Uncertainties after Averaging Trials If I conducted 6 trials for a lab, all with the uncertainty of (+/- 0.1 mL)... After averaging the 6 trials to plot the results, what is the new uncertainty?

In any case, you should calculate the mean $\overline X$, the standard deviation $\sigma$, and the standard error of the mean $\sigma_{\overline X}$ for the six trials.

However, the estimated uncertainty $u(y)$ of the final result $y$ depends on what is meant by “all with the uncertainty of $\pm0.1\ \mathrm{ml}$”, i.e. $u(x)=0.1\ \mathrm{ml}$.

For example, if this uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by the unknown error of a pipette and the same pipette was used for all six trials, each of the six individual results includes the same systematic error. You can neither eliminate this error nor reduce the caused uncertainty by making additional trials. In this case, you should combine (by using the rules for error propagation) the calculated standard error $\sigma_{\overline X}$ and the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. Therefore, the estimated uncertainty $u(y)$ of the final result cannot be smaller than $u(x)=0.1\ \mathrm{ml}$.

If the uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by a random error of each individual trial, it is already included in the observed standard deviation $\sigma$  and thus already taken into account for the calculated standard error $\sigma_{\overline X}$. Therefore, the observed standard deviation $\sigma$ should be greater than (or at least equal to) the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. In this case, you may assume that the estimated uncertainty of the final result $u(y)$ is approximately equal to the calculated standard error $\sigma_{\overline X}$.


Confused about parity of atom and chirality OK. I'm working in computational chemistry and the problem that bothers me is:

Given a molecule, I know that it has a central asymmetric atom with some parity which is either "Anticlockwise" or "AntiAntiClockwise" (or a really stupid way to say clockwise). I have the molecule drawn on a 2D canvas (they are in one plane).

For this example you could imagine 4 atoms. I have to specify which atoms lie in a different plane (3D perspective) and I can't seem to understand why is the parity important in that case? I'm using the SMILES O[C@@](F)(Cl)Br notation.

I'm really confused about this concept and it seems like the internet doesn't offer me a lot of information about it. So please tell me how is parity related to the plane the atom lies in? How can I map the edges and dashes to the SMILES  of this structure?

This is really a cheminformatics problem. It's a reflection of the chirality or atomic stereochemistry about the atom you mention.
There are multiple ways of specifying chiral information in 2D and 3D.
As mentioned by Dave SMILES uses a clockwise and counter-clockwise from left to right in the string.
In 2D, we usually use wedge and hash notation. The problem is that some users use wedges and hashes for graphical appearance and not for stereochemical information (e.g., pseudo-3D look). So file formats also add atom parity to indicate the actual stereochemistry of atoms.
The "parity" idea has to do with the relationship between the atoms as they are ordered in the file, vs. the stereochemistry around the central atom. As indicated in the other answer, these are enantiomers, even though they use the same "@@" SMILES stereochemical indicator:

$\ce{O[C$@@$](F)(Cl)Br}$
$\ce{O[C$@@$](F)(Br)Cl}$

Note that the atom ordering is different, so the "parity" will be different. For your example, $\ce{O[C$@@$](F)(Cl)Br}$ I get "1" for the MDL parity.
From the MDL specification (Appendix A: Stereo Notes):

Number the atoms surrounding the stereo center with 1, 2, 3, and 4 in order of increasing atom number (position in the atom block) (a hydrogen atom should be considered the highest numbered atom, in this case atom 4). View the center from a position such that the bond connecting the highest-numbered atom (4) projects behind the plane formed by atoms 1, 2, and 3.
Sighting towards atom number 4 through the plane (123), you see that the three remaining atoms can be arranged in either a clockwise or counterclockwise direction in ascending numerical order.

For more details, you might want to look at the Open Babel documentation, OpenSMILES or the MDL specification.


How to compare electrical conductivity of ionic liquids and aqueous electrolytes I am trying to compare the electrical conductivity of selected ionic liquids (ILs) and aqueous electrolytes.
By ionic liquid (IL), I mean a pure IL -- no water. Electrical conductivity of ILs is measured in units of Siemens per length. For example, the following table from Leys et al., J. Chem. Phys., 2008 gives the electrical conductivity of selected ILs in units of S/m:

On the other hand, aqueous electrolytes are obviously salts in water. I think, then, electrical conductivity of aqueous electrolytes are concentration-dependent. The following table in the CRC Handbook gives the electrical conductivity of selected aqueous electrolytes in units of S cm^2 / mol:

How can I compare the electrical conductivities of ionic liquids and aqueous electrolytes?
In other words, is there a way to relate ionic ($\sigma$ in S/m) and molar ($\Lambda$ in S cm^2 / mol) conductivities?

No, it is an apple to orange comparison. Ley et al. are changing the frequencies and in the molar conductivity measurements you do not know what was the current frequency. It is usually measured at 1 kHz if I remember correctly. Similarly, S/m cannot be changed to Scm^2/mol. You can indeed calculate the molarity of pure ionic liquids. The molarity of pure water is ~ 55 M. You should know the moles and the volume of the IL, but there is no point in calculating it.
You need to some heavy literature search. People have actually measured molar conductivity of ILs. Also note that many ILs are not water soluble so you have narrow down your choice and goals. Why are you comparing the conductivities to begin with? See this paper "On the concept of ionicity in ionic liquids" in Phys. Chem. Chem. Phys., 2009, 11, 4962–4967.
Also Walden plot might be useful for you.


Range of HMBC Coupling I need to determine the constitution of an organic molecule by several spectra like 1H NMR, 13C NMR, COSY, HMBC ,and HSQC.

Is it possible to get a HMBC crosspeak between the marked proton and carbon atom?

Typically, both two- and three-bond 1H–13C couplings appear in HMBC spectra, with three-bond couplings tending to be more prominent. The presence of other atoms in the path, e.g. oxygen, doesn't affect this. So, the three-bond coupling you have indicated is very likely to show up in the HMBC spectrum, as a cross peak at (f1, f2) ≈ (180 ppm, 4 ppm).

If all you are looking for is structural confirmation + assignment of 1H data (which would be typical if this was a product you expected from an experiment), I'd suggest that you don't necessarily need 2D data. The multiplet structures should contain more than enough information.


How to improve efficiency of the diamine synthesis from fluorescein? In this work, fluorescein was incorporated with an equal molar of 1-chloro-4-nitrobenzene and double potassium carbonate (molar ratio), the solvent of this reaction is dimethylacetamide (50 ml for 5 g of fluorescein).

The solution was stirred in at 110 °C for 24 h, after that cooled to the room temperature and precipitated in cold water, but a solid, similar to potassium carbonate, remains at the bottom of the reaction vessel, and the efficiency is very low (25%). I don't know the cause of this problem.

This is the structure that I expected for first stage:

In the next step the solid that was obtained from the previous stage, incorporated with 10% Pd/C and methanol. 30 eq hydrazine monohydrate was added dropwise and the solution refluxed at 50 °C for 12 h. After that the solution was filtered to separated catalyst and the crystals of diamine was obtained, but the efficiency of this stage is very low too.

What I think is happening is that you are not getting enough mixing between your reagents, the solid remaning is unreacted potassium carbonate. I think you can reduce the amount of potassium carbonate you are using - you only need 1 eq as it is dibasic, and use finely powdered if you have access to it. Run the reaction more dilute with a more powerful stirrer (have you an overhead stirrer?) so that material does not stick to the side and follow it by TLC. If this does not improve the outcome I would consider using a strong organic base such as DIPEA or DBU in place of the potassium carbonate so that everything is in solution.


Calculating the pH of a buffer made by a diprotic acid and its double salt The concentration of the diprotic acid (tartaric acid) will be constant at $\pu{ 0.1 M}$. I want to find the concentration of its double salt (potassium sodium tartrate) needed to create a buffer of $\pu{ pH }$=x (for example,x= 4.0). How would I do this? Do I have to use the Henderson-Hasselbalch equation? Make certain assumptions?

I tried to set up two equilibrium expressions with $K_\mathrm{a1}$ and $K_\mathrm{a2}$ as follows:

$$K_\mathrm{a1} =\frac{[\ce{C4H5O6-}][\ce{H+}]}{[\ce{C4H6O6}]}$$
and

$$K_\mathrm{a2} =\frac{[\ce{C4H4O6^{2-}}][\ce{H+}]}{[\ce{C4H5O6-}]}$$

Can I multiply $K_\mathrm{a1}$ and $K_\mathrm{a1}$ to eliminate$[\ce{C4H5O6-}]$, and then get the concentration of $\ce{C4H4O6^{2-}}$ necessary by plugging in $\pu{0.1M}$ for$[\ce{C4H6O6}]$ and the target $\pu{pH}$ in the appropriate form in $[\ce{H+}$]? I feel like I'm making some big assumptions here, though.

(Also, please forgive me; I don't know how to use LaTeX or Tex.)

If I try to make the assumption that the second dissociation does not occur as it is too weak (which it isn't) I wouldn't have the concentration of $\ce{C4H4O6^{2-}}$ in my equation, which is what I'm trying to find.

Can I multiply Ka1 and Ka1 to eliminate [$\ce{C4H5O6−}$], and then get the concentration of C4H4O62− necessary by plugging in 0.1 M for [C4H6O6] and the target pH in the appropriate form in [H+]

No, because $\ce{C4H5O6−}$ is a one of the major species. In fact, if you add the tartaric acid and its double salt at equimolar ratios, $\ce{C4H5O6−}$ will be the only major species. As a first approximation, you can let this reaction go to completion. If you have an excess of the acid, there will be a mixture of $\ce{C4H5O6−}$ and acid as major species. If you have the double salt in excess, there will be a mixture of the double salt and $\ce{C4H5O6−}$ as major species.

  Do I have to use the Henderson-Hasselbalch equation? Make certain assumptions?

If your ratio of reactants and products is such that you have a conjugate acid: base pair as the major species, you can use Henderson-Hasselbalch as the next step to estimate the pH. (If your major species are $\ce{C4H6O6}$ and $\ce{C4H5O6−}$, use $K_{a1}$, if they are $\ce{C4H5O6-}$ and $\ce{C4H4O6^2−}$ use $K_{a2}$.) In some cases, that's a good approximation, in others, not. You can test that by calculating the concentration of all species (based on the approximate pH) and then calculate Q for the two acid base equilibria to see how well they match K.


How to control the Friedel–Crafts alkylation reaction to methylbenzene as the major product? My question is due to a discrepancy I found in reactivity and organic conversions.
In reactivity of organic compounds it has been established that an alkyl group activates the benzene ring, and, hence for example, methylbenzene is more reactive than benzene.

Then consider the following situation. Suppose that there are two benzene molecules in a sample, and we convert one of them to toluene (methylbenzene) using Friedel–Crafts alkylation. Then we would see the following change in the mixture:

Now, consider we carry out alkylation once more. As toluene is more reactive than benzene, it will react first forming dimethylbenzene. By continuing this logic we would get hexamethylbenzene as the major product regardless of the amounts we add.

So, how is that we carry out conversions with Friedel–Crafts alkylation regarding toluene as the major product?

Feel free to correct me if my logic is incorrect and if so please specify what should be corrected.

The solution to this problem of having chance of poly-alkylation of Benzene rings by Friedel-Crafts alkylation has a general way, and the way is to first perform Friedel-Crafts Acylation and then reduction of the introduced Carbonyl carbon to saturated carbon atom.

Now, proceeding via this way helps to avoid all the problems that arise in a Friedel-Crafts alkylation. Firstly, after the first step, in the main product, we are having one Benzene ring attached to a carbonyl carbon , which deactivates the ring for participating in further electrophillic aromatic substitution and reduces any chances of poly-acylation which upon reduction also doesn't give any poly-alkylated product. Secondly, Friedel-Crafts acylation helps to reduce the possibility of rearrangement which occurs very often in the alkylation step. That's because the positive charge is not on any primary,secondary or tertiary carbon atom but on the acyl carbon which will not undergo any changes in its carbon skeleton structure. So, we can add the unrearranged carbon chain to the Benzene ring and reduce to get less-substituted carbon chains attached to the ring.

To solve your problem, I would rather prefer that you first react Benzene with a mixture of $\ce{CO}$ and $\ce{HCl}$ in presence of anhydrous $\ce{AlCl3}$ (Gatterman-Koch Aldehyde Synthesis) to get Benzaldehyde. Then you can perform a Clemmensen reduction ($\ce{Zn -Hg / conc. HCl}$) or a Wolff-Kishner reduction ($\ce{H2N-NH2/KOH}$) to get Toluene as the exclusive product.

Please note that, you can always control the reaction conditions to get Toluene by Friedel-Crafts alkylation only, still there remains a good amount of chance of getting di-substituted product as those kinds of controls need to be optimised and difficult to carry out and you have to perform purification techniques anyways. However, if you proceed via acylation and reduction, your side products will  be in lesser amount and product can be easily obtained.


Water and Ice - density We know that ice has a lower density than water despite both having the same [molecular] mass. I know that as water turns to ice, it expands. As far as I was taught, I know that it has something to do with hydrogen bonding. I would like to know in terms of hydrogen bonding, why ice actually expands?

The most straightforward answer is that water in the form of ice I (or $\ce{I_h}$) has an open hexagonal structure (see figure) whereas liquid water does not. In ice the oxygens are precisely tetrahedrally positioned where each oxygen is hydrogen bonded by four neighbouring oxygens with an O..O distance of approx 0.28 nm. The H atoms lie very closely along the O-O axes. Two of the H bonds are short, as in liquid water, and two are long. This is expected from residual entropy calculations and also from thermal and neutron diffraction experiments.

As this ice structure has a rather open lattice this is the reason its density is less at its freezing point than that of liquid water, rather than being due to the high density of water. One could argue that the reason that water density is a maximum at 4 $\ce{^0C}$ (277 K) is that the structure of  liquid water is changing from that of ice $\ce{I_h}$ to the short range structure of the liquid.

Liquid water is, of course, also highly hydrogen bonded but the extra energy it contains means that its structure is far less regular and so becomes slightly denser than ice. As the water molecules can move about more in the liquid than in ice, the hydrogen bonds will bend a little or break as molecules rotate. This will lead to a shorter O - O distance and hence slightly higher density in liquid than in the solid.

see also answers to this post Why really is ice less dense than water?

This figure shows the hexagonal ice structure; The O atoms are shown as dots and the lines are the (almost linear) O-H-O bonds.(Image from Murrell & Jenkins 'Properties of liquids & solutions')

(Ice also exhibits eight different phases depending on temperature  and pressure of which $\ce{I_h}$ is the one found at normal atmospheric pressure. In these other phases, angles are distorted from being tetrahedral)


Would 2 C₂ be 2 moles of diatomic carbon, or 4 moles of carbon? Would molecules like $\ce{C2, H2}$, etc., be considered 1 mole of diatomic [element] or 2 moles of [element]?

You can have 1 mole of $\ce{C_2}$ which contains 2 mole of $\ce{C}$ just as you can have for example 1 mole of $\ce{C_2H_5OH}$ which contains 2 mole of $\ce{C}$, 6 mole of $\ce{H}$ and 1 mole of $\ce{O}$.

The point is that you always have to mention whether you are indicating moles of molecule or moles of atoms. That is why a statement like: "I have 1 mole of oxygen" is ambiguous, because it could be referring to oxygen gas (i.e. $\ce{O_2}$) or to atomic oxygen (i.e. $\ce{O}$).


Effect of temperature on cooking I'm poaching some pears.  The recipe calls for bringing the liquid to a boil and adding the pears, then simmering until done, about 25 minutes.  It then calls for removing the pears from the liquid and letting both come to room temperature, then putting the liquid over the pears again and marinating overnight.  Separating the pears and the liquid will let the pears cool more quickly and presumably prevent overcooking.  I have heard a rule of thumb that every 10C in temperature doubles the reaction rate.  As the pears simmered for 25 minutes, it seems that the cooldown (even if they stayed in the liquid) will not add much to the reactions of cooking. Is there any sense to extracting the pears from the liquid?

(Note that I'm a rather bad cook, except in the lab!)

  Is there any sense to extracting the pears from the liquid?

Yes! Yomen Atassi has already explained that you actually have to consider two (distinct) processes in the recipe:

Cooking the pears for a short period of will yield the right consistency. Additionally, some transformations happening during the cooking will possible break up some cells, convert some ingredients to even more tasty ones, and maybe even facilitate the later take up of other of other compounds from the liquid. If you cook for too long, the pears will be sludge.
Marinating means that the compounds dissolved in the liquid will permeate the pear and add to the taste. Imagine that the pear acts like a spongue and sucks all the good stuff in from the marinade. This needs time.

With a two-step procedure given in the recipe, you should get the best of both worlds and I hope that the dessert was really tasty.


Stabilization of ipso and para position in Birch reduction by EWG of aromatic ring I'm trying to grasp the Birch reduction. In the lecture about it, it's said that the EWG group stabilizes ipso and para position through resonance. I tried to understand that by drawing resonances without success. What is wrong in it?

TL;DR: This question may not be readily answered with Lewis structures.

The problem is the electron is delocalized into an antibonding orbital.  By analogy, we can examine this concept with molecular oxygen.  The Lewis structure for molecular oxygen:

This clearly is a double bond, with all spins paired.  Now observe the molecular orbital diagram below:

The top energy levels clearly show oxygen with two unpaired electrons, which contradicts the Lewis structure completely.  There is no real way to simultaneously convey these two situations.

Similar diagrams exist and may be used for the aromatic system with an electron-withdrawing group exist, but they are rather complicated.


Polarity index vs. Dipole moment I'm looking to find whether water or methanol is more polar, and I'm getting conflicting answers.

My textbook says "the polarity of a bond is quantified by the size of its dipole moment." According to this table, at 20 °C, the dipole moments of methanol and water are 2.87 D and 1.87 D. This implies to me that methanol is more polar.

But there seems to be another measure called "polarity index," which I can't seem to find a definition for online. This table says the polarity indices of methanol and water are 10.2 and 5.1, which implies to me that actually water is more polar.

So what's the difference between the dipole moment and the polarity index? How is the latter defined? And is methanol or water more polar?

From Roger E. Schirmer "Modern Methods of Pharmaceutical Analysis":

  Solvents are generally ranked by polarity, but polarity is not a uniquely defined physical property of a substance. Hence the relative polarity of a solvent will be somewhat dependent on the method used to measure it....

  Solvent polarity is a complex function of many parameter in addition to adsorption energy. A more recent ranking of solvents by Snyder is based on a combination of parameter such as dipole moment, proton acceptor or donor properties, and dispersion force solvent...

  Snyder's polarity index ranks solvents according to a complex theoretical summation of these properties. As a rule, the higher the polarity index, the more polar the solvent.

Snyder's paper was published in   Journal of Chromatography A, Volume 92, Issue 2, 22 May 1974, Pages 223-230 . Take a look at it to see how the index was calculated.

There are different types of polarity indices. And each of them has different parameters and ways to calculate the polarity.

Unlike polarity, dipole moment is a physical property. Polarity indices often take dipole moments as a parameters when calculating the polarity of solvents.  The reason why there are polarity indices is because dipole moments alone couldn't explain the nature and interactions of solvents.

Part of your confusion stems from the repetition of the word "polar" to describe different phenomena. There is polarity of bonds and there is polarity of solvents. They are different things.


What would happen if I mixed a single water molecule in a beaker of hexane? What would happen if I mixed a single water molecule in a beaker of hexane? Would it sink or float?

How about 2 water molecules? 3? Etc?

In other words, how many water molecules does it take to be hydrogen bonding together to become more dense than hexane?

This is a clever question because the molecular weight of water is only 18.01528 g/mol while the molecular weight of hexane is 86.17536 g/mol. Naturally, you would expect the lower mass to rest on top, but water has a higher density... and we all know that water will sink! But there are other forces at work.

So, supposing that you neglect the fact that one molecule of water would be in a vapor-state, Google says the surface tension of Hexane @ 20 °C is 18.43 mN/m... so placing the water molecule ontop of the hexane would probably cause the molecule to float (like a feather on water).

However, if you suberged the molecule it should dissolve (Google says that the solubility of water in hexane is 0.01% at 20°C). Now, once the saturation point of hexane has been met, such that one more single molecule of water would cause a change to occur, that change would probably result in a condensation and precipitation of the dissolved water, because the reality is, that the hexane will become super-saturated, such that the extra water molecule will cause many water molecules to fall out of solution.

Understanding that you are transferring energy is key to understanding this. Consider pushing a ball over a hill, once the ball reaches the top and begins to fall, it falls into the valley. Once you push a boat to the tipping point, it falls to the bottom.

So the number of molecules necessary for hydrogen bonding together to become more dense than hexane would be the number of water molecules dissolved at the maximum supersaturation point (maybe 0.0105%, +1 molecule) minus the number of molecules at the saturation point (0.01%)

Now having a saturated solution, with water on the bottom of the container, adding another molecule of water would generally cause one of the other dissolved molecules to fall out solution (because there is an equilibrium... some of the water molecules at the bottom are dissolving, while others are falling out of solution).


Help me proving Dalton's law of partial pressure formula How do I prove

  $${{\displaystyle p_{i}=p_{\text{total}}y_{i}}}$$

where $y_i$ is the mole fraction of the $i^{th}$ component in the total mixture of $n$ components ?

Here is what I know

Mathematically, the pressure of a mixture of non-reactive gases can be defined as the summation:

${\displaystyle p_{\text{total}}=\sum _{i=1}^{n}p_{i}}$

or

${\displaystyle p_{\text{total}}=p_{1}+p_{2}+p_{3}+...p_{n}}$

And mole fraction $y_i= \dfrac{n_i}{n_{\text {total}}}$

I don't see how to relate , please help.

After having some discussion with @JohnRennie

Let $P_i=n_i \dfrac{RT}{V}$ then $P_i=\dfrac{n_i}{n_{\text{total}}}\dfrac{n_{\text{total}}RT}V$

$P_i=x_i P_{\text{total}}$

Where $x_i$ is mole fraction of $i^{\text{th}}$ particle.


How to calculate current density in an Electrolysis Process? I have an anode & cathode, both of $100 \times \pu{100\\mm}$ size. The process requires $\pu{450\\A/m2}$ current density. How do I reverse calculate this? Do I use the total surface area of both Cathode and Anode or just Anode?
Please help.

Thanks

Do I use the total surface area of both Cathode and Anode or just Anode?

You use the cross-section, just like it were a wire.

  Cross-section of both Anode and Cathode or just Anode?

Just one. Like in a wire, you would measure the cross-sectional area once, not cut the wire into many piece and add up all the areas.


Quantitative relationship between corrosiveness of acid and other properties Is the corrosive nature of acid relative to some substance quantifiable like acid dissociation constant ($K_\mathrm{a}$)? If yes, what can be used to describe the corrosive nature of an acid?

For example, it is known that citric acid is far less corrosive than, say, sulfuric acid. But, can it be quantified? Also, if corrosiveness can be quantified, then can the caustic nature of bases be quantified too?

In addition to the acid dissociation constant, some other factors that add to the corrosiveness of an acid are its ability to act as an oxidizing agent and its affinity for water (hence, ability to dehydrate material). Concentrated sulfuric acid has both these abilities in addition to simply being a strong acid. I do not know if these factors have been quantified, but they could be.

Mixing chromic oxide ($\ce{CrO3}$) with sulfuric acid makes it an even better oxidizing agent, especially when hot - so-called - "cleaning solution" is great in the lab for removing residues of tar from your glassware, but it is rather dangerously corrosive and has to be handled very carefully.


Enthalpy of dissolution of NaOH in small amount of water is there a way to calculate enthalpy of dissolution of NaOH in small amount of water?
I know that enthalpy of dissolution of sodium hydroxide in water is –44,5 kJ/mol. However it assumes the orders of magnitude higher amount of water than NaOH. (Which is not my case.) Is there any formula which does not omit the volume of solvent? (Or a tool which can answer questions such as: "How much heat is released if you make 60% solution of 80% solution?")
Thanks for help.

The dissolution enthalpy for concentrated final solutions highly depends on the dissociation and hydration degree, that varies a lot with the substance and concentration.
Therefore particular values for concentrated solutions have to be either measured experimentally, either calculated by some empirical formula, derived from experimental data ( if available ).


Compare stability of the given carbocation Give the order of stability of the given carbocations.

Usually stability is determined by studying effects taking place in the carbocation. All are resonating and there is no diff in inductive effect.So I want to know how the molecular weight might affect stability

The question is pretty easy though the INDUCTIVE and MESOMARIC effects are same but there is a diffrence in the HYPERCONJUGATIVE Effect as in HYPERCONJUGATIVE Effect the bond between $C-H$ breaks to provide stability thus any change in bond energy of the bond to be broken changes the strength of effect.

Now as the bond strength of $C-T$ bond if highest hence the HYPERCONJUGATIVE Effect of $\ce{CT3}$ decreases and hence the carbocation is least stable.And $\ce{CH3}$ is most stable

You can read more about Hyperconjugation in the following Site:
https://chemistry.tutorvista.com/inorganic-chemistry/hyperconjugation.html


Why does Dead Sea water feel greasy? The best explanation I found so far is this:

  These salts (magnesium, potassium, sodium, etc.) are responsible for the "greasy" feel of the water.

But it doesn’t say which salts exactly are responsible, and why it feels so different from seawater.

I don't know what you mean by the term "greasy", but here's a possible explanation:

When Dead Sea water evaporates on skin, then salts will precipitate when their maximum solubility is reached.

The mixture of tiny salt crystals formed, and water might be responsible for a greasy sensation.


Heat capacity of mixtures After reading this recent question I was interested in how to calculate the specific heat capacity of a mixture based on the specific heat capacities of its components.
According to this website the specific heat capacity of an ideal mixture is given by

$$C_{p(\mathrm{mixture})} = \sum\limits_iC_{p(i)}x_i$$
where $x_i$ is the mole fraction of each component.

Where does this come from and what are the assumptions involved in treating it as an ideal mixture?
What deviations from this are observed in non-ideal mixtures? Can you easily predict the deviations based on the mixture and/or predict the actual specific heat capacity?

Suppose you have $n_1$ mole of A and $n_2$ mole of B. If they are at temp. $T_1$ and you want to raise the temp. to $T_2$ then the amount of heat you need to add is
$$\Delta H~= n_1 \times c_{pA} \times \Delta T + n_2 \times c_{pB} \times \Delta T$$
$$\Delta H~= (n_1 \times c_{pA} + n_2 \times c_{pB}) \times \Delta T$$
$$\Delta H~= (n_1 +n_2)\times \frac{(n_1 \times c_{pA} + n_2 \times c_{pB})}{(n_1+n_2)} \times \Delta T$$
$$\Delta H~= n_{total} \times(y_1 \times c_{pA} + y_2 \times c_{pB}) \times \Delta T$$
$$\Delta H~= n_{total} \times c_{p,avg} \times \Delta T$$
So basically you are not losing any information in case of ideal situation where molecular interaction of two different species has no effect on each other's heat capacity. But from statistical mechanics we know,
$$C_v=\frac { <E^2>-<E>^2}{k_B \times T^2}$$
Now if we solve the schrodinger equation for individual species, we must include an extra potential energy term to account the interaction from other species, which in turn will result in fluctuation of $C_v$ from it's previous value.

Slightly off topic: When I see the question first things that came to my mind is kopp's rule. It's used to find heat capacity of molecule using atom's heat capacity or to find alloy's heat capacity using the constituents heat capacity.


How to refine manganese metal from manganese(IV) oxide found in batteries? I opened up some lantern batteries, and inside there's $\ce{MnO2}$ powder, carbon rods, and zinc casings, and other stuff that I don't know what it is exactly. I scraped out $\ce{MnO2}$-containing powder, but I would like elemental manganese from it.
How can I obtain manganese from manganese dioxide? How strong a reducing agent do I need? How hot does it need to get? Can I just add charcoal powder and smoke it with an oxyacetylene torch?

You may heat mixture of $\ce{MnO2}$ with charcoal to higher than $\pu{850 °C},$ in the absence of air. Manganese will be produced in the solid state by reduction of the oxide. But it is sensitive to air oxidation and easily reoxidized. So better use an excess of charcoal. In the industry, rough $\ce{MnO2}$ out of the mine is first reduced to $\ce{MnO}$ by $\ce{CO}$, then leached and reduced into $\ce{Mn}$ metal by a solution of $\ce{Fe^{2+}}$ in the absence of air. Metallic manganese is then dissolved in sulfuric acid and then reduced by electrolysis. See "Manganese metal" over Wikipedia.


What is the exact definition of the radial distribution function? I have been very confused by the radial distribution function which is often used in chemistry to predict the probability of finding an electron at a distance from the nucleus.

From Atkins' Physical Chemistry:

  Now consider the total probability of finding the electron anywhere between the two walls of a spherical shell of thickness $\mathrm dr$ at radius $r$. The... volume of the shell... is now $4\pi r^2 \mathrm dr$. ... The probability that the electron will be found within the inner and outer surfaces of the shell is the probability density at the radius $r$ multiplied by volume... i.e $|\psi|^2\times 4 \pi r^2 \mathrm dr$. $$P(r) = 4\pi r^2|\psi(r)|^2$$ The function $P(r)$ is called the radial distribution function (in this case, for an s orbital).

Immediately after this, I find:

  It is also possible to devise a more general expression which applies to orbitals that are not spherically symmetrical. ... $$P(r)=r^2R(r)^2$$

From Housecroft and Sharpe's Inorganic Chemistry:

  A useful way of depicting the probability density is to plot the radial distribution function ... $$\text{Radial distribution function} = 4\pi r^2R(r)^2$$

Again, from this page, the probability density turns out to be (for the 1s orbital) $$P(r)=4\pi r^2|R(r)|^2$$ However, the integral that the answerer calculated $\bigg(\int \limits _{\theta =0}^{\pi }\int \limits _{\varphi =0}^{2\pi }|Y(\theta ,\varphi)|^2\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi \bigg)$ to get $4\pi$ (for 1s orbital) is calculated to be 1 (for all orbitals) in Atkins', as "the spherical harmonics are normalized" (I don't know what that signifies).

My confusion is compounded by the fact that there is another function called the probability density which is $|\psi|^2$. This function is maximum at the nucleus and falls exponentially (for an 1s orbital). The radial distribution function, however, is maximum at a certain distance from the nucleus.

So, my questions are:

What is the physical significance of probability distribution and that of radial distribution function?

There is a similar question here but the answer does not satisfy me. If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?

What is the correct expression for calculating radial distribution function?

Which begs the question, what is the definition of radial distribution function?

The atomic orbitals (wavefunctions) $\psi(r,\theta,\phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(\theta,\phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.

As a concrete example, the $2\mathrm p_z$ orbital ($\{n,l,m\} = \{2,1,0\}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:

$$R_{2,1} = \sqrt{\frac{1}{24}}\left(\frac{Z}{a}\right)^{3/2}\rho\mathrm e^{-\rho/2} \qquad Y_{1,0} = \sqrt{\frac{3}{4\pi}}\cos\theta \tag{1}$$

where $a$ is a collection of constants ($a \equiv 4\pi\varepsilon_0\hbar^2/\mu e^2$), and $\rho$ is related to $r$ by another collection of constants ($\rho \equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2\mathrm p_z$ orbital.

The atomic orbitals themselves are normalised in the sense that

$$\int_{\text{all space}} |\psi(r,\theta,\phi)|^2 \mathrm dV = \int_0^{2\pi} \!\!\!\!\int_0^\pi\!\!\!\int_0^\infty |R(r)|^2|Y(\theta,\phi)|^2 r^2\sin\theta\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi = 1  \tag{2}$$

Conventionally, both the radial and angular wavefunctions are separately normalised. This means that

$$\int_0^\infty |R(r)|^2 r^2\mathrm dr = 1; \qquad \int_0^{2\pi}\!\!\!\!\int_0^\pi |Y(\theta,\phi)|^2 \sin\theta\,\mathrm d\theta\,\mathrm d\phi = 1  \tag{3}$$

(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)

Normally, since $\psi = \psi(r,\theta,\phi)$, the probability density $|\psi|^2$ depends on all three coordinates. The probability density (technically, $|\psi|^2\,\mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube frustum-like shape,† located at the point $(r,\theta,\phi)$. By summing these $|\psi|^2\,\mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.

However, often we are not interested in the angular dependency, and so what we can do is to sum up all these bits which have one value of $r$, but different values of $\theta$ and $\phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $\theta$ and $\phi$. By virtue of the angular wavefunction being normalised, the integral over $\theta$ and $\phi$ evaluates to $1$, and hence we have

$$|\psi|^2\,\mathrm dV = (|R|^2r^2\,\mathrm dr)\cdot (|Y|^2\sin\theta\,\mathrm d\theta\,\mathrm d\phi) \overset{\text{integrate over }\theta,\phi}{\longrightarrow} |R|^2r^2\,\mathrm dr \tag{4}$$

So, this quantity $|R|^2r^2\,\mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.

This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $\theta$ and $\phi$, it comes out to be $1$. The widely-cited alternative definition $4\pi r^2 |\psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have $\{l, m\} = \{0,0\}$ and

$$Y_{0,0} = \sqrt{\frac{1}{4\pi}}  \tag{5}$$

(you may check that this is normalised according to equation $(3)$.) Hence

$$|\psi|^2 = |R|^2|Y^2| = \frac{1}{4\pi}|R^2| \tag{6}$$

and it therefore follows that $|R|^2r^2 = 4\pi r^2|\psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4\pi r^2|\psi|^2$ is used (or $4\pi r^2 \rho$, where the probability density $\rho$ is simply $|\psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.

Based on the above arguments this is my take on the matter:

$|R^2|r^2$ is the best definition of the RDF;
$4\pi r^2|\psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);
I still don't fully understand why Housecroft has an additional factor of $4\pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4\pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $\sqrt{1/(4\pi)}$.

† Thanks to Ruslan for pointing this out! If you are curious as to what this looks like, there is a good diagram on LibreTexts, which I have re-uploaded to SE here. Note, however, that the convention in this diagram is $(\rho,\phi,\theta)$ instead of $(r,\theta,\phi)$.


Is (S)-3-hydroxypalmitoyl-CoA a hydroxy fatty acyl-CoA? Is there anything special about (S)-3-hydroxypalmitoyl-CoA comparing to those with different number of carbons, e.g. (S)-3-hydroxyoctanoyl-CoA, (S)-3-hydroxydecanoyl-CoA, (S)-3-hydroxylauroyl-CoA?

I am not a chemist at all; I am looking at the ChEBI ontology and wondering why (S)-3-hydroxyoctanoyl-CoA, (S)-3-hydroxydecanoyl-CoA, (S)-3-hydroxylauroyl-CoA are descendants of hydroxy fatty acyl-CoA, while (S)-3-hydroxypalmitoyl-CoA is not (it is only a descendant of long-chain fatty acyl-CoA).

This just looks like an arbitrary distinction they made in the ontology between 'short chain' acyl-CoAs and 'long chain'. As you can see, the 3-hydroxypalmitoyl chain is somewhat longer (16C) than the others. It stands to reason that the creators of the ontology have decided to subdivide hydroxy fatty acyl-CoAs into those with shorter hydroxyl fatty chains:

hydroxyoctanoyl:   8C
hydroxydecanoyl:  10C
hydroxylauroyl:   12C

and long:

hydroxypalmitoyl: 16C

As to where the dividing line is, the hydroxytetradecanoyl-CoA (14C) homologue is not 'long chain', whilst there is no entry for hydroxypentadecanoyl-CoA. The octadecanoyl homologue belongs to its own class (of 3-hydroxyoctadecanoyl-CoAs, of which there are only two conceivable enantiomeric members), and there does not appear to be icosanoyl homologue in the database. There may be other members of the 'long chain' family if you search around, who knows?

In short, it appears to be arbitrary.


Why is the addition of methanol during the prepartion of the liquid crystal cholesteryl benzoate necesary Here's one way to prepare cholesteryl benzoate crystals:

Step 1 Dissolve 1.0 g of cholesterol in 3 cm3
 of pyridine in a conical flask.

Step 2 Add 0.40 cm3
 of benzoyl chloride.

Step 3 Heat the mixture on a steam bath for about 10 minutes.

Step 4 Cool the mixture, and add 15 cm3
 of methanol.

Step 5 Collect the solid cholesteryl benzoate by suction filtration. Rinse the flask and the crude crystals with a little cold methanol.

Step 6 Recrystallize the cholesteryl benzoate using ethyl ethanoate as the solvent.

The methanol is said to be added to react with the benzoyl chloride. Why is this required?

Usually, in a preparation such as that of 1-bromobutane, some substance such as sodium carbonate is added in order to allow for separation of the aqueous substances from the organic layer. That makes sense. However, in the case of the above preparation, by adding methanol the substances within the conical flask go from benzoyl chloride, pyridine and cholesteryl benzoate to HCl, pyridine, methyl benzoate and cholesteryl benzoate. What's the point? Either way, both mixtures are going to eventually be separated from the cholesteryl benzoate by suction filtration.

Pyridinium hydrochloride and methyl benzoate are both soluble in methanol and thus will be washed through the filter leaving behind clean cholesteryl benzoate. If you don't use the methanol the solid you produce will be impure and sticky; the methyl benzoate and pyridinium hydrochloride will not pass through the filter but will bind to the product crystals.. Methanol also removes any excess of benzoyl chloride by reacting with it.


How can I interpret the colour of flames in terms of spectral series? I saw different colors of fire burning different metals. But how does this work? The Balmer series requires electrons to drop back down to n = 2. But, not all alkali metals have empty space in their second principle energy level. Does one electron, from the second principle energy level, just jump to somewhere of energy and come back down.

My description is kinda confusing, but all I'm  saying is, where does that space in the second energy level come from?

The Balmer series relates to the electronic transitions in the hydrogen spectrum. In other elements it only requires that the energy difference between different excited states corresponds to a wavelength in the visible spectrum. These will clearly differ depending on the effective nuclear charge of the elemnt concerned.


Is there a blanket term for all variants (both nuclides and ions) of a chemical element? I'm hoping to catalogue the various nuclides and oxidation states of an element but can't find a single term that describes them all, save for "chemical element variations". Is there a specific term used for this?

Yes, for example consider selenium. You can say "selenium species".


Adding ethanol to louche I'm watching MIT chemistry by Donald Sadoway. In one of his lectures devoted to solutions and phase separation, he performs experiments with absinthe.
First he mixes absinthe with 5 $\times$ water which turns into milky louche.

Then he adds some cognac and the mixture becomes transparent again. I don't grasp his explanation of this.

Here is a timestamped link to the video.
I'll also briefly summarize his argument below:

  ... and why did he add cognac? ... it's cause if you have a fat phase here and you've got an aqueous phase here and if you add alcohol, you got

  $(fat)\:\textrm{CH}_3 \textrm{CH}_2 \textrm{COOH} \: \dot{}\dot{}\:\textrm{OH}_2$

  ... this can bond to the water by hydrogen bond and this aliphatic tail can stab the fat and bring them into solution. This is why you have these recipes...

What does "stabbing the fat" mean?

Does it mean that the ethanol somehow makes the surface between fat and water disintegrate thus turning the mixture into a single phase solution?
Or maybe ethanol + water + fat somehow  combine to form a single molecule so that, again, there is a single phase solution now?

Absinthe is a strongly alcoholic beverage containing anethole, which is insoluble in water and very soluble in ethyl alcohol.  When water is added to the absinthe, the alcohol becomes too dilute to keep the anethole dissolved, and it appears milky. When more ethyl alcohol is added in the form of cognac, the solution again becomes strong enough in alcohol to dissolve the anethole and the solution becomes clear again.

The phrase "stab the fat" is a bit odd, but I'll do my best to explain. Anethole is a fat-like or fat-loving oil, meaning it dissolves well in other fatty sorts of compounds, but not in water (which is lipohpobic, or fat-avoiding/fat-fearing). Ethyl alcohol has two parts to it. One end of it (labeled (fat) in the chemical formula in the question) is lipophilic (attracted to fat) and the other end is hydrophilic (attracted to water). If there is enough alcohol in the mixture, the lipophilic end of the alcohol molecules can surround the anethole ("stabbing the fat") which leaves the other end of the alcohol free to be dissolved in the water (and also any excess hydrophilic portions of the alcohol).

I hope the "stab the fat" explanation was clear. Don't hesitate to ask for specific clarifications in the comments below.


Lambert-Beer's law coefficient While the Beer's law equation is $A=a \cdot b \cdot c$, according to my book, $a$ is a coefficient which is:

  a characteristic property of the analyte in a specific solvent, and its values will depend on the wavelength of absorption radiation.

This is quite unclear to me.

If we have for example  $a= 1.5$, what does that exactly tell us ? What do we know when this coefficient is given ? Or is this just a constant that helps us to solve the equation?

To start with, I recommend to reformulate Beer-Lambert's law in a way more frequently met in (analytical) chemistry, i.e.

$\mbox{ABS} = c \cdot{} l \cdot{} \varepsilon = \log_{10} \frac{I_0}{I}$

because then it more obvious that the (dimensionless) absorption (ABS) recorded is the product of analyte concentration ($c$), optical path length ($l$) across the sample investigated, and the molecular extinction coefficient ($\varepsilon{}$).  Absorption equally is defined as the logarithm (to base of ten) of the fraction of incoming light intensity ($I_0$) and light intensity that is transmitted across the sample ($I$).

Now to your question.  While concentration $c$ and optical path length $l$ may be changed (the later, for example, by exchanging a measurement cell of 1 cm path length to one of 2 mm or 5 cm -- depending on the setup of the spectrometer), the molecular extinction coefficient $\varepsilon$ is indeed more valuable than just assistance to solve the equation.  It relates to the ease the analyte may absorb irradiation.

UV-Vis spectroscopy investigates electronic transitions, prominently of $\pi{}$-systems (like of alkene double, and aromatic bonds), yet investigation of electronic transitions of $n$- (non-bonded, like of carbonyl-O electrons), $d$- (for example in transition metal complexes) and $\sigma{}$-electrons (in the short wavelength UV-range).  In principle, these transitions are discrete, i.e. only photons of this attuned energy ($h\nu{}$) that fits the energy gap between the electronic ground state and the excited state are absorbed.

An analyte molecule may possess more than one allowed electronic transitions.  They may occur simultaneously, with different efficiency.  For example, many organic dyes have a strong absorption in the visible range, like indigo has one around 600 nm.  Yet if you tune the wavelength of investigation to short wavelengths, below 250 nm, for example, absorption may still occur, too, yet to a lesser extent.  (This eventually contributes to processes that indigo stained cloths fade over time, for example.)

Side note:  The molecular extinction coefficient $\varepsilon$ depends on other, additional parameters than the "solvent" and "wavelength of investigation".  Varying the pH-value, ionic strength of the solution, presence/absence of other dissolved molecules than of your analyte and solvent may alter $\varepsilon{}$, too.  Do not forget, the concentration $c$ of your analyte may influence $\varepsilon{}$, too -- pointing to (concentration dependent)  potential dimerisation / aggregation of your analyte (for example charge transfer complexes).


Specific heat capacity and the periodic table Can someone explain me how can I use the periodic table in order to find the specific heat capacities of some elements? (for example, of Aluminium )
The question I have encountered is as follows:

Given the specific heat capacities of the following liquids, use the periodic table in order to find the specific heat capacities of the metals $Al, Fe, Cu, Au $ .
$ H_2 O - 4.18 J  /g \cdot C $, $ C_2 H_5 OH - 2.46 $ , $CCl_4 - 0.861 $ , $CCl_2 F_2 - 0.598 $ .

Either you have a periodic table that features the specific heat capacity of elements in their standard state at STP (standard temperature and pressure), in which case you merely have to look it up, or you don't, and there's nothing you can do.

Specific heat capacity of materials, even elemental solids, are physical properties resulting from complex behavior of the systems, and cannot be determined by simple laws or equations. They have to be determined experimentally, or by first principles calculations.

PS: water, methanol, carbon tetrachloride and dichlorodifluoromethane have got nothing to do with it.


What is the origin of the differences between the MO schemes of O₂ and N₂? Here are the MO schemes of $\ce{N2}$ (left) and $\ce{O2}$ (right).

Why is the $\sigma$-MO formed by the $p$ AOs energetically above the $\pi$-MO for $\ce{N2}$ but not for $\ce{O2}$?

Can it be explained this way:
In second period elements with more than a half filled $\ce{p}$ orbitals, the energy pattern of the MOs is regular.
In second period elements with less than or half filled $\ce{p}$ orbitals, as the $\ce{s}$ and $\ce{p}$ atomic orbitals have similar energies, so the formed $\ce{\sigma_{g}}$ MO would have similar energy as $\ce{\sigma_{u}}$, but because of electron cloud repulsion it has a big increase in energy, hence it is above the $\pi$ orbitals? And this phenomenom is known as sp mixing.

Or am I wrong somewhere? Is my reasoning corret?

This phenomenon is explained by s-p mixing. All the elements in the second period before oxygen have the difference in energy between the 2s and 2p orbital small enough, so that s-p mixing (combination) can occur lowering the energy of the σ(2s) and σ*(2s) and increasing the energy of the σ(2p) and σ*(2p) molecular orbitals. By moving towards right in a period, the s orbital gets more stabilized than the p orbital and the difference in their energies increases, making the s-p mixing for oxygen much smaller.


How can I calculate the standard reaction enthalpy if I lack one formation enthalpy? I have this reaction, for which I need the standard enthalpy of reaction:

$$\ce{CuO (s) + H2S (g) \rightarrow CuS (s) + H2O (g)}$$

This is the data from the NIST web book (standard formation enthalpies, $\mathrm{kJ}\cdot\mathrm{mol}^{-1}$):

$\ce{CuO (s)}$: -156.60
$\ce{H2S (g)}$: -20.5
$\ce{H2O (g)}$ -241.83

There are no data tabulated for $\ce{CuS (s)}$, so how can I compute the standard reaction enthalpy lacking this data?

The standard enthalpy of formation of $\ce{CuS}\,\mathrm{(s)}$ is -48.4 $\mathrm{kJ}\cdot\mathrm{mol}^{-1}$. See this reference  for your requested data and more.


Ionic compounds and van der Waals forces I'm having trouble understanding exactly what intermolecular forces occur in ionic compounds.

What I understand is that ionic compounds are held together by electrostatic interactions between opposite charged ions and that they aren't "finite" in the way covalent molecules are, and therefore they do not have intermolecular forces.

However, while my textbook says that only covalent molecules have van der Waals forces, my professor said that all compounds, even ionic compounds have London dispersion interactions.

Isn't London dispersion a van der Waals force? How does that work?

London dispersion forces are instantaneous dipole-induced dipole interactions. All compounds can have an instantaneous dipole moment, including ions, so London dispersion forces exist for all compounds.

Unlike molecules, ions are charged. The electrostatic interactions between charges are far stronger than the London dispersion forces between ions, so normally dispersion forces are ignored when considering ions.

Therefore, both your textbook and your professor are correct.


steam distillation of 4-bromoaniline , why render it basic prior In the multistage synthesis of 4-bromobenzenamine, why must the solution be rendered alkaline prior to the steam distillation after the hydrolysis of 4-bromo-N-ethanoylbenzenamine?

In principle, it is difficult to distill salts (ionic compounds). Therefore you first make sure that you have a free base in the solution. This is guaranteed in alkaline solution. For the steam distillation the compounds should be immiscible, which is another reason for having a free base. Let’s compare boiling points and solubility for aniline and its salt:

Aniline b.p. 184 °C, solubility in water: 36 g/L
Anilinium chloride b.p. 245 °C, solubility in water: 1070 g/L


When Iron reacts with any other species, how will we know that the reaction will be of Fe(II) or Fe(III) i.e. of ferrous or ferric? Like when iron reacts with water there forms iron oxide and hydrogen but how'll we come to know which of three equation will be right equation for reaction?
$$\ce{Fe + H2O → FeO + H2}$$
$$\ce{2Fe + 3H2O → Fe2O3 + 3H2}$$
$$\ce{3Fe + 4H2O → Fe3O4 + 4H2}$$

The stability of the various iron oxides depends on the oxygen fugacity (or oxygen partial pressure, which is a good approximation to fugacity in near-atmospheric pressures).
$\ce{Fe^3+}$ is the stable form of iron in atmospheric oxygen partial pressures, this in such conditions $\ce{Fe2O3}$ (hematite) will always be the stable product. The lower oxidation state oxides (such as magnetite $\ce{Fe3O4}$ and wüstite ~$\ce{FeO}$) can only stably form in deep earth environments, or in special controlled atmosphere reaction chambers.
That said, in normal conditions it is even more likely for iron hydroxides to form hydrated compounds, especially when water is a reactant, or is present. In these cases, you get a zoo of various Fe-oxides-hydroxides, some of which are amorphous (not crystalline), and others are metastable. These compounds are generally referred to as "rust".


Kinetic Theory of Gases and molecular velocities of gases The assumption that molecules collide with elasticity was postulated in the kinetic theory of gases.
So my question is molecules should have higher tendency to collide obliquely and would therefore result in change of velocity (as the vector components would cancel or add up the velocities of molecules, resulting in change of velocities of molecules) and energy states of electron, are the assumptions not very correct or am I wrong?

Inelastic collisions are such collisions, that dissipate mechanical energy to other energy forms. It may be e.g. acoustic energy, light, thermal energy ( in macro context only ) or internal energy of an object.

The change of the internal energy in molecular context can be creation/breaking of chemical bonds, or changes of electron energy states. In usual conditions of stable enough gases, these events are very negligible, as average kinetic energy is much smaller than the energy needed to such a change.

Be aware that even fully elastic collision generally leads to a change of the mechanical energy of an object, in whole, or its translational, vibrational and rotational parts.

Imagine 2 identical balls moving at the same speed on a colliding course. After the collision, they have still identical speed if and only if their collision was symmetric. Otherwise 1 ball will slow down and the other ball will speed up.

The justification of the kinetic gas theory is the best shown on very good agreement of prediction and observed behavior of permanent gasses at normal conditions.


Acid base titration curve Why we don't get sharp point at equivalence points in case of acid base titration other than the case of strong acid and strong base titration?

In the case of strong acid-strong base titration at equivalence point with the addition of very small amount of titrant (like a drop) the pH change is very large. That means the range over which indicator changes its color is likely to be passed in a drop of titrant, resulting in a sharp end point.

Whereas in the other cases the change of pH at equivalence point is less dramatic and there will be a smaller change in pH at equivalence point. So, the indicator will change its color slowly resulting in a wider end point.


Why is the active space for the dinitrogen cation defined this large in my example? I have a Molpro script which computes $\ce{N2^+}$ potential energies.

The significant part is the input for the wave function:

 {multi;
   occ,3,2,2,0,3,2,2,0;
   closed,1,0,0,0,1,0,0,0;
   wf,13,5,1;
   state,3;
 }

Why is the active space specified by OCC flag so large? It specifies 14 molecular orbitals, which means that both nitrogen should habe 7 atomic orbitals. Which atomic orbitals are additionally included?

The ground states of $\ce{N2^+}$ molecule has this electron configuration:

Thich can be specified as occ,3,1,1,0,2,0,0,0;.

What advantage does occ,3,2,2,0,3,2,2,0; have in the script and how can I understand and visualize those additional orbitals?

Explanation of OCC card

The OCC card in Molpro specifies symmetries of the orbitals in active space. In this case, they're specified in $D_\mathrm{}$ point group, according the following table:

$$D_\mathrm{2h}\\
\begin{array}{ccc} \hline
\text{No.} & \text{Name} & \text{Function}\\
1& \mathrm{A_g}    & s  \\
2& \mathrm{B_{3u}} & x  \\
3& \mathrm{B_{2u}} & y  \\
4& \mathrm{B_{1g}} & xy \\
5& \mathrm{B_{1u}} & z \\
6& \mathrm{B_{2g}} & xz \\
7& \mathrm{B_{3g}} & yz \\
8& \mathrm{A_u}    & xyz \\
\hline
\end{array}
$$

The active space is a truncation of the full CI space. Hence, including more virtual orbitals will lower the energy and eventually approach the FCI limit. The larger the active space, the more accurate the results.

The smaller active space will give you a good qualitative description.
But if your results turn out to be quantitatively wrong, increasing the active space is one way to improve them.

To get an idea about the higher orbitals, take a close look to the Molpro output. You might want to check the Print options, to get all desired energies and orbital coefficients. Furthermore, you can use for example put to create molden files and create 3D isosurface plots of all MOs.

Possible active spaces

The smallest meaningful would be to include the singly occupied $3\sigma_g$ orbital, together with its anti-bonding pendant $3\sigma_u$. Furthermore all the $1\pi$ orbitals should be included. This makes a CAS(5,6):

occ,   3,1,1,0,3,1,1,0
closed,2,0,0,0,2,0,0,0

Your occ,3,1,1,0,2,0,0,0 card actually excludes the anti-bonding $3\sigma_u$ and $1\pi_g$ orbitals, which corresponds to a ROHF calculation.

The next larger space would be the full valence active space, covering the whole $2s2p$ AO shell:

occ,   3,1,1,0,3,1,1,0
closed,1,0,0,0,1,0,0,0

Addtionally one could also include the $1s$ core orbitals with closed,0

The next thing to include would be the next higher lying virtual orbitals coming from the $3s3p$ AO shell: occ,5,2,2,0,5,2,2,0. After that one could go on with the $3d$ and $4s4p$ shells.

The occ,3,2,2,0,3,2,2,0 card would include the bonding $2\pi_u (3p_{x/y})$ and anti-bonding $2\pi_g (3p_{x/y})$ orbitals, but not the $4\sigma_{g/u} (3s)$ and $5\sigma_{g/u} (3p_z)$ which are similar in energy.


Does the water in hydrates affect final concentration? Say I have a hydrate, and I am going to dissolve it in water. Does the water in the hydrate affect the final concentration?

Here's an example: If I have $\pu{249.69 g}$ of $\ce{CuSO4·5H2O}$, and dissolve it in $\pu{200 mL}$ of water, will the concentration be $\pu{5 M}$?

Firstly, copper would have to be in the 4+ oxidation state for there to be $\ce{Cu(SO4)2.5H2O}$, so this isn't a realistic question.

But, no it won't be 5M, because you need to know the final volume of the solution to calculate molarity.  If you add 345.55 grams of just about anything I can think of to 200 mL of water, the volume will increase substantially.  So not only will the 90 grams of hydrate water add to the volume of the final solution, generally you need to consider how all added substances contribute the final volume.

In some cases, such as magnisium sulfate, a small amount added to water can actually decrease the volume.


What is the hybridization of the nitrogen in the azide ion? In the azide ion, there is a symmetrical arrangement of three nitrogen atoms. If I've been told to figure out the hybridization of each nitrogen atom, what's the best way to go about doing this? My initial guess was that the central nitrogen was sp hybridized and then the other two were sp2, but I can't seem to justify my guess.

Here are the resonance structures we can draw for the azide anion.

Resonance structure A has an octet on each nitrogen and just one unit of formal
charge on each nitrogen
Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens.  Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less.  In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B).  The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options.  For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene.  The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in.  This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know.  Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and C are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).


Why did J.J. Thomson observe protons in his cathode ray experiment? In this MIT lecture, at 7:22, the professor says that when J.J. Thomson added a positively charged plate on one side of the cathode ray and a negatively charged plate on the other side, he observed a large deflection towards the positive plate, and a small deflection towards the negative plate (see image below).

This observation implied that hydrogen gas consists of positively charged particles with lots of mass, and negatively charged particles with a small amount of mass.

However, according to Wikipedia, cathode rays are "streams of electrons observed in vacuum tubes." Thus, my question is: in J.J. Thomson's experiment, where are the protons (positively charged particles) coming from?

Thompson observed more than just cathode rays.

He experimented with different gases and different pressures in his tubes. When there is a significant amount of hydrogen you get ionisation of the gas and more than one type of charged particle can be generated and deflected (though different conditions are needed to see protons than electrons).

Part of the point of this was to distinguish between particles produced by the gas and particles produced when there was little or no gas. The point about cathode rays is that they can still be produced when there is virtually no gas in the tube by thermionic emission from the cathode. They have the same properties as one of the particles produced by ionising hydrogen which established some of the basis for saying that electrons are an important part of atoms.


How does the pressure change in the dimerization of nitrogen dioxide? say we have a reaction N2O4 reversible 2NO2

for this reaction,if initial number of moles of N2O4 is 'a' can we write

t = 0 a -

at equilbrium a-x 2x

similarly can we do the same for pressure? i.e.
if initial pressure is p,then can we say?

t=0           p -

at equilbrium p-y 2y

if yes why? and is x=y,i.e is the number of moles of x dissociated equal to decrease in pressure of reactant?
and is this method valid for all reversible reaction irrespective of the stoichiometric coefficient of reactants and products

If you assume ideal gas in a constant volume this works, but x is not y. 'a' and 'p' are related via the ideal gas law(a = n): P = n * R*T/V .

Take in mind that the pressure actually increases as the reaction progesses as more particles are formed.


Is dH(S,P) = dH(T,P) true (always)? I came across an interesting question with some physical chemistry students today. Based on the following steps, we're uncertain whether the statement in the title is/could be true. Assuming $dN = 0$,

Enthalpy's natural variables are Entropy and Pressure:

$$dH(S,p) = TdS + Vdp \space (1)$$

Enthalpy can be expressed as a total derivative of Temperature and Pressure:

$$ dH(T,p) = \frac{\partial H}{ \partial T}\vert_p \space dT + \frac{\partial H}{\partial p}\vert_T \space dp \space (2)$$

Total Enthalpy is $$H = U + pV = TS - pV + \sum_i \mu_i N_i + pV = TS + \sum_i \mu_i N_i \space (3)$$

Thus, taking partial derivatives in (2),
$$ dH(T,p) = SdT + 0 \space dp \space (4)$$

This means that, if $ dH(S,p) = dH(T,p)$,

$$ SdT = TdS + Vdp \space (5)$$ must be true. Can anyone see something wrong?

This seems to boil down to a math question: are two differentials of the same function (always) equal if expressed by different variables?

The reasoning stated is partially correct, but the final relation you arrived to is incorrect. I will try to explain why and write the N dependence explicitly for completeness. The crucial thing is that when one writes an expression such as $$\left(\frac{\partial H}{\partial T} \right)_{P,N}$$
what one really means is "take the partial derivative of $H$ written as a function of $T$, $P$ and $N$ with respect to $T$". When you take the partial derivatives in equation (2) then, you should take them considering $H$ as a function of $T$, $P$ and $N$. You have to consider then the expression
$H = H(T,P,N) = S(T,P,N)T + \mu(T,P) N$. If you differentiate that equation with respect to T and P the result is:

$$\left(\frac{\partial H}{\partial P} \right)_{T,N} = \left(\frac{\partial S}{\partial P} \right)_{T,N} T + \left(\frac{\partial \mu}{\partial P} \right)_{T} N~~~;~~~ \left(\frac{\partial H}{\partial T} \right)_{P,N} = \left(\frac{\partial S}{\partial T} \right)_{P,N} T + S + \left(\frac{\partial \mu}{\partial T} \right)_{P} N$$
If you replace those two relations in your equation (2) the result is:

$$ \mathrm{d}H = \left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T.$$

You can indeed equate this with your equation (1), which is what you ask in your main question. This is the same thing one ordinarily does when expressing a scalar as a function of different sets of coordinates, for instance $f = f(x,y) = x^2 + y^2$ and $f = f(r,\theta) = r^2$ this means, equating both, that $x^2 + y^2 = r^2$, which is a relationship that must hold if you want both $(x,y)$ and $(r,\theta)$ to refer to $f$ (this last sentence may be a bit tautological, I hope what it means is clear, note that it is certainly not "always" true that $f(x,y)$ and $f(r,\theta)$ are equal, since they are two different functions, albeit expressed with the same letter, for instance, $f(x=1,y=0) = 1$ but $f(r=2,\theta= \pi) = 4$, there must be a specific relation between the coordinates for these to be equal). If you do this you obtain:

$$\left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = T~\mathrm{d}S + V~\mathrm{d}P.$$

Note that you would get the same expression even if you considered processes in which $\mu \mathrm{d}N$ wasn't cero, cause both terms would cancel out. If one remembers that $T\mathrm{d}S = T\left(\frac{\partial S}{\partial P} \right)_{T,N}\mathrm{d}P + T\left(\frac{\partial S}{\partial T} \right)_{P,N}\mathrm{d}T$ then this simplifies to:

$$\left(\left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = V~\mathrm{d}P.$$

Dividing through by N:

$$\left(\frac{\partial \mu}{\partial P} \right)_{T} \mathrm{d}P + \;\left(\bar{S}+ \left(\frac{\partial \mu}{\partial T} \right)_{P} \right)~\mathrm{d}T = \bar{V}~\mathrm{d}P.$$

This is true if and only if:

$$\left(\frac{\partial \mu}{\partial P} \right)_{T} = \bar{V}~~;~~\left(\frac{\partial \mu}{\partial T} \right)_{P} = -\bar{S}.$$

These are correct relations and can also be deduced from the Gibbs-Duhem equation: $N\mathrm{d}\mu -V\mathrm{d}P + S\mathrm{d}T = 0$.


Why does the transverse magnetization in Fourier Transform NMR consist of Ix+iIy ($\hat{I}_{\!x}+\mathrm{i}\hat{I}_{\!y}$)? More specifically, the expression for the average expectation value of the transverse magnetization has the following form:

$$M^+(t)=N\gamma \hbar\,\text{Tr}(\hat{\sigma}(t)\hat{F}^+)$$

with $\hat{\sigma}(t)$ the density matrix. $\hat{F}^+$ is defined as:

$$\hat{F}^+ = \sum ^{K} _{k=1} \hat{I}^+_{\!k} = \sum^{K}_{k=1} (\hat{I}_{\!kx} + \mathrm{i}\hat{I}_{\!ky})$$

I don't really see how you intuitively obtain such an expression for the transverse magnetisation (in the syllabus it was posited, not derived). On the other hand, if you would take the complex modulus then you would find the magnitude of some 2D vector. It still does not motivate (enough) for me why it should look like this. Doesn't measurement happen in a single direction? Why is it not just

$$\sum^{K} _{k=1} \hat{I}_{\!kx}?$$

It also bears resemblance to the raising operator (which would explain the $+$), perhaps you measure how many quanta are absorbed, which corresponds to measuring the amount of spins that you raise to a higher energy level (but that's not what the raising operator signifies I think)? Or is it a phasor with a $90^\circ$ angle between both components?

As Mad Scientist said in the comments, measurement of the transverse signal in NMR occurs in both the $x$- and $y$-directions. This is known as quadrature detection. In practice this is not achieved with two detector coils, but rather with a "trick" which is described more fully in Keeler's Understanding NMR Spectroscopy, 1st ed., section 5.6.

Instead of having two separate quantities $M_x$ and $M_y$, it's easier to lump them into one complex quantity $M$:

$$M = M_x + \mathrm{i}M_y$$

For example, if we only have one spin with equilibrium magnetisation $M_0$ pointing along the $z$-axis, then following a $90^\circ (y)$ pulse (which rotates the magnetisation vector to $x$),

$$\begin{align}
M_x &= M_0 & M_y = 0
\end{align}$$

and after a period of free precession $t$

$$\begin{align}
M_x &= M_0\cos(\Omega t)\mathrm{e}^{-t/T_2} & M_y = M_0\sin(\Omega t)\mathrm{e}^{-t/T_2}
\end{align}$$

and $M = M_x + \mathrm{i}M_y = M_0\mathrm{e}^{\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2}$, where $T_2$ is the spin-spin relaxation time. Fourier transformation of this complex signal gives a spectrum $S(\omega) = A(\omega) - \mathrm{i}D(\omega)$. There is a constant of proportionality which doesn't matter and has been dropped. The real and complex components are known as the absorption and dispersion Lorentzians respectively:

$$\begin{align}
A(\omega) &= \frac{1/T_2}{(1/T_2)^2 + (\omega - \Omega)^2} & D(\omega) = \frac{\omega - \Omega}{(1/T_2)^2 + (\omega - \Omega)^2}
\end{align}$$

(Diagram from Keeler's Understanding NMR Spectroscopy, 2nd ed. (p 84).) The absorption component is usually what is shown in the actual NMR spectrum. There are some complexities because the spectrum has to be phase shifted, but the idea holds.

For a single spin it is very easy to use product operators instead of the vector model. After that, to obtain the expressions in your question, you just generalise to multiple spins. I might come back to this some time later, but for now, this will suffice.

The benefit of quadrature recording (which seems to have been cut from Keeler's 2nd ed., but remains in the 1st ed.) is that it allows us to find out the sign of the offset $\Omega$. Let's say that instead of having quadrature recording we only measure along the $x$-axis. Our signal is

$$\begin{align}
M &= M_x \\
&= M_0\cos(\Omega t)\mathrm{e}^{-t/T_2} \\
&= \frac{1}{2}(M_0\mathrm{e}^{\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2} + M_0\mathrm{e}^{-\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2})
\end{align}$$

It's relatively straightforward to show that for a function $M = M(t)$, if $\mathcal{F}[M(t)] = f(\omega)$:

$$\mathcal{F}[M^*(t)] = [f(-\omega)]^*$$

and using this, we can show that if we take the Fourier transform of the one-dimensional magnetisation, we get

$$\begin{align}
S(\omega) = \frac{1}{2}[A(\omega) - \mathrm{i}D(\omega) + A(-\omega) + \mathrm{i}D(-\omega)]
\end{align}$$

and if you take the full expression for this and replace $\Omega$ by $-\Omega$, you will find that $S(\omega)$ is the same regardless of the sign of the offset $\Omega$. This means that our spin of interest could have either a positive or a negative offset, relative to the spectrometer reference frequency $\omega_\mathrm{rf}$, but we would not be able to determine it.

In theory, if you could control the phase of the signal you are detecting, single detection would allow you to extract the sign of the offset. For example, in our example above, if you measure $M_y$ instead of $M_x$, you'll find that there is a difference when you replace $\Omega$ by $-\Omega$. Here's a schematic from Gunther's NMR Spectroscopy, 3rd ed. (p 265):

Here, the author Fourier transformed $M_x$ and $M_y$ separately before adding them up. Since the Fourier transform is linear, this has the same effect as adding $M_x$ and $M_y$ and Fourier transforming the sum. In any case, it shows that if the magnetisation vector starts off along the $y$-axis, you could, in theory, measure only the $x$-magnetisation to determine whether the offset is negative (case a) or positive (case b).

Such a spectrum would be very difficult to interpret, and would probably be impossible to analyse if positive and negative peaks overlapped. On top of that, it would also require you to know the exact phase difference between the initial signal and your single detection coil. I believe this is (due to several reasons) not experimentally possible, so unfortunately, there's no way to extract this information using single detection. (This is also part of the reason why NMR spectra have to be phase corrected – for the average NMR user the computer does this automatically.)

Now, we've established that you need quadrature detection to determine the sign of the offset. However, in the final NMR spectrum, we don't really look at offsets; we look at resonance frequencies. (To be precise, chemical shifts, but those are linearly related to resonance frequencies.) So, if you really wanted to, you could use single detection while placing the reference frequency at the far end of the spectrum. For example, if you want to measure peaks from $400.000$ to $400.001~\mathrm{MHz}$, you can place the reference frequency at $400.002~\mathrm{MHz}$. Then, by virtue of choosing an overly high value for $\omega_\mathrm{rf}$, you know that your offset $\Omega$ has to be negative.

However, a rf pulse centred at $400.002~\mathrm{MHz}$ would be less effective at exciting the spins (as it is further away from being on-resonance), and essentially you are throwing away half of the rf pulse (any components of the rf pulse with frequency greater than $400.002~\mathrm{MHz}$ are wasted). So, it's preferable to put $\omega_\mathrm{rf}$ right in the middle of the spectrum; but as said before, that would necessitate quadrature detection so that you can obtain information about the sign of the offset.


Colour changes in coordination complexes What causes coordination complexes to change colour as they are heated? Do the ligands detach from the central metal or do the electrons held in the d orbitals leave? If both happen, which happens first?

Specifically, I am trying to explain the colour changes observed when potassium ferrioxalate $(\ce{K3[Fe(C2O4)3]})$ was heated. The colour change is from green-yellow to brown/red.

Many things can happen when you heat a coordination complex: ligand loss, reactions in the ligand structure, loss of hydration water, isomerization, oxidation/reduction of the metal center etc.

In the specific case of potassium iron(III) oxalate trihydrate, as reported in this paper, the thermogravimetric analysis shows that the first structural change happens at 113 ºC with the loss of three water molecules (from the trihydrate salt), yielding potassium iron(III) oxalate, a yellow-orange solid.
When the temperature reaches 296 ºC (in air), iron(III) is reduced to iron(II) to form iron(II) oxalate, an orange solid.

Further heating leads to the decomposition of iron oxalate into FeO at around 470 ºC. FeO is a brown/red solid (probably the compound you mentioned in your question).


Diamonds into protective barrier Theoretically, is it possible to make a large, transparent piece of armor or a shield from a diamond?

Say, a piece of the said armor with a size  $7^\prime \times 7^\prime \times 6^{\prime\prime}$?

What kind of forces could this put up with?

I'm not a gun owner or military expert, but from what I understand the requirements for the armor material are more or less uniform; it should:

Be able to withstand an impact and quickly dissipate the kinetic energy from a projectile. Diamonds are are hard, but very fragile. They cannot redistribute incoming impact from a projectile in the opposite or perpendicular direction. Once they crack, there is no more protection, and they cannot self-heal.
Have low weight. The density of diamond is $\pu{3.51 g cm-3}$, which is nearly 2.5 times heavier than kevlar (ca. $\pu{1.44 g cm-3}$).
Be easily molded in desired form, and/or be flexible (relevant for body armor). Process of crystal growing is rarely precisely controlled and often requires mechanical post-processing, which is ineffective due to hardness of the diamond and can potentially weaken the structure.
Have adequate thermal interface. Due to exceptionally high phonon thermal conductivity, the person wearing this armor will suffer from cold in winter and from heat in summer time, not to mention what damage an opponent with a flamethrower can cause.
Be modular and easily repairable. Diamond plates can only be replaced, so one needs to carry a stockpile of the shards.
Be suitable for camouflage (absorb light and other EM radiation, optionally). Diamond surface is hard to modify, it has low adhesion, so there is no easy way to tarnish the object if the diamond layer is the outer one.
Be optically transparent and don't cast any distortions (relevant for the bullet-proof windows). Due to high refraction index it's a questionable choice for a transparent shield.
Be inexpensive in terms of production and maintenance. No comments here.

To wrap it all up, for now classical macro-scale diamonds seem to be a very expensive and ineffective material for use as an armor material due to their physical properties. However, at a nanoscale the situation can be different, as nanodiamonds show somewhat different properties, yet not nearly fully investigated for military applications.

Also, you might want to check this topic on Worldbuilding.SE: Is diamond armor better than traditional armor?


Mechanism of reaction of benzenediazonium chloride with ethanol to form benzene and acetaldehyde In my textbook, all that was told was that ethanol is a mild reducing agent and hence reduces BDC while oxidising itself. Is there any mechanism or am I just supposed to remember this like a redox reaction?
I did search for a mechanism but did not find anything. I have no idea as to how this reaction works. Can someone please point me in the right direction?

In reality, benzene and acetaldehyde are the minor product of the reaction of benzenediazonium chloride with ethanol. The major product is ethoxybemzene (phenetole). Peter Griess, who had discovered diazonium salts in 1858, had reported benzenediazonium salt (nitrate or sulfate) with ethanol undergoes aforementioned redox reaction in 1864 (Ref.1). However, by 1901, it was confirmed that the products from the redox reaction is minor (about ~9%) when compared to major product, phenetole (about ~64%) when benzenediazonium chloride was used. By 1956, Kelley (Ref.2) found that the yield is even low as 3% and his student, Miller (Ref.1) later confirmed this result.
Now, it is believed that the reaction follows the scheme $\bf{A}$ and $\bf{B}$ (Ref.1):
$$\ce{Ar-^+N#NX- + RCH2-OH -> Ar-O-CH2R + N2 + HX} \tag{$\bf{A}$}$$
$$\ce{Ar-^+N#NX- + RCH2-OH <=> Ar-N=N-O-CH2R HX \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -> ArH + RCHO + N2 + HX} \tag{$\bf{B}$}$$
where process $\bf{A}$ proceeding through a $\mathrm{S_N2}$ displacement of the diazonium groups while process $\bf{B}$ involves a homolytic fission of the bonds to produce radicals which lead ultimately to aromatic hydrocarbon and aldehyde.
Proposed mechanism:
$$\ce{Ar-N=N-X -> Ar^. + N2 + X^.} $$
$$\ce{Ar^. + R-CH2-OH -> ArH + R-^.CH-OH} $$
$$\ce{AR-^.CH-OH + X^. -> R-CHO + HX} $$
It was found that when oxygen is added to the system rate of reaction has not changed significantly. The propagation steps with $\ce{O2}$ has been suggested as:
$$\ce{AR-^.CH-OH + O2 -> R-CHO + H-O-O^.} $$
$$\ce{H-O-O^. + AR-CH2-OH  -> AR-^.CH-OH + H-O-O-H} $$
Ref.1 and Ref.2 have reported the first order kinetics of this reaction with $k = 0.96 \times 10^{-4}$ at $\pu{25 ^\circ C}$ and $k = 2.03 \times 10^{-4}$ at $\pu{30 ^\circ C}$, which are consistent with other published data.
Note: More insight on mechanism, read Ref.3.
References:

Robert Warren Miller, "The stoichiometry of the reaction of benzenediazonium chloride with ethanol," MS Thesis 1957, University of Arizona, Arizona (PDF).
A. E. Kelley, PhD Dissertation 1956, Purdue University, Indiana.
DeLos F. DeTar, Takuo Kosuge, "Mechanisms of Diazonium Salt Reactions. VI. The Reactions of Diazonium Salts with Alcohols under Acidic Conditions; Evidence for Hydride Transfer," J. Am. Chem. Soc. 1958, 80(22), 6072–6077 (https://doi.org/10.1021/ja01555a044).


Why does adipic acid give a cyclic ketone on heating while glutaric acid and succinic acid give a cyclic anhydride? Why does adipic acid give cyclic ketone on heating while glutaric and succinic acid give cyclic anhydride?

According to Blanc Rule for Dibasic Carboxylic acids, A cyclic Ketone is formed unless a five or six membered cyclic acid anhydride can be formed. This rule is valid for 1,4; 1,5; 1,6. dicarboxylic acids


What is the first organic compound to be discovered ? Urea or alloxan? It is widely accepted that urea is the first organic compound to be synthesized from inorganic chemicals. Wikipedia states that:

In 1828, the German chemist Friedrich Wöhler obtained urea artificially by treating silver cyanate with ammonium chloride.
$$\ce{AgNCO + NH4Cl → (H2N)2CO + AgCl}$$
This was the first time an organic compound was artificially synthesized from inorganic starting materials, without the involvement of living organisms. The results of this experiment implicitly discredited vitalism — the theory that the chemicals of living organisms are fundamentally different from those of inanimate matter.

But, alloxan (a not-so popular organic compound!) was isolated in 1818 by Brugnatelli (according to Wikipedia) 10 years before the discovery of urea!
So, is urea the first ever organic compound to be discovered or is it alloxan?
I also found a link which claimed that urea is not the first organic compound to be discovered.

The problem herein lies in the different possible definitions of organic chemistry.

The IUPAC defines an organic compound as a structure containing at least one carbon atom that can be named using the principles of organic nomenclature.

A less concise definition is

All compounds containing carbon and at least one other element that are not carbon monoxide, carbon dioxide or the latter’s inorganic salts.

Another one:

All compounds that contain at least one $\ce{C-H}$ bond.

Originally, the definition meant compounds isolated from organic materials i.e. organisms, living beings.

You see that the exact definition of organic compounds is difficult. But it was once very clear, historically: Anything not isolated from organisms or a derivative thereof was inorganic.
Note that isolation of organic compounds can be traced extremely far back in history: The ancient Egyptians isolated indigo as a colouring agent. However, they didn’t synthesise it and most attempts pre-19th century failed. Nobody questioned the possibility of isolating pure organic compounds from living matter, it was merely questioned whether they could be synthesised without involvement of organisms.
Alloxan was synthesised by oxidising uric acid with nitric acid as Wikipedia states. Thus, it was synthesised from an organic compound and would have been considered to contain enough vis vitalis even for further synthetic steps.
The compound your link states, 5-amino-3H-1,2,4-dithiazole-5-thione, is not described in any prime source as having been isolated from organisms. In fact, it is unknown to Wikipedia, so with all due skepticism, one can assume it to be of little to no relevance in living beings.
Compounds we today would call organic have been synthesised even earlier. Ethylene was first discovered in 1669 and it was successfully chlorinated to give 1,2-dichloroethane in 1794 — way before 1818, 1821 or 1828. Due to its name Dutch oil it is possible that it was considered an inorganic chemical. If it was synthesised from ethene that was, in turn, synthesised from ethanol, that would have just meant it carried over the vis vitalis from the original ethanol.
Wöhler’s discovery was that significant, because he used compounds that were (and are) unanimously declared inorganic (silver cyanate and ammonium chloride) and arrived directly at a compound that had previously been isolated from living material namely urea.


How do we know the natural abundance of isotopes on Earth? Without knowing the Average Atomic Mass or the percent abundance, how do we know that Protium is the most prevalent hydrogen isotope? What methods did scientists use to come to this conclusion? How did they reach such a conclusion?
I'm still starting out in learning chemistry and I felt like I hit a boulder. I know the formulas and how to get the natural abundance, but I haven't heard a concrete explanation of the concept. I know that it's impossible to survey all the hydrogen elements on Earth to see which isotope is abundant, so what method did scientists use?

Because we can sample the environment with great precision
You seem to assume that we can't know the exact makeup of a sample unless we already know the components. This is wrong. Mass spectroscopy can reliably measure the exact mass of every isotope in a sample and the relative abundance of each. If, for example, a sample of sea-water is measured, we can reliably tell the exact proportion of all the hydrogen isotopes and all the oxygen isotopes in it.
Given that his is possible, estimating the true natural abundance is a matter of statistical sampling. Ocean and river water is fairly well mixed so this is easier than it sounds (at least when there are no processes that radically alter the isotopic composition). So we have a lot of samples, for example, from different oceans and different locations in those oceans. Given that, we can be fairly confident that the isotopic composition doesn't vary a lot and, therefore, we can get a good estimate of the "natural" abundance of each isotope.
But the techniques we use are precise enough to notice that there are some (small) variations in that composition. Some processes do (very slightly) alter the isotopic ratios. Evaporation tends to happen slightly faster for lighter isotopes, for example, as do some biological processes. Some of those processes are temperature dependent leading to the ability to identify the ambient temperature preset when some fossils were alive by measuring the ratio of oxygen isotopes in them.
The key point is that we do have tools that can measure isotope ratios very precisely. So we know the ratios in any sample very precisely. With a very large number of samples, we can generate a reliable picture of the typical abundance of each isotope which is what we refer to as the natural abundance.


Is there any relation between resonance and smell of a hydrocarbon? I came across several examples (such as many benzene derivatives) of hydrocarbons having resonance that also seem to have a sharp odour. So, does resonance have an effect on smell?

I wish I could convey the spirit of Professor Louis Fieser on the development of organic chemistry. I couldn't find it in his textbooks, but in class, it went something like this:
Coal distillation led to a large number of different unsaturated compounds, most of which had strong odors, so they were classified as "aromatic". When more quantitative measurements (heats of combustion and heats of hydrogenation) were available, double bonds were investigated, and some compounds were found to be unusually stable. A single $C-C$ bond should have a certain energy, and so would a $C-H$ bond, and so also should a $C=C$ double bond, but some compounds gave surprising results.
Some molecules (like benzene) could be written with two positions for the double bonds - they could "resonate" between the two arrangements. Most of the compounds that had this sort of possibility were odorous (aromatic), so any compound with this sort of resonance possibility was called "aromatic", even if it had no appreciable odor. And they all had some sort of extra stability ("resonance energy") due to the double bond shifting.
Now while "resonance energy" ought to have some quantitative significance, it had some considerable variability, depending on what values you used to calculate the expected value for heat of hydrogenation or combustion. Another term popped into use: "aromaticity", which was never quantitatively defined, but was really a measure of the surprise at the difference between the calculated value and the observed value of the heats.
But I expect that odor is not a result of extra stability, but more likely a result of some reactivity of the multiple double bond system. And volatility.


Storing oxygen generated via water electrolysis I want to create a somewhat pure oxygen gas from electrolyzing water and using sodium bicarbonate as an electrolyte.

How can I harvest the oxygen once it is released from the water, and store it in a container without adding oxygen to the existing air in the container? Meaning that I want pure oxygen in the container and not normal air.

I think the simplest approach is to fill a test-tube with your electrolyte solution, seal the open end with your latex-glove covered thumb, then immerse the sealed end in your cell. If you can tilt your anode several degrees from the perpendicular, you should get a stream of oxygen bubbles that you can then catch with the test tube. You certainly won't get them all but it shouldn't be a problem to get 3/4 of them anyway. If you have trouble inverting the test tube without having a little air left, just slide some flexible tubing to the top and suction out the air. Take care to watch the level of your electrolyte as the solution that is displaced from your test tube will of course be filling your cell.

After you have collected all of the oxygen you need, seal the open end of the test tube with a piece cut from a latex glove and a rubber band. You will then have a test tube filled with oxygen gas and any residual electrolyte solution.


Is there a general method for working out electron configurations? For up to period 3, elements' electron configurations are relatively easy to work out until the transition metals begin. Then some elements may have $\ce{4s}$ orbitals which are preferred by electrons over $\ce{3d}$ orbitals. These small rules don't make sense all the time and only cover some elements. Is their any master method for working out the electron configurations of any given element?

There is a semi-empirical method to fill the atomic orbitals: atomic orbitals are filled  in the order of increasing n+l. If the sum n+l is equal for two atomic orbitals, they are filled in order of increasing n.
This gives the following order for filling the orbitals:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, 5g, 6f, 7d, 8p, and 9s