Water purification procedure that does not involve a flocculant of any kind? I have read about "electroflocculation" that uses less flocculant/alum, but is there a method – no matter how expensive or impractical – that uses no flocculant at all?

This of course depends on the scale and intent of the purification/treatment process, but flocculation is but one step in the water treatment process. So, I'm not sure whether you are asking if you can purify water without flocculation, or if you can get flocculation/aggregation to occur without the addition of floccing agents.

The simplest answer to water purification, on a small scale, is filtration and distillation. Flocculation serves to remove solid particles from the water which contribute to turbidity, but which may also include pathogens that are of real concern to human safety. On larger scales (household to community), incredibly effective methods for removing these types of contaminants include various forms of filtration. Some of the first successful water treatment plants on a large scale used sand filtration, and fast and slow sand filters are still used widely today. There are many commercially available drinking water filters for the removal of sediment, microbes and chemicals which are highly effective. Of course, on an industrial scale, filters of this nature would require frequent changing, and this is where sediment removal via flocculation is most effective in reducing costs. You could Google 'water purification in developing worlds' for a range of water purification methods that do not involve flocculation.

On the other hand, inducing flocculation without the addition of flocculant again depends on nature of the contaminant you wish to remove. The biofuel industries are pursuing this goal, developing methods of harvesting valuable microalgae stocks. Autoflocculation (or bioflocculation) of microalgae is known to occur as a function of pH, which can be controlled by restricting CO2 levels. Again a Google search on 'microalgae autoflocculation' will give you many hits. Another method is the light induced autoflocculation of microalgae. There is an interesting post on ResearchGate that may be of some interest to you if you are focused on flocculation of microalgae (if not, then ignore!)


how to obtain potential curves from database? I'm sorry if this is a really super basic question. I am looking at the potential curves of a diatomic molecule in a book , and to plot them. I google the molecule and got to the NIST website, there I get diatomic mass and constants for each state,  such as:

Te -     minimum electronic energy (cm-1)
ωe  - vibrational constant – first term (cm-1)
ωexe -     vibrational constant – second term (cm-1)
ωeye - vibrational constant – third term (cm-1)
Be -   rotational constant in equilibrium position (cm-1)
αe -   rotational constant – first term (cm-1)
γe -   rotation-vibration interaction constant (cm-1)
De -   centrifugal distortion constant (cm-1)
βe -   rotational constant – first term, centrifugal force (cm-1)
re -   internuclear distance (Å)
Trans. -   observed transition(s) corresponding to electronic state

Great, I should be able to use these to plot the potential energy curves for each electronic state somehow right? there is no formula i can see that here, is it just morse potential? what happens if the state is unbound?
Please help me understand how to plot the potential curves. Thanks!

The RKR method determines the classical turning points $r_{\pm}$ of the potential energy based on knowing experimentally determined spectroscopic constants. ( Rydberg, 1931, Z. Physik. v73,376, Klein, 1932 Z. Physik. v76, 226, Rees Proc. Phys. Soc.(Lond.) 1947, v59, 998.)
The energy equation takes the usual form. Parameters all in cm$^{-1}$, distances in m, reduced mass $\mu$ in u, $h$ in cm$^{-1}$s.
$$E_v=\omega_e(v+1/2) - x_e\omega_e(v+1/2)^2 + y_e\omega_e(v+1/2)^3 + z_e\omega_e(v+1/2)^4 +\cdots$$
The vibrational quantum number $v$ should be considered as a variable that has values from $-1/2$ to whatever positive value is required.
The derivative of the rotational energy wrt. $J$ at $J=0$  is also needed,
$$B'(v)= B_e-\alpha_e(v+1/2)+\gamma_e(v+1/2)^2+\cdots$$
The function representing the turning points is
$$r_\pm= \frac{f(v)}{2}\left (\sqrt{1+\frac{4}{f(v)g(v)}}\pm 1 \right)$$
where functions $f, \, g$ are integrals. These functions 'blow up' at the upper integration limit $x \to v$ so a correction is added to prevent this. The basic equation is
$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\int_{-1/2}^v\frac{dx}{ \sqrt{E(v)-E(x)}}$$
and the modified one
$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\left(\int_{-1/2}^{v-\delta}\frac{dx}{ \sqrt{E(v)-E(x)} }+ 2\sqrt{\frac{\delta}{Q_v}}\right)$$
where $\delta$ is small $\approx 10^{-5}$ and
$$Q_v=\omega_e - 2x_e\omega_e(v+1/2) + 3y_e\omega_e(v+1/2)^2 + 4z_e\omega_e(v+1/2)^3 +\cdots$$
The function $g$ with correction is
$$g(v)=\frac{4\pi\sqrt{2\mu}}{h} \left( \int_{-1/2}^{v-\delta} \frac{B'(x)}{\sqrt{E(v)-E(x)} }dx+2B'(v)\sqrt{\frac{\delta}{Q_v} } \right) $$
A set of values for CO is
$\mu = 6.85841,\; \omega_e= 2169.8135,\; x_e\omega_e=13.2883,\; y_e\omega_e=0.010511,\; z_e\omega_e= 0.000057$
$ B_e= 1.9312808,\; \alpha_e= 0.0175044,\; \gamma_e= 0.000000548$.
Plot energy $E(x)$(cm$^{-1}) $ vs $r_\pm(x)$ (pm) is shown. The value at $x=-1/2$ should be  $r_e=103.14$ pm the equilibrium bond length with energy of zero. As values are based on cm$^{-1}$ a scaling of $10^{14}$ is needed to go to pm.

I used Jupyter notebooks (via Anaconda) and python 3 to do the calculation using built in integrator from numpy/scipy and matplotlib to draw the graph. All are free and very easy to use.


Why does increasing orbital energy mean an electron can be more easily lost? Why is the higher the orbital energy, the easier the electron lost?
Also, if the distance between the electron and nucleus is getting father，the   electronegativity is getting weaker, the electron is getting easier to lost?
I feel confused about this two concepts.

Why is the higher the orbital energy, the easier the electron lost?

Orbital energies are measured relative to an electron at infinity (i.e. one which has been ionized), which corresponds to an energy of 0 eV. An electron which is lower in energy than this (i.e. bound to an atom) has a negative energy.

The first six energy levels for hydrogen are shown above. As you can see, the energies converge to 0 as n, the principal quantum number, goes to infinity. In order to ionize an electron it must be given enough energy to promote it to the n = ∞ energy level. Therefore the higher the energy of the electron, the less energy is required to ionize it.

Also, if the distance between the electron and nucleus is getting father，the electronegativity is getting weaker, the electron is getting easier to lost?

The most likely distance of the electron from the nucleus (the maximum in the orbital's radial distribution function) increases with n and so the electron is in a higher energy orbital and is more easily lost.


How did Halogens become known as Halogens? They are not the only elements that form salts! Having never given it a though before, I recently discovered (in a different context) that the prefix halo- actually means 'salt' or 'sea' and the suffix -gen means 'to form' or 'to generate'. So the Halogens are the elements that 'form salts'. But there are salts that do not have halogens in them, like $\ce{Na2S}$ or $\ce{(NH4)2SO4}$. Was it known at the time that other salts existed, and what other names for this group might have been considered?

On a side note, why aren't Group I metals called halogens? Doesn't it take two to tango (I mean, to form salts?)

This seems like a bit of a rhetorical question, so this isn't a terribly formal or authoritative answer, but anyhow - a lot of chemical nomenclature is like lava flow. It solidified and people just worked around it.

The halogens are so named because they have a rich chemistry of ionic compounds (fluorine through iodine, anyhow). However, both the halogens and the group I metals can form a wide range of things that aren't salts. The noble gases can form compounds with elements of low birth. The rare earth elements aren't particularly rare. Oxygen ('acid-former') is not a necessary component of acids. Technetium ('artificial' + ium) is produced in nature in significant quantities.

In addition, a lot of chemistry defies our human efforts to succintly categorise things, so at some point chemists have tried their best to find pragmatic general descriptors that unite groups of elements or molecules on the basis of the properties or constitution. The divide between organic and inorganic chemistry and the resulting exceptions and edge cases in classification (such as mellitic anhydride) is a good example of this. The map is not the territory.


How to assign the signals in the H-NMR of Aspirin? I need to assign the H-NMR of aspirin, below is aspirin with the H's labelled.

The peaks I have are:

2.30ppm   (I this is a singlet and would be F)
7.07ppm   (I think this is E)
7.29ppm   (I think this is D)
7.53ppm   (I think this is C)
8.05ppm   (I think this is B)
11.44ppm (this is a singlet and would be A)

For B,C,D,E I need to say what kind of splitting pattern there would be and how many coupling constants are present and there approximate value. I think I know the assignments of them but I don't know the splitting pattern or coupling constants.

Would E and B be doublet of doublets because they couple with D and C so they would have ortho and meta coupling?

Would D and C be coupling with each other and B and E so would they be doublet of doublets of doublets, with two ortho and one meta coupling?

Here is a link to the actual assigned proton nmr spectrum for acetylsalicylic acid.  Your peak assignments are correct.

  Would E and B be doublet of doublets because they couple with D and C
  so they would have ortho and meta coupling?

Yes, that is correct. Here is another link with more detail on the coupling constants (see p. 13 for example)

  Would D and C be coupling with each other and B and E so would they be
  doublet of doublets of doublets, with two ortho and one meta coupling?

Conceptually you are correct, however in the case where two coupling constants are very similar, the expected quartet collapses to a triplet. Since $\ce{J_{DC}}$ is very close in magnitude to $\ce{J_{BC}}$ the expected quartet appears as a triplet.  Therefore, proton C appears as a triplet from these two protons and then each of these lines is further split into a tight doublet by the smaller coupling to proton E.  The overall pattern appears as a triplet of doublets.  The same reasoning applies to proton D, and again a triplet of doublets results.


Simple Distillation of Unknowns Can an identification experiment (using acetonitrile, isopropyl, and cyclohexane) using simple distillation and boiling point be improved by using a digital thermometer?

Can simple distillation be improved in other ways, other than a more accurate thermometer? That's all I've come up with.

Ok, back to the start on all of this....

If you were looking for pure compounds or azeotropes what I'd want is a setup where a fractional distillation column was used with a very precise thermometer and a balance measuring the weight of the distillate. You could then nicely plot temperature vs weight.

Assuming that you just want to detect the pure compounds and/or azeotropes and not collect them.

Part of the problem here is that it takes some time to get the whole fractional distillation column up to temperature. So the temperature of the first drops is suspect.

If you're using a heating mantle you might be willing to boil the pot dry. But you wouldn't want to do that with a flame. So you might lose a bit on the back end of the distillation too.

The fractional distillation column doesn't have an infinite number of plates so the transitions from one compound/azeotrope to the next will be a bit "fuzzy." So if you were trying to collect "pure" compound/azeotrope fractions you'd avoid collecting the distillate from the transitions in those "pure" compound/azeotrope fractions.

Does this help?

EDIT

I guess I'd like two thermometers. One over the pot and one at top of the fractional distillation column.

Some sort of computer controlled temperature setting for the pot would be nice too.


Difference between titanium and titanium dioxide electrode I need an inert anode to perform some electrolysis.  I've tried graphite, which works to some extent, however I want to upgrade to titanium, which is pretty reasonably priced, and will perform better than the graphite.

My question is whether there is a difference between "Titanium oxide electrodes" and plain titanium.  Does a titanium dioxide electrode need to be specially treated to give it more inert characteristics, or does the oxide layer form naturally on the titanium?

I ask because I have seen this type of electrode classified as a Mixed Metal Oxide (MMO) electrode, but I have also seen some say that titanium is inert on its own.  I would like to now whether plain titanium is a viable electrode so I don't end up buying some that ends up corroding quickly during electrolysis!!

Any information in this subject would be greatly appreciated.

Thanks in advance!!

Used as an anode, a normal untreated piece of titanium metal will "anodize", forming a non-conductive titanium dioxide coating.  In my experience, the current through a large untreated titanium sheet anode drops below the milliamp level in under a minute.

But there are a variety of electrically conductive but oxidation resistant micron-scale coatings that can be formed onto titanium, including platinum (a "platinized titanium anode"); conductive oxides of common metals like Pb or Sn (lead dioxide or tin oxide); or conductive oxides of rare metals like Ir, Ru, or Ta (a "mixed metal oxide anode").  There are some decent electron microscope microstructure photos in Umicore's promotional catalog.

The advantage to using a "valve metal" like titanium or aluminum as the substrate on a coated anode is if part of the fancy conductive coating flakes off, the bare titanium just anodizes and stops conducting, rather than getting eaten away and dissolving into the electrolyte.


Converting SMILES string into valid sub-strings I have a SMILES string as follows.
smi= 'CCOC(=O)C1=C(COCCN)NC(C)=C(C(=O)OC)C1c1ccccc1Cl'

Is there anyway that I convert this SMILES string into valid sub-strings which can be later processed in RDKIT.
For instance
smi_sub1= 'C1c1ccccc1Cl'
smi_sub2= 'CCOC(=O)C1=C(COCCN)NC(C)=C(C(=O)OC)'
smi_sub3= 'C1=C(COCCN)NC(C)=C(C(=O)OC)'

These sub-strings should be subset of the main SMILES string and should also be valid from the SMILES syntax point of view.
So that when I use the function MolFromSmiles, It get processed without any error or none in return?
Mol_sub1 = Chem.MolFromSmiles(smi_sub1)
Mol_sub2 = Chem.MolFromSmiles(smi_sub2)
Mol_sub3 = Chem.MolFromSmiles(smi_sub3)

While this might work in some cases, I am not sure how well it will work in general. The problem here might be reflected in this statement:

These sub-strings should be subset of the main SMILES string

Not every substring will be a subgraph of the parent molecule. As mentioned in the comments, this seems essentially a fragmentation problem. Take this one of your examples:

smi_sub1= 'C1c1ccccc1Cl'

The reason this won't parse is that it starts with a 'C1' but has no corresponding 'C1' to finish the ring. So it should be 'Cc1ccccc1Cl', which is 1-chloro-2-methylbenzene.
Theoretically, you could take substrings, then post-process them to remove 'disconnected' ring numbers. However, it would make more sense to use proper fragmentation on the molecular graph, and then you would have correct subgraphs anyway, without the need to convert them from SMILES.


Why do gases only transmit sound waves whose wavelength is long relative to the mean free path? Also, when is this fact useful in practice?

Since sound waves are longitudinal waves (generally in air) there needs to be some kind of interaction between neighboring "air" molecules (a mixture consisting mostly of $\ce{N2}$, $\ce{O2}$, $\ce{CO2}$ and $\ce{Ar}$). You can see that this is just not possible if the mean free path of the "air" molecules is larger than the desired wavelength. How would the "air" molecules interact on such distance if there are no "air" molecules to interact with? This interaction (i.e. sound of a particular wavelength) will only be possible if there are sufficiently many "air" molecules nearby to interact with, or formulated otherwise: if the mean free path is sufficiently much smaller than the desired wavelength of the sound. Of course, it is possible to make sound waves with a larger wavelength than the length of the mean free path, just not the converse.

One application I can think of lies more in Physics but whatever. In the days before Einsteins theory of special relativity was developed/accepted, there was a common consensus, which was supported by Maxwell himself, that light was travelling in a special reference frame called "The Aether". In this reference frame the speed of light equaled $c$. We now know of course that light can travel through vacuum and that there exists no such special reference frame as The Aether. There were various problems with this Aether theory, most famously the Michelson-Morley experiment that suggested that either the Aether did not exist or was moving for a weird reason with the exact same trajectory and speed as earth in space. Another problem, related to your question, was that the speed of waves in a medium is dependent on the particle density of that medium. And since (at least some) electromagnetic waves have an incredible small wavelength (the $\lambda$ of visible light for example was known), the density of this Aether medium had to be extremely high, yet it had to also weight almost nothing and be able to penetrate trough everything!

From this and other facts Einstein must have concluded that the Aether theory was in fact invalid. This idea ultimately lead to the theory of special and general relativity, both of which have many spectacular and far reaching consequences.


Number of grams of CO2 released from a carbonated water The partial pressure of $\ce{CO2}$ gas in a bottle of carbonated water is $3.84\mathrm{~atm}$ at $25\mathrm{~^\circ C}$.  How much $\ce{CO2}$ gas (in g) will be released from $0.5\mathrm{~L}$ of the carbonated water when the partial pressure of $\ce{CO2}$ is lowered to $1.04\mathrm{~atm}$?  $25\mathrm{~^\circ C}$, the Henry’s law constant for $\ce{CO2}$ dissolved in water is $1.65 \times 10^3\mathrm{~atm}$, and the density of water is $1.0\mathrm{~g/cm^3}$.

First, I found the number of moles of $\ce{CO2}$ using this formula:

$$n = \frac{PV}{k}$$

So the answer that I found which is $0.000351$, can I multiply it by a molar mass of a $\ce{CO2}$ to get the number of grams inside the bottle? This approach seems too easy for me, but at the same time it makes sense.

This shows part of the problem.

Henry's law as given is

$K_{AW}^{px} = $1.65*10^3$ = \frac{p_{\ce{CO2(gas)atm}}}{[\ce{CO2(aq)}]}$

where:
$K_{AW}^{px}$ atm = aqueous-phase mixing ratio of $\ce{CO2}$ and the units are $atm$

$[\ce{CO2{(aq)}}]$ = the concentration of $\ce{CO2{(aq)}}$ in $mole/L$

$p_{\ce{CO2(gas)atm}}]$ = the partial pressure of $\ce{CO2{(gas)}}$ in $atm$

So rearranging...

$[\ce{CO2(aq)}] =\frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}$

The moles of $\ce{CO2}$, $m_{\ce{CO2}}$, is given by concentration, $[\ce{CO2}]$, times the volume, $v$ in $L$.

$m_{\ce{CO2(aq)}} = [\ce{CO2(aq)}]v =  \frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}v$


Why is density of water maximum at 4 degree celsius? Why is density of water maximum at 4 degree celsius?

We were having a discussion to this question in class and I could gather the following points.

1) $H_2O$ exists in a cage like structure in ice form, due to extensive hydrogen bonding.

2) density=mass/ volume : We are not changing the mass here, therefore increase in volume will directly correspond to decrease in density and vice versa.

2) This cage like structure can trap gases and result in increase in volume. Therefore, forming of this cage like structure will result in increase in volume and breaking of this structure will result in decrease in volume.

3) At 273K , this cage like structure of ice exists as shown. On heating, we find that molecules gain kinetic energy and this cage like structure breaks. Due to this breaking, the volume decreases. At 277 K or (4 degree celsius), it so happens that the entire cage like structure is broken which results in decrease in volume resulting in maximum density.
Therefore, density of water is maximum at this temperature.

4) At this stage, the cage has been entirely broken. Now, the molecules form this cage like structure again due to which volume again starts to increase as a result of which density decreases.

Here's what I couldn't understand:

  1) I cannot understand why trapping of gases results in increase of
  volume. The gases are trapped in the free space inside the cage.
  Whether that free space is occupied by gases or whether it remains
  unoccupied shouldn't result in increase of volume. Then why it's said
  that trapping of gases inside the cage like structure formed by ice
  results in increase in volume?

  2) I am not sure I heard the 4th point right because it clearly
  doesn't make sense. If at 277K, the cage like structure broke, why it
  will reform on further heating. On heating, we are giving the
  molecules more kinetic energy.They should roam about and not get tied
  down to a cage like structure. Why that cage like structure is being
  formed again after 4 degree celsius?

The answer appears to be fairly complicated, but fundamentally it results from a competition between two different options for water molecules interacting.

One place to start would be The structure of water; from ambient to deeply supercooled, Lars G.M. Pettersson and Anders Nilsson, J. Non-Crystalline Solids 407 399-417 (2015) where they discuss the 'inhomogeneous structure hypothesis'. At its heart they define high density liquid (HDL) and low density liquid (LDL) (not connected to cholesterol!) structures. LDL is tetrahedral, HDL is more distorted. LDL is directional H-bonded, while HDL is more van der Waals influenced. LDL minimized enthaly, HDL maximizes entropy. These are all consistent with a variety of experimental and simulation work given in the paper.

A recent paper applying DFT to the problem, Interstitial Voids and Resultant Density of Liquid Water: A First-Principles Molecular Dynamics Study, Sohag Biswas et al., ACS Omega 3 2010-2017 (2018), suggests a similar general picture. Again, the tradeoff of hydrogen vs van der Waals bonding appears to be important. Different exchange/correlation terms provide somewhat different answers. One suggested reason for the HDL structure is that, by incorporating an 'interstitial' water molecule into the LDL, one disrupts the directional hydrogen bonds and enhances the breaking and making of them, blurring out the structure (while being higher density).

So, it isn't a simple one-phase picture, but still seems to be a tradeoff of entropy and enthalpy of different configurations in the water.


How did early chemists determine the major product of a solution? For example, how did they determine using evidence that when 2-chloropropane reacted with ethoxide, a strong base, formed propene as the major product and 2-ethoxypropane as the minor product?

Could someone explain how did chemists determine the major products without NMR Spectroscopy? I have no lab experience and do not have access to them either.

In absence of currently available (or, in principle available) spectroscopies like NMR, IR, Raman, UV-Vis; and mass spectrometry our ancestors tried first to separate the products obtained by sequential recrystallisation.  Or by extraction under acidic, neutral, basic conditions (including back extraction).

They then recorded other physical properties of the purified products, if accessible, which then were compared with reference data.  For example, melting point and crystal shape if the products were solid, otherwise boiling point, refractive indices of the products obtained.  Then, they knew a whole array of pretests to check for the presence / absence of elements in the products formed, like Beilstein test for halogens and Lasaigne test for nitrogen.  They would determine the molecular weight by means of cryoscopy or ebullioscopy.

From a chemical point of view, it was common to subject the products to combustion analysis to determine the molecular formula.  (One of the instruments for this, the  Kaliapparat is still present in the logo of The American Chemical Society.)  Till today, there are journals preferring this over a mass spectroscopic determination of the molecular weight.

They knew about reactions that would yield derivatives, for example from the reaction with picric acid (affording pircrates), or with toluenesulfonic acid (yielding amides); often crystalline with a sharp melting point, easing to compare with reference data.  If not, other reactions like the haloform reaction would sequentially degrade the products into fragments, and then these products obtained would be characterised by physical or chemical means.

And obviously, the balance was much more "sacred" than today, where often we "just" weight-in reactants, and weight-out products.


What is the role of ammonium chloride in the workup of a Grignard reaction? In the following Grignard reaction, why is aqueous ammonium chloride used to get to the products? I don't see how it participated in or modified the reaction from what is normally expected.

Ammonium chloride ($\ce{NH4Cl}$) is the work-up reagent that quenches the magnesium alkoxide product of the Grignard addition. It is the reagent of choice as it is a proton source without being acidic; acidic conditions could result in protonation of the tertiary alcohol product and elimination to the alkene. It also ensures that all inorganic salts of Mg will extract into the aqueous phase.


Do high-energy chemical bonds indicate strong chemical bonds? In my head it seems reasonable that a high-energy bond would indicate that the bond is strong, however a question got me thinking.
Take for example lead azide, a very explosive compound that is often used as a detonator or fuse. It has a decent amount of energy stored in its bonds while being very sensitive to heat, friction, shock, etc.
This indicates to me that the lead azide has a low activation energy and therefore weak bonds, yet still plenty of energy available to be released.

There are two factors here to consider.
First, the energy released by a given reaction is (partly) due to the difference between the bond strength in the product vs reactant, not the absolute bond strength, unless the reaction involves just breaking a bond and not forming a new one.
Second, even if your reaction is just a bond breaking, there are multiple ways to break a bond. Reported bond enthalpies are usually defined in terms of the energy required for homolytic bond breaking, ie one electron stays on each component. For example,
$$\ce{HCl -> H. + Cl.}$$
By that measure, the H-Cl bond is quite strong, but we know that in a polar solvent that mitigates the negative effect of charge separation, the H-Cl will readily undergo heterolytic bond cleavage, where both electrons stay on one atom:
$$\ce{HCl -> H+ + Cl-}$$
In that context, we might consider the H-Cl bond to be very weak. (Note, though, that I'm also overlooking the amount of "bonding" between solvent and ions in this case, which means that it's not an example of a pure bond breaking reaction.)
All of which is to say that before you can make a statement about bond energy and bond "strength" you need to define both terms more specifically.


Why is this stereocenter by the Br an R? I thought it was S? Why is this stereocenter by the Br an R? I thought it was S? For problem b in the following image:

Both structure (a) and structure (b) have misassigned chirality centers. There should be only one $R$ assignment over all four structures. The correct assignments are as follows:

  Structure (a)

The misassigned chirality center is the one in the center. The carbon atom with the OH group has the highest priority (one C-O bond beats any number of C-C or C-N bonds). The second highest priority is the pyrrole ring (one C-N bond and one C=C bond beats one C-N, one C-C, and one C-H). The carbon atom with the primary amine is the third priority. The groups are arranged counter-clockwise and the lowest priority (H) is in the back.

  Structure (b)

The misassigned chirality center is the one with the bromo group. The bromine atom is the highest priority. The carbon atom bonded to chlorine is second, and the carbon atom bonded to fluorine is third. The groups are arranged clockwise, but the lowest priority (H) is forward.


Number of stereoisomers and symmetry of pentane-2,3,4-triol The answer is apparently 4, but the solution does not make sense.

I can only identify 2 chirality centers (the central carbon cannot be a chirality center) so I thought 2^2 = 4 stereoisomer possibilities, but because of the existence of a plane of symmetry, there is a meso compound: thus there should only be 3 stereoisomers.

However, the solution claims that initially, we can deduce 3 chirality centers and claim 8 stereoisomers, then eliminate some of them through the observation of plains of symmetry (there is only one as far as I can see either way). This solution does not make sense, but when I checked by looking at all of the structures, it seemed true. Can someone help me what I am missing?

It is easier to compare molecules is you rotate them all into a comparable position.  Say 1 with the central C-3 hydroxyl group out of the plane of the screen/page like in 1, 4, 5 and 7.

So when we rotate a molecule by 180 degrees along a vertical axis in the plane of the paper, something going into the plane of the page on one end off the molecule rotates to become something coming out of the plane of the page on the other end of the molecule.  Also central group flips from into to out of the page and vice-versa.

Hence we can tell that actually pairs (1,2), (3,4), (5,8) and (6,7) are the same molecule, just rotated by 180 degrees.

The absolute positions of the three centres doesn't matter, only the relative orientation of the end two to the central site.


Would decomposing an object into molecules be an environmental problem? In a thought experiment, a friend proposed to me a machine capable of decomposing any object (e.g., a brick) into its molecules. I'm not a chemist or physicist, but my initial reaction was that this was certainly an environmental issue if used in a larger scale.

Is that right? Or is a molecule so basic that there would be no health concern from such a (theoretical) decomposition.

EDIT @Aesin:
My thinking was:

objects can't disappear; the brick will be breathed in, get into
water etc.
if we start decomposing more advanced objects, we
get all kinds of harmful elements in the air (mercury etc.)
possible radiation from decomposition

As for the purpose: Let's assume it would be used for general waste destruction. Let's assume energy is not a concern (thanks for mentioning it!)

Decomposing something into 'molecules' is immediately indicative of a non-chemistry background! To try to address your question a bit more directly, I can see a couple of ways of looking at the thought experiment:

1) It decomposes everything to simplest possible component stable molecular structures.

In the case of organic material, plastics etc, you will probably end up with a preponderance of various gases and water, as largely this disassembly is equivalent to complete combustion (CO₂, H₂O, NH₃) likely being the main resulting candidates. These would float out into the atmosphere unless captured.

In the case of inorganic material, e.g. a computer motherboard you will end up with many molecules of various metals, most of which would rapidly oxidise due to their very high surface area (being unimolecular), and various silicates and ceramic unit cells. All of these would be incredibly fine aerosols or dissolved in atmospheric water, similarly rapidly dispersing unless contained.

Results would likely include enhancing the greenhouse effect (on balance as some compounds will diminish it), inducing acid rain (sulfates, nitrates), causing algal blooms/biome damage.

If you had storage for different outputs then you would be producing several industrially useful compounds and if you partnered it with metal reclamation facilities to deal with all the metal oxides it might be a useful way of recycling, energy no object.

2) It decomposes and sorts everything by chemical element.

Assuming you have a nice little storage container for each element (reasonable given you are sorting into the simplest molecule of a pure element), you would end up with various cylinders of oxygen, hydrogen, nitrogen, lumps of metal, lumps/powders of inorganic materials like phosphorus, sulfur, silicon. This would be potentially useful, but incredibly energetically expensive even beyond the separation system outlined by Uncle Al, so would be considerable only in the case of having essentially limitless free energy (many nuclear fusion plants/orbital solar collectors/dyson sphere type energy availability I suspect).

Issues here would relate to various elements being highly reactive on contact with air (or rather the oxygen/water in it). Obvious examples include allotropes of phosphorus and the rare earth metals.

Other issues unaddressed include fractional distillation and cracking of small molecule gases into smallest molecule gases, how to prevent reaction inside the vessel i.e. actually separate things rather than just provide them with huge amounts of energy to react with each other etc


Order of electrophilic substitution What will be the order of reactivity towards electrophilic substitution in case of the following compounds:

benzene, ethyl benzene, isopropyl benzene, tert-butyl benzene

The answer at the end of the book says that ethyl benzene will be most reactive (the book doesn't explain the cause though it's a MCQ) but as according to inductive effect I think the tert-butyl group will have more +I effect on the ring. Doesn't it?

In your series, all of the alkyl benzenes will have roughly the same +I inductive effect.  Where they differ is with regard to the resonance effect.  Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.

There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation.  In isopropyl benzene there is only one and there are none in *tert-*butylbenzene.  Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.

Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:

benzene=1, toluene=24, tert-butylbenzene=15.7 (reference, page 1060)

Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest.  The expected order in your series would therefore be:

ethylbenzene > isopropylbenzene > tert-butylbenzene > benzene


clock reaction dilutions for KI Stock solution 0.1M KI solution 100mL

Volume of KI
30
10.5
5.25

Volume of water
0
19.5
14.75

concentration of KI
0.1
0.05
0.025

How did they calculate these numbers?
 I tried c1 v1 = c2 v2
For example for 0.05M:
(0.1)(v1) = (0.05)(30/1000)
v1= 0.015L = 15mL, what does this 15mL represent. How did they calculate the volume of KI and water.

I want to dilute KI and I want the final solution to be 30mL. And the concentrations that I need are shown above (0.1, 0.05, 0.025, 0.0125).

Please help

The formula

$$c_1\cdot V_1=c_2\cdot V_2$$

is correct.

Since your stock solution has a concentration of $c=0.1\ \mathrm{mol\ l^{-1}}$, the value for the initial concentration $c_1$ is always

$$c_1=0.1\ \mathrm{mol\ l^{-1}}$$

Since you want the final volume of each solution to be $V=30\ \mathrm{ml}$, the value for the final volume $V_2$ is always

$$V_2=30\ \mathrm{ml}$$

You want four individual solutions (labelled $\mathrm a$, $\mathrm b$, $\mathrm c$, and $\mathrm d$) with four different final concentrations ($0.1\ \mathrm{mol\ l^{-1}}$, $0.05\ \mathrm{mol\ l^{-1}}$, $0.025\ \mathrm{mol\ l^{-1}}$, $0.0125\ \mathrm{mol\ l^{-1}}$. These are the individual values for the final concentration $c_2$:

$$\begin{align}
c_{2,\mathrm{a}}&=0.1\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{b}}&=0.05\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{c}}&=0.025\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{d}}&=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$

Now you can rearrange the above-mentioned equation to solve for the initial volume of the stock solution $V_1$:

$$\begin{alignat}{2}
V_1&=\frac{c_2\cdot V_2}{c_1}\\[6pt]
V_{1,\mathrm a}&=\frac{0.1\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=30\ \mathrm{ml}\\[6pt]
V_{1,\mathrm b}&=\frac{0.05\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=15\ \mathrm{ml}\\[6pt]
V_{1,\mathrm c}&=\frac{0.025\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=7.5\ \mathrm{ml}\\[6pt]
V_{1,\mathrm d}&=\frac{0.0125\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=3.75\ \mathrm{ml}
\end{alignat}$$

You measure these volumes of the stock solution and transfer them into four separate vessels.

Finally, you have to fill up each solution with pure water to the final volume of $V_2=30\ \mathrm{ml}$. The required additional volume of water $V_{\ce{H2O}}$ is approximately

$$\begin{alignat}{2}
V_{\ce{H2O}}&\approx V_2-V_1\\[6pt]
V_{\ce{H2O},\mathrm a}&\approx30\ \mathrm{ml}-30\ \mathrm{ml}&&=0\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm b}&\approx30\ \mathrm{ml}-15\ \mathrm{ml}&&=15\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm c}&\approx30\ \mathrm{ml}-7.5\ \mathrm{ml}&&=22.5\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm d}&\approx30\ \mathrm{ml}-3.75\ \mathrm{ml}&&=26.25\ \mathrm{ml}
\end{alignat}$$


Percentage ionic character when electronegativity is given What is the ionic character of a bond, $\ce{A-B}$, in terms of the electronegativities of $\ce{A}$ and $\ce{B}$ ($\chi_\ce{A}$ and $\chi_\ce{B}$)?

I have been taught that the percentage ionic character is:

$$ \frac{\text{observed value of ionic character}}{\text{calculated value of character}} $$

but I can't understand how electronegativity is used here. I couldn't find anything on the internet either.

Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference $\Delta \chi$.

Percent ionic character $= (1-e^{-(\Delta \chi/2)^2} )\times 100$

But I'd like to correct the definition of percent ionic character in your question using dipole moment  $\mu$ (not Observed value of ionic character):

Percent ionic character = $\Large\frac{\mu_{\text{observed}}} {\mu_{\text{calculated}  }}$ $\times 100 \%$

Where $\mu_{\text{calculated}}$ is calculated assuming a 100% ionic bond.

For more details please see this page.


Why is ammonium chloride rather than an alkylammonium chloride salt the product of nucleophilic substitution of a chloroalkane by ammonia? My textbook said that when an excess of ethanolic ammonia is heated with a haloalkane ($\ce{R-X}$) one should give the products as being $\ce{R-NH2}$ and $\ce{NH4Cl}$ as opposed to $\ce{RNH3Cl}$. I understand that in reality an equilibrium will exist between the alkylammonium and ammonium cations, but I thought amines were stronger bases than ammonia, so I find it odd that I would be asked to give the ammonium salt and an amine rather than the alkylammonium salt as the major product of this reaction. Have I misunderstood something?

You have actually answered the question right in the first sentence of the description.

You are right in saying that $\ce{RNH_3Cl}$ will be formed.

The alkyl halide, $\ce{R-X}$, is the alkylating agent. The reaction medium is basic in nature, so I'd expect the $\ce{S_N2}$ mechanism. So,

$\ce{H3N +RX\to H3NR+X-}$

We know that an excess of ammonia has been used. This simply means that the forward reaction is favoured:

$\ce{H3NR+X- +NH3\rightleftharpoons RNH2 + NH4+X-}$

It's all Le Chatelier's once again - makes extraction simpler, doesn't it?


In Lassaigne's test, the organic compound is first fused with Sodium Metal instead of any other Group 1 Metal, Why is this so? I know that Sodium has high polarising power than potassium so it can convert the respective functional groups more readily than Potassium and both Lithium and Sodium salts are soluble in water. This indicates that both Sodium and Lithium can be used but this is not the case as in each case the organic compound is first fused with Sodium to make respective ionic compounds and then identification tests are done. Why is this so? Why any other metal is not used?

You asked this historical question before (check Sci math history which is pretty good). For historical searches, Google scholar works the best. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=lassaigne+test&btnG=&oq=lassaign

The very first article is very useful on the history of sodium fusion test.
"A lost centenary: Lassaigne's test for nitrogen. The identification of nitrogen, sulfur, and halogens in organic compounds" J. Chem. Educ., 1945, 22 (5), p 212

The first few lines of the article clearly mention that Lassaigne used potassium rather than sodium. He must have the same question his mind which you are asking now. Surprisingly, it was another person named Jacobsen who suggested using sodium. The original test indeed used potassium.

Why sodium is used today in some teaching labs? Lower price and less reactivity than potassium. Essentially nobody uses this historical test in analytical laboratories anymore. Today, an very small quantity of organic compound is burned and the gaseous products are analyzed by gas chromatography.


Finding the value of pressure equilibrium constant Question:

  A rigid vessel containing only $\ce{NO2(g)}$ is heated to $\pu{337^{ \circ}C}$  and allowed to come to equilibrium according to the following reaction:
  $$\ce{2NO2 <=> 2NO + O2}$$

  The density of the resulting mixture is measured to be $\pu{0.52 g/L}$ at a total pressure of $\pu{ 0.75 atm}$. What is the value of $K_\mathrm p$?

The answer key says it should be $\pu{0.65 atm}$, but I can't get that number. I first tried to find mols using the given density and molar masses, and then finding the partial pressure of each component using the mol fractions, but it doesn't work out.

The density is given.From here you will get a relation between initial concentration and volume.From $K_c$ you will get relation between volume,extent of dissociation and initial concentration.From $K_p$ you will get relation between pressure,extent of dissociation and initial concentration.After that use the formulas that Klaus Warzecha gave.I think that will be enough to solve this.


Is ethanol an acid or a base or amphoteric? Is ethanol considered to be an acid or a base? Or does it act as an amphoteric species? Is acidity and basicity a relative concept? Also, can someone please explain this on the basis of the Brønsted–Lowry acids/bases theory which is based on the protonic concept; will it donate or accept $\ce{H+}$ ions? I'm unsure which acid base theory should I follow in this case, any help will be appreciated.

Because we are talking about Brønsted–Lowry acids and bases, we are only concerned with whether or not a molecule has (or the degree to which it has) the ability to donate a proton (acid) or the ability to accept a proton (base) or to do either (amphoteric).

The case of ethanol is rather simple to evaluate. Firstly, the hydroxyl proton could be donated to a strong base, although this only happens to a minimal degree in an aqueous solution as water is a much stronger acid than ethanol. Still, ethanol has the ability to act as an acid because of the ability to donate it's hydroxyl proton.

However, aqueous solutions of ethanol are slightly basic. This is is because the oxygen in ethanol has lone electron pairs capable of accepting protons, and thus ethanol can act as a weak base.

Summary - TL/DR:
As ethanol has the ability to both donate and accept protons, it should be considered an amphoteric compound with respect to Brønsted–Lowry acid-base theory.


Does Gibbs free energy of formation always equal zero for elements in standard state? I was looking through the appendix in my chemistry textbook when I noticed that every compound that had a $\Delta H_f$ of zero also had a $\Delta G_f$ of zero. So of course the compounds that have a $\Delta H_f$ of zero are the elements in their standard states but why does that make their $\Delta G_f$ zero? Does every compound that has a $\Delta H_f$ of zero also have a $\Delta G_f$ of zero? What exactly is $\Delta G_f$?

Does Gibbs free energy of formation always equal zero for elements in standard state?

No.  Elements occur in different allotropes.

$\Delta H_f^o$ and $\Delta G_f^o$ are defined to be zero at 298K, 1 bar for the lowest energy allotrope, with the exception that the values for white phosphorous are defined to be zero even though it is not the lowest energy allotrope.

The other allotropes do not have zero $\Delta H_f^o$ and $\Delta G_f^o$ even though they are still elements.

  So of course the compounds that have a $\Delta H_f$ of zero are the elements in their standard states but why does that make their $\Delta G_f$ zero?

$\Delta H_f = 0$ does not imply $\Delta G_f = 0$

The two values are independently defined to be zero for a particular allotrope as explained above.

  Does every compound that has a $\Delta H_f$ of zero also have a $\Delta G_f$ of zero?

If a compound (other than by definition) happens to have $\Delta H_f = 0$, then $\Delta G_f$ could be any value, it is not necessarily zero.

  What exactly is $\Delta G_f$

It is the change in Gibbs energy upon forming a compound in its standard state from its constituent elements in their standard states at 298K.


Bomb Calorimeter vs Cup Calorimeter In the textbook,it is said that bomb calorimeter measures the constant volume heat ,while the cup calorimeter measure Constant pressure Heat.I was wondering, why cup calorimeter is considered constant pressure,even though it also does not allow a gas to expand (it has a fixed volume).So if not then that means cup calorimeter allows some air to get in so that pressure is constant ?(I.e the system is not closed )?

Even though the pressure remains constant, heat can get in (or out) of the system. From the cup calorimeter, we are trying to find $\Delta H$ which is equal to the heat evolved from the system at constant pressure. We measure this heat and our job is done.

$$\Delta H=q_p$$

If we want to calculate $\Delta U$, we just to find the heat evolved from the bomb calorimeter, which can be done by measuring the change in calorimeter's final and initial temperature given that the heat capacity of the calorimeter is known.

$$\Delta U=q_v=nS\Delta T$$


Which definition of van der Waals forces is correct? I've been told by my teacher that van der Waals forces is the formal name for intermolecular forces. But on the internet, many people are saying van der Waals forces is another name for London dispersion forces. If van der Waals forces is another name for dispersion forces, why is it named after another scientist?

van der Waals forces is not another name for London dispersion forces. In his answer, airhuff has given IUPAC definition for van der Waals forces.
London dispersion forces are the intermolecular forces that occur between atoms and between nonpolar molecules as a result of the motion of electrons in those atoms and molecules. They are induced dipole-induced dipole interaction, better described in following diagram:

These inter-atomic or inter-molecular forces are the weakest of van der Waals intermolecular forces. London dispersion forces got its name due to its discoverer, Fritz London (Ref.1). You can find some details of the distance dependence on London forces here.
References:

Fritz London, "The general theory of molecular forces," Trans. Faraday Soc. 1937, 33, 8b-26 (https://doi.org/10.1039/TF937330008B).


Unique calling code for a chemical reaction I'm writing a script and would like to use a "Gödel numbering" to search for specific reactions.

Is there any used procedure to assign an symbol (integer, possibly) to a chemical reaction? E.g.
$$CH_4+4O_2=CO_2+2H_2O\ \ \ \Longrightarrow \ \ \  121231412$$

I don't know about Godel numbering specifically (though this is interesting, my being a math guy at heart), but there are existing textual systems for encoding molecules (not reactions that I know of).  From these, you may be able to encode reactions.

Two examples are InChI and SMILES.  These are Wikipedia links.


Does the Lewis acid have to match the halogen in electrophilic halogenation of benzene $\ce{FeBr3}$ catalyzes bromination and $\ce{AlCl3}$ chlorination. I'd like to know whether bromination and chlorination can be done in sequence without changing the catalyst? So - will using the $\ce{FeBr3}$ for chlorination and $\ce{AlCl3}$ for bromination work? Is it less efficient or slower or is there another reason?

The reason you use the respective halides for halogenation is that the halogen and Lewis acid can inter-react to form a chemical equilibrium if you don't.

$$\ce{FeBr3 + Cl2 <=> Cl+ + [FeBr3Cl]- <=> Br+ + [FeBr2Cl2]- }\\
\ce{AlCl3 + Br2 <=> Br+ + [AlBrCl3]- <=> Cl+ + [AlBr2Cl2]- }$$

Given these equilibrium reactions you can see that now you have two electrophile species which means your halogenation is not as controlled. It will work and halogenation will still occur, but now and you will have bromination and chlorination occurring in competition with each other when you only wanted one or the other.


Why are there no alkaline foods? It takes some work to find a list of the actual pH of various foods.  (Searches are confounded by a popular "alkaline diet" theory that the pH effect on a human body of consuming various foods is not a function of the pH of the food itself.)
It appears that almost all food has pH < 7.0 (most much lower), and I can't find any with pH > 7.5.
Is there a biochemical reason that there is no significantly alkaline food?

Most people find alkaline taste bitter and "bad tasting". There are some exceptions; shellfish, particularly conch, may have a pH of more than 8.0. This is also the approximate pH of bile, which tastes particularly foul when retching.
Crackers made with baking powder or baking soda also may be somewhat alkaline, as is egg yolk.
However, your question is valid... why are the vast majority of foods neutral or somewhat acidic? Why do most enjoy the taste of grapes, plums and citrus fruits, which may have pH below 3.0? Does it have to do with metabolism? With the preservative effects of low pH?


Why do some old masters paintings deteriorate faster than others? Specifically, Wikipedia states that Sir Joshua Reynolds' "The Age of Innocence" was painted by the artist over another one of his paintings because the latter was "had suffered some paint losses," and goes on to say that "Innocence itself has deteriorated since 1859." Why?

Some pictures where lead-based paint was used tend to deteriorate due to formation of black PbS. However, I have found that the cause of this is the artist's 'defective' technique, which caused flaking of paint. It's the matter of technique, not chemistry.


Does copper react with HCl when current is present? Copper does not react with hydrochloric acid, according to many sources.  However, I found an experiment posted in a video, which I will try to repeat, that shows a copper pipe dissolving into HCl and plating a nail when the copper pipe and the nail are connected to a 12-volt battery, becoming electrodes.  No other chemicals, according to the video, are present.  How does the electrical current cause the copper to react with the HCl when otherwise it would not?

A reaction between copper metal and hydrochloric acid would go something like this: $$\ce{Cu + 2HCl -> CuCl2 + H2}$$
...which we can write as: $$\ce{Cu + 2H+ -> Cu^2+ + H2}$$
If we break this up into half reactions, we get: $$\ce{Cu -> 2e- + Cu} ~~~E°=-0.34~V$$ $$\ce{2H+ + 2e- -> H2} ~~~E°=0.00 ~V$$
If we add the voltages together, we get the cell potential of the combined reaction is $-0.34+0.00=-0.34~V$. Since the cell potential is negative, the reaction is not spontaneous, so the reaction is not going to occur without any artificial interference ($\Delta G°=-nFE°_{cell}$).

However, if you connect a battery (galvanic cell) to this reaction, you can supply the necessary electrons for this reaction to occur (electrolytic cell). Since you are connecting a 12 V battery, there is more than enough voltage to cause this reaction to occur.


Substance of higher latent heat of melting than water, similar melting point I saw ingeniously simple "cooler box" yesterday - a double-walled plastic box with some liquid between the walls. You freeze it, and it keeps temperature near the freezing point of that liquid, until the liquid melts, while you can store dry foods inside.

Now, the efficiency of such a box will depend on volume of the liquid, to a lesser degree - how far below its melting point it was frozen, and on latent heat of its melting. (and insulation of course, but that's an entirely different matter.)

While I suspect the liquid used in the box I saw was just water with a little coloring added, I wonder if I could do better than that. Is there a substance that can be reasonably obtained by a DIY tinkerer, that has a latent heat of melting significantly higher than water, and a melting point at normal pressure somewhere "within reason" - achievable by normal kitchen freezer, and not above "food inside will spoil"? (bonus points for it not being overly dangerous...) Or maybe is there some way to increase latent heat of melting water ice by adding something to it?

Indeed, it will be hard to find a substitute for water. You might want to look into how ice packs work.

Some of the things that are added to water in such packs include substances to prevent bacterial growth and to prevent the water to become fully liquid. The water thus become a gel.

  Gel packs may be made by adding hydroxyethyl cellulose (Cellusize) or vinyl-coated silica gel.

One factor that you might consider, is that if there is more mass (for example, dissolving something in the water) will provide more substance for the any heat to go to, so that the temperature rise will be slower.

Another, is hydrogen bonds, which are very present in water molecules and which explains its high specific heats (latent heat and heat capacity).

Unfortunately, adding in stuff to the water will tend to disrupt the hydrogen bonds. I haven't tried it, nor can find something to support my next sentence, but I would expect ammonia to be a possible additive. It has a higher specific heat capacity than water and dissolves easily as well since it does not disrupt hydrogen bonds that much due to its molecular structure.

Glucose is also a potential additive since it doesn't disrupt the hydrogen bonds much, but its specific heat is apparently a bit lower than ammonia. I found a site where I could input some parameters and it outputs the heat capacity of the sugar water and the outputs often say that the specific heat capacity is lower than that of ammonia alone (from the same site that michielm mentioned, ammonia has a specific heat capacity of a bit less than 4.6 at $0^oC$).


Molarity of hydrogen gas in standard hydrogen electrode I am unable to understand the folowing statement of Wikipedia it is about standard hydrogen electrode :

  The concentration of both the reduced form and oxidised form is maintained at unity. That implies that the pressure of hydrogen gas is 1 bar.

But why if I apply ideal gas equation then I am getting molarity of nearly 0.004M .this is condradiction to the statement in Wikipedia. N/V=P/RT THEREFORE M=N/V~0.004

The standard hydrogen electrode is a "reference electrode." It is supposed to be used with ultra-low currents.

In operation, hydrogen gas is bubbled through a 1 N acid solution, but there is little hydrogen gas in the solution. In order to get the hydrogen gas to bubble through the solution, a pressure of a bit more than 1 bar must be used.


Logical way to compare dipole moment I have to find the decreasing order of dipole moment of $\ce{CHCl3}$, $\ce{CH3Cl}$, $\ce{CH2Cl2}$, and $\ce{CCl4}$. Now I know that in $\ce{CCl4}$ there will be no dipole moment. Now to compare the other three compounds I don't have logical reasoning to compare the dipole moments but I have the data to compare but in the examination we won't be having these data so I was looking for any logical reasoning to compare the dipole moments of the remaining three compounds. Thank you

Look at dipole moment as a vector with the head of vector pointing towards the more electronegative atom. Ccl4 has 4 vectors pointing outwards but these vectors cancel, hence 0 dipole. Chcl3, 1 vector from H towards c, 3 vectors from c to each cl atom. Note they have tetrahedral geometry. The net vector would be along one of cl atoms but as other cl atoms have components of their vectors opposite to the victorious cl atom, dipole would be reduced. Ch2cl2, net vector would be somewhere between the 2 c==>cl vectors but there are not much opposite components of vectors as compared to chcl3, it has greater dipole than the former. Ch3cl, net dipole is along the cl atom and further each H==>C dipole is increasing the magnitude of the vector. These problems are based on visualization and vectors help you visualize better.


How exactly does the oxidation of TEMPO by NaOCl/KBr work? I would like to understand the exact process by which TEMPO is oxidized to the nitrosonium cation by NaOCl, as in the part of the mechanism shown in the diagram (1) surrounded by the red box:

The confusion for me is that TEMPO clearly needs to donate 1 electron to $\ce{OCl-}$ in order to form the neutral, nonradical nitrosonium cation, but we also get a $\ce{Cl-}$ anion, so there must also be a 2 electron process involving the $\ce{Br-}$ "co-catalyst" coming into play. How do I reconcile these 1 electron and 2 electron processes so as to draw a formally appropriate mechanism for this part of the reaction?

What exactly happens to the $\ce{OCl-}$ oxygen, and how does it happen?

(1) Org. Synth. 2005, 81, 195. DOI:10.15227/orgsyn.081.0195.

I would propose the following mechanism:

Note that the oxygen radical anion isn't that bad as a leaving group as the Cl-O bond is weak. It can participate in an electron transfer with chloride anion, or rather the bromide at the beginning of the reaction as it is more abundant (I have written the protonated form - hydroxyl radical as it is less likely that two anions will come close enough for electron transfer):

The bromine radical can in turn react with another TEMPO molecule. That is why two equivalents of TEMPO react with one equivalent of hypochlorite.

Keep in mind that there are countless possibilities and deviations from the mechanism I proposed.


Why can mercury oxide be decomposed into mercury and oxygen upon heating? As I am a grade 9 student, please use simple English that does not involve many technical words in Chemistry. Thank you.

I have learnt from my teachers' handout that physical methods of separation include heating, and that a compound can only be broken down into elements by chemical methods such as electrolysis.

But why can mercury oxide (a compound) be decomposed into mercury and oxygen upon heating (physical method)?

Or is it simply because heating is a chemical method? (that my teacher probably was mistaken)

In simple terms, heating is indeed a physical method. You might have heard of heat being, in fact, molecules/ions moving faster. There’s nothing in there per se that would make it a chemical method.

But. The atomic particles that are moving quicker can bump into each other with more energy (because they are faster — think of a car crash which is worse at high speeds); this can be understood as activation energy, i.e. energy that allows reactions to proceed.

So by adding enough heat energy, you are inducing a chemical change which leads to the (thermal) decomposition of mercury oxide into its elements.


Shape of sp³ hybrid orbital I have a doubt regarding the shape of the $\mathrm{sp^3}$ hybridised orbital. I looked up for it on stackexchange and found a post where the guy said that where ever the orbitals interfere with opposite signs of the wavefunction they cancel out and hence we get a small lobe .
He further said that when the right side of the p orbital overlaps with s orbital( i have linked the post) they both add up.

But the $\mathrm{s}$ orbital wavefunction can also have negative sign (in the post the guy has assumed that s orbital wavefunction is always positive) . So in all four interference combinations are possible : ++ , +- , -+ , - - . Hence the shape should be different  but its not.

I am really confused about this.
Any help is appreciated.

Thanks in Advance.
Link to original post : https://chemistry.stackexchange.com/a/15909

Don't put too much trust in the absolute signs of wavefunction, for they all are arbitrary anyway. Look at it this way: an s orbital has one sign (*). One lobe of p orbital has the same sign, so when they add up, it grows huge. Another lobe inevitably has the opposite sign, so when they interfere, it is reduced to a tiny pig tail.

(*) This is not quite true if we consider radial nodes. That's when things get really hairy.


What is the difference between "spontaneous" and "energetically favourable"? I read somewhere that the difference between energetically favorable reactions and spontaneous reactions is that energetically favorable reactions are ones where energy is released, i.e., $\Delta H<0$, whereas spontaneous reactions are ones where the change in Gibbs free energy is negative, i.e., $\Delta G<0$. Thus it is possible for a reaction to be energetically favorable but not spontaneous if $\Delta H$ is negative, and yet $\Delta G$ is positive because $\Delta S$ is negative and the product of it and T is large enough to compensate for $\Delta H$ as per this equation: $\Delta G = \Delta H - T(\Delta S)$.

My questions are:

1) Is this true?

2) If so, what would be an example of a reaction that is energetically favorable but not spontaneous?

Here is a very simple example of an exothermic, nonspontaneous process at room temperature:

$$\ce{H2O_{(l)} -> H2O_{(s)}}$$

You need to extract heat from water to freeze it, but it is obvious that at $25\ ^{\circ}\mathrm{C}$, this does not happen spontaneously. Only if you decrease the temperature to decrease the contribution of the negative $\Delta S$ term will this reaction occur spontaneously.


Why are non-spontaneous reactions driven in the presence of energy? Here is my understanding. If you have an endergonic reaction (m) with $\Delta G_m=-m$ and you couple it to an exergonic reaction (n) with $\Delta G_n=n$, where $n>m$, then the endergonic reaction can proceed as the overall $\Delta G$ for the system is negative. I understand why having the endergonic reaction occurring is not a violation of the second law of thermodynamics and why it is made possible in that sense.

I was recently studying electrolysis and they talked about hooking up an electrochemical cell to a power supply and using electrical energy to drive a nonspontaneous redox reaction. If you include the power supply in the system, then the $\Delta G$ is negative, as the power supply gives a lot of energy.

This is my confusion: if the redox reaction is nonspontaneous (let’s call it R1), then it must be spontaneous if the opposite direction (let’s call it R2). However, in the presence of electrical energy… R1 is made spontaneous? Therefore it can occur? But what of R2? Shouldn’t that reaction be even more spontaneous, as it were? Shouldn’t the system try to minimize its energy by preferentially allowing R2 to proceed, even in the presence of an energy source? Thus, why would R1 occur? Why does shoving energy into a system cause a nonspotaneous reaction to occur when it would be more favorable to continue having it occur in the opposite direction?

I would also appreciate it if anyone could extend this to general energy coupling, say with hydrolyzing ATP and a nonspontaneous biological reaction. Why is the original spontaneous reaction not made more spontaneous?

You are incorrect in your assumption that the addition of a power source makes both the forward and the reverse redox reactions more spontaneous. An electrochemical cell operates on the same principle as a battery. On the anode of the cell, you will find a compound that is being oxidized, or losing electrons. These electrons then flow through some sort of a wire to the cathode, where the electrons are being used to reduce another compound. The electricity will flow in the "spontaneous" direction, in other words, from the compound that prefers to be oxidized to the compound that prefers to be reduced. The "voltage" across an electrochemical cell is determined by adding up how badly the reducing agent wants to get oxidized ($E°_{ox} = -E°_{red}$) and how badly the oxidizing agent want to be reduced ($E°_{red}$), and is a measure of how much the reaction wants to occur.

Now, lets try adding a battery in between the anode and the cathode. If the battery is added such that its anode points towards the cell's cathode, the battery will add to the voltage of the cell, causing the effective voltage for the reaction to increase. On the other hand, if the battery is placed such that its anode points towards the cell's anode, it will be creating a voltage in the other direction. When this happens, it becomes a question of whichever voltage is stronger. If the battery has a stronger voltage than the cell, electrons will flow in the opposite direction. This is what occurs in electrolysis, causing the originally non spontaneous process to become spontaneous. This results in charging the electrochemical cell. On the other hand, if the battery has a smaller voltage than the cell, the electrons will flow in the original direction - charging the battery as they pass through it.

While a power source can be used to either make a spontaneous process more spontaneous or to reverse the flow of electrons, it cannot do both at the same time.


How to find the concentrations of the acid-base equilibrium between sulfuric acid and sulfur dioxide for a known pH using Octave? The system is an aqueous solution of sulfuric acid and sulfur dioxide. I know how much sulfur there is and I know the pH.  The system is in equilibrium.  I would like to use Octave to solve the system.

I set up the following equations, ignoring $\ce{SO3^2-}$ because the pH is around 1.5.

By definition:
$$ \mathrm{pH} = -\log [\ce{H+}]$$

Conservation of mass:
$$[\ce{H2SO4}] + [\ce{H2SO3}] + [\ce{HSO4-}] + [\ce{SO4^2-}] + [\ce{HSO3-}] = M $$

Henderson-Hasselbalch equations for the system:
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{HSO3-}) + \log \frac{[\ce{HSO3-}]}{[\ce{H2SO3}]}\\
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{SO4^2-}) + \log  \frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}\\
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{HSO4-}) + \log \frac{[\ce{HSO4-}]}{[\ce{H2SO4}]}\\
\end{align}

I then defined the solution vector to have the following order:
$$ x = [ [\ce{H+}], [\ce{HSO4-}], [\ce{SO4^2-}], [\ce{HSO3-}], [\ce{H2SO4}], [\ce{H2SO3}] ]$$

and my code for Octave is attached.  For some reason I always get negative values for some of the results.

function y = phCalc(x)
   pH = 1.87;
   pKa1 = -5;   % this is a guess to drive H2SO4 to ionization in this system
   pKa2 = 1.987;% from a textbook
   pKa3 = 1.857;% from a text book
   M = 0.0012 / 32 * 1000 ; % total sulfur moles per volume
   y(1) = 10^(-pH) - x(1);
   y(2) = x(5) * 10^(pH-pKa1) - x(2);
   y(3) = x(2) * 10^(pH-pKa2) - x(3);
   y(4) = x(1) - x(2) - x(3) - x(4);
   y(5) = x(2) + x(3) + x(4) + x(5) + x(6) - M;
   y(6) = x(4) / 10^(pH-pKa3) - x(6);
endfunction

Your code seems to be calculating the differences between a guessed concentration vector x and a calculated concentration vector.  These differences in concentrations between guessed and calculated are y.  For example, the guessed pH is apparently x(1), and the calculated pH is 10^(-pH).  The difference between them is y(1).

Thus, its no surprise that the y values are sometimes negative.  To solve the problem you need to feed your function to a numerical solver such as fsolve.  That function will take your guessed solution x, other parameters such as pH, the various pKas, and M, and your function phCalc as input, and return the value for x which makes all the y variables zero.  That solution is the one you want.

On the chemistry side, you mention sulfur dioxide, but I don't see any equations for the solubility of sulfur dioxide gas in water.  Additionally, sulfur dioxide is reduced relative to sulfuric acid, so there would need to be an additional reaction, and parameter (extent of reaction or degree of reduction or something) that describes the relative amounts of reduced sulfur (as sulfur dioxide, bisulfite, or sulfite) vs. oxidized sulfur (as sulfate, bisulfate, sulfuric acid, or sulfur trioxide).

There is also the issue of charge balance: what concentration of inert cations (e.g. sodium or potassium or whatever) are you assuming?  Charge balance must hold in bulk aqueous solution.  If you don't have any sodium / potassium etc., then you should include the self ionization of water as an equation, hydroxide as a species, and y(7) = x(1) - x(2) - 2*x(3) - x(4) - x(OH) as an equation.  If you do have sodium/potassium/whatever at significant concentration, you can neglect hydroxide and just go with the equation y(7) = x(1) - x(2) - 2*x(3) - x(4) + x(Na) where x(Na) is the cation concentration, which I have assumed is sodium.

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Easy-to-do experiments that clearly show outstanding properties of graphene By chance, we received some monolayer graphene sheets (20 cm x 20 cm) and mixed coper-graphene wires for free at our university. I would like to prepare some very-easy-to-do experiments for the chemistry students that show the awesome properties of graphene, and which could be carried out with very simple lab equipment or even home experiments. What are some good experiments we could do?

You may want to post this in the Physics forum, as many of the outstanding properties of graphene are expected to be physical.  See this Periodic Video: Graphene

Given this, two things to try.  First, graphene should conduct electricity, like graphite but not like carbon.  Second, graphene may dissolve in nonpolar solvents like carbon tetrachloride, similar to buckyballs, to produce colorful solutions.  See this Periodic Video: BuckyBall

Both these experiments should be checked in advance: I'm extrapolating from other experiments, but have not done these myself.


Molar concentration of water and pKa value By defintion, $\mathrm pK_\mathrm a$ value is defined as
$$\mathrm pK_\mathrm a = -\lg ([\ce{H2O}] K_\mathrm{eq})$$

Consider the reaction:
$$\ce{HCl + H2O <=> H3O+ + Cl-}$$

The Eq constant of $\ce{HCl}$ is defined as

$$K_\mathrm{eq} = \frac{[\ce{H3O+}][\ce{Cl-}]}{[\ce{H2O}][\ce{HCl}]}$$

Why is molar concentration of water fixed at 55.5? Wouldn't it depend on the extent of ionization?

For most practical purposes, we assume that the aqueous solution is dilute and that the concentration of water is constant, that is, $[\ce{H2O}] \approx 55~\mathrm{M}$. This is the standard that is used to define the ionization constant of water, $K_\mathrm{w}$.

If the equation $1\times 10^{-14}~\mathrm{M^2} = [\ce{H3O+}][\ce{OH-}]$ didn't hold (at $25~\mathrm{^\circ C}$) then we would have to come up with other equations for acid-base equilibrium. Even for strong acid/base dissociations we approximate the equilibrium equation with full dissociation (i.e. one forward arrow).

I recommend reading pp 12-15 of the following if you want a more theoretical treatment of strong acid/base dissociation where activities are used: http://www.chem1.com/acad/pdf/c1xacid2.pdf


How does change in entropy influence the tendency of a reaction to occur? I know that entropy is the measure of the randomness of particles within a system, but how would it affect the tendency of the reaction? Would it just increase or decrease the chance of the reaction occurring, based on what change in enthalpy is?

Entropy ($S$) is a measure of the disorder in a system.

In a closed system, entropy always increases over time. In an open system, energy can be added to a system to cause a decrease in entropy, but this is not necessarily a spontaneous reaction.

A reaction will be spontaneous if the change in free energy, $\Delta G$, is negative.

Quantitatively,
if entropy increases, and the reaction enthalpy is exothermic $(\Delta H<0)$ or weakly endothermic ($\Delta H>0$ and $\Delta H < T\Delta S$), the reaction becomes spontaneous according to the free energy equation.

Since the observable universe is essentially a closed system, entropy will continue to increase until the universe essentially ceases to exist.

This way an increase or decrease(sometimes) in the randomness can cause a reaction to occur.

Here is a table relating to spontaneity and signs of $\Delta G$:

Image Source


How does sulfuric acid catalyze the reaction of potassium chlorate and sucrose? Part of the total reaction is mentioned here:

http://chemistry.about.com/od/demonstrationsexperiments/ht/instantfire.htm

$\ce{2KClO3(s)} + \text{heat} \rightarrow 2\ce{KCl(s)} + 3\ce{O2(g)}$

The rest is a mystery. (And where does the heat come from?)

I regret to say that I have delayed acquisition of organic chemistry principles that might have answered this question already but can't help but wonder why, right now.

I feel that once the role of sulfuric acid is cleared up, the reaction with sucrose not described above will explain itself.

(And no I am going to run around burning things down with this knowledge.)

I think the sulfuric acid isn't really a catalyst here -- I think you use the heat from the reaction between sulfuric acid and sucrose (which will start happily at room temperature) to start the reaction between the potassium chlorate and the sucrose (which usually won't).

ETA: Once you've done that, it's pretty much a straight-up messy combustion reaction, only with the oxygen being provided from a solid source (thermal decomposition of chlorate to chloride and oxygen as the reaction you mentioned shows) rather than anything mechanistically clean. Compare, for example, this fairly typical demonstration just using hot chlorate and a sugar source: Oxidation of Sugar or Gummi Bear with Potassium Chlorate.


Why are there as many molecular orbitals as atomic orbitals? So molecular orbitals are represented as a weighted sum over atomic orbitals. Of course I reckon that the total sum of the weights per atomic orbital has to equal 1, when summed over all molecular orbitals. But why then should n atomic orbitals results in exactly n molecular orbitals? Is there a fundamental reason for this I am not seeing?

The orbitals, both atomic and molecular, must be orthogonal, that is, they have no net overlap or correlation:  equal constructive and destructive interference. You can think of them as forming the basis set of an n-dimensional vector space. If you think of the atomic orbitals as the standard basis...

Example for a 4-dimensional space:

$\langle 1,0,0,0 \rangle$,
$\langle 0,1,0,0 \rangle$,
$\langle 0,0,1,0 \rangle$,
$\langle 0,0,0,1 \rangle$

...Then the molecular orbitals are just a rotated basis. In other words, they take linear combinations of the atomic orbitals to span the same space.
Ex:
$\langle \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \rangle$,
$\langle \frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2} \rangle$,
$\langle \frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2} \rangle$, $\langle \frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2} \rangle$

The same n-dimensional space requires at least the same number (n) of vectors to span it. This is exactly the same phenomenon as rehybridising individual s and p orbitals to form sp$^3$ orbitals.

You can find explicit formulae for the orbital vectors with the actual numbers (one possible representation), along with fantastic illustrations here Darmstadt, LCAO Theory.

There are two normalisation conditions when combining atomic orbitals to make either hybridised orbitals or molecular orbitals. The first is that every atomic orbital must be used completely. That is, the sum of its contributions to all the molecular orbitals must be 1. The other is that each of the new orbitals cannot have more than 100% orbital. Although I think that it is this second part that you were questioning the reasoning for.


Coordination complex - nitrito-N & nitrito-O I am currently learning the coordination compounds and encountered a new isomerism called linkage isomerism.
I am thinking that if I want to get some $\ce{[Co(NO2)6]^3-}$, I should have to dissolve $\ce{CoCl3}$ in water to get $\ce{[Co(H2O)6]^3+}$, then add some concentrated $\ce{HNO2}$ to displace the six water molecules away (although I don't know why $\ce{NO2-}$ is a stronger ligand than $\ce{H2O}$).
Questions:

Will I get a 50-50 mixture of nitrito-N & nitrito-O co-coordination complexes? Or will there be a chance of getting random combinations of nitrito-N & nitrito-O coordinate bonds to a $\ce{Co^3+}$ (e.g. 4 nitrito-O + 2 nitrito-N)?
If I just want to get nitrito-N complex, are there ways to prevent the formation of nitrito-O complex? Or the mixture must form and then we need to do some other reactions to get rid of nitrito-O complexes?

Linkage isomerism is of major interest in the field of inorganic chemistry. Linkage isomerism occurs when an ambidentate ligand such as $\ce{NO2, SCN,}$ etc. binds to the metal center. Nitro-nitrito metal complexes can be regarded as good examples of linkage isomerism. It must be noted that the $\ce{NO2}$ group can be coordinated to the metal atom in different ways such as nitro (via the nitrogen), nitrito (via oxygen), chelating via both oxygens (nitrito-O,O′), and bridging nitro, usually via nitrogen and oxygen atoms (Ref.1):

Source of diagram: Modified from Ref.1
Amongst various coordination modes, above diagram shows not only $\ce{NO2}$ is capable of having the wide range of binding modes in mononuclear metal complexes, but also its potential tendency to act as a bidentate ligand in some cases.

Will I get a 50-50 mixture of nitrito-N & nitrito-O co-coordination complexes? Or will there be a chance of getting random combinations of nitrito-N & nitrito-O coordinate bonds to a Co3+ (e.g. 4 nitrito-O + 2 nitrito-N)?

Not necessarily! It is not easy (at least not as easy as OP described it) to synthesize specific linkage isomer. It has been shown that the coordination mode of nitrite ion depends on its nature and the stereochemical environment which surrounds the metal ion (Ref.1 & references there in).

If I just want to get nitrito-N complex, are there ways to prevent the formation of nitrito-O complex? Or the mixture must form and then we need to do some other reactions to get rid of nitrito-O complexes?

These specific compound have been synthesized. Thus, there should be ways to such synthesis. However, I'm not qualified to give you a definitive answer for that question. Inorganic coordination chemistry expert would be the most qualified one. Yet, references in here including Ref.1-3, and references there in would give you better view. For instance, Ref.3 has described a ways to synthesize $\ce{Cu(NO2)2}$ and $\ce{Mg(NO2)2}$ complexes. Both metal showed nitrito-O,O′ (as bidentate) coordination with different stabilities.

References:

Hoda Pasdar, Bahare Hedayati Saghavaz, Naser Foroughifar, and Mehran Davallo, "Synthesis, Characterization and Antibacterial Activity of Novel 1,3-Diethyl-1,3-bis(4-nitrophenyl)urea and Its Metal(II) Complexes," Molecules 2017, 22(12), 2125 (12 pages) (DOI: https://doi.org/10.3390/molecules22122125).
Andrew J. Timmons and Mark D. Symes, "Converting between the oxides of nitrogen using metal–ligand coordination complexes," Chem. Soc. Rev. 2015, 44, 6708-6722 (DOI: 10.1039/C5CS00269A).
Andrew J. Timmons and Mark D. Symes, "Coordination of $\ce{NO2}$ to $\ce{Cu}$ and $\ce{Mg}$ in $\ce{M(NO2)2}$ Complexes. A Theoretical Study," Chem. Soc. Rev. 2015, 44, 6708-6722 (DOI: 10.1039/C5CS00269A).


How is chemical energy stored in carbon I'm fairly new to chemistry, and I have a hard time understanding how chemical energy is stored in carbon (I'm aware that the question can be generalised to 'how is chemical energy stored', but I'm trying to understand it in the example of carbon). In the example in my book, carbon and oxygen react to form carbon dioxide, releasing energy in the form of warmth in the process.

My book says that the energy released in this process mainly comes from carbon, which is originally the solar energy stored by plants. I have to tried to look for explanations how that energy is stored (not how the process of photosynthesis works, but in what form or what is the meaning of the 'stored energy'), but I still can't really wrap my head around it. The best answer so far was What is the nature of chemical energy?

If I understand the answer there correctly, the bond potential energy of a carbon dioxide molecule is lower than the combined bond potential energy of a carbon atom and an oxygen molecule. Given that, there are still two things I don't understand:

Why is the bond potential energy in the carbon atom and/or oxygen molecule higher than that of a dioxide molecule? How can that explained? Is this directly related to the chemical traits of the elements carbon and oxygen and the compound carbon dioxide?
Why, or better said how is said bond potential energy converted to warmth in the process?

I'm aware that there might be concepts in physics here I don't know or understand to understand a possible answer, and I would be thankful for any pointers to what subjects I should be first reading to understand such an answer.

Edit: I would like to quote here something I came across on wikipedia. I'm not providing it as an answer nor claiming it is correct, just trying to illustrate how confusing it is to find a satisfying answer for this matter. This comes from https://en.wikipedia.org/wiki/Chemical_bond#Overview_of_main_types_of_chemical_bonds:

  In the simplest view of a covalent bond, one or more electrons (often a pair of electrons) are drawn into the space between the two atomic nuclei. Energy is released by bond formation. This is not as a reduction in potential energy (sic), because the attraction of the two electrons to the two protons is offset by the electron-electron and proton-proton repulsions. Instead, the release of energy (and hence stability of the bond) arises from the reduction in kinetic energy due to the electrons being in a more spatially distributed (i.e. longer de Broglie wavelength) orbital compared with each electron being confined closer to its respective nucleus.

Italic and bold added by me. Note that wikipedia itself has a '[clarification needed]' tagging in the middle of that text, right after 'two protons', which I removed.

So if I understand that correctly and assuming that info is correct, any formation of a covalent bond would result in energy being released. But that still doesn't explain why my book is claiming the energy is particularly stored in carbon (and not in oxygen), and why it is released as warmth. In addition, the explanation on wikipedia would contradict what I referred to earlier, namely that the energy released comes from differences in bond potential energy.

One of the most important issues in understanding bond energy in chemistry is the question: energy relative to what?

The formation of bonds from (mostly hypothetical) atomic elements releases energy (or they wouldn't be bonded.) Solid carbon (graphite or diamond) has less energy than a cloud of carbon atoms so it could be said graphite has less energy relative to a cloud of atomic carbon atoms. We would describe this energy as "bond energy". But this is the energy released by the hypothetical process of forming lists of C-C bonds from isolated carbon atoms.

But there is another subtle complication if we want to understand reactions. sometimes, when chemists write a reaction scheme they don't specify all the detail. So C +O2->CO2 doesn't actually mean atomic carbon atoms plus oxygen molecules give carbon dioxide. Most chemists would assume that the carbon starts in the state it is normally found as in laboratories: graphite (so not isolated atoms but a molecular solid containing lost of C-C bonds). This terminology seems to have confused some people. But chemists don't like to waste space in formulae redundantly specifying things most of their colleagues will already know. The default state of an ingredient in a reaction is assumed unless specified otherwise and this usually means the form the compound or element takes in a bottle on the lab shelf.

So how can we say that carbon stores energy? Well we have to ask the question, relative to what? In the case of the question the answer is relative to carbon dioxide gas. The point here is that the double bonds in CO2 are stronger than the combined bonds in oxygen gas and graphite (so they are lower energy) This means that the reaction between carbon and oxygen releases energy (it takes less energy to break all the O-O and C-C bonds than is released when new C=O bonds are formed. So elemental carbon "stores" energy relative to carbon dioxide.

The claim that carbon "stores" solar energy absorbed by plants is a little simplistic (mainly because plants don't store carbon but usually store more complicated molecules like sugars or poly-sugars like lignin or cellulose). Dead plants are converted to more carbon like compounds (oil is mostly hydrocarbons, but coal is mostly carbon) under some geological conditions. What plants actually do at the start of this process is to convert carbon dioxide into sugars using energy input from sunlight. They do this to store energy and to grow by converting the sugars into the structures of their leaves, stems and trunks.

We can reverse that process by burning the plants or their degradation produce like oil and coal in air to reproduce the carbon dioxide and release energy. In this highly simplified sense, carbon is a store of "solar" energy.

But the key to understanding what is meant is to understand the chemical reaction involved in the creation or burning of the carbon. Then you can understand what energy is stored and how it is stored.


What is the number of nodes in the molecular orbitals of cyclic H3? A quick google search does reveal the molecular orbitals of this cyclic H3.

However, one source mentions there are are two nodes; while two of my professors said there's one node. Which is correct?

This seems to be a related question - I am not sure. There also is a question on linear H3+ - not cyclic.

In linear $\ce{H_3}$ you have one MO with zero nodes, one with one node and one with two nodes.  When you bend the molecule around to make a cycle, the two nodes in that last orbital merge to make a second orbital with one node.  The two orbitals of cyclic $\ce{H_3}$ with one node are, of course, degenerate when the ring has threefold rotational symmetry.


Multiple solutions in the balancing of redox reactions While looking at the reaction between potassium permanganate and hydrogen peroxide in the presence of sulfuric acid a friend and I were attempting to find a general equation to yield the coefficients of the balanced reaction depending on varying levels of acidity. I solved using half reactions and came up with equations for each coefficient in terms of the mole amounts for the manganese(II) ions and manganese(IV) oxide produced. My friend however, went about the problem by setting up a system of equations from the initial skeleton equation and yielded answers which depended on 3 variables.

From this, his method gave the answer

$$\ce{H2O2 + 4MnO4- + 6H+ -> 4O2 + Mn^2+ + 3MnO2 + 4H2O}$$

as a possibility, while my method yielded

$$\ce{7H2O2 + 4MnO4- + 6H+ -> 7O2 + Mn^2+ + 3MnO2 + 10H2O}$$

from the same ratio of manganese(IV) oxide to manganese(II). His reaction appears to balance correctly. Is there something he is missing that prevents that reaction from occurring with his coefficients, or does the half reaction method hide some possible answers?

The first thing that you should realise is that under specific conditions (here: acidic ones) only one product is formed in this reaction. Permanganate is not reduced to both manganese(IV) oxide and manganese(II) cations, it will either cleanly give manganese(IV) oxide (if the acid concentration is very low) or cleanly give manganese(II) cations (if the acid concentration is sufficient). The only way to generate both is to have sufficient acid at first which is subsequently used up by the reaction and drops below a threshold. Thus, if you were titrating and notice the dark precipitate of $\ce{MnO2}$, immediately stop, throw away the analysis and start again as it has become worthless.

Therefore, there is only one simplest solution to the problem, depending on which half-reaction you prefer:

$$\begin{align}\ce{MnO4- + 8 H+ + 5 e- &-> Mn^2+ + 4 H2O}\tag{Red1}\\[0.6em]
\ce{H2O2 &-> O2 + 2 H+ + 2e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 6 H+ + 5 H2O2 &-> 2 Mn^2+ + 8 H2O + 5 O2}\tag{Redox1}\\[1.6em]
\ce{MnO4- + 3 e- + 4 H+ &-> MnO2 + 2 H2O}\tag{Red2}\\[0.6em]
\ce{H2O2 &-> O2 + 2 H+ + 2 e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 2 H+ + 3 H2O2 &-> 2 MnO2 + 4 H2O}\tag{Redox2}\end{align}$$


Molecular knots absolute configuration How is the absolute configuration of chiral molecular knots determined? What rules should be applied? For centrochiral compounds, the CIP rules are applied, for axially chiral is described here and for planar chiral here. I can't find the system for assigning R/S (or possibly some other stereodescriptors) for molecular knots, e.g  the attached knot doesn't contain an Sn improper axis, therefore it is chiral.

I guess that the IUPAC Gold Book rules for helicity apply here.

  The chirality of a helical, propeller or screw-shaped molecular entity. A right-handed helix is described as P (or plus), a left-handed one as M (or minus).

The above knotane is right-handed, so it should be P.

For more complex molecular knots such simple rules most likely will fail.


What is the hybridisation of hydrogen in methane? I am trying to understand hybridisation. In methane carbon has $\mathrm{sp^3}$ hybridisation, but what is the hybridisation of hydrogen? Is it $\mathrm{sp^3}$? If yes then why?

Hydrogen does not hybridise, as it only has one $s$ orbital. Hybridisation is the "mixing of $s$ and $p$ orbitals; here is a good explanation (for ethane, but it explains the general theory as well).


Which bonds need to be broken to form a liquid? I'm confused whether it's the intermolecular forces, or the bonds that are being broken when a substance changes state? Does this differ among different types of compounds (eg. covalent, ionic, covalent network solid? What is actually happening when they change state?

It could be either chemical bonds or weaker intermolecular forces that change during a change of state. Water is held together by hydrogen bonds and make its boiling point higher than for analogous compounds.

Sulfur changes both ways with temperature:

  Sulfur forms polyatomic molecules with different chemical formulas, the best-known allotrope being octasulfur, cyclo-S8... At 95.2 °C (203.4 °F), below its melting temperature, cyclo-octasulfur changes from α-octasulfur to the β-polymorph. The structure of the S8 ring is virtually unchanged by this phase change, which affects the intermolecular interactions. Between its melting and boiling temperatures, octasulfur changes its allotrope again, turning from β-octasulfur to γ-sulfur, again accompanied by a lower density but increased viscosity due to the formation of polymers. At higher temperatures, the viscosity decreases as depolymerization occurs. Source

Metallic bonds are delocalized, so in melting, there is no real change in bonding (though on boiling, a metallic vapor is quite different).

One could say, though, that this is simply a matter of definition: chemical bonds act over shorter distances, and intermolecular forces over longer, but they are not intrinsically different, the polar or hydrogen bond being an example of one that bridges these realms.


Trivial name of a compound vs its chemical formula I had a question like below and I've had trouble with similar problems before and how to go about solving for the chemical formula.

  What is the chemical formula for lithium aluminum hydride?

The first thing that comes to mind, we have one lithium, one aluminum and one hydride. Then why is the chemical formula $\ce{LiAlH4}$? Where did the four $\ce{H}$ atoms come from? Earlier on in the course we had stuff like

  What is the chemical formula for di-arsenic trioxide?

And that one is easy, I have 2 arsenic and 3 oxygen. Given the full name I can easily deduce the formula (or the other way around). But what do you do in situations where you don't get "all the information". Do I go about thinking up how the structure works and see that it somehow logically follows that it needs to be four $\ce{H}$?

I know most basic compounds, e.g. sulphate is $\ce{SO4^2-}$, but I feel like I can't memorize some of the things. There has to be a more general way to do this. It feels like and I must have missed it. Essentially, the full systematic names are no problem, but the trivial names are giving me a rough time and I'm scared if they come up during my finals next week. Any tips regarding that?

No you are absolutely right here, but even sulphate or nitrate is wrong. If you take a closer look at what a sulphate essentially is, you will see the following: It is a separate unit, it has a central atom and ligands and the whole unit is negatively charged. If this was a metal we would say it's a complex and name it with some sort of -ate in the name. And this is exactly what you see here. So tetraoxosulfate in short is sulfate. Whit this shorter notation you are just not able to guess it unless it is so common that you should know about it. For lithium aluminium hydride it's the same thing, it should be called lithium tetrahydroaluminate.

Now you may ask yourself how are you supposed to work with these problems? In all scientific cases just google it. That's what we do as well, if this is class or homework related, then your teacher will only give you certain examples. If he doesn't, talk to him about it.


Reaction in Zinc-Carbon-Ammonium Chloride battery I made a dry cell battery from zinc (Zn), carbon (C) and ammonium chloride (NH4Cl) which produced a current of 0.4 amps and voltage of 1.1V. However when I could not find the redox reaction that was occurring.

What is the chemical reaction happening in the cell?

Surprisingly this battery uses Zn without any oxidant like $\ce{MnO_2}$ and it  produces a voltage as high as $1.1$ V. In order to establish its characteristics, the two compartments should be studied independently.
First, the anode, with zinc. $\ce{Zn}$ should not produce simple $\ce{Zn^{2+}}$ ions, as there are ammonia and ammonium chloride in the solution. So the anode reaction is probably :  $$\ce{Zn  + 4 NH_4^+ ->  [Zn(NH_3)_4]^{2+}  +  4 H+  +  2 e-}\label{rxn:zn4}\tag{1}$$ The reduction potential for this anodic equation is not given in the tables. But it may be obtained from the next equation, whose standard reduction potential is known to be $\ce{E_2° = - 1.03 V}$.  $$\ce{Zn  + 4 NH3 ->  [Zn(NH_3)_4]^{2+}  +  2 e-}\label{rxn:zn}\tag{2}$$  In a $1$ M solution of $\ce{NH4Cl}$, $\ce{[NH_3]}$ = $2.4 ·10^{-5}$ M. So the corresponding reduction potential of ($2$) is : $$\ce{E_2 = - 1.03 \mathrm{V} + 0.0296\mathrm{V}· log (2.4 ·10^{-5})^4} = - 1.03~ \mathrm{V} - 0.54~ \mathrm{V}  = -1.57 ~\mathrm{V}\label{rxn:nh}\tag{3}$$
Now the cathodic reaction should be studied. Suppose the cathode reaction is  : $$\ce{2 H2O  + 2 e- -> 2 OH- + H2}\label{rxn:z}\tag{4}$$ with a redox potential equal to  $- 0.83$ V. Couping these two expressions ($2$) and ($3$) are never sufficient to get $-1.1$ V as the overall voltage. So, as the ammonium chloride solution is acidic, the cathode reaction must rather be : $$\ce{2 H+ + 2 e- ->  H2}\label{rxn:znn}\tag{5}$$ In a $1$ M  $\ce{NH4Cl}$ solution, the pH is $4.62$, so the corresponding potential of the cathode is E$\ce{_5}$ = - $0.059$ pH = - $0.059·4.62$ V = - $0.27$ V.
To summarize, the anode reaction is ($1$) or ($2$), at will, and the cathode rection is ($5$). As a consequence, the theoretical voltage of the cell is : $\ce{E_5}$ - $\ce{E_2}$ = - $0.27$ V + $1.57$ V = $1.3$ V, if the concentration of ammonium chloride is $1$ M.
Ref.: Standard electrode potentials have been taken from "Chemistry Data Book" from J. G. Stark and H. G. Wallace, John Murray, London, $1994$, Table $43$, p. $70$.


Does CO2 dissolution in different salt concentrations affects the amount of carbonic acid formed? We have a school experiment, where I must report on the following question:

  Does salinity affect the solubility of $\ce{CO2}$ in saltwater?

So we selected four different concentrations of salt (in the range of 1-4 %) in 100 mL of water and added 5 g of $\ce{CO2}$ and measured the time it took to dissolve completely. We then used titration to figure out the amount of $\ce{CO2}$ (reacting with water to give weak carbonic acid) that is dissolved.

My questions are:

Will the amount of $\ce{CO2}$ that reacts with water depend upon the salt concentration?
Will the amount of titrant used to figure out the amount of $\ce{CO2}$ dissolved also change depending on the salt concentrations?

100 mL of water and added 5 g of $\ce{CO2}$ and measured the time it took to dissolve completely

This statement doesn't make sense.  Firstly, if you know it dissolved completely, why try to determine how much dissolved?  Secondly, the only way that much CO2 would dissolve is if you did the experiment at high pressure.

If you want literature data to compare your experiment to, see Fig. 6c or equation 6 of Behaviour of the CO2-H2O system