Why is nucleus charge squared in Rydbergs formula? According to coulumbs law, the energy changes linearly with charge. So why is the nucleus charge squared in this formula? Is there a way to derive it?

I was suspecting it might be because it pulls the electron closer too.

Derivation comes from the formula for electrostatic potential energy.

The electrostatic potential between the point charges of an electron, and a proton would be represented by the formula

$$U_E(r) = k_e \frac{q_eQ_p}{r}$$

However, in the derivation that we are looking at a difference of energy is from an emission or absorption, such that the energy concerned is

$$E_i - E_f$$

The subtraction appears in the Rydberg formula from the distance in the form

$$\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$

However, the terms other than the radius are multiplied together and come outside the parentheses.  Even though we are looking at only a single electron, the term for the charge in the nucleus is squared because of this. This can be seen in the Bohr derivation of the Rydberg formula.

$$E = h\nu = E_i - E_f = \frac {m_eq_e^2q_Z^2} {8h^2\epsilon_0} \left( \frac {1} {n_f^2} - \frac {1} {n_i^2} \right)$$

This shows the origin of the squared term for the nuclear charge which correlates with the Rydberg formula for a hydrogen-like element since all terms outside the parentheses other than charge are constant.

$$\frac {1} {\lambda_{vac}} = RZ^2\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$


Friedel Craft Alkylation Why do we get isoproyl benzene on treating benzene with 1-chloropropane instead of n-propyl benzene?

Is it because of Markovnikov's rule?

First AlCl3 converts 1-chloropropane into propane with a positive charge on terminal carbon

Since secondary carbocation is more stable than primary carbocation it undergoes rearrangement and positive charge shifts to secondary carbonation atom.

Hence we get isopropyl benzene


Determining reaction order in equilibrium reactions Am hoping that someone can help me with the below question, and that I am asking in the right place. Am new here, so please forgive any missteps.

Given a reaction like this:

$$\ce{A + B <=> C + D},$$

is the differential/initial rates method of determining a reaction order valid? That is, the method which determines $b$ in $r = k C_\ce{B}^b$ as the slope in a $\ln({\mathrm{d}C_\ce{C}}/{\mathrm{d}t})$ vs $\ln(C_\ce{B})$ plot.

Basically, I wonder about three points here:

I have only seen examples for this method on single forward reactions, so is it valid when there is a reverse reaction?
Am I correct in thinking that I can use measurements of $C$ for $\ln(\text{rate})$, even though I'm interested in the reaction order of $\ce{B}$, because the rate can be written in terms of any of the reactants or products?
How crucial is it that the concentration (rate) measurements must come from initial times? If the change in all concentrations is very small, is it then not valid to use even a later values?

The difficulty with problems in kinetics is that you can envision a pool of intermediates such that, while B is consumed, you don't see significant formation of C, so that C is a poor reporter of the consumption rate of B. Concepts such as intermediates, autocatalysis, feedback, steady-states, lag times can be important here. Things can get complicated pretty quickly.

The initial forward rate, in the absence of a significant accumulation of reaction intermediates, can for simplicity be written as

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

Again, this assumes a negligible accumulation of intermediates whose reverse reaction may lead back to formation of B, as in a process under steady state.

For this to be useful in the determination of $k_\mathrm{f}$ it helps to satisfy two conditions:

$k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b \gg k_\mathrm{rev}[intermediates]^m$
$[\ce{A}]^a\approx \text{const}$

These conditions are essential to guarantee a decay of the concentration of $\ce{B}$ with a rate constant $k_\mathrm{fwd}[\ce{A}]^a$. The answer to your second question is no unless the mechanism is simple and conditions 1 and 2 hold (small amount of product), that is, if the reverse reactions from intermediates or products are not important.

So, it is not essential that the measurements come from the beginning as long as the mechanism is elementary (no rate limiting steps leading to accumulation of intermediates), the reverse reaction is insignificant in rate and the concentration of A greatly exceeds that of B throughout your measurement time.


Basicity of boric acid My book defines the basicity of the acid as the number of $\ce{H^+}$ ions furnished by one mole of the acid in solution. Well, I know that $\ce{B(OH)_3}$ is a weak Lewis acid and it accepts an $\ce{OH^-}$ to form an anion - thereby increasing the net $\ce{H^+}$ concentration in the solution.

So, what should exactly be its basicity? Should I just assume its basicity to be one as well - although it doesn't satisfy the definition?

Boric acid does not dissociate in aqueous solution, rather it is acidic due to its interaction with water molecules (i.e accepts $\ce{OH-}$) to form the tetrahydroxyborate ion:

$\ce{B(OH)3 + H2O -> [B(OH)4]- + H+}$

$\ce{B(OH)3}$ does not act as a proton donor but behaves as a Lewis acid i.e. it accepts a pair of electrons from $\ce{OH-}$ ion and in this process $\ce{H+}$ ions are furnished.

I understand monobasic means there should be 1 replaceable hydrogen atom in the molecule, but $\ce{B(OH)3}$ in itself actually doesn't contain any replaceable hydrogen atom; rather, on interaction with water (i.e after accepting a $\ce{OH-}$ ion) a hydrogen ion is liberated. But if you see the after effect, say adding a monobasic acid (which holds the definition correctly) like $\ce{HCl}$ to water, per molecule it liberates one $\ce{H+}$ ion, similarly each $\ce{B(OH)3}$ also liberates one $\ce{H+}$ after it accepts $\ce{OH-}$ from water. So only looking at the after effects/ final result, both $\ce{HCl}$ and $\ce{B(OH)3}$ liberate 1 hydrogen per molecule though the process through which it was liberated is different.

So considering the after effect we consider $\ce{B(OH)3}$ as an acid and as it finally liberates 1 electron per molecule (though after accepting $\ce{OH-}$), we still can consider it to be a monobasic Lewis acid. At least, Wikipedia calls it a monobasic Lewis acid.


Inductive effect in Organic Chemistry Why does $\ce{O-}$ show +I effect?
On the other hand why does $\ce{OH-}$ show -I effect?
both oxygens are electron rich and are negatively charged. my teacher says that charge on the species is a headache for the species ,so according to me both the oxygens should show +I effect, that is donating electrons tendency. Plus they have lone pairs too.
How can we extend the similar explanation for the halides? they have 3 lone pairs so they should also not take up more electrons.
If carbon is double bonded to oxygen then is there any inductive effect taking place over there?

Organic is confusing me. Help would be appreciated.

Lone pairs can sometimes be delocalized. An acceptable rule of thumb is that if an atom possesses lone pairs and is connected to an sp2 atom, then the two atoms participate in resonance. So consider phenol. Oxygen has lone pairs. It's connected to an sp2 carbon. And so yes, the oxygen can participate in resonance with the sp2 carbon it's connected to, and also the other sp2 carbons in the aromatic ring. We can ascertain the fact that every atom is phenol is sp2 by noting that phenol is planar.

In general, a negative formal charge makes delocalization a (more) favorable move as this decreases the electron (charge) density on one atom. Spreading out electron density between or among multiple atoms generally lessens reactivity.

Now, your question is why would oxygen exhibit a +I or electron-donating effect despite its high electronegativity. Well, it can participate in resonance donation as explained above; see phenol and its conjugate base, the phenoxide ion, for examples. Examining both in terms of electrophilic aromatic substitution also reveals that the oxygen atom in both is strongly electron donating. True, oxygen can withdraw electron density via sigma bonds (inductive effect) but doing so is unfavorable especially in the phenoxide ion in which the oxygen already bears a negative formal charge. As your teacher explained, bearing charge isn't something that is very favorable.

And in general, you're going to find that resonance effects outweigh inductive effects. Hence the reason why an sp2 oxygen connected to another sp2 atom generally exhibits a +I effect. Consider dimethyl malonate and methyl acetoacetate . Which is more acidic? From inductive effects alone we'd say dimethyl malonate because the oxygen is highly electronegative as opposed to an sp3 carbon and should withdraw more electron density via the inductive effect. From a resonance-standpoint we'd say the opposite; the oxygen in the ester donates electron density via resonance more than a methyl group can inductively donate electron density. And you'd have to realize that the resonance-standpoint is more important in evaluating the acidities of the two weak acids.

A common exception to resonance>inductive effect in organic chemistry is an sp2 atom connected to a halogen. Yes, the halogen has lone pairs and these can be delocalized into the adjacent p-orbital. But the halogens (except possibly fluorine) are much bigger than many other atoms, and therefore the magnitude of resonance donation is limited. So the great electronegativity of the halogens trumps their magnitude resonance donation, and we can see this in the context of electrophilic aromatic substitution again; halogens are ortho-para directors (delocalize the lone pairs in the Lewis structures to see why) but are also mild deactivators.


Is so called "Polymerized LiNb (Lithium Niobocene)" a possible compound? The so called "Polymerized LiNb (Lithium Niobocene)" is a fictional compound.

I heard about this substance being used for an interesting purpose. It reacts to mechanical stress or pressure by creating an electric charge (piezoelectricity) and, conversely, deforms, as if external pressure was applied, when voltage is applied. Apparently it was far more "piezoelectric" than any other similar compounds, i.e. it deformed far more drastically than other piezoelectric materials.

Does such a compound, or similar compounds, exist?

Materials such as Terfenol-D and Galfenol are magnetostrictive ferrous alloys that expand and contract in the presence of magnetic fields, as well as creating magnetic fields when acted on by a physical force.  Both these materials were developed at Ames Lab (formerly the Naval Ordnance Laboratory, that's why the material names end with -NOL).  Terfenol-D has the highest magnetostriction of any Fe-alloy, but the high rare earth content or Terbium and Dysprosium limits its commercial applications.

The "polymer" moniker for Lithium polymer batteries does no refer to the lithium metal or oxides, it only speak to the polymer electrolyte or a polymer casing.  It just a marketing term, no a chemical designation

So on to your question... Polymer, no.  Monomer, no.  Niobium has an extremely high melting point (nearly 2500C/4500F).  It is a very difficult material to melt.  Lithium boils around 1300C.  From a metallurgical processing point of view, you would need a chamber pressurized to several MPa with an inert gas to prevent the lithium from volatilizing as you attempt to alloy it with niobium.  You would also need some powerful magnetic stirring to attempt to keep the mix homogeneous since the lithium would literally float out of the solution.


Stabilisation of phenoxide ion Which one of the following substituents at para-position is least effective in stabilizing the phenoxide ion $\ce{PhO-}$
(a)$\ce{-CH3}$
(b)$\ce{-OCH3}$
(c)$\ce{-COCH3}$
(d)$\ce{-CH2OH}$

According to me it should be $\ce{-OCH3}$ because of its mesomeric effect which will counter the mesomeric effect of $\ce{O-}$.
$\ce{-CH3}$ has a positive inductive effect but, the mesomeric effect of $\ce{-OCH3}$ should be stronger and destabilize the phenoxide ion to a greater extent

Which one of the following substituents at para-position is least
  effective in stabilizing the phenoxide ion $\ce{PhO^{−}}$

Yes, the p-methoxy group is the least stabilizing of the choices. I'm not quite sure what you mean by

  its mesomeric effect which will counter the mesomeric effect of $\ce{O^{−}}$
  .

so let me add a few words of explanation for the choice.

In the figure below, I've drawn a few of the resonance structures possible for p-methoxyphenoxide (A-C).  The methoxy group donates electrons through resonance and resonance structure B illustrates this effect.  However, note that in resonance structure B we have placed two negative charges close to each other - this is very destabilizing.  Therefore, as the methoxy group donates electrons through resonance it destabilizes the system. In resonance structure C, the electron withdrawing inductive effect of the methoxy group is stabilizing.  However the resonance effect outweighs the inductive effect and overall the p-methoxy group is strongly destabilizing in the case of phenoxide.

Resonance structure D illustrates the p-methyl case.  Methyl is very mildly electron releasing both through resonance and inductively, so it also destabilizes the phenoxide anion, but since its effect is weak, it is not as destabilizing as the p-methoxy substituent.

Finally, notice in the p-acetyl case we can draw resonance structure E which actually stabilizes the phenoxide anion via resonance.  The p-acetyl group is also inductively stabilizing in the phenoxide example.


What is the proof of complete ionic dissociation in water? Is there any way to prove that ionic compounds, for instance $\ce{NaCl}$, completely dissociate into singular $\ce{Na+}$ and $\ce{Cl-}$ ions within water by hard experimental evidence?

Besides adhering to the perfect solvation model in theory, I am interested in definitive proof of monoatomic solvation in practice.

How do we know that the ionic crystal lattice isn't only partially dissolved? When the superficial reaction between water and the ionic compound takes place, could it stop at say 10-100 atom microcluster lattices ($\ce{Na/Cl/Na/Cl}$ etc), giving the appearance by standard means that it is completely dissolved?

I am only interested in experimental means of verification, not concepts taken from books.

One method is by making electrical conductivity measurements on aqueous solutions of salts. If we take four strong electrolytes:

Sodium chloride
Potassium chloride
Sodium perchlorate
Potassium perchlorate

We can measure the electrical conductivities at very low concentrations which allows us to obtain the limiting specific conductivties of the salts (extrapolation to zero concentration).

We will then find that the sum of the limiting conductivity of sodium chloride and potassium perchlorate will be the same as that of potassium chloride and sodium perchlorate.

This is because the limiting conductivity of a strong electrolyte salt is the sum of the contributions of the cations and anions. This allows us in an indirect way to show that some salts can totally dissociate when very dilute.


Determine direction of mass transfer For part a) I don't really know what to look at. Diffusion occurs from low to high concentrations so by looking at $t_1$ I would say that the direction of the mass transfer is from phase II (organic phase) to phase I (aqueous phase) since the organic phase starts at a lower concentration than the aqueous phase.

But apparently it is supposed to be the opposite. And the answer is the same at $t_2$.

Could anyone please explain how to determine the direction of the mass transfer and how it can be the same at $t_1$ and $t_2$. Thank you!

You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.

The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.

So you should follow these rules:

When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).

In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.


Ionic Radius of Iodine in High Temperature Form The ionic radius of iodine in the high temperature form can be determined once the unit cell is known, but how can I calculate the radius of $\ce{I}$ given that the cubic lattice parameter is 6.10 Å.

Considering by cubic lattice you mean a Simple Cubic structure, you could do the following:

Reference image1
Reference image2

(could not put it here because of copyright)

In a cubic lattice the atoms touch each other along the edges(see image1 and image2), hence the edge length/parameter will be equal to twice the radius (see image2).
$$\therefore\ \ce{Edge\ Length(a)=2\times radius(r)}$$
$$\therefore\ a \ = \ 2\times r$$

This method can be adopted for any lattice, you only need to find the axis along which the atoms touch each other(image1 bottom row).

If they exhibited an fcc close packing structure, you would notice that the atoms touch each other along a face-diagonal.

$$
\begin{align}
\therefore\ce{Length\ of\ face\ diagonal&\ =\ radius\ of\ top\ left\ atom +\\
                                        &\quad\ diameter\ of\ atom\ at\ face-center + \\
                                        &\quad\ radius\ of\ bottom\ left\ atom}\\
\therefore\ce{Length\ of\ face\ diagonal&\ =\ r\ + ( 2\times r ) + r}\\
\therefore\ce{\sqrt{2}\times a&\ =\ 4\times r}\\
\therefore\ce{a&\ = 2\sqrt{2}\times r}
\end{align}
$$

Similarly for bcc structure the atoms touch each other along the body-diagonal
$$\therefore\ce{\sqrt{3}\times a\ =\ 4\times r}$$


pH level of fabric with regard to inhibiting fungal growth Someone on another forum made the point that doing laundry with sodium hypochlorite (bleach) will kill any fungus such as athlete's foot fungus, but that after the fabric dries, the pH level of the fabric is conducive to more fungal growth, since bleach is a base.  The person made the point to substitute bleach with an acid like lemon juice, or to first do the laundry with bleach and then a second time with lemon juice.  Exact quote below:

  "Bleach removes fungus and mold but also promotes it’s later development. I had jock itch on and off for years until I stopped using bleach in my wash. Once I stopped using bleach it never returned. Fungus thrives in high PH environments caused by Bases like bleach. My jock itch returned for a short time when my wife used bleach again. If you must use bleach you should neutralize it with a post acid rinse like vinegar or lemon juice."

Is there any scientific rationale to backup this statement?  Is lowering the pH of fabric with an acidic substance a workable means of preventing fungal growth?  Or does it not matter- in other words, washing your clothes with bleach should neither promote nor inhibit future fungal growth (emphasis: athlete's foot fungus)

Hypochlorite bleach is a potent biocide, so throughly washing clothes with it should kill all fungi. The pH part is true, bleach is alkaline, but so are all common detergents. But this should not even matter, if the clothes are properly rinsed with clean water, that should remove almost all of the bleach and restore the pH to close to neutral.
What I think happened, is that his clothes were not rinsed well enough, so traces of the bleach and the detergents stayed in the clothes, causing or exacerbating skin problems. Some people have skin especially sensitive to bleach and allergic reactions to detergents can happen to.
So I would recommend washing with bleach, but making sure the clothes are well rinsed. You may need to set your washing machine on an extra rinse cycle to get the desired effect. Using lemon juice should not be necessary.


What is the "mechanism" of the decomposition of hydrogen peroxide? In my classwork on the decomposition reaction of hydrogen peroxide by catalase, I wrote the following equation:
$$\ce{2H2O2 -> 2H2O + O2}$$
However, my teacher said that I should go further in depth and write down the (quote unquote) "mechanism" of this reaction. I'm not sure what he is talking about, can anyone help me?

EDIT: I am an IB student studying chemistry in Standard Level so the answer to the above problem should remain within the scope of the IB syllabus. Here's more information from my teacher:

“Although we show reactions in one neat equation, the majority of reactions occur in a series of steps. (…) There’s three step to that (the one above) reaction and catalase will act on one of those steps to reduce the activation energy and therefore make the reaction faster”.

Below is one way to write it that shows what the enzyme does and avoids the uncertainty of the details:

$$\ce{E + H2O2 -> E-O + H2O}\tag{1}$$
$$\ce{E-O + H2O2 -> E-O2 + H2O}\tag{2}$$
$$\ce{E-O2  -> E + O2}\tag{3}$$

In words, the enzyme binds to the first molecule of hydrogen peroxide, release one molecule of water and holds on to the other oxygen atom (via the iron bound to heme bound to the enzyme). Then, the second molecule of hydrogen peroxide binds to the enzyme, which again release water and holds on to the second oxygen atom. Finally, the enzyme releases the two oxygen atoms in the form of the dioxygen molecule.

Alternatively, you could write the following three steps, giving even less information about what is happening in the enzyme:

$$\ce{E + 2H2O2 -> [E\cdot H2O2\cdot H2O2]}\tag{A}$$
$$\ce{[E\cdot H2O2\cdot H2O2]-> [E\cdot H2O\cdot H2O\cdot O2]}\tag{B}$$
$$\ce{[E\cdot H2O\cdot H2O\cdot O2] -> E + 2H2O + 2O2}\tag{C}$$

In words, the enzyme first binds to the substrates, then catalyzes the reaction, and finally releases the products. While this in general might be a good way of describing the action of an enzyme, it is not quite correct for catalase, which releases one product before the second substrate binds. However, this might be the answer you are looking for at the IB level. I would ask the teacher for further clarification.


In aminobenzoic acid, do we have three different pKa's? In para-aminobenzoic acid, how can I know whether the amino group loses a proton or not when $\mathrm{pH}$ exceeds $8.5$? I have only found two values for $\mathrm{p}K_\mathrm{a}$ but I never found info about the pKa (in this case it would be the third $\mathrm{p}K_\mathrm{a}$ at $\mathrm{pH}>8.5$) of the amino group.

In any case I guess at basic $\mathrm{pH}$, negative charge over $\ce{NH2}$ would be stabilized by aromatic ring and $\ce{COO-}$.

The two acidity constants in 4-aminobenzoic acid are due to the loss of $\ce{H^+}$ from the protonated form of 4-aminobenzoic acid (the 4-carboxyphenylammonium cation), transforming it into 4-aminobenzoic acid, and then a further loss of $\ce{H^+}$ turning it into the 4-aminobenzoate anion.

The first two ionizations have a $\mathrm{p}K_\mathrm{a}$ of around 5, while the next ionization quite likely has a $\mathrm{p}K_\mathrm{a}$ well below ~25 (compare with other substituted anilines), so you won't find any appreciable amount of deprotonated nitrogen atoms in an aqueous solution. Groups with N−H bonds are far less acidic than their O−H counterparts due to the lower electronegativity of the nitrogen atom, which is less capable of stabilizing the extra negative charge.


What is the flow of electrons, cations, and anions in a voltaic cell (please see image attached)? This is an IB question. The mark scheme says that the correct answer is A and only A. But I believe B is also correct. The following is my thought process. Since A is more reactive, it undergoes oxidation to form A2+ ions, while B2+ ions undergo reduction and form B. From this we can deduce that A is the anode and B is the cathode. According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image). Therefore, A and B would both be correct. Please can someone confirm my thinking or point out any mistakes. Thank you

According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image).

The idea of the salt bridge is to prevent electrolytes mixing while providing ion flow. When you have a high concentration of inert ions in the salt bridge, cations in the salt bridge will flow into B, and anions in the salt bridge will flow into A. The solution in the salt bridge is not mixed, so it would take a very long time for the electrolyte in A to diffuse all the way to B.

To conclude, ions flow out of the salt bridge into the solutions A and B, but virtually no ion flows all the way through the salt bridge. This tells you that the salt bridge will be depleted after a certain amount of current flows through the wire. Accordingly, electrochemical cells used in batteries usually don't have a salt bridge but a semipermeable membrane instead, or a clever choice of half reactions where mixing of the electrolytes is not a problem (lead acid battery for example, where the only soluble species, sulfuric acid, is common to both half reactions).


Number of stereoisomers of trisubstituted cyclopentane When 4-methylcyclopent-1-ene reacts with bromine, then 1,2-dibromo-4-methylcyclopentane forms. And I am stuck on determining the amount of stereoisomers that can form as a result.

1,2-dibromo-4-methylcyclopentane has two chiral carbon atoms, those of bromine. And an internal mirror plane, we should have 3 optical isomers.

But I am confused upon drawing the following:

1 and 4 appear to be the same, 2 and 3 appear to be the same.

5 and 7 appear to be the same, 6 and 8 appear to be the same.

This gives me 4 isomers. So I draw them different again:

The puckered structure leaves me doubting myself: while I think that the internal mirror plane causes there to be only 2 different cis structures instead of 4 (or instead of 1 if it had been 1,2-dibromo-4,4-dimethylcyclopentane), I am not sure if the trans structures are actually the same.

Almost everything that needs to be said has been said. But I do have a few comments about configurations. In enantiomeric, trans-dibromides 1 and 2, the methyl groups are chirotopic and non-stereogenic. Inversion of the stereochemistry of the methyl group in either 1 or 2 returns the same structure, respectively. Thus, there is no designation for stereochemistry at the C4 methyl center. On the other hand, inversion of the stereochemistry of the methyl group in 3 gives 4 and the same operation on 4 produces 3. Therefore, the methyl center in achiral isomers 3 and 4 is achirotopic and stereogenic. Accordingly, the C4 centers have lowercase descriptors, r and s, respectively. This site might be of help.


What is the optimal voltage for storing a 3.7 V lithium polymer battery? According to this website, storing a lithium polymer (LiPo) battery/cell at around 40% capacity would be more beneficial for the lifespan of the LiPo cell comparing storing the battery at 100% capacity. The author also stated that "it is best to store LiPo batteries at 3.6 V to 3.8 V. This is applicable for standard LiPo batteries that hold 4.2 V per cell when fully charged."
However, the instruction of an RC helicopter (with a LiPo battery (3.7 V (4.2 V when fully charged), 180 mAh capacity)) I bought stated that if I am not using the model, I need to make sure that its voltage is at least 3.9 V at storage.
The manufacturer, upon inquiry, told me that I should store the battery fully charged and only recharge it after the voltage drops below 3.9 V. I.e., my correspondent's answer and the instruction of the helicopter is quite different from the website mentioned before.
Who is the right one? The website or the instruction and the manufacturer, according to the working principle of LiPo battery?
Some other questions relevant for the maintenance of the battery (question added that were not asked originally):
Would the charging frequency be a determinant for battery lifespan? Usually how fast would the battery lose voltage by self-discharge (is it dependent on the voltage the battery has)? Also, when charging, does the charge in the LiPo cell build up linearly (so that I will know when is 40% charge reached)? Would it be better if I do not charge the battery fully even right before using the battery in the RC helicopter?
There is a undervoltage cut-off for my RC helicopter. What is usually the percentage of the charge at which the undervoltage cut-off is reached?
A/two question(s) out of curiosity:
Would the time needed for charging a completely new and empty cell to full for the very first time be different from the time needed for charging an already-used empty cell (maybe not completely empty, since the RC helicopter has undervoltage protection) to full? What is the difference between these two charging process?

The $\pu{3.6-3.8 V}$ range is a good general choice, but it may be battery-specific. The particular voltage for 40% charge may differ for different cell technologies, e.g. various deviations of electrode materials and due to cell aging.
The optimal storage conditions, as you mentioned, are more often expressed as charge/capacity % ratio. Usually, the optimal storage conditions are said to have about 40% of the maximal charge. For longer term storage, it is advised to be kept in a fridge, but not to get below
$\pu{0 ^{\circ}C}$. Note that by the semi-empirical van 't Hoff's rule(*) chemical reactions (= battery deterioration) speed up typically 2-4 times per temperature increase by $\pu{10 ^{\circ}C}$.
Lithium-ion/lithium polymer (Li-ion/LiPo) cells reportedly lose typically 20% of capacity per year when fully charged, but just 4% at 40% charged state.
For practical purposes, sometimes a trade off is made between battery lifetime and battery readiness.

Some background about Li-Ion/LiPo cell chemistry:
The negative anode is formed by lithium graphite intercalate $\ce{Li_\mathrm{x}C6}$, x<=1
The positive cathode is formed most commonly by $\ce{Li_xCoO2}$
The electrolyte is most commonly formed by $\ce{LiPF6}$ dissolved in mixture of dialkylcarbonates $\ce{R1-O-CO-O-R2}$ (**)
About electrode reactions from Li-Ion wikipedia article:

The positive electrode (cathode) half-reaction in the lithium-doped cobalt oxide substrate is
$$\ce {CoO2 + Li+ + e- <=> LiCoO2}$$
The negative electrode (anode) half-reaction for the graphite is
$$\ce{LiC6 <=> C6 + Li+ + e^-}$$
The full reaction (left to right: discharging, right to left: charging) being
$$\ce {LiC6 + CoO2 <=> C6 + LiCoO2}$$
The overall reaction has its limits. Overdischarging supersaturates lithium cobalt oxide, leading to the production of lithium oxide, possibly by the following irreversible reaction:
$$\ce {Li+ + e^- + LiCoO2 -> Li2O + CoO}$$
Overcharging up to 5.2 volts leads to the synthesis of cobalt(IV) oxide, as evidenced by x-ray diffraction:
$$\ce {LiCoO2 -> Li+ + CoO2 + e^-}$$
In a lithium-ion battery, the lithium ions are transported to and from the positive or negative electrodes by oxidizing the transition metal, cobalt ($\ce{Co}$), in $\ce{Li_{1-x}CoO2}$ from $\ce{Co^{III}}$ to $\ce{Co^{IV}}$ during charge, and reducing from from $\ce{Co^{IV}}$ to $\ce{Co^{III}}$  during discharge. The cobalt electrode reaction is only reversible for $x < 0.5$ (x in mole units), limiting the depth of discharge allowable. This chemistry was used in the Li-ion cells developed by Sony in 1990.

In addition to the above, there is another deteriorating process, related to keeping Li-Ion/LiPo cell long time at fully charged state ( typical for laptops permanently connected to AC, unless special coutermeasures are involved ). There is a side tendency of lithium ions in the highly charged cell state to form metallic lithium at surface of graphite-lithium intercalate. Normally, this is reversible. But there is ongoing slow side reaction with the dialkyl carbonate solvent:

The Reaction of Lithium with Dimethyl Carbonate and Diethyl Carbonate in Ultrahigh Vacuum Studied by X-ray Photoemission Spectroscopy
Abstract: The reaction of dimethyl carbonate (DMC) and diethyl carbonate (DEC) with clean metallic lithium in ultrahigh vacuum was studied by the use of X-ray photoelectron spectroscopy with the temperature-programmed reaction methodology. Both molecules are of interest as solvents in ambient-temperature lithium batteries. The solvent molecules were condensed onto the surface of an evaporated lithium film at $\pu{120 K}$, and spectra were collected as the sample was warmed in ca. $\pu{25 K}$ to $\pu{30 K}$ increments. The reaction of either DMC or DEC with lithium was initiated at $\pu{180 K}$, a temperature much lower than their bulk melting temperatures, producing lithium methyl carbonate, methyllithium and lithium ethyl carbonate, and ethyllithium, respectively. At temperatures greater than $\ce{270−300 K}$, the lithium alkyl carbonates start to decompose with $\ce{Li2O}$, elemental carbon, and alkyllithium as products on the surface. Both DMC and DEC are more reactive toward metallic Li than another carbonate solvent, propylene carbonate, which we have studied by the same methodology. Because methyl and ethyllithium are highly soluble in the parent solvent, electrodeposited $\ce{Li}$ is predicted to have poor stability in an electrolyte composed of either DMC or DEC.

(*) Not to be confused with the van't Hoff's equation - the van't Hoff's rule is about temperature dependency of kinetic rate, the van't Hoff's equation is about temperature dependency of equilibrium constant. But both are related.
(**)( dimethyl carbonate, diethyl carbonate, methylethyl carbonate, ethylen carbonate, propylene carbonate )
---- Feedback to comments
Being fully charged helps to prevent deep discharges, what is not the same as saying it is optimal storage condition. Note that storing cells fully discharge is worse than fully charged.
About half charge is optimal for lifetime and full charge is optimal for the time before needed recharging. But self discharging is not dramatic. LiPo have reportedly self-discharge rate approximately 5% per month.
This discharging is probably not linear, and its rate would be decreasing. Also, consider it as illustrative number only. It may differ for different cell technologies and age. Discharge cutoff cuts the connection before a real 0% charge. So the effective 0% charge > real 0% charge. Generally self discharging rate may increase with battery age, but it can be an individual number.
The usual mode of Li-ion/LiPo charging is by a constant current, until the threshold voltage is reached. Since then, charging switches to the constant voltage charging, until the charging current drops below a threshold value. The point of mode switching is typically between 70 (old cell)-80(new cell)% of the full actual capacity.
All that means the charging current is constant until the charge mode switching, and then it is asymptotically decreasing toward zero, until the current threshold cutoff stops the charging.
For lifetime, more shallow charging cycles is better than fewer deep charging cycles. The former has bigger total provided energy value for the cell lifetime. E.g., just illustratively, if a cell lifetime is 300 0-100% cycles, it has lifetime e.g. 800 30-80% cycles, giving 4/3 times bigger total charge. And it goes further for even more shallow cycles. But, unfortunately, shallow charging strategy is seldom practical.
So, if you know you are going to use the battery shortly after charging, it is the best to fully charge it, or at least to 90%. If you do know it will not be used for some time, it is better to charge the about discharged battery just partially, e.g. just 30-40% of the typical full charging time, depending on the residual charge.
The charging time depends on the battery state. A new one may have bigger capacity and generally different remaining charge. Note that new batteries are shipped partially charged, never empty. So bigger free capacity = longer charging time. The same free capacity, but older battery = longer charging time, because the second stage kicks in sooner due higher internal battery resistance.
The discharging cut-off voltage is vendor and technology dependent. It is usually somewhere between 3.0-3.5 V. Note that voltage as a function of charge is strongly nonlinear. More discharged = steeper voltage fall. The same vendor and technology dependency is for the cutoff charging voltage. The percentage itself is hard to tell, but estimated from the voltage curve just few percents of the capacity, perhaps 2-3%.


What is the LUMO of a molecule with an alkene and a carbonyl? In a molecule that has both an alkene (C=C) and carbonyl (C=O) that are NOT conjugated, such as pent-4-en-2-one, how do you determine which π* anti-bonding orbital is the LUMO?
Essentially, I'm trying to reconcile two statements that I thought were true but apparently contradict here:

Increased electronegativity of an atom lowers the energy of its molecular orbitals. This lead me to think that π* C=O is the LUMO because the higher electronegativity of oxygen would lower the energy of the anti-bonding orbital (as well as the bonding orbital).
The lower the energy of the bonding MO, the higher the energy its anti-bonding MO is. This leads me to think that π* C=C would be the LUMO because π C=C is the highest energy bonding orbital, so π* C=C would be the lowest energy anti-bonding orbital.

Any help you could provide would be GREATLY appreciated!!

I think here you are missing the point on what exactly a HOMO and a LUMO are. They are descriptions of the energy splitting that occurs when quantum wave functions constructively, or destructively, interfere with eachother when in proximity. Because this system is not conjugated, the alkene and the carbonyl with both INDIVIDUALLY have their own HOMO and their own LUMO. In the case of both of these systems, the HOMO will be the pi bonding orbital and the LUMO will be the pi* anti-bonding orbital. Now, when it comes to which of these LUMOs will be lower in energy, it honestly will depend on the compound. Electronegative atoms, electron withdrawing groups, etc. narrow the energy gap between bonding and antibonding orbitals, making the LUMOs more reactive. So in this particular example, the carbonyl will probably have a larger pi* coefficient and thus be lower in energy. For more complex systems, reactivities need computer algorithms, clever thinking, or just experimentation to determine which energies are lower.


How can I clean stained garden rocks? I have some white garden pebbles which  are now  yellow with trace of rust. I have tried to soak them in bleach diluted with water without success.  Could I receive suggestions of an appropriate material to remove the rust?

The iron oxide from rust is not soluble in high pH (basic) solutions such as bleach. If you want to remove the the rust, you need an acid such as muriatic acid (wear PPE, follow label instructions). I would recommend an oxalic acid based product such as Bar Keepers Best Friend (I have no affiliation).  In my (pseudoscience) experience, oxalic acid its more effective at sequestering rust (when scrubbed) than anything else I've used including nitric, muriatic, and sulfuric acid. I would be cautious if the white rocks are marble as acid can disslove marble into oblivion.


Free radical substitution - bromination of ethane I am new to this service so please overlook any mistakes.

My question is with regards to free radical substitution. I understand the three step process of the bromination of ethane and what would occur if chlorine was in excess.

I do not however understand what would happen if ethane was in excess. What would occur and what would be the major product produced?

If there is an excess of ethane, you would be getting the monosubstituted bromoethane, $\ce{CH3CH2Br}$, as the main product.

When ethane, $\ce{CH3CH3}$, is in excess, for any bromine radical formed, most of the reacting particles around it would be an unreacted ethane molecule (high concentration of $\ce{CH3CH3}$), this means that there is a greater chance of a bromine radical colliding with and reacting with $\ce{CH3CH3}$ to form the monosubstituted bromoethane.

This is in contrast with when bromine is in excess, where the excess of highly reactive bromine radicals mean that most of them react fully with the available ethane molecules to form fully substituted $\ce{CBr3CBr3}$.


What is the difference between K and Kp in the equilibrium equation? In this equilibrium equation, $$K = K_p\cdot (RT)^{-\Delta n}$$ what does $K$ represent in comparison to $K_p$? It seems to me that they both are equilibrium constants, so how are they different?

The equilibrium constant is defined by the expression
$$K_x=\prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}}.\tag1$$

For $x$ you can substitute a number of quantities, most commonly when working with solutions is (amount) concentration $c$ and when working with gasses often the partial pressure $p$ is used. In the latter case the equilibrium constants can be related via the ideal gas law $$pV = n\mathcal{R}T.\tag2$$

Substituting this into $(1)$ gives you
$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag3$$

This can be transformed into
$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}} \prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{3a}$$

We know that the concentration is defined as
$$c_{\ce{B}}=\frac{n_{\ce{B}}}{V}\tag4$$
and from $(1)$ we can set up the concentration based equilibrium constant
$$K_c=\prod_{\ce{B}} c_{\ce{B}}^{\nu_{\ce{B}}}
     =\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}}.\tag5$$

Substituting $5$ in $3$ we find
$$K_p=K_c\prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{6}$$

From the associativity of multiplication it follows therefore
$$K_p=K_c\cdot\left(\mathcal{R}T\right)^{\sum_\ce{B}\nu_{\ce{B}}}.\tag{7}$$

This is often abbreviated with $K=K_c$ and $\sum_\ce{B}\nu_{\ce{B}}=\Delta n$ and therefore
$$K = K_p\cdot (\mathcal{R}T)^{-\Delta n}.\tag{7a}$$

Associative law of multiplications for exponents:
\begin{align}
  x^{n+m} &= x^n\cdot x^m\\
  x^{\sum d} &= \prod_d x^d
\end{align}


Why is this chair conformation the most stable and which leaving group is best? I'm confused about how this chair conformation was constructed and why that particular Cl leaves. Why is one of the Cl's drawn axial while the others are drawn equatorial (it seems like the others could have been drawn axial as well)? And, to reiterate, why is it that Cl that's favored to leave over the other two?

In cyclohexane, the equatorial position is energetically favored over the axial position. So the lowest energy conformer is the one where the most substituents are in equatorial position. (Or rather: Where you minimize the energy according to the A Value).

The axial Cl is favored as leaving group because of the elimination reaction mechanism. Normally, we see $\beta_{1,2}$ elimination in an anti-fashion. Which is what is proposed by your image.


Stability of resonance structure My book states- " the one that has ...less separation of opposite charges ,...., more dispersal of charge is more stable than others"

Aren't 'less separation ' and 'more dispersal' opposite to each other?
Please explain with an example.

Re-read the statement carefully. It says that the separation of opposite charges ("charge separation") destabilizes the resonating structures. The second phrase ("charge delocalisation") refers to the delocalisation of the same charge around a resonating structure.

For example, here's an example of how "charge delocalisation" increases stability:

(source)
Notice the same positive charge is delocalised throughout the conjugated system. It reduces - the unit positive charge on a single carbon atom - to a partial positive charge over several carbon atoms in the resonance hybrid, thus increasing the stability of the molecule.

Here's an example of "charge separation":

which destabilises the molecule. Remember that uncharged structures are more stable than charged structures.

I hope it helps!


Do all decomposition reactions involve heat? When water decomposes into Hydrogen and Oxygen as in 2H2O—-> 2H2+O2 does this involve heat? Is it always implied that heat is involved in a decomposition reaction or does this just spontaneously occur? Chemistry 1st semester so excuse the fundamental nature of the question. Thanks

Plainly put: no, decomposition reaction does not always have to entail providing heat. A decomposition reaction (in simple language) is when one compound breaks into multiple other compounds, like so:

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

As a general rule, if the product side is relatively more stable, it will decompose without you having the to provide extra heat. In the case of the reaction you gave, $\ce{H2O}$ is more stable than $\ce{H2}$ and $\ce{O2}$, and thus you will have to provide heat for the breaking of bonds.


Why can potassium hydroxide not be used to remove atmospheric carbon dioxide? As we know that $\ce{KOH}$ absorbs $\ce{CO2}$, then why can't it be used to remove the excessive $\ce{CO_2}$ present in the atmosphere?

$\ce{KOH}$ is produced by electrolysis of aqueous $\ce{KCl}$ solutions. Taking into account that the electrical energy still mostly comes from power plants that burn fossile fuels, it doesn't look like the most clever idea to release $\ce{CO2}$, just in order to capture it from the atmosphere again.

While it doesn't make sense to tackle the global $\ce{CO2}$ problem with $\ce{KOH}$ or other hydroxides, they do nevertheless bind $\ce{CO2}$ just as you said. Consequently, these hydroxides were and are still used for the individual protection against $\ce{CO2}$ (self)poisoning in emergency cases and/or in confined spaces. Rebreathers used by fire brigades, divers or miners are typical examples. Initially, $\ce{KOH}$ was indeed used, but nowadays, these $\ce{CO2}$ filters mostly contain mixtures of sodium hydroxide and calcium hydroxide.


Hybridization sp3 and sp2 When we calculate the electrons and orbital numbers in $\ce{CH4}$ $sp^3$ hybridization we have taken 3 electrons from $2p$ of $\ce{C}$ and 1 electron from $2s$ of $\ce{C}$ but in $\ce{C2H4}$ we have $2sp^2$ (2 electrons from $2p$ and 1 from $2s$).

Why do $\ce{C}$'s atomic orbitals make $2sp^2$ in $\ce{C2H4}$ and $sp^3$ in $\ce{CH4}$?
How we associate the numbers of electrons from $2p$ and $2s$?

This can be explained in terms of orbital overlap and therefore bond strength.

Consider a single carbon in an ethene molecule with two sp3 hybridised carbon atoms. The orbitals on the carbon repel each other and arrange themselves into a tetrahedron. Two of the four sp3 hybrids are used to form sigma bonds with hydrogen atoms. The carbon is then left to share its remaining two sp3 hybridized orbitals with its neighbouring carbon. Draw this system and you can see that two sigma bonds can be formed, though they are "bent" as the sp3 orbitals on the carbon are at a 109.5 degree angle to one another.

Now consider a single sp2 hybridised carbon in an ethene molecules with two sp2 hybridised carbon atoms. Again two orbitals are used to bond with hydrogen. This time we have an sp2 orbital pointing directly at the neighbouring carbon with which to form a strong sigma bond. The remaining p orbital which is orthogonal to the C-C sp bond can also forms a C-C pi bonding interaction.

Draw these orbital systems and compare.

The sp2 hybridised system constructs a strong linear sigma bond.
The sp2 hybridised system better shields the nuclei of the bonding atoms.
The sp2 hybridised system provides more orbital overlap, leading to stronger bonding and lower free energy.


Kinetic energy of electrons emitted in Photoelectron Spectroscopy In multiple sources (example), I found the information that the kinetic energy $E_k$ measured using (Ultraviolet / X-ray) Photoelectron Spectroscopy is given by:

$E_k = h\nu - B.E._F - \phi_{spec}$

where $h\nu$ is the energy carried by the incident photon, $B.E._F$ the energy difference between the energy level of the electron and the Fermi energy and $\phi_{spec}$ the work function of the spectrometer.

As far as I understood the measurement process, the actually measured quantity in Photoelectron Spectroscopy is the kinetic energy $E_k$ of the electron. This energy is measured by the radius of the electron in a homogeneous electric field. Thus, the kinetic energy of the electron after the emission would $E_k = h\nu - B.E._F - \phi_{sample}$, this independent of the work function of the spectrometer. So where does this difference come from?

From Wikipedia (emphasis mine):

  Because the energy of an X-ray with particular wavelength is known
  (for aluminum Kα X-rays, Ephoton = 1486.7 eV), and because the emitted
  electrons' kinetic energies are measured, the electron binding energy
  of each of the emitted electrons can be determined by using an
  equation that is based on the work of Ernest Rutherford (1914):

  \begin{equation}
     E_\text{binding} = E_\text{photon} - \left(E_\text{kinetic} + \phi\right)   \end{equation}

  where $E_\text{binding}$ is the binding energy (BE) of the electron, $E_\text{photon}$ is
  the energy of the X-ray photons being used, $E_\text{kinetic}$ is the kinetic
  energy of the electron as measured by the instrument and $\phi$ is the
  work function dependent on both the spectrometer & the material. This
  equation is essentially a conservation of energy equation. The work
  function term $\phi$ is an adjustable instrumental correction factor
  that accounts for the few eV of kinetic energy given up by the
  photoelectron as it becomes absorbed by the instrument's detector. It
  is a constant that rarely needs to be adjusted in practice.

So, it is the absorbtion of the photoelectron by the detector of the spectrometer that causes the work function term. The work function of the measured sample should be part of $E_\text{binding}$ as this is the ionisation energy of the electron. However, I think the equation you quote might have it slightly wrong because if $B.E._F$ is the energy difference between the energy level of the electron and the Fermi energy then there is indeed some portion of the work function of the sample missing since $B.E._F$ is not equal to the ionisation energy.


Cannizzaro or aldol condensation? The reaction between p-methylbenzaldehyde and $\ce{NaOH}$ is an example of:
  A) Cannizzaro reaction
  B) Aldol condensation
  C) Hydrolysis
  D) Haloform reaction

I selected B, aldol condensation, as the given compound is an aldehyde and it reacts with $\ce{NaOH}$. But the given answer is A. Why is it so? The $\ce{NaOH}$ is not mentioned to be concentrated.

The aldol condensation needs $\alpha$-hydrogens next to the carbonyl group in order to proceed. Check out this reaction equation from the Wikipedia article:

One of these hydrogen atoms is removed to form the enolate. See this mechanism (also from the Wikipedia article).

Now, look at the structure of p-methylbenzaldehyde. This molecule does not have any $\alpha$-enolizable hydrogen atoms. The $\alpha$-position is an $sp^2$-hybridized carbon atom in the aromatic ring. The aldol condensation is thus not possible.

We can also rule out the haloform reaction because there are no halogen compounds given. There are also no hydrolyzable functional groups (halides, ethers, esters, anhydrides, amides, nitriles, etc.).

The only remaining possibility is the Cannizzarro reaction, which requires a benzaldehyde derivative or other aldehyde lacking $\alpha$-enolizable hydrogen atoms.


Why is it that protons and electrons undergo the same amount of deflection in an electric field if they have the same energy? The question I have is with respect to this diagram. Which depicts a stream of protons and electrons entering a proton field with equal energy.

Why is it that in the case where the energy of protons and electrons are equal, their angle of deflection is exactly the same? However when speed is kept the same, the deflection of electrons is greater. Why is this so?

Now that's a mildly non-trivial observation. Why would they be equal, really?
Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the acceleration $a={qE\over m}$. By the time the particle reaches the right side of the picture, it gets deflected by $y={at^2\over2}={1\over2}{qE\over m}\left({L\over v}\right)^2={1\over2}{qEL^2\over mv^2}$.
That pretty much sums it up. $E$ and $L$ are the same for everyone, $q$ is the same in absolute value for electron and proton, so the only difference is $mv^2$. Now, considering your two situations:

Same speed: same $v$, but a proton is slightly (that is, a thousand-and-something times) heavier, hence lower deflection.
Same energy: same $mv^2$, same deflection (just to the opposite sides).

Things get hairy in relativity, but that's another story.
So it goes.


Can hydrogen and oxygen gasses be turned back into water? I electrolized water today, and got hydrogen. And I have a bottle of oxygen. Can I combine those to get myself water? I know $\ce{H2O}$ is super pure water, but I just want to try. Is this too idiotic to ask or can it be done?

When hydrogen burns, pure water is created; the reaction is exactly the reverse of the electrolysis and shown in the equation below.

$$\ce{2 H2 + O2 -> 2 H2O}$$

If you just have hydrogen, then it will burn with a flame as you might know from a methane gas flame. However, once you have a certain mixture of hydrogen and oxygen gas in your container, the reaction becomes explosive. If you have the exact stoichiometric mixture of $2:1$, the reaction is a very strong (and loud) explosion; a tiny amount created a louder bang in Munich’s chemistry lecture hall than all other explosions the experimentors showed on that day.

The fact that hydrogen burns to give water and only pure water (vapour) was one of the main points used to try and sell hydrogen-powered cars, since water vapour is environmentally friendly and does not cause global heating like $\ce{CO2}$ does.

The electrolysis/explosion difference shows how different conditions for forwards and backwards reaction can be.

It bears repeating: do not attempt the combustion at home unless you know exactly what you are doing!


How to remove wash anthracene oil/creostae oil from stainless steel tanker? I once loaded anthracene oil/creosate Oil in my stainless steel tanker and I had to transport it to some factory. The material was inside the tanker for throughout the journey that was of around 8 days.

The material got unloaded at the said factory. However I had another order to load Formaldehyde from the same city. I took my truck to the usual tanker service station to wash the tank with Detergents/water/diesel/scrubbing and everything which is quiet a normal procedure after and before any other liquid chemical.

However I have never had trouble with washing any other material as much as I am having with anthracene oil/creosote oil. It just doesn't seem to come off , the servicing person had severe respiratory problem when he went inside the tank.

Even after washing with several detergents, fast water streams and other things... the chemical is just not leaving the surface of my tank. One more thing, I think the service would have been much easier if the fumes/gas inside the tanker weren't so strong and scorching to the skin of the service person. He straight away told us to never come back with the same material.

Kindly let me know how to make sure to remove the essence of the gas/fumes left in the tank due to 8 days of material in the tank, how to remove the material from the surface of the tank and how to make this process well by using any other chemicals.

Anthracene oil (CAS registry number: [90640-80-5]): Both European Chemical agency (ECHA) and US Toxicology Date Network (TOXNET) classify anthracene oil as:

A complex combination of polycyclic aromatic hydrocarbons obtained from coal tar having an approximate distillation range of $\pu{300 ^{\circ}C}$ to $\pu{400 ^{\circ}C}$ ($\pu{572 ^{\circ}F}$ to $\pu{752 ^{\circ}F}$). Composed primarily of phenanthrene, anthracene, and carbazole.

Anthracene oil  is also classified as a carcinogen the risks to human health and to the environment as well (though the relevant symbol is not issued).

Creosote oil (CAS registry number: [8001-58-9]): Creosote oil is a synonym for coal tar creosotes. It is a distillation product of coal tar. Creosote oil is an oily liquid consistency and range of color from yellowish-dark green to brown.  It is a mixed compound, composed primarily of aromatic hydrocarbons, anthracene, naphthalene, and phenanthrene derivatives. At least 75% of the coal tar creosote mixture is polyaromatic hydrocarbons.
Creosote oil is heavier than water (specific gravity: 1.07-1.08) and has a continuous boiling range beginning at about $\pu{200 ^{\circ}C}$  (range: $\pu{194\!-\! 400 ^{\circ}C}$).  They are slightly soluble in water, but miscible with alcohol and ether (Ref.1).
Koppers' Safety Data Sheet has following warnings for creosote oil:

Hazard Statements:

Causes eye irritation.
May cause cancer.
May cause respiratory irritation (No wonder your servicing person had severe respiratory problem when he went inside the tank).
Very toxic to aquatic life.
Toxic to aquatic life with long lasting effects.

It also listed following precautionary statements for prevention:

Avoid breathing dust/fume/gas/mist/vapors/spray.
Wash thoroughly after handling.
Wear protective gloves/protective clothing/eye protection/face protection.
Obtain special instructions before use.
Do not handle until all safety precautions have been read and understood.
Use only outdoors or in a well-ventilated area.
Avoid release to the environment.

I put all this effort to show that how dangerous your situation is and how ignorant you have been. Your first mistake is not receiving all special instructions for handling these kind of chemicals from responsible people (see above bold-face precaution). In your case, I assumed the manufacturer of the chemicals should provide these details. What you should have done now is go back to the manufacturer, and request the instruction for clean-up procedure safely. And be aware that whoever the person following that instructions, should wear with all personal protective equipment (PPE) required, specially suitable breathing equipment.
Alternatively, since the both oils are viscous, you may slant your tanks backward to make them drain out to safety container (for long time) before starting to wash them (assuming drainage is in the back). Once drained, a person wearing PPE can go inside and wipe out inner walls with cloths soaked with alcohol several times, since both oils are miscible in alcohol. Also, kerosene or gasoline or even coconut oil may work as well (please dispose those cloths accordingly). Finally, you should use appropriate detergent (or soap) with water to remove trace of remaining hazard chemicals (if there any) and washing organic material used.
References:

B. F Hibbs, J. Georges (Agency for Toxic Substances & Disease Registry), Toxicological Profile for Creosote (updates):  Wood Creosote, Coal Tar Cresote, Coal Tar, Coal Tar Pitch, and Coal Tar Pitch Volatiles; Forwarded by Dr. D. Satcher; Prepared for US Department of Health & Human Services, 1996.


1H splitting pattern of benzyl CH2 protons During my PhD, I prepared the two compounds below (there was actually a slightly longer chain but it's not inherently relevant).

When looking at the 1H NMR spectra of these compounds, something slightly weird occurred.

In one of them, the CH2 group of the benzyl appeared as a singlet (s, 2H). In the other, the CH2 group of the benzyl was observed as a multiplet (an AB quartet like structure, to be specific).

Is there any rationale for why the two cases would be different, given the distal nature of the closest stereocentre (i.e. why are the two CH2 protons diastereotopic to begin with given how distal they are)?

This is a bit speculative. I'm still thinking about it, and maybe this shouldn't even been an answer. I drew out the left structure and the enantiomer of the right structure because I want the configuration at the carbon with the OMOM group to be the same.

I think you're calling these the 1,2-anti-1,4-anti and 1,2-syn-1,4-syn, respectively. The main focus here was on the conformation about the 1,2-carbon-carbon axis. I put the two hydrogens anti to each other. In the anti-anti case, the alkyl chain extends off away from the benzyl moiety while in the syn-syn case, the alkyl chain is pointed more towrads the benzyl moiety.

I'm not in love with this explanation. It's a bit speculative, but I used the tools I have and at least, it's consistent with your observation about splitting.


Prins reaction mechanism I am trying to figure out the product for the following reaction:

Its a Prins Reaction. My proposed mechanism for it is:

In my second step, that particular carbocation is formed since $\ce{Ph}$ is an electron-withdrawing group.

Mechanism looks neat enough, but the reaction given in Wiley's Solomon and Fryhle (Page 379) is:

I'd like to know why my mechanism is "wrong"? I think there's problem in the carbocation generation step but isn't the carbocation that I have formed more stable?

Your generation of electrophile is correct the subsequent primary carbocation formation by the attack of electrophile is incorrect.

  In my second step, that particular carbocation is formed since $\ce{Ph}$ is an electron-withdrawing group.

The benzene ring  stabilizes the carbocation through mesomeric effect.

The electrophile will attack at $\alpha$ position rather than $\beta$


Do non-ergodic chemical reactions exist? Do reactions whose final outcome is not even theoretically predictable by any means possible exist in Nature or does the outcome of all possible chemical reactions can (at least in theory) be predicted if we could get enough information about the conditions the reaction system is in?

My question isn't about a well-defined reaction system whose behavior is unknown to a certain degree but about the possibility in principle to discover a chemical reaction or a system of chemical reactions able to produce compounds which can't be predicted even in theory, even if we manage to acquire all the possible information about the system that can be acquired by all experimental means possible? What I want to ask is whether we can, at least in theory, know all possible products a reaction system can generate or do we "hit a wall" limiting our ability to predict certain chemical reactions over a certain limit?

I know very well the difference between a stochastic and a deterministic process, but what I'm asking isn't about stochasticity. It is about the possible predictability of all chemical reactions that could exist. As far as I know if we use the concept of stochasticity we may be able to derive possible, albeit only probable predictions about the state the system can be in. Therefore we can at least compute some probabilities different compounds can have to emerge under certain conditions. What I'm asking about is the validity of the concept of ergodicity, therefore I want to know is it possible to know the end state and all the reaction products of all possible chemical reactions, or there are some which could generate products that are impossible to predict in both theory and reality? Do reactions generating such unpredictable behavior exist in the real world or are they possible only in theory?

The time point of the radioactive decay of a single atom cannot be predicted, even given all experimentally available information, and is not even predictable theoretically by any possible means.

Just isolate a single atom and observe its radioactive decay, and you have the purely random system you strive for. Could also take more than one atom and observe fluctuations with Geiger counter.


Kinetic vs thermodynamic control of HBr to 1,3-Butadiene The author claims that the transition state leading to the formation of the 1,2 product from the allylic carbocation is lower in energy since the carbon (which is attached to Br in the 1,2 product) is more electrophilic.

This makes sense to me, however, doesn't the transition state represent the energies of both, the reactants, and the products? In which case the thermodynamically favored product (1,4 product) should also have a lower energy transition state.

A transition state does not ‘represent the energies of […] reactants and […] products’. A transition state is a certain arrangement of atoms somewhere between the most stable atom arrangement the reactants can have and the most stable atom arrangement the products can have. The key characteristic of a transition state is that it is directly en route from reactant to product and that it is the arrangement with the highest internal energy.

For any arrangement of atoms you can calculate a single point energy: reactants, products and the transition state. The energy difference between reactants and the transition state is colloquially known as activiation energy although that may not always be correct. It is, however, important to realise that each arrangement has its own energy that cannot a priori be deduced from any other arrangement. Thus, a transition state’s energy is not directly related to the energies of reactants or products (with the exception that by definition it must be higher in energy than both).

While it is a general rule that the transition state leading to the thermodynamically more stable products has a lower energy than any other transition state, this is not true for all cases. Indeed, for any reaction in which one can discuss kinetic (lower energy transition state) versus thermodynamic (lower energy products) control, the transition state energies must be ‘reversed’ with respect to the product energies.


Is there a way to faciliate remembering which are alpha/beta hemiacetals of sugars? I understood the concept of hemiacetals quite well, but have considerable trouble remembering which are called $\alpha$ or $\beta$ respectively.
Is there some concept/mnemonic/common reaction/something else to faciliate remembering this convention?

When you draw the Haworth projection, the alpha anomer is written with the -OH group pointing downwards (see the image), when you write the symbol alpha, you start by writing it downwards. For the beta anomer the -OH group is written upwards and when you write the symbol beta, you start writing it upwards. This mnemonic applies only to D-monosaccharides. For L-monosaccharides the opposite applies.

Actually, a better definition is that the alpha isomer has the -OH group pointing in the direction opposite to the -CH2OH group (on carbon 5), the opposite applies for the beta isomer, therefore for L-monosaccharides the -CH2OH group should point downwards and the alpha isomer will have the -OH group pointing upwards.


Increasing the length of a carbon chain in an alkane What would be the best reaction to increase the carbon chain length in an alkane? Preferably, it would be by one at a time; however, if that is not possible, just generally increasing the total length of the carbon chain (such as producing pentane from propane) would be fine.

Moreover, is there a way that I can add a methyl group to a compound in the form $\ce{R-CH3}$?

One method could be metathesis of alkanes. There is a nice paper about it published by J. M. Basset et al., SCIENCE 1997, 276, 99-102, "Metathesis of Alkanes Catalyzed by Silica-Supported Transition Metal Hydrides" and for sure more recent ones have been published since then.

  Abstract
  The silica-supported transition metal hydrides $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$ and $\ce{(≡Si-O-)xM-H (M, chromium or tungsten)}$ catalyze the metathesis reaction of linear or branched alkanes into the next higher and lower alkanes at moderate temperature ($\pu{25^\circ}$ to $\pu{200^\circ C}$). With $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$, ethane was transformed at room temperature into an equimolar mixture of propane and methane. Higher and lower homologs were obtained from propane, butane, and pentane as well as from branched alkanes such as isobutane and isopentane. The mechanism of the step leading to carbon-carbon bond cleavage and formation likely involves a four-centered transition state between a tantalum-alkyl intermediate and a carbon-carbon bond of a second molecule of alkane.


adiabatic expansion vs Joule-Thomson Effect Source: The1stLawofThermodynamicsLesson4

I am looking for difference between adiabatic expansion and Joule-Thomson effect. Looking at the setup demonstrating Joule-Thomson effect, I find no essential difference between the set-up for the two.

What is the feature of the setup for Joule-Thomson effect that makes the process isenthalpic, as oppose to adiabatic expansion? (I am not looking for proof initial H= final H)

What is the use of the throttle/ porous plug? If it is the throttle that makes the process isenthalpic, what is the physical reason behind it?

From another source,

  The gas is first compressed isothermally, then expanded isothermally
  (may be at different temperature as initial temperature)

I am not sure what it means. Does it mean that its temperature jumps abruptly from initial temperature to final temperature? It has only two temperature throughout the process? How is this possible? I expect the temperature to change continuously, because the volume of the gas changes throughout the process.

In the Joule-Thompson experiment, there is major viscous dissipation of mechanical energy (in this case pressure) to internal energy occurring within the pores of the plug.  This is analogous to Poiseuille flow of a viscous fluid in a tube.  If the tube is adiabatic, the fluid flowing through a tube experiences a temperature rise as a result of the viscous "heating."  The work to push the fluid through the plug is totally dissipated and converted to increased internal energy of the gas.  The reason that viscous effects are so important in flow through a plug is that the pores are very small.  So flowing the gas through the plug is analogous to flowing the gas through an array of very tiny tubes in parallel.

Ordinarily when a gas expands adiabatically, it cools.  But, for flow of an ideal gas through a porous plug, the adiabatic cooling is exactly offset by the viscous heating.  So there is no temperature change.  And, as a result, there is no enthalpy change.


onset of oxidation for cyclic voltammetry I have plotted a cyclic voltagram for my different polymer blends and I am trying to work out the onset of oxidation. picture below is my graph

from this website: https://www.ossila.com/pages/cyclic-voltammetry
There is a diagram which shows how the straight line should be placed so that it intersects the horizontal line, and thus the onset of oxidation can be found

However I am being told to draw my lines like this:

The procedure I am being told to do, doesnt make sense to me, I dont think the data from that would be useful. I would only get values for the Eonset that are very similar if I were to do it the way I'm being told. From my data, theres a clear difference in oxidation slope, so wouldn't it make sense to plot the diagonal line against that slope, rather than a small section near the beginning? From my 5 cyclic voltagrams, it would make much more sense if I were to do it as the website has told me to. There are more pictures on google which do it that way too

Please don't apply the reversible CV diagram on something which is clearly a non-classical cyclic voltammogram. When you experimentally don't get the shape as predicted by the theory, it means it that particular theoretical model does not apply in your case.
Look at three different types of voltammograms in this picture. Does your CV look like a reversible one? If not, then you were comparing an apple with an orange.

There is a very nice article which is relevant to you case:
Practical Aspects of Cyclic Voltammetry: How to Estimate Reduction Potentials When Irreversibility Prevails,  Journal of The Electrochemical Society, 2019,  166(5):H3175-H3187
This figure is relevant:

Examples of cyclic voltammograms showing chemically reversible and irreversible oxidation with designated potentials: E a = anodic potential, E c = cathodic potential, E (i) = inflection-point potential, E (p/2) = half-peak potential, E (e) = edge potential.


Cubane systematic name Why is cubane called pentacyclo[$4$.$2$.$0$.$0^{2,5}$.$0^{3,8}$.$0^{4,7}$]octane?

I am especially intrigued by the pentacyclo part.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), polycyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system as follows.

The main ring of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible (here: 8 atoms).

The main bridge is the bridge that includes as many of the atoms as possible that are not included in the main ring (here: 0 atoms).

Bridges other than the main bridge are called secondary bridges:

  P-23.2.6.1 Naming polycyclic alicyclic hydrocarbons

  Rings not designated by the bicyclic system described above (…) are defined by citing the number of atoms in each secondary bridge as an arabic number. The locants of the two attachment points of each secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower first) separated by a comma. The numbers indicating independent secondary bridges (bridges that connect atoms of the bicyclic system) are cited
  in decreasing order. (…)

  P-23.2.6.1.1 The prefixes ‘tricyclo’, ‘tetracyclo’, etc., in place of ‘bicyclo’, indicate the number of rings in the polyalicyclic system. The number of rings is equal to the number of bond cuts necessary to transform the polycyclic system into an acyclic skeleton, unbranched or branched.

Therefore, the correct systematic name is ‘pentacyclo[4.2.0.02,5.03,8.04,7]octane’. However, the retained name ‘cubane’ is used in general nomenclature and as preferred IUPAC name (PIN).


Aromaticity of different compounds I'm new here and I'm still learning, I understand that there four rules for aromaticity:

cyclic
conjugated double bonding (every atom in the ring should be sp2)
huckle number should be 2, 6, 10, 14, 18, ..18
planar ring

The 2nd question should be 4 electrons and is antiaromatic, but how about the first question?

My answer would be to refuse to answer this case.
The 4n+2 rule applies only if the electrons are conjugated in a cycle -- nothing more, nothing less.  If you have pendant pi bonds in the conjugated system, like the carbonyl group in cytosine and many similarly structured compounds, you don't have the right conditions to apply the 4n+2 rule at all.
Interior conjugated atoms, as in pyrene, also mean you don't have specifically conjugated cycle.


What is exactly ortho effect in Reimmer-Timmer reaction? In the Reimmer-Timmer reaction cho group is added to only at ortho position not at para position , I got the answer for this as ortho effect. So what is exactly ortho effect.

I think that the coordination step is responsible for the ortho selectivity. The phenoxide anion has a large negative charge on the oxygen atom, so the electron-poor carbene will tend to go there and the conjugated attack will result in the ortho product.


What is the major product obtained on acidification of substituted epoxide? The problem

Source : MS Couhan (problems in organic chemistry . Chapter: Alcohols , phenols and ethers).
My Thoughts

If I proceed along $\ce{path 1}$ , $\ce1a$ is formed . It is unstable due to inductive withdrawing effect of $\ce{C=O}$ neighboring to it. Eventually $\ce2$ is obtained.
However, along $\ce{path 2}$ relatively more stable $\ce{3a}$ (compared to $\ce{1a}$) is formed. $\ce3a$ on ring expansion gives $\ce{3b}$ , leading to $\ce{4}$  ($\ce{3b}$ is relatively unstable compared to $\ce{1b}$).
My question
Which of the two paths should I choose that would lead to a major product ?

The reaction of epoxyketone 1 affords diketone 2 under photochemical and thermal conditions ostensibly through the diradical.1 Boron trifluoride provides the diketone 3 through acyl bond migration and not via alkyl group migration as in your intermediate 3a. I have not located a proton-catalyzed reaction. Unfortunately, diketone 3 was not one of your choices.

1) John R. Williams, George M. Sarkisian, James Quigley, Aaron Hasiuk, and Ruth VanderVennen, J. Org. Chem., 1974, 39, 1028.


How to show/represent an ionic bond visually? How to represent/ show an ionic bond visually?
For example in NaCl

Some examples are as follows,
$$
\ce{[Na]+ [Cl]-} \, , \\
\ce{[Mg]^2+ 2[Cl]-} \, , \\
\ce{3[Mg]^2+ 2[N]^3-} \, .
$$
Hopefully you get the idea, so a general rule is:
$$
\text{#ions}[\text{element symbol}]^{\text{magnitude of cation charge}} \text{#ions}[\text{element symbol}]^{\text{magnitude of anion charge}} \, .
$$


Why does nucleophilic substitution of a haloalkane with ammonia produce ammonium, not a hydrogen halide? For example, $\ce{CH3CHBrCH3}$ reacts with $\ce{:NH_3}$ to make $\ce{CH3CH(NH3^+)CH3 +:Br-}$,$$\ce{CH3CHBrCH3 \rightarrow CH3CH(NH3^+)CH3 +:Br-}$$then reacting again with $\ce{:NH_3}$  to make $\ce{CH3CH(NH2)CH3 +:Br^- +NH4^+}$:

$$\ce{NH_3 + CH3CH(NH3^+)CH3 +:Br- \rightarrow CH3CH(NH2)CH3 +:Br^- +NH4^+} $$

Question: why does the last step not produce $\ce{CH3CH(NH2)CH3 +HBr}$? That is, why is Ammonia a better nucleophile than the halide ion, despite having a lower partial negative charge (the halide has a $-1$ charge)?

Edit: a suggestion made by a teacher is that my proposed step does in fact happen, it's just that it occurs as an intermediary to the last step.  Apparently, it's very favourable to have a Hydrogen Halide and Ammonia acid-base reaction. Is he correct, and if so, why is this?

The answer to this is quite similar to one I gave recently for another set of reactions with hydrogen halides as possible products.

Your proposed last step is:

$$\ce{CH3CH(NH3^+)CH3 +Br- \rightarrow CH3CH(NH2)CH3 +HBr}$$

Here, you have a weak acid $\ce{(CH3CH(NH3^+)CH3)}$ and a very weak base $\ce{(Br^{-})}$ reacting with each other to turn into a weak base $\ce{(CH3CH(NH2)CH3)}$ and a very strong acid $\ce{(HBr^)}$. This is highly unfavourable from a thermodynamic sense, because you're trying to inject a lot of chemical potential into the molecules at no cost. Though your teacher suggests it might happen, I wouldn't be very comfortable saying that, if only for the fact that there's another, easy option.

It is possible that the previous argument is slightly circular, so it's best to explain why $\ce{HBr}$ is such a strong acid / $\ce{Br^-}$ is such a weak base. As you mention, $\ce{Br^-}$ has a negative charge, and one might expect that to mean it is a better base. However, that affirmation is only useful as a rule of thumb when comparing substances of different structure, being strictly valid solely when looking at a single sequences of related structures (such as $\ce{PO_4^{3-}}$/$\ce{HPO_4^{2-}}$/$\ce{H_2PO_4^{-}}$). The strength of a base is determined by several factors, such as charge localization, bond strength between base and a proton, steric hindrance, and strength of interaction of the protonated/deprotonated species with the solvent (if it exists). There are very strong neutral bases and very weak negatively charged bases ($\ce{SbF_6^-}$) known.

$\ce{Br^-}$ is a poor base because, while it has a net negative charge, that extra electron is bound to a somewhat electronegative atom, thereby stabilizing the charge. Not only that, but the bromide anion is quite large, so the charge is "spread out" over a significant volume and would attract a proton less, in some sense. Meanwhile, though $\ce{NH_3}$ has no extra charge, and even though the central nitrogen atom is more electronegative than bromine, the non-bonding pair of valence electrons in $\ce{NH_3}$ is highly localized, especially due to the fact that second period p-block elements are anomalously small. This makes $\ce{NH_3}$ an attractive target for a proton.


Meaning of Flammability Limits and reaction in Air of Ammonia I have read on Wikipedia that the flammable limits of gaseous ammonia are between 15–25% and I have also read that the autoignition temperature is $651\rm~^\circ C$.  And that the combustion of ammonia in air is $$\ce{4NH3 + 3O2 -> 2N2 + 6H2O}$$  My understanding of this is that with a 20% air to $\ce{NH3}$ that I could hold a flame.  Outside of that range, I cannot hold a flame.

My question is if I am at 30% air and above $651\rm~^\circ C$ will Ammonia still combust/react with Air or will it stay $\ce{NH3}$?

My follow up question would be that when if I sprayed ammonia into a heated chamber (above $650\rm~^\circ C$) would flames/the exothermic reaction only happen when the ammonia vapor and the air reach that 15–25% range as the vapor propagates into the air and not react outside that range?

Chemical reaction would proceed in an air/ammonia mix, even outside the concentration range at which the reaction is self-sustaining, particularly if there were a catalyst to help. It is just that at the specified concentration range, when it reaches $651^{\circ}\mathrm{C}$ at one point in the mixture, the reaction proceeds fast enough to heat the rest of the mixture to that ignition temperature.

This is similar to an Otto cycle spark-ignition engine: with the correct mixture, the flame front propagates from the spark plug throughout the cylinder.

In the Ostwald process, a $1:10$ ration of $\ce{NH3}$ and air can sustain the reaction at the surface of a catalyst.

In a similar fashion, hot copper can catalyze the oxidation of acetone, methanol or ethanol, providing a surface where the reactants can combine fast enough to keep the surface hot.


How does the partial pressure of gas C change with the change in volume of the container? Suppose I have the equilibrium in a closed container:
$$\ce{3A(s) + 4B(g) -> 4C(g) + D(s)}$$

What happens to the partial pressure of C if the volume of the container is halved?

I was pondering about this question, because by the gas equilibrium constant
$$K_\mathrm{p} = \frac{[C]^4}{[B]^4}$$
 the partial pressure of each should remain equal to each other. But using an ice table, $K_\mathrm{p} = \frac{[C]+x}{[B]-x}$, or you can flip it around. Either way, $x$ should equal 0 by this equation for them to remain the same, which means that the partial pressures don't change, which then means that the gases are turning into the solids. But by the chemical equation 1 mole of one gas will produce 1 mole of the other gas, so how does the gas even turn into the solid? So wouldn't the partial pressures of both increase?

Another way to think about it is to imagine a container with only B and C. That means a reaction cannot even occur, so when volume decreases, the partial pressures increase.

This is my thought on the problem. Can someone please explain to me what really happens?

The equilibrium constant $K_P$ is independent of pressure but does depend on temperature which we assume to be constant. The equilibrium constant $K_P$ is the ratio of partial pressures $p_i$ of the gaseous substances raised to the stoichiometric powers, $K_P= p_C^4/p_B^4$ and is defined at a standard state of $1$ atm. pressure.

Writing the equilibrium constant for the reaction $\ce{mB <=>lC}$ in terms of the mole fraction $x_i=n_i/n$ for n total moles gives $K_x= x_C^l/x_B^m$ where l and m are constants, both four in your question. The mole fraction i of a species behaving as a perfect gas can also be written  $x_i=p_i/P$ where $p_i$ is the partial pressure and P the total pressure.

Rewriting the equilibrium constant in terms of partial pressures of B and C gives
 $$K_x=\frac{p_C^l}{p_B^m}P^{m-l}= K_PP^{\Delta n}$$
where $\Delta n$ is the change in the number of moles of the gases. Thus unless the change in the number of moles is zero, which it is in your example, the mole fractions of the components  will change with change in total pressure even though $K_P$ is constant. If $\Delta n > 0$ the pressure will reduce the mole fraction of products.


What happens to the solute when a solution freezes? I came across a situation in which my chemistry textbook gives phase diagram of a pure solvent and a solution with the same solvent.I am considering 2 different , say beakers, one containing some amount of pure solvent , and the other contains a solution formed by a non volatile solute and the same amount of solvent. In this, the author has shown the part of the graph corresponding to solid solvent , and solid form of solution to be identical. It's natural for me to think that a solid obtained by freezing a solution would be different from that obtained by freezing a pure solvent. So they must not have identical graphs. Is my argument correct ? Or is it like this- when the solution is frozen , since the solvent molecules are brought together for making a solid, we simply don't have solvent molecules to solvate  our solute, and the solid is more or like a frozen pure* solvent with solute particles trapped in it?

If you were to freeze the solution in a normal manner, using something similar to a home freezer, a more and more concentrated solution will form as the majority of the solvent freezes. The solute will be found in a small amount of solution that did not freeze, since the solute allows for freezing point depression. So, if you put both of your beakers into the freezer, one beaker will have a completely frozen block of solvent and the other will have a frozen block of solvent with some volume of highly concentrated solution.


Difference between solubility and dissolution? How is solubility different from dissolution? And how are they both different from solvation?

Solvation [...] is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. As ions dissolve in a solvent they spread out and become surrounded by solvent molecules.
  [...]

  Dissolution is a kinetic process, and is quantified by its rate. Solubility quantifies the dynamic equilibrium state achieved when the rate of dissolution equals the rate of precipitation.

Quote from Wikipedia.


Are phospholipids capable of forming automatically circles or vesicles? Sometimes scientist think that life began with membrames made of a kind of phospholipids. There wasn't any DNA needed for because those molecules would automatically form vesicles like cells. But is this really possible and how do those molecules form circles and vesicles?

Phospholipids structure can be described as a charged phosphate head group, with long hydrophobic chains coming off it. When you put these molecules in water, like oil drops, the oils spontaneously conglomerate into sheets on the surface of the water.

How, depending on concentration and size of the phospholipids you can get from sheet like membranes to spherical droplets. In either case the spontaneous formation of sheets or micelles (spheres) results from the entropically driven hydrophobic effect.

The simple way to understand this hydrophobic effect and entropy, is that when two hydrophobic tails between two phospholipids are in close proximity, the hydrophobic tails are able to interact via van der Waals forces. But even more important is that water molecules that surround each hydrophobic region of the molecule, are excluded between them. This means that the entropy of water increases a lot more compared to the loss of entropy from having the phospholipids come together and stay together.

This is a pretty fundamental concept to how proteins fold. Look up hydrophobic effect on google for nice, diagramatic explanation.

Or check intro biochemistry textbook like Nelson and Cox (2012) principles of biochemistry textbook.


Why doesn't this reaction happen? In the synthesis of DMAD, we eliminate two bromides from dibromosuccinic acid in basic environment to get acetylenedicarboxlyic acid (Step 2).

However, this reaction could also happen:

The reaction in Organic Syntheses (http://www.orgsyn.org/demo.aspx?prep=cv2p0010) went with a yield of 72-88 %, which means that these pathways probably do happen, but only in a minority of cases.

Why is that so, given that the elimination of CO2 is entropically favoured?

This elimination is unfavorable because C--C bonds are generally quite stable. Decarboxylation would involve the formation of an unstable carbanion before proceeding to the elimination as you have outlined. (Note, this is more favorable under different functional group conditions, namely when a neighboring carbonyl can stabilize the carbanion through the formation of an enol  https://www.masterorganicchemistry.com/tips/decarboxylation/ ) So, the formation of the alkyne is favored here over the elimination of CO2.


How to selectively convert benzene to p-nitrotoluene? I converted benzene into toluene by Friedel-Crafts alkylation reaction.
Then by nitration I got two compounds o-nitrotoluene and p-nitrotoluene.

Is it possible to get only p-nitrotoluene?

I'd doubt that exclusive nitration of toluene in para position is possible in homogenous solution.

But imagine to stuff a toluene molecule into a very narrow "tube". This tube will block any attack at C-2 (and C-3), while a nitrating agent may approach C-4.

Exactly this has been done in the vapour phase nitration of toluene with nitric acid over a zeolite (ZSM-5). See: Kalbasi, R.; Ghiaci, M.; Massah, A. Highly selective vapor phase nitration of toluene to 4-nitro toluene using modified and unmodified H3PO4/ZSM-5. Applied Catalysis A: General 2009, 353 (1), 1–8. DOI: 10.1016/j.apcata.2008.10.013.


Are these chemicals suitable for disinfecting my water tank? This is the same question that I asked before, only for another product this time. I have more detailed information about this product, including a data sheet.

There is an 8000 liter tank containing water for household use like showers and washing dishes and clothes (but not drinking or cooking). For more details, read here or here.

I need to combat harmful organisms and microorganisms in the tank, but I live in a remote rural area and can't just buy any chemicals I want. Thus, I am asking whether this product will work.

Here are the ingredients of the product:

  Water
  Sodium Hypochlorite
  Sodium Chloride
  Sodium Carbonate
  Sodium Chlorate
  Sodium Hydroxide
  Sodium Polyacrylate

Here is a full data sheet.

If this product is suitable, what is a good amount and frequency?

If this product is not suitable, which ingredient is the culprit?

When you start asking questions about chemistry you are only left with more questions... in general terms of safety, using household bleach to create a concentration of 1 - 3 ppm sodium hypochlorite is relatively safe (especially compared with sewage). In specific terms, the answer is two-fold; you have to consider the amount of exposure and the component's toxicity (the difference between medicine and poison is the dose).

It is important to note that the SDS for household bleach only lists sodium hypochlorite in the hazardous ingredients section. This directly implies that the only chemical of significant hazard in the bottle is sodium hypochlorite, and the rest of the ingredients are not present in sufficient amounts to cause issues.

First let's look at what thise things are:

Water : you know
Sodium Hypochlorite : bleach- household bleach is made from chlorine, caustic soda, and water and DC current (electrolysis).
Sodium Chloride : table salt, leftover from production or evaporation of water and chlorine.
Sodium Carbonate : washing soda- kind of like baking soda but about 40% stronger. This may be considered a buffer or stabilizer or byproduct.
Sodium Chlorate : when bleach is cooked/boiled (and dried), the leftover crusty stuff is salt and sodium chlorate... so this would be a byproduct, probably only in a very tiny amount.
Sodium Hydroxide : lye (caustic soda, NaOH) - household bleach solutions are typically stabilized by lye as part of the manufacturing process. Lye is used for raising the pH of drinking water and for making many things including soap.
Sodium Polyacrylate : thickening agent - It may be used in food, toothpaste, and shampoo.

APPROXIMATE Household Bleach Amounts:

Water : 93%
Sodium Hypochlorite : 5.5%
Sodium Chloride : 0 - 2%
Sodium Carbonate : guestimating 0 - 0.2%
Sodium Chlorate : guestimating ~ 0 - 0.2%
Sodium Hydroxide : since the pH of bleach is about 11, the amount of sodium hydroxide must be less than 0.001 molarity, so less than 0.004%
Sodium Polyacrylate :  based on the viscosity of bleach and my personal experience with this thickening agent, about 0.5%.

Now lets look at the LD50 (a general measure of toxicity). The LD50 is the amount of material necessary to kill 50% of the test subject (usually rats). For these chemicals I listed acute (hard and heavy) oral dosage. Chronic exposure may be more relevant because you will not be drinking this, but chronic exposure data is also far more difficult to measure accurately, and with respect to all of these chemicals, chronic effects are generally nil, so I wouldn't have anything to say.

Water : greater than 90,000mg/kg, rat (yet drowning causes more deaths than all of the following chemicals combined)
Sodium Hypochlorite : 8910 mg/kg, rat
Sodium Chloride : 12357 mg/kg, human
Sodium Carbonate : 4090 mg/kg, rat
Sodium Chlorate : 1,200- 7,000 mg/kg, rat
Sodium Hydroxide : 100mg/kg, rat
Sodium Polyacrylate :  greater than 2000 mg/kg, rat. When something is above 2000 mg/kg the exact lethal dose is often not pursued.


Electron Affinity trends Imagine an arrangement of atomic orbitals in an alternate universe, in which the s subshell contains 2 orbitals instead of 1, and the p subshell contains 4 orbitals rather than 3. In which groups would you expect the anomalously low electron affinities to occur?

My thoughts are groups 3 and 7, due to the fact that they are "one away" from filling up their subshell. The groups would look like:

s: _ _ _ _  ... p: _ _ _ _ _ _ _ _

However, the answer says 4 and 8. Is there something I'm missing?

Edit: in terms of electron configuration blanks, it would be
s: _ _ ... p: _ _ _ _

After rediscovering my own question, I'm posting it as an answer for anyone else to reference or to correct.

Electron affinity (EA) refers to "the energy released when an atom in the gas phase accepts an electron." (Burdge, Atoms First, p.118)

Lower EA means that the process of accepting an electron is not energetically favorable. In other words, lower EA = less likely to accept an electron.

Groups 4 and 8, in this case, already have filled subshells. Adding an electron would place it in a new subshell that is higher in energy. The EA would be lower for these groups because of the "reluctance" of the atom to accept an additional electron.

Groups 3 and 7 have the anomalously high EAs, which might have been where the confusion was.


How does the surface tension of water decrease with an increase in the concentration of NaCl? Can anyone help me with the theory of surface tension? My chemistry experiment is related to the surface tension and concentration of NaCl. I am confused about my hypothesis. On the internet, I found that increasing salt concentration also increases the surface tension of the water. But in my experiment, surface tension is decreasing with the increase of NaCl. So can anyone help me with the theory behind this?
With the help of needle and water drops, I was able to calculate the value of surface tension of water for each concentration of NaCl.

The surface excess of a salt (NaCl) is negative because salts prefer to be hydrated.  So there is less salt at the inferface than in the bulk phase.  The Gibbs adsorption equation says the change is surface tension is the opposite of the surface excess - so for a salt solution, the change is positive and the surface tension increases.


Why is the pKa (COOH) of alanine and glycine similar but the pKa (NH3) different? Alanine is heavier with a ($\ce{-CH3}$) than glycine which has an ($\ce{-H}$). What makes them have different $\mathrm{p}K_\text{a}(\ce{-NH3+})$ $(\mathrm{p}K_\text{a}: \text{Ala}=9.69, \text{Gly}=9.60)$ but similar $\mathrm{p}K_\text{a}(\ce{-COOH})$ $(\mathrm{p}K_\text{a}=2.34)$ when the rest of the amino acid is similar? Why does changing a hydrogen for a methyl group affect the $\ce{-NH3+}$ but not the $\ce{-COOH}$?

Also there are three amino acids (glycine, leucine, aspartine) which have same $\mathrm{p}K_\text{a}(\ce{-NH3+})$ but different $\mathrm{p}K_\text{a}(\ce{-COOH})$.

I found the following pKa values for glycine and alanine from the CRC Handbook of Chemistry and Physics via Sigma-Aldrich:

Glycine $\ce{COOH}$: 2.34
Alanine $\ce{COOH}$: 2.34

Glycine $\ce{NH3+}$: 9.60
Alanine $\ce{NH3+}$: 9.69

First, although the values for the carboxylic acid are identical as reported, you would see a difference if you found a value reported with enough precision. The carboxylic acid is not immune to changes in the structure of the rest of the molecule.

Second, the difference between 9.60 and 9.69 is pretty small. In most practical situations, the two amino acids would behave identically with regard to the $\ce{NH3+}$ pKa.

The fact that the magnitude of the difference is larger for the $\ce{NH3+}$ pKa values can likely be attributed to proximity. In this case, the relevant proton is 2 bonds away from the $\alpha$ carbon, whereas the proton in the $\ce{COOH}$ group is 3 bonds away.