How can I understand the Lewis acid-base interaction of iron(III) and water? The interaction between $\ce{H+}$ and $\ce{NH3}$ as respectively a Lewis acid and a Lewis base is clear for me. $\ce{NH3}$ has a lone electron pair and $\ce{H+}$ has no electrons, or, saying politely, an incomplete duet, and can use nitrogen's lone pair to complete its duet, and a dative (coordinate) bond is formed in the process.

However when I think about $\ce{[Fe(H2O)6]^3+}$, I don't see where these six lone pairs of electrons (which are coming from water molecules) can go in the iron ion. The electron configuration of $\ce{Fe^3+}$ is: $\mathrm{1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^5}$. It has five 3d electrons, and all (namely, 2) of its 4s electrons are gone. All five 3d electrons have the same spin, and they occupy five d orbitals. OK, I do remember that 3d orbitals are funny and in the transition metals 3d orbitals get divided into low and high energy level 3d orbitals (hence the color of transitional metal solutions).

But even with this, how do six water molecules position their lone pairs there? We have space for one pair of electrons (where 4s electron left) and then space for 5 more electrons, so in total, for 7 electrons. 6 molecules of water would give us at least six pairs of electrons, i.e. 12 electrons. So we are 5 places short to place all these electrons.

So the question is: how are they placed?

So, effectively, the question is: How are electrons from the covalent bond between ligand and central metal atom or ion placed with regard to the electron configuration?

$\ce{Fe^{3+}}$ has the condensed electron configuration of [Ar]$3d^5$. When the ligands approach the iron ion from the X, Y, and Z, the ion forms six bonds. It would be incorrect to state that the remaining 3d electrons are involved in the bonding with the ligands. Instead, the iron ion uses 6 orbitals from the 4s, 4p, and 4d orbitals to accept the lone pairs of electrons from the ligands. However, before they are used, the orbitals are hybridized in order to create six orbitals of equal energy.

If you want to dig deeper into the subject, I highly recommend the lecture notes from this webpage here.


Does every solute spend at least $t_M$ in the mobile phase? In 'Fundamentals of Analytical Chemistry' 9th edition by Skoog, West, Holler and Crouch, there is the following passage (page 865).

  The dead time (void time), $t_M$, is the time it takes for an unretained species to pass through a chromatographic column. All components spend at least this amount of time in the mobile phase.

Apologies if it seems a silly question, but how does that make sense? Wouldn't all components spend exactly that amount in the mobile phase? That is, of course, assuming a constant flow rate for the mobile phase. Is this where I have gone wrong?

Dead time ($t_{\bf M}$, also called holdup time) is the time it takes for the mobile phase (eluent) to traverse one length of the column, for a given flow rate. Because moieties in the eluent do not interact chemically with the stationary phase, $t_{\bf M}$ is therefore a lower bound on retention times for the species that do interact (chemically) with the stationary phase.

Simply put, anything that does interact with the immobile phase cannot elute faster that the pure mobile phase: retention times will be greater than $t_{\bf M}$.


Nucleophilicty and Basicity of OH and F Why is $\ce{OH-}$ more basic than $\ce{F-}$?

From what I understand, $\ce{O}$ has larger radius so it should be more stable (meaning less basic) than $\ce{F}$. When they form anions, since $\ce{O^{2-}}$ is larger, it can distribute its electrons over larger volume than $\ce{F-}$, meaning $\ce{OH-}$ should be less basic. Is this because of partial negtaive charge on oxygen?

  In a polar aprotic solution, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

I learned that basicity (not nucleophilicity) is about stability. So can this be explained by electron density? Is explaining it with polarizability right?

  In a lecture, my teacher said $\ce{RO-}$ is relatively stable because of the high electronegativity of $\ce{O}$.

I think it should be unstable because of high electron density caused by high electronegativity. What's right?

  When people say something is stable, are they referring to its acid/base characteristics, or can stability also be referring to nucleophilicity/electrophilicity?

Why is $\ce{OH-}$ more basic than $\ce{F-}$?

To answer this question, let us consider the $\mathrm{p}K_\mathrm{a}$'s of the conjugate acids of these anions: $\ce{H2O}$ and $\ce{HF}$. $$\mathrm{p}K_\mathrm{a}(\ce{H2O})=15.7\ \\  \mathrm{p}K_\mathrm{a}(\ce{HF})=3.17$$

You didn't need me to tell you that $\ce{HF}$ is a stronger acid than water though, but it is -- $10^{12.53}$ times stronger to be exact. This means that this is also how many times stronger $\ce{OH-}$ is than $\ce{F-}$

  In a polar aprotic solution, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

Small, electron-dense ions are always great nucleophiles, except when the solvent crowds the nucleophile with hydrogen bonds, as in the case of $\ce{F-}$. It is so heavily surrounded by solvent molecules in polar protic solvents that it can't make an attack very well. $\ce{I-}$ is not as heavily solvated in these solvents, and it is very easily polarizable, making it a better nucleophile in these solvents.

In polar aprotic solvents, the small $\ce{F-}$ ion can easily slip between solvent molecules and make attacks. Now in comparison, $\ce{I-}$ is a much poorer nucleophile.

  In a lecture, my teacher said $\ce{RO-}$ is relatively stable because of the high electronegativity of $\ce{O}$.

Again, we can answer this by looking at $\mathrm{p}K_\mathrm{a}$'s. The $\mathrm{p}K_\mathrm{a}$ of ethanol, for example, is $15.9$! Only $10^{0.2}=1.58$ times weaker an acid than water! That means that ethoxide will deprotonate everything $\ce{OH-}$ will and then a little more, so it's not really that much more unstable.

  When people say something is stable, are they referring to its acid/base characteristics, or can stability also be referring to nucleophilicity/electrophilicity?

The two are pretty well related. You have exceptions like how $\ce{F-}$ is a poor nucleophile in water despite being a good base, but as a rule of thumb, I would say people mean acid/base reactivity when they talk about "stability."


When Z>1 are repulsive forces dominant or are attractive forces dominant? According to PVreal = ZRT

If z >1 repulsive forces are dominant or are attractive forces?
Different teachers are telling contradictory answers and after referring a few books, I have become more confused.

Points for ATTRACTIVE FORCES are dominant:

At high pressure the molecules do not strike the walls with full impact as they are dragged back by attractive forces.
When molecules are farther away from each other(as Vreal > Videal) the attractive forces are dominant.
Repulsive force are dominant when he molecules are almost in contact .

Points for REPULSIVE FORCES are dominant:

At high pressure the molecules are close each other that attractive forces are negligible but molecular size is significant.Thus repulsive forces are significant
As volume of real gas is more than volume of ideal gas , the real gas is more difficult to compress.Thus repulsive forces are dominating.

There was a question where P = 12atm T = 350K and Z = 1.2(after calculation). It was given that repulsive force are dominant.But my teacher said it was wrong and it should be attractive forces.

Your arguments for attractive forces are not quite correct, particularly the first one, the other two are partly correct. The points for repulsive forces are more accurate but not completely correct.

The intermolecular potential energy is fixed by the type of molecule and varies with separation between molecules (e.g. a Lennard-Jones potential). The potential well formed by the interaction is short-range. It is effectively zero at about four to six molecular diameters decreases to negative values as the molecules approach  one another (negative energy is attractive) but rapidly becomes positive ( repulsive) when the molecules are effectively in contact.The potential is unaffected by temperature and pressure.

The ratio $p\overline{V}/RT$ is called the compressibility ratio Z and is unity for an ideal gas. ($\overline V$ is the molar volume $V/n$). Ideal gases have no intermolecular potential, or size. Real gases differ primarily in their size and magnitude of their intermolecular potential energy and so deviations from $Z=1$ are to be expected. At all temperatures and pressures the molecules have an attractive potential that tries to pair up molecules, and repulsion on collision when the charge on the electrons in each molecule force them to repel one another.

It is the competition between attraction, repulsion and kinetic energy at a give temperature that determines the shape of the $Z ~ vs. p$ plot.

For molecules at low temperature the line of the graph $Z= p\overline{V}/RT$ vs. p dips down further below $Z=1$ than it does for the same molecules at higher temperatures. This is because at low temperatures the  molecules have less kinetic energy ( so move more slowly) and are more influenced by attractive forces than at higher temperatures. This makes the actual volume of gas smaller than the ideal (as a larger fraction of molecules are paired up), which in turn makes Z smaller.

Similarly, at a given temperature, a larger negative Z is seen in gases with a larger attractive interaction, say $\ce{CO2,  CH4 or ethene}$ compared to those with smaller attraction, e.g. $\ce{He or H2}$ .

At high temperature the molecules have more kinetic energy and so attractive interactions are relatively less important, and repulsion more so as the kinetic energy is larger. This makes the actual volume greater than the ideal and $Z>1$.


Why do different elements form different types of carbides? What property of the elements make them form different types of carbides like:

$\ce{Be}$ and $\ce{Al}$ - $\ce{Be2C}$ and $\ce{Al4C3}$ (Methanides) contains $\ce{C^4-}$ ion

$\ce{Na}$ and $\ce{Ca}$ - $\ce{Na2C2}$ and $\ce{CaC2}$ (acetylides) contains $\ce{C2^{2-}}$ ion

$\ce{Li}$ and $\ce{Mg}$ - $\ce{Li4C3}$ and $\ce{Mg2C3}$ (sesquicarbides) contains $\ce{C3^4-}$ ion.

Boron - $\ce{B4C}$ (covalent carbides)

Titanium and tungsten - $\ce{TiC}$ and $\ce{WC}$  (interstitial carbides)

According to wikipedia,

  Carbides can be generally classified by chemical bonding type as
  follows:

  salt-like carbides
  covalent carbides
  interstitial carbides
  "intermediate" transition metal carbides.

This type of classification is based on the electronegativity of the element to which the carbon is bonded.

According to this site:-

  The most electropositive metals form ionic or
  saltlike carbides, the transition metals in the middle of the periodic
  table tend to form what are called interstitial carbides, and the
  nonmetals of electronegativity similar to that of carbon form covalent
  or molecular carbides.

Actually credits goes to @jatin  who gave me an important hint:-

  You might wanna consider diagonal relationship which accounts for
  similarities between those particular metals.

As you can see that Li and Mg, Na and Ca and Be and Al all form diagonal element pairs. So, they are expected to have similar properties. So, considering their electronegativity, the diagonal element pairs have almost same value (though Li and Mg electronegativity is quite different). So they form similar type of carbide.

(original picture)

As for the intestitial carbides, @bon said that

  Insterstitial carbides only form with large cations (Ti and W), as you might
  expect if the anion (very small) to fit in the spaces in the cation crystal
  structure.


At which temperature does carbon monoxide explode? I had doubt whether carbon monoxide explodes or not. I checked out this resource and found out that carbon monoxide explodes when heated. But they didn't mention the temperature at which it explodes. So my question is that at which temperature does carbon monoxide explode and also at which pressure?

At which temperature does carbon monoxide explode?

At a wide range of temperatures.

If mixed with air in a proportion between 12 and 75% it is explosive (assuming a there is an ignition source) - I think this is at standard temperature and pressure - 273K, 0.1 MPa.

Note that 1% of CO in air will cause death within 3 minutes due to its toxic effects.


Strong and Weak Acids/Bases I'm so confused... $\ce{HClO}$ is a weak acid. then why is its conjugate base, $\ce{ClO-}$, a weak base in water? shouldn't they be inversely proportional? shouldn't $\ce{ClO-}$,  be a strong base?

If $\ce{HClO}$ is a weak acid, that means it does not readily give up a proton and has a strong pull on them. So when it becomes a conjugate base, $\ce{ClO-}$, shouldn't it readily take protons and therefore be a strong base in water?

ALSO,
when an acid is dissolved in water, and some of that acid dissociates into $\ce{H3O+}$, is the Kw of water negligible and you only consider the Ka when finding pH? But that confuses me because I thought you use 14, the exponent of Kw when finding pH.

$K_\mathrm a\times K_\mathrm b = K_\mathrm w$.  Some teachers tell their students that the conjugate base of a weak acid is strong but it's not true.  What is true is that the stronger the acid, the weaker the conjugate base and vice versa.

Yes to your second question.  When calculating the pH of a solution, the hydronium ion concentration is usually controlled by the strongest acid in the solution.  This is the case in the type of problems found in lower level chem classes.  In a higher level class problems with acids close enough in strength so that both acids matter may be encountered.


Why does the solubility of Group II hydroxides increase and the solubility of sulphates decrease down the group? I know that, solubility of alkaline earth metal hydroxides increases down the group and solubility of alkaline earth metal sulfates decreases down the group.

I wish to understand the reason for these trends.

Okay, for this scenario, the two concepts that play a key role are the lattice enthalpy and the hydration enthalpy.
Lattice enthalpy is the energy released when one mole of a compound is formed from its constituent gaseous ions under standard conditions. Lattice enthalpy is dependent on the length of the ionic bond between the cation and the anion in the compound. If we look at Coulomb's law, we find that the attractive force between two bodies with different charges is inversely proportional to the square of the distance apart between their centres of mass, i.e. $F\propto 1/r^{2}$. A shorter bond would require more heat energy to be broken. Therefore, the shorter the bond, the higher the lattice enthalpy.

Hydration enthalpy is the energy released when one mole of gaseous ions is completely hydrated by water molecules under standard conditions. If two ions of different sizes were surrounded by an equal number of water molecules each, the smaller ion would be more hydrated compared to the larger one due to the size differential. Therefore, the smaller the ion, the higher the hydration enthalpy.

Let's deal with $\ce{OH-}$ first. While progressing down Group 2, the size of the cation increases since the number of shells increases to accommodate to the extra electrons. Due to this, the distance between the centres of mass of the cation and the anion increases and by definition, the lattice enthalpy decreases. The hydration enthalpy also decreases since the size of the cation increases. However, due to the square factor, the lattice enthalpy decreases faster than the hydration enthalpy. This is why the solubility of Group 2 hydroxides increases while progressing down the group.

Now let's look at $\ce{SO4^2-}$. The same thing applies to the cation while progressing down the group. However, the $\ce{SO4^2-}$ ion is shaped in such a way that the change in the distance between the centres of mass of the ions is far less significant compared to that in the case of Group 2 compounds containing $\ce{OH-}$ ions. Therefore, the lattice enthalpy decreases much slower when progressing down the group compared to the case of Group 2 $\ce{OH-}$ compounds. Again, the hydration enthalpy decreases the same way as it does in the case of Group 2 cations bonded to OH⁻ ions. Since the hydration enthalpy decreases faster than the lattice enthalpy in the case of Group 2 sulphates, the solubility of Group 2 sulphates decreases while progressing down the group.


Is a diamond a single molecule? Assuming a perfect diamond with no impurities. Would this diamond considered to be a single large molecule?
Browsing the interweb I found several opinions about this but did not find a clear Yes or No.

From what if gathered so far I'm pretty sure a diamond is a monocrystalline solid. Since all its parts are connected with covalent bonds I would assume that it should be considered a molecule even though it has no fixed number of atoms in its molecule formula.

Diamond is a covalent network solid, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).

Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc. - network solids form one of the two main classes of macromolecules, the other being polymers. You'll note that the Wikipedia article on macromolecules seems to imply that 'macromolecule' and 'polymer' are synonymous. They are not, at least not to those chemists working in the field.


Understanding the hybridization of cyclohexane? I'm having quite a fair bit of trouble comprehending the hybridization of cyclohexane. I understand that this is a ringed structure with single C-C bonds, therefore all the carbons should have sp3 hybridization right?

Apparently however, this is incorrect, but I am not sure why. I'm reading about how it not planar like benzene is, and adopts a chair/boat conformation which I am having trouble visualizing as well.

If someone could provide some insight on this topic, that would be great!

I'm having quite a fair bit of trouble comprehending the hybridization
  of cyclohexane.

Atoms can be hybridized. Not entire molecules. Just atoms.

  I understand that this is a ringed structure with single C-C bonds,
  therefore all the carbons should have sp3 hybridization right?

Yes, all the carbons in cyclohexane can be said to be sp3 hybridized.

  I'm reading about how it not planar like benzene is, and adopts a chair/boat conformation which I am having trouble visualizing as well.

In benzene all the carbons are said to be sp2 hybridized because benzene is flat. Carbons which are sp2 are in the same plane with their ligands. Sp3 carbons are not flat with respect to all their ligands.

  why are the sp3 carbons not planar (not flat)?

You seem to be adept at identifying sp3 carbons. So I won't go over that.

Within the framework of VSEPR theory -

1) Atoms are made of charged particles, namely, protons and electrons.

2) Protons are at the core (nucleus); electrons are distributed around the protons.

3) Electrons repel each other (like charges repel).

So if we have 4 atoms arranged around a central carbon atom the lowest energy conformation or the configuration which will keep each of the 4 atoms the furthest apart is tetrahedral. The angle between the four ligands in a perfect tetrahedron is about 109.5 degrees. If we constrain ourselves to 2 dimensions and force all the atoms to exist in the same plane then the best angle of separation we can achieve is 90 degrees (imagine carbon to be at the center of a square with each corner of the square representing an atom). Clearly, 109 degrees > 90 degrees so the tetrahedral configuration is favored.


How does carbonic acid cause acid rain when Kb of bicarbonate is greater than Ka? How does carbonic acid cause acid rain when $K_b$ of bicarbonate is greater than $K_a$? Should it not create an alkaline solution?

$K_a = 4.8 \times 10^{-11}\ (mol/L)$. $K_b = 2.3 \times 10^{-8}\  (mol/L)$. These are the values for $\ce{HCO3-}$.

These numbers are from a school book that I read, but it's not in English. It's called "Kjemi 1" by Harald Brandt

What is the ${K_a}$  of carbonic acid? Note that sources differ in their ${K_a}$ values, and especially for carbonic acid, since there are two kinds - a pseudo-carbonic acid/hydrated carbon dioxide and the real thing (which exists in equilibrium with hydrated carbon dioxide but in a small concentration - about 4% of what what appears to be carbonic acid is true carbonic acid, with the rest simply being $\ce{H2O*CO_2}$.

Either way, I find that the ${K_a}$ of the mixed carbonic acid is about $4.2 \times 10^{-7}$, which is greater than $1.0  \times 10^{-7}$, and this implies that a solution of carbonic acid alone should be acidic no matter what. This suggests to me that your numbers are wrong; would you mind sharing your numbers and their source if possible?

EDIT: I see that you have updated your numbers. Nonetheless, I believe that your ${K_a}$ for carbonic acid is wrong; that number looks suspiciously like the ${K_a}$ instead for hydrogen carbonate ion (or the bicarbonate ion). According to Wikipedia, the ${pKa}$ of carbonic acid, is 6.3 (and this is taking into account any aqueous carbon dioxide). Convert this to a ${K_a}$ value and we get about $5.0 \times 10^{-7}$. This is in-line with the value I obtained from a copy of Daniel C. Harris' Qualitative Chemical Analysis.

EDIT 2: I think you've realized your mistake; as you say, the values are for $\ce{HCO_3^-}$, which is the hydrogen carbonate ion.


What defines basis sets in computational chemistry? I have a simple question about the Linear Combination of Atomic Orbitals (LCAO):

$$
\psi_{i} = \sum_{r} c_{ir} \phi_{r}
$$

where

$\{\psi\}$: Set of molecular orbitals
$\{c\}$: Set of coefficients
$\{\phi\}$: Set of atomic orbitals

When we talk about basis sets, do we mean here the coefficients or the atomic functions?

In LCAO, it is the set of atomic orbitals (AOs) that is the basis, and the coefficients are the basis expansion coefficients. For example, take the hydrogen molecule, with 1 atomic orbital on each atom:

The lower energy MO, the bonding one, will be (excluding normalization):

$$
\psi_{\mathrm{1s}} = \phi_{\text{left}} + \phi_{\text{right}}
$$

and the higher energy MO, the antibonding one, will be (excluding normalization):

$$
\psi_{\mathrm{1s}^{*}} = \phi_{\text{left}} - \phi_{\text{right}}
$$

This means that for the bonding MO, $c_{\text{left}} = 1$ and $c_{\text{right}} = 1$, while for the antibonding MO, $c_{\text{left}} = 1$ and $c_{\text{right}} = -1$. If your basis was different (say, the atomic orbitals were of different radial extents for the two atoms), then the coefficients would be different to accommodate that. They aren't part of the basis set.

Since you tagged this with computational-chemistry, I assume you also want to know about basis sets in computational chemistry. Conceptually, they are identical to LCAO-MO bases, but what may be confusing is that each atomic orbital itself may be composed of multiple functions (called primitive functions), rather than just the single function for each AO as seen above. This leads to another set of coefficients, called contraction coefficients, describing how the primitive functions are linearly combined to form a contracted function.

An "implementation" detail is that the functional form of AOs is usually Gaussian functions, which have a parameter in the exponent (here, $\alpha$):

$$
\phi_{r}(x) = e^{-\alpha x^2}
$$

This means that to fully define an atomic orbital basis composed of contracted Gaussian-type orbitals (CGTOs), one needs both the contraction coefficients and exponents. They usually look like the following. For each angular momentum type, there may be one or more contracted functions, each composed of one or more primitives. For example, in the hydrogen definition below, there are 2 functions to describe s orbitals, one of which has three primitives and the other is not contracted. The contraction coefficients are in the first column and the exponents in the second column.

****
H     0
S   3   1.00
     13.0107010              0.19682158E-01
      1.9622572              0.13796524
      0.44453796             0.47831935
S   1   1.00
      0.12194962             1.0000000
****
C     0
S   5   1.00
   1238.4016938              0.54568832082E-02
    186.29004992             0.40638409211E-01
     42.251176346            0.18025593888
     11.676557932            0.46315121755
      3.5930506482           0.44087173314
S   1   1.00
      0.40245147363          1.0000000
S   1   1.00
      0.13090182668          1.0000000
P   3   1.00
      9.4680970621           0.38387871728E-01
      2.0103545142           0.21117025112
      0.54771004707          0.51328172114
P   1   1.00
      0.15268613795          1.0000000
D   1   1.00
      0.8000000              1.0000000
****

The last note is that when performing a Hartree-Fock or similar calculation, the contraction coefficients and the exponents don't change. That would mean the basis set changes. The basis set is fixed, it is the MO coefficients that change, the same coefficients that appear in the LCAO-MO equations.

Image taken from here. Basis set is Def2-SV(P).


How to purify sodium chloride table salt that contains sodium ferrocyanide I wish to obtain pure sodium chloride crystals from ordinary table salt.  But all proprietary table salts seem to contain the anti-caking agent sodium ferrocyanide which I should like to remove.

Don't.  Ferrocyanide is a good way of tying down cyanide ions so they don't do their usual bad things to you.  Trying to purify the salt could release the cyanide ions.


What is the name of this 3 open tube and where can I purchase one? I would like to do some plasma activated water tests.

Does anyone know the name of this three open tube along with where I can purchase one?

The screenshot came from this video if you need a better angle to look at it.  Here's the video, the screenshot came from
YouTube

The flask pictured in your question is simply a three necked round bottom flask. They're available commercially in a variety of sizes with different configurations of ground glass joints (personally, I've always found it easier to buy single neck round bottom flasks and then get our glass blower to append the joints where I want/need them).


Drawing and naming structures I am trying to learn how to draw and name skeletal chemical structures and would like some opinions and help if you may.

In the second (bottom) structure I can see that there are 6 carbons so hex would appear in the name of the structure somewhere, and I can also see that it is part of the methyl group but I have also seen that it looks very similar to 3-methyl-2-pentene which has thrown me a bit, could somebody walk me through this one step by step? Also this structure is a resonance structure if I am correct?

The other one (top) I am completely lost with, I know there is a carbon ring so that is 6 carbons plus the three which are stated, 9 in total would mean 'mon' would have to appear in the name and is also part of the methyl group, how would I go about naming this structure?

Much appreciated!

At first you need to identify, if there are any well know functional groups.

https://en.wikipedia.org/wiki/Functional_group

If there are any, then they will be used to create the final name, and this is done according with the rules of IUPAC nomenclature.

https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry

In your first example.
There is a benzene ring, which is a very common structure in organic chemistry. There are other more complex derivatives of it too, for this example or similar looking molecules the name would refer to either benzene or phenol functional group, this is a little nuanced.

So, in this case we know it is a benzene we also know, there are 3 methanes attached to it or more specifically methane derived groups called methyl groups.(Methane is alkane and a functional group of alkanes is called an alk(-yl) group).

3 methane groups = trimethyl

The last thing is to say where they are attached. This can be found by counting around C atoms on the ring 1, 2 and 3. Ring has no beginning and no end so the starting point is arbitrary, by convention the first group has number 1 and the rest follows.

1,2,3-Trimethylbenzenes

The second molecule.
Here, the idea is to pick the back bone of the molecule and then attach to it smaller and smaller parts like a tree with its branches.
So the trunk of the tree would be 5 C atoms so pentane.
Here the functional groups attached is also methyl group on 3th atom of the backbone pentane, so that would give 3-methylpentane
but..
there is still a double bond that has to be included in the name which  changes pentane to pentene, and lastly its nice to say where is it exactly therefore pen-2-ene

Putting it all together gives us,
3-Methylpent-2-ene


How do proton exchange membrane electrolysers work? I've seen lots of videos, animations, and a few Wikipedia pages, but I'm still trying to figure out a few things.

Is an electrolyte needed to make the water conductive like an alkaline electrolyser?
Does the cathode remain dry? If so, how do the protons get across the gap from the membrane to the cathode?
Are platinum cathode and iridium anode catalysts required for the reaction, or just needed to speed it up a small amount? Why are these specific materials used?

PEM process allows to increase the density of current. the distance between the cathode and anode is relatively small. The polymer has acidic residues that can transfer protons. You can use pure liquid water or even water vapor (Energy Environ. Sci., 2011, 4, 2993).

Cathode and anode remain wet. They touch the PEM.

Pt and Ir are not required, but they worth the money. PEM itself is relatively expensive, so there is not much point to save on Pt or Ir. However they can be replaced with a cheaper material. But then efficiency drop to alkaline electrolysis method.


Why can't helium be solidified at 'ordinary' pressures? According to the UC Davis ChemWiki Chemistry of Helium, helium has a comparatively unusual property, specifically:

  Helium is the only element that cannot be solidified by lowering the temperature at ordinary pressures.

'Ordinary' referring to standard air pressure (1 atmosphere).  In order to solidify, there needs to be a corresponding pressure increase, with a projected density of:

  $0.187 \pm 0.009~\mathrm{g~mL^{-1}}$ at $0~\mathrm{K}$ and $25~\mathrm{bar}$.

So why can't helium be solidified at 'ordinary' pressures?

When the temperature of helium gas is decreased to about 5.2 K, a phase transition to ordinary liquid helium ($\ce{He}$ I) occurs. The behavior of this liquid phase is normal and identical to any other liquid phase.

As temperature is decreased more (at moderate pressures), helium does not solidify. In fact, it undergoes a phase change to a second liquid phase known as $\ce{He}$ II at a temperature of approximately 2.17 K. As shown in the phase diagram below, the $\ce{He}$ II phase persists until absolute zero. Due to quantum effects, $\ce{He}$ II has remarkably unique properties. It is considered a superfluid.

For more details, please click here. Pressure is required to produce solid helium (25 atmospheres or more), because of the small mass and extremely weak forces between the helium atoms.


Enthalpy change of reaction Suppose you have the following equation:

$$\ce{2NH4SCN(s) + Ba(OH)2(s) ->Ba(SCN)2(s) + 2H2O(l) + 2NH3(g)}$$

You initiate this reaction using 15.22g of the $\ce{NH4SCN}$ (and excess of the barium hydroxide) in a boiling tube resting in 50g of water. The temperature rises by 11 degrees. Therefore energy = $4.2 \times 11 \times 50 = 2.31kJ$ and moles = $\frac{15.22}{76.1} = 0.2mol$

I would then assume that it would just be $\frac{2.31}{0.2} = 11.55kJmol^{-1}$

But the actual answer is 23.1 (double my answer) which I am guessing is because there are 2 moles of $\ce{2NH4SCN}$. I do not understand why this is the case, because you are using 0.2 moles and the energy raised is 2.31 so why the multiplication by 2?

The reason lies with the definition of the Heat of Reaction. It states that it is the amount of heat when 1 mole of the product is formed. As you noted, since 2 moles of $\ce{NH4SCN}$ is needed for each mole of product, then you need to multiply by 2.

Added:
If you change the number of moles of $\ce{NH4SCN}$ to 0.1 moles (to remove the stoichiometric 2), then your calculations would give you the answer.


What is the 'ancestral term' in the LS coupling scheme? On the physics.nist.gov site: http://physics.nist.gov/Pubs/AtSpec/node06.html

The table titled "Atomic structural hierarchy in LS coupling and names for the groups of all transitions between structural entities" contains the hierarchy of atomic LS coupling.  But there is a rather interesting footnote in the table: The letter $\gamma$ represents any additional quantum numbers, such as ancestral terms, necessary to specify a particular term.

What's ancestral term?  Is there an example of this in the literature?

"Ancestral Terms" are terms such as parent and grandparent terms that couple to form the final term.

For example, if you see $3d^7(^4F)4s4p(^3P°) ^6F°_{9/2}$

this means 7 3d electons couple to form a $^4F$ grandparent term

the 4s and 4p electrons couple to form a $^3P°$ parent term and

the grandparent and parent terms couple to form the final $^6F°$ term.

The above is one of several example given at the NIST page Notations for Different Coupling Schemes


Does freezing water do work? My brother and I were talking about how he needed to take his beers out of the freezer because they would burst open. This made me think of something no professor ever addressed... Obviously work is occurring on the beer can, but oxymoronically it's due to the removal of heat energy. The secret I imagine is in the phase change itself, which likely takes its own energy (which would normally be given off as heat) and uses it to expand (do work).

My questions extend on this idea...

What if we replace the beer can with a 5 inch thick steel capsule. Does the capsule prevent the water from freezing because there isn't enough room to expand?
What about if you take even more energy away? Would absolute zero finally break the capsule?
Can a strong enough container simply prevent water from phase changing to a solid altogether, regardless of temperature?
Normally increasing the pressure of a substance will cause it to freeze (in chemistry class my teacher made water boil at room temperature with a vacuum pump removing pressure). With water though, it expands while freezing... So paradoxically if you have water in a strong capsule and keep increasing the pressure (instead of temperature), will the water eventually freeze or simply remain very cold liquid?

Thanks!

Long story short, it goes as follows:

Yes, the freezing water does perform some work. You put a load on top of a bucketful of water, and as it freezes, it will lift the load all right.
Yes, a capsule would prevent the water from freezing, at least for a while (provided that the capsule is immensely strong). Then some of the beer (water?) will freeze. Then... see the next item.
No, the absolute zero would not finally break the capsule. Rather, it would turn water into some alternative kind of ice which is more dense than water. BTW, you won't have to get precisely to absolute zero to achieve that.
True, normally increasing the pressure of a substance will cause it to crystallize ("freeze"). Water is different in this regard, which is quite rare, but by no means unique.


Unfavorable reactions How can you make an unfavorable reaction go backwards?

From my understanding if we wanted a reaction at equilibrium to go backwards (i.e. form reactants from products), we would add more product into the solution.

But I'm not sure how the reaction being an unfavorable one changes the above statement.

After making the reaction go backwards, we have changed the reaction rate going backwards correct?

If the process we're discussing is an equilibrium, and if the forward reaction is "unfavorable" (I take that to mean that the products are of higher energy then the reactants), then there are still ways to shift the equilibrium in whichever direction you want.  For example

A <-> B + C

using LeChatelier's principle, if we increase the pressure in our vessel, we will push the equilibrium back towards A, the opposite will occur if we decrease the pressure.  Temperature, catalysts, reagent concentrations, etc. can all affect the position of an equilibrium.


Can we combine two reactions and then calculate the equilibrium concentrations? At a certain temperature, $\ce{N2O5}$ dissociates as:
  \begin{align}
\ce{N2O5 (g) &<=> N2O3 (g) + O2 (g)} & K_1 &= 4.5
\end{align}
  At the same time, $\ce{N2O3}$ also dissociates as:
  \begin{align}
\ce{N2O3 (g) &<=> N2O (g) + O2 (g)}  & K_2 &= \mathrm{?}
\end{align}

  When 4 moles of $\ce{N2O5}$ are heated in a flask of $\pu{1 L}$ volume, it has been observed that the concentration of $\ce{O2}$ is $\pu{4.5 M}$ at
  equilibrium. What are the equilibrium concentrations of the other products?

The combined reaction:

$$\ce{ N2O5(g) <=> N2O(g)  + 2O2(g)}$$
$$K_\text{combined} = K_1 \times K_2 = 4.5 K_2$$

There are 4 moles of $\ce{N2O5}$ initially. If $x$ moles of $\ce{N2O5}$ dissociate to give $x$ moles of $\ce{N2O}$ and $2x$  moles of $\ce{O2}$, then plugging in the values in the $K_c$ expression for this reaction we get
$$4.5 K_2 = \frac{4x^3}{4-x},$$
where
$$x = \frac{4.5}{2 \pu{mol}}.$$

Solving for $K_2$, I get $K_2=\frac{81}{14}$, which is incorrect.

Is it allowed to first combine the reactions and then use the combined reaction?

Karsten's insight is correct. Your mistake is in assuming that you have either $\ce{N2O5}$ or $\ce{N2O}$ and $\ce{O2}$.

There are four concentrations of interest: $\ce{[N2O5]}$, $\ce{[N2O3]}$, $\ce{[N2O]}$, and $\ce{[O2]}$.

How can you relate these variables?

Well, you can apply conservation of matter in terms of oxygen and nitrogen.

For nitrogen:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$

For oxygen:

$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2 \ce{[O2]} = 5\times(4\ \mathrm{M})\tag{2}$$

Then, there are the equilibrium expressions:

$$K_{1}=4.5=\frac{\ce{[N2O3][O2]}}{\ce{[N2O5]}}\tag{3}$$

$$K_{2}=\frac{\ce{[N2O][O2]}}{\ce{[N2O3]}}\tag{4}$$

If you substitute $\ce{[O2]} = 4.5\ \mathrm{M}$ into (2), (3), and (4), you end up with:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$
$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2\times(4.5\ \mathrm{M}) = 5\times(4\ \mathrm{M})\tag{2'}$$
$$4.5=\frac{\ce{[N2O3]\times(4.5\ \mathrm{M})}}{\ce{[N2O5]}}\tag{3'}$$
$$K_{2}=\frac{\ce{[N2O]\times(4.5\ \mathrm{M})}}{\ce{[N2O3]}}\tag{4'}$$

That's 4 equations and 4 unknowns, and it's all algebra from there.

Assuming I did that correctly, I got $$K_{2} = \frac{45}{7}$$


What does the peak separation of carbon dioxide IR rotational-vibrational spectrum corresponds to and why? Spectrum of $\ce{CO2}$ at $2349\space \mathrm{cm^{-1}}$   region

For some other molecules (for example $\ce{CO}$) the peak separation corresponds to $2B$ (where $B$ is the rotational constant wavenumber)
However for $\ce{CO2}$ when I calculated the peak separation and compared it to literature value, it seems approximately to equal to $4B$? Any idea why is this?

Also, what is the explanation for $\ce{CO2}$ spectrum being asymmetric? and why does the $P$ branch have larger peak separations than the $R$ branch?

(In this answer I am using the symbol $\require{begingroup} \begingroup \newcommand{\tb}[0]{\tilde{B}} \tb$ to represent a wavenumber instead of $B$, which I choose to define as an energy – not that we will be using it.)

  Why is the peak separation $4\tb$ instead of $2\tb$?

This has to do with nuclear spin statistics. Because oxygen-16 is a boson (spin 0), the wavefunction of the entire molecule (which comprises translational, rotational, vibrational, electronic, and nuclear spin components) must be symmetric upon exchange of the two oxygen-16 atoms.

The translational component is symmetric, as usual. The electronic ground state is ${}^1\Sigma_\mathrm{g}$, which is totally symmetric. The nuclear spin component must be symmetric by virtue of the fact that both oxygen atoms are spin 0; this means that there is only one possible spin state that they can be in (cf. hydrogen-1 has a spin of 1/2 and two possible spin states). Consequently, exchange of the oxygen atoms leads to no change in the nuclear spin wavefunction.

As a result, this means that the vibrational component times the rotational component must be symmetric; or in other words, the vibrational and rotational components must have the same parity.

Since the peak structure at $\pu{2349 cm-1}$ arises due to the antisymmetric stretch, the vibrational ground state $v = 0$ is symmetric with respect to exchange of oxygen, and must therefore be paired with rotational levels with even $J$, which are likewise symmetric. In a similar manner, the vibrational excited state $v = 1$ is antisymmetric, and must be paired with odd $J$ levels.

In $\ce{CO}$ where there are no nuclear spin statistics at work, the following transitions are all allowed in the $R$-branch ($\Delta J = +1$):
$$\begin{align}
(v,J) = (0,0) &\to (1,1) & \text{line 1}\\
(0,1) &\to (1,2) & \text{line 2}\\
(0,2) &\to (1,3) & \text{line 3}\\
(0,3) &\to (1,4) & \text{line 4}\\
\end{align}$$

and so on. However, in $\ce{CO2}$ the state $(v,J) = (0,1)$ is not permissible, and hence line 2 does not appear in the spectrum. Similarly, neither does line 4, or line 6, and so on; the separation between adjacent lines is therefore doubled.

  Why are the peak separations in the $P$-branch larger than in the $R$-branch?

This is because the rotational constant $\tb_0$ in the vibrational ground state is different from the rotational constant $\tb_1$ in the first vibrational excited state. The rotational energy levels (in wavenumber units) are therefore by:

$$\tilde{G}(v,J) = v\omega_0 + \tb_vJ(J+1)$$

where $\omega_0$ is the wavenumber of the pure vibrational transition (i.e. the band origin). For the transition $(v,J) = (0,J) \to (1,J-1)$ in the $P$-branch, the associated energy is therefore

$$\tilde{P}(J) = \omega_0 + \tb_1J(J-1) - \tb_0J(J+1)$$

The separation between adjacent lines in the $\ce{CO2}$ spectrum is then

$$\begin{align}
\tilde{P}(J) - \tilde{P}(J+2) &= \tb_1J(J-1) - \tb_0J(J+1) - \tb_1(J+1)(J+2) + \tb_0(J+2)(J+3) \\
&= 4J(\tb_0 - \tb_1) + 6\tb_0 - 2\tb_1
\end{align}$$

You can easily verify that if $\tb_0 = \tb_1 = \tb$, then this simplifies directly to $4\tb$, as expected.

However, because $\tb_0 \neq \tb_1$, the gap between adjacent lines acquires a $J$-dependency, because the term $(\tb_0 - \tb_1)$ is nonzero. In the case of the asymmetric stretch, it turns out that $\tb_0 > \tb_1$ (can you figure out why? It's best explained using the moment of inertia $I$, since $\tb = \hbar/8\pi cI$). Therefore, as $J$ increases (i.e. as we move farther left along the spectrum), the separation between lines increases.

An exactly analogous calculation will explain why the lines in the $R$-branch tend to "bunch":

$$\tilde{R}(J+2) - \tilde{R}(J) = 4J(\tb_1 - \tb_0) + 6\tb_1 - 2\tb_0$$

$\endgroup$


Determine which orbitals will form hybrids with one another I've been teaching myself chemistry, so any help is greatly appreciated.  I've been reading an online tutorial that claims the two orbitals that merge in Aluminum trihydride are  1 orbital of 2s and 2 orbitals of 2p.  This confuses me, because I would expect the orbitals of the electron configuration would form a hybrid with one another, that is, Al electron configuration is [Ne]3s^2 3p^1.  Therefore, why do the lower orbitals form a hybrid? Also, what is the rule in general to determine which orbitals will form hybrids?

Let's take your questions in reverse order.

  what is the rule in general to determine which orbitals will form
  hybrids?

The "general" rule is that you mix one atomic orbital for each bond you want to form.  In methane, carbon has 4 valence electrons and is forming 4 bonds so it mixes 4 atomic orbitals (1 2s and 3 2p) generating 4 $\ce{sp^3}$ hybrid orbitals, and it places 1 electron in each orbital.

In the case of aluminum trihydride, aluminum has 3 valence electrons and will form 3 bonds, one to each of the 3 hydrogen atoms.  Therefore, aluminum must mix 3 atomic orbitals (1 3s and 2 3p) generating 3 $\ce{sp^2}$ orbitals and places 1 of its electrons in each of these orbitals.

  why do the lower orbitals form a hybrid?

A 2s orbital is relatively close in energy to a 2p orbital and the same applies for a 3s and 3p orbital.  The closer in energy 2 orbitals are, the more effective is their mixing.  Therefore, 3s and 3p orbitals can mix (form new hybrid orbitals) effectively.  The reason orbitals mix in the first place is to form new orbitals that have greater directionality and can therefore form stronger bonds.  After all, we could form 3 bonds to aluminum trihydride by just bonding to the 3s orbital and separately to 2 3p orbitals.  Two of these bonds would be oriented 90 degrees from one another and the third bond would be fluxional, moving around the surface of a (3s) sphere. But hybridizing the aluminum so that we can form 3 $\ce{sp^2}$ orbitals and use them for bonding with hydrogen produces a more stable molecule and one with the observed geometry (3 bonds in a plane with a 120 degree angle between them).  If hybridization doesn't produce stronger bonds and a molecule that is more stable, then the molecule wouldn't hybridize. $\ce{PH3}$ is an example of such a molecule, the central phosphorous atom remains unhybridized.


How to I identify the diene and dienophile through retrosynthesis of a given compound? I'm stuck while finding what to use to make this compound:

My first attempt was to use a triple bond as dienophile and a ring as my diene, as shown below -

But I'm not sure if this is right.

The $\ce{CO3}$ substituents are questionable. Let's assume that this is $\ce{CO2Me}$ instead!

Since this is homework, I won't provide a full solution but:

Note that double bond patterns may isomerize (in the direction of conjugation).
Alkynes can serve as dienophiles too!


Do covalent salts produce sodium carbonate extract? In case of qualitative analysis of anions, sodium carbonate extract is prepared for insoluble salts.

It is done by heating (for 10 minutes) an aqueous solution of a mixture of the solid salt and sodium carbonate in 1:4 ratio (by mass). As far as I understand, the principle behind preparing sodium carbonate extract is to prepare a clear solution containing the anion by precipitating the cation as insoluble oxide or carbonate.

But is this process applicable for salts that do not ionize at all in water, or covalent salts like $\ce{SnCl2}$, $\ce{SbCl3}$ etc.? I have never tried this myself.

Soda extract is done to remove multivalent cations from your sample solution, by precipitating them as (mixed) carbonates and hydroxides.

Antimony or tin chlorides are well soluble even in water, and even if the solids have a strong covalent character, they are still ionic in solution.

The sulfates of barium, strontium and calcium are however very badly soluble, so it doesn't work for those. You can however do it in the melt, instead of solution. After cooling and powdering that melt, you add water. What dissolves are sodium sulfate and soda. Add $\ce{BaCl2}$ to detect the sulfate. The residue are the alkaline earth metal carbonates, which dissolve in $\ce{HCl}$.

(Please check your textbook/lab manual for the actual procedure. I have Jander, Blasius, Einführung in das anorganisch-chemische Praktikum. Perhaps someone else can recommend a good English book.)


Finding reaction rate constant for second order equation in Wolfram Mathematica I’m working in Mathematica for my master’s project and I can’t work out how to find $k$ in the second order rate equations where I have a concentration vs time graph for $\ce{AHA}:$
$$\mathrm{rate} = k[\ce{AHA}][\ce{H}].$$
I have the starting concentrations for $\ce{AHA}$ and $\ce{H}.$

The general second order equation is
$$\frac{\mathrm d[A]}{\mathrm dt}=-k[\ce{A}][\ce{B}]\tag{1}$$
with $[\ce{A}]_0,$ $[\ce{B}]_0$ the initial amounts. At a time $t$ there are $[\ce{A}]_0-x$, and $[\ce{B}]_0-x$ remaining if amount $x$ reacts. Thus
$$\frac{\mathrm dx}{\mathrm dt}=k([\ce{A}]_0-x)([\ce{B}]_0-x).\tag{2}$$
You should be able to integrate this from $0\to x$ by using partial fractions or directly using computer algebra. The result is
$$\ln\left(\frac{[\ce{A}]_0-x}{[\ce{B}]_0-x}\right)=k([\ce{A}]_0-[\ce{B}]_0)t+\ln\left(\frac{[\ce{A}]_0}{[\ce{B}]_0}\right),\tag{3}$$
so that you can plot (as Maurice has pointed out) the left hand side vs $t$ and obtain the rate constant $k$ from the slope $k([\ce{A}]_0-[\ce{B}]_0)$ in units $\pu{dm^3 mol^-1 s^-1}.$


Do all adiabatic processes involve a change in temperature? I learnt that adiabatic processes involve no heat exchange between the system and the environment. Does this mean that all adiabatic processes involve some changes in temperature then?

Thanks a lot!

The first law of thermodynamics says,
$$\Delta U = Q - W$$

By definition, no heat is exchanged between the system and the surroundings in an adiabatic process. Therefore, the 1st law reduces to,

$$\Delta U = -W$$

In an adiabatic process, $PV^\gamma =const. $ has to be followed. If you change pressure, the above equation implies that volume changes. If the volume changes, there will be some work done which in turn will change the internal energy of the system. Therefore, the temperature changes.

You can also come to the same conclusion using ideal gas equation.

We know the ideal gas equation,

$$PV = nRT$$

If there should be no change in temperature, $PV$ should remain constant.

But in case of an adiabatic process, we know that
$$PV^\gamma =constant$$

So if you change one of those parameters, the product $PV$ will obviously change (so as to keep $PV^\gamma$ constant). Hence, temperature always changes in an adiabatic process.

*I have ignored ridiculously insane cases where matter can exit/enter the system through an adiabatic wall (which isn't possible). If you could discover a wall that allows matter to move across and does not allow heat to conduct through it, you deserve a nobel prize.

*Free expansion and mixing gases are two processes where the process is adiabatic and there is no temperature change.


Nucleophilic aromatic substitution on pyrimidines: C2 vs C4 selectivity [There are many similar questions on Chemistry.SE (e.g. 1, 2) and a quick search will reveal all of them, but nothing on this in particular.]

The reaction of 2,4-halopyrimidines with nucleophiles generally leads to selective displacement of the halide at the 4-position:1

In their textbook, Joule and Mills don't explain why, though, simply saying that it "is the normal situation for nucleophilic – 4-chloro > 2-chloro".

I don't see any obvious difference in the stability of the anionic intermediates, either (the so-called Meisenheimer complexes).

What can explain the regioselectivity of the above reaction?

Reference

Joule, J. A.; Mills, K. Heterocyclic Chemistry, 5th ed.; Wiley: Chichester, U.K., 2010, p 257.

An explanation can be provided by frontier molecular orbital theory.  The LUMO coefficient is higher on the C4 vs C2.  This is a general effect meaning that an EWG para to the leaving group is more reactive than ortho.  See 4-halopyridines vs 2-halo and 4-nitrobenznes vs 2-nitro.  There could also be an effect from the repulsion between the nitrogen lone pairs and the approaching electron rich nucleophile, which would be greater in proximity to 2 nitrogens rather than 1.


Is water a possible fuel for jet engines? Jet engines can run on almost any fuel, and the operating temperatures of modern jet engines' hottest sections are anywhere between 3000 and 3150 degrees F (1648 and 1732 degrees Celsius). Does that mean that a hydrogen on-demand system could work on modern jets?

Water is pumped and heated first by the exhaust section, then directed towards the hotter sections of the engine (when hot enough to not cause cooling and lower engine efficiency) where it's broken down into hydrogen and oxygen at a heat above 1472 degrees F (800 degrees Celsius), then those gases are pumped into the engine for combustion.

The advantages are that firstly, water is abundant and therefore cheap. Even sea water could be used because at those temperatures it's easy to design a system that would get rid of the impurities that would otherwise corrode critical engine parts.

Secondly, it would save on manufacturing costs given that non-heat critical parts in the exhaust section would not need to be made of sophisticated and expensive materials and alloys given the cooling effect of water.

Thirdly, the costs of the fuel weight would be reduced given that the energy density of hydrogen is twice that of fossil fuels, so less would need to be carried. And most importantly, the environment problem would be solved in aviation given that there would be little or no carbon dioxide emissions.

It can't work because of the fundamental thermodynamics

What you are proposing is, basically, the plane carries water; the water is broken down into its components, hydrogen and oxygen; the components are recombined by burning them as fuel. Burning hydrogen and oxygen is a perfectly good way to create a lot of heat. But it doesn't much matter how you break the water apart into hydrogen and oxygen, the thermodynamics of the reaction won't work.

The problem is simple: you need to have a source of energy to split the water apart. In chemistry we know the energy levels of the reactants and the products and we can work out whether energy is released or stored in a reaction. Burning hydrogen and oxygen releases a lot of energy, but by the rule of thermodynamics, breaking water apart to its components requires the input of exactly the same amount of energy. You can't get round this. Worse, in the real world, there are losses at every conversion step so you can't even break even (ain't things unfair!)

In your plane you could, in principle, split water and burn it in the engine for propulsion. But you would need to have some other source of the vast amount of energy required to split the water. That implies both another fuel and another engine. In reality they would vastly outweigh any imagined savings in weight and cost.

Even if you could build some sort of SciFi engine that both splits water and then burns it again you would still be nowhere: the entire output of the burning would be required for the splitting with nothing left over to generate thrust (and in the real world there are losses so your plane would rapidly drop out of the sky).


Why does ester hydrolysis occur in basic medium Ester hydrolysis occurs in basic medium and this is the medium which is usually preferred for the same. But I am facing a very fundamental doubt and not able to understand the reason.
Ester Hydrolysis goes like this:

$\ce{RCOOR' + OH- <=> RCOOH + OR'-}$

Now according to me $\ce{OR'-}$ should be a better base because of +I inductive effect of alkyl group. If it is a better base then why does it not attack the acid back again !? In short, why does the equilibrium not reverse ?

In basic medium, there is a huge amount of $\ce{OH-}$ and they will deprotonate the carboxylic acid as soon as it forms.
$$\ce{RCOOR' + OH- <=> RCOOH + R'O-}$$
$$\ce{R'O- + H2O <=> R'OH + OH-}$$
$$\ce{RCOOH + OH- -> RCOO- + H2O}$$
The last deprotonation step is almost irreversible and consumes the base. This is why, an excess of base is usually added for ester hydrolysis.
The $\ce{RCOO-}$ is a negative ion, so it repels any nucleophile that comes close to it, so nucleophilic attack cannot happen, and the reverse reaction is prevented (as Maurice has mentioned in the comments). In essence it is this last step that drives the reaction to the product side.


What does an endpoint saturated signal mean. How does it relate to equilibrium constant? I have had a question that has been bugging me for a couple of years in my graduate school career, and I was wondering if perhaps someone could help explain it to me. Given an endpoint assay like this one I've shown below (an ELISA assay, but I assume it could work for any endpoint), how does the endpoint signal (the signal at the highest concentration) relate to binding? And how does $EC_{50}$ (the concentration between the maximum and minimum signal) relate to the equilibrium constant?

For years, biologists have used the $EC_{50}$ to give a rough approximation for the equilibrium binding constant $K_d$. However, it always bothered me that all curves features on the graph could produce nearly the same $EC_{50}$, but with different levels of maximum absorbance. For instance two of the curves are saturated at around 1.2 and the others are around 2.5.

Do all of these curves have roughly the same $K_d$, or is it worrisome that we don't get equivalence signal for the maximum concentrations?

$EC_{50}$ is concentration of a bioactive molecule which gives half the maximum response. You should note that the response is bound by a maximum which means this dose-response curve follows a saturation kinetics.

Saturation kinetics can be modelled in many ways and Hill and Michaelis-Menten kinetics are the most commonly used models in biochemistry. Again, a point to note is that many mathematical functions/models do not explain the underlying mechanism. They just mimic the behaviour of the system.

Lets assume that the antibody-response follows a two-step kinetics like Michaelis-Menten.

$$\ce{Ab + Ag <=>[k_a][k_d] AbAg ->[k_{res}] Response }$$

Lets call $R_{max}$ as the maximum response rate. Like in the case of MM kinetics, the response rate $R$ would be (see the wiki page for the derivation):

$$R=R_{max}\frac{[Ag]}{K_m+[Ag]}$$

Here $K_m$ is the concentration of antibody which would give half the maximum response rate (half response rate times times a sampling time would give you the $EC_{50}$).

Now, what $K_m$ really is depends on the approximation of the model:

Equilibrium model: this assumes that the binding and unbinding reactions are much faster than the response reaction and are hence at equilibrium (wrt response dynamics). In this case $K_m$ would be same as the dissociation constant.
Quasi Steady State model: this assumes that enzyme-substrate complex is at a quasi steady state (QSS) i.e. it doesn't change with time. In the case of enzyme substrate reactions, the substrate gets consumed and so the QSS would result from both the binding-unbinding reactions and conversion reaction. In this case $K_m=\dfrac{k_d+k_{res}}{k_a}$. However in this case the antibody is not consumed to give the response. So QSS would be equivalent to equilibrium approximation.

This would also apply for the Hill kinetics in which response can be described as:

$$R=R_{max}\frac{[Ag]^n}{{K_m}^n+[Ag]^n}$$

where $n$ is an integer which represents co-operativity in binding i.e. one bound antibody enhances the binding of the next (total number of binding sites is the upper bound of $n$).

Note that equilibrium model may not be applicable for drugs that can get consumed in a reaction. Also, not all drug induced reaction would follow this simplified model. There are other models for co-operative binding too.

Therefore, before concluding anything about kinetics from an empirical measure like $EC_{50}$, we need to understand (or at least assume) the underlying model.


How to predict precipitate upon evaporation from calcium ions and either chloride, carbonate, or nitrate counterions? An aqueous solution containing $\ce{Ca^2+}$, $\ce{Cl^-}$, $\ce{CO3^2-}$, and $\ce{NO3^-}$ is allowed to evaporate. Which compound will precipitate first?

The correct answer is $\ce{CaCO3}$. I had guessed that, and I know nitrates are soluble, so that wouldn't precipitate, but why the $\ce{CO3^2-}$ and not the $\ce{Cl-}$? Is it because of the equal and opposite charges between them? Does electronegativity play a role?

This question is testing your knowledge of general solubility trends, which largely need to be memorized. Here is a fairly comprehensive list, and there are many others scattered through the internet.

You are correct that nearly all nitrates are soluble, so that would be a bad guess.

The halides are generally pretty soluble. There are more exceptions than for nitrates, but as a rule halide salts tend to be soluble.

Carbonates however are frequently insoluble. The Group I cations tend to be an exception, but not the Group II cations, like $\ce{Ca^2+}$.

So, the general trend in solubility for the three anions in your question is $\ce{NO3- > Cl- > CO3^2-}$, so you would expect $\ce{CaCO3}$ to precipitate first.

We can verify this against the known solubilities of each of the salts at room temperature:

$\begin{array}{rr}
\pu{\ce{Ca(NO3)2}}: & \pu{1212 g/L }\\
\pu{\ce{CaCl2}}: & \pu{75g/L}\\
\pu{\ce{CaCO3}}: & \pu{0.01g/L}
\end{array}$

So, the quantitative solubility data confirms the original prediction and $\ce{CaCO3}$ will clearly precipitate first.


Green kerosene in copper container goes colorless I recently got hold of an old blowlamp that had not been touched for more than 30 years.
The fuel tank of the blowlamp is made of brass.
I poured the old kerosene from the blowlamp into a clean polyethene flask, the kerosene was clear and quite green in color. The flask got about half full.

The next day the kerosene had turned colorless and there was a brown  precipitate. The flask was not shaken or subjected to strong light.

What happened here? Something that had been stable for years suddenly reacting. If it is a reaction with oxygen, does oxygen diffuse that easily through kerosene?

Was the green substance an inorganic copper salt dissolved in the kerosene or was it some kind of organic "copper kerosenate"?

The brass is a metal mixture of cooper and zinc. Copper compounds are known to form blue and green salts.

Examples: Copper Chloride

I believe that green color it's a copper compound and the copper reacted with oxygen (air) and the preciptate is copper oxide.

Also, the oxygen solubility in hidrocarbons it's around 60/70 ppm¹.
¹Corrosion for everybody - Alec Groissman


Racemization of biphenyl compounds How does the effect of heat and effect of substituents affect racemization of biphenyl compounds?

Preparation: Take the parent compound per se, biphenyl.  The bond keeping the two phenyl rings together leads to one carbon atom we call C1, the other C1'.  Now if you label the other carbon atoms in each  phenyl ring seperately in clockwise direction, you'll get C2, C3, C4, C5, C6 in one cycle; and C2', C3', C4', C5', C6' in the second.  (I haven't a credit to upload a picture yet.)

Now: The larger, the more sterically demanding substitutents on C2, C6, C2' and C6' are (condition), the higher the energetical barrier is to interconvert the conformers into each other.  Eventually, this may lead to atropisomerism, and the isomers are atropisomers.

Note, please, sometimes there may be already a preference for one or the other conformation.  If, for example, you substitute C2 and C2' of biphenyl by nitrogen, free molecules of this compound 2,2'-bipyridine energetically favour a conformer with the nitrogens as far as possible separated (similar to a s-trans conformation).  Offering a cation to bind, to complex to, however may lead to a preference towards the s-cis-like conformation.


Nucleophiles and bases (elimination and substitution) Is it safe to say that all elimination reactions (E1 and E2) involve bases (since in elimination reactions, a hydrogen is being removed, and thus something that accepts a $\ce{H+}$ is called a base)

Whereas, In substitution reactions, bases are not involved.

To summarise:

substitution reactions- can involve bases/nucleophiles/both
elimination reactions- must involve a base, but the base can also be a nucleophile (i.e. elimination reactions cannot occur if you only have a nucleophile)

There are two possible definitions of base:

Substance which accepts a proton throughout the course of the reaction (catalyticly or stoichiometricly). This is basically the Brønsted definition. To accept a proton, the molecule in question needs a lone pair whose energy level and orientation is in a good range for a proton to attach (e.g: ammonia: good; phosphane: bad). We could also extend the definition Lewis-style.
Substance which generally may accept a proton in reactions. This is still basically the Brønsted definition only without the requirement of it actually happening. The molecule needs a lone pair whose energy level and orientation is in a good range for a proton to attach even if it doesn’t happen. We could also extend it to a Lewis-style definition.

Applying the first definition, nucleophilic substitution reactions usually do not require bases. However, the molecules that attack are generally bases: A nucleophile generally needs an accessable lone pair, as would a base.

Consider the case of an alcoholate. If it is methanolate, it is small and can attack nucleophilicly. However, we also know that it can act as a base to be protonated to generate methanol. Depending on the substrate, it would choose to do substitution or elimination reactions.

If the alcoholate is tert-butanolate, we have a large molecule. Technically, it still could attack as a nucleophile. However, it is usually too big to attack and rather abstracts protons to allow elimination reactions.

You cannot really draw a line. Both are bases, both could be nucleophiles.


Conversion of propane to methane I had a question in my textbook to convert propane to methane. The basic idea was chlorination in presence of sunlight. However, two products are obtained in this case - namely 2-chloropropane and 1-chloropropane. So it didn't work out at all.

I came up with another idea i.e. dehydrogenation, but it has not been taught to us at all. In addition the dehydrogenation reactions I searched on google were too complex. So, I hope there's another method for this.

Can anybody help me to sort this out?

This probably defies the purpose, but still. Two-step process with steam reforming and Fischer–Tropsch synthesis

$\ce{C3H8 +O2 +H2O -> CO + H2 + CO2}$

This step is usually performed with methane over nickel catalyst, but should work with propane as well. Remaining water is condensed and carbon dioxide is separated using triethanolamine as absorbent. Then a reverse reaction is performed

$\ce{CO + 3 H2 -> CH4 + H2O}$

depending on catalyst, and exact conditions it is possible to get methanol, linear alkane/alkene mix or methane. AFAIK, $\ce{Ni}$ catalysts favor methane formation, while cobalt and iron favor linear alkanes and terminal linear alkenes.


Chemicals that form coloured complexes with Pb2+ I’m currently conducting a photo-spectrometric analysis of lead concentration in water and was looking into using Eriochrome Black t as an indicator for Pb2+ (due to the complex forming a reddish colour). Because Eriochrome is relatively expensive, would there be any alternate chemicals that can complex lead in solution of the form Pb(OH)NO3, such that colour is present. This is because the source of lead is from a dissociation reaction of lead (II) nitrate when it is dissolved in water. The resulting lead is complexed.

All the best!

In the old book from Hopkin and Williams, Organic Reagents for Mineral Analysis. London (1933), the following information can be found in the Chapter "Diphenylthiocarbazone, or Dithizone".

Diphenylthiocarbazone is a black powder, insoluble in water, but soluble in organic solvants, giving green solutions. Its main application is its reaction with lead. Lead ions produce a red color to its CCl4 solution. To 10 mL Lead aqueous solution, 1 mL KCN 10%, and two drops of ammonia are added. Add 2 mL of a dilute solution of diphenylthiocarbazone in CCl4 (50 mg/L). Stir. A reddish color must appear if the sample contains 0.001 mg Pb. A similar color appears without lead, due to the solubility of the reagent in CCl4. Addition of KCN removes totally any excess of reagent in the aqueous phase.

The only drawback about this method is the necessity to use KCN, which is forbidden today. I have found in the same book a small article about benzidine, where they say that lead solutions give a blue color when mixed with benzidine, if a solution 0,05% benzidine in acetic acid 10% is used.

Lead can also be detected and quantitatively determined by 8-hydroxyquinolein. But the book mentions only the following bibliographic  references :
V. Marsson, L. Haase, Ueber die Bestimmung des Bleis mit Hilfe von o-Oxychinolein, Chem. Ztg, 52, 993, (1928)
V. Hovorka, Sur le dosage du plomb au moyen d'ortho-oxyquinoléine, Coll. Czech. Chem. Comm, 9, 191 (1937)


Preventing the gas when Hydrogen peroxide is added to a detergent In organic detergents, when I add Hydrogen peroxide to increase the stain removing feature a gas is produced.

How should I prevent the gas production process?

While hydrogen peroxide is fairly stable if pure in water solution, it can quickly break down to water plus oxygen in the presence of a catalyst such as transition metal ions--probably what is causing the gas evolution you observe. One approach you might try is to test other types of detergent - perhaps you can find one that does not catalyze breakdown of peroxide. Liquid facial soap or glycerine soap--or even dishwashing liquid--might be worth a try.

Another approach would be to mix the peroxide in right before use rather than try to store the mixture--the more dilute, the better.


Draw a diagram to show the shape of Chloromethane and explain why it has this shape. I know it has this shape, but I don't know how to explain it. Any help is appreciated, thanks.

Carbon has in his outer shell 4 electrons and needs 4 more electrons to get the full shell like the noble gas Ne.

Chlorine has in his outer shell 7 electrons and needs one electron to get the full shell like the noble gas Ar.

I give you an answer because the question about the tetrahedral structure of Chloromethane contradicts the perceptions of physics about a s- orbital with 2 electrons and 3 p-orbitals with 2 electrons in each p-orbital. To avoid the contradiction between the claimed s- and p-orbitals and the empirical founded tetrahedral structure the sp-hybridizations were postulated. The carbon in Chloromethane has four sp3-hybridizations with all atoms around.

If you have a look around you will see that the hybridizations are the norm. Let the s- and p-orbitals be a toy for physicists.

What you have to check is, always to get pairs of electrons. That electrons like to be pairwise was expressed by the Pauli exclusion principle. It is a principle and not an explanation why this happens this way in atoms. But if you need an imagination of how the electrons form pairs you could think for yourself of the electrons as tiny bar magnets (they indeed have a magnetic dipole moment) and two of them form a magnetic pair. This pairs around the carbon will be in some equilibration to each other. Since chlorine has a lot of other electron pairs it pushes the 3 hydrogen atoms. So you get this shape:

All images from Wikipedia


Lewis structures of molecules containing oxygen As far as I know, when writing Lewis structures for molecules, we need to consider the octet rule and the valence electrons each atom has.

But, if we take for example, the molecule $ Cl O_2^- $ , we have the following Lewis structure: (as it appears here )
O - Cl - O with 6 valence electrons surrounding each oxygen atom, and 4 valence electrons surrounding the chlorine atom.

My problem is:
The oxygen atom has 6 valence electrons. By forming a chemical bond with the chlorine atom, it shares one electron, which leaves it with a total of 5 non-bonded electrons. So, how can it be that in the Lewis structure of this molecule, it has 6 non-bonded electrons and one shared electron? (or a total of 7 electrons! )

If you consider the $\ce{ClO2-}$ as coming together from neutral atoms (plus the extra electron), all the bonding electrons "come from" the chlorine. This is commonly called a 'dative covalent bond' (or just 'dative bond' for short), and is sometimes drawn like this, instead of just a line:

As this image suggests, you can think of it as an atom with a lone pair bonding with an atom with no spare valence electron.


Will Avogadro's number change if definition of atomic mass unit is changed? I know that the relation between atomic mass unit and Avogadro's number is a reciprocal relation.
My question is if the definition of atomic mass unit is changed, suppose instead of 1/12th mass of carbon-12 isotope, we consider 1/24th mass of carbon-12 isotope, will the Avogadro's number change?
And if Avogadro's number changes will that change the mass of a mole of atoms of any element?

Simple and actual answer:
The current definition of Avogadro's number doesn't depend on the unit of mass. They are completely independent quantities. The Avogadro's number is currently simply defined as $\pu{6.02214076e23}$. It's just a number which is important enough to get a special name.
The answer you are probably looking for:
Historically the Avogadro's number was defined in such a manner that it did depend on the unit of mass.
At that time the Avogadro's number was defined as the number of atoms present in $\pu{12 g}$ of $\ce{^{12}C}$ isotope of carbon.
Now as you can see even if we defined an atomic mass unit as $1/24$ of the mass of an $\ce{^{12}C}$ atom instead of $1/12$, it won't make any difference to the Avogadro's number as it was defined on the basis of grams of carbon and not atomic mass units of carbon.
Now coming to your second question of whether this would have changed the mass of a mole of a substance. Yes it would have changed that, all the molar masses would have been double of as known currently.
Note:
All this discussion is just about how someone wrote a definition and the loopholes someone can detect from it using language.
For example if we really would have defined an atomic mass unit in this different way, we would have also defined Avogadro's number as number of atoms in $\pu{24 g}$ of $\ce{^{12}C}$. Even $\ce{^{12}C}$ would have been written as $\ce{^{24}C}$!
The takeaway is that the scientific idea and origin of all these units was(and is) always there  but it's just the nuance of defining it in such a way that it doesn't create such problems was the problem. That's the only reason all the standard units were restandardized recently. It's a sad reality that schools are still teaching the old definitions even after we have said them a good bye.


Why does the trimethylammonium cation have a larger -I effect than the ammonium cation? Why does $\ce{N(CH_3)_3^+}$ have a larger $-I$ effect than $\ce{NH_3^+}$?

Since the methyl group is good at donating electrons it would stabilize the charge on nitrogen atom, decreasing its potential to withdraw electrons hence its $-I$ effect.

I can't figure out why its the other way around. Can someone help me understand this?

$\ce{N(CH3)3+}$ has more -I than $\ce{NH3+}$
, this irony happens because if you consider
$\ce{N-CH3}$ bond and $\ce{N-H}$ bond,  which is more polar?
Of course $\ce{N-H}$ bond will be more polar  due to more difference in electronegativity  value,  so that implies electron density will be more on $\ce{N}$ in $\ce{NH3+}$ than $\ce{N}$ on $\ce{N(CH3)3+}$.
And therefore since electron density is less on $\ce{N}$ of $\ce{N(CH3)3+}$ , it will have greater -I effect.


What nomenclature do names like fumarate and malate derive from? When reviewing the citric acid cycle, I find names like glutarate and succinate to be helpful in thinking about the structure of the molecules because I was taught about the sequential dicarboxylic acid naming scheme: oxalic (2 carbon), malonic (3C), succinic (4C), glutaric (5C) etc.
But I often struggle with fumarate and malate because I have not learned the numbering or nomenclature system that these names are derived from. Can someone point me in the right direction so that I can learn the background behind these names?
Knowing that will be helpful to my recall of the molecular structure. Then I can stop memorizing and start understanding.

Unfortunately, you will have to memorize the structures and the names because there is no chemical structure hidden in those names.
From Elsevier's Dictionary of Chemoetymology

fumaric acid $\ce{C4H4O4}$, derived from the genus name Fumaria (fumitory), from fumus (Latin:
smoke)

Fumaria plant

malic acid $\ce{C4H6O5}$, derived from malum (Latin: apple)

The same goes for all other common names like oxalic acid, acetic acid etc.


Is the axial position more suited for a lone pair or an electronegative element? In one question, I found that it is given $\ce{XeF2}$ and $\ce{CO2}$ have the same shape.

$\ce{CO2}$ is linear. In $\ce{XeF2}$, if the fluorine is placed in an axial position, then only shape is linear. But in most compounds, lone pairs are placed in an axial position.

Is the lone pair or the electronegative element more appropriate for placement in an axial position?

So in $\ce{XeF2}$ there will be three lone pairs on the xenon, and the xenon will have two covalent bonds to fluorines. The xenon is of course the central atom, on to that we apply VSEPR-theory. We count five electron dense areas (two bonds and three lone pairs) and so VSEPR-theory predicts a trigonal bipyramidal shape for $\ce{XeF2}$

The lone pairs want to stay as far away from each other as possible (in VSEPR). Indeed would there have been just two lone pairs putting them axially would have introduced the maximal possible angle of 180°. Instead we have three lone pairs, so the equatorial positions introduce the best possible angle of 120°. Compare this to the 90° angle introduced if we had put two lone pairs axially and just one equatorial.

What is left now are the axial positions to which we allocate the fluorines, obtaining a linear molecule. In this case the VSEPR prediction has been confirmed experimentally.


How to calculate the osmotic pressure of a sucrose solution? The boiling point of an aqueous solution containing sucrose, $\ce{C12H22O11}$, is $101.45~^\circ\mathrm{C}$. Calculate the osmotic pressure of this solution at $35~^\circ\mathrm{C}$, at which the solution density is $\pu{1.036 g/mL}$. The boiling point elevation constant is $k_b = 0.512~^\circ\pu{C\cdot m^{-1}}$.
  The expected answer is
  $\Pi = 37.6~\mathrm{atm}$.

I tried it this way:

I used $\Delta T_b = k_b \cdot m$ and substituted $T_b = 101.45$ and $k_b = 0.512$
and got $m = 198.144~\mathrm{mol/kg}$.
Density equal to $\pu{1.036 g/ml}$ equals $\pu{1.036 kg/L}$, which means $\pu{1.036 kg}$ for $\pu{1 L}$ and $\pu{1 kg} = \pu{0.965 L}$.
$M = \pu{198.144 mol}/\pu{0.965 L} = \pu{205.33 mol/L}$
The osmotic pressure equation $\Pi = MRT = 205.33 \cdot 0.08206 \cdot (35+273)
  = \pu{640 atm}$.

Where have I gone wrong?

I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg

This should be the first clue that something went wrong.  198 mol of sucrose is about 67 kilograms of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose.  That can't be right.  (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)

I think you forgot to use the change in boiling point temperature in your $\Delta T_b = k_b m$ equation.  The change is only 1.45 °C, not 101.45 °C.

What happens if you plug in that value and redo your calculations?

Update: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.

  Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.

This is true, but keep in mind the definition of molality: it is moles of solute per kg of solvent.  However the density is obviously referring to the kg of total solution.  So, it is inappropriate to find the molarity by simply dividing the molality by the density.  As I described in a past answer on chem.se, if the molarity is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density.  So you will have to invert this relationship to find the molarity from the molality.


Why do the first and second laws of thermodynamics not contradict each other? We're learning about entropy right now in general chemistry, and I'm trying to understand something.

From the fact that entropy can be directly compared with enthalpy, free energy, electromotive force, etc., and its being measured in units of $\pu{J//mol*K}$ – including joules, the units of energy – it would seem that entropy is some form of energy. By the first law of thermodynamics, then, entropy can't be created nor destroyed, only transformed and transferred. From Gibbs' equation we see that entropy is being transformed out of the molecular potential energy, and from Clausius' equations we see that entropy can be transferred between a system and its surroundings, with $\Delta S_\mathrm{system}=-\Delta S_\mathrm{surroundings}$. Since $\Delta S_\mathrm{universe}=\Delta S_\mathrm{system}+\Delta S_\mathrm{surroundings}$, it follows that $\Delta S_\mathrm{universe}=0$ – as one would expect from the first law.

Yet the second law of thermodynamics states that $\Delta S_\mathrm{universe}>0$.

If entropy is a form of energy, then how can universal entropy tend to increase?
If entropy is not a form of energy, then how can it be compared with actual forms of energy and measured in energy units?

It is something of a historical accident that entropy has units of J/K.  It came out of the fact that the connection between heat, temperature, and energy was not obvious to early scientists, and so they effectively picked different units for measuring temperature and for measuring energy.

In the more modern statistical interpretation of entropy, the entropy of a system is simply a number.  Specifically, if the number of microstates associated with a given macrostate is $\Omega$, then $S = k \ln \Omega$.  The number of microstates ($\Omega$) is just a number, without any units, and therefore so is $\ln \Omega$.  You can see that we actually have to insert Boltzmann's constant, with its units of J/K, to make the units "come out right".

Arguably, a more natural way to define entropy would be to just make it it a dimensionless quantity:  $S = \ln \Omega$, without the factor of $k$.1  This would be equivalent to measuring temperature in units where $k$ is equal to 1 exactly, rather than defining our unit of temperature such that $k= 1.380649 \times 10^{-23}$ J/K exactly.
If we did this, we'd be effectively measuring temperature in units of energy as well;  for example, in an ideal monatomic gas with a "temperature of 1 J", the average KE of each molecule would be $\frac{3}{2}$ J.  Quantities such as Helmholtz free energy would still have units of energy, since we'd still define $F = U - TS$, with $T$ having units of energy and $S$ being dimensionless.

Of course, in this parallel universe where entropy is defined as a dimensionless number, another gen. chem. student would be asking why temperature is not the same thing as energy, even though they're measured in the same units.  But that's another question and another answer.

1 In fact, entropy is defined exactly this way in information theory, since there's not really a notion of energy (or temperature) to speak of in such contexts.


Chemical to dissolve rubber/latex glove I seem to have a drain pipe in my flat blocked by what seems to be a latex/rubber glove. I am unable to pull it out — is there a chemical that can reliably dissolve the glove and at the same time not damage the PVC pipe itself?

This is really a plumbing issue much more than a chemical dissolution issue. The problem is uncertain: "seems" to be a latex rubber glove; complete blockage.
I'm trying to envision how a rubber glove completely blocks a drain. Perhaps by partially blocking the drain, then catching hair and other particles - but on what? Perhaps on something else sticking into the drain - a root? Some other partial blockage?
Whatever; let's imagine pouring some solvent into the drain. For now, let's forget the PVC pipe: it is much thicker than the glove and will not dissolve before the glove does. As soon as the glove dissolves, the built-up head of solvent should flush away and prolonged rinsing with water will pretty much clean off the PVC pipe.
Unless the glove is caught on some projection and merely softens, while the 5 gallons of solvent drips away over a few minutes, leaving the glove softened and intermixed with the other captured trash. Now the problem has intensified. As solvent dissipates, the glove will become the adhesive that binds the trash together and to the PVC. But then the solution will be obvious: discover where the blockage is (with your snake), then remove that portion of the pipe and replace it. And don't do that ever again.
But if you insist on a chemical solution, I suggest air: pump up the pipe with a tire pump or one of those air blasters for toilet blockages and push the blockage down to larger pipe. It would also be helpful to load the blocked pipe up to your drain opening with a solution of dish detergent for lubrication and to decrease the volume that you have to pressurize.
If you must get more aggressive, there are tables of glove resistance (https://www.augusta.edu/services/ehs/chemsafe/PDF%20files/gloveselechart.pdf) online, but it would be easier to do your own experimenting. Since you have suspected it is a rubber glove, you might have other rubber gloves available that you could put into various solutions for a few hours or days to see how the glove behaves. But I think the glove is only part of the problem, and a good commercial drain cleaner would attack what I suspect is the helper-blocker, and dissolve that, and alkali isn't too gentle with latex rubber anyway (although the chart suggests natural rubber withstands alkali fairly well - except for ammonia). Also bleach is tough on natural rubber. Of course, YOU MUST NOT EVEN THINK OF MIXING AMMONIA WITH BLEACH!!!


Are there any substituents other than halogens which are ortho-para directors but ring deactivators? I was reading about electrophilic aromatic substitution reactions, and in every book, I found that the only example given for ortho-para directors - which deactivate the ring - were halogens. So are there any other substituents, other than halogens, which are ortho-para directors but ring deactivators? If so, do they do it like halogens?

The nitroso $\ce{-NO}$ group is deactivating as well as ortho para directing just like the halogens in electrophilic aromatic substitution.

For reasons of this behaviour, refer to this question on the website.

Here are more examples from Peter Skyes' Guidebook to Mechanism in Organic Chemistry: (least deactivating to most deactivating)

  $\ce{-F}<\ce{-Cl,-Br, -I}<\ce{CH2Cl < -CH=CH-COOH < -CH=CH-NO2}$

For all these groups $\ce{-I}$ effect is more dominating than $\ce{+M}$ effect. Hence they deactivate the ring. However, when the arenium ion is formed they stabilise the ring through the $\ce{+M}$ effect (hyperconjugation in case of $\ce{CH2-Cl}$)

The book doesn't mention the nitroso group in the order but since it has very weak +M we should expect it to be after $\ce{CH2Cl}$

$\ce{Cl-, Br- , I-}$ are almost equally deactivating therefore they have been separated by a comma.


extracting parameters from .SKF files , I have a question about interpreting .SKF files, I am using these parameters to model defective graphene,

when I use parameter set given by dftbBaby/C-C (I think this is from Hotbit)
or the parameter set given by the following paper:

Porezag, D., Frauenheim, T., Köhler, T., Seifert, G., & Kaschner, R. (1995). Construction of tight-binding-like potentials on the basis of density-functional theory: Application to carbon. Physical Review B, 51(19), 12947.

(both of which give approximately $H_{pp\pi} ~ -3 eV$ at ~1.4 Angstrom) then after performing slater-koster transformations and embedding the H0 and S values into the secular equation, I get the correct band structure.

however, for other parameterizations, for example say dftb/mio, the value I get for $H_{pp\pi}$ from interpolation of hopping integrals is much smaller ~ 2eV and gives wrong band structure:

here for sure it is me who is doing sth wrong because using the same parameter set DFTB+ gives the correct band structure,

I am not sure whether I am not interpreting the SKF files properly (in a naive attempt the band structure can be corrected by increasing the value of r0 given in the SKF file which is equal to shifting the curves downwards) or is there any other step I need to perform before embedding the H and S values into the secular equation ?

Thank you for your help

I figured that I was using the wrong format, the detailed format of .SKF files is given here

The correct extracted parameterization profile is given below:


How can a group be both a good nucleophile and a good leaving group? I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms, and I have a few questions.

My book has a couple of tables, one lists a bunch of substances grouped as good, moderate and poor nucleophiles. Among the good are $\ce{Br-}$ and $\ce{I-}$. So a couple of halogens. I'm figuring chlorine is in that group too.

Then it has a table showing ability to function as leaving group. Iodine, bromine and chlorine anions are at the top of that list too.

My question is: How can these be quite attracted to the positive charge AND quite easy to remove at the same time?

I'm not entirely sure of this, comments appreciated

Leaving group tendency is a thermodynamic variable. So, it depends more on the equilibrium and energies of the reaction. We calculate it from the basicity of the group (weak base -> more stable while solvated -> better leaving group).

Now, iodide is a weaker base than $\ce{OH-}$, since its conjugate acid $\ce{HI}$ is a stronger acid than $\ce{H2O}$. So, iodine is a better leaving group.

On the other hand, nucleophilic tendency is basically "how willing to sacrifice a lone pair is the group?".  It's more dependent on forces, and less on energy, as it is a kinetic phenomenon. The outer shell electrons in $\ce{I-}$ are more loosely bonded to it than those in oxygen in $\ce{OH-}$. So, it is a better nucleophile.

Another way to look at it would be via the "accessibility" of the group in solution. This basically attempts to explain the difference between a nucleophile and a leaving group. In a leaving group, you have to "give up" the ion to the solution. This means that solvation doesn't really matter here -- once you've given it up, it's gone. The process will have a similar rate for different ions. However, when looking at the reverse process, you need to check how solvated the ion is. If the ion is surrounded by water molecules (like $\ce{F-}$, it will be much harder for the electrophile to "extract" it. On the other hand, an ion like $\ce{I-}$ is less solvated, and thus more "accessible".


How was Mendeleev able to develop his table? I asked my general chem instructor this and he didn't know:

Mendeleev was able to group elements and discovered a bunch of them and put them in the correct order in the mid-1800s.  What equipment did he use to do this?  Was he just weighing stuff and guessing?

Specifically, how did he figure out atomic weight?

This quote from the wikipedia article on History of Chemistry might provide some insight into the determination of atomic mass:

  Chemists in different countries developed several different incompatible atomistic systems. A paper that suggested a way out of this difficult situation was published as early as 1811 by the Italian physicist Amedeo Avogadro (1776-1856), who hypothesized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, from which it followed that relative molecular weights of any two gases are the same as the ratio of the densities of the two gases under the same conditions of temperature and pressure.

So this implies that you could take equal volumes of different gases at the same temperature and pressure, then weigh them and take the ratios. This would allow you to get relative masses for $\ce{H2}$, $\ce{N2}$, $\ce{O2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$. Other gases could be used, but they wouldn't be as useful here. You could then determine that $\ce{H2}$ is the lightest gas, and that $\ce{O2}$ is 16 times heavier, $\ce{N2}$ is 14 times heavier, and so on. You don't even need to know that these are diatomic gases in order to make these measurements. Once you had these determined you could move on to gases like $\ce{CO2}$, or $\ce{H2S}$. Once you knew the mass of the gas molecule and what elements were in it, you could calculate how much of each element must be there, and by subtracting the mass of the known element, the remaining mass must be the atomic mass for the other element. As you accumulated more data you could approach substances that don't easily form gases, such as $\ce{NaCl}$. If you know the mass for $\ce{Cl}$ you could determine the mass for $\ce{Na}$.

What I suggest might not be the way atomic masses were determined, but it would probably work. Looking at this image of Mendeleev's original table shows they already had pretty good estimates of atomic mass by 1869.


Why does radium have a higher first ionisation energy than barium? I'm wondering why radium appears to buck the general trend that first ionisation energies decrease as you move down a group in the periodic table: barium (the group 2 element preceding it) has a first ionisation energy of $\pu{502.9 kJ/mol}$, whereas radium has a slightly higher first I.E. of $\pu{509.3 kJ/mol}$ (from the Wikipedia, although my textbook agrees).

Is there any explanation for this at present? (I imagine that quantum mechanics may be involved in some way, but I'm not entirely sure how)

There are most likely other examples of the trend being broken, but this is the only one I've come across so far and I'm curious to know why this is the case.

I think it's also important to mention relativistic effects here.  They already start becoming quite visible after $Z=70$, and $\ce{Ra}$ lies a good bit after that.

In very heavy atoms, the electrons of the $\ce{1s}$ orbital (actually, all orbitals with some electron density close to the nucleus, but the $\ce{1s}$ orbital happens to be the closest and therefore most affected) are subjected to very high effective nuclear charges, compressing the orbitals into a very small region of space. This in turn forces the innermost electrons' momenta to be very high, via the uncertainty principle (or in a classical picture, the electrons need to orbit the nucleus very quickly in order to avoid falling in). The momenta are so high, in fact, that special relativity corrections become appreciable, so that the actual, relativistically corrected momenta, ($p_{\text{relativistic}}=\gamma p_{\text{classical}}$) are somewhat higher than the approximate classical momenta. Again via the uncertainty principle, this causes a relativistic contraction of the $\ce{1s}$ orbital (and other orbitals with electron density close to the nucleus, especially $\ce{ns}$ and $\ce{np}$ orbitals).

The relativistic contraction of the innermost orbitals creates a cascade of electron shielding changes among the rest of the orbitals. The final result is that all $\ce{ns}$ orbitals are contracted, getting closer to the nucleus and becoming shifted down in energy. This is relevant to the question because the $\ce{7s}$ valence electrons in $\ce{Ra}$ are more attracted to the nucleus than one would expect from a simple trend analysis, since they rarely take into account the increase of relativistic effects as one goes down the periodic table.

Thus, the first (and second) ionization energy of $\ce{Ra}$ becomes higher than expected, to the point that there's actually a upward blip in the downward trend. Eka-radium ($Z=120$) would have far stronger relativistic effects, and can be expected to have a significantly higher ionization energy compared to $\ce{Ra}$. In fact, relativistic effects will conspire to make the group 2 metals slightly more noble! Though the periodic table becomes such a mess near the super heavy elements that it's hard to say whether it'll be a clearly visible trend, or just one effect to be combined with several others.


Calcium carbonate and hydrogen peroxide reaction In application of hydrogen peroxide to soil and groundwater to remediate petroleum contamination, how will the naturally occurring limestone react with the peroxide, thereby reducing the effectiveness of the application.  Would calcium peroxide be more suitable.

The effect of aqueous hydrogen peroxide on natural limestone has been examined in a different context (removal of moss and algae from the walls of frequently visited and therefore illuminated show-caves).

The results were published in Environ Pollut., 2003, 122, 417-422 (PubMed).

The authors exposed etatons of limestone to various aqueous solutions for about four months and examined possible damages using a scanning electron microscope. They found:

  Based on SEM study, the polished surfaces of the carbonate etalons did not show any changes after 135-day exposure to rainwater, 15% peroxide, and 30% peroxide, respectively. No etched pits or any other visible
  effects of dissolution [...] were detected.


Why does the colour of the same compound (e.g. magnesium nitride) differ? When I am searching the colour of $\ce{Mg3N2}$ I came across photos showing different colours: greenish-yellow on Wikimedia,
and some other just yellow on some Google Images results.

I am eager to know what causes some samples to be bit greenish.

Minor differences in color are likely either due to impurities or due to intrinsic crystal defects such as f-centers. For example, $\ce{ZnO}$ is often a shade of yellow, rather than pure white, due to those defects. In alkali halides, h-centers are similar defects giving rise to colors. You can easily create color centers in $\ce{NaCl}$ with a spark coil in just a few minutes.

Another possibility is that there are mixtures of different allotropes, or polymorphic forms. An extreme example is transparent diamond vs. black graphite. BTW, diamonds having varied colors due to impurities (e.g. nitrogen) and to intrinsic defects are useful in studying quantum phenomena, as well as gems.