Are there any trends for electrical conductivity? Which element has the greatest electrical conductivity?
  (A) $\ce{As}$
  (B) $\ce{Ge}$
  (C) $\ce{P}$
  (D) $\ce{Sn}$

I am unsure of how to approach this problem. Is there some sort of trend for electrical conductivity or some other method to determine which of these answers the question?

I am pretty sure it is not $\ce{Ge}$ as that is used for semi-conductors, and am pretty sure it is not $\ce{P}$ because that is a non-metal but I don't know how to narrow down between the others.

I am unsure of how to approach this problem.

No, you are not! You have already reduced the number of candidates by good reasoning!

At this point, one

could have a look at conductivity tables or remember in which period the metal blocks start for every main group or
remember typical electricity-related uses of $\ce{As}$ and $\ce{Sn}$

Did you ever form a conducting connection between electronic components (or just some wires) by soldering? $\Leftarrow$

(Do you recall that $\ce{GaAs}$ is used as a semiconductor material in thin film solar cells?)


Trying to displace iron from ferrous sulphate heptahydrate using aluminium foil but nothing has happened I started by dissolving the sulphate into water and filtering it as I noticed there was a dirty residue lying at the bottom. (I am using ferrous sulphate that is commonly used in gardening.) I then poured this solution into a jar along with torn pieces of tin foil (tin foil being a common name; the foil is made of aluminium). At the start the solution was fairly clear and greenish in colour. A few days later there is no sign of iron collecting at the bottom of the jar and there has been a slight change in colour. The solution is now slightly cloudier and looks more like apple juice. The orange colour makes me think there must be iron oxide being produced. I need to know whether my method for displacing the iron is correct and if elementary iron will be produced.

You are correct: Commercially available metal foil is typically made from aluminum, even if it's commonly called 'tinfoil' sometimes. As you say, technically, tin foil is made of tin; aluminum foil is made of aluminum.

Jannis Andreska is also correct: The passivating oxide on the aluminum is likely preventing the iron ions from reaching the $\ce{Al^0}$ and swiping their electrons.

So, you are correct again: The yellow/orange color you're observing is probably oxidation of the $\ce{Fe^{2+}}$ to $\ce{Fe^{3+}}$, by $\ce{O2}$ in the air.

One possible approach to breaking the oxide would be to connect the foil to the negative terminal of, say, a AA battery, with the positive terminal connected to a junk stainless steel spoon, also submerged.  That might not be enough voltage to break down the oxide film, though; on the other hand, a higher voltage might lead to corrosion of the spoon and the release of iron and other metals into your solution.  May or may not be a problem for your experiment?

Regardless, I would recommend adding some table salt $\left(\ce{NaCl}\right)$ to your solution, both to increase its conductivity and also to introduce chloride, which may help to attack the oxide on the aluminum, in tandem with the applied potential.


How to remove paraffin wax drop from cam lense glass? When I put a my cellphone on a table for charging, some drops of paraffin wax were laying there which I couldn't see.  After some time during charging, heat melted the paraffin wax and it stuck on the rear camera lens of my Lenovo k6 Power.

I tried to wipe the lens with a special cloth especially used for cleaning my glasses, but it couldn't remove all the scratches/stains.

Please suggest any solutions.

The first thing I would try is to re-heat the phone then try to wipe the paraffin off with your lens cloth. Paraffin wax typically melts between 46 and 68 °C (115 and 154 °F)$^1$. Assuming you don't have a good way to accurately heat the camera to this temperature, I would just repeat what you did to melt the wax in the first place; let the camera discharge it's battery and then recharge it. Once it has heated up, try wiping it off with the lens cloth.

Unfortunately, I have a feeling the above procedure alone may not be sufficient to remove all of the wax. It's worth trying because it's simple and safe for your phone. If that procedure does not remove all of the wax, the next thing I would try is using a solvent in addition to the heating. The wax should dissolve in diethyl ether$^2$, which is the primary component of most engine starting fluid products. You should be able to go to any auto parts store and purchase a can of starting fluid who's primary ingredient is diethyl ether (sometimes just called ether).

The first thing you will want to do with the ether is to verify that it does not attack the material holding/surrounding your lens. If the area around the lens is all metal and glass, you are safe and can skip this step. If the area around the lens is plastic (unlikely) then you will want to test a small area of the plastic to be sure it will not be dissolved or distorted in any way by the ether. Simply soak a corner of the lens cloth with the ether and touch it to the plastic. You have to be fast because ether evaporates very quickly. You might want to use a magnifying glass to look at the area that you whetted with the ether. If there is any change in the color or texture (you should feel the area with your finger), then you have to decide whether you want to proceed and risk ruining your camera. Again, it is unlikely that this will be a problem. I looked at some images of your phone model, and while there are some plastic parts to the construction, the lens assembly appears to be just metal and glass.

If you decide to proceed with cleaning the lens with ether, it should be a pretty simple, straight forward process. I would heat the phone by charging it as discussed above, soak a portion of the lens cloth with the ether-based starter fluid, and gently wipe the remaining wax from the lens. The lens glass and any metal surrounding it are impervious to the ether (and any other ingredients of the starter fluid), but avoid contact with any plastic components.

Given your problem as you've stated it and the make of your phone, I'm pretty confident that your camera lens can be thoroughly freed of the paraffin wax. Don't hesitate to ask for clarifications of anything I've said in the comments, and best of luck!

1) Kaye, George William Clarkson; Laby,Thomas Howell. "Mechanical properties of materials". Kaye and Laby Tables of Physical and Chemical Constants. National Physical Laboratory. Retrieved 25 October 2013.

2) Seager, Spencer L.; Slabaugh, Michael. "Alkane reactions". Chemistry for Today: General, Organic, and Biochemistry. Belmont, CA: Cengage. p. 364. ISBN 978-0-538-73332-8.


How ph affects aesculune fluorescence? I am facing a ween problem. I am trying to find out, how would look like the graph for fluroscence of aesculine (intensity of fluoroscence on Y axis and pH on X axis).
I know, that increasing the pH causes a corresponding increase in the maximum fluorescence intensity from 77.74 (units) in the acidic sample to 146.13 at neutral pH and 232.69 at alkaline pH.
But, can I apply this on aesculine?

You should investigate several pH and study the effect of time on the fluorescence spectra. One cannot quote single values without mentioning the wavelengths of excitation and emission.
pH can alter the state of  ionization and it can also cause hydrolysis of the sugar unit.
You need to collect three different types of spectra at each pH, say from pH 2 to 12.
At each pH, first measure the absorption spectrum at those pH values. This is the easiest experiment. Do you see a difference in absorption spectra at different pH values? Note the trends.  If yes, pH is affecting either the structure (decomposing) or the state of ionization.
Next, collect the excitation spectrum at a fixed emission wavelength (you have to estimate that). The excitation spectra should look similar to the absorption spectrum in shape.
Once you know the correct excitation wavelength at each pH, collect proper emission spectrum by using the lamba max of excitation.
So the answer is that be ready to see the changes in fluorescence spectra since there is no "the" fluorescence spectrum of a given molecule in a solution. You have to specify the solvent, pH, and excitation wavelength for experimental measurements.


How to quantify amount of pigments in plant leaves? Using thin-layer chromatography, it is possible to see what pigments plant leaves contain. For example, I can test whether there are chlorophyll a or b, carotenoids, or anthocyanins.

What kind of technique I would need to quantify the amount of these pigments in the leaves?

Once you've found a suitable mobile phase in analytical TLC, you could upscale the separation and transfer the results to preparative TLC or column chromatography.

In the case of column chromatography, collect the eluted material in test tubes, check the content by analytical TLC and/or UV-VIS spectroscopy, unite identical fractions, evaporate the solvent and determine the weight of the respective fractions.

In the case of preparative TLC, scratch the separated bands off the plate, elute with a solvent of your choice and proceed as above.

If necessary, repeat the chromatographic separations for the different compound classes with other solvent mixtures.


Alloys/metallurgy : promising elements for for Sn-based solders I am trying to research potential modifications to lead-free solders (intended to be used in electronics). Base alloy is eutectic $\ce{Bi_{58}Sn_{42}}$, and I am going to add various combinations of a bit of copper, silver, gold, platinum, gallium, lithium, and indium.

Is there something there which is known to not work? I am wondering this because I see that gold, platinum, gallium, and lithium are not typically used as dopants for electronic soldering alloys. Should I expect a lot of problems attributed to intermetallic compounds? (I know there are issues due to a Sn-Cu reaction, but nothing about the rest).

Is it theoretically possible to find a combination with significantly high copper content (> 10-20%) while still having low melting temperature (< 250 $^{\circ}\mathrm{C}$)?

Lithium might be tricky to alloy in an oxidizing atmosphere (e.g. air), and Pt might take some effort to form an alloy, because of its high m.p., which, however, is still less than than the boiling points of Bi or Sn. Other than that, there should be little trouble forming alloys.

You can search on line for "phase diagram alloy components" to find the melting point for at least some of the alloys. E.G. an online search for

phase diagram "copper-indium" alloy

turned up this link at Springer-Verlag GmbH, with the diagram below.

That said, I would expect the "10-20 % copper" requirement precludes melting point less than 300 C, even with low-melting In, Hg, Ga etc.

BTW, I would not use "dopant" for a component of an alloy that will not appreciably change its electronic properties; it is just another component, be it minor, of a ternary, quaternary etc. alloy.


What are the products formed when phenol is reacted with benzyl bromide? What are the products formed when phenol is reacted with $\ce{PhCH_2Br}$ (benzyl bromide)?

I can think of two ways in which the reaction would proceed. One would be the formation of an ether through SN2 mechanism. The other is formation of 4-benzylphenol (Friedel–Crafts alkylation)

Which one of the two products is likely to be formed?
If both are formed, is there any preference for one being the major product?

I believe the SN2 mechanism will predominate. The benzyl bromide is not sufficiently electrophilic to perform a Friedel-Crafts Alkylation, as that often requires the use of a metal catalyst.


Conversion of methylcyclopentanol to cyclopentanone How to convert methylcyclopentanol to cyclopentanone?

I tried eliminating the alcohol group as $\ce{H2O},$ but that gives Saytzeff product. If only I can get the Hofmann product (methylene cyclopentane), I would subject it to ozonolysis and get cyclopentanone.

I don't know exactly what to use as the reagent to get that Hofmann product. Could there be other methods for the conversion?

If you leave it to E1, you will not get the selectivity you need.

The difference between the methyl group and the methylene unit is steric bulk. You can select for deprotonation of the methyl for elimination via a bulkier base in an E2 reaction.

In addition, you will need to convert the substrate so that it has a good leaving group. For an alcohol, this is typically accomplished by first converting to a tosylate or a mesylate.


Why amino acids (Zwitterion) become either negative or positive at low and high pH solutions? The amino acids are Zwitterions. In neutral pH, an Amino acid's amino group has a postive charge and Carboxyl group has negative charge. They cancel each others charge thanks to the $Hydrogen$ that's roaming between the two groups.

But, I am not able to understand why Amino acids have positive charges at a low pH and negative at high pH? Given amino acids are Zwitterions, how can both terminals have the same charge (either positive or negative) at low and high pH?

Can someone explain this from general chemistry and pH perspective (acid/base)?

PS: first posted in Biology site but community recommended it to be moved here.

Firstly a note about terminology. The word "terminus" is reserved for the N- or C-termini of a polypeptide chain. For a free amino acid, you should refer to the carboxyl and amino groups as the $\alpha$-$\ce{COOH}$ and $\alpha$-$\ce{NH2}$ groups respectively.

Anyway, the -$\ce{COOH}$ group is acidic; above a certain pH, typically around 2, it can be deprotonated to form a negatively charged -$\ce{COO-}$ group. On the other hand, the -$\ce{NH2}$ group is basic; below a certain pH, typically around 9, it can be protonated to form a positively charged -$\ce{NH3+}$ group. I made a short table describing the protonation states of the two groups.

$$\begin{array}{c|c|c|c}
\textbf{pH} & \textbf{Carboxyl group exists as} & \textbf{Amino group exists as} & \textbf{Net charge} \\
\hline
<2 & \ce{-COOH}\text{ (neutral)} & \ce{-NH3+}\text{ (positively charged)} & +1 \\
>2 & \ce{-COO^-}\text{ (negatively charged)} & \ce{-NH3+}\text{ (positively charged)} & 0 \\
>9 & \ce{-COO^-}\text{ (negatively charged)} & \ce{-NH2}\text{ (neutral)} & -1 \\
\end{array}
$$

For more information you can read Wikipedia or any biochemistry textbook - the first few chapters will often be devoted to discussing chemistry like this.


Timescales in Chemical Analysis Has anyone seen a comparative list of typical run times of the techniques used in analytical chemistry? I am making a list of rough timescales of analytical techniques (for a slide). Assume that the sample preparation is aleady done.

For example:
Volumetric titration may take up to 10 minutes

Gravimetry can take several hours to a day

NMR- several minutes

Electron spin resonance ?

X-ray photoelectron spectroscopy: survey spectrum for surface analysis?

Auger electron spectroscopy for surface analysis

X-ray crystallography

Inductively coupled plasma: approximately a minute to read one sample

FTIR- minutes

Raman spectroscopy- typically a minute

HPLC- seconds to hours

GC/ Supercritical Fluid Chromatography/ Capillary electrophoresis
 - seconds to hours

Mass spec - less than a minute

Ion mobility spectrometry- milliseconds

Proton NMR (1D) acquisition time can vary by orders of magnitude depending on concentration (sample amount), instrument sensitivity and relaxation times. For a standard experiment it is on the order of minutes.


If Alkynes can be reduced to Alkenes using Sodium in liquid ammonia, why can't Alkenes be reduced to Alkanes? Firstly, I would like to ask why the alkyne reacts at all with the solvated electron? Surely the alkyne group is extremely nucleophilic (more so than the alkene). This, to me, seems to suggest that an alkene would be more likely to react in this manner than the alkyne. However, this is clearly not the case; why?

As shown in your diagram, when the initial electron is transferred to the alkyne a radical anion is formed; the transferred electron pairs with an electron from the alkyne to form an anion at one end of the molecule. These two electrons are in an $\ce{sp^2}$ orbital.  If you try to repeat this process with an alkene, you wind up with the paired electrons in an $\ce{sp^3}$ orbital.  An $\ce{sp^2}$ is lower in energy than an $\ce{sp^3}$ orbital since it contains more s character and an s orbital is lower in energy than a p orbital.  The electron solvated in ammonia has enough energy to transfer into a lower energy $\ce{sp^2}$ orbital, but not enough energy to transfer into a higher energy $\ce{sp^3}$ orbital.


If capsaicin is insoluble in water, what makes chilli water taste spicy? I have always been told that the 'heat' from chillies came from capsaicin which is oil based. Recently I began wondering how environmentally friendly chilli based insecticide/repellent sprays could work if they are water based. A chemist friend confirmed for me that capsaicin is not soluble in water.

I minced a chilli and added this to a coffee mug of boiling water, left overnight then strained through a coffee filter. The resulting chilli water is definitely spicy, but why? What has dissolved into the water (or is the mixture some sort of emulsion of tiny oil droplets?).

You got me curious, so I poked around a bit on this.

First there is insoluble and then there is insoluble. The Wikipedia article on capsaicin lists its solubility as $\pu{0.0013 g}/\pu{100 mL}$ which is $13$ parts per million. So capsaicin is "relatively insoluble", but not wholly so. Second the Wikipedia article also points out that capsaicin itself is just one of the various capsaicinoid chemicals which give peppers their "hot" taste.

At the end of the (old) Wikipedia article on the Scoville scale, a taste test for hotness, it states

  [...] A measurement of one part capsaicin per million corresponds to about 16 Scoville units. [...]

So, 13 pmm multiplied by 16 Scoville units per ppm is about 200 Scoville units.

A table on hot sauces and peppers lists the Original TABASCO® brand Pepper Sauce from McIlhenny Company as being 2,500 to 5,000 Scoville units. So 200 Scoville units would be about one tenth strength sauce. That should still be enough to tingle your tongue.


In halogenation of ethene using bromine for example, why does the lone pair of bromine attack the C atom? In the presence of the electron rich (pi bond) of ethene, the bromine molecule becomes a temporary induced dipole, one end being electron deficient and acting as the electrophile. (The pi bond in) ethene attacks the electrophile.

At the same time, one of the lone pair of electrons in bromine attacks one of the C atom. Why does this happen?

Normally, I'm familiar with the same process, but with the carbocation being formed without the lone pair of electrons of bromine attacking the other C atom:

Does this have something to do with resonance as shown by the diagram below?

There are too many bond lines to explain the reaction; the question suggests that we have to explain the bond lines!

Let's start from scratch, with a different viewpoint. Ethylene isn't the attacker; it doesn't burst into flame in oxygen or do anything aggressive. It may be reactive, but in a "normal" range. Bromine, on the other hand, is a potent oxidizer and does go around interacting with relatively inert materials - it's a powerful bleach.

So, when bromine comes near the ethylene pi bond, it attacks the electron pair and grabs it, but the second bromine steals it away and departs as Br-. After all, bromine is more electronegative than carbon; the first bromine is just a link in the chain of events that allows the second bromine to escape with an electron. The first bromine winds up stuck to the C-C double bond; the two carbons and the bromine share the positive charge. The bromine certainly gets the lion's share of electron density, leaving the carbons with most of the positive charge. That's why another bromine can attack the carbons from the rear to give a dibromide. (If a Br- attacked the bromine attached to the double bond, it would just regenerate the reactants.) I bet that the second bromine to bond to the ethylene is not from the same bromine molecule as the first bromine. (But since all bromine molecules look the same, I can't prove it.) Doing the bromination in a solution with an overwhelming abundance of another nucleophile would probably give a mixed product.

A C-C double bond is about 1.34 Angstroms long; a bromine atom (in Br2) is about 2.29 Angstroms in diameter. Rather than envisioning the cation as a triangle of three grapefruit, I see it as a cantaloupe sitting on two large oranges. Just consider the empty p orbital of bromine and the pi lobe of the carbons; no need or benefit from making a three-membered ring.


Why is hexafluoridocobaltate(III) the only common high-spin cobalt(III) complex? Almost all $\ce{Co^3+} (\mathrm{d^6})$ complexes are low spin, including $\ce{[Co(H2O)6]^3+, [Co(ox)3]^3–, [Co(NH3)6]^3+}$ and $\ce{[Co(CN)6]^3–}$. The only common high-spin cobalt(III) complex is $\ce{[CoF6]3–}$. Why is the hexafluorido complex in a high-spin state?

This is a question that the crystal field theory cannot answer because of its inherent shortcomings.

The crystal field theory assumes point-type charges approaching the central metal from the coordinate axes leading to the destabilisation of those $\mathrm d$ orbitals that do not feature a nodal plane in the coordinate axes: $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$. The others have zero contribution in all $x,y$ and $z$ direction and are thus not affected by point-sized ligands approaching from there. As simple as this model is, it only gives a crude approximation of why field splitting occurs and which orbitals are likely to end up higher in energy. For everything that involves quantitative discussions, the full molecular orbital approach is needed.

Below you will find the fullest molecular orbital scheme of an $\ce{[ML6]}$ complex including σ and π interactions. A simplified version without the π interactions can be found here; you should attempt to understand it before adding the π interactions to the picture. In this question, however, we need to understand the π effects and thus I cannot leave them out.

Scheme 1: Molecular orbital scheme of an $\ce{[ML6]}$ complex featuring both σ and π interactions.

The system we are looking at — $\ce{[CoF6]^3-}$ contains the π-basic fluoride ions (meaning that their $\mathrm p$ orbitals that can interact with the metal in a π symmetric manner, are fully populated). The interaction of the $\mathrm{t_{2g}}$ fluorine group orbital with the metal’s $\mathrm{t_{2g}}$ orbitals will create a bonding-antibonding interaction. With fluoride most likely having lower-lying orbitals, this interaction should raise the energy levels of cobalt’s $\mathrm{t_{2g}}$ orbitals which thereby turn into $\mathrm{t_{2g}{}^*}$ orbitals.

In this case, fluorido is the only strongly π-basic ligand you are considering in your comparison. And chloride does not form a hexachlorido complex like fluoride would but a tetrahedral tetrachloridocobaltate(III) for steric reasons. Thus, we don’t know whether the fact that fluoride is π-basic enough to raise the $\mathrm{t_{2g}}$ levels high enough so spin pairing does not occur is because of its π basicity alone or whether further elements come into play. Oxalato, ammin and aqua ligands have only weak π-basic effects while cyanido, if it displays π effects, is a π acid.


How to balance this redox reaction taking place in a basic medium? How to balance this reaction which is taking place in a basic medium?
  $$\ce{Al + MnO4- -> MnO2 + Al(OH)4-}$$

My Attempt

Reduction half reaction:

$$
\begin{align}
\ce{4 H+ + MnO4- + 3 e- &-> MnO2 + 2 H2O}\\
\ce{28 H+ + 7 MnO4- + 21 e- &-> 7 MnO2 + 14 H2O}
\end{align}
$$

Oxidation half reaction:

$$
\begin{align}
\ce{Al + 4 H2O &-> Al(OH)4- + 8 H+ +7 e-}\\
\ce{3 Al + 12 H2O &-> 3 Al(OH)4- + 24 H+ + 21 e-}
\end{align}
$$

Adding half reactions:

$$
\ce{3 Al + 7 MnO4- + 28 H+ + 12 H2O + 21 e- -> 7 MnO2 + 3 Al(OH)4- + 14 H2O + 24 H+ + 21 e-}
$$

Adding $\ce{OH-}$ and eliminating $\ce{e-}$ and formed $\ce{H2O}$.

Final equation:

$$
\ce{3 Al + 7 MnO4- + 2 H2O -> 7 MnO2 + 3 Al(OH)4- + 4(OH)-}
$$

Please tell if I am wrong.

Writing an answer since I can't comment yet.

You are using the ion-electron method for basic medium, but are balancing using $\ce{H+}$ and later adding $\ce{OH-}$? Sounds suspicious to me, but the outcome must be the same.

Generally, I add $\ce{OH-}$ ions on the oxygen deficient side and then add water to compensate the hydrogen deficiency on the other side.

Note: your first half-reaction balancing the conversion of permanaganate to $\ce{MnO2}$ must be regarded incorrect, as $\ce{MnO2}$, being soluble in acidic conditions (presence of $\ce{H+}$, as the equation indicates!) would yield to further reduction, and end up as $\ce{Mn^2+}$.

Here's how I balanced it:

Oxidation half

$$
\begin{align}
\ce{Al &-> Al(OH)4-}\\
\ce{Al + 4OH- &-> Al(OH)4- + 3e-}
\end{align}
$$

Reduction half:

$$
\begin{align}
\ce{MnO4- &-> MnO2}\\
\ce{MnO4- &-> MnO2 + 2OH-}\\
\ce{MnO4- + 2 H2O &-> MnO2 + 4 OH- + 3 e-}
\end{align}
$$

Adding:

$$
\ce{Al + MnO4- + 2 H2O -> Al(OH)4- + MnO2}
$$

Your method is fine as well. The only mistake is the wrong $\ce{H+}$ coefficient in $\ce{Al}$ oxidation half-reaction, which will be 4 instead of 8.

P. S. (only concern): note the differences in half reactions in basic and acidic media. Conclusion: when products differ in acidic/basic conditions, it is best not to balance in the asker's method, as it cannot be considered entirely correct. Nothing more, nothing less.


Why is the order of reflection simplified to n = 1? My professor told us that for the sake of simplicity, we will assume that the order of reflection for Braggs' Law is always equal to 1 but failed to explain why. Is it because d(h,k,l) for materials has been determined using an order of reflection 1?

The common form of the Bragg equation is $\lambda=d_{hkl}\sin(\theta)$ but in crystallography it is used as $\displaystyle \lambda =\frac{2d_{hkl}}{n}\sin(\theta)=2d_{nh,nk,nl}\sin(\theta)$ where the indices are divided through by $n$ to give $\lambda=d_{hkl}\sin(\theta)$. The indices $hkl$ have to be integers (from derivation via Laue equations) but can also be multiples of integers, i.e. they can have a common factor. When the $hkl$ do have a common factor they are not strictly lattice planes but planes parallel to $(h/n, k/n,l/n)$ with a spacing $1/n$ of the lattice. Suppose a plane is (210) then each lattice plane passes through a lattice point (atom). If the indices were (420) then not all planes are equivalent as only half of the points (atoms) pass through a lattice plane so they are not equivalent. Thus sticking with $n=1$ ensures we always use for example, (210), and not (420) or (630) etc., and so always calculate the correct value for $d_{hkl}$


Is this amide compound electrophilic enough to be attacked by a cyclopentadiene in base? I am trying to make a fulvene molecule with dimethylamino and phenyl substituents. My worry is that the amide group donates electron density into the carbonyl carbon making it less electrophilic.

Although I have no prove for this, I am almost completely certain, that this will not work.

If you look closely at the cyclopentadienyl anion, what you would like to utilise as the nucleophile, you will notice that it is and aromatic system. (It even completely obeys the Hückel rules.) There are no electrons in an orbital of suitable symmetry, that would allow attacking the π* C-O bond.
In the other molecule, as you have noted, the electrons of the nitrogen will be interacting with the π carbonyl bond. Moreover, this part of the π system is also overlapping with the aromatic system of the phenyl moiety. That will certainly remove quite a bit of charge density from the carbonyl carbon. A nucleophilic attack at that position seems highly unlikely.


Do all flames require the same minimum concentration of oxygen to sustain themsleves? Assume you have burned some substance A in an airtight container until the fire goes out due to insufficient oxygen. You then place some burning substance B in the container through an airlock. Are there substances A and B such that the substance B would continue to burn for a significant amount of time using the oxygen left behind after substance A burned out?

In other words, do all materials require the same concentration of oxygen to sustain a flame?

No, different substances will cease combustion at different concentrations of oxygen. The concentration needed can be described by the limiting oxygen index. This is the reason that the use of pure compressed oxygen can be so hazardous; not only will materials burn that would not have burned before, but now materials that do burn require a lower temperature/energy input to ignite.

You can understand this conceptually when you consider where the heat goes. On combustion in air some of the heat must be used to heat the $\pu{79\!\%}$ of the air that isn't oxygen and that heat is wasted in the inert gas and the heat is diluted. That means only ~$\pu{20\!\%}$ of the heat goes into activation energy to propogate the combustion. When the oxygen is more concentrated, less heat goes into inert gases and more goes into oxygen to propagate the combustion and vise-versa.


What metals aren't dissolved in/attacked by aqua regia? Aqua regia (Latin: Royal Water) is one of the strongest acids known in Chemistry, and is capable of dissolving gold and platinum.

My copy of the Oxford science dictionary goes on to say (under the entry: Aqua regia) that metallic silver does not dissolve in aqua regia. Morever it does not  mention any other examples of aqua regia-resistant metals. Further down, it mentions that silver's invulnerability to aqua regia is due to the formation of a protective silver chloride coating on the metal, which serves to protect the metal from further decomposition.

However, this Wikipedia article claims:

  ...aqua regia does not dissolve or corrode silver...

This, I find contradictory to the dictionary's "formation of silver chloride" claim.

So,

What metals (elemental, forget alloys) are neither attacked by nor dissolved in (freshly prepared) aqua regia?
What makes those metals that don't dissolve or corrode in aqua regia so impervious to the acid?
Does silver metal actually develop a silver chloride layer on exposure to aqua regia? If so, would that mean the Wikipedia article is incorrect?

Keep in mind

The answer will depend upon the reaction conditions. Most importantly,

physical state of the metal: porosity, degree of comminution;
temperature;
mechanical aggravation of metal surface during reaction.

Often times a chemistry text mentions that no reaction occurs. The reaction might still happen. It is just that for the specified parameters the process is meaningless and negligible.

Short overview of aqua regia

Aqua regia is the $3:1$ volumetric mixture of $\ce{HCl}$ and $\ce{HNO3}$. Its additional reactive power draws from monochlorine created in situ.$^{[1]\ [2]}$

$$\ce{HNO3 + 3HCl -> Cl2 + NOCl + H2O\\ NOCl -> Cl + NO}$$

  Which metals are impervious to $3:1\ \ce{HCl/HNO3}$?

Almost every metal will react with aqua regia provided certain criteria are met.$^{[1]\ [2]}$ The closest you will probably get is ruthenium $\ce{Ru}$, and perhaps osmium $\ce{Os}$. To the best of my knowledge, $\ce{Ru}$ will not react with aqua regia in a meaningful way even if aqua regia is boiling.$^{[2]}$ The difference with $\ce{Os}$ is that powdered osmium is attacked by boiling aqua regia.$^{[1]\ [2]}$

$$\ce{Ru + HNO3 + HCl $\kern.6em\not\kern -.6em \longrightarrow$}$$

$$\ce{\underset{powder}{Os} + $\underbrace{\mathrm{HNO_3}}_{\text{boiling}}$ -> OsO4 + N_xO_y + H2O \\ OsO4 + 2H2O <=> H2[OsO4(OH)2] \\ OsO4 + HCl ->OsO2Cl2 + Cl2 + 2H2O\\ OsO2Cl2 + HCl ->OsCl4 + Cl2 + 2H2O\\ 2OsO2Cl2 + H2O <=> OsO2 + H2[OsO2Cl4] \\ 3OsCl4 + 2H2O <=> OsO2 + 2H2[OsCl6]\\ OsO2 + 6HCl <=> H2[OsCl6] + 2H2O}$$

Brief discussion about the list provided in the comments

Titanium $\ce{Ti}$ does react, and does so at room temperature.

$$\ce{3Ti + $\underbrace{\mathrm{12HCl + 4HNO_3}}_{\text{room temperature}}$ -> 3TiCl4 + 4NO + 8H2O}$$

Rhenium $\ce{Re}$ reacts slowly at room temperature $\ce{->HReO4}$. This will further react with $\ce{HCl -> ReCl4 + Cl2}$.$^{[2]}$

Hafnium $\ce{Hf}$ does react at room temperature. The reaction is slower than with titanium; overall equation is identical.$^{[2]}$

Tantalum $\ce{Ta}$ reacts when aqua regia is heated to $150\ ^{\circ}\mathrm{C}$. Rhodium $\ce{Rh}$ reacts in a grinded state. As a large compact piece, iridium $\ce{Ir}$ is affected over temperatures of $100\ ^{\circ}\mathrm{C}$. Niobium $\ce{Nb}$ is inert at room temperatures.$^{[2]}$

Summary: ruthenium $\ce{Ru}$ is your best bet.

  What makes metals $\ce{Ru}$ and $\ce{Os}$ so stable in aqua regia?

The nobility of these metals is not the best explanation. As you correctly pointed out, $\ce{Pt}$ and $\ce{Au}$ react fine. This is direct evidence that for other metals a protective layer should form. The layer varies from metal to metal, but usually is either an oxide (or oxide hydrate), or a chloride.

Effectiveness of mechanical aggravation also points to stable, non-reactive compound formation on the metal's surface.

For ruthenium, as of now I am unsure what this precipitate could be. If anyone has a reference, please edit or leave a comment.$^\text{[reference needed]}$

  What happens with silver?

Silver and aqua regia react very poorly, and for a short amount of time.$^{[2]}$ The culprit is $\ce{AgCl}$ ($K_s = 1.8 \cdot 10^{-10}$)$^{[2]}$. A slow reaction might still take place due to complexation.$^{[2]}$

Surprisingly, silver reacts with $\ce{HBr}$!$^{[2]}$ Its solubility product is even worse, $K_s = 5.0 \cdot 10^{-13}$.$^{[3]}$ My guess is that this layer is not as dense as $\ce{AgCl}$ but this still needs verifying.$^\text{[citation needed]}$

References

(In progress)

$[1]$ N. N. Ahmetov. Anorgaaniline keemia. (1974)

$[2]$ H. Karik, Kalle Truus. Elementide keemia. (2003)

$[3]$ Skoog, West, Holler, Crouch. Fundamentals of Analytical Chemistry. 9th edition. (2014)


Lowest Michaelis constant Km I want to find a lower limit of the Michaelis constant for some evaluations of Michaelis-Menten enzyme kinetics.
What is the lowest $K_m$ you ever encountered? Is there a theoretical limit?

Take this generic reaction representing the Michaelis-Menten enzyme kinetic:

You can derive the Michaelis-Manten constant $K_m$:
 (for derivation see https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics)

The quasi-steady state hypothesis was used, therefore the rate of formation and breakdown of $ES$ are the same; this is a good approximation if the enzyme concentration is much less than the substrate concentration $[S]$ or $K_m$ or both. That means that $K_m$ can be extremely small (almost zero) and the Michaelis-Menten equation still be valid proven that the enzyme concentration is much less than  $[S]$.
$K_m$ is very small if:
$k_r$ and $k_{cat}$ are both very small compared to $k_f$, thus if the complex $ES$ is formed very quickly compared to its breakdown.
In this case the reaction proceeds at its maximum velocity $V_{max}=k_{cat}*E_0$, with $E_0=[ES]+[E]$:

About the theoretical limit on how small $K_m$ can be there are two considerations:
 First, $k_f$ cannot be cannot be faster than the diffusion-controlled encounter of an enzyme and its substrate. This means that $k_f$ cannot be higher than $10^9 s^{-1} M^{-1}$  (ref.https://www.ncbi.nlm.nih.gov/books/NBK22430/#:~:text=This%20rate%20cannot%20be%20faster,s%2D1%20M%2D1.)
Second, $k_{cat}$ cannot be too small or the product $P$ formed is produced at a rate not suitable for the cell survival.
Generally, I haven't encountered $K_m$ smaller than $10^{-6}M$ in biological enzymes (e.g Triosephosphate isomerase, TPI) but possibly there are. In TPI, $ k_f$ ∼ $10^9 s^{-1} M^{-1}$, therefore  $k_r$  ∼ $k_{cat}$  ∼ $10^{3}s^{-1}$ that is still quite high. It would be very interesting to know if there are enzymes with an even lower $K_m$. Very intresting question.


Quantum mechanical explanation for Le Chatelier's principle? We know from basic chemistry that a reaction reaches equilibrium as described by Le Châtelier's principle. What's the quantum mechanical explanation to this principle? How a change in temperature/concentration causes an equilibrium to shift, from the point of view of wave functions?

The main thing that classical and quantum statistical mechanics have in common is probability.  In classical stat mech the equilibrium value of an observable is given by
$$\langle A \rangle = \sum_i^N p_iA_i$$
where $A_i$ is the value of $A$ in microstate $i$ and $p_i$ is the probability of microstate $i$.  The expression for $p_i$ at equilibrium depends on the ensemble (microcanonical, canonical, etc).  One way to define equilibrium is that the probabilities are constant
$$\frac{\partial p_i(t)}{\partial t}=0$$
So if one perturbs the system so that this equation doesn't hold the $p_i$'s will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.

In quantum stat mech the equilibrium value of an observable is given by
$$ \langle A \rangle = \tfrac{1}{N} \sum_\alpha^N \langle \Psi^{\alpha}| \hat{A} |\Psi^{\alpha}\rangle$$
where $\Psi^{\alpha}$ is the wavefunction for microstate $\alpha$

The wavefunctions are typically expanded in an orthonormal basis ($\phi$'s)
$$ \langle A \rangle = \tfrac{1}{N} \sum_{k,l} \rho_{l,k}\langle \phi_l| \hat{A} |\phi_k\rangle $$
where the density matrix is can be defined by a density operator
$$ \rho_{l,k} = \langle \phi_k| \hat{\rho} |\phi_l\rangle $$

The equilibrium is then defined as
$$\frac{\partial \hat{\rho}(t)}{\partial t}=0$$

So if one perturbs the system so that this equation doesn't hold the $\hat{\rho}$ will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $\Psi^{\alpha}$ contributions to  $\hat{\rho}$ increase of decrease.

The time dependence of $\hat{\rho}$ can be calculated by the time-dependent Schrödinger equation.  So Le Chatelier's principle is a heuristic way of predicting roughly how the time-dependent wavefunctions will evolve at the simulation approaches equilibrium.

To get a more physical picture one can introduce the Born-Oppenheimer approximation in which case the numerical simulation of the time-dependent Schrödinger equation becomes a classical MD simulation (with a QM energy function) from which the time dependence of the probabilities $p_i$ can be computed and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.


Among hydrogen, helium and carbon dioxide, which gas would behave most like ideal gas and why? I was wondering which gas would behave most ideally out of $\ce{H2},$ $\ce{He},$ $\ce{CO2}.$ All gasses are in the same condition.

I know the answer is either $\ce{H2}$ or $\ce{He}$ because of London dispersion forces. I was wondering which one it was and why.

In the simplest model, a gas is called ideal when its particles are point-like (no volume) and have no interactions. Real gases behave like ideal gases at low pressure (where the particle volume is neglible compared to the total volume) and high temperature (where condensed phases, i.e. interatomic or intermolecular interactions are disfavored).

The size-comparison between helium and dihydrogen is straightforward: Dihydrogen is larger. As for the strength of interparticle interactions, we can compare normal boiling points: Helium's is 4 Kelvin and dihydrogen's is 24 Kelvin.

So this would suggest that helium is "more ideal" as it has the lower boiling point and the smaller size.


Hydrolysis of epoxide There are two possibilities:

Oxonium ion formation
Cleaving of epoxide

Firstly, I don't know if epoxide is more reactive than carbonyl group.
If epoxide reacts with the proton,

But the given answer is:

In the above mechanism, an unstable carbocation is formed because its near a carbonyl group which is electron withdrawing:

I would go with your approach but your final product is unstable due to high repulsion's by $\pi$ electrons.

Hence your final product will react further to form a more thermodynamically feasible product.

NOTES:
In second structure two anti-periplanar migrations are possible and the one which leads to the more stable carbonation is favored (or consider the better migrating group based on electronegativity). Third structure rotates about spiral carbon due to repulsion's.


How can I predict if a reaction will occur between any two (or more) substances? I am new to chemistry and I find it fascinating. I am trying to learn about chemical reactions and I was wondering if there was an easy way to quickly tell if any combination of chemical substances would produce a reaction and what product(s) if any might be formed.

For example, if I pick any two random substances $\ce{A}$ and $\ce{B}$, can I determine if a reaction will occur and predict the products?

$$\ce{A + B -> \ ?}$$

More specifically, let's say I just learned that chlorine bleach (sodium hypochlorite) can be made by a reaction of sodium hydroxide and chlorine with sodium chloride as a byproduct:

$$\ce{NaOH(aq) + Cl2(g) -> NaOCl(aq) + NaCl(aq)}$$

Is there a way that I could have predicted this reaction (and any other) before I learned about it? I do not want to memorize the outcome of every combination o that I can answer questions about chemical reactions. I am hoping there is a short list of simple rules that govern all chemical reactions that I can commit to memory and then apply to any combination of substances. I might also like to be able to develop a simple computer program built around an algorithm for these reactivity rules that can sample databases of known substances and predict new reactions.

Can I predict the products of any chemical reaction?

In theory, yes!

Every substance has characteristic reactivity behavior. Likewise pairs and sets of substances have characteristic behavior. For example, the following combinations of substances only have one likely outcome each:

$$
\ce{HCl + NaOH -> NaCl + H2O} \\[2ex]
\ce{CH3CH2CH2OH->[$1.$ (COCl)2, (CH3)2SO][$2.$ Et3N] CH3CH2CHO}
$$

However, it is not a problem suited to brute force or exhaustive approaches

There are millions or perhaps billions of known or possible substances. Let's take the lower estimate of 1 million substances. There are $999\,999\,000\,000$ possible pairwise combinations. Any brute force method (in other words a database that has an answer for all possible combinations) would be large and potentially resource prohibitive. Likewise you would not want to memorize the nearly 1 trillion combinations.

If more substances are given, the combination space gets bigger. In the second example reaction above, there are four substances combined: $\ce{CH3CH2CH2OH}$, $\ce{(COCl)2}$, $\ce{(CH3)2SO}$, and $\ce{Et3N}$. Pulling four substances at random from the substance space generates a reaction space on the order of $1\times 10^{24}$ possible combinations. And that does not factor in order of addition. In the second reaction above, there is an implied order of addition:

$\ce{CH3CH2CH2OH}$
$\ce{(COCl)2}$, $\ce{(CH3)2SO}$
$\ce{Et3N}$

However, there are $4!=24$ different orders of addition for four substances, some of which might not generate the same result. Our reaction space is up to $24\times 10^{24}$, a bewildering number of combinations. And this space does not include other variables, like time, temperature, irradiation, agitation, concentration, pressure, control of environment, etc. If each reaction in the space could somehow be stored for as little as 100 kB of memory, then the whole space of combinations up to 4 substances would require $2.4 \times 10^{27}$ bytes of data, or $2.4\times 10^7$ ZB (zettabytes) or $2.4\times 10^4$ trillion terabytes. The total digital data generated by the human species was estimated recently (Nov. 2015) to be 4.4 ZB. We need $5.5\times 10^5$ times more data in the world to hold such a database. And that does not even count the program written to search it or the humans needed to populate it, the bandwidth required to access it, or the time investment of any of these steps.

In practice, it can be manageable!

Even though the reaction space is bewilderingly huge, chemistry is an orderly predictable business. Folks in the natural product total synthesis world do not resort to random combinations and alchemical mumbo jumbo. They can predict with some certainty what type of reactions do what to which substances and then act on that prediction.

When we learn chemistry, we are taught to recognize if a molecule belongs to a certain class with characteristic behavior. In the first example above, we can identify $\ce{HCl}$ as an acid and $\ce{NaOH}$ as a base, and then predict an outcome that is common to all acid-base reactions. In the second example above, we are taught to recognize $\ce{CH3CH2CH2OH}$ as a primary alcohol and the reagents given as an oxidant. The outcome is an aldehyde.

These examples are simple ones in which the molecules easily fit into one class predominantly. More complex molecules may belong to many categories. Organic chemistry calls these categories “Functional Groups”. The ability to predict synthetic outcomes then begins and ends with identifying functional groups within a compound's structure. For example, even though the following compound has a more complex structure, it contains a primary alcohol, which will be oxidized to an aldehyde using the same reagents presented above. We can also be reasonably confident that no unpleasant side reactions will occur.

If the reagents in the previous reaction had been $\ce{LiAlH4}$ followed by $\ce{H3O+}$, then more than one outcome is possible since more than one functional group in the starting compound will react. Controlling the reaction to give one of the possible outcomes is possible, but requires further careful thought.

There are rules, but they are not few in number. There are too many classes of compounds to list here. Likewise even one class, like primary alcohols (an hydroxyl group at the end of a hydrocarbon chain) has too many characteristic reactions to list here. If there are 30 classes of compounds (an underestimate) and 30 types of reactions (an underestimate), then there are 900 reaction types (an underestimate). The number of viable reaction types is more manageable than the total reaction space, but would still be difficult to commit to memory quickly. And new reaction types are being discovered all the time.

Folks who learn how to analyze combinations of compounds spend years taking courses and reading books and research articles to accumulate the knowledge and wisdom necessary. It can be done. Computer programs can be (and have been) designed to do the same analysis, but they were designed by people who learned all of the characteristic combinations. There is no shortcut.


Why does CH3CH(OH)CH2CH3 give iodoform test? The iodoform test is given by compounds containing $\ce{-(C=O)CH3}$ group or compounds which can be oxidized to such.
$$\ce{ CH3CH(OH)CH2CH3}$$

Is a 2 degree alchohol and even though it were to be oxidized it would not contain a $\ce{-(C=O)CH3}$ group but according to a question asked by JEE in 1997 it as a matter of fact does give the test.

HOW?

The substrate you have provided (2-butanol) would indeed give a positive test.

The iodoform test is performed in presence of iodine and potassium or sodium hydroxide, which first oxidizes the aforesaid compound to 2-Butanone.

Mechanism (taken from this Wikipedia page) which shows $\ce{BrO-}$ instead of $\ce{IO-}$, but would be executed in a similar fashion:

Thus 2-Butanone responds to iodoform test successfully.

References:

Ueber Entstehung von Jodoform und Anwendung dieser Reaction in der chemischen Analyse, Lieben, Ann. Spl. Bd. 7, 218 1870


Reorganization during ionisation for d block elements This is a quote from my textbook:

  The irregular trend in the first ionisation enthalpy of 3d lmetals,can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d orbitals. So the unipositive ions have dn configuration with no 4s electrons.There is thus a reorganization energy accompanying ionisation with some gains in exchange energy as number of electrons increases and from transference of s electrons into d orbitals.However the value of Cr is lower because of the absence of any change in d configuration.

What I basically understand from this is that during first ionisation, electron is lost from $\mathrm{4s}$ orbital because it have higher energy and forms uni-positive ion and the remaining one electron is transferred back into $\mathrm{3d}$ lower energy orbital as energy gap between $\mathrm{3d}$ and $\mathrm{4s}$ lowers. This transition of electrons into lower orbital gives a net decrease in ionisation enthalpy (where energy is released as reorganization energy). But I don't understand what they mentioned about $\ce{Cr}$.

https://www.chemguide.co.uk/atoms/properties/3d4sproblem.html

In this site when they explain about Vanadium electronic configuration it has been mentioned that Vanadium in its unipositive ion state have 4 electrons in $\mathrm{3d}$ rather than 3 electrons in $\mathrm{3d}$ and 1 electron in $\mathrm{4s}$ which supports the idea that electrons is transferred from $\mathrm{4s}$ to $\mathrm{3d}$.

Is it true that during ionisation there is a transference of electron from $\mathrm{4s}$ to $\mathrm{3d}$ orbital?Why the text mention reorganization and exchange enthalpy when talking about ionisation?

Text source:NCERT textbook ($\mathrm{d}$- and $\mathrm{f}$-blocks, page 215)

As your textbook notes, the relative energy of atomic orbitals is not fixed as you start ionizing your atom. As you ionize, all orbitals drop in energy. However, higher angular momentum orbitals drop more than lower ones, and specifically in your case, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ orbitals. So consider vanadium, the atom has a $\mathrm{[Ar]4s^23d^3}$ configuration, but after you ionize it, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ ones and you get $\mathrm{[Ar]3d^4}$ as the electron configuration of the $\ce{V+}$ ion. So your measured ionization energy includes both the energy it takes to remove an electron from $\ce{V}$ and also some reorganization energy. You know there is some reorganization energy since two electrons (both $\mathrm{4s}$ ones) have "moved" but you only removed one.

Now, what about $\ce{Cr}$? Recall that $\ce{Cr}$ is a bit special and has a $\mathrm{[Ar]3d^54s^1}$ configuration when neutral. After we remove an electron (in this case, a $\mathrm{4s}$ one), we have an $\mathrm{[Ar]3d^5}$ configuration for $\ce{Cr+}$. Note that no electrons had to move around besides the one we removed. So there is no reorganization energy in the case of $\ce{Cr}$, as your textbook notes.


Hess's Law/Enthalpy of reaction The problem:

  Chloroform, $\ce{CHCl3}$, is formed from methane and chlorine in the
  following reaction:

  $$\ce{CH4(g) + 3Cl2(g) ->3HCl(g) + CHCl3(g)}$$

  Calculate $\Delta rH^{\circ}$, the enthalpy of formation for this
  reaction, using the enthalpies of formation of $\ce{CO2(g)}$
  (-393.509kJ/mol-rxn), $\ce{H2O(l)}$ (-285.83kJ/mol-rxn) and
  $\ce{CHCl3(g)}$ (-103.1kJ/mol-rxn) and the enthalpy changes for the
  following reactions:

  $$\ce{(1) \ \ CH4(g) + 2O2(g) -> 2H2O(l) + CO2(g) \ \ \ \ \Delta
 rH^{\circ}=-890.4kJ/mol-rxn \\ (2) \ \ 2HCl(g) -> H2(g) + Cl2(g) \ \ \ \ \Delta
 rH^{\circ}=+184.6kJ/mol-rxn }$$

My attempt at the problem:

Given the hint in the problem statement and by looking at the two given reaction equations i assumed that i needed to create another equation. I created an equation for $\ce{CHCl3(g)}$ from $\ce{CO2(g)}$ and $\ce{H2O(l)}$ and calculated it's $\Delta rH^{\circ}$ from the enthalpies of formation as shown below (Let's call this equation 3):

$$\ce{2CO2(g) + 5H2(g) + 3Cl2(g) -> 2CHCl3(g) + 4H2O(l)}$$

$\Delta rH^{\circ} = -562.502kJ/mol$

Here's where i run into difficulties: I can see that in the 'target reaction' we have HCl on the right and CHCl3 on the right, so it stands to reason that equation 3 will stand as it is presented but equation 2 will switch direction (and the sign of the enthalpy of reaction swop too). I don't want to switch the direction of equation 1 around because it has CH4 on the left and in the target reaction CH4 is on the left. We could multiply equation 1 by 2 and equation 2 by 3 in order to have equal amounts of chlorines and Co2's so we can cancel them, however we are now left with hydrogens on the left and H2O's on the right with no way to cancel them. Hydrogen and water obviously don't appear in the target reaction so i have gone wrong somewhere along the line.

Am i missing a reaction equation? Is there something i can do to cancel the unwanted reactants in the reaction equations? Any help would be greatly appreciated.

If you add up the enthalpies of the following reactions with the given factor and divide the result by two you will get the desired result.

$$\ce{CH4 + 2 O2 -> 2 H2O + CO2} \tag{· 2}$$
$$\ce{2 C + H2 + 3 Cl2 -> 2 CHCl3} \tag{· 1}$$
$$\ce{C + O2 -> CO2} \tag{· -2}$$
$$\ce{2H2 + O2 -> 2 H2O} \tag{· -2}$$
$$\ce{2 HCl -> H2 + Cl2} \tag{· -3}$$

$$\ce{2 CH4 + 6 Cl2 -> 6 HCl + 2 CHCl3}$$


Would there be a difference if deuterium is embedded instead of protium (regular hydrogen) in acids? So instead of regular hydrogen, it would be a deuterium (still a Hydrogen). For example, instead of $\ce{HCl}$ it would be $\ce{DCl}$ where D is a deuterium.

Yes, there would be a difference. The acids would be weaker.  Why? Because the heavier isotope lowers the zero-point energy (the lowest possible energy) in the acid molecule, making the acid's bond to the D stronger than it would have been to the H. Acids in $\rm D_2O$ are weaker than they are in ordinary water, too.

Any reaction that involves dissociation of these acids will run more slowly, too. This is called "the kinetic isotope effect". You can read more about that here: http://en.wikipedia.org/wiki/Kinetic_isotope_effect

You might want to ask further questions: why is the zero point energy lower for a bond to D than a bond to H? How exactly does the lower zero point energy make the bond stronger? How exactly does it affect the rates of reactions that include acid dissociation?

Ask away!

If you'd like to see how substituting a D for an H in acids affects the acid dissociation constant for particular acids, here's a good reference:

R. A. Robinson, M. Paabo, R. G. Bates, "Deuterium Isotope Effect on the Dissociation of Weak Acids in Water and Deuterium Oxide", J. Res. Nat. Bureau. Standards, 73A(3), 1969, pp 299-308.


RO- as leaving group in saponification process If RO- is a stronger base than OH- (and leaving groups are weak bases), why is RO- a leaving group here?

$\ce{OH-}$ of course can, and will, be a leaving group here. The reaction is in equilibrium, $\ce{OH-}$ adds and leaves very many times, but that is a dead end as it does not drive the reaction forward. $\ce{RO-}$ can also be the leaving group - the difference between the basicity of $\ce{OH-}$ and $\ce{RO-}$ is not so great that it does not happen. Even if it happens one time in ten thousand, when it does the cascade it sets off makes the reaction irreversible so the reaction proceeds, if slowly, and the salt of the carboxylate is produced.


What exactly is a 'chiral centre' in a carbon compound? The typical progression (in books, YouTube videos) of explanation of chiral carbons in a compound is as follows:

Chiral compounds are non-superimposable by their mirror-image;
It is easily seen that only carbon atoms with all substituents different are chiral;
In a carbon compound with more than one atom, (2.) is exactly how you check for chirality;
It isn't necessary for a compound to be chiral despite having chiral carbons.

It seems to me that something between (2) and (3) is being eaten up. How does a chiral carbon in an chiral compound show the non-superimposability property? How does it satisfy the basic definition of chirality?

Skipping axial and helical chirality altogether, no to point #2: there are examples of atom centred chirality without a carbon atom in the centre.  The careful oxidation of thioethers may yield chiral sulfoxides:

(Han et al., Chem. Soc. Rev. 2018, 47, 1307-1350, doi 10.1039/C6CS00703A).
Here, chiral requires just the two carbon substitutents to differ, because the third is oxygen, and the fourth the lone electron pair.  Thus, applying CIP rules as usual where the lone electron pair has the lowest priority, DMSO is not chiral, contrary to methyl phenylsulfoxide which is.
Just as with stereogenic centres around carbon, deprotonations on the adjacent prochiral carbon atom may proceed with preference for one hydrogen over the other:

(Solladié, Synthesis 1981, 185-196, doi 10.1055/s-1981-29378).


Degeneracy of orbitals? Why is that in an external magnetic field(uniform) the degeneracy of d,f orbitals is lost but the degeneracy of p orbitals remain intact if the main cause of losing degeneracy is the difference in energy of differently oriented orbitals in the magnetic field?

My textbook states that the energy of the p orbitals will split in a magnetic field. (Physical Chemistry, McQuarrie, p.210)


Method and Tips for Inorganic Qualitative Analysis In a few weeks time, I will be doing an inorganic qualitative analysis test which is very pressed for time (a majority of people don't finish). The form of the test is most likely going to be that there will be around 9 unknown solutions (each contains 1 cation and 1 anion) which I have to identify. I will be given a list of all the possible cations and anions (there will be around 20 cations and  20 anions) and also we will have pH paper and a heat gun.

I wondering if anyone could suggest any tips on how to go about determining the composition of each unknown solution.

Currently, my method is to do it systematically by making a table and reacting all the solutions together. However, this can very time-consuming and probably not the most efficient way to identify the solutions.

Also, is there any distinctive reactions or properties that we will help me identify certain ions. For example, antimony and sulfide will give a unique peachy pink precipitate, or adding water to bismuth may cause a precipitate (as acid is usually required to make bismuth go into solution so by adding water, the acid is diluted and the bismuth precipitates out.)

This is a rough draft of the complete answer. A lot more will be added to the answer very soon. At present, it includes the common tests for sulfite and bisulfite ions. Please do not downvote just because it is not a complete answer yet.

Identification Of Sulfite Anion: ($\ce{SO3^{2-}}$)

All sulfites are insoluble except "group IA sulfites " and$\ce{(NH4)2SO4}$. Sulfite ppt. is always white in color.

Test With Acid:

$\ce{SO3^{2-} + H^{+} -> SO2 (g)}  $

Sulfur dioxide gas is evolved which gives lime water test as well as baryta water test:

$\ce{SO2 + Ca(OH)2 -> Ba(HSO3)2}[\text{soluble}]$

$\ce{SO2 + Ba(OH)2 -> Ca(HSO3)2}[\text{soluble}]$

Test with $\ce{BaCl2/CaCl2}$:

$\ce{SO3^{2-} + CaCl2 -> CaSO3 + 2Cl^{-}}$

$\ce{SO3^{2-} + BaCl2 -> BaSO3 + 2Cl^{-}}$

Note that $\ce{BaSO3}$ is slowly oxidized by atmospheric oxygen to form heavy white ppt of $\ce{BaSO4}$. Similarly:

$\ce{BaSO3 + Br2 -> BaSO4 + Br-}$

$\ce{BaSO3 + H2O2 -> BaSO4 + H2O}$

$\ce{BaSO3 + HNO3 -> BaSO4 + NO}$

Test With $\ce{Pb(NO3)2}$:

$\ce{Pb(NO3)2 + SO3^{2-} -> PbSO3 [\text{white ppt.}]}$

On boiling $\ce{PbSO3}$ in presence of $\ce{O2}$, $\ce{PbSO4}$ is produced.

(On boiling $\ce{PbS2O3}$ we get $\ce{PbS}[\text{black ppt}]$ and $\ce{H2SO4}$. Hence, this fact can be used to distinguish sulfites from thiosulfates).

Test With $\ce{AgNO3}$:

Schiff's Test/Fuschin's Test/Magenta Reagent Test:

Both $\ce{SO3^{2-}}$ and $\ce{SO2}$ are decolorized in this test.

Distinguishing b/w $\ce{SO3^{2-}}$ and $\ce{CO3^{2-}}$ ion:

Some Important Redox Reactions:

$\ce{MnO4- + SO2/H2S -> Mn^{2+} + SO4^{2-}/S}$

$\ce{Cr2O7- + SO2/H2S -> Cr^{3+} + SO4^{2-}/S}$

$\ce{Br2 + H2O + SO2/H2S -> Br- + SO4^{2-}/S}$

$\ce{Fe^{3+} + SO2/H2S -> Fe^{2+} + SO4^{2-}/S}$

$\ce{IO3- + SO2/H2S -> I2 [\text{gives deep blue color in starch}] + SO4^{2-}/S}$

$\ce{SO2}$ can either be oxidized to $\ce{SO4^{2-}}$ or reduced to $\ce{S^{2-}}$. But, $\ce{H2S}$ can only be oxidized to $\ce{S}$.

Notable Reaction (Comproportionation) : $\ce{\ce{SO4^{2-}} + \ce{H2S} -> S + H2O}$

Points to remember:

$\ce{CO2/CO/CO3^{2-}/HCO3^{-}}$ do not show any change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$
$\ce{S^{2-},SO3^{2-}/HSO3^{-}}$ do show change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$.
$\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O}$ $ [\text{no visible change is observed}]$

Dvarda Alloy ($\ce{Zn}/\ce{Al}/\ce{Cu}$):

$\ce{Zn + NaOH -> Na2ZnO2 + 2H}$

In presence of dvarda alloy the following reactions can take place:

$\ce{SO3^{2-} -> S^{2-}}$

$\ce{SO2 -> H2S}$

$\ce{NO2^{-} -> NH3}$

$\ce{NO^{3-} -> NH3}$

$\ce{Zn}$ reacts with dil. $\ce{H2SO4}$ to produce $\ce{H2}$ gas which can also reduce $\ce{SO4^{2-}}$ to $\ce{S^{2-}}$.

Test with Sodium Nitroprusside:

$\ce{Na2[Fe(CN)5NO] + ZnSO4 -> Zn[Fe(CN)5NO] [\text{salmon pink ppt }]}$

$\ce{Zn[Fe(CN)5NO] + \text{moist} SO2 -> \text{red color compound}}$

$\ce{S2O3^{2-}}$ and $\ce{S^{2-}}$ ions also give the test. So they can be removed on treating with $\ce{HgCl2}$ as follows:

$\ce{Hg^{2+} + S^{2-} -> HgS}$

$\ce{Hg^{2+} + S2O3^{2-} -> HgS2O3}$

$\ce{HgS2O3 [\text{on heating}]-> HgS [\text{black ppt}] + H2SO4}$

Identification Of Bisulfite Anion ($\ce{HSO3^{-}}$):

Test With Acid:

$\ce{HSO3^{-} + H^{+} -> SO2 (g)}$

Test With $\ce{H2O2}$:

$\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O} $

$\ce{HSO3^{-} + H2O2 -> H^{+} + SO4^{2-} + H2O} $ (acidic)


In this specific case is the proton considered product or catalyst? Assuming we have a reaction of

$$\ce{CO2 + H2O ->  HCO3- + \color{red}{H+}}$$

then is the proton (in red) is considered a product or a catalyst?

I read the Wikipedia article titled Product (chemistry), but the issue is still not clear to me. The article gives the following illustration:

In my original example above, I assume that the proton is considered a product since the hydrogen was originally part of the water molecule.

But in this case (see here, at 6:45) the hydrogen proton of the product of the reaction goes to the hemoglobin and kicks off an oxygen molecule.

As you've illustrated with your figure from the Wikipedia article, a catalyst appears as the same compound on each side of a chemical equation. The short answer to this part of your question is that since the proton only appears on one side of the equation, it cannot be a catalyst. A hydrogen atom bound to a water molecule is not the same thing as a proton. Rather, it is the result, or product, of a chemical reaction. Also note that the reaction given in your question would be more correctly expressed as:

$$\ce{CO2 + H2O <=> H2CO3 <=> HCO3− + \color{red}{H+}}$$

In the second part of the question, that regarding the video excerpt, the proton from your reaction actually becomes a reactant in a second completely different reaction. Here, the proton reacts with hemoglobin resulting in two new products: the protonated hemoglobin an an oxygen molecule.

I hope that I have addressed the crux of your question. Please don't hesitate to ask for any clarifications in the comments below.


What is a good method to measure the redox potential of a cellular system? I know how to measure the concentration of free thiols, which is reflective of the redox-potential of cellular compartment such as the cytoplasm or a lysosomes.

What other methods exist?

There are many complexes of d-elements, that can reversibly lose/gain one electron, changing their color in process. They are known as Redox Indicators. Some of them are stable enough to be used in vivo.


What determines malleability of metals? I'm designing an aircraft, and certain sections need to hold under a lot of stress. I know that it is the ways the metal crystallizes that effects the mallebility, but does the distance between the atoms/ crystal arrangement also effect malleability?

In general, malleability is determined by how easily metal crystals can be moved around. Anything that gets in the way, such as another substance blocking crystal movement, makes the metal less malleable.

For example, pure iron is malleable, but add some carbon to molten iron, and the carbon precipitates out on cooling, blocking crystal movement and creating much harder and more brittle carbon steel. Think of it as a two-phase material.


Raman Scattered (inelastic) photons rescattering outside of capture area Imagine that you are focusing the raman laser right at the center of the vial sample with water (or any liquid) using 532nm laser.

(from "Photometric Standards for Raman Spectroscopy)

For raman 180 degree backscattering sample geometry, can the inelastic scattering photons upon travel to the spectrometer becomes elastically scattered on way to it by the thick liquid sample such that it is scattered outside the area?

If yes, does it mean it is better not to focus so deep into the sample so the return path would be less (that is focusing the raman laser just after the glass on the liquid sample)? The side effect of this is getting some raman noises from the glass itself.

Yes, of course the Raman scattering can suffer from additional scattering on its way through the sample, just like it will generally be attenuated on its way to the detector by any element in its path (mirrors, lenses, glass surfaces, etc.).

If scattering is a problem or not very much depends on your sample. The ideal sample will not scatter your excitation laser or Raman scattering much, like for instance a clear solution, and then I would say the influence on the measurement is negligable. If you have something opaque, such as an emulsion, additional scattering will be relevant and you would want to measure near the sample surface.

If you have problems with a glass-affected background you could switch to quartz glass or use a confocal Raman setup, if available.

And also depending on your sample, absorption might be something you should consider as well.

By the way: I would probably rather go for a planar glass surface in order to avoid refraction as it occurs on a curved glass surface.

Edit due to comments:
Pure water does neither absorb nor scatter light with a wavelength around 532 nm much. If it would, you would be able to see this with your eyes. Wikipedia claims that a fraction of 10-7 of the incoming light is scattered inelastically. Although this should be only used as a rule of thumb, this tells you that a loss of Raman intensity due to "secondary" Raman scattering is not a practical problem you would usually be able to observe. Also, I would not be worried about elastic scattering although it is much more relevant than inelastic scattering (Wikipedia claims 0.1% to 0.01%).

I do not know your Raman setup in detail. But if you have anything like a confocal aperture (pinhole, slit...) on the path to the detector, refraction on the container/water surface will definitely reduce your signal intensity with increasing penetration depth, especially if you do not measure with an immersion objective lens. This is because your focus is severely distorted especially in length when focussing deeper into the sample, it will become much longer. Your confocal aperture then blocks part of your signal because it originates from a part of the sample which is far from the focal plane. This is descibed in detail for example in this paper, and also in this paper. However, the papers are behind a paywall.


Will a copper anode and hydride ion cell work? Can a galvanic cell with the overall reaction
$$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$ work?

$\Delta G$ of this reaction is negative (because $E^{\circ}$ of the reaction is $\mathrm{0.34~V}$ as the standard electrode potential of copper will be $\mathrm{-0.34~V}$ as it is acting as an anode). This shows the cell is spontaneous;

However, I have read that $\ce{Cu(s)}$ cannot reduce $\ce{H+}$ so I was wondering if this cell would work?

TL;DR - What you know from the activity series of metals is still true. You made a mistake in your interpretation of standard electrode potential. The rationale follows:

Standard electrode potentials are written in the reduction direction:

$$\ce{M}^{x+} + x\ce{e- -> M(s)}$$

Standard electrode potentials are measured against the standard hydrogen electrode:

$$\ce{2H+ + 2e- -> H2(g)}$$

Thus for the $\ce{Cu(s)|Cu^{2+}}$ electrode, $E^\circ$ is defined for the followed redox reaction (everything at 1 molar, 1 bar, and in standard states):

$$\ce{Cu^2+ + H2 -> Cu(s) + 2H+}\ \ \ E^\circ = +0.34$$

For the oxidation of copper with acid, the electrode potential would be negative in the direction you wrote it, which translates to a positive $\Delta G^\circ$, since:
$$\Delta G^\circ = - nFE^\circ$$


What fluorescent materials absorb visible light and emit UVC light? What fluorescent materials, if any, exhibit anti-stokes shift such that they absorb visible light and emit ultraviolet light in the range of 250-285 nm? I'm interested in particular in materials that are efficient in their conversion and not difficult to acquire.

It is not possible in a single molecule as energy conservation will forbid it. There is not enough thermal energy, $k_BT \approx 210 $ wavenumbers, visible to uv thousands of wavenumbers.  It is sort of possible using triplet-triplet annihilation ($\ce{T + T \to S^* + S}$ ) but this needs significant initial excitation and long lived triplets, which means de-oxygenated solution, and produces the same singlets as were excited initially. There are also inorganic up-conversion materials, for example some lanthanides, such as ytterbium based ones will convert near ir light to visible light.


Convert .car file PBC to basis vectors? I have a .car file that contains dimensions (periodic boundary conditions) for the unit-cell of a hexagonal crystal material I'm working with:

PBC   27.1979   27.1979   15.4999   90.0000   90.0000  120.0000 (P1)

That's the a, b, c lengths and $\alpha$, $\beta$, $\gamma$ angles.

I need to convert this information into basis vectors in cartesian space, e.g.

basis1 27.1979 ?       0.0
 basis2 0.0     27.1979 0.0
 basis3 0.0     0.0     15.4999

This may be better suited as a math question but it's computational-chemistry specific.

I found the way to convert the hexagonal parameters:

I have a,b,c and $\alpha$, $\beta$, $\gamma$. The solution in basis vectors is:

basis1 = (a, 0, 0)
basis2 = (xy, yx, 0)
basis3 = (xz, yz, c)

where
\begin{align}
xy &= b\cdot\cos(\gamma)\\
yx &= b\cdot\sin(\gamma)\\
xz &= c\cdot\cos(\beta)\\
yz &= \frac{b\cdot c\cdot \cos(\alpha) - xy \cdot xz}{\sqrt{b^2 - xy^2}}\\
\end{align}

Thus my solution was:

basis1 = (27.1979,  0,       0     )
basis2 = (-13.5990, 23.5541, 0     )
basis3 = (0,        0,       15.499)


Why doesn't N2O give a positive brown ring test? A common nitrate test, known as the brown ring test can be
performed by adding iron(II) sulfate to a solution of a nitrate, then
slowly adding concentrated sulfuric acid such that the acid forms a
layer below the aqueous solution. A brown ring will form at the
junction of the two layers, indicating the presence of the nitrate ion.

I know for sure that certain oxides of nitrogen, such as $\ce{NO}$ and $\ce{NO2}$ test positively in the brown ring test for nitrates, but I had never found a reliable source stating the same for the other oxides of nitrogen.
But one of the questions in a class test I took was:

Which one of the following does not give the Brown Ring test?

$\ce{N2O}$
$\ce{NO}$
$\ce{N2O3}$
$\ce{N2O5}$

The answer was given as $\ce{N2O}$ and I am unable to identify the logic behind the answer. Could it be because the nitrogen compound is reduced in the process of forming the brown complex with iron, while $\ce{N2O}$ already has too small an oxidation state to be reduced to $\ce{NO+}$?

The brown ring nitrate test is based on forming ferrous nitrosyl cation, with nitrous oxide formed from nitrates by iron oxidation in acidic solution:

$$\begin{align}
\ce{3 [Fe(H2O)6]^2+ + NO3- + 4 H+ &->3 [Fe(H2O)6]^3+ + NO + 2 H2O} \\
\ce{[Fe(H2O)6]^2+ + NO &->3 [Fe(H2O)(NO)]^2+} + H2O\\
\end{align}$$

Positive results are given also by nitrites and nitrogen oxides able to form nitric or nitrous acid, directly by hydration, or indirectly by disproportionation.

$$\begin{align}
\ce{ [Fe(H2O)6]^2+ + NO2- + 2 H+ &-> [Fe(H2O)6]^3+ + NO +  H2O} \\
\ce{N2O3 + H2O &-> 2 HNO2}\\
\ce{2 NO2 + H2O &-> HNO2 + HNO3}\\
\ce{N2O5 + H2O &-> 2 HNO3}\\
\end{align}$$

Therefore, $\ce{NO}$ and all higher nitrogen oxides oxidizing $\ce{Fe^2+}$, forming $\ce{NO}$, react positively in the brown ring test.

$\ce{N2O}$ cannot do that, because it cannot form $\ce{NO}$ at given conditions.


How is NH4Cl a salt? So what I know is, a salt is an ionic compound between a metal and a non-metal which exchange electrons. Like in $\ce{NaCl}$, $\ce{Na}$ is the metal and $\ce{Cl}$ is the non-metal, $\ce{Cl}$ hogs up the valence electron of $\ce{Na}$ and so the atom becomes charged.

But in the salt $\ce{NH4Cl}$, neither nitrogen, hydrogen nor chlorine are metals. All of them are non-metals. How can it be a salt?

This is not the correct definition of salt. The IUPAC definition of salt is "a chemical compound consisting of an assembly of cations and anions". In fact there are many salts which do not fit the definition you give.


Why do atoms stay nearer to each other even when delocalized electron moves away When the delocalized electron moves from between two atoms to other then why yet the measurement of distance between those older two atoms is less than normal single bond.

There is a significant quantum probability a delocalized electron is there, what in average attracts atoms together, compared to the state there was no such an electron.

E.g. for benzene, the distance is shorter than for a single bond, but longer than for a double bond, being effectively 1.5 bond.


Confused on normallity So normality is 'equivalents' over liters solution right? I'm having a hard time understanding exactly what an equivalent is. At first from one thing I read i thought it was just the H released from acids in the reactions but then something else started talking about ti with compounds without H, and in double/single displacement reactions. From part of what I read it sounds like a single would have 1 eq and 2 for double, but what found to read was pretty brief so I'm not fully sure. Can someone please explain these a bit better?

Normality = Equivalents of solute/liters of solution

Equivalents are determined by the number of electrons exchanged (or protons in water solutions, or yet other ions in different solvents). For example, if HCl is used to neutralize an alkali, then one mole of HCl will neutralize one mole of NaOH, but it takes two moles of HCl to neutralize one mole of Ca(OH)2 or three moles of Al(OH)3, because they have, respectively, two and three active OH- groups.

There are some pretty clear explanations in more detail at Molarity, Molality and Normality and at Normality.

And then there's Mel Brooks Abby Normal...


Chemical desalination Is there a way to remove salt from water, even just partially, using chemical or filtration methods prior to a reverse osmosis?

The problems is that salt is very hard on a reverse osmosis membrane which is expensive. Is there some way to attract some of the salt out of the water before using the membrane?

One possibility I have heard of is decanoic acid which can be used to sequester salt.

The method you mentioned is known as directional solvent extraction and seems quite straight forward and efficient. It is a relatively new method of desalination. Soybean oil, hexanoic acid, decanoic acid, and octanoic acid can be used (the last two are most effective) [ source: https://dspace.mit.edu/handle/1721.1/78539 ]. This video shows the setup used.


How to calculate bicarbonate and carbonate from total alkalinity It's been a long time since I did my chemistry classes and I'm currently trying to analyze groundwater samples for hydrogeology purposes. I would like to evaluate carbonate and bicarbonate concentration from groundwater samples, but I only have values of total alkalinity as $\ce{CaCO3}$, $\mathrm{pH}$, and temperature. Is it possible? Some of the $\mathrm{pH}$ values are above 8.3.

I know that:

If $\mathrm{pH}$ is above 8.3: $[\mathrm{alk}_{tot}]=[\ce{HCO3-}]+2[\ce{CO3^2-}]+[\ce{OH-}]-[\ce{H+}]$
If $\mathrm{pH}$ is under 8.3: $[\mathrm{alk}_{tot}]=[\ce{HCO3-}]+[\ce{OH-}]-[\ce{H+}]$

With the $\mathrm{pH}$, I can find calculate $[\ce{OH-}]$ and $[\ce{H+}]$. But how can I calculate $[\ce{HCO3-}]$ and $[\ce{CO3^2-}]$?

Thank you!

If I understood your question correctly, you have solutions where you know there is a given amount of calcium carbonate dissolved, and would like to know the distribution of this carbonate between all the species present. This proportion is commonly refered as the alpha($\alpha$) for a given species, that varies from 0 to 1(0% - 100%). From your question, I can make some assumptions:

As a groundwater sample, any solids dissolved are very diluted, so we don't need to worry about ionic strength;
pH is not fixed;
Temperature is not fixed, but I will assume its close to room temperature;
As other components are not mentioned, I will assume all carbonate comes from calcium carbonate.

Carbonic acid, $\ce{H2CO3}$, has two ionizable hydrogens, so it may assume three forms: The free acid itself, bicarbonate ion, $\ce{HCO3-}$(first-stage ionized form) and carbonate ion $\ce{CO3^2+}$(second-stage ionized form). The respective proportions in comparison with the total concentration of calcium carbonate dissolved are $\alpha0$, $\alpha1$ and $\alpha2$. They must sum to 1(100%), as in chemical reactions matter is neither created or destroyed, only changing between forms. When the calcium carbonate dissolves, a equilibrium is established between its three forms, expressed by the respective equilibrium equations:

First stage:
$$\ce{H2O + H2CO3 <=> H3O+ + HCO3-}$$
$$K1 = \frac{\ce{[H3O+][HCO3-]}}{\ce{[H2CO3]}} \approx 4.47*10^-7 $$

Second stage:
$$\ce{H2O + HCO3- <=> H3O+ + CO3^2-}$$
$$K2 = \frac{\ce{[H3O+][CO3^2-]}}{\ce{[HCO3-]}} \approx 4.69*10^-11 $$

You can also write a equation for the overrall reaction, by sum of each stage (and multiplication of the respective equilibrium constants):
$$\ce{2H2O + H2CO3 <=> 2H3O+ + CO3^2-}$$
$$K1K2 = \frac{\ce{[H3O+]^2[CO3^2-]}}{\ce{[H2CO3]}}$$

Analysing our system, to give a full treatment, if we know the solution pH, we can calculate $\ce{[H3O+]}$. So we are left with three unknown variables, $\ce{[H2CO3]}$, $\ce{[HCO3-]}$ and $\ce{[CO3^2+]}$. But so far we have only two independent mathematical equations, for K1 and K2 (the overrall equation does't count as independent, as it's only the merging together of the other two). To solve it, we need at least one more independent equation, to match the number of unknows. What we need is the equation for the material balance of the system. As we assumed all carbonate came from calcium carbonate, we can write:
$$Cs = \ce{[CaCO3]} = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$

Where Cs here stands for the known concentration of the salt, calcium carbonate. Now we can start replacing values taken from the equilibrium expressions into the material balance, isolating each unknow. For the bicarbonate, for example:
$$Cs = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$
$$Cs = \ce{\frac{[HCO3-][H3O+]}{K1} + [HCO3-] + \frac{K2[HCO3-]}{[H3O+]}}$$
$$Cs = \ce{\frac{[HCO3-][H3O+]^2 + K1[HCO3-][H3O+] + K1K2[HCO3-]}{K1[H3O+]}}$$
$$\frac{\ce{[HCO3-]}}{Cs} = \ce{\frac{K1[H3O+]}{[H3O+]^2 + K1[H3O+] + K1K2}} = \alpha1$$

So we got the expression for $\alpha1$, that has a curious structure: a fraction, where the denominator is a polynomial of degree 2, and the numerator its middle term. The same procedure can be repeated to find the expressions for the alphas  of the other dissolved species. For sake of brevity, I won't do it, but the final result will be:
$$\alpha0 = \frac{\ce{[H2CO3]}}{Cs} = \ce{\frac{[H3O+]^2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$
$$\alpha2 = \frac{\ce{[CO3^2-]}}{Cs} = \ce{\frac{K1K2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$

Note that a interesting pattern emerges. The expressions for the remaining two species have the same structure, just changing the term that goes in the numerator.

With the expressions for all species, it's helpful to use a spreadsheet to automate the calculations for a entire range of pH values, to grasp in a visual way what happens with carbonates as pH changes. I did just that, look at the results (here the spreadsheet, to whomever wants to download and play with it):

We see that in lower pH the predominant form for carbonate is the free carbonic acid. A bit over 6 bicarbonate ion takes over, and reigns up to pH a bit over 10, from where fully ionized carbonate ion takes over. For a given pH, the concentration of each species can be computed multiplying the respective $\alpha$ by the concentration of total calcium carbonate originally present.

The plot that looks like a "XX" also allows us to see a interesting property of carbonates. Nowhere in the plot you will find a pH value where we have the three species all in significant amounts. In the lower pH region you can find both bicarbonate and carbonic acid. But carbonate only shows up when carbonic acid goes away. It's like the unconfortable situation where you have two close friends who both hate each other. In a given moment I can see you in a room talking with either friend, but I will never see you three in the same room, or both friends of yours. The dividing line is close to the pH 8.6 you mentioned in your question. However, that sad situation has a upside. It makes the problem easier to calculate.

The full treatment I gave to this problem was indeed overkill. If I have three species, but only two show up together at any given time, I can "forget" I'm dealing with a diprotic acid. I need only to see the dividing line I've found, around pH 8.6. If I'm above it, free carbonic acid concentration is zero, and I have to deal only with the pair bicarbonate/carbonate, pretending the bicarbonate anion is just a monoprotic acid. From the equilibrium, we have:
$$\ce{[H3O+]} = \frac{\ce{K2[HCO3-]}}{\ce{[CO3^2-]}}$$

Or in logarithimic form:
$$pH = pK2 + log(\frac{\ce{[HCO3-]}}{[CO3^2-]})$$

This is the old Henderson–Hasselbalch equation you surely heard about before. As we know the pH and K2, we can calculate the ratio between carbonate and bicarbonate.

In the other side, if I'm below my dividing line near 8.6, carbonate ion concentration is zero, now I have to deal only with the pair carbonic acid/bicarbonate, pretending carbonic acid is just other monoprotic acid. From the equilibrium, we have:
$$\ce{[H3O+]} = \frac{\ce{K1[H2CO3]}}{\ce{[HCO3-]}}$$

Or in logarithimic form:
$$pH = pK1 + log(\frac{\ce{[H2CO3]}}{[HCO3-]})$$

As we know the pH and K1, we can calculate the ratio between carbonic acid and bicarbonate.

If you want to study in depth such calculations, I recommend this book: Butler, James N. Ionic Equilibrium: Solubility and PH Calculations. John Wiley & Sons, 1998.


Selectivity of diazo-coupling with p-amino phenol I came across this question and I've been thinking about it since, without being able to find a convincing major product. This question is related to one I found on this site, Diazo Coupling reaction with para-substituted phenol?, but it doesn't totally clear my doubt, and hence I specifically posted this reaction.

My thoughts were, since benzene diazonium is a weak electrophile, it would prefer to attack the most nucleophilic or electron rich site, and $\ce{-NH2}$ being better nucleophile than $\ce{-OH}$, I feel that would be more preferred site of coupling. Also I found out that aniline reacts with diazonium in basic medium to form diazo-aminobenzene which later rearranges to form more stable amino azobenzene; only to support my opinion. Thus my product was:

But the actual answer given was ortho coupling w.r.t. to $\ce{-OH}$, and it supports the product shown in the previously mentioned question, and @orthocresol commented that it occurs due to formation of phenoxide in basic medium:

But still it wasn't very convincing for me, as no one posted a solid answer to the question, neither are their any suitable articles I found on the internet that compare the reactivities of different functional groups towards diazonium.
I would be very obliged if someone could provide a deeper insight to the matter, and clear my misconceptions(if any), and a critical view about the barriers or aids in formation of the products.

I think the right way to approach this question isn't "we don't know the product, let's try to reason it out from first principles"; rather it should be more like "we found that this is the product when we ran the reaction in the lab, how can we rationalise this observation using a model of nucleophilicity that we know?"

So, in that previous comment which you quote, I did not mean to say "phenol will be deprotonated, hence this product will definitely be formed"; rather I meant something more like "if this is indeed the correct product, then a possible reason is because the phenol is deprotonated, making it (or the positions ortho to it) more nucleophilic than the aniline".

That said, I did just look it up, and there are indeed some references from the primary literature which support the claim that the product above is formed.

So, since we have experimental evidence (and insofar as this evidence is sound), you should not doubt what the product is; the only question left is why it is formed. And so I offer you the same explanation: that phenol is deprotonated under the basic conditions, and that phenoxide is more nucleophilic than aniline (at the ortho positions).

As for why phenoxide is more nucleophilic than aniline: I could offer you some wishy-washy explanations, like how a negative charge makes the oxygen atom more electron-donating towards the ring (and that outweighs the natural nucleophilicity ranking of N > O), but I don't think there is a simple one-line answer which is sufficiently rigorous for me to accept. I tried to look for some other evidence, but it is very difficult to compare C-reactivity of phenoxide vs aniline, since most EAS reactions take place in acidic media. There are some studies which conclusively show that phenoxide-O is more nucleophilic than aniline-N, but that isn't what we're looking for either.

So... all in all, it is better to use this experimental observation (that the azo coupling occurs ortho to oxygen) as proof that phenoxide-C is more nucleophilic than aniline-C, rather than to try to do it the other way round, i.e. "phenoxide-C is more nucleophilic than aniline-C, hence it must be true that this reaction will form this product". After all, chemistry is first and foremost an experimental science.

It's a subtle difference, but adopting the correct scientific approach will be better for you as a chemist in the future.


Calculate the root-mean-squared speed of free electrons in the Sun's corona given pressure, temperature and mass of an electron $m_\text{e}=9.1\times10^{-31}\ \mathrm{kg}$, $T=2\times10^{6}\ \mathrm{K}$ and $p=0.030\ \mathrm{Pa}$ the solution is given as: $5.4\times10^6\ \mathrm{ms^{-1}}$ but I got $9.5\times10^6\ \mathrm{ms^{-1}}$ using the following method (which does not use the value given for pressure. I simply plugged the numbers into the equation below:

$$v_\text{rms}=\sqrt{\frac{3K_\text{B}T}{m}}$$

Where have I gone wrong?

I'll go out on a limb here and say your answer is right and theirs is wrong.

Note that the answers differ by a factor of the square root of 3, so I'm guessing they forgot to include the factor of 3.

There would be a small correction for relativistic effects, but not enough to explain the differences in the two answers.


What is the qualitative definition of enthalpy in beginner terms? So in my coursebook it says enthalpy is a property that accounts for the increase in internal energy and the accompanying work done by the system and that its equal to the sum of the internal energy $U$ and product of pressure $P$ and volume $V$. Also on the net, mostly on the internet it says:

  A thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume.

And where it explains in other words I don't quite understand, e.g. as on Wikipedia. So please I would like anyone of you to explain (in beginner terms) that what is the meaning of enthalpy and heat content as used in the definition above.

In my judgment, it's not worth spending much of your valuable time trying to assign physical significance to the physical property we call Enthalpy.  Enthalpy is just a convenient parameter to work with when we are doing certain types of thermodynamic analysis.  One situation which it is particularly useful for is in applying the first law to open systems in which material is entering and/or leaving the system.  Another is where changes in a closed system are occurring at constant pressure.


What is the difference between non-adiabatic coupling and diabatic coupling? The diabatic couplings are the off-diagonal terms of wave functions written in a diabatic basis. There is a need to use the diabatic basis when there is a breakdown of the Born-Oppenheimer approximation. Then a transformation is done from the adiabatic to diabatic basis. But what does non-adiabatic coupling mean?

The reason I am asking is that most quantum chemistry softwares (I use Molpro) calculate the non-adiabatic coupling. Then some procedure has to be used to convert them to diabatic coupling. But what is the difference between the two and why do I need to use the diabatic coupling?

The difference is in the representation: adiabatic representation, being exact, allows you to make adiabatic approximation. If you do not want to have adiabatic approximation, there will be off-diagonal matrix elements in kinetic energy, so the coupling goes through kinetic energy. In diabatic representation, the kinetic energy is diagonal, and the couplings come from potential.
As was pointed out above, diabatic coupling allows you to apply Landau-Zener rule and estimate probabilities of transition.


Does evaporation reverse salt hydrolysis? It has been a long time since I studied organic chemistry, but one thing I do remember is that when we needed cyanide salts we were told not to bother keeping any remaining solution because the salts hydrolyze very quickly.

But can't hydrolyzed salts be fully recovered by simply evaporating the water?

E.g., we don't worry about evaporating NaCl from a distilled water solution and finding anything other than NaCl.  So I'm wondering if the lab instructions to neutralize and dispose of cyanide salt solution was just based on expediency.  Or: can something unsafe happen during the evaporation of a hydrolyzed salt solution?

Simple cyanide salts

In aqueous solution, simple cyanide salts more or less completely dissociate into their constituent ions.  E.g. for potassium cyanide:

$\ce{{KCN}(s) + H2O(l) -> K+(aq) + CN- (aq)}$

Those solutions are relatively stable.  There is an acid-base equilibrium between $\ce{CN-}$ and $\ce{HCN}$, but in unbuffered or in alkaline aqueous solution this equilibrium will be almost completely on the left:

$\ce{CN- + H2O <<=> HCN + OH-}$

Over very long times, if the solution is left in an open container, some of the tiny, minute amount of $\ce{HCN}$ formed in this way could evaporate, leading to a very slow loss of cyanide from the solution.  If stored closed this will not be a problem.

(Mixing simple cyanide salts with an acid will result in an immediate and dangerous release of gaseous $\ce{HCN}$, so in this question I assume that no acids are involved.)

Thus, for simple cyanide salt solutions stored appropriately, yes you should indeed be able to recover all of the initially added the solid salt after boiling off or otherwise removing any water.

Cyanide complexes of transition metals

Salts such as potassium ferricyanide $\ce{K3Fe(CN)6}$ consist of potassium cations and complexed ferricyanide anions, $\ce{Fe(CN)6^{~3-}}$.  They are soluble in water as well, but unlike their simple cyanide anions, they can be hydrolyzed.

The hydrolysis is catalyzed by UV light and results in the formation of free $\ce{CN-}$ anions as well as iron hydr(oxide) solid precipitates.

See this book for more information.


How to know which atom is the central atom in VSEPR theory? How to find the central atom? What is the central atom in $\ce{PCl3}$
and  how to find the shape using VSEPR theory?

You know that phosphorus has 5 valence electrons and that chlorine has 7 valence electrons ... and that you somehow have to combine them.

As chlorine is more electronegative (Allred-Rochow: Cl 2.83, P 2.06) you know that each chlorine will attract one electron from the phosphorus to get the noble gas configuration with 8 electrons in the valence shell.

Now that in turn means that there will be one single bond between the phosphorus and each chlorine, which automatically makes phosphorus the center.


Effect of dilution on weak acids pH My textbook has a question which says what would happen to the pH of a weak acid if it is diluted with water. The answer is its pH would decrease.
My teacher’s argument is that it would indeed decrease because water ionizes to hydronium ions which would increase the concentration of H+ ions and therefore, decrease the pH.
My argument is that it would increase because when the acid is diluted, the concentration of H+ ions will decrease and hence, the pH will increase. Further details in the attached photo.
My father, who is a chemist, argues that it would decrease then increase. He says that it varies upon the amount of water added and that it is not a direct relation. Instead, it would form a graph similar to the letter V.
I am honestly lost. What would exactly happen?

Whatever the nature of the acids, weak or strong, dilution of an acidic solution produces an increase of pH. The dissociation coefficient may increase, but the pH will always increase. The expression "like a Y letter" is pure fantasy.
Your calculations about acetic acid are correct. Just a small remark : in the second part, you say :"if  you double the volume". It should be stated that this operation is made by adding pure water. In this case, the total concentration of acid is divided by 2. Surprisingly, the concentration of $\ce{H+}$ is not divided by $2$ : $\ce{[H^+]}$ = $2.99 ·10^{-3} mol L^{-1}$. So it has been divided by $√2$ (and not by $2$).
Now, let's speak about the number of moles. During the dilution, the total number of moles of acetic acid has not changed, but the total number of moles of $\ce{H^+}$ has slightly increased during the dilution, which means that the acid is more dissociated in the diluted solution.


Can two atoms with the same amount of electrons but differing electronic configurations be considered isoelectronic? Can two atoms with the same amount of electrons but differing electronic configurations be considered isoelectronic?

For example, would we consider yttrium(I) $(\ce{Y+})$ with configuration: $\ce{[Kr] 5s^1 4d^1}$ and strontium $(\ce{Sr^0})$ with the configuration $\ce{[Kr] 5s^2}$ to be isoelectronic?

$(\ce{Y+})$ with configuration: $\ce{[Kr] 5s^1 4d^1}$ and strontium $(\ce{Sr^0})$ with the configuration $\ce{[Kr] 5s^2}$ are not isoelectronic. In order to be isoelectronic the two atoms would have to have the same electronic structure as well as the same number of valence electrons.

IUPAC Goldbook

  isoelectronic - Two or more molecular entities are described as isoelectronic if they have the same number of valence electrons and the same structure, i.e. number and connectivity of atoms, but differ in some of the elements involved.


What is the best approach to balance this equation? Why I cannot do it efficiently? What is the best approach to balance this equation?
$$\ce{CO2 + H2O <--> C2H2 + O2}$$

I did it at first this way and I was not successful:
\begin{align}
\ce{4CO2 + H2O &<--> C2H2 + 3O2}\\
\ce{2CO2 + 6H2O &<--> 6C2H2 + 5O2}\\
\ce{12CO2 + 6H2O &<--> 6C2H2 + 5O2}\\
\end{align}

Should I start from right side or left? They say start from the most complicated molecule? If it is like that, then what is the most complicated one and why?

This may be a bit dumb method, but sometimes I do use it as it is pretty straightforward and doesn't require any guessing. I setup a few equations and try solving over integers.
 $$\ce{aCO2 + bH2O <--> cC2H2 + dO2}$$

Balancing $\ce{C}$ atoms both sides,
$$a = 2c\\$$
Balancing $\ce{H}$ atoms both sides,
$$2b=2c\\$$
Balancing $\ce{O}$ atoms both sides,

$$2a+b=2d$$

Solving gives this. Not hard to see that $a=4$ does the job.