Balancing redox equation So I had these two reactions that confused me .. I tried the standard way of balancing it (Putting oxygen number everywhere, +1,+2,-2 etc.) and then go on with it but it didn't work

So anyways here are the 2 reactions:
$$\ce{3 Cu + 8HNO3-> 3Cu(NO3)2 + 2NO + 4H2O}$$
$$\ce{3Ag + 4HNO3 -> 3AgNO3 + NO + 2H2O}$$

I've no idea how do you get those coefficients (numbers).. Please, is there an easy way to do it (perhaps it has to do with electrons added/given? ) and if so - where do I start?

There are a few basic principles that are useful to keep in mind when you want to balance a redox equation.

1) OILRIG - Oxidation Is Loss, Reduction Is Gain (of electrons)

Let's look at the basics of the first equation:  $$\ce{Cu + HNO3-> Cu(NO3)2} + ?$$

On the left hand side we have Cu(0) metal and N(V), on the right hand side we have Cu(II) ion and N(V), so from OILRIG we know that the copper is being oxidised... but we know from OILRIG we have to put the electrons from the Cu(0) somewhere...

2) Hydrogen ions are always +1, and oxygen ions are always (there are a few exceptions but this is a good general rule) -2

... So the only thing that can change is the nitrogen. The nitrogen must be being reduced.

Nitrogen actually has a large number of possible oxidation states, taking any value between -3 and +5, so how do we decide which one it goes to in this case?  This depends on the temperature and the concentration of the acid. We can probably assume room temperature and bench concentration of acid (otherwise conditions need to be specified), and that the nitrogen isn't going to gain 5 electrons and become nitrogen gas, which means that we are going to form an oxide. Of the oxides, the most thermally stable is NO (nitrogen is in oxidation state +2 here).

What does that look like now?

$$\ce{Cu + HNO3-> Cu(NO3)2 + NO}$$

To balance the electron transfers, we need 3 Cu(O) going to Cu(II) to donate 6 electrons for 2 N(V) to go to N(II). Balancing for electrons and nitrogens, we now have:

$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO}$$

Ok, now we need to look at balancing oxygens and hydrogens.

3) Use water if you can to balance oxygen and hydrogen. If not, start looking at hydroxide or hydrogen ions.

Here we need to somehow account for 8 hydrogens and 8x3 - 3x3x2 + 2 = 4 oxygens... That's 4 water molecules, so we finally have the answer you provided at the start!

$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO + 4H2O}$$


What's the order of oxidising strength of the oxyacids of Chlorine? What I have been told is that the acidity of the four oxyacids of Chlorine increase in the order $\ce{HOCl} < \ce{HClO2} < \ce{HClO3} < \ce{HClO4}$.

I have also been told that the oxidising strength goes in the reverse order. In my knowledge acidic strength should increase with oxidising strength. Where am I going wrong?

For your reference, the Ka values of the acids are as follows:

$\ce{HOCl} = 2.9 \times 10^{-8}$
$\ce{HClO2} = 1.1 \times 10^{-2}$
$\ce{HClO3} = 5.0 \times 10^2$
$\ce{HClO4} = 1 \times 10^3$

For many acids, the proton $\ce{H+}$ is the oxidizing agent. For example, in the oxidation of zinc with $\ce{HCl}$, the oxidizing agent is $\ce{H+}$.

$$\ce{Zn + 2H+ -> Zn^{2+} + H2}$$

However, for some acids, the conjugate base (anion) is also an oxidant (and usually a more powerful oxidant than $\ce{H+}$. The oxyacids of chlorine are all oxidizing agents through their anions, as is nitric acid and some others:

$$\ce{Zn + 2H+ + 2OCl- -> Zn^{2+} + H2O + 2Cl-}$$

In the above example, hypochlorite is the oxidizing agent. For the oxyacids of chlorine and others, the strength of the oxidizing agent is independent of the strength of the acid.


Can I compare DFT calculations with different grids? When doing DFT calculations, some integrations are commonly done numerically on grids. [In fact, more than a single grid may be used at the same time for different integrals, e.g. approximations such as the resolution of identity (RIJCOSX, RIJK, etc., see e.g.  J. Chem. Phys. 118, 9136 (2003)) use grid schemes too (I believe some programs call this approximation density fitting).]

There's the downside that those grid schemes introduce a source of error. In fact, I find that only by increasing the quality of the grid I can remove some imaginary frequencies. Can I rigorously compare energy values among calculations that used slightly different grid schemes?

I think yes and I reasoned as follows. Since I'm interested in energy differences, let's imagine two structures with true energies $E^*_1$ and $E^*_2$. Both energies are approximated by the calculated energies $E_i = E^*_i + \epsilon_i$, where $\epsilon_i$ is the error introduced by calculations (including grids, etc.). Now the energy difference $\Delta E^* = E^*_2 - E^*_1$ is approximated by $\Delta E = E_2 - E_1 = \Delta E^* + \Delta \epsilon$, where $\Delta \epsilon = \epsilon_2 - \epsilon_1$. Since grid schemes introduce errors that are smaller than other sources of error (e.g. implicit solvation), everything should be fine as long as $\Delta \epsilon$ is acceptably small for the particular application at hand.

Does this seem reasonable? If not, what's wrong with the above?

Yes, your error analysis is valid for energy differences. However, I believe it is also valid for the absolute error $\epsilon_i$ of any calculated quantity, not just the error $\Delta\epsilon$ of an energy difference.

In any numerical computation, the key thing for situations like this is to ensure that these "structural" sources of error have been reduced to a magnitude that don't affect your results. In this case, this is achieved when you reach a grid quality $Q$ where the numerical results no longer change when you further increase to $Q+\delta Q$.

To be clear: in this case elimination of the imaginary frequencies is not a good criterion for a sufficiently large $Q$.  The proper stopping point for refinement of $Q$ is when the results no longer change appreciably.

Note that, as you allude to implicitly in the question, this situation is different than the one of trying to compare, e.g., absolute energies from computations run with different methods or at different levels of theory.  In this case, there are fundamental/theoretical reasons why those comparisons cannot be meaningfully made.


If PQ, T2U2, RS2 and VW4 have the same Ksp, which has the lowest solubility in moles of solute per litre? If the solubility products of the following ionic compounds all have the same numerical value, and all solubilities have values between zero and one, which one would have the lowest solubility in moles of solute per litre?

$\ce{PQ}$

$\ce{RS2}$

$\ce{T2U2}$

$\ce{VW_4}$

Their chemical equations and $\ce{Ksp}$ values would be:

$\ce{PQ <-> P +Q}$  and $\ce{Ksp=[P][Q]=[P][P]=[P]^2}$
$\ce{RS2  <-> R +2S}$ and  $\ce{Ksp=[R][2S]^2=[R][2R]^2=4[R]^3}$
$\ce{T2U2  <-> 2T +2U}$ and $\ce{Ksp=[2T]^2[2U]^2=[2T]^2[2T]^2=16[T]^4}$ (although the textbook says $\ce{[T]^4}$ which I don't understand)
$\ce{VW4  <-> V + 4W}$ and $\ce{Ksp=[V]^2[4W]^4=[W]^2[4W]^4=256[W]^5}$ (although the textbook says $\ce{256[V]^5}$ but I am pretty sure its equivalent whether it is written in terms of V or W)

FYI my thought process here is that:

Based on the chemical equation RS2⟷R+2S
The concentration of one product (S) is twice that of (R).
So if there is X moles of RS2, and there is X moles of R and 2X moles of S
Hence in the Ksp=[A]^n⋅[B]^m , A=X and B=2X

Now the textbook says

  As $\ce{[P][P]=4[R]^3=[T]^4=256[V]^5}$
  [P] must be the lowest because $\ce{[P]^2}$  has the same value as  $\ce{4[R]^3 and [T]^4 and 256[V]^5}$

I can't get my head around this. I would have thought this proves
$\ce{[P]}$ has the highest value not the lowest. Why is [P] the LOWEST?

E.g. noting that all values are between $0$ and $1$ (as defined in the question)

$\ce{[P]^2=256[V]^5}$  (the other two options are in the middle so don't matter)

If both sides equaled e.g. $0.25$ (which is between $0$ and $1$)

$\ce{[P]^2=0.25 and 256[V]^5=0.25}$

${[P]=0.5}$ and ${[V]=(0.25/256)}^{1/5} = 0.25$

Showing [P]>[V], so [P] (moles/L) is the greatest solubility, not the lowest.

Is anyone able to spot my error?

Question and answer as shown in book below (I have typed up the relevant parts above but sometimes I interpret parts of the textbook question incorrectly (making my stack exchange question invalid or misleading):

Easier just to migrate my comment into an answer. Your assumption
$$\ce{[PQ]} > \ce{[VW4]}$$ is not valid based on
$$\ce{[PQ]^{2}} = 256\ce{[VW4]}^{5}$$

For $y = a^{x}$, if $a>1$, larger $x$ leads to larger $y$. But for $a<1$, larger $x$ leads to smaller $y$, and the question indicates that solubilities are between zero and one, i.e., small.

If you plug in $K_{\mathrm{sp}}= 10^{-10}$, you get $\ce{[PQ]}=10^{-5}$ and $\ce{[VW4]} = 3.3\cdot 10^{-3}$.

As I said in my comment, I believe that this is a very poorly designed question: the constraints are too broad and there is a typo in the key. Someone was very sloppy...


Concept of "non-superimposable mirror image" in chirality I didn't understand the fact that chiral objects are non-superimposable on their mirror images. I mean, if you put your right hand (apparently an achiral object) in front of a mirror, the mirror image will be like your left hand, but, when you make your hand touch the mirror, isn't it superimposable on its mirror image (and hence achiral)? Doesn't that mean that everything is achiral?

Your hands are chiral, that is why you need two different leather gloves, one that only fits your right hand, and one that only fits your left hand.  If your hands were superimposable, then you would only need one kind of glove and it would fit both hands.


Correct way of drawing D-configuration with this stereochemistry? Hello! I have completely memorized how to draw all 20 amino acids in the way seen in the picture (middle, between the "ball-and-stick" and fisher projection)

By my understanding this is in the L configuration, what is the correct way of drawing them in D- configuration? Is there a simple trick? Does anyone have an image of them all drawn in D-config in the same way?

Is it basically having the COO- and NH3+ switch places. Like here:

Thanks!

You don't have to memorize the whole structures. Easier is to memorize the general structure including the stereochemistry, and the residues, R (except from proline). The Fisher projection is useful, but on your depiction, it's not in the usual orientation. The carbon chain should be vertical, carboxy group at the upper side, residue at the bottom. Hydrogen and amino group (pointing towards observer) horizontal.

(non-ionized forms)

One can easily memorize that for L- amino acids, amino group is on the left, for D- amino acids the opposite, which is the case for your orientation as well (and don't forget to invert chiral residues for D- AA's, as Zhe has already noted in the discussion; it is the case for isoleucine and threonine).


A compound that only absorbs UV-A radiation As a similar idea to the question A compound that absorbs all visible light, I am curious if there is a compound (or compounds) that absorb UV-A radiation (shown below)?

However, preferable the compound would have relatively good transmission in the UV-B and Visible wavelengths.

Image source

Yes, there are many such compounds.  "The evolution of UVA protection" is a nice article detailing the progress on UVA blockers.  I'm sure you're aware that  sun care is a big business ($8.7 billion USD worldwide in 2013), so there is no shortage of chemists working in the area. In fact, there is enough (medical and clinical) focus that UVA (315-400 nm) has been subdivided into UVA1 (340-400 nm) and UVA2 (315-340 nm)

Here is a figure from the above link showing the UV spectra of some compounds.  The top part shows UVA absorbers and the bottom half shows broad-spectrum UV absorbers.

and a table with the names of the compounds

BMBM was an early UVA absorber (solid blue line), but was photochemically unstable (dashed blue line), in other words, its blocking power deteriorated with UV exposure time.  More recent UVA absorbers, like DHHB are photostable.

UVA absorbers belong to many different structural classes and contain many different types of functional groups.  Here are few examples.

BMBM

DHHB

TDSA


Lab goggle recommendations? I'm taking my first chemistry class with a laboratory component this quarter. My first laboratory session was earlier this evening (observing simple metathesis reactions). While I found the lab work to be fairly straightforward, the fact that the interior surface of my safety goggles accumulated a considerable amount of condensation proved to be a moderate hinderance.

I'm wondering if anybody has recommendations for lab goggles? Particularly ones which may avoid this problem?

Note: it is school policy that we wear lab goggles with side shields, not lab glasses.

Any insights here would be appreciated!

I don't have specific product recommendations (and I think they're discouraged in general on SE), but these are the general principles:

You'll only get condensation on goggles (or anything else) if they're cooler than the dewpoint of the air contacting them. If your goggles are cold, you'll get condensation.

If you're sweaty, it'll be worse.

If the goggles are glass or another high-thermal-mass material, they'll stay cold longer, making the problem worse.

If the goggles are ventilated, humid air can escape from the space between your goggles and your face, making the problem better.

Does the problem only arise after you've been working in the lab for a while, or does it happen as soon as you put the goggles on? If the latter, you might be able to fix it by just putting the goggles in a warm interior pocket for a few minutes before you start work.

There are also spray-on anti-fog products for goggles. If you can't find one of those, try rinsing the goggle interior surfaces with soapy water, then not rinsing the soap off. The soapy film left on the goggles won't prevent condensation, but it will slow the formation of droplets, which will slow the fogging effect.


Does supersaturated caffeine polymerize? I tried preparing a supersaturated solution of caffeine by mixing anhydrous caffeine powder with water in a thin Turkish tea glass and heating it up over a candle (desperate times...).

This worked as expected and a lot more caffeine dissolved into the clear solution after I slowly swung it back and forth over the flame. The water didn't boil, no visible vapors formed on the surface, and I was holding the glass with my fingers, so I would say that the temperature was well below 100°C.

After the solution cooled down, instead of crystals precipitating, I found the glass filled with a fibrous cotton-like white mass. I pulled it out (it mostly stuck together) and dried it up on sheets of paper. Now I have white flakes which are flexible instead of brittle (still feel very much like cotton).

This would seem an awful lot like some polymer forming - except the substance tastes a lot like crystalline caffeine, and melts and evaporates much like caffeine - with the vapors having the same distinctive acrid (but not burnt) smell.

I'm pretty sure the answer to the polymer question is no, but in case I'm right I'd be very curious as to what's actually happening.

Google Images "caffeine crystals"  They are beautiful fibrous masses.  The solvent can influence crystal shape by making various faces grow faster or slower.  Benzil slowly crystallized from hot optically resolved pinene is slender needles.  Crystallized from room temp supersaturated mixed xylenes with a needle seed crystal, it is single crystal blocks.


How can we determine the Acidity or Base only from information molecular formula provides?No concentration given How can we determine the Acidity or the base of a solution just from the information the molecular formula provides, like $\ce{KCl}$, $\ce{NH4Cl}$, $\ce{MgSO4}$, $\ce{NaCO3}$, like which ones have a pH of less or more than 7, if I was given the concentration , I would have used $-\log{\ce{[H3O+]}}$ formula, but without it, how can I determine the pH levels?

You cannot determine exact pH levels without knowing specific equilibrium constants (or equivalent values) and concentration of the substances.

Determining whether something is acidic is relatively straight forward. Inorganic compounds like the one you've presented can be broken down into cations and anions.

For the anions, you should consider:

Is this anion acidic ($\ce{H2PO4-}$)
Is this anion basic ($\ce{OH-}, \ce{HCO3-}$)

For the cations, you should consider:

Is this cation the hydrogen ion ($\ce{H+}$)
Is this cation a Brønsted-Lowry acid ($\ce{NH4+}$)
Is this cation Lewis acidic ($\ce{Al^{3+}}$)

For compounds where ions are both acidic and basic, the exact balance matters. Then, you're crossing into the complicated regime where things are interesting.


What is the justification for Hückel's rule? I know that experimentally, the rule has been verified but I am having trouble understanding why it works. I see no reason why equivalent cyclic structures can undergo resonance and be just as stable. Any clarification would be nice.

The justification comes from molecular orbital theory. I'm going to analyze the MO diagrams for benzene (a molecule with $4n+2$ π electrons) and an MO diagram for cyclobutadiene (a molecule with $4n$ π electrons). These results generalize for all planar cyclic molecules with π orbitals on every atom.

The above π MO diagram of benzene may be obtained through Hückel theory (see also: Pi molecular orbitals of polyenes). Here, the "zero" of energy (referred to as $\alpha$ in Hückel theory) refers to the energy of an isolated carbon 2p orbital that does not participate in bonding. Hence, any orbitals below this energy are bonding in nature, and orbitals above it are antibonding.

In the benzene MO diagram above, you'll notice that all the molecular orbitals that are being filled by the π electrons are more stable than leaving the carbons unbonded. On top of that, it can be shown that the stabilisation derived from delocalised π bonding in benzene is greater than the stabilisation derived from simply forming three localised C–C π bonds, in conjugation with one another, but not delocalised over the entire ring (e.g. in hexa-1,3,5-triene). This is the condition known as aromaticity.

Let's look at the MO diagram for a hypothetical perfectly square cyclobutadiene, for contrast.

As with all molecular orbitals, the most stable orbital is filled first. However, problems start to emerge when going up to next two orbitals. First, unlike the next orbitals in sequence for benzene, these are formally nonbonding within simple Hückel theory.

It can be shown that this situation is less stable than the case for an alternative version of cyclobutadiene, where the double bonds are localised between two adjacent carbons. In practice, cyclobutadiene therefore undergoes a distortion away from a square geometry to a rectangular geometry, which has longer C–C bond lengths for the single bonds and shorter bond lengths for the now localised double bonds (see also: Cyclobutadiene - Jahn–Teller effect or not?). This is the condition known as antiaromaticity.


How does a diatomic molecule vibrate in response to heat? Do you know of any applet/animation that shows how a diatomic molecule like $\ce{O2}$ or $\ce{N2}$ vibrate in response to/ to produce heat?

Can it vibrate only on one axis, the one joining the two atoms? What happens to the electrons during the vibration?

The diatomic can vibrate only by extending and contracting the bond. The more energy that is added the bigger the bond excursion. (More vibrational levels are excited and have bigger excursions). The bond only extends by a few % of its length in lower vibrational levels so its pretty hard to see. If heaps & heaps of energy are added (temp say $1000$ K) then in a real molecule the bond can break as it is actually anharmonic in its motion. When this happens the atoms fly apart.

Additionally the molecule can also rotate while it is vibrating. Rotation is a slower motion than vibration so many vibrational periods occur within a $360$ degree rotation.

At any given constant temperature in a sample of molecules the energy moves between vibrations, rotations and translations so that each type of energy is in equilibrium.

Some of the energy can  also be transferred as  radiation, in the infra-red (for vibrations hundreds to a few thousand $\pu{cm^{-1}}$) and microwave region ( in the range few $\pu{cm^{-1}}$) for rotational transitions but only if the diatomic molecule has a permanent dipole, i.e. is not homonuclear. If these molecules are in equilibrium with surrounding then these surroundings must also emit/absorb radiation to keep the total energy constant. If you were to suddenly isolate your block of (heteronuclear diatomic) molecules, say in space, then the radiation emitted would over a few hours remove energy so that the molecules will end up in their zero point levels.

The figure shows the potential energy profile of HI. The the average length for the for the lowest level (quantum number $0$) is approx $106.4 $ pm and the chnage due to vibration is about $\pm 10 $ pm and clearly greater for higher vibrational levels. The vibrational period is $\approx 14 $ fs.


Why is the alpha particle called a particle when it is made of four particles? We know the alpha particle is the nucleus of helium. It contains four subatomic particles - two protons and two neutrons. The protons and neutrons are further made of particles called up and down quarks. So why call an alpha particle a particle. And similarly, why do we say electrons, protons, neutrons etc. are particles, even though, they are made of particles?

I can understand the case of electrons, protons, and neutrons as when they were discovered and named, quarks were not discovered. But the helium nucleus is a different case.

I reckon talking about an English word's meaning changes in the frame of reference. We do not call a quark an object, although, an object could be any real or virtual thing. So in chemistry, I don't think calling an alpha particle a particle is a good practice. I do understand there are dust particles and all but we mention them only while talking about the macroscopic world. Not so in the case of chemistry. I hope you get my point.

Actually, the alpha particle contains two protons and two neutrons. It is what is emitted from the nucleus under alpha decay, giving off what was first classified as alpha radiation by Rutherford at the end of the 19th century (very late 1800's).

In 1932 the neutron was discovered. Only then did it become clear that the nuclei of atoms contain more than just some positively charged particles (which we know as protons).

So there is this historic reason for the naming of the alpha particle.

"Particle"

The word particle does not only pertain to elementary particles (and in [quantum] chemistry, we stop at the level of nuclei as particles, everything that goes deeper is nuclear physics). There are smoke particles, dust particles, etc.

There is also the wave-particle duality. In principle, since we can observe this "alpha radiation" from radioactive processes, there must be a particle that belongs to the wave. With electromagnetic radiation, this is the photon, and in alpha radiation it is the alpha particle.


If the ferric ion is more stable than the ferrous ion, then why is it readily reduced to the latter? The standard reduction potential value for the ferric ion:
\begin{align}
\ce{Fe^3+ + e- &<=> Fe^2+} & E^\circ &= \pu{0.77 V}
\end{align}
The value is positive, hence it means that ferric ion is readily reduced to ferrous ion, i.e. ferrous ion must be more stable than ferric ion.
But in a book I have also seen that ferric ion is more stable than ferrous ion as it has stable $\mathrm d^5$ configuration.
I am little bit confused that if ferric is more stable, then why ferric is readily reduced to ferrous. Please explain it in detail.

As already pointed out, the notion of stability needs to be defined much more accurately.

The fact that the standard reduction potential is positive has no direct bearing on the gas-phase stabilities of the two ions. The only thing that the standard reduction potential actually reflects, is that the standard Gibbs free energy of the following reaction

$$\ce{Fe^3+ (aq) + 1/2H2 (g) -> Fe^2+ (aq) + H+ (aq)}$$

is negative.

So, for example, Curt already hinted in the comments that under an atmosphere of oxygen, $\ce{Fe^3+}$ would be more stable. This is simply because the reaction

$$\ce{4Fe^2+(aq) + O2 (g) + 4H+ (aq) -> 4Fe^3+ (aq) + H2O (l)}$$

(or some variation of it) has a negative Gibbs free energy. That is very easy to prove, since we know that $E^\circ(\ce{O2}/\ce{H2O}) = +1.23~\mathrm{V}$ and $E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77~\mathrm{V}$.

In addition if you were to carry out the reaction in base, the stability of $\ce{Fe(III)}$ relative to $\ce{Fe(II)}$ will be increased greatly, since any $\ce{Fe^3+}$ will be precipitated as $\ce{Fe(OH)3}$.

Remember that a standard reduction potential is measured against an arbitrarily determined zero. If, instead of the standard hydrogen electrode, we chose to use the $\ce{O2/H2O}$ reaction as the standard, then $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ would have a negative value.

Let's also get this myth out of the way. $\ce{Fe^2+}$ does not, under any circumstances, have a $\mathrm{(3d)^5(4s)^1}$ configuration. Period.

Ok, with all of that settled, let's answer the question you were actually trying to ask. Yes, $\ce{Fe^3+}$ is a decently good oxidant. There are plenty of half-reactions that have $E^\circ$ below the $\pu{0.77 V}$ of $\ce{Fe^3+/Fe^2+}$. If you want to use $\ce{H+/H2O}$ as a reference point (which is perfectly fine, but you have to make it very clear that this is a relative stability, not an absolute stability), then yes, $\ce{Fe^3+}$ is a better oxidant than $\ce{H+}$.

The reason behind that has nothing to do with $\mathrm{d}^5$ stability. It is simply because $\ce{Fe^3+}$ has a pretty high charge and would quite like having another electron so that it doesn't have quite so high a charge.

More formally, you could construct a cycle for the half-reaction $\ce{Fe^3+ + e- -> Fe^2+}$, neglecting entropy for the time being.

$$\require{AMScd}
\begin{CD}
\ce{Fe^3+ (aq) + e-} @>\large\Delta H^\circ>> \ce{Fe^2+ (aq)} \\
@V\large-\Delta_\mathrm{hyd}H(\ce{Fe^3+})VV & @AA\large\Delta_\mathrm{hyd}H(\ce{Fe^2+})A \\
\ce{Fe^3+ (g) + e-} @>>\large -I_3(\ce{Fe})> \ce{Fe^2+ (g)} \\
\end{CD}
$$

The quantity $\Delta_\mathrm{hyd}H(\ce{Fe^2+}) - \Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is positive since $\Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is more negative than $\Delta_\mathrm{hyd}H(\ce{Fe^2+})$. So, the hydration enthalpy terms disfavour the reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$.

However, the third ionisation energy of Fe is pretty large, and it turns out that in the 3d block, the third ionisation energy is the major factor that controls the reduction potential of $\ce{M^3+}$ to $\ce{M^2+}$. In particular, it is quite large because Fe is the 6th element in the 3d-block; you should know that $Z_\mathrm{eff}$, as well as ionisation energies, steadily increase across the 3d block due to the poor shielding provided by 3d electrons.

Put simply, the large value of $I_3$ makes it difficult to oxidise $\ce{Fe^2+}$ to $\ce{Fe^3+}$.

The $\mathrm{d^5}$ configuration of $\ce{Fe^3+}$ does have an impact. It can explain why $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ is relatively small compared to its neighbours: $E^\circ(\ce{Mn^3+}/\ce{Mn^2+})$ and $E^\circ(\ce{Co^3+}/\ce{Co^2+})$ are both much larger than $\pu{0.77 V}$. However, this "half-filled shell stabilisation" (more properly phrased in terms of exchange energy) is not quite a large enough factor to make $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ negative.

The question that I linked above has slightly more information.


How does evaporation even possible? Suppose the pressure is 1 atm and we put water (at 25 °C) in an open container. After some time the water will disappear as the water evaporates. We know that water at 25 °C and 1 atm has a vapour pressure. And this vapour pressure is obviously less than the atmospheric pressure or external pressure as no boiling is occurring at this pressure, temperature. If they were equal, boiling would occur which is clearly not happening.
So, if the vapour pressure is less than the atmospheric pressure, then how are are liquid molecules converting into gaseous molecules? I mean the atmospheric pressure above the liquid is exerting a greater force on the liquid and it's vapour pressure is less than the atmospheric pressure.
Then how are the liquid molecules overcoming the atmospheric pressure and changing into gaseous molecules if their vapour pressure is less than the atmospheric pressure? They can only change into gaseous molecules when they can overcome the atmospheric pressure which can occur when their vapour pressure equals atmospheric pressure.

I remember my days when I started confusing myself in the exact same way. The issue I believe is that you haven't been explained the details of how all this happens.
Evaporation:
All molecules in a liquid have various amount of energy. But the ones at the top, the ones that form the surface - they have the maximum energy in them because they are present there by overcoming any forces of attraction. (The other way to imagine this would be, the layer of molecules at the surface will have mostly downward forces of attraction on them compared to the distributed force of attraction from all directions on a bulk molecule.) Now these surface molecules have the tendency to escape. These molecules will individually leave the liquid by getting rid of the forces of attraction because it is easier for them compared to bulk molecules. And when individual molecules leave the liquid, it is called evaporation. Although this is a form of vapourisation, it doesn't need a constant temperature where the change of state occurs. They are just individual molecules that leave the liquid and escape and cause a vapour over the liquid.
Boiling:
But in boiling, a lot of constant heat supply, constant temperature, constant pressure. All this is required. And when this is obtained, the liquid changes its state and becomes a gas. This gas forms bubbles and escapes from the bulk of the liquid as well and not only from the surface as in case of evaporation.


Will piperine endure heating for several days of soxhlet extractions? I'm extracting piperine for use with curcuminoid supplements using a soxhlet apparatus and 91% isopropyl alcohol. My first extraction is right about done but I still need to do two more because of the limited size of my soxhlet.

My question is, instead of distilling off and reusing the distilled IPA could I just use the same solvent that contains the black pepper extract for three subsequent extractions?

I tried googling but could not determine if piperine is stable enough.

So, I promised to answer this question based on my experiences if I'd remember. And I did. Here we go:

I did three soxhlet extractions, each with 175 grams of black peppercorns.

The first extraction took 6 hours and the cycles were quite slow due to the vapor pathways of the apparatus not being insulated.
The second extraction took 2 hours, again with an uninsulated apparatus.
The third extraction also took 2 hours, but the cycles were much faster because I used aluminum foil for insulating the apparatus.

Each extraction used 400 ml of 99% isopropyl alcohol of which 200 ml was distilled off after extraction. Something to note is that this reclamation for the first run took at least two times as long as it did for the two later ones.

The yields were as follows:
- Run 1: 8,6 grams of piperine
- Run 2: 6,8 grams of piperine
- Run 3: 7,5 grams of piperine

From this I concluded that with sufficient insulation, something like 3-4 hours could potentially provide the optimal efficiency and that the yields of piperine aren't significantly affected by prolonged periods of heating in an alcoholic solution.

I hope this write-up will help or be of interest to someone!


Why does pyrite form cubic crystals? Pyrite ($\ce{FeS2}$) forms cubic crystals, like these ones:

I know that the crystal structure is primitive cubic, but I don’t see how cubic structure on the molecular level translates to cubic structure on much larger scales. If you are given some number of cubes, there are many different ways to assemble the cubes face to face, and only one of these ways is to create a large cube. So why is it that the pyrite in this rock forms nearly perfect cubes?

Pyrite can actually come in a range of crystalline morphologies, as seen below:

Image source: Modelling nanoscale $\ce{FeS2}$ formation in sulfur rich conditions (J. Mater. Chem. 2009, 19 (21), 3389), who state that the shape is dependent on the concentration of sulphur during its formation.

Modelling of cubic pyrite crystals, as in the picture in your question, are explained in the article Modeling the Shape of Ions in Pyrite-Type Crystals (Crystals 2014, 4 (3), 390–403.), modelling the following structure:

From the caption, the blue spheres represent $\ce{Fe}$ and the ellipsoids represent $\ce{S}$. The shapes of the constituent chemicals used in this diagram is based on their modelling, specifically one of the conclusions the author makes is that:

  It turns out for $p$ valence shell ions that an isotropic ionic radius only occurs on cubic lattice sites. For all other site symmetries, however, two or three radial parameters will apply. Appropriate geometric shapes for $p$ valence shell ions are thus given by ellipsoids instead of spheres.

The $p$ valence ion in this case being sulphide.  With this model in mind, the explanation provided in the caption for this diagram states:

  The mesh inscribed into the sulfur ions now reveals the ellipsoidal compression along <111> directions. In this model, the number of contact points of $\ce{S}$ and $\ce{Fe}$ ions become four and six, respectively, and the ellipsoidal deformation is concluded to enable a stable packing.

The molecular shape does not necessarily mean that the pyrite crystal has a cubic shape, however if the conditions are right (pressure and temperature), and the composition is 'correct' - if the formation is in a iron rich environment (as per the first reference), then this molecular shape could express as a cubic crystal.


Thermodynamics of photochemical reactions Suppose I have a photochemical reaction in gas phase, such as
$$
\ce{CO2 + {$h\nu$} -> CO + O}.
$$
I would like to work through the thermodynamics of such a reaction and understand the meaning of every term.

For a non-photochemical reaction, such as $\ce{CO + O -> CO2}$, I have a good understanding of how the change in Gibbs energy breaks down into enthalpy, internal entropy and a log-concentration term corresponding to the entropy of mixing. If I ignore the photon in the first reaction above, I can regard it as simply the reverse of this reaction, which due to its photochemical driving force is able to reduce the Gibbs energy by the same amount, $\Delta G$. All of this is unproblematic.

However, $h\nu$ itself is an energy change, which I can calculate, assuming I know the frequency $\nu$ of the absorbed light. I would like to know how to think about the thermodynamics of the whole system, including the coupling of the chemistry to the radiation field. In particular, my questions are:

How does the energy $h\nu$ relate to the other terms with energy units, such as $\Delta H$ and $\Delta G^0$?
I know that while a single photon has no internal entropy, a beam of light of a particular frequency can be thought of as having a temperature and an entropy. How should I think about the role of radiation entropy in the thermodynamics of photochemistry, and in particular, how can I do second-law calculations for photochemical reactions?

$h\nu$ is not a thermal energy so thermodynamic considerations do not apply. For example, you cannot predict the probability of being in an electronic excited state from equilibrium thermodynamics.  The main reason is that the excitation is not a result of (classical) thermal energy transfer, but rather the adsorption of a photon which changes the quantum state.

With a constant light source

$$\ce{CO2} + h\nu \ce{-> CO + C}$$

will reach a steady state with constant concentrations that will be different from the thermal equilibrium concentrations.  However, you can not extract a standard free energy difference by
$$ \Delta G^\circ=-RT\ln \left( \frac{p_{\ce{CO}}p_{\ce{O}}}{p_{\ce{CO2}}} \right) $$
because the products and reactant are not in thermal equilibrium.


What is special about the molecule luciferin that it can emit light? A classic example of bioluminescence; the protein luciferin is oxidized by the enzyme luciferase, releasing energy as light.
What is it that gives luciferin this ability? Why can't just about any protein display bioluminescence?

Are there more chemical reactions where electromagnetic waves are produced? Also, are there any chemical reactions which produce high energy electromagnetic waves, such as X-Rays?

[...] the protein luciferin is oxidized by the enzyme luciferase

Luciferin is not a protein, but a benzothiazole with a thiazole attached to the carbon atom between nitrogen an sulfur.

Upon oxidation at the thiazole ring, an oxetanone is formed. This four-membered ring breaks, releases carbon dioxide and turns the thiazole into a thiazolone. The resulting molecule is called oxy-luciferine.

Amazingly, the oxy-luciferine generated through this reaction is in an excited state! While typically excited states are the result of the absorption of a photon, here excited state is generated chemically!

Anyway, the fate of this excited state is as to be expected: it relaxes to the ground state by emitting a photon.

In general, these reactions are known as chemoluminescence.

Another known example is the peroxyoxalate reaction of glow sticks.

Phosphorescence at ~450 nm has been observed from the comproportionation of cyclohexanoxy radicals in polymer matrices, when hydrogen atom atom transfer between two radicals produce cyclohexanol and cyclohexanone (in an excited triplet state).


Why is the oxidation of sulfur dioxide not termed as heat of formation of sulfur trioxide? According to me, the following reaction's heat should be termed as heat of formation of sulfur trioxide:

$$\ce{SO2 + 1/2 O2 -> SO3}$$

However, in my textbook, they have that given it is not the formation reaction. Why is it so? And which is the formation reaction then?

Heat of formation is the enthalpy difference to the elements in standard state as defined by IUPAC, which is in most cases their most thermodynamically stable form at STP.

It is completely irrelevant if the respective chemical equation makes any sense, i.e. it would work out in a laboratory. However high the number of intermediate steps to reach the desired product is, energy is always conserved.


How is seawater alkaline? If when water and carbon dioxide react they form carbonic acid, carbonate, and bicarbonate, how does seawater still have a pH of around 8? Doesn't a compound need an hydroxide ion to be a base?

The short answer to the title question is the ubiquitous presence of geologic and biogenic calcium carbonate $(\mathrm{p}K_\mathrm{a} = 9)$.  The oceans as a whole can largely be thought of as residing on beds of calcium carbonate. The following are some excerpts from this Wikipedia page:

Eggshells, snail shells and  most seashells are predominantly calcium carbonate.
Carbonate is found frequently in geologic settings and constitute an enormous carbon reservoir. Calcium carbonate occurs as aragonite, calcite and dolomite. The carbonate minerals form the rock types: limestone, chalk, marble, travertine, tufa, and others.
Calcium carbonate contributors, including plankton (such as coccoliths and planktic foraminifera), coralline algae, sponges, brachiopods, echinoderms, bryozoa and mollusks, are typically found in shallow water environments where sunlight and filterable food are more abundant. Cold-water carbonates do exist at higher latitudes but have a very slow growth rate.

Regarding your question

Doesn't a compound need an hydroxide ion to be a base?

Here you go:
$$\ce{CO3^2- (aq) + H2O (l) <=> HCO3- (aq) + OH-(aq)}$$
Although calcium carbonate is only sparingly soluble $(K_\mathrm{sp} = 3.3×10^{−9})$, as stated above the oceans are simply rife with it.


Why do larger alkyl groups show a greater inductive (+I) effect? I have been taught that an alkyl group shows a +I (positive inductive) effect. Why do they do so? I know that hyperconjugation can explain how alkyl groups could stabilize a carbocation (better than a hydrogen). But, how does a longer chain alkyl group (or a bulkier alkyl group) show more +I effect than a smaller one?

Qualitatively:

If a methyl group is more electron donating than a hydrogen, then an ethyl group would be more donating that a methyl group by the same reason.

The effect drops off with distance, and generally a pentyl group is considered no more donating than a butyl group.

As for why a methyl group is more electron donating than a hydrogen, the carbon has a higher electronegativity than hydrogen (slightly), polarizing the bond (slightly).  This increased electron density near the carbon, which increases (slightly) its electron donation (I+) ability.


Raman spectra in Gaussian I want to simulate Raman spectra of benzene in Gaussian 09.

Calculated spectra for a single ring Benzene ($\ce{C6H6}$) in Gaussian differs about $\pu{40 cm^{-1}}$ for each peak, compared with experimental Raman's databases.

How can I solve this problem?
Gaussian file created with GaussView6:
MP2 config

DFT config:

Parameters: (for optimization)

Thank you for attaching your input file.
The first place to start for improving these calculations is your choice of basis set. 6-31G is a rather small basis set and is likely a source of significant error here. If you're fond of using the Pople type basis then you would be better off using something like 6-31++G* so that you include polarization and diffuse functions. With your current basis I suspect that you are missing important polarization of the atoms upon distortion.

If you really wanted a very accurate answer I would recommend the following scheme. Optimize your geometry at the DFT level and then use that geometry as a starting place for a optimization and frequency calculation at a higher level of theory. For example use the cc-pvtz basis set (a correlation consistent basis set that's more efficient generally than the Pople type) and CCSD(T). This will provide a very accurate gas phase raman spectrum.

Edit:
The only additional Gaussian specific parameter you should need is the Geom=Checkpoint command for the CCSD(T) optimization and frequency run. This will use your existing checkpoint file as a geometry starting point.
The other keywords (basis and method) are exactly what they should be cc-pvdz and ccsd(t). For more information on basis keywords I recommend the users manual.


What are the effects of adding water to a system at equilibrium? Given the reaction $\ce{NaCl~(s) <=> Na+~(aq) + Cl^{-}~(aq)}$. $\ce{NaCl}$ is added to water and equilibrium is reached.

Does the reaction shift to the left/right/not shift at all if more water is then added?

It seems to me that adding water would decrease the concentrations of $\ce{Na+}$ and $\ce{Cl-}$, and therefore the reaction would shift to the right. Is this correct?

The reaction $$\ce{NaCl~(s) <=>[\ce{H2O}] Na+~(aq) + Cl^{-}~(aq)}$$ is at equilibrium when approximately $359~\mathrm{g/L}$ sodium chloride are dissolved in water, i.e. the solution is saturated, or $c(\ce{Na+}) = c(\ce{Cl-}) = c_\mathrm{max}$.

If less $\ce{NaCl}$ is dissolved, it is only one phase, which is in equilibrium with itself and all solid salt is dissolved, therefore the equilibrium would be entirely on the right side. At that point adding water would not change the equilibrium at all, it would only dilute the whole system.

When the solution is saturated, there is still solid salt present, the equilibrium is not entirely on the right side. Adding water will disturb the equilibrium, i.e. $c_\mathrm{present}<c_\mathrm{max}$. Therefore more of the excess $\ce{NaCl}$ will dissolve, until the maximum concentration is reached again. If the solution is still saturated after that, nothing will have changed in your solution except for the volume.
If your solution is not saturated anymore, the equilibrium is entirely on the right side and the concentrations will decrease.


How can benzaldehyde have a pKa of 14.9? There are numerous websites out there which claim that the pKa of benzaldehyde – C6H5CHO – is 14.90. (Just do a Google search for benzaldehyde pka to see what I mean.) This doesn't make sense chemically, as there are no protons in benzaldehyde which would be so acidic.

However, it doesn't seem to have been made up. The CRC Handbook of Chemistry and Physics lists the same value of 14.90 for the pKa of benzaldehyde, at $\pu{25 ^\circ C}$ in aqueous solution. But I can't find this information in the primary literature.

If this value refers to the acidity of the hydrate, does it include the equilibrium constant for hydrate formation? i.e. is this Ka the equilibrium constant for the reaction:

$$\ce{PhCHO + 2H2O <=> H3O+ + PhCH(OH)O-}$$

or is it just

$$\ce{PhCH(OH)2 + H2O <=> H3O+ + PhCH(OH)O-}?$$

I would appreciate some kind of definitive evidence. Using chemical reasoning and intuition is great, but if this value is to be of any use to anybody, then we need to know exactly what it describes.

Significant amount of geminal diol of benzaldehyde exists in an aqueous solution of benzaldehyde at 25 °C because $\mathrm{p}K_{\text{hyd}} = 2$ (Ref. 1)

The $\mathrm{p}K_{\mathrm a}$ of benzyl alcohol is listed as 15.40 (Wikipedia). Thus, one can reasonably assume that the given value of $\mathrm{p}K_{\mathrm a}$ 14.9 represents a composite equilibrium constant for the hydration of benzaldehyde and dissociation of the geminal diol thus formed.
Late addition to the answer:
In his paper, "Acidity constants of benzimidazolium ketone and pyridinium aldehyde hydrates" (Ref.2), Terence C. Owen states that:

It is known that the acidity constants of gem-diols typically are about 2.5 units lower than those of the corresponding monohydric alcohols.

When do some extensive literature survey, one can find few examples to backup that statement. For example, $\mathrm{p}K_{\mathrm a}$ of methanol is reported as 15.7 while that of formaldehyde hydrate is 13.3, between which $\Delta\mathrm{p}K_{\mathrm a} = 2.5$ (Ref.2). Interestingly, $\mathrm{p}K_{\mathrm a}$ of 1,1,1,3,3,3-hexafluoropropan-2-ol is reported as 9.22 while that of hexafluoroacetone hydrate is 6.45 where $\Delta\mathrm{p}K_{\mathrm a} = 2.77$ (Ref.3). However, $\mathrm{p}K_{\mathrm a}$ of 2,2,2-trifluoroethanol is reported as 12.37 (Ref.4) while that of 2,2,2-trifluoroethanal hydrate is 10.05 (Ref.3) where the difference is < 2.5 ($\Delta\mathrm{p}K_{\mathrm a} = 2.33$).
Thus, we can conclude that the $\mathrm{p}K_{\mathrm a}$ of benzaldehyde is derived from its hydrate (gem-diol).
Also see: Yoshiro Ogata and Atsushi Kawasaki, In The Chemistry of Carbonyl
Group, Volume 2; Jacob Zabicky, Ed.; John Wiley & Sons Ltd.: New York, NY, 1970, Chapter 1: Equilibrium additions to carbonyl compounds, pp 1–69 (https://doi.org/10.1002/9780470771228.ch1).

Late Edit:
It is also noteworthy that $\mathrm{p}K_{\mathrm a1}$ of the cyclic hydrate of hthalaldehyde is reported to be $11.50 \pm 0.10$ at $\pu{25 ^\circ C}$ (Ref. 5):


Phenyl vs phenol functional group? Phenyl is a functional group with an aromatic ring bonded to another group. And, phenol is a molecule that is just a phenyl bonded to a hydroxyl group.

However, some sources consider phenol itself a functional group. Are they both functional groups, or is phenol just a molecule?

Phenol is both a molecule and a functional group (or at least a class of molecules). IUPAC has the definitive answer on the matter:

Phenols:

  Compounds having one or more hydroxy groups attached to a benzene or other arene ring, e.g. 2-naphthol


Precipitation titration with Volhard method In the Vollhard method Chloride is precipitated from a solution using escess silver ions.
$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)}}$$
After precipitation the solution is filtered and ammonium thiocyanate is added to the silver supernate with a small amount of ferric ammonium sulfate as an indicator where thiocyanate reacts preferentially with the silver:

$$\ce{Ag+_{(aq)} + SCN-_{(aq)} -> AgSCN_{(aq)}}$$

Once the all of the silver has reacted with thiocyanate the $\ce{Fe^3+}$ will then react to form $\ce{Fe[SCN]^2+}$ which produces a dark red color and indicates an equivalent point. and the chloride concentration is calculated by difference.

$$\ce{Fe^3+_{(aq)} + SCN-_{(aq)} -> Fe[SCN]^2+_{(aq)}}$$

Since the silver in the precipitate, why will you have to filter $\ce{AgCl}$ before you can start back titration instead of treating the solution? Would filtering also necessary for $\ce{AgBr}$ and $\ce{AgI}$?

Yes filtering is necessary as thiocyanate can complex the solid silver chloride per:

$$\ce{AgX_{(s)} + SCN-_{(aq)} -> Ag[SCN]X-_{(aq)}\tag{X = Cl, Br, I}}$$

This additional reaction requires more thiocyanate for back titration and thus gives a lower chloride concentration than is true.


Electrolytic refining of precious metals — why the anode rather than the cathode? In electrolytic refining, metals like platinum and gold are deposited as anode mud.

Why aren't they deposited on the cathode since they have higher reduction potential?

For these metals to deposit, they would have to be present in solution such that they can come into contact with the cathode and be reduced. However, this would first require them to be oxidized, so that they dissolve in solution. Unless you are actually electrolytically refining for gold, the voltage will not be enough to oxidize it.


I'm pretty sure this doesn't follow Le Chatelier's Principle The likelihood is that I'm misunderstanding what's going on here.

Consider the reaction $\ce{A <=> B}$, where $K_\mathrm{c}=1$.

Initially, the system at equilibrium, where $[\ce A]=\pu{1M}$ and $[\ce B]=\pu{1M}$.
The system is then disturbed to increase $[\ce A]$ by $\pu{1M}$. Instantaneously, $[\ce A]=\pu{2M}$ and $[\ce{B}]=\pu{1M}$. Le Chatelier's principle suggests that the equilibrium position moves to the right to minimise the increase in $[\ce A]$.
The system then returns to equilibrium, such that $[\ce A]=\pu{1.5M}$ and $[\ce{B}]=\pu{1.5M}$.

Clearly, the relative amounts of $[\ce A]$ and $[\ce B]$ have not changed, so the equilibrium position has not changed, which is in direct contradiction with Le Chatelier's principle.

I think my misunderstanding may lie in one of the following:

My definition of equilibrium position (the relative amounts of reactants and products at equilibrium).
Between which numbers the equilibrium position is considered to move, let me elaborate on this: the relative amount of $[\ce A]$ and $[\ce B]$ have not changed between 1 and 3 ($1/1=1.5/1.5$), but they have changed between 2 and 3 ($1/2\ne 1.5/1.5$), so I see how the equilibrium position could be said to have moved right from step 2 to step 3 as the amount of B increases and the amount of A decreases.

Le Chatelier's Principle does not directly say what happens to concentration ratios.  Nor does it directly compare conditions after a disturbance with those before.  It says what happens only after a disturbance has been applied.

Here, the disturbance is adding one mole of A per liter.  When equilibrium is re-established you have only an additional one-half mole of A per liter, the rest having reacted to form B.  The initial addition of A has been partially offset, which is exactly what Le Chatelier's Principle says.


Interchange of axial and equatorial groups in Me2NPF4 The $\ce{^{31}P}$ NMR spectrum of $\ce{Me2NPF4}$ when recorded at room temperature consists of a quintet with intensities $1:4:6:4:1$. However, at $-100~\mathrm{^\circ C}$, the spectrum is as shown below:

Propose a structure for $\ce{Me2NPF4}$ that is consistent with these observations.

The phosphorus has five groups surrounding it so it must have some sort of five-coordinate geometry. The low temperature spectrum shows a triplet of triplets with similar coupling constants, which is what you would expect from a trigonal bipyramidal structure with two axial and two equatorial fluorines. On this basis I assigned the structure as below:

However, I don't understand why all four fluorines are equivalent at higher temperatures. Does this mean that they are constantly interchanging and if so, what is the mechanism for this?

Obviously, in the structure you drew you have two sets of two fluorines which are non-identical: The axial ones that form a 4-electron-3-centre bond with phosphorus and the equatorial ones that are bound by the classical 2-electron-2-centre bond. And also obviously, at higher temperatures this difference vanishes so the fluorines must somehow interconvert into each other (that is the easier explanation when compared to a different structure and is thus preferred by Occam’s razor) rapidly on the NMR-timescale. You noticed this already.

The mechanism for this is called Berry pseudorotation, and it involes a square pyramidal transition state as shown in this image taken from the Wikipedia page:

(The image is drawn with pentacarbonyliron but just exchange iron with phosphorus, four carbonyl groups with fluorine and the fifth with dimethylamine and voilà.)

Since this mechanism involves a significant change in the binding situation (we no longer can write two clearly 2-e-2-c and one 4-e-3-c bond(s)), we need a certain activation energy — that is usually supplied thermally. At $-100~\mathrm{^\circ C}$, there is not enough thermal energy to induce this change; at room temperature there is.

Ben Norris mentioned another important observation in a comment: There is not $^3J_\ce{PH}$ coupling visible. I suspect that this is due to the spectrum actually being $\ce{^{31}P\{^1H\}}$ — a proton-decoupled phosphorus spectrum.


Determine the equilibrium constant for this reaction The question is:

  Ions $\ce{B}$ and $\ce{C}$ react to form complex $\ce{BC}$. If $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{B}$ is combined with $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{C}$, and our initial concentration of BC is $0$, yet the end in equilibrium for complex $\ce{BC}$ is $\pu{0.00450 mol}$ determine the equilibrium constant for this reaction.

Firstly, I converted the total volume $\pu{15.0 mL + 15.0 mL = 30 mL}$ to L and I got $\pu{0.03L}$

Secondly, I used the given $\pu{0.00450 mol}$ and divided it by $\pu{0.03 L}$ and I got $\pu{0.15 M}$ for BC

Thirdly, I did the ICE chart

After that, I used the quadratic formula to solve for $x$ and calculate the concentration for each substance, as well as getting the $K_\mathrm{c}$ at equilibrium

I got $1.53$ for $K_\mathrm{c}$ but I am not sure if the whole process is correctly done. Please point out my mistakes

You made a slight error in calculations for B and C. While those concentrations are correct, that is 1 M at 15 mL. When doing RICE tables and equilibrium calculations, we can use concentrations, however I prefer to convert concentration into moles.

We are trying to find $K_c$, and we know that equal moles of the reactant react to give us 0.0045 mol of BC.

$$\ce{B + C <=> BC}$$

\begin{array} {|c|c|c|c|}
\hline
\text{Initial conc.} & \text{0.015 mol} & \text{0.015 mol} & \text{0 mol}\\
\hline
\text{Change conc.} & -x & -x & +x\\
\hline
\text{End conc.} & \text{0.015 mol} - x & \text{0.015 mol} - x &  0 + x \\
\hline
\end{array}

We know that our change in equilibrium is 0.00450 mol, as that is the end amount of product that we have in our equilibrium calculations.

Accounting for the concentrations, we divide our number of moles by the total volume to get mol/L. 0.5 mol/L, 0.5 mol/L, .15 mol/L.

Assuming that our reaction is at equilibrium:

$$K_c = \frac{[\text{0} + \text{.15 mol/L}]}{[\text{0.5 mol/L} - \text{.15 mol/L}][\text{0.5 mol/L} - \text{.15 mol/L}]} $$

$$K_c = \frac{[\text{.15 mol/L}]}{\text{[0.35 mol/L]}^2} $$

$$ K_c = 1.22 $$

Thanks @Nicolau for pointing out the error in the question and reviewing my work.


PE expression for vibrational spectra of molecules using Newton's second law I am currently studying the textbook Infrared and Raman Spectroscopy: Principles and Spectral Interpretation, second edition, by Peter J. Larkin. Section 10. Calculating the Vibrational Spectra of Molecules, says the following:

In the case of vibrational spectroscopy, the polyatomic molecule is considered to oscillate with a small amplitude about the equilibrium position and the PE expression is expanded in a Taylor series and takes the form:
$$V = V_0 + \sum_{i = 1}^{3N} \left( \dfrac{\partial{V}}{\partial{q_i}} \right)_e d q_i + \dfrac{1}{2} \sum_{i = 1}^{3N}\sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e d q_i d q_j + \dots$$
The above expression is expressed in internal coordinates, $q_i$ and $q_j$ which are directly connected to the internal bond lengths and angles. The above expression is simplified since:

The first term $V_0 = 0$ since the vibrational energy is chosen as vibrating atoms about the equilibrium position.
At the minimum energy configuration the first derivative is zero of definition.
Since the harmonic approximation is used all terms in the Taylor expansion greater than 2 can be neglected.
This leaves only the second term in the PE expression for $V$. Using Newton's second law the above is expressed as
$$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$

Even after taking the points above into account, I don't see how we get from the first equation to the second. I'm assuming that $V$ is the potential energy (PE), right? And Newton's second law in differential equation form is $F = m \dfrac{dV}{dt}$. So how does this and points 1., 2., and 3. then lead to $$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$?
I would greatly appreciate it if people would please take the time to explain this.

It's probably simplest to start with the familiar example of the 1D harmonic oscillator. Here
$$F=-\frac{dV}{dx}=-\frac{d\left(\frac{1}{2}kx^2\right)}{dx}=-kx\tag{1}$$
which is often called Hooke's law, $F=-kx$, where x is the displacement from the equilibrium position, and the potential is assumed quadratic (parabolic) in the position coordinate. The derivation above requires the substitution $V(x) = \frac12 kx^2$, which can be obtained by trimming a Taylor expansion of the potential about the energy minimum (assumed = 0) at the quadratic term, and the force constant is defined as
$$k=\left(\frac{d^2V}{dx^2} \right)_e\tag{2}$$
Also,
$$F=ma=m\frac{d^2x}{dt^2}\tag{3}$$
If we transform to mass-weighted coordinates (or equivalently set m=1), we therefore can write that
$$\frac{d^2x}{dt^2} = -\frac{dV}{dx}=-\left(\frac{d^2V}{dx^2} \right)_e x\tag{4}$$
The derivation in your book does essentially the same thing. Note that since the potential is assumed quadratic and the other listed conditions are imposed, the potential can be written as
$$V = \frac12 \sum_{i = 1}^{3N}\sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j \tag{5}$$
The double summation can be collapsed into a single one when you take the derivative with respect to a spatial coordinate, say $q_k$:
$$\begin{align} \dfrac{d^2 q_k}{d t^2}  &=-\left( \dfrac{\partial{V}}{\partial{q_k}} \right)  \\&=-\left(\dfrac{\partial{}}{\partial{q_k}} \left[  \frac12 q_k^2  \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e   + \frac12 q_k\sum_{j \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e  q_j  + \frac12 q_k \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i  + \frac12 \sum_{i \neq k }^{3N}\sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j  \right] \right)  \\ &= -\left( \left[ \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e  q_k  + \frac12 \sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e  q_j  + \frac12 \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i    \right] \right)    \\ &= - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k} \partial{q_j}} \right)_e q_j \end{align}$$


What can I use to remove print from hard surfaces? I have a lot of profile products with the logotype of the company handing them out printed on them. In some situations I don't like to show these logos. However, it is stupid to bin working products for this reason so I would like to remove the print on them.

What solvent can be used for this purpose? What solvent is used in this type of colours?

Currently there are two items I want to remove the printing from:

A clear plastic bottle in BPA free tritan.
A tool in, I think, aluminium (metal anyway) with a surface similar to a Macbook Pro.

Tritan is a co-polyester. It is prone to stress cracking in the presence of a number of common solvents - cyclohexanone and tetrahydrofuran are used as surface bonding agents.

Acetone can induce stress cracking, as can IPA.

I would suggest overprinting if you want to disguise the logo.

As for the aluminium item - take your choice - cyclo, THF, acetone, xylene - lots of options.


Is axis of symmetry considered a true symmetry? In most of the books when i read about optical activity of a compound this is what they say:

A compound that does not posses any kind of symmetry is optically active. In other words chiral compounds without having any kind of of symmetry are optically active.

However, this holds true when we refer to plane of symmetry or centre of symmetry, because molecules having axis of symmetry or alternate axis of symmetry are still optically active.

For example this compound possesses axis of symmetry but is still optically active:

And therefore i am totally confused about relation of optical activity with symmetry.

Thanks for answering the question

A molecule is chiral (1, 2) if it does not have an improper axis of rotation. Since your drawing does not seem to show the meso form of 2,3-dichlorobutane, it is optically active.


In a mass spectrum, are there peaks corresponding to a combination of 13C and bromine isotopes? I just began studying mass spectroscopy and I don't understand why certain peaks are not represented in the mass spectrum graph. Like let's say you have methyl or ethyl bromide. Usually, there are two equal molecular ion peaks at the right-most side, which indicate a $1:1$ ratio of $\ce{^{79}Br}$ and $\ce{^{81}Br}$ isotopes.

But this is only when you assume all carbon atom are $\ce{^{12}C}$. Why doesn't the graph show peaks with combination of $\ce{^{13}C}$ and bromine isotopes?

None of the examples online indicate this.

It does appear. The graphs you're looking at online probably just dumbed it down.

Here's some proof if you like, a compound synthesised by yours truly, with molecular formula $\ce{C7H7BrN2O2}$. I clipped the relevant part of the mass spectrum I got. This was done with ammonia gas CI, so instead of the $\ce{M^{+.}}$ radical cation we expect to see the $\ce{[M + H]+}$ quasi-molecular ion. We expect the following monoisotopic masses (I used this website to calculate these):

$$\begin{array}{ccc} \hline
\text{Molecular ion} & m/z & \text{Predicted relative abundance} \\ \hline
\ce{[C7H7^{79}BrN2O2 + H]+} & 230.9769 & 100.00 \\
\ce{[C6^{13}CH7^{79}BrN2O2 + H]+} & 231.9798 & 8.48 \\
\ce{[C7H7^{81}BrN2O2 + H]+} & 232.9749 & 98.01 \\
\ce{[C6^{13}CH7^{81}BrN2O2 + H]+} & 233.9777 & 8.29 \\ \hline
\end{array}$$


Why is reduction potential of water ( 2H₂O + 2e⁻ → H₂ + 2OH⁻ ) is -0.83 insted of 0, as in the case of (2H⁺ → H₂)? I came across this value of reduction potential of water while solving a question. Also, I noticed that, for example

$$\ce{Ag+ → Ag}$$

has a fixed standard potential, and the same is used when we do it for a reaction like

$$\ce{AgCl → Ag + Cl-}$$

Then why here in the reduction of water the value differs from the $\ce{H+ → 1/2 H2}$ reaction?

Also, while doing electrolysis of water, which reaction actually happens at the cathode:

$$\ce{2 H2O  + 2 e- → H2 + 2 OH-}\quad\text{or}\quad\ce{2 H+ → H2}~?$$

Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1,

If we consider reactions

$$\begin{align}\ce{
2 H+ + 2e-    &<=> H2 \\
2 H2O + 2 e-     &<=>  2 OH- + H2 \\
}\end{align}$$

their standard redox potentials imply
$\rm{p}H=0$ for the former, $\rm{p}H=14$ for the latter.

For the former reaction and $\rm{p}H=14$, $E=-0.83\ \rm{V}$. As for $\ce{H+/H2}$ potential, at $\pu{p_{\ce{H_2}} = 1 atm}$

$$E_\ce{H2\ /H+}  = E^{\circ}_\ce{H2\ /H+} + 0.059 \log [\ce{H+}] = E^{\circ} - 0.059\  \rm{p}H$$

Thermodynamically, the equations they are equivalent, but kinetically, the latter reaction has in strongly alkaline environment the advantage of many concentration orders. And vice versa.

Elements can have multiple standard redox potentials, as such a potential can be expressed for any redox reaction (even if for some only formally from the thermodynamic data).

For example, some standard redox potentials for $\ce{Ag}$ are:

$$
\begin{array}{lcll}
\ce{Ag+ + e- &<=>& Ag(s)};\ & E^\circ_{red} = \pu{0.80 V}\\
\ce{AgCl(s) + e- &<=>& Ag(s) + Cl-(aq)};\ & E^\circ_{red} =\pu{ 0.22 V}
\end{array}
$$

(There are more; see the Standard electrode potential data table).

The latter can be calculated from the former using
$$E_{\ce{Ag/Ag+}} = E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+}$$

For the $\ce{Ag / AgCl / Cl-}$  system, we have to include into potential calculations the $\ce{AgCl}$ solubility constant.

$$\begin{align}
K_{\rm s, AgCl} &= c_{\ce{Ag+}}\cdot c_{\ce{Cl-}} \\[1em]
E_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+} =\\
  &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} - 0.059 \log {c_\ce{Cl-}}= \\
  &= E^{\circ}_{\ce{Ag/AgCl/Cl-}} - 0.059 \log {c_\ce{Cl-}} \\
E^{\circ}_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} =\\
 &= 0.80 + 0.059\log(1.77\cdot10^{−10}) = \\
 & = 0.22
\end{align}$$

Note that the $\ce{Ag/AgCl/Cl-}$ electrode is often used as the secondary standard of the electrode potential, aside of the calomel electrode $\ce{Hg/Hg2Cl2/Cl-}$, because maintaining of the primary potential standard electrode $\ce{Pt-H2/H+}$ is not convenient. The $\ce{Ag/AgCl/Cl-}$ electrode is used also as the electrode to measure $c_{\ce{Cl-}}$.


Is a carbon chiral it two of its groups are cis - trans isomers? For example:

In this image, the two groups sticking to a carbon are cis-trans isomers. So does that make the carbon chiral, exhibiting optical properties?

What if the two groups were both having the same configuration(both cis or both trans), would it then be achiral?

Thanks for any help :D

Any time all four groups bonded to carbon are distinct, including any sort of isomeric groups, you have a chiral center.  Comments associated with the related question referenced by @Loong indicate that the cis vs trans scenario is even included in the $R$ vs. $S$ naming convention.  This case is an $S$ configuration as the cis group takes second place and is counterclockwise from the hydroxyl group in the appropriate view.


Why is the first energy of ionization of oxygen lesser than that of nitrogen? The following question arises from a question I found in my book.

  Experimentally it has been determined that the value of the first
  energy of ionization of oxygen is lesser than the first energy of
  ionization of nitrogen. Select the statement which explains this:

  The higher electronegativity of oxygen
  The lesser electron affinity of nitrogen
  The lesser radius of oxygen
  The higher stability of nitrogen
  The higher electron affinity of nitrogen

According to my book the answer is the fourth option, but I couldn't find a way to justify this. What sort of explanation can do this?.

My guess is that nitrogen forms a triple bond with itself which is higher than the oxygen and this makes the molecule more stable but I don't know if this can be translated into a higher energy of ionization nitrogen rather than the oxygen which does have a sigma and pi bonds. Am I right?

Can someone please illustrate this or bring some sort of diagram with some values to justify this?. It would help a lot if an answer could help me why should I discard the second option?.

Doesn't nitrogen has smaller electron affinity than the oxygen? Wouldn't it mean that having a smaller electron affinity cause it to be easier to strip out an electron from its outer shell rather than the opposite as the question indicates?

Two electrons of the oxygen molecule are located on a $\pi $* level. They are relatively easy to ionize. It is not the case in $N_2$ : all electrons are located in binding orbitals.


Why does the second electron affinity has an opposite sign of the first one? Many first electron affinities are positive, indicating a favourable process, but the corresponding second electron affinities are negative. For example, the first and second electron affinities of oxygen are $+141$ and $-780~\mathrm{kJ~mol^{-1}}$ respectively. Why is this so?

Many non-metals like those in Group 16 and 17 have first electron affinities that are negative.  In other words, the following reaction is exothermic and energy is released.

$$\ce{X + e^{−} -> X^{−} + ~energy}$$

By gaining an electron the group 17 atoms attain a complete octet of electrons and become stabilized.  By gaining an electron the Group 16 atoms are just one electron away from having a filled shell.  Further in both the Group 16 and 17 cases, the protons in the nucleus are not fully screened by the outer electrons, so adding one more electron to form an anion is electrostatically favorable.

However, if we try to add a second electron the reaction becomes endothermic; adding a second electron is unfavorable and energy must be added (endothermic).

$$\ce{X^{-} + e^{−} + ~energy~ -> X^{2−}}$$

It is difficult to add a second electron primarily because of the increased electron-electron repulsion that occurs with the second electron.

Since adding the first electron is exothermic, while adding the second electron is endothermic, the sign reverses when we measure the energy added or released by these two processes.


What are the plastic connectors called (for glassware)? I can't remember for the life of me, and google isn't making it easy. e.g. the things labeled 24/40 etc., as in this picture:

Those are called Keck Clips. 24/40 is the size of the joint.


Is there anything that can react with molecular gaseous nitrogen? I am looking for something to react with nitrogen in the air. Is there any substance that can react fast with molecular nitrogen at room temperature?

From Wikipedia

  Elemental lithium reacts ("burns") in an atmosphere of $\ce{N2}$ to give lithium nitride:
  $$\ce{6Li + N2 → 2 Li3N}$$

  Magnesium also burns in nitrogen, forming magnesium nitride.
  $$\ce{3 Mg + N2 → Mg3N2}$$


Are labile tetrahedral [Mabcd] complexes optically active? In the above mentioned complex, there is a chiral centre at the metal, and hence it seems the compound must be optically active.

But my course notes say that the ligands keep on changing their their positions at a very fast rate and as a result such complexes get converted into their mirror images and are hence optically inactive.

I am unable to reconcile my understanding of the stereochemistry with this statement in the notes. How can such complexes be optically inactive?

In a labile complex, ligands are detaching and attaching to the metal centre at extremely high rates. This means the solution will racemise quickly.  Therefore, even if the complex is truly optically active, you have equal proportions of each enantiomer and thus equal rotations of the polarised light plane - in opposite directions.
The end result is that rotation cancels and so none is observed.


Tiffeneau–Demjanov rearrangement products With reference to the paper McCasland, G. E. Pinacolic Rearrangements of Epimeric Aminocyclanols. J. Am. Chem. Soc. 1951, 73 (5), 2293–2295 DOI: 10.1021/ja01149a110. (ResearchGate link). It states that

  cis-2-Aminocyclohexanol on treatment with nitrous acid undergoes a pinacol rearrangement, yielding both cyclohexanone and cyclopentylmethanal as products.

However, it goes on to state that the trans epimer would specifically yield the latter as the major product. Can someone explain this (preferably via mechanism)?

Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for the migrating group to be anti to the leaving group ($\ce{N2+}$), for orbital overlap reasons.

(apologies for hand-drawing - doing this on my phone)

You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the $\ce{OH}$ is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when $\ce{OH}$ is equatorial, the $\ce{C-C}$ bond is anti and the cyclopentane product is formed.

Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the $\ce{OH}$ and $\ce{N2}$ are axial, there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the $\ce{N2}$, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.


Surface Tension and Evaporation Why does acetone evaporate faster than ethanol yet still has a higher surface tension?

I would expect it to have a lower surface tension... I got the data from surface-tension.de.

Surface tension and evaporation are not as linked with each other as posed by your question. They are two distinct characteristics of liquids which generally follow direct relationship but not always. There are several other factors which you should not ignore:

Surface tension is the inherent force of liquid described by its tendency to minimize its surface area. A liquid with high molecular attraction takes account for a higher surface tension. It's temperature dependent and does not have similar slopes for all liquids. The case you provided is for standard temperature i.e. 20 C so you may get a contrary result at some other temperature. Compare: http://www.ddbst.com/en/EED/PCP/SFT_C11.php and http://www.ddbst.com/en/EED/PCP/SFT_C4.php2.
Evaporation usually deals with heat capacity, thermal energy required to overcome inter-molecular attractions and other thermal properties. Usually higher surface tension means energy required to overcome inter-molecular attractions is higher hence it may also have a higher value for the liquid having higher surface tension but it is not necessary. So liquids resembling each other in thermal properties usually form intersecting surface tension curves. i.e. one have a larger value till intersecting temperature and later on second has more. Ethanol and propanone is also such borderline case to be compared.


Can Aquatabs be used to clean swimming pool water? Can Aquatabs be used to prepare/chlorinate water for swimming?  They are used  to clean water for drinking so is it safe to assume that if used in proper quantity they can also clean water for swimming?

We are currently at a place where there is no pool and the temperatures are soaring, our 3 year old son has a small pool which we take along on picnics.  I was not able to find pool chlorine somewhere over here but we do have adequate quantity of aquatabs with us.

I tested this and the water smelled/tasted just like it does in a properly chlorinated pool but want to be sure if they are adequate.

Just in case, the label says

  Each tablet contains 500mg Sodium Dichloroisocyanurate (300mg Free available
  chlorine)

I am not really worried about maintaining a pH level over the matter of days, its more of a one time thing then we can change water.

Aquatabs can certainly be used as a disinfectant in a small swimming pool like you've described. The active ingredient has many uses, including drinking water treatment as you've pointed out, but it is also used as a disinfectant in swimming pools.

According to this Wikipedia article regarding the active ingredient of Aquatabs, sodium dichloroisocyanurate:

  As a disinfectant, it is used to sterilize drinking water, swimming pools, tableware and air, fight against infectious diseases as routine disinfection.

A study was done[1] comparing the effectiveness of sodium hypochlorite (common household bleach, also used for pool chlorination) with sodium dichloroisocyanurate. From the abstract:

  A comparison of commercial sodium hypochlorite (NaOCl) and sodium dichloroisocyanurate (NaDCC) products was made. Solutions of NaOCl and NaDCC containing the same levels of available chlorine (av.Cl) exhibited very similar bactericidal activities, despite significant differences in pH.

The ideal level of active chlorine for a swimming pool at neutral pH is about 0.5 to 1.5 mg of active chlorine per liter of pool water. So, if you aim for the middle of this range, you would want to add one Aquatab tablet for every 300 liters of water in your pool. Also, note that chlorine is less than half as effective at a pH of 8 as it is at a pH of 7, so I would pay some attention to the pH anyway.

D. Coates, J. Hosp. Infect. 1985, 6 (1),31-40. A comparison of sodium hypochlorite and sodium dichloroisocyanurate products. DOI: 10.1016/S0195-6701(85)80015-3


What is the reason for phase separation? I have seen phase separation as a phenomenon taking place. However I do not understand why this phenomenon occurs as a counter example to the natural laws of diffusion along the concentration gradient. During phase separation the richer side becomes richer with the atoms which it already has and becomes more and more deficient with the poorer atoms and vice-versa. Similar things will occur in the opposite case as well. How can phase separation be explained in detail, for example with factors like the chemical potential?How do you quantify whether 2 liquid interactions will be unfavorable or not?

The change in Gibbs energy has to be negative for a process to happen (at constant pressure in the absence of work). Gibbs energy is a combination of entropy and enthalpy.

  I do not understand why this phenomenon occurs as a counter example to the natural laws of diffusion along the concentration gradient.

Diffusion is driven by entropy changes, specifically the entropy of mixing (in an ideal solution, the enthalpy of mixing is zero, so the entropy term is the only consideration). The entropy goes up when a solute moves from high to low concentration.

Phase separation occurs when the enthalpy term dominates. If substances A and B are initially mixed and then undergo phase separation, some change occurred (change of temperature, added third component, etc.) so that now phase separation more exothermic (lower energy for [A to interact with A and B interact with B] than [A interacting with B]). Under these circumstances, phase separation can happen.


How come water would not affect the results of a titration? We were told to rinse the buret with the base so that the concentration won't be messed up, but when it comes to the flask that is going to hold the acid, there is no need to rinse, because it would not affect the result if it has water in it.

Why is that? Wouldn't the acid be diluted a little bit and that throws off the accuracy?

The volume and concentration of base equals the number of moles of base. This is use to determine the number of moles of acid in the flask. Since you are interested in the number of moles of acid in the flask, why would adding water (which doesn't have any acid) change the number of moles of acid?


Titration calculations - working out cation so I was doing this question and I have some questions about what I did. Firstly am I correct to say that the concordant results = 24.1 cm^3 because there are 2 of these values. From this I then calculate that M = 32.9 however this can not be true, so what have I done wrong?

You must have forgotten a factor $2$ somewhere.
You are right is thinking that $24.10$ mL is needed for the final titration. But now let's redo the calculations.
$24.10$ mL $\ce{NaOH} ~0.25$ M contains $24.10·0.25 = 6.025$ millimole $\ce{NaOH}$. Obviously this $\ce{NaOH}$ has neutralized the same amount $6.025$ millimole $\ce{HCl}$. As a consequence, there was $10$ times more $\ce{HCl}$ in the $250$ mL flask. This is $0.06025$ mol $\ce{HCl}$, and was the residual amount of $\ce{HCl}$ remaining after reaction of HCl plus the unknown carbonate. It also means that after adding $0.100$ mole $\ce{HCl}$ in the initial flask, the difference $0.100 - 0.06025 = 0.03975$ mol $\ce{HCl}$ had been used to destroy the carbonate.
But whatever the formula of the initial carbonate, $2~\ce{H+}$ are needed to react with  $1~ \ce{CO3^{2-}}$. So the amount of unknown $\ce{XCO3}$ or $\ce{Y2CO3}$ carbonate in the original powder is half the amount of $\ce{HCl}$ and it is : $0.03975/2$ mol = $0.019875$ mol $\ce{XCO3}$.
As the mass of this carbonate is $2.50$ g, its molar mass is: M = m/n = $2.50/0.019875 = 125.8$ g/mol. As the molar mass of $\ce{CO3}$ is $60$ g/mol, it means that the molar mass of $\ce{X}$ is $125.8 - 60 = 65.8$ g/mol.
This final value may be twice the atomic mass of a monovalent cation, or the atomic mass of a bivalent cation. As there are no monovalent cation with an atomic mass of about $32.9$, the unknown element must be a bivalent cation. And the only element with such an atomic mass is Zinc : A(Zn) = $65.39$
So the unknown carbonate is $\ce{ZnCO3}$


How Does a Precipitate Get a Colour? If you can use precipitate colours to determine an anion, what makes the precipitate have that particular colour?

Like for example what in sodium sulfate or barium chloride makes a white precipitate when these two combine?

The 'white colour' is due to the fact that the product does not absorb light in the visible part of the spectrum but forms as many fine crystals which in themselves scatter room lights. So the 'whiteness' is due to the scattering of light back towards your eyes after the light is repeatedly reflected of the many facets of the many crystals in the ppt. If you use red light to illuminate then the 'white colour' will become red. If the ppt is coloured then absorption of light occurs at some wavelengths as well as scattering at all wavelengths.

At the surface of a crystal some small fraction of light is reflected, but with millions of randomly oriented crystals numerous reflections occur. The amount of reflection at each surface depends on the difference in refractive index between the crystal and air for the wavelength of light used. If you could add a solvent say dichloromethane which has  a high refractive index and will probably not dissolve the ppt then the crystals can be made to almost disappear as scattering is now far less.  You can do the same thing with frosted glass beads which, when in air, scatter a lot but in water become virtually invisible.


Can you make R-Candy without heating the propellant mix? I'm wondering if there's any way you can make "Rocket Candy" (or R-candy) fuel without cooking it. I'm building a model rocket that I'm hoping will get about 3 km on my first build (It's not going to be super big), and I thought R-candy was a good idea. I am just getting into this stuff, but I'm doing a ton of research and trying to be safe.
In every recipe I've seen, it says that you need to "cook it" on a stove or use some other heat source. I'm wondering if you can avoid using a stove, oven, etc.
If there's no way to make R-candy without cooking it, is there some other fuel I can make from household materials?
P.S. The ammonium–aluminum fuel isn't an option for me.

First of all, when it comes to something that may impact your future well-being in a bad way, do a thorough research on the topic. Wikipedia is a good starting point as it highlights the role of homogenization in the fuel–oxidizer mix preparation:

There are many different methods for preparation of a sugar-based rocket propellant. Dry compression does not require heating, only the grinding of the components and then packing into the motor. However, this method is not recommended for serious experimenting.[2] Dry heating does not actually melt the $\ce{KNO3},$ but it melts the sugar and then the $\ce{KNO3}$ grains become suspended in the sugar.

Heating assures homogenization and packing of the fuel–oxidizer mix to create constant burn rate. If you decide to heat the mix, it is imperative that you look up the properties of all components, especially melting points. For example, in the most common nitrate–sorbitol mix sorbitol (m.p. 95 °C) would act as a flux for potassium nitrate (m.p. 334 °C). Make sure you are using a well-functioning hot plate with a temperature controller and not open flames, otherwise the mix will ignite.
Make sure the powders are dry before you mix them. Otherwise trapped water inside molten sorbitol (remember, m.p. 95 °C and both sorbitol and nitrates are hygroscopic) will create steam pockets which may cause local explosion with the stuff dispersed around and potentially catching fire from the hot plate. Needless to say, the mix must be carefully stirred and slowly heated until fully homogenized; only then it can be poured into the firm template or the rocket body. The mix is going to be at about 150 °C and sticky, so wear heat-resisting protective gloves.
If a hot plate is not an option, you may opt out for the worse alternative of grinding the powders and pressing into shape. First, you have to produce fine powders of the fuel and oxidizer separately. Normally, separated sorbitol and potassium nitrates should not ignite because of friction, but work with small batches at a time regardless. Second, they must be thoroughly mixed and pressed to assure good contact between particles. Work with dry components in small quantities without producing too much dust flying around to prevent dust explosion.
General notes:

Work with small portions of materials at a time in a well-ventilated area using PPE.
Check your equipment thoroughly and make sure there are no sources of sparks and open flames nearby. Molten mix will ignite when overheated, and the dust is explosive.
Remaining water will quench burning due to its tremendous heat capacity and may cause unstable engine operation with a non-zero chance of explosion, so I'd avoid adding water at all regardless what preparation method you pick.


What is the strongest acid? According to Wikipedia, $\ce{HeH+}$ and fluoroantimonic acid are the strongest.

According to Nature, Carborane acid is the strongest, but Wikipedia says fluoroantimonic acid is stronger.

Links:

$\ce{HeH+}$ / Wikipedia - stating HeH+ the second most acidic
$\ce{HeH+}$ / Wikipedia 2nd source - stating $\ce{HeH+}$ the most acidic (see in 'Notes' section)
Fluoroantimonic Acid / Wikipedia - stating it is the strongest acid (super_acid)
Carborane Acid / nature.com - According to Nature and even Google itself, it is the strongest acid created but according to wikipedia Flouroantimonic acid is stronger.

So, which is the strongest ?

The problem with this question is that the exact answer is "it depends..."

First off, it depends on your definition of acidity and how you measure it. Everyone seems to be using Brønstead acids (i.e. $\ce{H+}$ donors). I see two different measures in other answers:

Proton affinity: This is a gas-phase measurement of $\ce{A^{−} + H^{+} -> HA}$
Hammett acidity ($H_0$): This is a solution measurement, given by $\mathrm pK_{\ce{BH^+}} - \log\left(\frac{[\ce{BH^+}]}{[\ce{B}]}\right)$

Secondly, it depends on medium as mentioned by LDC3.

Leveling effect: The solvent leveling effect reflects the lowest possible $\mathrm pK_\mathrm a$ in a particular solvent, based on the basicity of the conjugate base. So you need $\ce{HF}$ or fluorosulfuric acid to reach low $\mathrm pK_\mathrm a$.

So the problem in my mind is that carboranes and fluoroantimonic acid are solution measurements, but $\ce{HeH+}$ is a gas-phase measurement. It does have the highest gas-phase proton affinity. But I'd put my money on things I can use in lab.

Incidentally, the Reed group prepared the fluorinated carborane acid this year: Angew. Chem. Int. Ed. 2014, 53 (4), 1131–1134. So that compound, $\ce{H(CHB11F11)}$ wins the crown for strongest solution-phase Brønstead acid (for the moment, at least).

There's also a nice review article "Myths about the Proton. The Nature of H+ in Condensed Media" by the same group in Acc. Chem. Res. 2013, 46 (11), 2567–2575.


Identifying the molecular structure of compound with sum formula C8H14O3 formula using NMR and IR spectrometry data I am looking to identify the molecular structure of a compound with
the sum formula $\ce{C8H14O3}$ using the NMR and IR spectrometry data below.

I was able to figure out the molecule as

My friend asked me why I didn't consider the following molecule:

I couldn't answer that. What inferences from the IR & NMR data can I observe, to explain why the structure my friend suggested is wrong?
Source for these IR & NMR Spectra: https://www.orgchemboulder.com/Spectroscopy/Problems/7.shtml

Compare the quartet at $\pu{4.2ppm}$ and the singlet at $\pu{3.5ppm}$. These quite obviously would, in your friend’s structure, correspond to the protons on either side of the ether-oxygen. However, we should expect the methylene singlet to be more deshielded than the ethyl group’s methylene: It is squashed between an electron-withdrawing $\ce{OR}$ and an electron-withdrawing $\ce{C=O}$. The ethyl group’s methylene is between an electron-withdrawing $\ce{OR}$ and a mostly electron-neutral $\ce{CH3}$ group. Therefore, the order of peaks we observe in the spectrum does not make sense when we consider your friend’s proposal.
Thereafter, one might also argue that a quartet at $\pu{4.2ppm}$ is so obviously an ethyl ester as likely every organic chemist has had way to many NMR spectra after a column that contained a peak exactly there. An ethyl-ether is more shielded (diethyl ether’s methylene proton has a chemical shift of $\pu{3.5ppm}$.


Abnormal deviation of CO2 from ideal gas behaviour Why does carbon dioxide show larger abnormal deviation from ideal gas behaviour than other gases. Why does carbon dioxide first show negative deviation and then positive?

When the compressibility factor z for CO2 is expressed in terms of reduced pressure and reduced temperature (i.e., using the law of corresponding states), the PVT behavior of CO2 is not very different from that of other gases.  The reduced temperature and reduced pressure are defined as the actual pressure and temperature normalized in terms of the critical pressure and critical temperature, respectively.  The real question is, "what causes the critical temperature and critical pressure of CO2 to be what they are?"


Explanation of the Selection rule in light of IR spectroscopy I am in mathematics, but I am writing a paper on molecular vibrations (because you can use representation theory to handle this).

I want to prove the selection-rule, but I also have to give a presentation of my intermediate results. Can someone give a little (maybe trivial) explanation (in one or maximal two small sentences) of what the selection-rule does? (for example for the IR-spectrum). This is for my presentation if someone doesn't know something about the mathematical and chemical background.

The operator for interaction with infrared light is the permanent dipole of the molecule, $\mu$.  I don't know anything about representation theory, but in group theory, you decompose the molecule by its symmetry elements to find the eigenmodes.  The infrared active modes are those modes in which the permanent dipole changes over the course of a vibration.  An excellent reference is Symmetry and Spectroscopy by Harris and Bertolucci which should be available in your library.


Surface area: what is the difference? What are the differences between:

van der waals surface area
solvent accessible surface area
topographical polar surface area

What is the significance of these differences, and what are the differences in how each of the three is used?

A Van-der-Waals surface Area (VdW) is a surface constructed by spheres around atoms with the radius of each sphere being the van-der-Waals radius of the respective atom.

The Solvent accessible surface area (SAS) is an surface normally being constructed by "rolling" a sphere the size of the vdw-radius of the solvent molecule around the solute and taking the middle of the solvent sphere as the surface of the solute.

Picture is taken from the original GEPOL publication of Pascual-Ahuir and Silla J. Comp. Chem., 1990, Vol. 11, No. 9, 1047-1060

So the difference is just the size. Both have the same questionable physical meaning. They are used for example in continuum solvation models like COSMO/PCM.

I used the SAS for Frozen-Density-Embedding-models to close unphysical holes (in PCM the surface constitutes an infinite continuum of solvent, so there would be an infinite volume of solvent between two subsystem molecules) between molecules of different subsystems in a QM/QM/Cont. multilayer method. Nonetheless, they are both only viable for qualitative results.

For the PSA, just read the wikipedia article.

The PSA is only the surface of the polar atoms of a molecule and it is used in medicine to calculate permeability of cell membranes of molecules in the body.


Corrosive? Aluminium Oxide? I moved into a house that has a light fitting that has been made out of an old food 'tin' can.

The light hadn't worked since moving in. On trying to remove it the glass of the bulb pulled straight out into my hand. There is a white powder on the bulb and also the wire towards the bulb.

It also looks in a structure like mould or mildew, as the 'white' is also up the wire and on the beam that the light lead is hanging from.

What is the powder? Is it dangerous, as in corrosive to other materials as it fell onto the floor and myself? Or is it possible that it is more a mould?

IMO this might be tungsten(VI) oxide, $\ce{WO3}$, which is fairly non-toxic, unless you consume it spoonwise.

From the look of this DIY setup I would doubt the quality of the light bulb itself was superb, and most likely one wouldn't use a fancy lamp anyway. So, the glass enclosure could be breached and some point, and the inner inert atmosphere (Argon/Nitrogen) was substituted with air. When hot tungsten filament oxidizes, it produces tungsten oxides which evaporates/scatters from the filament as white smoke:

$$\ce{2W + 3O2 -> 2WO3}$$

Actually, bulky $\ce{WO3}$ is yellow, but being dispersed in a form of tiny crystals, tugsten(IV) oxide has pale-yellow or white color depending on the fraction.

Anyway, I would get rid of this can enclosure and buy a decent light source, mainly from an aesthetic point of view.

P. S. Here is a funky experiment showing what could happen in your case: https://www.thenakedscientists.com/get-naked/experiments/light-bulbs-liquid-nitrogen


Desorption instead of adsorption from solution using activated charcoal I was just trying to see the effect of concentration on adsorption in solutions, i.e. I was trying to verify the Freundlich isotherm. I used oxalic acid and a batch of homemade activated charcoal (using this site).

First I prepared solutions of different concentrations. Then I added the activated charcoal, stirred it for 5 minutes and filtered the solution. When I titrated the filtrate using KMnO4, after calculating the concentration, I found that the concentration of the solution had been increased!

I tried the same process with different concentrations of oxalic acid solution and the result was the same. Can anyone tell me where did I go wrong?

Note: I used the same batch of activated charcoal in all the experiments. (Even then, how did the concentration of the solution increase? Shouldn't it be the same or lower than the initial concentration?)

For titrating oxalic acid, I heated the solution to about 60 degrees C.( After filtering the solution)

In such experiments you typically run a blank. So add the homemade activated charcoal to just some water, filter, then titrate that filtrate as a blank.

There are two possibilities for the increase.

    (1) Your batch of homemade activated charcoal was contaminated with who knows what soluble compound. (This seems the most likely...)

    (2) Quite possible that some of the charcoal particulates are small enough to not be trapped by the filter.

Another way to run the experiment would be to add the homemade activated charcoal to just some water, filter, then wash the wet charcoal into the oxalic acid solutions (this would remove soluble compounds from the charcoal). Then dilute all the solutions to some co0nstant volume, filter, and titrate the filtrates. (You should still run a blank.)


What is the mechanism for the oxidation of a lactol to a lactone? Recently, I came to know that lactols can be converted to lactones but I wasn't able to find the suitable reagent and the reaction mechanism. Can someone elaborate the same?

Methyl(trifluoromethyl)dioxirane (TFDO) in 1,1,1‐trifluoro‐2‐propanone (TFP)may be able to do your sought oxidation (see image [1] below): In there, cyclic ethers (shown example is tetrahydropyran) yield lactones by subsequent oxidation of a stable cyclic hemiacetal intermediate such as lactol in your case.

It is considered as selective and reactive oxidizing agent, capable of epoxidation of unreactive alkenes and arenes, oxyfunctionalization of alkanes, and oxidation of alcohols and ethers in following reference: Encyclopedia of Reagents for Organic Synthesis (1; March 2016); 11 pages. (https://doi.org/10.1002/047084289X.rm267.pub3)

You may also read: Oxidations by methyl(trifluoromethyl)dioxirane. 5. Conversion of alcohols into carbonyl compounds: J. Am. Chem. Soc., 1991, 113(6), 2205–2208 (DOI: 10.1021/ja00006a044).