Are carbocation intermediates in equilibrium? So for the reaction of $\ce{H_2O}$ with an alkyl ethene, in the presence of $\ce{D^+}$ ions, forms two possible carbocations:

$\ce{ R-CHD-CH2^+}$ and $\ce{R-CH^+-CH2D}$

The major product formed is the secondary carbocation as it is stabler than the primary carbocation.

So my questions are:

Are the two carbocations in an equilibrium or are they formed in
specific, fixed quantities?
If they are in equilibrium, i.e., the primary and secondary
carbocation, why and how would the secondary carbocation revert back
to being a less stable primary carbocation?
Also, is the hydride shift that converts the primary to a secondary
carbocation the only shift that can occur; even if after the shift,
a more stable carbocation can be formed (depending on the attached
alkyl group)?

Are the two carbocations in an equilibrium or are they formed in specific, fixed quantities?

$\ce{R-CH=CH2 + H^+ <=> R-CH2-CH2^+ + RCH^+-CH3 <=> alcohol~products}$

Yes, the two carbocations formed are in equilibrium. Initially the two carbocations will be formed in a ratio dependent upon the difference in activation energies leading to them (kinetic control).  However, since all of the steps in this reaction involve equilibria, after a while when equilibrium is established several things can happen

many carbocations will revert back to starting material which can then be reprotonated and reform the carbocations
or alcoholic product can be reprotonated, eliminate water and reform the carbocations
or the carbocations themselves can equilibrate directly by hydrogen shift

Once the equilibrium is established, the ratio of carbocations formed will be dependent upon the energy difference between the carbocations themselves (thermodynamic control).  Often, but not always, the relative stabilities of the transition states leading to the carbocations matches the relative stabilities of the carbocation intermediates themselves.

  If they are in equilibrium, i.e., the primary and secondary carbocation, why and how would the secondary carbocation revert back
  to being a less stable primary carbocation?

why - that's what happens in equibria; even though phenol is much, much more stable than its enol counterpart, there will be an exceedingly small amount of the enol present at equilibrium

how - by any of the 3 routes mentioned in the bullets above

  Also, is the hydride shift that converts the primary to a secondary carbocation the only shift that can occur; even if after the shift, a
  more stable carbocation can be formed (depending on the attached alkyl
  group)?

Other groups like alkyl or phenyl can shift too, but hydrogen shift is usually faster than the others.


Solubility of lead cyanide I read that lead(II) iodide is soluble in excess KI solution due to complexation, but lead(II) cyanide does not dissolve in excess cyanide solution. This seems wrong since cyanide is a much better ligand that iodide, so shouldn't it be able to form a complex more easily?

Cyanide is generally better than iodide in transition metal complexes because it acts as a pi-acceptor. It has empty orbitals that overlap with the partially filled d-orbitals of the metal, lowering the energy of these electrons.

Iodide is a sigma-donor, meaning that it donates electron density to empty metallic orbitals. The effect is small compared to a pi-acceptor in a transition metal, because the major contribution to stability is the energy of d-orbital electrons.

However in Pb(II) complexes, the d-orbitals are full already, so here the sigma-bonding mechanism for stabilisation is key. As cyanide isn't as good of a donor as iodide, then iodide forms a complex whereas cyanide doesn't.


Is the dissolution of sodium acetate trihydate endothermic? Sodium acetate trihydrate dissolves in water to its constituent ions:

$$\ce{NaOAc.3H2O (s) ->[H2O] Na+ (aq) + OAc- (aq) + 3H2O (l)}$$

The crystallisation of sodium acetate from a supersaturated solution is well-known to be exothermic. Since dissolution is the reverse of crystallisation, should the enthalpy of dissolution be positive (i.e. an endothermic process)?

The process of dissolving sodium acetate trihydrate $(\ce{NaC2H3O2.3H2O})$ in water is endothermic. The molar enthalpy of solution at $T=25\ \mathrm{^\circ C}$ is $\Delta_\text{sol}H^\circ=19.66\ \mathrm{kJ\ mol^{-1}}$.*

Accordingly, since crystallization is the reverse process of dissolution, the crystallization of sodium acetate trihydrate from aqueous solutions is exothermic. For practical purposes, the enthalpy of solution with a reverse sign is taken as enthalpy of crystallization.

The enthalpy of solution mainly depends on two energy contributions: lattice energy and hydration energy. Lattice energy is the energy released when the crystal lattice of an ionic compounds is formed. Conversely, energy equal to the lattice energy has to be supplied to break up the crystal lattice; i.e. this process is endothermic. Hydration energy is released when water molecules hydrate the ions; i.e. when water molecules surround the ions and new attractions form between water molecules and ions. This process is exothermic.

Therefore, whether the process of dissolving a salt in water is exothermic or endothermic depends on the relative sizes of the lattice energy and the hydration energy. If the lattice energy is greater than the hydration energy, the process of dissolving the salt in water is endothermic. Conversely, if the lattice energy is smaller than the hydration energy, the process of dissolving the salt in water is exothermic.

For anhydrous sodium acetate $(\ce{NaC2H3O2})$, the lattice energy is actually smaller than the hydration energy. The molar enthalpy of solution is negative; at $T=25\ \mathrm{^\circ C}$ it is $\Delta_\text{sol}H^\circ=-17.32\ \mathrm{kJ\ mol^{-1}}$.* Thus, the process of dissolving anhydrous sodium acetate in water is exothermic.

However, hydration of ions can also occur in crystalline solids. Sodium acetate trihydrate $(\ce{NaC2H3O2.3H2O})$ contains three water molecules for each $\ce{NaC2H3O2}$ unit in the crystal. The corresponding amount of hydration energy is already released when crystalline anhydrous sodium acetate absorbs water and is converted to crystalline sodium acetate trihydrate. Therefore, the additional amount of hydration energy that is released when sodium acetate trihydrate is dissolved in water and completely hydrated is smaller than the total hydration energy that is released when anhydrous sodium acetate is dissolved in water. Hence, the resulting value for the molar enthalpy of solution is larger for sodium acetate trihydrate than for anhydrous sodium acetate.

* “Enthalpy of Solution of Electrolytes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


Contribution of Tetrahedral and Octahedral voids in HCP I cannot find anywhere what the contribution of atoms situated at octahedral or tetrahedral voids in a HCP unit cell would be. I need to know this to be able to calculate the formula of a compound with HCP unit cell, when the positions of the atoms in the lattice are given.
What I mean by 'contribution' is how much of the atom in that lattice point belongs to one unit cell. Eg:- in CCP packing the contribution of atoms at the corners is 1/8 th.
I know that in a CCP unit cell the octahedral voids are there at the body center and the edge centers- so the contribution of an atom from both locations is different- 1 and 1/4 th. So I'm not sure whether the positions of OTs or TVs in a hcp unit cell are different (for the atoms of the same type) if so could someone tell how the contribution would be?

I cannot find anywhere..

Couldn't find? You can figure it out on your own, with me, right now.

Disclaimer: By "hexagonal unit cell" I assume you mean hexagonal prism, which comprises of 3 primitive hexagonal Bravais lattice.

Octahedral Voids (OhV)

The one on the left displays octahedral voids(visibly surrounded by 6 atoms).
The OhV is at the center of the highlighted octahedron, and lies on the plane containing four of the six lattice points in the octahedron.
Note that these lattice points belong to the unit cell, and the OhV must hence lie inside the unit cell.

This is the same for the other 5, amounting to a total of 6 OhV per HCP unit cell.

Being completely inside, its contribution is taken as 1.

BONUS: If you remember, No. of atoms per unit cell is 6, which
tallies with the no. of OhV given by the formula $N_{OhV}=N$ where
$N$ represents no. of atoms per unit cell.

Tetrahedral Voids (ThV)

I'm going to be frank: it IS not as easy as locating OhV.
The figure above absolutely freaked me out when I tried to visualize ThV's. I'm not keen on recommending that.
Instead, this 3D rendition should help immensely.

Type 1 ThV

The Type 1 Thv's are the ones inside in unit cell completely(so again, contribution=1).
Notice that I have marked 2 tetrahedrons. Also note that each tetrahedron we form will tell us about the ThV its existence creates.

RED: Note that there will be another 2 like this one in the lower half. That makes a total of 6 Red-type ThV in the entire HCP unit cell.
GREEN: Note that there will be a total of 2 of the green-type ThV, the other one belonging to the inverted tetrahedron just below it.

That gives us a total of $6+2=8$ Type 1 ThV.

Type 2

Let's do a little recap to see which ThVs have we counted so far(displayed in red)

I constructed a 3D model of 3 HCPs, which is required to visualize the remaining.

Now I will mark the ThV forming lattice points in red.

As you can see, lattice points from 3 unit cells are involved.

A quick analysis tells us that there will be 2 such Tetrahedrons for every vertical edge.
A quicker analysis tells us that since each ThV so formed is present in 3 HCP unit cells, or is "shared" by them, Its contribution must be $\frac 13$.

We count the number of such tetrahedrons to be 12 per unit cell; with contributions, ThV Type 2 will be $12*\frac 13$ $=4$.

Ultimately, we get:

ThV type 1= 8

ThV type 2= 4

Total ThV= 12.

Which brilliantly tallies with the formula for number of ThV($N_{ThV}=2N$).

I hope managed to enlighten you.

TL;DR

OhV contribution is 1, no issues there.
ThV does not have a single "contribution", but rather could be 1 or
$\frac 13$ depending on which ThV you're talking about.


Mechanism of a cyclization reaction - why does an enol ether form? I was asked to predict compound A from the follwing organic scheme:

Looking at the formation of A, my prediction was that since TsOH is a strong acid, it would dehydrate the alcohols (like sulfuric acid does). However, looking at the next reaction of A, it seems to me that what happened was that the ketone formed its enol and did an aldol condensation with the aldehyde. But then, if this was the case, why do we have to add TsOH to get to A, then add $\ce{H3O+}$ to get to the final product when we could've gotten to the final product from the start with an aldol condensation?

I was told that the structure for A is actually:

I was not expecting this at all, and I'm not sure how this was formed. I can see that the enol ether in A is hydrolyzed in acid to get the final product, but the formation of A seems strange. What is the mechanism for the formation of A and why were my predictions wrong?

@Carbenoid: A year has passed since you requested a mechanism from @Mithoron. The statutes of limitation have run out. Your comment about why not do an aldol condensation (in base) is well taken. But the question was what is the structure of compound A.  I have no problem with the dienol ether 5 (aka A) in spite of many people's aversion to forming 8-membered rings.
Full stereochemistry of structure 1 could have been informative but not essential. I will forgo the mechanism of the acid-catalyzed aldol condensation and defer to @ringo's contribution. The bold bonds in structure 2 form a rigid framework so that the degrees of freedom are minimized in the formation of the hemiketal 3. Effectively, the cyclization is that of a much smaller ring. This ring closure is mechanistically more sound than the SN2 displacement with an enol particularly one with an (E)-geometry when at least the (Z)-geometry would be required. The red arrows terminate with dienolether 5. Aqueous acid and microscopic reversibility via the blue arrows return dienolether 5 to structure 2. In fact, structures 2 and 3 may well be in equilibrium with each other.


When does a pinacol undergo dehydration or rearrangement? What happens when the above pinacol is treated with -
(a) conc. sulphuric acid
(b) dil. sulphuric acid

I'm confused between pinacol rearrangement and dehydration (E1 or E2?), as both possibilities exist. I think the concentration of the reagent might have a role to play, but I'm not sure. Please help.

P.S. It'd be great if you could quote data and sources with your answer.

When you treat pinacol with sulfuric acid, an E1 elimination will take place ( having a carbocation as intermediate ) and it forms pinacolone ( 3,3-dimethyl-2-butanone ). I am not aware of how concentrated the acid should be. Do you require a photo depicting the mechanism ?


Why diethyl ether does not absorb in UV region? Explain why diethyl ether does not have a UV spectrum, even though  it has lone-pair electrons.

Can we say that π* molecular orbital does not exist because there are no π-bonded electrons in diethyl ether?

We can measure the UV-Vis spectra for π → π* or n → π* electronic transitions (other transitions are forbidden according to selection rule) [1, pp. 577–578]. But in diethyl ether, being a saturated compound, it has no π-bonds. This successfully dismisses the idea of having π* orbital. And, as there is no π* orbital, we can not have a transition to the π* orbital resulting in no UV-Vis absorption spectra.
Reference

Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed.; Cengage Learning: Stamford, CT, 2015. ISBN 978-1-285-46012-3. (PDF)


qualitatively: Why are thioesters more reactive than regular esters? In ochem students are taught that shorter bonds (like alkynes) have more energy than longer bonds such as alkanes. However in biochem we are taught that thioesters are very energetic bonds and release alot of energy upon breaking.

Is there a specific mechanism that explains the relationships between bond energy and length?

background: a university student with a chem minor

http://guweb2.gonzaga.edu/faculty/cronk/CHEM440pub/thioesters.html

"thioester cannot make as much of a contribution because the required overlap between carbon's 2p orbital and the sulfur 3p orbital is ineffective. The large difference in energy between these two orbitals means they cannot "mix" as well as the 2p orbitals of oxygen and carbon."


Conversion of 4-nitrotoluene to 2-bromobenzoic acid In the above conversion why doesn't bromine add to methyl group? Is the first reaction following free radical or polar reaction (Friedel-Crafts). If it's following free radical mechanism then Br should add to CH3, if it's F-C substitution then why there is no catalyst when NO2 is deactivating group?

What would be the change if I first do seconds step (i.e reduction to amine) and then bromination? Would it be wrong to do bromination at last?

Is the intermediate (2-bromo-4-nitrotoluene) the only major product of first step?

Sorry but I would like to answer your questions from last to first,

(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place

(2) -COOH  is deactivating meta directing group (due to presence of  >C=O) so friedel-crafts bromination would take place at meta position.

(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2

(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced  one of the -H at ortho (to -CH3) of benzene.

(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)


Enthalpy, heat of solution test: Calcium chloride vs magnesium chloride in water I did a school experiment to measure the heat of solution (enthalpy) of various salts dissolved in water.
My results are completely different to the teacher's answer.
Specifically my test showed Calcium Chloride increasing the temperature of the cup of water by 6 degrees Celcius, while the Magnesium Chloride only increased the temperature of the same volume of water by 1 degree Celcius.
The teacher's answer shows Magnesium Chloride with an enthalpy more than double (twice as negative) as that of Calcium Chloride.
I'm happy to be wrong about this but could someone point me to an "official" answer so I can conclude that the experiment was done incorrectly.

It may be caused by the real content of hydrate water in the used calcium and magnesium chlorides. As there is possible disagreement between declared and true water content, due long term reaching equilibrium with air water vapour. I have once seen a glass bottle with reportedly solid $\ce{LiCl}$, sitting on an shelf for years. The content was a clear liquid of LiCl solution.(*)
Both chlorides form multiple hydrates, stable at different humidity or temperature conditions. Generally, with increasing of crystal/hydrate water content, the dissolution enthalpy raises for both salts from negative to positive values. E.g. anhydrous $\ce{CaCl2}$ warms up water when dissolved, $\ce{CaCl2 . 6 H2O}$ cools water down significantly. Note that both anhydrous chlorides are highly hygroscopic.
I suggest to make repetition of the experiments with salts of different origin and short history. Or, heating them well in an oven to get anhydrous stuff.

(*) This hygroscopicity of $\ce{LiCl(s)}$ is used in some humidity meters, measuring equilibrium electric resistence of a soaked fabric.


Comparing the bond strength between a polar bond and non-polar bond According to NCERT class 12 part 2, $\ce{ICl}$ has lesser bond enthalpy than $\ce{I2}$. I have read that non polar covalent bonds are weaker than polar covalent bonds as in polar covalent bonds partial opposite charges develop on the bonding elements which leads to extra electrostatic interaction between them which is absent in non polar bonds. So according to what I read the answer should have been opposite. Please let me know where this logic is wrong. Also using this logic I understood that ionic bond is stronger than covalent bond.

In polar covalent bonds, say $A–X$, assume electronegativity of $X$ is greater than that of $A$ then their is polarisation of bond $A–X$, $X$ gains partial negative charge $\delta–$ and $A$ gains partial positive charge $\delta+$ but this charge is very less (as electrons of sigma bond are held tightly) to account for any significant increase in bond strength of $A–X$ bond but the $–X$ group in $A–X$ molecule tends to attract $A–$ group of another $A–X$ molecule in this way there is electrostatic interaction between two different $A–X$ molecules giving rise to intermolecular forces (Dipole-Dipole interaction, hydrogen bonding etc).

DO NOT CONFUSE
FORCES BETWEEN TWO DIFFERENT MOLECULES AND COVALENT BONDS WHICH EXIST IN A SINGLE MOLECULE. THEY ARE TWO COMPLETELY DIFFERENT THINGS

However, attraction or repulsion may exist between two parts of a single molecule (generally happens for organic  molecules) but doesn't significantly affects bond strength between them rather they tends to change orientation of those groups or may rotate about the sigma bond.

And for $ICl$ and $I_2$, bond of iodine and chlorine is weak due to polarisation. And bond between iodine and iodine is strong covalent bond. I hope you know that polarisation of electron cloud of sigma bond decrease overlapping between orbitals.


Basis sets that can be used for macrocyclic complexes What could be the possible basis set that can be used for a metal base macrocyclic complex in gaussian09?

I am using transition metals such as $\ce{Ni}$, $\ce{Co}$, etc.  Also, I have tried using DFT and HF theory for optimization but I am not getting appropriate results.

I am basically looking for binding of metal with the ligand in these coordinate complexes on basis of which i can use them for a particular application.
The basis set that I am using right now for DFT is Opt=(Cartesian,maxcycles=500)  b3lyp/gen  nosymm   pseudo=read SCF=(conver=5,vshift=400,maxcycles=1000)

Edit: Earlier I was using a specific basis set for transition metals and now I am using lanl2 as a single basis set for the DFT calculation. I am also using hf cal in which the sto3g basis set is used but iam getting two consistent errors in optimization which are convergence failure and number of step exceed error processed link 9999.

In order to check on the availability of basis sets you may want to look at the Basis Set Exchange. After determining which basis set suits your problem (this will likely involve some literature search, Nickel, DFT, basis set, in Web of Science should do the trick), you can enter basis sets into Gaussian using the Gen keyword. BSE will export any selected basis sets in Gaussian format, so that shouldn't be a problem.


Why does the extent of racemisation in an SN1 reaction depend on the solvent used? Why is it that the extent of racemisation is greater in an $S_N 1$ reaction caried out using a polar protic solvent but the racemisation is not as extensive if an aprotic solvent is used although the carbocation intermediate is common in both?
eg:- 2-Bromobutane is solvolysed with $\ce{H2O}$ and $\ce{CH3OH}$ but the racemisation is 98% in water and 73% in methanol

Secondary haloalkanes can undergo substitution by both the $S_N 1$ and $S_N 2$ routes. In general these two mechanisms compete with each other. The following diagram is a More-O'Ferrall-Jencks Plot showing the possible mechanism space for a nucleophilic substitution reaction. The reactants are in the upper left. The products are in the lower right. The horizontal axis represents the extent to which the C-LG bond has broken. The vertical axis represents the degree to which the C-Nu bond has formed. The $S_N 1$ pathway is represented by traveling along the perimeter of the plot first to the right and then down. The $S_N 2$ pathway is actually the red diagonal line represented both processes happening simultaneously. There is a third pathway (down and then right), representing nucleophilic attack and then loss of leaving groups. $S_N$ reactions at $sp^2$ centers often follow this route.

Very few reactions stick completely to one of these three extreme paths. For $sp^3$ centers, the mechanism sticks to the upper right of the diagonal depending on a variety of factors like the polarity of the solvent, the nucleophilicity of the nucleophile, the strength of the C-LG bond, steric hindrance at the subsitution center, and how stable the carbocation would be if it formed. The arrows on the plot above suggest the directions that the transition state would move if the nucleophilic attack step became more favored (stronger nucleophile, less polar solvent, less steric hindrance). The end position is not possible for this particular reaction, but such is the result of reusing and image from somewhere else.

In your case, you can tell how close the actual reaction pathway sticks to one of the idealized pathways. Or, you can think of these numbers as the relative likelihood that a substrate will undergo solvolysis by $S_N 1$ or $S_N 2$ mechanisms in that solvent.

In your case, methanol $\ce{CH3OH}$ is less polar than water $\ce{H2O}$, so the $S_N 1$ pathway is less favorable in methanol than in water. However, the $S_N 1$ pathway in methanol is still way more favorable than the $S_N 2$ pathway in methanol (by 73:27).


Order of magnitude of ions in water My book says water has approximately $\pu{3.35E28 molecules/m^3}$  and that only a small part of water molecules are dissociated to $\ce{H+}$ and $\ce{OH-}$ ions and that the approximate amount of ions of each sign is $\pu{6E19 ions/m^3}.$

I am trying to prove the last part. I thought I could used the ionic product of water

$$[\ce{H+}][\ce{OH-}] = \pu{1E-14},$$

so that in neutral conditions

$$[\ce{H+}] = \pu{1E-7 ions/L} = \pu{1E-4 ions/m^3}.$$

What am I doing wrong or how should I proceed?

You are right: 1 cubic meter of water contains $\pu{1E-4}$ moles of $\ce{H+}$ ions. But one mole is $\pu{6E23}$ atoms or ions.

So, $\pu{1E-4 mol}$ contains $\pu{1E-4}\times\pu{6E23} = \pu{6E19 ions}.$
Here we are.


What is the colloidal osmotic pressure? The oncotic pressure or colloidal osmotic pressure is the osmotic pressure developed due to the presence of colloids in a solution. But since the colloids are not true solution, why should the colloids be termed as solutes soluble in the solvent and consequently capable of producing any osmotic pressure? Should they not behave like insoluble substances who do not affect the osmolarity of the solution? And theses colloidal osmotic pressure plays a very important role in trans-capillary transfer dynamics.

If the colloids do affect the osmotic properties, is the expression to measure the colloidal osmotic pressure same as that for true solutions(vant-hoff equation)?

I think, first I should clarify what causes the osmotic pressure:
Osmosis occurs when two solutions of different concentrations are separated by a membrane which will selectively allow some species, e.g. the solvent, through it but not others, e.g. the solute.
So, there is a concentration gradient between the two solutions which would lead to a diffusion from the side with the higher to the side with the lower concentration.
But since the solute cannot pass through the membrane the diffusive flow cannot even the concentration differences out and thus constantly presses against the membrane exerting a force on it which leads to a pressure.
The pressure originating from this tendency for osmotic flow to occur is called the osmotic pressure $\Pi$ of the solution.

So, the question is: Is the theory behind diffusion applicable to colloidal solutions?
To reason about this, I will first address the matter of whether colloids can be considered "real" solutes or not. Colloidal particles are usually some polymer particle coated with a tenside layer. So, it is the molecules of those tenside layers that interact with the solvent molecules and they are chosen in such a way that the colloidal particles neither lump together (coalesce) nor sink to the ground or float to the top - that keeps the colloidal solution stable. So, a colloidal particle is surrounded by solvent molecules which solvate the tenside layer. Of course, the situation is a bit different from a "normal" solution since the colloidal particle is a lot bigger than, say, an alcohol molecule and thus the the ratio between solute particle size and the size of its solvation sphere is much smaller. But, in essence, a colloidal particle and an alcohol molecule are solvated in pretty much the same way.
Furthermore it can also be assumed that interactions (e.g. electrostatic forces) between the colloidal particles are weak.
This means that colloidal particles are thus not subject to external forces which influence their movement. They simply move through the solution by Brownian motion. This is important because then their motion can mathematically be described by a stochastic process which is the basis for the thermodynamic description of diffusion.
And, as I described above, the reason for diffusion to happen is the same as the origin of the osmotic pressure.

So, to answer your last question: Yes, I think, the description for osmotic pressure caused by colloidal particles should be the same as for "normal" solutes.


Is it valid to use Gibbs Free Energy for gases? This has confused me for a long time now. Why do we compute Gibbs free energy when we figure out the spontaneity of a reaction involving gases? In my mind, I envision a reaction involving gases to be always in a closed container, which would clearly change the pressure of the gas, but it seems to be that constant pressure is a requirement for using Gibbs free energy.

Now if it is that we are supposed to assume that the reaction involving gases does not occur in a closed container, why do we use molar concentration of gases when computing equilibrium constants? In this situation, it seems to be that volume of gases will change.

I'm suspecting that most of my confusion is on what constant pressure actually means in a reaction, so as a third question, what does constant pressure mean when we are referring to a reaction?

The Gibbs Free Energy has been found to be the most convenient thermodynamic function to work with when determining the equilibrium constant for a reaction or in establishing the chemical equilibrium state for a chemically reacting system involving multiple reactions.  The condition for equilibrium of these is that, at a given temperature and pressure, the Gibbs Free Energy is minimized with respect to changes in extent of reaction.  This applies to whatever temperature and pressure are currently present in the reactor, irrespective of whether they have been changing. For a mixture of ideal gases, it is possible to show that this Free Energy minimization is equivalent to requiring that the sum of the free energies of the pure reactants at pure gas pressures identical to their partial pressures in the reactor are equal to the sum of the free energies of the pure products at pure gas pressures identical to their partial pressures in the reactor.


What is the canonical reference for studying stereochemistry? I am well versed with all the named reactions and all of organic chemistry.  I am quite weak in stereochemistry.

What canonical text can I use to study this topic and connect with other parts of the chemistry?

The classical book on stereochemistry is Eliel: Stereochemistry of Organic Compounds, first edition appeared 1962.

You will find overwhelming amount of information there. I would recommend it, regardless your age and level. It is so well written, that it is worth at least of fast reading over most important parts.
You should find it in any reasonable library, no need to get the latest edition, even the older ones should cover what you need.


In an experiment testing catalyst influence on reaction, is NaOH a reactant or a catalyst? I have recently completed an experiment with hydrogen peroxide (5%), 0.1M KI and 1M NaOH.

Is the sodium hydroxide a reactant or a catalyst in this experiment?

In basic solution the half reactions are:

Reduction:

$\ce{H2O2 + 2 e- -> 2 OH-}$

Oxidation:

$\ce{I- + 6 OH- -> IO3- + 3 H2O + 6 e-}$

So the overall reaction is

$\ce{3 H2O2 + I- -> IO3- + 3 H2O}$

thus $\ce{OH-}$ is overall neither a reactant nor a product. Hence it is either a catalyst, an inhibitor, or a spectator ion.


Gravimetric analysis In a gravimetric determination of phosphorus an aqueous solution of dihydrogen phosphate ion is treated with the mixture of ammonium and magnesium ions to precipitate magnesium Ammonium Phosphate. This is heated and decomposed to magnesium pyrophosphate which is weighed. A solution of $\ce{NaH2PO4}$ yielded 1.054 gram of $\ce{Mg2P2O7}$. What weight of $\ce{NaH2PO4}$ was originally present?

I attempted the question and my answer was coming out to be 1.32 gm where as the real answer is 1.14 gm

P.S: I wanted to post the pic of my solution so that i could be corrected  , but just cant figure out how to upload the photo.

And also can somebody give the answer without using POAC

The amount of $\ce{Mg2P2O7}$ in 1.054 gram of a pure sample is:
$$ n = \frac{\pu{1.054g}}{\pu{222.57g mol-1}} = \pu{0.004736 mol}$$

There are two atoms of $\ce{P}$ in $\ce{Mg2P2O7}$, but only one in $\ce{NaH2PO4}$. Without considering the two reactions done in detail, we can infer a 2:1 ratio:

$$\ce{2 NaH2PO4 + ... -> Mg2P2O7 + ...}$$

Thus, the mass of $\ce{NaH2PO4}$ is
$$  m_\ce{NaH2PO4} = 2 \cdot \pu{0.004736 mol} \cdot \pu{120.0 g mol-1}= \pu{1.137g} $$


Time dependence of integrated rate equations Is it possible for an integrated rate equation to not be dependent on time? What law or principle says so in chemical kinetics?

What if an integrated rate equation obtained only depends on the concentrations of the reactants? Is it possible? Can you refer me to books and references?

By definition, an integrated rate equation shows the change in concentration as a function of the progress of the reaction (time). So I don't understand why professor insists that an integrated rate equation can exist without having the variable time in it.

From a mathematical standpoint: If the integrated rate law is independent of time, then the rate of change of the concentration must equal zero (by the fundamental theorem of calculus). So you're looking for a reaction in which the concentration of the species you want in the rate law (presumably the reactant) does not change appreciably.

As others pointed out, this is a common case for intermediates under the steady-state approximation (rate of production = rate of consumption, net rate zero). But that's just for intermediates. For a substrate, you would need the substrate concentration to be essentially constant. That's probably not a "reaction" that will be interesting to study. A trivial solution, if you will.


Is water the most common multielemental compound? Water is a very common substance, not just on Earth, but also in space. Although water usually occurs in space as a solid and occasionally as a gas and relatively rarely as a liquid, the most common compound is H₂. I believe based on the abundance of elements in the universe (ignoring helium because it is a noble gas) that water is the most common multielemental (having more than one type of element in its structure) compound. Are there more common ones, or is water the most common one?

You are indeed correct that the elemental molecule $\ce{H2}$ is the most abundant molecular species in the universe by far.[1]. In order to answer your question

Is water the most common multielemental compound?

I only focus on neutral species, of which around $50\%$ are in molecular, i.e. multiatomic form.[2]
In this category, second only in total abundance to molecular hydrogen $\ce{H2}$ is carbon monoxide $\ce{CO}$.
This has become evident during the extensive work that has been conducted in the last few decades as large-scale $\ce{CO}$ surveys of the Milky Way galaxy and other galaxies have been performed, further corroborating $\ce{CO}$ as the second most abundant neutral species and technically[3] as most abundant interstellar compound.
To conclude, it might be interesting to realize how substantial the difference in mass is still between the two most abundant molecular species in interstellar space:[2]

Estimates of the galactic mass of $\ce{H2}$ and $\ce{CO}$ can now be given with reasonable accuracy $m(\ce{H2}) = 2 \cdot 10^9 \; M_\odot$, and $m(\ce{CO}) = 2 \cdot 10^6 \; M_\odot$.

So in order to estimate $m(\ce{H2})$ which constitutes the bulk of the gas mass, carbon monoxide $\ce{CO}$ is typically used as tracer or proxy for $\ce{H2}$, as further reading I suggest a very informative post on the physics.SE: Why is $\ce{CO}$ a good tracer for $\ce{H2}$? How are those molecules correlated?
The mass unit $M_\odot$ is commonly used in astrophysics and represents the mass of the Sun, i.e. $M_\odot \cong \pu{1.98E30 kg}$.
The water molecule $\ce{H2O}$ is probably next, i.e. in third place, of the most abundant molecular (neutral) species in the universe which can be directly underpinned by the fact that it made up of the first and third most abundant elements in the universe, hydrogen $\ce{H}$ and oxygen $\ce{O}$*[4]. But I am not aware of any reputable source that explicitly states or confirms this assumption.
Footnotes & References

vide, e.g. Why is $\ce{CO}$ a good tracer for $\ce{H2}$? How are those molecules correlated? or Wikipedia: List of interstellar and circumstellar molecules respectively

Winnewisser, G. The chemistry of interstellar molecules. Topics in Current Chemistry,Cosmo- and Geochemistry 1981, 39–71. DOI: 10.1007/3-540-10920-X_14.

vide ut supra, comment by Karsten Theis:

$\ce{H2}$ is not a compound, it is an element. It is a molecule rather than an atom, though.

van Dishoeck, E. F.; Herbst, E.; Neufeld, D. A. Interstellar Water Chemistry: From Laboratory to Observations. Chem. Rev. 2013, 113 (12), 9043–9085. DOI: 10.1021/cr4003177.


Why is MP2 failing to calculate proper IR frequencies for [Co(CO)4]- and [Fe(CO)4]2-? I was doing some IR calculations on different 3d transition metal carbonyl complexes during some practical courses where we were introduced to effective core potentials. While doing this we found that MP2 has some problem with the calculation of at least the asymmetrical $\nu_\text{CO}$ band (that was the only one we were interested in).

We made the calculations for $\ce{[Ti(CO)6]^2-}$, $\ce{[V(CO)6]-}$, $\ce{[Cr(CO)6]}$, $\ce{[Mn(CO)6]+}$ in octahedral symmetry and $\ce{[Fe(CO)4]^2-}$, $\ce{[Co(CO)4]-}$ and $\ce{[Ni(CO)4]}$ in tetrahedral symmetry. The method was MP2, as stated in the title, and the basis set was 6-31G(d). The program we used was Gaussian$~$09 Rev. A.02 but that was not the problem, as I did calculations with other programs with similar results. All optimized structures are considered to be true minima as there either is no imaginary frequency or it is calculated to be $\ge -50~\mathrm{cm^{-1}}$ with was stated to be a problem of the insufficient DFT grid but not further questioned.

The following graphic shows that most of the calculated values are acceptable and near their experimental values but two of them, namely $\ce{[Co(CO)4]-}$ and $\ce{[Fe(CO)4]^2-}$ differ significantly from the experimental values by nearly $300\ldots1000~\mathrm{cm^{-1}}$.

$\hskip1.4in$

After this I varied some parameters of the calculation that could influence this deviation as was:

additional diffuse functions for the better description of ions
tight optimization criteria (T) as this might reduce the imaginary frequency
no frozen core (F) just to try it
solvents: PCM(DCM), PCM(Heptane) as experimental values aren't taken in gas phase
I also optfreq'ed with B3LYP/6-31G(d)/MDF10 (B) to see if it's the method itself

The results can be seen in the following table:
$$\begin{array}{cccccccc}
\hline
\text{Basis~Set} & \mathrm{Solv.} & \mathrm{d(MC)} & \mathrm{d(CO)} & \nu_\text{CO}(calc.) & \mathrm{Int.} & \nu_\text{CO}(exp.) & \Delta\nu_\text{CO}\\
 &  & \mathrm{Ang.} & \mathrm{Ang.} & \mathrm{cm^{-1}} & \mathrm{km\, mol^{-1}} & \mathrm{cm^{-1}} & \mathrm{cm^{-1}}\\
\hline
\mathrm{E}/6{-}31\mathrm{G}^\ast & & 1.802 & 1.184 & 2830 & 2.0\cdot10^5 & 1890 & 940\\
\mathrm{EFT}/6{-}31\mathrm{+G}^\ast & & 1.799& 1.187& 2966& 2.7\cdot10^5 & 1890 & 1076\\
\mathrm{E}/6{-}31\mathrm{+G}^\ast & & 1.800 & 1.187 & 2996 & 2.9\cdot10^5 & 1890 & 1106\\
\mathrm{E}/6{-}31\mathrm{+G}^\ast & \mathrm{DCM} & 1.795 & 1.186 & 4671 & 0.0 & 1890 & 2781\\
\mathrm{E}/6{-}31\mathrm{G}^\ast & \mathrm{Hept.} & 1.799 & 1.184 & 3203 & 4.3\cdot10^5 & 1890 & 1223\\
\hline
\mathrm{BE}/6{-}31\mathrm{G}^\ast & & 1.767 & 1.169 & 1978 & 1322 & 1890 & 88\\
\hline
\end{array}$$

What can be seen:

Within MP2 there is no significant effect on $\ce{M-C}$- or $\ce{C=O}$-bond lengths.
The diffuse functions shift the frequency to higher wave numbers and
the impact of the solvent is big to huge. As without the solvents the
deviation was calculated to be somewhere around
$+1000~\mathrm{cm^{-1}}$ it is now at $+2800~\mathrm{cm^{-1}}$!
Also the intensities are horribly wrong ... they are somewhere around $2$
to $4\cdot10^5~\mathrm{km\, mol^{-1}}$ instead of somewhere around
1000 or in the special case of DCM they are completely zero.
B3LYP predicts everything within common error ranges

What are your ideas why MP2 fails with this special task?

Some months later but here we go...

I agree with most of the comments that multiconfigurational effects should be present for this system, and maybe they are very strong. But I tend to think that this is not the cause of the huge difference with experimental values. The fact that B3LYP returns reasonable values should be taking into account. Surely it "could be" just a coincidence but let's face it, I is very unprovable that all the results are reasonable due to coincidence (it is very simple to check it out, just try some others well proved functionals).

My guess is that you found the correct nuclear coordinates for the minimum of the PES, but the SCF part of the calculation have serious problems and it converged to some unreasonable electronic "state". This would imply that the MP2 results are ridiculous.
As HF and B3LYP are very different in the SCF part, B3LYP converged correctly.

You could try different initial guesses (Hückel, PA, a previous converged calculation) and different methods (like RHF, OHF, UHF etc.). Other basis sets can also help (a little larger if possible). I would try to avoid effective core potentials! (I understand that this can be a problem for your teaching purposes). Convergence method is of course of main importance (consider quadratic or very damped convergence) . I found almost impossible that calculating inner electrons correlation helps in this case.

Of course that the refinements that you tried are useful, but I think that just to improve quantitatively the results.

I hope that the ideas above help you. I would like to know if you effectively solved your problem.

Regards


Difference between vapor-phase transport and vapor-phase deposition? I am currently studying the textbook Physics of Photonics Devices, Second Edition, by Shun Lien Chuang. In a section discussing The Invention of Semiconductor Lasers, the author says the following:

  At the Solid State Device Research Conference in July 1962, an MIT Lincoln Laboratory group and RCA Laboratories reported extremely high efficiency (85% to 100%) electroluminescence from GaAs diffused junction diodes. Semiconductor lasers were invented during September to October 1962 by four groups within 30 days [9-12] (see the review article by Dupuis in Ref. 13). They were led by Robert N. Hall of General Electric Research Development Center, Schenectady, New York; Nick Holonyak Jr. of General Electric, Syracuse, New York; Marshall I. Nathan of the IBM Research Laboratory, Yorktown Heights, New York; and Robert Rediker
  of the MIT Lincoln Laboratory, Lexington, Massachusetts. Among the four groups, only Holonyak's laser diodes and light-emitting diodes (LEDs) were created from single-crystal GaAs$_{x}$P$_{1 - x}$ alloy material grown by vapor-phase transport and were the only devices emitting in the visible region.

I had already read about vapor-phase deposition in a textbook on MEMS (microelectromechanical systems), but I was not familiar with vapor-phase transport. Reading about these two methods, they sound very similar, and I'm unsure if they're just different names for the same thing, or whether there is some difference between them. I would greatly appreciate it if someone would please take the time to explain the difference between, if any.

The paper of interest is C.M. Wolfe et al., "Growth and Dislocation Structure of Single-Crystal Ga(As$_{1-x}$P$_{x}$)", Journal of Applied Physics 36(12) 3790 (1965).

The key paragraph is:

  The GaAsP of this investigation has been prepared
  by a closed-tube method utilizing a quartz ampoule
  which is cleaned and baked-out under vacuum near
  the softening point of quartz. The halide is introduced
  by subliming a metallic halide, such as PbCI2, PbI2,
  NH4CI, NH4I, etc.,ro into the reaction vessel. Typically,
  2.2 g of source GaAs and GaP, 4-8 mg of excess P,
  40-160 mg of halide, and a dopant are placed in an
  8-cm$^{3}$ ampoule as in Fig. 1. This vessel is then evacuated
  to 10$^{-6}$ Torr, sealed, and placed in a single-zone furnace
  at 950°C with a gradient of 5°-15° along the tube.
  Commercially available GaAs is used with GaP grown
  from high-purity elements by a method previously
  described.

So, yes, it is a vapor deposition technique, but uses a thermal gradient to evaporate from one place and redeposit in another place, controlled by the temperature.


Can a polymer have different resonance structures? Several organic compounds have resonance structures, but can a polymer have resonance structures?

There's no reason why the structural subunits of polymers could not exhibit resonance, provided there are conjugated $\pi$-bonds present. A polymer such as kevlar, for example, certainly benefits from resonance stabilization within each monomeric unit.

Whether there exist any polymers in which the resonance is truly distributed over the entire molecule (if that's what you're really asking), however, I'm unsure. From the perspective of valence bond theory, the electron delocalization required for resonance necessitates a planar geometry with unhybridized $p$ orbitals that align and overlap to form $\pi$-bonds perpendicular to the plane of the $\sigma$-bond skeleton of the molecule. In molecular orbital theory, the sort of extensive electron delocalization that resonance implies requires a bonding molecular orbital, distributed over all participating atoms, constructed from a set of symmetrical, in-phase $p$ orbitals, with all planes of symmetry present in the original molecule(s) conserved in the generation of new molecular orbitals.

Given the rigidity of those criteria, I sincerely doubt that polymers of any great length could exist in which the resonance is distributed over the entire polymer chain. However, that could be just a failure of imagination or a lack of knowledge on my part, and I hope somebody else weighs in with a more conclusive answer.


Does mass equal to moles times molar mass take into account coefficients? I have an equation $$\ce{ 2NaN3 -> 2Na + 3N2}$$ I've calculated there are $\pu{4.4 moles}$ of $\ce{N2}$ (using the data given in the question). I then did $\large{\frac{4.4}{3 \cdot 2}}$ to get number of moles required of $\ce{NaN3}$ (although if this is wrong then correct me). And now I want to calculate the mass of $\ce{NaN3}$ used (knowing there are $\pu{3 moles}$).

Where I have $\pu{ 3 moles}$ of $\ce{2NaN3}$ and I want to calculate mass do I use the $\mathrm{M_r }$of $\ce{M_r(NaN3)\cdot 2}$ or just a single mole of it.

Eg:
 $\text{mass = moles} \cdot  \mathrm{M_r} = 3 \cdot 65 $ or $\text{mass = }~3 \cdot 130$?

What do you mean by $\pu{3 moles}$ of $\ce{ 2NaN3}$? The coefficient (the 2) is the number of moles of that molecule that you have.

If you're looking at an equation with $\ce{ 2NaN3}$ written in it, and you want that 3 times, then you're going to have $\pu{6 moles}$ of $\ce{NaN3}$, so $\text{mass}~= ~6 \cdot 65$ .


Are higher branches of mathematics used in sub-fields of chemistry? I'm wondering is there any branches of Chemistry that make use of manifolds, Riemann geometry, tensors etc. So far I have not come across anything but hoped that the wider community would know of something?

Forgive me if this is too vague a question but I'm just looking for some topics I can research myself.

We use tensors in my research lab to analyze solid state NMR.
In my research laboratory we use manifolds all the time for stimulating annealing calculations.

There are several research groups in physical chemistry which utilize what you are describing, but it will never be an undergraduate topic. One particular group that comes to mind that uses this sort of stuff a lot is the Hirata group at the University of Illinois. They actually made a piece of software called the Tensor Contraction Engine. http://www.scs.illinois.edu/~sohirata/software.html


Acid dissociation constant and equilibrium constant What is the relationship between acid dissociation constant and equilibrium constant for an acid's reaction with water?
If we have this reaction:

$\ce{CH3COOH + H2O → CH3COO- + H3O+}$

I would calculate the equilibrium constant to 0,005241 M at 25 celcius

Acid dissociation constant for acetic acid is as known 1,75 × 10$^{−5}$. How can this be true? The equilibrium constant I have calculate is about 300 times larger than the acid dissociation constant!

An acid dissociation constant is a particular example of an equilibrium constant.
For the reaction:
$$\ce{CH3COOH +H2O <=>CH3COO- +H3O+}$$
The equilibrium constant would be:
$$\newcommand{\ac}[1]{\text{activity of }\ce{#1}}{\bf K_{eq}}=\frac{\ac{CH3COO-}\cdot\ac{H3O+}}{\ac{CH3COOH}\cdot\ac{H2O}}$$
And the acid dissociation constant would be:
$${\bf K_a}=\ce{\frac{[CH3COO-][H3O+]}{[CH3COOH]}}$$
Now the relation between both would be:
$$\newcommand{\acc}[1]{\gamma_{\ce{#1}}}{\bf K_{eq}}={\bf K_a}\underbrace{\frac{\acc{CH3COO-}\acc{H3O+}}{\acc{CH3COOH}\acc{H2O}}}_{\Gamma_0}\frac1{\ce{[H2O]}}$$
Now:
$${\bf K_a}={\bf K_{eq}}[\ce{H2O}]\Gamma_0^{-1}$$
So you're missing some factors.They're not exactly the same.


How can I estimate how much humidity is present in a NaOH granulate? I have a container with NaOH granulate. I want to use it for making soap again, as I did in the past.

When I checked it I saw big clumps because the seal was not good and humidity got in since last time. Yesterday I used it anyway, increasing the amount arbitrarily by 10%, and still was not enough.

I have the following tools:

scale (sensitivity 1 gram),
arbitrary glass or plastic volume meter calibrated with said scale,
temperature probe (up to 70C with 0.5C tolerance and 0.1C sensitivity),
white vinegar,
concentrated lemon juice,
concentrated HCl (I think 36%).

More common "kitchen" chemicals may be available, if you ask in the comments.

How can I estimate how much humidity is present in the NaOH granulate using household items/chemicals?

A measurement performed directly on the NaOH solution used for the soap (around 300 g NaOH and 750 g water) is acceptable.

For a rough estimate, you could do the following:

Weight a sample of it and heat it above 100°C for a while (say, one hour?).
Weight the sample afterwards.

The difference should be the amount of water given away and is an estimate (from below) of how much water there was in the sample.

But be aware that you may have some sodium carbonate depending on storage conditions.


After opening, when is a can of Diet Coke most acidic? (Based on a true story) Situation: Clogged toilet, no plunger.

Goal: Trying to use the acidity from a can of Diet Coke as toilet drain cleaner.
Question: How can I maximize the acidity of a Diet Coke?

Question 1: Should I heat up the Coke (using a microwave)?

Is an acidic solution more acidic when it is hot or cold?

Question 2: Would heating the Coke make the Coke less soluble to Carbon Dioxide (which is acidic)?

Is a solution more soluble to gases when it is hot or cold?

Question 3: What is the loss of carbon dioxide (acidity) during the time taken to microwave the soda?

Coke (or any carbonated drink) is most acidic when it is just opened. The acidity comes primarily due to phosphoric acid. Citric acid if present also adds some acidity. Carbon dioxide (which is present as dissolved form) will increase acidity though by  small amount. Carbonic acid, formed upon dissolution of $CO_{2}$ is a weak acid. Its contribution to acidity, when phosphoric acid is present would be negligible.

Facts

Solubility of a gas is more in cold liquids.
Carbon dioxide dissolves in water. The dissolved $CO_{2}$ is now in equilibrium with carbonic acid. The so formed carbonic acid is much less in concentration than $CO_{2}$ itself.

Considering small effect on acidity from carbonic acid:
 Just opened coke can will have higher carbonic acid (whatever be its amount), due to higher (relatively) dissolved $CO_{2}$. As temperature increases $CO_{2}$  fizzles out to air, and dissolved $CO_{2}$ is reduced  so is carbonic acid.

PS- Microwaving coke can will be dangerous. There is a hazard of explosion and severely damaging your microwave oven of-course !

I am not sure if cleaning with coke will help in your situation.


50-90'C color changing chemical to use as warning marker for hot electronics? I am testing electronics with many potentially unstable contacts, and I wish to monitor if they rise above 50 to 90 degrees Celsius, i.e. using carbon paper that would darken progressively from 50 to 90.

The substance should stay a specific warning color if it goes too high and not change hues again same as thermochromic pigment, so that I can detect high temps.

I hope you know a chemistry trick that could help? Thank you.

There are commercial  sources for thermochromic permanent-change ink, but you might also try thermal printer paper. You might need to check if there is a time-dependence on the color change, as well as temperature dependence. The printer paper is available at Amazon, Staples and elsewhere for less than US$20 / package, but you might just test a cash-register receipt, many of which are on thermal paper. According to this PDF, most start to darken at 79 C.


If caffeine is an alkaloid, why is coffee acidic? I have just put together on my mind these two facts: caffeine is an alkaloid and brewed coffee is slightly acidic (pH = 5).

My Biology teacher and my Chemistry teacher could not elaborate satisfying answers, but my Physics teacher said it could be because of substances to the industrial coffee powder (hillbillies in Brazil, for instance, do not display gastritis attained from too much coffee).

Coffee contains hundreds, if not thousands, of other compounds in addition to caffeine.  Included among these other compounds are many acids.  Many small, organic acids such as  citric, malic, lactic, pyruvic and acetic acid are present, but both quinic acid and chlorogenic acid (and their derivatives) are usually present in even higher concentration.  Phosphoric acid, an inorganic acid, is also present. The exact concentration of these various acids depends upon processing variables such as roast conditions and grind size.  Here is a link to an interesting, one-page discussion of the subject.
quinic acid

chlorogenic acid

Edit: In Season 3, episode 6 of "Breaking Bad", we meet Walt's new lab assistant Gale, who is brewing coffee using a lab glassware set-up.  Walt says that it is the best coffee he has ever tasted and Gale responds by mentioning "quinic acid" as something to be considered in brewing great coffee.


What is meaning of unit (mg-Pt/cm^2) in relation to fuel cells? I do not understand what the meaning is of the unit

mg-$\ce{Pt}$/cm$^2$

which I have come across in reading about the theory of fuel cells.

In simple terms, what does this unit mean?

The unit in question means milligrams of platinum per square centimeter.

In your application - fuel cells, and specifically the membranes used to fabricate them - it describes how much of the metal is present per square centimeter of membrane.


Does Nitrogen dioxide change colour if we put it in a constant temperature of e.g. 100 kelvin? The changing of the colour of $\ce{NO2}$ is because of a change in temperature, resp. heating and cooling?

The changing color of $\ce{NO_2}$ is because it exits in equilibrium with $\ce{N_2O_4}$ as shown in the following equation $$\ce{2 NO_2 <=> N_2O_4} + \mathrm{58~ kJ/m}$$ $\ce{NO_2}$ is a reddish-brown colored gas, while the $\ce{N_2O_4}$ is colorless.  Considering Le Chatelier's principle,

Taking heat out of the equilibrium will push it to the right
Putting heat in, will push it to the left
Increasing pressure will push it to the right

As we shift the equilibrium we change the concentration of the components, and consequently the color of the gaseous mixture.

Back to your question, in the solid state (which is where it would be at 100°K), the equilibrium is shifted completely to the right.  So if we had a tube of gaseous $\ce{NO_2}$ and $\ce{N_2O_4}$ at room temperature it would have a distinct red-brown color.  If we then plunged it into a bath at 100°K, it would solidify to $\ce{N_2O_4}$ and appear colorless.


Weird Wikipedia Section on Oxidizing Behavior of Nitric and Sulfuric Acids It has been my understanding that when nitric, perchloric, or sulfuric acids act as oxidizing agents, they must exist as whole $\ce{HNO_3}$ or $\ce{HClO_4}$ or $\ce{H_2SO_4}$ molecules that are still protonated. For example, putting copper in dilute sulfuric acid does nothing, but using concentrated sulfuric acid does cause a redox reaction. This agrees with the fact that only concentrated forms of these acids can oxidize; only concentrated forms have undissociated/whole molecules. A dilute solution would contain mostly hydronium and the conjugate base of the strong acid, thus not be able to oxidize. However, the "General Properties" section of this Wikipedia article flat out contradict that.
It says,

However, even dilute nitric acid can oxidize copper to $\ce{Cu^{2+}}$ ions, with the nitrate ions acting as the effective oxidant. (Emphasis: nitrate ions).

Then below it says that concentration is a factor for sulfuric acid. Is something amiss here?

UPDATE: I now embrace what user @matt_black is saying, and by now
I'm not even surprised that something like this could happen. Chemical guidelines are just guidelines. However, I'm now even more curious as to this departure from the trend. If he or someone else would like to give an explanation of the mechanism of reactions involving oxidation (of copper, for instance) by $\ce{HNO_3}$ and $\ce{H_2SO_4}$ that accounts for this anomaly, that would be great.

It is chemistry that is being weird not Wikipedia

The trouble with simple explanations of oxidising acids is that there is no simple explanation that covers all the cases. The specific reactions leading to oxidation are complicated and have a variety of different mechanisms.

So copper in sulphuric acid can be oxidised by a reaction that involves SO2 production. That has a completely different mechanism to the reaction with nitric acid and a different concentration dependence.

And, given the variety of reactions, the need to have protonated molecules of the acid is not a good generalisation: it depends on the specific reaction that is happening.

The oxidising reactions depend on the substrate being oxidised as well as the acid. Which mechanism will dominate isn't easy to determine without trying the reaction. So don't trust simple generalisations.


Definitive Guide To Removing Stains From Baby Clothes I'd like someone to explain, scientifically, what makes some baby clothes' stains so hard to remove, and what the right way to treat the stains is, chemistry-wise. A lot of internet sites have solutions (pardon the pun) that are 'my experience' and anecdotal, which is better than nothing, but not always transferable (and sometimes may damage clothes).

So yeah... how do you get out the hard to remove orange stains (e.g. from iron, or carrot/sweet potato soup)? How about other things? Why? I'd love someone to make this the definitive guide to removing stains from baby clothes, if you can :D. (Feel free to expand beyond my stated question above so there's just a one-stop reference question here.)

Wash the garment in warm water with an enzyme detergent and oxygen bleach combo, such as Bio Kleen Oxygen Bleach Plus. To get rid of baby food stains, dunk the clothing in cold water, then soak the stain for at least 15 minutes in equal parts water and rubbing alcohol, which is safe for most fabrics when diluted.


Calculating laser wavelength/power to cause emission of light in a gas? It's been a long, long time since I've had to worry about electron excitation levels and band gaps. Please forgive (and correct!) my terminology and misconceptions.

I've become interested in volumetric displays. In particular I'm looking at a pair of lasers scanning a volume of gas such that the intersection of the lasers would supply enough energy to cause the gas to emit visible light. Since this is the start of my research I would like to investigate a number of gasses in order to determine a number of gasses which might be suitable for this application. I have, however, forgotten how to determine the amount of energy needed to cause an outer shell electron to emit a photon, and then trying to back-calculate the required wavelength(s) and power requirements of the laser emitters.

I'm looking for some information on how to identify the required information for a given gas and how to calculate the energy required to cause visible light emission, and also information on how the wavelength of laser light impacts the ability of the gas to absorb the energy with the intended effect of emitting visible light. Links to introductory texts are always helpful, as are answers which help me get the terminology correct so that I can improve my searches for this information.

Two-photon excitation using femtosecond lasers that emit in the infrared range is a technique used in living cell imaging via fluorescence microscopy. Advantages of the technique, such as the minimized light scattering (~ 1/$\lambda^4$) using long wavelength for excitation and the fluorescent dyes used are discussed here.

For large volumetric displays, a different technique was/is under development in a joint project of the Japanese National Institute of Advanced Industrial Science and Technology (AIST), the Keio University and an industrial partner.

Here, fluorescence does not play a role and special gases are not needed either.

The images can be displayed in air and are formed - pixel by pixel - by plasma dots generated in the focal point of pulsed infrared lasers with a pulse width of 1 ns and a repitition rate of 100 Hz.


Is a transition metal with 7 d electons ever diamagnetic? A transition metal with 7 d electrons forms a complex ion with cyanide. I know that cyanide is a high field ligand, which means the electrons will pair before moving to the next energy level, but wouldn't there still be one unpaired electron? And wouldn't that make it paramagnetic? I was told that my prediction was wrong and that it was in fact diamagnetic, but I don't understand why.

As I noted in my comment above, I think the question is whether there are 7d electrons in the atom or in the ion.

For example, if you take 3 electrons from 7, you get 4d electrons in the $\ce{M^{3+}}$ ion which obviously could (and should) be diamagnetic.

This is all, of course, a guess given what little you posted.

Let's say I had a similar question in my class. Students might then assume this $\ce{M^{3+}}$ is octahedral. (Personally, I'd make it a bit more obvious in the question.)

Given that the ligand is high-field, you get an initial MO diagram like this:

But... we're not done. This electron configuration is subject to a weak Jahn-Teller distortion. So you'd end up with an MO diagram like this:

That, of course, is diamagnetic.


Why is absorbance in UV-VIS spectrometer negative even after calibration? I was collecting data to plot a calibration curve using pure beta-carotene sample. Prepared 5 different concentrations of beta-carotene dissolved in methanol. Pure methanol was used as blank. However, data collected for all 5 concentrations observed was negative (between -0.005 and -0.010) at 610nm ( orange region along the spectrum). Please advice.

Possible causes include:

If it is a dispersive instrument (simultaneous recording of blank and sample in two cells): is the logic about which one is sample and blank set correctly?  At least some spectrometers by PerkinElmer and Shimadzu allow to swap how front and rear path are managed by the software.
How long did you run the spectrometer to warm up?  Maybe the instrument wasn't yet stable, elder spectrometers need 10mn+ to warm up all their internals.
Check the lamp for its run time and compare with the specifications by the manufacture.  Depending on the setup, the instrument's software may tell you this, or / and the lamps may have a counter (the little box between the plug and the mount of the lamp shown in the picture below).  If the lamp's intensity over time is not stable any more while recording sequentially blanks and samples, you will have a hard time to record faithfully spectra.  This cause typically remains unnoticed with mono-beam instruments less well maintained and run until the lamp «breaks» (multiple times older than the recommended run time).

(image credit)

Check your instrument with a known standard.  This either may be a liquid solution of a known concentration (e.g., by SigmaAldrich) to purchase, or already sealed into a cell, e.g.

(image credit)

or solid samples of metals deposit on glass or quartz

(image credit)

Don't forget that groups active in materials science / spectroscpy / optics may have such references to calibrate their instruments regularly and may be happy to share with you how to use these accurately, too.


Determining the most stable enolic form of 1-phenylbutane-1,3-dione? Question

Which form of 1-phenylbutane-1,3-dione is more stable?

Answer

3-hydroxy-1-phenylbut-2-en-1-one

My question
I've been told that between the above two, 3-hydroxy-1-phenylbut-2-en-1-one is more stable because the double bond is in the middle which is unclear to me.
Could someone explain the above statement?

In opposition to the line of thought presented, I argue that the 4-hydroxy isomer is more stable than the 3-hydroxyl form for that the conjugated $\pi$-electron system is larger, than in the former.
This is based on multiple assumptions.

For one, given a molecular framework which may have multiple $\ce{C=C}$ bonds, if other parameters are not altered, the system with double bonds in conjugation is more stable than the one with isolated double bonds. As an example, the computed free enthalpy of formation, $\Delta{}_RG$ about 1,5-hexadiene (two isolated $\ce{C=C}$ bonds) is $\pu{92.70 kJ/mol}$ (link to computation on MolCalc.org), the about 1,3-pentadiene (two $\ce{C=C}$ bonds in conjugation) $\pu{86.57 kJ/mol}$ (link to computation).  And the less endothermic a compound, the more thermodynamically stable.

For the other, I said the 3-hydroxyl isomer would contain a less extended conjugated $\pi$-system, than the 4HP isomer.  Elements to participate in this system are the benzene ring, the carbonyl group, and the simple $\ce{C=C}$ bond, present in both.  If you imagine a water wave running from one to the other other end of the $\pi$-system, $\ce{C=O}$ only is tangent in 3HP, while in 4HP the wave runs all across these bonds.  Thus, equally a reason why the 4HP isomer is more favourable, than 3HP.
For a raw simplification, you may consider the benzene ring as "super atom" and count the longest pathway accessible. 3HP: 4 atoms, 4HP: 5 atoms participate in the $\pi$ system; again energetically more favourable for 4HP, than for 3HP.

Sadly, the engines to run MolCalc.org's web site constrain the computations to molecules with 10 non-H atoms maximum.  Thus, contrasting to earlier posts by others (e.g., this by Ben Norris), I'm not able to support my arguments by a similar computation for your substrates (12 non-H atoms).


Problem digesting sample for Ti analysis using ICP-OES I'm having trouble digesting a paper sample for $\ce{Ti}$ analysis using our ICP-OES. I've been digesting my ash samples with 50% $\ce{H2SO4}$, but I'm still getting lower $\ce{TiO2}$ results than the target. What is the best way to digest paper sample with $\ce{TiO2}$?

If you want to stay with a sulfuric acid type of digestion, there is this(1) paper that describes a microwave digestion using copper(II) and hydrogen peroxide to facilitate the digestion.

From my personal experience, if your lab is properly equiped, my first choice would be an $\ce{HF}$ digestion or a nitric/perchloric acid digestion.

1)"A rapid and simple microwave-assisted digestion procedure for
spectrophotometric determination of titanium dioxide
photocatalyst on activated carbon", ScienceDirect, Talanta 71 (2007) 1867–1872


Creating a 10 ppb Lead Nitrate Solution I’m trying to make a 10 ppb Lead Nitrate solution as part of a concentration calibration step for a Beer Lambert’s Law experiment. I plan on performing a dilution in three steps. The calculations have been evaluated using:

$c_{1}v_{1}=c_{2}v_{2}$

Step 1: Creating a 1000 ppm (=1000 mg/L) stock solution (let’s call it A) by dissolving 1 gram (1000 mg) of Lead Nitrate in 1 L of distilled water [creating 1L of stock solution A]

Step 2: Creating an even more dilute  stock solution B of concentration 1 ppm (=1 mg/L = 1000 ppb) by diluting 1 mL of stock solution A with 1999 mL of distilled water. [creating 2L of stock solution B]

Step 3: Creating the desired solution C of concentration 10 ppb by diluting 0.25 mL of stock solution B with 24.75 mL of distilled water. [creating 25 mL of desired solution C)

Does this seem like an adequate method to obtain the desired solution? Should I add 1% v/v of Nitric Acid between every dilution step as a stabilizer to keep the lead ions in solution for as long as possible?

I only plan on using these solutions for about 10 days, so I’m not too concerned about the longevity of the solutions.

Your dilution approach can be heavily improved.

Rule no. 1: 2+2 $\neq$4 in solution preparation from an analytical chemist's perspective. An extreme example is that you add 50 mL of water+50 mL of ethanol $\neq$ 100 mL solution. If you add a solid NaCl to 1 L of water, the volume may not remain 1 L. It may decrease!

Analytical dilutions are always prepared in highest quality volumetric glassware (Called class A glassware), which comes in fixed volumes. You might have 2 mL, 5 mL, 10 mL, 25 mL, 50 mL, 1000 mL flasks in the university. Similarly, pipettes comes in fixed volumes. The graded ones (so called Mohr's) are not that great.

With these limitations, you design dilutions, in such a way that volumes are not fractional so a volume of 24.75 mL for dilution is a no-no.

The dilution relation is handy $$\mathrm{C_iV_i=C_fV_f}$$, where i and f indicate initial and final concentrations (C) and volumes (V).

So think about it now in order to prepare 1000 mg/mL you would like to dilute it 1000-fold in steps to make it a 1 ppm solution or 1000 ppb $\ce{Pb(NO3)2}$. From there you would like to dilute it 100-fold to make it 10 ppb.

The first question you would like to think now, is what volumetric glassware is available.


What is the definition of current efficiency? What is the exact definition of current efficiency of an electrolytic cell? I've looked online and there is no suitable answer to it.

I thought current efficiency would be the amount of electrons that actually participate in a desired reaction. For example, if we are electrolyzing a an aqueous solution of $\ce{FeCl3}$ and we have a $\pu{10A}$ current running for, say, $\pu{100 s}$, at $60\%$ current efficiency, two possible products are $\ce{H2}$ gas and reduced $\ce{Fe}$ metal. Current efficiency of $60\%$  means that half of the electrons that pass through this circuit would reduce the $\ce{Fe}$ instead of reducing $\ce{H2}$.

However, there is another way to think about it, in that it means 60% of the theoretical mass yield of $\ce{Fe}$ comes out of the reaction instead of my original definition, which is different because the reduction of iron (III) requires 3 electrons rather than only 2 electrons for the reduction of $\ce{H2O}$/$\ce{H+}$ ions. So what is the actual definition?

The theoretical definition is the ratio of electrons used to make the useful product out of the total electrons that were supplied. The problem in the chemistry literature with efficiency definitions, however, is that you will more than likely find that different people will use different definitions for one reason or another (easier to calculate, some type of comparison in their work, or simply because it sounds or looks better). The right way to deal with efficiencies, then, would be to define how the efficiency is calculated in order for the reader to know what you are talking about. If this is an academic setting, I would stick to the theoretical definition, but ask your professor as to what their favorite definition would be.


Where can we buy chlorinated paraffin for lab application? Any idea where to buy chlorinated paraffin for lab application? We need very viscous analog of $\ce{CHCl_3}$.

Should we think about synthesizing it?

I found a lot of places where I could buy 5 tons:) But we need only few liters for the rest of our life.

Sigma-Aldrich is a global supplier of purified chemicals for research users.

They have a whole page of the haloalkane compounds they sell (in small quantities such as you desire).  You might be interested in compounds such as trichloropropane, pentachloroethane, etc.

A reason it might be difficult to get the higher-molecular weight compounds is that they are toxic and difficult to dispose of properly.  In the US, their use is subject to increasing regulatory requirements.


How to accurately determine mass transfer using balances? I had troubles the other day using a balance that goes to the tenth of a milligram place (i.e. .0005 vs .001 g), when trying to determine how much solid I transferred. I would take some amount of mg, record it, and then set it aside. I'd take the receiving flask, tare it, and add the solid to said flask. The # I would get would either be wildly lower or slightly lower. I am very sure I am not mispouring anything.

So I am wondering, is it possible these sensitive balances cannot handle things with larger masses, or if the receiving flask is not perfectly flat, would that mess up the mass? Similarly, would the location where I put the flask or weigh paper on the balance matter (e.g. dead center, to the left, right, etc.)? Is there another thing I could do? technically I could take the difference in weigh paper with solid vs without, but that doesn't tell me what got into the receiving flask.

Accurate weighing requires you to be aware of enormous amount of factors. It is not possible to cover them all in one answer. For starters, where do you put your balances? Are they levelled? Are there any heat or vibration sources nearby? Are there any sudden air flows? Is your mains power reliable? Were the balances calibrated properly? Is the ambient temperature or humidity too high or too low? Do they change much? Is there much static electricity (depends on the materials you are working with and on the ambient humidity)? Do you use gloves? How much do you weigh? I mean, not you weigh, but isn't it something less than the minimum weight requirement for your balances? Et cetera, et cetera.

Returning to your questions: in general, sensitive balances can handle large masses which are within the specification. However, the measurement uncertainty depends on the mass. Then, if the receiving flask is not perfectly flat, there should not be any problem unless the flask is wobbly or unsteady. The location where you put the flask should be as close to the center of the weighing pan as possible. It matters.

As for the other things you could do, I'd like to recommend you to check out some weighing practice manuals. You can start with the balances manufacturers - Mettler has brief Good Weighing Practice whitepapers, maybe others like Sartorius have something similar. Analytical chemistry handbooks should have something to offer. Then, check out the manual for your balances. Aren't you trying to weigh something more accurately than the accuracy of the balances (remember, accuracy $\ne$ resolution)? Then, if you don't, just ensure that you are following the recommended weighing practices as closely as humanly possible.


What is a hydrogen bond? I was doing a little research on the Chemistry of water, and it said that water molecules have a hydrogen bond between them. What is it, and what does it do to water?

Firstly you should consider this simple model of atoms: There is a compact nucleus and smeared out electron clouds. In H this cloud is simply spherical. In many other atoms (e.g. O, C, N, S, etc.) the clouds are usually arranged like this (I did not find english site for some reason) (imagine tying 4 ballons together, forming a tetrahedron). This also explains the general geometry of water (and ice).

As you know in water the hydrogen is covalently bonded to oxygen. But due to the different properties of O and H the electrons of these bonds are not distributed symmetrically but much nearer the oxygen, leaving the hydrogen slightly positively charged. (Note that different elements cause different levels of asymmetry, a higher difference in electronegativity corresponds to a stronger asymmetry. This is why hydrocarbons ( C-H-bonds are rather apolar) are not very soluble in water.)

Now these positive H nuclei can bond to "full" electron clouds (e.g. the two of oxygen in water that are not bonded to hydrogen)

For something to be suluble in water you need to have many of those polar bonds and "free electron clouds", so e.g. sugars are easily soluble but e.g. alkanes are not.

These illustrations hopefully make now more sense. Sadly I could not find an english text about the tetrahedron model, which I consider didactically very useful.

Edit: Note that although hydrogen bonds are due to the electrostatic attraction, they are quite distinct from ionic bonds in the aspect that they have a definitive orientation, i.e. in ice the bonds still form tetrahedrons and not a cubic lattice or something like that.
I do not know a very good explanation but I think this is simply because the scale is much smaller; if the angle changes just a bit, the negatively charged part of the hydrogen atom interferes already.


Is crushing an avocado a chemical or physical change? Applying pressure can result in chemical changes, does this hold true for an avocado?

I don't think so.  A chemical change implies changing molecular-level structure usually accompanied by transferring electrons.  When you squish an avocado you're only rearranging the fats, carbohydrates, and fiber in it - I don't think a chemical reaction occurs.

If you left it out, and it turned brown (oxidized), that would be a chemical change!


Understanding dissolution of CO₂ in Selexol and Rectisol I'm trying to understand pre-combusiton carbon capture processes as described in this presentation.

I'm told that solvents such as chilled methanol ($\ce{CH3OH}$) and dimethyl ethers of polyethylene glycol ($\ce{C_{2n}H_{4n + 2}O_{n + 1}}$) can dissolve $\ce{CO2}$ from a gasified coal fuel (before combustion) under the action of increased pressure.  I'm struggling in understanding how the increased pressure induces the dissolution of $\ce{CO2}$ from the gasified coal.  I'm also a bit rusty with my chemistry, by saying $\ce{CO2}$ is dissolved by the solvents, what is really going on, chemically?

Simply, $\ce{CO_{2}}$ is dissolved in those solvents in a similar way to gases dissolved in water. The molecules do not react with the solvent but mix. The higher the pressure of the gas you pump into the liquid, the more that dissolves in it (rather than turning into bubbles and escaping). In fact, amount of gases dissolved in liquids can be measured both as a concentration (mol/L) or as a partial pressure (Pa).

Henry's law states that:

  "At a constant temperature, the amount of a given gas that dissolves
  in a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid."

The formula is:

$p = k_{H}c$

where $p$ is partial pressure of the gas in solution, $c$ is its concentration and $k_{H}$ is a constant based on what the solute is (in this case $\ce{CO_{2}}$), what the solvent is (eg methanol) and the temperature.

The harder you push, the tighter you can pack it in.

Here is a picture:

In contrast, with the amine-based solvents, the $\ce{CO_{2}}$ actually reacts, which is why a degasification process needs to happen to release it.


What is negative adsorption? In adsorption, the concentration of adsorbate is greater on the surface. So, how is it possible that the solvent will be more on the surface and the solute (adsorbate) in the bulk… or does solute not always mean adsorbate?

Solute does not mean adsorbate always.An Adsorbate is a substance that is adsorbed  on a surface of another substance. In a dilute solution, negative adsorption occurs as solvent is adsorbed by the adsorbent and not solute. In positive adsorption, solute is adsorbed.


Reaction of Metallic Tin with NH4Cl and treatment with Sulfur Metallic $\ce{Sn}$ is reacted with $\ce{NH_4Cl}$ and the resulting solution is treated with Sulfur.

What is the gaseous product?
What is the color of final precipitate?
The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple    coloured solution. What is the composition of the solution?

I do realise the answer to (3) will be the purple of Cassius i.e colloidal gold in $\ce{Sn(OH)2}$ (as per Vogel's Qualitative Inorganic Analysis textbook fifth edition). But I can't figure out the answer to the first two questions (I tried Googling with no fruitful search).

Note: this is an incomplete answer because question is unclear till now

  Metallic $\ce{Sn}$ is reacted with $\ce{NH4Cl}$..

$$\ce{2NH4Cl + Sn → SnCl2 (white ppt.) + 2NH3 (both aq and g.) + H2 ^}$$

Ammonium chloride react with tin to produce tin(II) chloride, ammonia
and hydrogen. (source)

Some tin(II) chloride will react with aqueous ammonia to form tin (II) hydroxide and amm. chloride.(source)

$$\ce{SnCl2 + 2(NH3•H2O) or NH3(aq) → Sn(OH)2 or SnO.H2O(white ppt.)  + 2NH4Cl (white ppt.)}$$

  ..and the resulting solution is treated with Sulfur.

I am suspecting that is hydrogen sulfide and not sulfur.(Are you sure?)

$$\ce{SnCl2 + H2S → SnS(dark brown ppt.) + HCl}$$

Tin(II) chloride react with hydrogen sulfide to produce tin(II) sulfide and hydrogen chloride. Tin(II) chloride - concentrated solution. Hydrogen sulfide - saturated solution. (source)

So, answer to your questions:

What is the gaseous product?

  Hydrogen, Some ammonia gas.

What is the color of final precipitate?

  Stannous sulfide ($\ce{SnS}$)- dark brown ppt.

The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple coloured solution. What is the composition of the solution?

  $$\ce{2[AuCl4]- + 4Sn^2+ + 2H2O -> [2Au + Sn(OH)2)](purple ppt.) +
3Sn^4+ + 2H+ + 8Cl-}$$

  (Vogel)

  Note: $\ce{AuCl3}$ exist as $\ce{AuCl4-}$ ions in solution.


How to predict the 31P NMR spectra? What can I predict about the $\ce{^{31}P}$ spectra of $\ce{TiP2O7}$?
To get the required information about the crystal structure of the compound

Here is a screenshot from the ICSD website:

Since there is only one oxidation state (+5) for $\ce{P}$, there will be only one peak. And from the Wyckoff symbol, 4c, the intensity of the signal is 4, i.e. 4 $\ce{P}$ atoms in the unit cell.
I can't go any further to complete the task and extract all the relevant information. I'm using SIMPSON software to simulate the spectra. Can somebody walk me step by step through this?

If the question you were asked to answer was literally "what can you predict about the $\ce{^31P}$ NMR spectrum of $\ce{TiP2O7}$ (from the crystal structure), then I believe you have vastly overcomplicated the problem.

This is a question about solid-state NMR (as $\ce{TiP2O7}$ is a solid).

In solid-state NMR of crystalline compounds, the chemical shift depends on the exact arrangement of atoms within the crystal unit cell. The number of signals (for each element) depends on how many non-equivalent atoms of a particular element there are within one unit cell.

From the screenshot, it says that "formula units per cell" = 4. As the formula is $\ce{TiP2O7}$, this indicates there are 4*2 = 8 different phosphorus atoms in the unit cell.

So the answer is:
You expect to observe 8 phosphorus signals in the $\ce{^31P}$ NMR spectrum.

There is no need to simulate the NMR spectrum with simpson. In fact, it is not possible to simulate the spectrum using simpson with only the crystal unit cell parameters. In order to set up a simulation in simpson, you actually have to supply the chemical shift and the interactions between the sites. You can calculate the interaction between the sites from the distances specified in the unit cell, but there is no information about chemical shifts!


Nanocar racing competition and the race tracks used In the Nanocar race competition the molecules are raced on a gold surface.
Is there a specific property of gold that it forms a good "racetrack"? On what criteria does it compare with other metal surfaces?

While I can't find specific justification for the surface, the first competition involved both gold and silver "racetracks."

Drivers gear up for world’s first nanocar race
How to build and race a fast nanocar

The competition involves propulsion and imaging using STM so the substrate must be conductive. Gold (and silver to a lesser degree) is particularly useful, since it won't easily oxidize and is easy to purify.

I've seen several comments about having 'tracks' in the gold surface, which suggests a well-defined crystal face with reconstruction, e.g. Au(100):

(Image from Wikipedia)


Is PbO ionic compound? In IIT-JAM 2018, There was one question, which one of the following oxides are ionic?
There was one option PbO also. According to the official answer key, PbO isn't ionic.

Although I know this distinction between ionic and covalent isn't defined sharply. But Pb is a metal and Oxygen is a non-metal. So according to me, it should be Ionic. I tried to find on the internet I couldn't able to find out about PbO, although PbO2 is defined in between of ionic and covalent according to this paper. PbO should be more ionic than PbO2 since oxidation state of the Pb is lower, it will be less electronegative.

Even $\ce{PbCl2}$  is considered as Ionic according to many books, where Cl is less electronegative and much bigger in size as compared to O, So At least PbO should have more ionic character than $\ce{PbCl2}$  according to the Fazan's rule.

So should I challenge the official answer key? If yes, then on what basis? Which reference should I send to them?

Iono-Covalent Character of the Metal–Oxygen Bonds in Oxides: A Comparison of Experimental and Theoretical Data

If you want a reference to challenge the question, try page 248 of Advanced Study Guide Chemistry.  There is a table on that page that says the PbO is "ionic".


Where can I obtain tables of P-V-T/compression factor data? Some recent research I'm dealing with has required me to look for tables of experimental pressure/volume/temperature values (equivalently, a table of compressibility/compression factors in terms of any two of pressure, volume, or temperature) for various substances. Thus far, I've managed to obtain compression factor tables for a short list of substances from Perry's Chemical Engineers' Handbook.

I am wondering, however, as to where I should be looking for more extensive tables (e.g. tables with finer sampling), as well as tables for substances that were not included in that handbook. What books/websites/journal articles should I be looking at?

My attempts at searching on the Internet have not been very fruitful; I would appreciate knowing what keywords should I be using with the usual search engines to get better results.

One place to look for simple fluids is http://webbook.nist.gov/chemistry/fluid/.  But there are only a limited number of compounds treated.

Be warned that the data in the NIST Webbook is smoothed.  Collecting data of this sort is very difficult.  Many different sets of observations are combined and fitted to an appropriate equation.  What is at the URL given above is computed from that equation.

If you are looking for overall compression factors, this will do very nicely.  If you want five figure precision in compression factors, well, this won't do.

Another place to check is the  which is a primary source of raw data.


Reasons for solid or liquid soap Soap is made by a saponification reaction, where a fat reacts with hydroxide ions to form a surfactant and glycerol.

To make a solid soap $\ce{NaOH}$ is used, while $\ce{KOH}$ is used for liquid soaps.

I don't understand why the alkali metal has such a great impact on the state of matter. Usually the argumentation is based on intramolecular interactions, such as Van-der-Waals forces or hydrogen bonds, but if the same fat is used once with $\ce{NaOH}$ and once with $\ce{KOH}$ the resulting surfactants are basically the same, so the interactions shouldn't differ too much.

The only reason I could think of is the size of the alkali metal. Potassium has an atomic radius of 231 pm which is quite a bit more than the radius of sodium, being 186 pm. But why the atomic radius should have an impact on the state of matter of the soap is still unclear to me. Maybe I'm also completely wrong with this assumption.

A while ago, this question was already asked here on this forum, but I wonder if the given explanation is the only reason for the different state of matter. It's absolutely true, that the reactivity of the alkali metals increases from top to bottom, but can this solely explain the phenomenon? @rch provides the solubility of $\ce{NaOH}$ and $\ce{KOH}$ to back up his answer, but I don't think that this is sufficient. You cannot simply change the hydroxide ion by a fatty acid and assume, that there are no substantial amendments in the reaction behavior, can you?

Does anyone can explain this in more detail?

I don't know if anyone is still looking for the answer but here I go anyway because I spent 40 minutes researching this for an assignment
TL;DR - The better solubility of potassium salts is the key factor, but not in the way one would initially suspect. Industrial processes and the efficiency of large-scale soap making explains the choice.
The reason why potassium-based soaps are 'soft' or liquid is to do with the manufacturing process and the difficulty in separating them (R-COO-K) from the glycerin/glycerol after reaction with the soap lye (The alkaline salt solution used to make the soap e.g. KOH). The better solubility of the potassium salts contributes to this significantly.
Glycerin/glycerol has a lot of good properties for making a liquid soap too. It's colourless, viscous, non-toxic to humans and even has some antiviral and antimicrobial properties. It makes sense to utilise these and make the soap liquid instead of trying to get it out. Also it is often cost ineffective to dissolve out the glycerin with large amounts of expensive KCl or other K salt. Comparing this to sodium-based soaps, it takes smaller amounts of cheaper Na salts to separate the by-products.
It all comes down to money I guess.
Here are my sources, read them if you want to know more:

https://pubs.acs.org/doi/pdf/10.1021/ed025p379
http://www.madehow.com/Volume-2/Soap.html
https://pubchem.ncbi.nlm.nih.gov/compound/Glycerol


Can I get an accurate temperature reading from a flask with a built-in thermometer well? I recently purchased a flask with a built in thermometer well, but I have discovered that simply setting a rod thermometer in the well does not produce a very accurate reading. I was wondering if there was maybe a specific method for measuring temperature in wells that I am not aware of. Here are some pictures of the flask:

The thermal conductivity of air isn't very high, when compared to organic solvents or silicon oils used for heating baths. Consequently, one doesn't place thermometers in the vapour phase above the liquid when monitoring or controlling temperatures. (That is, of course, except for distillations, where you want to measure exactly that - the temperature of the vapour phase in the head.)

In your case, it might help to fill the well with the same silicon oil that you use for the heating bath. In order to keep the rod thermometer in place (rather than letting it float), you might want to use a Quickfit adapter (similar to this one) that fits the joint of the well.


Is it possible to compress an atom to infinite density? I know, by nature, normally atoms are stable. But I wonder, if it is possible to compress an atom, say the Hydrogen atom, to infinite density by applying external force? After all, electrons and protons would attract each other by electrostatic forces and nuclear forces are also there at least up-to a certain distance...

Has it been done?

The reason I thought of it, was because I was trying to find the ratio of the amount vacuum to the amount of matter in the whole universe...

We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near $10^{-33}$ centimeters and $10^{-43}$ seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about $10^{-5}$ GRAMS or $10^{19}$ Billion Electron Volts (BeV or GeV). It is a simple matter to construct a 'Planck density' by computing $10^{-5}$ grams / $\text{(} 10^{-33} \text{centimeters)}^{3} = 10^{94}$ grams/cc. Many physicists think that this represents the highest density for any natural compression mechanism, because at higher densities, space-time itself transmutes into some type of quantum mechanical system for which 'density' is not a meaningful concept. So, if fundamental particles are truly without any structure larger than the Planck scale, then they can be packed together into a 'system' with a density of $10^{94}$ grams; the highest achievable density for energy/mass in this physical universe. Even within black holes, the famous 'Singularity' may not have infinite density, but merely a finite density near the Planck value. This is how 'quantum gravity' is thought to regulate, or eliminate real physical singularities; something which ordinary general relativity is not able to do.

I read it in a book Elegant Universe, by Brian Greene. This book has a whole seperate chapter for this.


Effect of concentration on rate How do I solve this question?

My thinking is this: since O2 is not a reactant in any step, and since the reaction rate depends only on the reactants, increasing the concentration of O2 does not affect the rate of the reaction.

Is my thinking correct?

Question is solved. Poutnik taught me to do it using steady state approximation :)

Just adding the image for somebody who might find this in future.


Why are oxime geometrical isomers stable? 231st page of Clayden OC 2nd edition:

  Unlike the geometrical isomers of alkenes, however, those of an imine
  usually interconvert quite rapidly at room temperature. The
  geometrical isomers of oximes on the other hand are stable and can
  even be separated.

Why are oxime geometrical isomers much more stable than the geometrical isomers of immines?

It is the same reasoning that is used to predict relative barrier heights in amines.

In the example pictured below, the imine starts with the nitrogen lone pair in "more or less" an $\ce{sp^2}$ orbital.  In the transition state for interconversion, the lone pair is in a $\ce{p}$ orbital.

The term "more or less" is important.  If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit more p character than $\ce{sp^2}$ (say $\ce{sp^{2.2}}$ for example), then it is already closer to the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be reduced.  On the other hand, If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit less p character (more $\ce{s}$ character) than $\ce{sp^2}$ (say $\ce{sp^{1.8}}$ for example), then it is further from the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be greater.

In the case of oximes ($\ce{X=OH}$) or other molecules where the substituent on nitrogen is electronegative (electron withdrawing), the electron density in the $\ce{N-X}$ bond is shifted towards the electronegative atom.  Consequently there is less electron density in the nitrogen portion of this bond.  As a result the nitrogen rehybridizes. Since $\ce{s}$ orbitals are lower in energy than $\ce{p}$ orbitals, orbitals with increased $\ce{s}$ character are better able to stabilize electrons. Since there is less electron density in the orbital nitrogen is using for this bond, $\ce{s}$ character is removed from the orbital.  This "extra" $\ce{s}$ character is used to stabilize the orbital containing the lone pair of electrons.  The lone pair orbital now has more $\ce{s}$ character and is further away from the $\ce{p}$ hybridized transition state - the barrier to reach this transition state increases.  If our substituent is an alkyl group (X=R, an imine), the carbon is less electronegative then nitrogen, there will be increased electron density in the nitrogen orbital used to form the $\ce{N-R}$ bond.  The nitrogen will hybridize such that there is more $\ce{s}$ character in this orbital so as to better stabilize these electrons.  As a consequence, there will less $\ce{s}$ character (more $\ce{p}$ character) in the lone pair orbital.  The lone pair orbital is therefore closer to the $\ce{p}$ hybridized transition state and less energy will be required to reach the transition state.

As a result of this reasoning $\ce{NF3}$ has a higher inversion barrier then $\ce{NH3}$ and oximes have a higher barrier to inversion than imines.


How to compare SN2 rate of cyclopropane with bridged cyclic compound? How to compare the SN2 rate of iodocyclopropane and 1-iodobicyclo[2.2.1]heptane?

Backside attack is must, but as the cyclopropane ring is planar it's quite impossible for back side attack. Then the bridge compound is non planar, we know in bridge compound both SN1 and SN2 substitution are bad, so I just confused in these two.

Concerning the rates of SN2 reaction, the rate of reaction for cyclopropyl iodide will be relatively higher than 1-iodonorbornane (name of given bridged compound).

The cyclopropyl iodide (same as 1-iodocyclopropane) has one hydrogen atom substituted by the iodine atom. If you look into the proper stereochemistry of the compound, this $\ce{I}$ atom can be above the plane of the cyclopropane ring or below the plane. Angle between the above-planar atom and below-planar atom is approximately 115 degrees. So, the attacking nucleophile can attack from above the plane if $\ce{I}$ is below the plane, or vice-versa. This attack would be best if leaving group $\ce{I^-}$ and the nucleophile are in almost 180 degrees position. Here the repulsion due to lesser bond angle and with the hydrogen will decrease the rate, but SN2 is still possible to a low extent in the cyclopropane ring.

Now to the bridged compound. Firstly, there is very very little chance for backside attack. Secondly, for occurrence of SN2, there should be a Walden inversion of configuration. But in the briged bicyclo compound, the ring is in a very compact structure, and if it tries to undergo any inversion of configuration, a huge strain will be created in the bicyclo-compound, which is thermodynamically, very unstable. So, the rate of SN2 is almost nil for 1-iodo norborane.