axial in that position is downward), the middle axial methyl is upward, and the right equatorial methyl  is downward (as axial in that position is upward). This doesn't match the Haworth.

While you are correct that B should be the most stable conformer of trimethylcylcohexane, the question asked which one best represents the one shown in the Haworth projection. Answer choice B has R stereochemistry on carbon B (see below), while carbon B has S stereochemistry in the original projection. As such, the initially drawn compound is different from the answer choice B. On the other hand, answer choice A has the correct stereochemistry, and also has the maximum number of possible methyl groups in equatorial positions, causing it to be the best depiction of the Haworth projection.

Note: Instead of finding the absolute stereochemistry by using R/S notation, you could also notice that the Haworth projection and answer choice B have different relative stereo chemistry. In the original molecule, carbons B and C are cis, while in answer choice B, these carbons are trans.


Dangers of using activated carbon past its lifespan Activated carbon (AC) is one of the most widely used media for water filtration. It's effective at removing chlorine taste and odor, and it's recognized as being safe.

However, most activated carbon filters have an indicated lifespan of about 4-6 months, after which they need to be replaced. But is there any risk to using an AC filter past its lifespan other than the fact that it won't effectively filter contaminants anymore?

I've read in passing that an AC filter past its lifespan can actually release contaminants in water, because some contaminants in the influent water may be preferentially adsorbed by the AC, replacing existing contaminants previously trapped in the AC and causing them to be released in the effluent water. Is this something that has been explored in research?

In particular, has it been shown that using a saturated AC filter can cause the levels of certain contaminants — e.g. a trihalomethane such as bromoform, or perhaps fluoride or lead — to rise to a level that is well above the EPA maximum contaminant level?

Every adsorbent has a saturation capacity depending on its active surface area. The idea to replace carbon filter is not only due to the saturation capacity but imagine fungi, bacteria may be adsorbed as well. It is worth replacing.


Role of catalyst in elementary reactions towards rate law We do know that the concentration of a catalyst might affect the rate of reaction. But what if the catalyst is included in the rate law? For example,
$$\ce{O3 + M <=>[$k_1$][$k_{-1}$] O2 + O + M},\tag{1}$$
where $\ce{M}$ is the catalyst, when I am writing the rate change of $\ce{O},$ do I write
$$\frac{\mathrm d[\ce{O2}]}{\mathrm dt} = k_1 [\ce{O3}][\ce{M}] - k_{-1}[\ce{O2][O][M}],\tag{2}$$
or
$$\frac{\mathrm d[\ce{O2}]}{\mathrm dt} = k_1 [\ce{O3}] - k_{-1}[\ce{O2][O}]?\tag{3}$$
Which is correct?

The catalyst uses a different reaction path with the zero net catalyst consumption, compared to not catalyzed reaction.
E.g. instead of $\ce{A + B <=> C}$, there is $\ce{A + M <=> AM}$, $\ce{AM + B <=> M + C}$.
If there is a homogenous catalyst, then its concentration in rate equations is processed by the same way as the reactants or products. But the different reaction path, compared to non catalyzed reaction, remains.
So it would not be $$\ce{O3 + M <=> O2 + O + M}$$
but rather e.g. :
$$\ce{O3 + M <=> O2 + MO}$$
and
$$\ce{MO <=> O + M}$$

OR:
$$\ce{O3 + M <=> O + MO2}$$
and
$$\ce{MO2 <=> O2 + M}$$
So you need to solve the kinetics of the system of linked reactions, considering catalyst form concentrations in steady states.

If it is a heterogenous contact catalysis, then the amount and activity of catalyst is implicitly involved in the catalyst-dependent kinetic constant:
$$\frac{\mathrm d[\ce{O2}]}{\mathrm{dt}} = k_{1,\mathrm{M}} [\ce{O3}] - k_{-1,\mathrm{M}}[\ce{[O2][O]}$$
where the ratio $\frac{k_{1,\mathrm{M}}}{k_{-1,\mathrm{M}}}$ remains constant, independent of the catalyst activity.


Is it necessary to add prefixes of halogens (more than 1) to the iupac name alphabetically? If I draw structure of 2-bromo-4-chloropentane and
4-chloro-2-bromopentane I get same result, when I try to draw structure. So why is it necessary to name prefixes alphabetically. Can we write iupac name not alphabetically.

With the preferred IUPAC name (PIN), there is exactly one way that you expect to find a compound written.

Assume you want to buy the compound from the chemical supplier of your choice but don’t have a CAS number at hand. What do you do? You search for the compound in the catalogue. In the old days, these were always paper books that you could use as a weapon but nowadays electronic catalogues exist to. Still, if you’re looking for 1-bromo-3-chloropentane but it is entered into the catalogue as 3-chloro-1-bromopentane, you’re going to have a very hard time searching for it.

There are also other reasons why having one defined standard way of naming compounds is strongly preferred over random but clear.


Simple experiments involving enantiomer synthesis I want to do an experiment to investigate factors affecting the synthesis of chiral molecules. The only problem is, my institution doesn't have much money, and it seems like the chemicals required for any experiment involving chiral molecules are very expensive!

My initial research found an experiment based on the Bürgi-Dunitz trajectory (see Addition to Carbonyl Double Bonds), but with the use of L-Selectride it isn't the cheapest. I then researched the Diels-Alder reaction, but the conjugated dienes are also relatively expensive, or are only available in large quantities!

I am only an amateur in the field of chemistry, and much of the interesting material is very hard to find unless one knows the specific terms to use. Now I come to you for help - are there any cheap experiments involving the synthesis of enantiomers? Other than cost, the constraints are:

Equipment: standard laboratory equipment is available, but nothing too fancy
Aim: the experiment needs to investigate something, for example factors affecting the synthesis of L and D enantiomers. Two different chiral auxilliaries or catalysts would do well here.

If you have some way of measuring the enantiomeric excess of the product mixture, I suggest trying proline catalysis of an aldol reaction. The materials are cheap / readily available, and the reaction setup is straightforward. Proline is probably the cheapest chiral catalyst available.

Here is a procedure for an organic chemistry lab run at Rutgers that might get you started:
http://chem.rutgers.edu/sites/default/files/coursefiles/Courses_f09/310/Procedure_Proline_Catalysis.pdf


OD660 readings of yeast I am working on a yeast fermentation project, and I am getting final OD660 reading of 1.85 for different media: yeast extracts+cane molasses, nitrogen source+cane molasses, and glucose+cane molasses at different concentrations. The final OD reading is the same for all media. Is it normal? Are my readings correct?

Many inexpensive spectrophotometers cannot measure absorbances over 1.5 or so reliably. It's possible that you're out of the measurement range of your instrument. Try diluting your samples with water first. If you dilute 1:1, then the absorbance value of your sample is double what you measure (assuming the Beer-Lambert law is valid in this situation)

Also, I imagine molasses is very absorbent at that wavelength so it might be obscuring any change from the yeast and it may not be possible to measure the yeast growth by absorbance, with such a matrix.


Should the volume of a solute be factored into the volume of a solvent when preparing a substance of a certain molarity? Upon being asked to prepare a solution of a given molarity from a solution with high concentration I was confused whether or not I should factor in the volume taken from the high concentration substance into the volume of the solvent. I think that an example could illustrate what I am asking more appropriately. Lets say I have to prepare a 5 M substance in 50 ml of water. The substance is in a container that is 10 M so I must calculate the number of moles and then the amount of solute to extract from the 10 M container.  $$M = {m\over{L}} $$ $$5 M = {m\over{.05}} $$ $$moles = .25$$ given that the solution's concentration is 10 M $$10M={.25\over{L}}$$ $$L = .025$$ Now in order to achieve the 5 M in the 50 ml of water, do I pour 25 ml of water and then the 25 ml, which then adds to 50 ml? or do I pour the 50 ml of water and then the 25ml resulting in a total volume of 75 ml?

The only volume that is important in the denominator of your equation is the total volume of solution; neither the volume of solute or solvent are explicitly relevant.

Pragmatically, there is not likely to be much of a volume change when taking a concentrated solution and diluting it.

So with reference to your question, you should add enough solvent to get the proper final volume.  Explicitly, add the 25 mL of stock solution and 'dilute to mark'-- that is add enough solvent so the total solution is 50 mL.  The added solvent might not be exactly 25 mL to make this final volume.


What is the 3D shape of an ethane molecule? I am currently teaching a course on groups, and want to give my students "real life" examples of symmetry groups to think about. The ethane molecule would be one such.  But, the on-line sources don't all agree on the exact shape. I'm fairly sure ethane looks like either this

or this.

I oriented the molecules in such a way that the carbon atoms are on top of each other. The difference between the two choices is how the groups of three hydrogen atoms relate to each other. In the latter alternative the molecule is its own mirror image with respect to the plane bisecting the line segment joining the carbon atoms. Meaning that the hydrogen atoms also pair up in a way that they are on top of each other. However, in the former version this is not the case, and you would need to rotate one triple of hydrogen atoms by 60 degrees to move them on top of each other.
To turn this into a homework problem I would like to know which is correct?
Wikipedia suggests that the one on the top is correct. However, the picture below also appears in some places. Like here.
My theory: Whatever little I know about physics/chemistry suggests to me that the top one is more likely. The reason being that the similarly charged hydrogen atoms repel each other (possibly this is only a dipole or a higher order effect?). Therefore it stands to reason that one triple of hydrogen atoms is aligned in a way that they use the "in-between" angles.
Nitpick: The "correct" answer probably is that neither of these two images is quite right. You see, I generated them using the tetrahedron (=methane) angle of 109.5 degrees everywhere (simpler for me to do it that way). But Wikipedia says that the angle between the carbon-carbon bond and a carbon-hydrogen bond is slightly larger, 111.17 degrees. So the hydrogen atoms at the top should be pushed a teensy bit further up and similarly the ones at the bottom a little further down. A bigger similar effect is that in the actual molecule the distance between the carbon atoms is larger than the carbon-hydrogen distance. Neither of these has an effect on the collection of symmetries the molecule has, but the answer to my main question does have an impact. That's why I am asking.
If somebody is inclined to elaborate on the reason(s) why we don't have that methane angle everywhere, I will be educated further :-)

You need to know the terms "staggered" and "eclipsed", because that's what we call these two conformations. Now to the point: the real ethane molecule rotates and changes shape all the time, so both geometries are possible, albeit the top one is more likely, because it is the energy minimum. To make the molecule stop in the staggered conformation, you have to freeze it to nearly absolute zero, or maybe put some bulky groups in place of hydrogens.
This, BTW, is the reason why I dislike to use ethane molecule as an example when I teach space groups to kids. Better use something rigid.
So it goes.


Is the ring or the chain chosen as the parent structure in octan-2-ylbenzene? How would you name the above compound according to IUPAC rules? I see some sources say it is 2-phenyloctane, but ChemDraw and a few other sources say it is 2-octanylbenzene.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): “octan-2-ylbenzene” (preferred IUPAC name; ring preferred to chain)

Method (2): “2-phenyloctane” (chain has greater number of skeletal atoms)

The function Convert Structure to Name of ChemDraw correctly yields “octan-2-ylbenzene”.


How to compare stability between 5‐methylcyclohexa‐1,3‐diene and 3‐methylenecyclohexene? Problem

Compare stability of 1 and 2:

Answer

 1 is more stable than 2.

My approach

Both have the same number of π-bonds.
Both have one resonance structure.
Both are non-aromatic.
In 1 there are three α-hydrogens, whereas in 2 there are four α-hydrogens.

As extent of hyper-conjugation is more in 2 than in 1, making 2 more stable, which contradicts the answer.

As far as I have read, in general, exocyclic double bonds (which are outside the ring) are stable for 3 or 4 numbered rings while endocyclic double bonds (which are inside the ring) are stable in 5 or 6 numbered rings.
As you note all the factors, more or less all factors are same in both cases, so it becomes a little handwavy thing.
Additional Reading:

Goumans, T.; Ehlers, A.; Lammertsma, K.; Würthwein, E. endo/exo Preferences for Double Bonds in Three-Membered Rings Including Phosphorus Compounds. Eur. J. Org. Chem. 2003, 2003 (15), 2941–2946. DOI: 10.1002/ejoc.200300120.

Fleck, B. Notes - Stability and Ease of Formation of Endo and Exo Double Bonds in 5- and 6- Ring Systems. J. Org. Chem. 1957, 22 (4), 439–439. DOI: 10.1021/jo01355a023.

Exocyclic vs. endocyclic double bonds in E1 elimination (although this one is not much relevant here).


Hexagonal water, alkaline ionized water I am chemical engineering and i care my nutrition and health. in my opinion, it is so important that what we drink or eat daily.
 Respectfully, I have a question about alkaline ionized water bottle, Which is sold commercially. I have some information about production of electrolyzed water technology. as i study in some articles, hard water (water with minerals such as calcium bicarbonate) electrolyzed in a electrolysis cell and Hydroxide is produced. As i knew, calcium and magnesium hydroxide are insoluble in water. But we don't see any white residue in commercial alkaline ionized water bottle. These particles (as an alkaline agents) are probably filtered, my question is " if calcium and magnesium hydroxide are filtered, why this water have pH 9 or high??"
thank you
I'd be happy and it is my pleasure if you tell me this.

As Waylander states, calcium and magnesium hydroxides do have some solubility in water.  In the case of magnesium hydroxide you can demonstrate this by simply buying some milk of magnesia and applying red litmus paper to it; the paper turns blue like there were an alkali.

Simply dissolving either calcium or magnesium hydroxide into water would raise the pH beyond the target (you get 10 for magnesium, 12 for calcium).  Additional calcium or magnesium salts may be introduced to shift the dissolution equilibrium for the sparingly soluble hydroxide, and thus control the pH.

Here and here the use of lime and magnesia for water treatment is mentioned.  Both work by making the water slightly alkaline so as to displace heavy metals as less soluble hydroxides.


Reactivity and its meaning When we say that “potassium” is reactive, are we referring to a potassium atom or bulk potassium as in the solid metal? In other words does “potassium” refer to a type of atom or an agglomeration of potassium atoms?

Reactivity is a vague concept in chemistry. It is rather confusing than useful. Reactivity involves both thermodynamic and kinetic factors, it predicts whether a chemical compound will undergo a chemical reaction (either by itself or with other compounds) or not and if it does how fast is the reaction going to completion i.e., rate of reaction.

Reactivity is also used to compare or study rates of two or more chemical reaction. For example when we say reactivity of HBr is more than reactivity of HCl - here we refer to how fast the compund will give acid-base reaction.

Reactivity is sometimes referred to bulk and sometimes to a single molecule but mostly it is referred as bulk, but it is not predictable.

In pure compounds, reactivity is regulated by the physical properties of the sample. For instance, grinding a sample to a higher specific surface area increases its reactivity. In impure compounds, the reactivity is also affected by the inclusion of contaminants. In crystalline compounds, the crystalline form can also affect reactivity. However, in all cases, reactivity is primarily due to the sub-atomic properties of the compound.


Anti-Solvent Precipitation Of Lycopene I have a question about anti-solvent precipitation in organic chemistry:

Is it possible to precipitate lycopene or other carotenes using this method?

I have been extracting carotenes like lycopene by Soxhlet method using acetone as solvent. I was wondering if it's possible to precipitate these compounds using the anti-solvent method.

If it's possible which anti-solvent would be more accurate, in the documents that I have been reading , it was mentioned methanol and other alcohols but I'm not sure if the acetone that I used in the extracting part would cause any interference.

If you have a solution, so long as you don't have too much of your "good", more solubilising solvent,  you should be able to precipitate your terpene.  Maybe not crystallise, but at least bring it out of solution. Lower alcohols and water act as anti-solvents for long chain hydrocarbons like lycopene.
If this is for lab purification scale, another option is to simply evaporate your solvent and dissolve it in your crystallisation solvent of choice. For these longer chain polyenes,  it may require a two solvent system anyway to crystallise the carotene of choice.
Going back to one of your comments, the acetone will not interfere so long as it is the minimum amount. With the Soxhlet method you use, presumably you extract unwanted compounds also, so your experience of how easily these dissolve/precipitate will affect your choice of isolation method and specifically, the anti-solvent choice.


What is the "curse of dimensionality" in molecular dynamics? The 'curse of dimensionality' or the 'bottleneck problem' in molecular dynamics is explained in page 5 and 6 of Ab Initio Molecular Dynamics: Basic Theory and Advanced Methods by Dominic Marx and Jurg Hutter .

They proved that Car-Parrinello MD outperforms classical MD with a computational advantage growing as $\sim 10^N$ with system size, which I am unable to understand, especially the last paragraph of page 5.

To be more specific, I am copying the relevant contents:

  In the case of using classical mechanics to describe the dynamics - which is the focus of the present book - the limiting step for large systems is the first one, why should this be so? There are $3N-6$ internal degrees of freedom that span the global potential energy surface of an unconstrained $N$-body system.  Using, for simplicity, 10 discretization points per coordinate implies that of the order of $10^{3N-6}$ electronic structure calculations are needed in order to map such a global potential energy surface. Thus, the computational workload for the first step in the approach outlined above grows roughly like ∼$10^N$ with increasing system size ∼$N$. This is what might be called the curse of dimensionality or dimensionality bottleneck of calculations that rely on global
  potential energy surfaces.

What is needed in ab initio molecular dynamics instead? I am confused about the parts placed in bold:

  Suppose that a useful trajectory consists of about $10^M$ molecular dynamics steps, i.e. $10^M$ electronic structure calculations are needed to generate one trajectory. Furthermore, it is assumed that $10^n$ independent trajectories are necessary in order to average over different initial conditions so that $10^{M+n}$ ab initio molecular dynamics steps are required in total. Finally, it is assumed that each single-point electronic structure calculation needed to devise the global potential energy surface and one ab initio molecular dynamics time step require roughly the same amount of cpu time. Based on this truly simplistic order of magnitude estimate, the advantage of ab initio molecular dynamics vs.  calculations relying on the computation of a global potential energy surface amounts to about $10^{3N-6-M-n}$. The crucial point is that for a given statistical accuracy (that is for $M$ and $n$ fixed and independent of $N$) and for a given electronic structure method, the computational advantage of ``on-the-fly” approaches grows like ∼$10^N$ with system size. Thus, Car–Parrinello methods always outperform the traditional three-step approaches if the system is sufficiently large and complex. Conversely, computing global potential energy surfaces beforehand and running many classical trajectories afterwards without much additional cost always pays off for a given system size $N$ like ∼$10^{M+n}$ if the system is small enough so that a global potential energy surface can be computed and parameterized.

How do the bolded points prove that fact that "computational advantage of 'on-the-fly' approaches grows like ∼$10^N$ with system size" over classical MD?

It's certainly more of a "back of the envelope" calculation than a rigorous proof. The main point is, that for a certain system, the cost of obtaining a set of trajectories with a given statistical accuracy is fixed with ab initio MD, as only the dynamically relevant parts of the PES are calculated.

With what is here called "classical MD" (but also applies to "quantum dynamics" and "wave packet" methods), the whole PES is computed in advance, and this cost is system dependent (scales with $3N - 6$ for nonlinear systems); the individual trajectories over this PES are then calculated comparably quickly.

The ratio of the two scalings is $\frac{10^{3N-6}}{10^{M+n}}$, increasing with system size (making ab initio MD favourable) and decreasing with required statistical accuracy (making classical MD favourable).


De-excitation of a molecule In our chemistry lecture today on UV/vis-spectroscopy, we discussed the origin of colour in conjugated molecules due to electronic transitions from lower-energy molecular orbitals to higher-energy molecular orbitals. One of the examples we talked about was the origin of the orange colour of carrot. The lecturer mentioned that carrots appear orange as it contains conjugated molecules that absorbs blue light, resulting in the complementary orange colour being emitted. This blue light is used to excite electrons from π MOs to π* MOs in the molecule. One of my classmates asked what happens to the excited electron when it falls back to the ground state, wouldn't it cause blue-light photons to be emitted?

To this, the lecturer replied that perhaps, there is de-excitation of the electrons through other pathways, such as through molecular vibrations (i.e. kinetic energy of the molecules). However, he also was not so sure about the matter. Thus, I would like to seek clarification on how these excited electrons de-excite after initial excitation.

After radiation is absorbed and the electron is at an excited state in the molecule there are several pathways for de-excitation to occur (see fig.). The pathway of choice depends on its rate, ie how fast it can happen. It turns out that the fastest de-excitation pathways are radiationless (wavy arrows in fig) such as internal conversion (IC) that happens through vibrational relaxation of the molecule and external conversion (EC) where the molecule returns to its ground state through collision with solvent molecules. This is why matter usually doenst emit light after absorbing it. There are instances however where this is actually possible because it happens that the radiationless pathways are not the most efficient way for the molecule to return to its ground state and that is when fluorescence (FL) and phosphorescence (Ph) occur.


Energy levels transitions of an electron? I have an example that tells me that the energy transition between $n=1$ and $n=3$ is $12.1\ \mathrm{eV}$ using the following formula:

$$E= -RH (\frac{1}{m^2} - \frac{1}{n^2})$$

and that the conversion from $12.1\ \mathrm{eV}$ gives a wavelength of $102.6~\mathrm{nm}$.

I understand that $-RH$ is the Rydberg's constant that corresponds to $13.60\ \mathrm{eV}$ and $m$ is the inner orbit and $n$ the outer orbit (but I might be wrong with the latter). So,

$$E= -13.60\ \mathrm{eV}\ (\frac{1}{1^2} - \frac{1}{3^2})$$

$$E= -13.60\ \mathrm{eV} \cdot 0.83$$

$$E= -11.288$$

I don't understand how they get the first value for the electron Volts and then convert these into wavelength. Additionally, my task is to find the energy level for the 4th orbit in the Lyman's series.

If I remember correctly, you can use the equation:

$$E = \frac{hc}{\lambda}$$

and

$$\nu = \frac{c}{\lambda}$$

to solve for the frequency, wavelength, or energy of the photon. Note that $E$ is energy, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength, and $\nu$ is frequency.

The energy, (actually 11.97eV not 11.2eV)converted into joules (2.719 $*10^{-18}$ J) and then put into the first equation:

1.918 $×10^{-18}$ J = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{\lambda}$

then lambda can be solved for:

$\lambda$ = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{1.918×10^{-18} J}$

$\lambda$ = $1.036×10^{-7} m $

= 103.6 nanometers.

This solves for your wavelength like you asked for.


Which of the following compounds possess more than one percent enol content? Which of the following compounds possess more than one percent enol content?

I understood that stability of enols are dependent on the hydrogen bonding in the compounds. But here, what is the significance of more than one percent enol content?

All the compounds that have a relation $\alpha-\beta$ between the carbonyl groups. The hydrogen between two $\ce{CO}$ is very acid because two electrowithdrawing remove electron density from the $\ce{C-O}$ bond. Therefore for this kind of compound the most stable for is the enol. In Organic Chemistry this is very useful in creating new carbon-carbon bonds in mild conditions like in aldoholic reactions, Michael reactions and so on. See the very good book of Marc Loudon, Organic Chemistry.


Is water committed to vapor state at boiling point or after? Quite early into my studies, I learnt about phase changes with this graph.

I’m just going to focus on vaporization/condensation. I learnt that the plateau was caused because all the heat was going into breaking the bonds between the molecules, freeing them from a liquid state to a gaseous state. Thus, no energy was going into raising the kinetic energy of the substance or the temperature.

Later, I learnt about thermodynamics and Gibbs free energy. Spontaneity is dependent on temperature—some processes are spontaneous at some temperatures, and some are not. For a temperature above 100 °C, the boiling of water is spontaneous. For a temperature below 100 °C, the condensation of water is spontaneous. At 100 °C, both states coexist in equilibrium (i.e. $\Delta G = 0$). It is only when you raise/decrease the temperature ever so slightly that water is committed to either state.

I fail to reconcile this with what I learnt. At 100 °C, energy is going into breaking intermolecular bonds. That suggests to me that at 100 °C, water is being committed to vapor. According to the thermodynamic interpretation, at 100 °C, both liquid and gas coexist in equilibrium, and water is not committed to vapor until the temperature exceeds 100 °C.

Can anyone clear up my confusion?

At equilibrium, not only are molecular bonds of the liquid breaking to form vapor.  There are also molecular bonds of the vapor reforming to produce liquid.  At equilibrium, the rates of the two processes are equal.  But, if you add heat to the liquid, the rate of breaking bonds exceeds to rate of reforming bonds (i.e., the temperature and vapor pressure of the liquid increases slightly), and some of the liquid changes to vapor.  But once you stop adding heat, equilibrium is re-established.


What is the origin of the maximum in rotational heat capacity? The graph of rotational heat capacity above shows a small maximum before approaching the equipartition value. What is the origin/physical explanation of this maximum?

Most heat capacities go through a maximum as the temperature increases. $C_V = \left( \frac{dU}{dT} \right)_V$ so a maximum in $C_V$ corresponds to a minimum in $\left( \frac{dT}{dU} \right)_V$, i.e. the point where the temperature changes very little as energy is being supplied to the system.  At this point (most of) the energy is being used to excite molecules from the ground state to the excited state, rather than increasing the kinetic energy (i.e. the temperature) of the molecules.

I have made video about it here.

For a more quantitative explanation, consider a two-state system with a singly degenerate ground-state and a excited state with degeneracy $g$, that is $\varepsilon$ high in energy than the ground state.  The heat capacity can be written as
$$C_V = \frac{\varepsilon^2}{RT^2}p_0p_1 $$
where $p_0$ and $p_1$ is the probability of being in the ground and excited state, respectively
$$ p_0 = \frac{1}{1+ge^{-\varepsilon/kT}}$$
$$ p_1 = \frac{ge^{-\varepsilon/kT}}{1+ge^{-\varepsilon/kT}}$$

Now some plots for $N_A \varepsilon$ = 2.5 kJ/mol and $g$ = 100

The first plot shows $p_0p_1$ (green), $C_V$ (red) and  $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis).  This shows that the general shape of $C_V$ as a function of temperature comes from $p_0p_1$

The second plot shows $p_0$ (red), $p_1$ (green), $p_0p_1$ (yellow) and $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis) - right $y$-axis.  This shows that the shape of  $p_0p_1$ come from the shape of  $p_0$ and $p_1$, and that the peak occurs where  the ground and exited states are equally likely ($p_0 = p_1$).  The peak in $C_V$ occurs at slightly lower temperature due to the $\frac{\varepsilon^2}{RT^2}$ term, but ultimately the peak in $C_V$ derives from the peak in $p_0p_1$.


Book about biochemistry for physicist I want to understand the basics of biochemistry, from a physicist point of view. I am looking for a textbook that discusses things like the covalent bond and the oxidation number, starting from first principles (quantum mechanics), and builds up to explain things like why ATP is an energetic currency of metabolism, the organization of glycolysis, and oxidative phosphorylation. Of course I do not expect to have detailed mathematical derivations of all of these. But I do want to end up with a firm grasp of the physical principles involved.

Maybe there is single textbook covering these areas, so I would also appreciate suggestions to build a "reading-list", which can include classic papers.

Thanks.

Your question will probably get closed because any answer is going to be an opinion. Instead of recommending a book-I'll shed a little light on what your journey will be like.

The Subject(s)

Your pursuit to describe high level biological functionalities using first principals covers many fields. I don't think you'll find a single book that explains it all. Instead, you're going to have to learn from a number of different sources.

Physical Chemistry - This will give an overview of electronic structure using quantum mechanics
Physical Organic Chemistry - Learn how to apply physical chemistry to organic molecules.
Bio-Chemistry - Apply organic chemistry to biological processes.
Physical Bio-Chemistry Combines physical chemistry with bio chemistry.

You're really after point 4, but it will need to learn a fair amount before getting there.

The Reality

The reality of things is that we don't need to dig down to first principals to describe many biological processes. Chemistry is able to abstract enough of the nitty-gritty away to come up with easier & faster high level concepts.

Take programming for example, web developers don't need to write websites in assembly language because someone has gone ahead and abstracted it all into a nice user friendly language. You could spend a lifetime coding a site like stackexchange in assembly-or you could use a higher level language and do it in a few months.

Just how you won't find a book on website design using assembly language (first principals), you probably won't be finding one describing biology using pure quantum mechanics.

Further Steps

Assess your current knowledge of physical and (general) chemistry. Once you know which level you're at, go to different university sites and check which books the professors are using to teach. If you find a course that is covering the topics that you're interested in-contact the TA or professor and ask for recommendations; I'm sure they'd be happy to help.


Can a first order reaction change to a second order reaction under specific temperatures? I'm studying basic reaction behavior. My question is:

Consider a Continuous Stirred Tank Reactor (CSTR) where an exothermic, first-order, irreversible reaction takes place:

$$
\ce{A ->[k_1] B}
$$

Is there any possibility or (does it make sense in chemistry) that under high and/or low reaction temperature the reaction will switch to a second order reaction? i.e.

$$
\ce{A + A ->[k_2] C}
$$

If yes, can you give an example? Please consider answering in layman's terms.

It's theoretically possible, though I don't know if there exists specific examples.

The first thing to recognize is that reactions don't really "switch" in an on/off fashion (at least not based on temperature). If there's two physically possible reactions, they both occur under all conditions ... theoretically. The main reason you see one and not the other is due to the comparative rates of reaction - one reaction might be so slow that no appreciable amount of product forms in reasonable timeframes.

So while the rates of reactions change based on temperature, as Zhe mentions, the way that rate changes in most cases (the Arrhenius equation) means that the two reactions aren't going to swap positions based on those considerations alone.

However, kinetics is not the only knob we're playing with here, when we change temperatures. We're also messing with the thermodynamics. Specifically, temperature is a factor in the free energy equation

$$\ce{\Delta G = \Delta H - T \Delta S }$$

If the two reactions have different $\Delta H$ (enthalpy) and $\Delta S$ (entropy) values, then we can change the spontaneity of the reactions. At low temperatures, something with a very favorable enthalpy will be spontaneous, even with a disfavorable entropy. Conversely, at high temperatures, something with a very favorable entropy will be spontaneous, even with a disfavorable enthalpy.

So if you have a system which can undergo first-order and second order reactions (for example, a molecule which has ) and choose your reactant such such that the first order reaction has a more favorable enthalpy and a less favorable entropy than the second order reaction (or vice versa), there will be a temperature where the thermodynamically favored product will go from that of the first-order reaction to that of the second order reaction. Whether this would be enough to manifest itself as an appreciable difference in product depends on the particular numbers.

While we're on the discussion, there is another technique related to controlling product outcome, that exploits the balance between the two factors already discussed. That's using thermodynamic versus kinetic reaction control. In this setup, you have one reaction which provides a more stable product, but due to things like steric hindrance the reaction rate is lower.  The other product forms faster, but the product is less stable. If you set up reaction conditions such as short reaction times at high temperature where the speed of the reaction is the dominant factor, the faster-made "kinetically controlled" product dominates. However, if you use conditions such as long reaction times at low temperatures, where the stability of the products is the dominant factor, the more stable "thermodynamicly controlled" product dominates.


Alphanumeric preference between hexyl and heptyl Our chemistry teacher taught us that hexyl (five letters) is preferred to heptyl (six letters) as hexyl has lesser number of letters. After searching many sources I could not find any such rule.
Which one would be preferred while writing IUPAC names?
For example, will it be 8-heptyl-10-hexylheptadecane or 10-heptyl-8-hexylheptadecane?

According to the IUPAC nomenclature, preference for placing prefixes is given in accordance with alphabetical order (just like in a standard English dictionary). It has nothing to do with the number of letters in a prefix. Since "P" in heptyl comes before "X" in hexyl, heptyl will be placed before hexyl in IUPAC nomenclature.
Hence, the correct name will be the one where heptyl is placed before hexyl (8-heptyl-10-hexylheptadecane).


Can water volume reduce with introduction of another chemical Can a chemical be introduced to a volume of water which can actually reduce the volume of water in that container?

Yes, at least if only the total volume should decrease:

Density of water: $\pu{1000 kg/m^3}$
Density of ethanol: $\pu{789 kg/m^3}$
Expected density of 52 Vol.% ethanol solution: $\pu{890 kg/m^3}$
True density of 52 Vol.% ethanol solution: $\pu{909 kg/m^3}$
All values are for the same conditions.

The volume of every ethanol-water solution is smaller than the sum of the original volumes of water and ethanol. For example if you add $\pu{52 mL}$ of ethanol to $\pu{48 mL}$ water you will get a solution with an volume of approximately $\pu{97.9 mL}$.
This effect is known as volume contraction and depends on the mixing ratio of both components in a nonlinear way. Its reason is the formation of additional bonding forces (e.g., hydrogen bridges) between the molecules, which results in a higher density.

Note: I don´t think that there is something that can increase density of water in such a way that the resulting total volume is actually smaller than the volume of water used.


How much water would one loose if one burns hydrogen and oxygen? By using electrolysis one can separate hydrogen and oxygen, but when burning hydrogen and oxygen together steam is released, therefore creating water. My question is:

  how much of the original quantity of water would be lost after this
  process?

Theoretically, none.

Actually, since the gases are produced in a damp environment, there would be a lot of water vapor in the gases. You could easily remove most of the water vapor, but some (maybe as much as 1%) would still be in the gases.

I think it would be best to elaborate on what you mean as water being lost?


Mendeleev's predictions While studying about Mendeleev's periodic table in my textbook( class 11 NCERT chemistry), I have studied that Mendeleev has predicted some elements like Eka-boron(scandium), Eka-Aluminium(Gallium),Eka-Silicon(Germanium)etc. He also predicted their physical properties which were most accurate. One such element about germanium is given in the below link.

How could he predict such accurate values of such physical properties?

Well, guess what, even I study the exact same textbook (class 11 NCERT, from India). I had asked my teacher the exact same doubt, and the answer was:

Due to its proximity with neighbouring substances, like Arsenic (which was already discovered by then - see here), Mendeleev was able to accurately calculate the average values of density and melting point of Germanium, or eka-Silicon.
The formula of its Oxide and Chloride was predicted with respect to the formula of the Oxides and Chlorides formed by other elements of the group IVA (as per old IUPAC numbering), such as $\ce{CO2}$, $\ce{SiO2}$, $\ce{CCl4}$ and $\ce{SiCl4}$. Their densities were calculated using the knowledge of Chlorine's and Oxygen's atomic weight, and the predicted density and/or atomic mass of the new element.

Although I don't know how the boiling point of a substance can be predicted, my best guess for it is that it was approximated using the boiling point of other similar compounds near it in the Periodic table. For more information on how Boiling point can be possibly approximated, here is a link.


Why a TS must have only one negative frequency The vibrational frequencies are related to second derivatives; a minimum will have only positive frequencies while transition state should have one negative frequency. The vibrational analysis must be performed at the the optimized geometry.

  Reproduced from ucsb.com

When finding a minimum (local or global), the vibrational frequencies should all be positive. This makes sense (at least looking at it simplistically) as if we travel in any direction we're always going up (else we're not in a minimum).

When looking for a transition state, the common goal is to end up with one (and exactly one) negative frequency, with all others being positive. This makes less sense to me, as, from a transition state there are (in 2D space) two possible ways to go the wrong way (down in energy) - clearly one is more favourable as it leads to a lower energy intermediate/product, but none the less, the other pathway exists.

Is my analysis too basic (i.e. thinking about a reaction profile curve is too simplistic), or is this okay but i'm missing some fundamental transition state theory?

Well, let's just look at the simple case you mention. If we are at a transition state, then we are at a saddle point on the potential energy surface (PES). As you say, there are two directions where we go down in energy and two where we go up in energy which are along or perpendicular to the gradient vector.

Note that to confirm we are at a saddle point, in mathematics one could do the second-derivative test. Or, in chemistry (and in math) in order to do this, we construct the Hessian matrix and find its determinant. So, let our PES be given by a function $f(x,y)$. Then, the Hessian takes the form,
$$
\begin{bmatrix}
f_{xx} & f_{xy} \\
f_{xy} & f_{yy} \\
\end{bmatrix}
$$
where $f_{xx}$ denotes the second partial derivative of $f$, both with respect to $x$ and likewise for the rest of the terms. Notice that $f_{xy}=f_{yx}$, so this matrix is symmetric and hence all its eigenvalues are real. Also note that sometimes people will write this in terms of the force constants instead of the second-derivatives because a harmonic force-constant is a second derivative.

So, if we take the determinant of this matrix, its sign tells us about whether the function is concave up or concave down at the point we are interested in. This determinant is given by,
$$
f_{xx}f_{yy}-(f_{xy})^2.
$$
The sign of this actually tells us information about the eigenvalues of the matrix because the determinant of a matrix is equal to the product of its eigenvalues.
Notice that in order to find the eigenvalues of this matrix we would solve the following equation:
$$
(f_{xx}-\lambda)(f_{yy}-\lambda)-(f_{xy})^2=0
$$
Or, after rearranging,
$$
\lambda^2-\lambda(f_{xx}+f_{yy})+f_{xx}f_{yy}-f_{xy}^2=0
$$

These eigenvalues can be related to the frequencies of the vibrations if we first mass-weighted each second-derivative. All I want to show here is that for this case there are only two eigenvalues, and these eigenvalues do not have to do with the number of directions to move towards a minimum. One of the vibrations you can imagine as being the mode which in two directions have an increase in energy. So, along this mode, the molecule stays at the transition state and hence there is a physical frequency associated with this motion. If the second derivative test is worth anything, then the overall sign of this determinant will be negative at a transition state and hence there will be one and only one negative eigenvalue. Note that this only works at critical points, so that's why at the transition state there must be only one negative imaginary frequency (or negative eigenvalue).

Because for most every molecule we have more than two modes to worry about, you could imagine splitting a much larger matrix into many 2-d matrices and looking at two modes at a time. From this idea it is clear that a transition state need not be limited to having only one negative frequency. Rather, at second-order saddle points, (i.e. a saddle point where following the eigenvector corresponding to a negative eigenvalue leads you to another saddle point) there will be two negative imaginary eigenvalues.

To summarize:

The number of imaginary frequencies is the number of negative eigenvalues of the Hessian matrix (the matrix of all possible mixed-second-partial derivatives)
There can be more than one negative frequency at a higher-order saddle point, but physically speaking these are much less common than transition states with only one imaginary frequency
One cannot determine the number of imaginary frequencies by thinking about the number of paths to a minimum. You could think about the number of modes which do not kick you off of a saddle point if you prefer this.

Edit:

I mistakenly referred to negative frequencies which is not correct (as alluded to by a commenter). The eigenvalues of this matrix correspond to the force constants (second derivatives with respect to displacement coordinates), so one gets negative force constants, not negative frequencies at a saddle point. Because the force constant for one mode is negative at a transition state, the "frequency" of this mode is evaluated as $\nu=\sqrt{\frac{k}{\mu}}$ as usual, and the frequency is imaginary. So, one is looking for either one negative force constant or one imaginary frequency, not a negative frequency. Sorry I missed that.


How to calculate the concentration of all relevant species in a buffer of a given pH? You prepare $1.0~\mathrm{L}$ of a $0.25~\mathrm{M}$ acetic acid solution with a final $\ce{pH}$ of $6.0$. What are the molar concentrations of all relevant acetic acid species ($[\ce{HA}]$ and $[\ce{A-}]$) given that the $K_a$ for acetic acid is $1.74 \cdot 10^{-5}~\mathrm{M}$?

I am getting confused with this problem.

Since the $\ce{pH}$ is given, I know what the $[\ce{H^+}]$ is.
So now when I try to do the ICE table

$$
\begin{array}{l|ccc}
               & \ce{HA} &  \ce{H+}  &  \ce{A-} \\\hline
\text{Initial} & 0.25    &  0        &  0       \\
\text{Change}  &  -x     &  +x       &  +x      \\
\text{End}     & 0.25 -x &  +x       & +x       \\\hline
\end{array}$$

And from here I begin to assume $[\ce{H^+}] = [\ce{A^-}]$, which I am not sure about.
Then I set $x = 10^{-6.0} = 1\cdot 10^{-6}$ so I get $[\ce{A^-}] = 1\cdot 10^{-6}~\mathrm{M}$ and $[\ce{HA}] = 0.24999~\mathrm{M}$ which I think is incorrect, and to even further ensure my that it's incorrect, when I attempt to check the $K_a$ value with this, it does not match.

My second approach:
$[\ce{HA}] = 0.25~\mathrm{M}$

In this I determine the $\mathrm{p}K_a$ from the $K_a$ which turns out to be $4.759$, which indicates that there should be more $[\ce{A-}]$ than $[\ce{HA}]$.

I now use the Henderson–Hasselbalch equation:
\begin{align}
    6.0   &= 4.759 + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\\
   17.40  &= \frac{[\ce{A-}]}{[\ce{HA}]}\\
   17.40  &= \frac{[\ce{A^-}]}{0.25~\mathrm{M}}\\
[\ce{A-}] &= 4.35~\mathrm{M}\\
\end{align}

I feel more confident about my second answer.

Can someone please help me out with this particular problem and perhaps tell me procedure I should use as well as what the correct answer should come out to be and why?

The value given for the pH of the solution seems to be incorrect*.

As the reaction is $\ce{HA <=> H+ + A-}$, your initial assumption that $[\ce{H+}]=[\ce{A-}]$ is correct.

From your chart,

\begin{array}{l|ccc}
               & \ce{HA} &  \ce{H+}  &  \ce{A-} \\\hline
\text{Initial} & 0.25    &  0        &  0       \\
\text{Change}  &  -x     &  +x       &  +x      \\
\text{End}     & 0.25 -x &  +x       & +x       \\\hline
\end{array}

it follows that
$$K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.25 -x} = 1.74\cdot10^{-5}.$$

We can solve this for $x$:
\begin{align}
x^2 &= 1.74\cdot 10^{-5}(0.25-x)\\
x^2 + 1.74\cdot10^{-5} x - 4.35\cdot10^{-6} &= 0\\
x &= 0.00207698~\mathrm{M}\\
\end{align}

Thus, $[\ce{HA}] = 0.25~\mathrm{M} - 0.00207698~\mathrm{M} = 0.24792302~\mathrm{M}$ and $[\ce{H+}]=[\ce{A-}] = 0.00207698~\mathrm{M}$.

Also note that in your second attempt, the value $[\ce{A-}] = 4.35~\mathrm{M}$ makes no sense, as the amount of $\ce{A-}$ present should be in a 1:1 ratio to the amount of $\ce{HA}$ consumed in the reaction, and $[\ce{HA}] < 4.35~\mathrm{M}$.

The given $\ce{pH}$ is likely incorrect because the concentration of $[\ce{H+}]$ ions for the given molarity and concentration does not match with the given $\ce{pH}$, and because the given $K_a$ is close to values of $K_a$ provided online.


Where do I go to get details on setting up different instruments? Say I want to set up a system to preform cyclic voltammetry. How do I know what instruments to get/ electrochemical cells/ etc? Is there a resource for other setups as well, like EIS?

Your question is very broad and hard to answer without pointing out multiple books on the topic. Here is a good one to start: Lab Techniques in Electroanalytical Chemistry, Edited By P. Kissinger and W. Heineman

However, your specific experiment will depend on the type of your sample and what information you are looking for. The best place to start is to look for papers in your field and read their experimental sections.

Instrument manufacturers also write application notes specifically for this reason. You may find useful information on their websites as well. Here is an example : Application Notes Regarding EIS from Gamry

Full disclosure: I used to work for Gamry Instruments.


Democritus vs John Dalton atomic theory What is the difference between atomic theory given by Democritus and John Dalton? Because I have read in many books and figured out that both theories explain the same i.e atom is an indivisible particle and all elements differ in shape and weight of their atoms. So basically what is the difference in both the theories?

Demokrit was a philosopher, while Dalton was arguing on the basis of (somewhat limited, but still) solid, quantitative evidence. I'm not knocking Demokrit (!), but his was not a scientific statement, let alone a theory. You cannot compare the two, or at least you should not try to compare them from the view of natural science.

Both came to a similar conclusion, but they were reasoning on a completely different basis. Of course, Dalton had the advantage that the idea was already there and known to him.

In Dalton's time, the modern concept of elements was already known and widely accepted. Iron, oxygen, etc., not fire, water and stone. You could say that he knew what he was talking about, regarding atoms. And Dalton and his contemporaries had the numbers they measured.

Demokrit, on the other hand, had no idea about quantitative elemental analysis. Which ultimately was the downfall of his idea: He had no numbers that others could have proved and extended, or disproved. Only when the chemists at Dalton's time were able to make quantitative measurements the atomic theory could become useful. Demokrit had probably theorised that atoms of the same kind had the same weight (we don't know exactly, because we have very little original records from him), but Dalton now had solid evidence, that only made sense if that was indeed the case.


Oscillation of particle in harmonic oscillator In harmonic oscillator, as vibration quantum number increases then does the distance of oscillation of particle from the mean position where there is minimum potential energy increases?
If yes, then the expression of potential energy $(kx^2/2)$ which we derived from Taylor expansion of potential energy at mean position should not hold as we neglect high powers terms $(x-x_0)^3$ etc.but as x increases so we can't neglect high power terms. Then the expression of quantized energy $(n+1/2)\hbar\omega$ which we get by solving Schrödinger equation also doesn't hold.

The wave function diagram also shows that the distance from mean position increases as vibration quantum number increases.

Please tell me where is flaw in my argument.

Classically speaking, the (idealised) harmonic oscillator is defined by Hooke's law. Quoting from Wikipedia:

In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force $F$ proportional to the displacement $x$:
$$\vec{F} = -k\vec{x}$$
where $k$ is a positive constant. If $F$ is the only force acting on the system, the system is called a simple harmonic oscillator.

In one dimension we can drop the vectors and write this as $F = -kx$. But the force is also the (negative) derivative of the potential, i.e. $F = -\mathrm dV/\mathrm dx$, and integrating this to get $V = kx^2/2 + C$ is trivial. The constant of integration $C$ is arbitrary, but it doesn't fundamentally change the physics (it just moves the energy scale), so we can ditch it by setting it to zero.
If you had a potential energy with higher-order terms in $x$ (i.e. $x^3$ and above), then the derivative with respect to $x$ would contain terms that are $x^2$ and above. This directly contradicts the definition of the harmonic oscillator. It could be some other kind of oscillator, but its motion would not be called simple harmonic.
Consequently, for this particular oscillator, trying to perform a Taylor expansion is a little pointless: there are no terms in $x^3$ or above, so all the higher-order derivatives $\mathrm d^3V/\mathrm dx^3, \cdots$ are all zero, and all those terms in the Taylor expansion are zero. Of course, in the general case, Taylor analysis of a different oscillator could be very useful. Just not here.
The potential energy of the quantum harmonic oscillator is directly inherited from the classical simple harmonic oscillator, by replacing the classical quantities $x$ and $p~ (= mv)$ with their quantum analogues $\hat{x}$ and $\hat{p} ~(= -\mathrm i\hbar(\partial/\partial x))$. So given that the classical potential energy is $V = kx^2/2$, it's not too much of a stretch to get the quantum potential energy $V = kx^2/2$.


Discovery of optical activity of enantiomers - where did the idea of using polarized light came from? According to Wikipedia, optical activity was first observed in 1811 in quartz. Louis Pasteur's contribution is also mentioned there and in other sources as a major advancement in establishing optical activity as a distinct characteristic of a compound, specifically as a differentiating factor between two enantiomers - a difference that became a standard descriptive characteristic.

My question is how did they even think of that? Why would Pasteur, Arago, Herschel, and other chemists would think of testing the effect of polarized light on chemical compounds? (vice versa actually, chemical compounds affect polarized light, but I assume they did not know that prior to that discovery...). Was it a standard lab procedure at the time?

Testing and determining physical characteristics (such as melting point, boiling point, etc.) seem reasonable and pretty straightforward, but I am wondering how did polarized light entered the chemistry lab as an identifying procedure.

It wasn’t a standard lab procedure to test the effects of polarised light on a certain compound/solution/whatnot. However, back in the day it was standard lab procedure to re-run a certain experiment with practically everything your lab had: water, acids, solvents, table salt, cat hair, human hair, dust, fabric, … You would note if you observed an interesting effect and, where applicable, if that interesting effect would only occur in some cases.

So once you are aware that something like Iceland spar produced interesting phenomena (even though you might not be able to explain these phenomena just yet) you would start trying out all sorts of things to test what this interesting optical phenomenon is capable of doing. By chance, you would run across something that you would be able to observe as cancelling of light, even though all components of the system would let light through — and then you would attempt at understanding what you just observed.

They say science is merely grown men playing around for a reason.


What would be the reaction if melting iron is put in normal water? What would be the reaction if melting iron is put in normal water? Will water be chemically changed?

Unlike the other metals we usually see in the reactivity series, such as calcium, magnesium, zinc, etc., hot iron does not form iron hydroxide, instead it forms iron oxide. The reaction is:

$\ce{3Fe + 4H2O <=> FeO.Fe2O3 + 4H2}$

Iron is similar to magnesium and zinc as they react only with hot water (not cold water). But the reaction of iron with hot water is less vigorous than that of magnesium and zinc.


How do I go about balancing these equations? I have to use oxidization number approach.

$\ce{Co^2+ (aq) + H2SO3(aq) -> Co^3+(aq) + S2O3^2- (aq)}$

This is my approach so far:

The change in Co is 1 electron, the change is S is 2 electrons, so CO must be multiplied by two

$\ce{2 Co^2 + H2SO3(aq) -> 2 Co^3+ + S2O3^2-}$

I balance the O's, then balance the H's and I get

$\ce{2 Co^2+ + 2H2SO3(aq) + 2H^+ -> 2 Co^3+ + S2O3^2- + 3H2O(l)}$

I did it the half reaction way as well, I got the same answer but the coefficients on the Co's were 4. I suspect my problem is my assumptions about the oxidization number change at the beginning.

Thanks in advance

I agree with you, the right equilibration of the redox reaction is the following:
$$\ce{4 Co^2+(aq) + 2H2SO3(aq) + 2H^+ (aq)-> 4 Co^3+(aq) + S2O3^2- (aq) + 3H2O(l)}$$

The problem is  in your assumptions about the oxidization number change at the beginning, and more precisely about sulfur.
The oxidation number of  sulfur in $\ce{H2SO3}$ is four. But in thiosulfate ion, you have two inequivalent sulfur atoms:
Please see:
 Oxidation states of the sulfur atoms in the thiosulfate ion
The oxidation number is $\ce{+IV}$ for the central one, while it's zero for the terminal one.

In sum, the variation in the oxidation number for sulfure is four, not two.
(As if you have two sulfur atoms with oxidation number of four in $\ce{2H2SO3}$. One of them changes its oxidation number to zero. while the other one stays intact).


Hydrogen Bonding in Water Phase Changes How is the disparity between the heat of fusion and the heat of vaporization for $\ce{H2O}$ best explained?

A) It takes more hydrogen bonds for water to fuse than it does to vaporize.

B) Water molecules are moving farther apart during fusion than during vaporization.

C) Water molecules are moving closer together during fusion and farther apart during vaporization.

D) Vaporization occurs at a higher kinetic energy than fusion.

E) More hydrogen bonds are broken during vaporization.

At first, I tried this problem via Process of Elimination. I looked at (B) and immediately knew that that fact was false, because water molecules move closer together during fusion, and farther during vaporization. Next, I saw (C), and though it was true, it had no involvement with the problem. (How far the water molecules move does not have anything with the disparity between heat of fusion and heat of vaporization). Then, I saw (D), and though it was also true (vaporization occurs at a higher temperature than fusion, and kinetic energy is directly proportional to temperature), I felt that it did not relate to the problem. So I have ruled out 3 of the 5 answers by now. It was the options (A) and (E) that stumped me. I did not know the relationship between the temperature and Hydrogen bonds in water, and conducted some research on the internet.

I found a useful link http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html , and saw that answer choice (A) is in fact true, if you look at the graph towards the end of the page. It has more hydrogen bonds in the solid state than in the vapor/gas state. So, I thought the answer was (A). However, I looked in the answer key provided with this question, and the answer was (E): more hydrogen bonds are broken during vaporization. I don't really see how this is, and would like assistance and answers why. Also, I would greatly appreciate an explanation for why the number of bonds increases when temperature is low, and decreases when temperature is higher. Even an explanation on how hydrogen bond numbers (approximately) change as phase changes would be great.

The molar heat of fusion for water is 6.02 kJ/mol (http://www.chemteam.info/Thermochem/Molar-Heat-Fusion.html), while the molar heat of vaporization for water is 40.7 kJ/mol (http://www.chemteam.info/Thermochem/Molar-Heat-Vaporization.html). That is equal to 6 times difference, even a little more. I suggest this energy is used to break existing hydrogen bonds and barely for anything else. The energy of hydrogen bonds is "6-30 kJ/mol" according to this source: http://book.bionumbers.org/what-is-the-energy-of-a-hydrogen-bond/.
Let's consider for simplicity, that breaking 1 mole of hydrogen bonds will require 6.02kJ of energy.
So, when water melts, there is maximum one mole of hydrogen bonds is broken per one mole of water. That is totally different for vaporization process: during vaporization of 1 mole of water it looks like 6 moles of hydrogen bonds are broken.
This invariably leads one to answer E:
More hydrogen bonds are broken during vaporization.


How to calculate the charge on silicates? An isolated $\ce{SiO4^4-}$ ion has $-4$ charge. These discrete tetrahedral units form the ortho-silicates. As these tetrahedral units combine, they form various types of silicates.
How do we calculate the total charge on various forms of silicates?
Examples: $\ce{Si4O11^6-}$ has $-6$ charge and so does $\ce{Si2O7^6-}.$ But I am not able to understand how to calculate this charge. Is there any general rule?

In this answer, the basic concept of how to understand what figure they are and how much charge they have would be discussed.
As OP has well stated, the basic unit of any silicate is the $\ce{SiO4^4-}$ tetrahedron which has four oxygens that are connected to one silicon atom.  Each unshared oxygen provides -1 charge and each shared oxygen contributes 0.5 oxygens to one tetrahedral unit.
Now, depending on how these units interconnect and how oxygens are shared, we get multiple different structures. I shall discuss three of them here since that should be enough to get the idea across.
Type 1: Chain Silicates

In this type of silicates, as you can see above, the tetrahedron shares one oxygen on either side and so the basic unit here would be the $\ce{SiO3^2-}$ unit, since two oxygens are being shared.
Since two oxygens are shared per tetrahedral unit, the basic unit would have 3 (2 unshared + 0.5 * 2 shared) oxygens and 2- charge.
Type 2: Pyro silicates

Here, the one oxygen in the middle is shared by the two tetrahedral units. This means that there is a total of 6 free oxygen atoms that are only bonded to the silicon with a single bond. Therefore the structure arises as $\ce{Si2O7^6-}$.
Since one oxygen is shared per tetrahedral unit, the basic unit would have 3.5 (3 unshared + 0.5 * 1 shared) oxygens and 3- charge. However since the values need to be integers $\ce{SiO_{3.5}^3-}$ would be written as $\ce{Si2O7^6-}$
Type 3: 3D Silicates
In this case, all the oxygen atoms are shared leading to the formation of $\ce{(SiO2)_n}$ molecules.
Further Reading
https://www.adichemistry.com/inorganic/p-block/group-14/silicates/silicates-1.html