How can we rationalise the amount of crystal water in coordination complexes? I was wondering why specific number of water molecules are attached to salts like $\ce{CuSO4.5H2O}$? Searching gave me a satisfying picture :

I think $\ce{FeSO4.7H2O}$
 should have a similar structure with $6$ water molecules in coordination sphere. But what about $\ce{MgSO4.7H2O}$? Or $\ce{Na2CO3.10H2O}$? How do we explain that?

I have given two answers: one regarding the hydration of metal cations in solution and one regarding salts of metal cations and their hydration extents.

When in solution, the answer lies with size. Think of a BB gun pellet and a basketball. How many basketballs can you stick to the sides of a BB gun pellet? 2, maximum. That's why some metal cations are hydrated with fewer waters than others; these metal cations tend to be the smaller ones.

Also, it is not $\ce{FeSO4}$ that is hydrated but rather just $\ce{Fe^{2+}}$. Iron sulfate is soluble in water so there shouldn't be unionized iron sulfate in solution; in addition, if the iron sulfate were unionized, it would be hard to coordinate water molecules to the metal cation - it's charge density is diluted, so to speak, by the presence of the doubly negative sulfate anion. So the correct formula should be $\ce{[Fe(OH_{2})_6]^{2+}}$.

When not in solution - i.e about the salt examples you posted: outside of water these salts can coordinate that many water molecules likely because there is no water to hydrolyze the metal-sulfate bond. By having the sulfate part of the molecule stick around, this likely facilitates more hydrogen bonding with water molecules - note that the sulfate anion is polarized by the water molecules; the negative charge density resides overwhelmingly on the oxygen molecules.

Consider the above picture of ferric sulfate. Note that there are four oxygens with partial negative charges on the sulfate ion. At least some of these partial negative charges facilitate hydrogen bonding in the solid-state. Also note the charge on the iron atom; this positive charge density also facilitates hydrogen bonding to water molecules.

In addition, we can consider calcium sulfate dihydrate - gyspum, drywall, sheet rock - the stuff your walls are likely made of. From here we clearly see that two of the oxygens on the sulfate anion are available for association with water molecules. Hence the existence of $\ce{CaSO_4*2H_2O}$


What is the main difference between Quantum Clusters and Nanoclusters? I know that Quantum Dots have high quantum yield and ranges from 2nm-10nm, and Nanoclusters have high fluorescence and ranges from 1nm-10nm. But the difference between them is that Quantum Dots exhibits red shift as the size gets larger while vice versa for Nanoclusters. My guess is Quantum Clusters is the combination of both, size is less than 2nm and exhibits red shift as size gets larger. Is this correct? Are there more things that distinguish them to each other?

Short answer: "quantum clusters" is not widespread and likely confusing. I'd suggest "small nanoparticles" or perhaps "metal nano clusters."

I can't comment on optical shifts, but many of these properties are distinct to the particular material and the onset of quantized energy levels in that material. So I doubt you can simply combine properties of quantum dots and nanoparticles.

I asked a colleague who synthesizes gold nanoparticles. She hasn’t heard anyone use the term "quantum clusters" and thinks people would find it very confusing.

Typically the usage:

Quantum Dots refer to semiconductor nanoparticles, e.g., $\ce{CdS}$ or $\ce{PbSe}$ etc.
Nanoparticles refer to nanometer-sized particles of basically anything (proteins, polymers, metals, etc.)
Clusters refer to very small particles (i.e., <1-2 nm diameter) with well-defined numbers of atoms (i.e., a few hundred at most). There is an IUPAC Gold Book definition:

  A number of metal centres grouped close together which can have direct metal bonding interactions or interactions through a bridging ligand, but are not necessarily held together by these interactions.

For reference, Wikipedia suggests nanoclusters are aggregates of nanoparticles. I don't think that's in common use, but certainly it suggests your suggested term would be very confusing.


How can we prove the inequality TdS > dU + pdV? I know that for reversible process
$$T\,\mathrm dS = \mathrm dU + p\,\mathrm dV,$$
but since all the terms in the given equation are state variables so the above equation should also be true for irreversible processes but in my textbook for irreversible process the inequality is used for irreversible process.

For a closed system with a single phase of a pure substance, when only expansion work is possible
$$\mathrm dU= -\mathrm p\,\mathrm dV + T\,\mathrm dS, \label{eqn:1}\tag{1}$$
where $p$ (the state variable) is the internal pressure and $T$ (the state variable) is the internal temperature of the system.
The first law, however, says more generally that
$$\mathrm dU = \mathrm dw + \mathrm dq. \label{eqn:2}\tag{2}$$
Equation \eqref{eqn:2} was derived by applying equation \eqref{eqn:1} to a reversible process.
For the reversible process, thermal and mechanical equilibrium between internal and external pressure means that $p_\mathrm{ext} = p$ and $T_\mathrm{ext} = T,$ and, combined with the second law, that
$$\mathrm dw = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV \tag{3}$$
and
$$\mathrm dq = -T_\mathrm{ext}\,\mathrm dS_\mathrm{ext} = T\,\mathrm dS. \tag{4}$$
More generally, the second law says that
$$\mathrm dS_\mathrm{tot} = \mathrm dS_\mathrm{sys} + \mathrm dS_{surr} \ge 0 \tag{5}$$
the equality holding for a reversible process. Dropping the label for the system
$$\mathrm dS \ge -\mathrm dS_\mathrm{surr} \quad\leftrightarrow\quad \mathrm dS \ge \frac{\mathrm dq}{T_\mathrm{surr}}. \tag{6}$$
Therefore for an isothermal process (thermal equilibrium condition applies)
$$\mathrm dS \ge \frac{\mathrm dq}{T}. \tag{7}$$
The inequality applies for instance to an isothermal free expansion.


Which is the most acidic hydrogen in vitamin C I thought it would be the lower first $\ce{-OH}$ because it's close to the double bond with oxygen, so I thought there would be more electron withdrawal from the oxygen atom due to its electronegativity.But the correct answer is actually the second lower $\ce{-OH}$, which I really don't understand

My attempt to the question:

The correct anion is stabilized by the mighty resonance.

That's about it.

Hydrogen bonding is irrelevant, since it affects both variants the same way.


Alternative software for processing Molden files? I am trying to salvage old data, and one of my output files is in Molden format.

I tried downloading Molden at its homesite but the link does not seem to work.
Is there another way to access Molden format files?

Thank you very much.

I am fairly sure Molden can be downloaded. However, programs like JMol, and Chemcraft are usually cross-compatiblr with molden.


Partial Atomic Charge in GAMESS In GAMESS, a calculation of optimisation gives data about partial atomic charges (like Mulliken or Lowdin charges). Can you please let me know how to calculate the other types of partial atomic charges (like Hirshfeld, Maslen or Politzer charges)?

FYI: I have used B3LYP and MP2 for optimization.

Unfortunately GAMESS does not support any other than the named population analyses.

You will need to use a different program. For that matter I suggest MultiWFN, which comes with a variety of different population analyses and is still in active development. From the website:

  Population analysis. Hirshfeld, VDD, Mulliken, Löwdin, Modified MPA (including three methods: SCPA, Stout & Politzer, Bickelhaupt), Becke, ADCH (Atomic dipole moment corrected Hirshfeld), CHELPG, Merz-Kollmann and AIM methods are supported.

The currently developed version MultiWFN 3.4 should be able to even read GAMESS output directly as stated in the update log:

  GAMESS-US output file now can be used as input file (not comprehensively tested, currently only single point task at HF/DFT level is formally supported). The suffix of output file should be changed to .gms so that Multiwfn can properly recognize it

As a workaround for older versions, and a reliable way to import this data, you can export a .wfn file with GAMESS. I will give a brief summary for this, since I am still using MultiWFN 3.3.8. From the corresponding manual:

  For example, the wavefunction represented by GTFs is enough for Hirshfeld population, so you can use .fch/.molden/.31~.40/.wfn/.wfx file as input, but .pdb, .xyz, .chg, .cub and .grd files do not carry any wavefunction information hence cannot be used.

As I have previously stated, .wfn files are technically outdated, especially when it comes to larger basis sets. However, for the time being, they are our only option to make it work.

I am using GAMESS version % DEC 2014 (R1) and the following input to demonstrate the procedure. I am using a Linux environment, so you might have to adjust this guide accordingly.

 $CONTRL SCFTYP=RHF            ! Restricted calculation.
         RUNTYP=OPTIMIZE       ! Geometry optimisation.
         COORD=ZMT             ! Z-matix specifies coordinates.
         AIMPAC=.TRUE.         ! Requests wfn file to be written.
         $END
 $BASIS  GBASIS=STO NGAUSS=3   ! STO-3G minimal basis
         $END
 $GUESS  GUESS=HUCKEL          ! Start with Hückel MO.
         $END
! Following group contains molecule specification.
! Blank lines are important.
 $DATA
Water
Cnv  2

O
H  1 rOH
H  1 rOH  2 aHOH

rOH=1.09
aHOH=110.0
 $END

Note that the indentations before the keywords are important. ! indicates a comment. Any additional output will be written to the punch file. This is dependent on how you set up GAMESS, but it normally is located in ~/scr/<rootfilename>.dat.

To extract the .wfn file you can use the following little script, or extract it manually from the .dat file.

#Script to extract wfn file from GAMESS dat (punch) file
#! /bin/bash

[ ! -z $1 ] && inputfile="$1" || exit 1
outputfile="${inputfile%.*}.wfn"

startpattern="----- TOP OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"
endpattern="----- END OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"

sed "/$startpattern/,/$endpattern/!d;//d" "$inputfile" > "$outputfile"

This should give you the following .wfn file for analysis:

Water
GAUSSIAN              5 MOL ORBITALS     21 PRIMITIVES        3 NUCLEI
  O    1    (CENTRE  1)   0.00000000  0.00000000 -0.14560943  CHARGE =  8.0
  H    2    (CENTRE  2)  -1.43257773  0.00000000  1.05592484  CHARGE =  1.0
  H    3    (CENTRE  3)   1.43257773  0.00000000  1.05592484  CHARGE =  1.0
CENTRE ASSIGNMENTS    1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  2  2  2  3  3
CENTRE ASSIGNMENTS    3
TYPE ASSIGNMENTS      1  1  1  1  1  1  2  2  2  3  3  3  4  4  4  1  1  1  1  1
TYPE ASSIGNMENTS      1
EXPONENTS  1.3070932E+02 2.3808866E+01 6.4436083E+00 5.0331513E+00 1.1695961E+00
EXPONENTS  3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01 5.0331513E+00
EXPONENTS  1.1695961E+00 3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01
EXPONENTS  3.4252509E+00 6.2391373E-01 1.6885540E-01 3.4252509E+00 6.2391373E-01
EXPONENTS  1.6885540E-01
MO  1                     OCC NO =   2.00000000 ORB. ENERGY = -20.25158070
  4.22735195E+00  4.08851077E+00  1.27421022E+00 -6.18804850E-03  8.27701479E-03
  6.24678653E-03  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  6.97684832E-03  4.38722945E-03  6.95010854E-04
 -1.54630881E-03 -1.49552256E-03 -4.66089059E-04 -1.54630881E-03 -1.49552256E-03
 -4.66089059E-04
MO  2                     OCC NO =   2.00000000 ORB. ENERGY =  -1.25754111
 -9.93961353E-01 -9.61316148E-01 -2.99600254E-01 -2.02173727E-01  2.70423693E-01
  2.04092795E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  2.05800516E-01  1.29412887E-01  2.05011755E-02
  4.30889918E-02  4.16737970E-02  1.29879022E-02  4.30889918E-02  4.16737970E-02
  1.29879022E-02
MO  3                     OCC NO =   2.00000000 ORB. ENERGY =  -0.59384261
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  1.02654522E+00  6.45519180E-01  1.02261083E-01  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
 -1.24405583E-01 -1.20319664E-01 -3.74983838E-02  1.24405583E-01  1.20319664E-01
  3.74983838E-02
MO  4                     OCC NO =   2.00000000 ORB. ENERGY =  -0.45973131
 -4.42352599E-01 -4.27824176E-01 -1.33334109E-01 -1.28843153E-01  1.72338126E-01
  1.30066155E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00 -1.26642108E+00 -7.96359559E-01 -1.26156734E-01
 -8.17262213E-02 -7.90420432E-02 -2.46339524E-02 -8.17262213E-02 -7.90420432E-02
 -2.46339524E-02
MO  5                     OCC NO =   2.00000000 ORB. ENERGY =  -0.39261686
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  1.67545020E+00
  1.05356804E+00  1.66902879E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00
END DATA
RHF      ENERGY =      -74.9659012162   VIRIAL(-V/T)  =   2.00600297

You can now use this file and load it with MultiWFN. After the start-up process you should find the key values printed to the screen like this (I abridged it a little bit):

 Multiwfn -- A Multifunctional Wavefunction Analyzer (for Linux)
 Version 3.3.8, release date: 2015-Dec-1
[...]
 System energy:     -74.965901216200 Hartree,   Virial ratio:  2.00600297
 Total/Alpha/Beta electrons:     10.0000      5.0000      5.0000
 Net charge:     0.00000    Expected multiplicity:    1
 The number of orbitals:     5,  Atoms:      3,  GTFs:     21
 This is restricted close-shell single-determinant wavefunction
 Title line of this file: Water
 Formula: H2 O1
 Molecule weight:        18.01528

 Loaded water-sto3g.wfn successfully!

                    ------------ Main function menu ------------
[...]
 7 Population analysis
[...]

You may now choose 7 and enter the following menu:

                 ============== Population analysis ==============
 0 Return
 1 Hirshfeld population
 2 Voronoi deformation density (VDD) population
 10 Becke atomic charge with atomic dipole moment correction
 11 Atomic dipole corrected Hirshfeld population (ADCH)
 12 CHELPG ESP fitting charge
 13 Merz-Kollmann (MK) ESP fitting charge
 14 AIM charge

Follow the guidance of the manual and the on-screen instructions to produce the required charges. For comparison here are the computed Hirshfeld charges, written to water-sto3g.chg with the following command sequence 7; 1; 1; y.

  O     0.000000    0.000000   -0.077053   -0.296017
  H    -0.758088    0.000000    0.558771    0.147959
  H     0.758088    0.000000    0.558771    0.147959

The element identifier is followed by the Cartesian coordinates, which are followed by the charges.

In principle all other implemented charges are equally as easy to compute. Keep in mind, that basin based charges (e.g. AIM) need more computing power and depend heavily on molecule and basis set size.

Apart from this procedure, you can obtain natural charges with the NBO6 program, which conveniently interfaces with GAMESS. Unfortunately you have to buy it.


Combustion of paraffin Find the heat of combustion for 1 mol of $\ce {C21H44}$ in the reaction where some wax is oxidized to elemental C (soot) and water vapor. The Heat of Formation of $\ce {C21H44}$ is -476 kJ/mol.

When it says some, does that necessarily change the molar ratio?
I thought the balanced equation would be:

$\ce {C21H44 + 11O2 -> 21C (graphite) + 22H2O}$

$\ce {O2}$ and elemental carbon are zero for heat of formation

22(-241.818 kJ/mol) - (-476 kJ/mol) = -4843.996 kJ/mol

Is this right?

Thanks

The question is ambiguous, unfortunately.  There are two possibilities that I can see.  Either the question:

is asking about a reaction in which some quantity of wax undergoes the reaction you give, in which case you are 100% correct.
is asking about a reaction in which part of the paraffin is being oxidized to C and part is being oxidized to CO2, the usual reaction that occurs.  Though that is more likely to occur in real life, in my opinion, it also is impossible to solve without further information.  The best you could do would be to set a lower limit (the value you've already found) and an upper limit, the heat of reaction for $\ce{C21H44 + 32 O2 -> 21 CO2 + 22H2O}$.  (I guess that would be an upper limit and a lower, numerically, since the values are both negative, but this reaction would release far more heat.)


How can value of acid dissociation constant is vary with concentration of acid? First of all I have learned that $K_\mathrm{a} = \frac{[\ce{H+}] [\ce{A-}]} {[\ce{HA}]}$ . Also I have been informed that $K_\mathrm{a}$ value is independent of concentration.

So let us take an example. Suppose 1 mole of acid $\ce{HA}$ dissociates to form 0.9 moles of $\ce{H+}$ ions and 0.9 mole of $\ce{A-}$ ions and 0.1 mole of undissociated $\ce{HA}$. So, the $K_\mathrm{a}$ value will be $K_\mathrm{a} = \frac{[0.9] [0.9]} {[0.1]}=8.1$.

But if we take 2 moles of same acid then, 1.8 moles of H+ ions and 1.8 moles of A- ions will be formed leaving 0.2 moles of undissociated HA. This time $K_\mathrm{a} = \frac{[1.8] [1.8]} {[0.2]}=16.2$.

So we can see that $K_\mathrm{a}$ has increased with increase in concentration. Can you tell me where my assumption is wrong?

Before we begin, note that $K_a$ is defined in terms of concentration and not amount. Furthermore, $K_a$ is a constant so it must be the same provided the conditions, other than concentration, are the same.

The nature of equilibrium is such that it does not always give the same absolute reaction (i.e. a fixed percentage of reaction, or the reaction does not proceed to the same "position" each time). This is unlike a fixed forward reaction which always ends at 90% completion. In other words, although the 1 molar solution of HA gave 0.9M of $H^+$ and $A^-$, the 2 molar solution would only give out...

$$ K_a = 8.1 = x^2/(2-x) $$

where $x$ is the number of moles of $H^+$ produced. When you solve this, you would get $x=1.66M$


What is the mechanism of endothermic dissolution? My real question is in fact "How can dissolution happen if it's endothermic?" However, there
have been many questions regarding this exact topic and the answers about Gibbs free energy
determining spontaneity of the reaction doesn't help my intuition at all. The majority of answers
list the increase of entropy as the cause of endothermic dissolution, but in my opinion, entropy
doesn't make sense as the cause.
Let's take a similar problem, namely the melting of a solid. If we place a solid with lower
temperature into surroundings with higher temperature (enough for the solid to reach its melting
point), it's not that liquification would cause the temperature of the surroundings to drop, but
rather that the transfer of heat from the surroundings to the solid would cause liquification. The
acquired heat allows the solid to increase its entropy, not the other way around.
So, if increased entropy is the result of endothermic dissolution, what is the cause? Or, what is
the mechanism that allows entropy to be increased?
Let's take the dissolution of NaCl in water as an example. This process is endothermic by around
4 kJ/mol. My guess is that, since some ions in the solid have higher kinetic energy than average, this energy
will be sufficient to overcome the 4 kJ/mol barrier and leave the solid, taking some of its thermal
energy away from it. As a result, the temperature of the remaining solid will drop a little. Since
the temperature of the solid is now slightly lower than the temperature of the surrounding water,
heat will be transferred from the water to the solid until the temperatures are equal. This will
result in lowering the temperature of water and in the increase of entropy.
Is my thinking correct? If not, why?

Since the temperature of the solid is now slightly lower than the temperature of the surrounding water, heat will be transferred from the water to the solid until the temperatures are equal

It is fine to think that way as a rough draft in rationalizing this process. The only issue is that temperature is a macroscopic quantity, and mechanism tends to be a particular (molecular) concept. So you could say that the individual solid particles (e.g. ions) will have slightly lower kinetic energy than average, and collisions with individual solvent molecules will balance this out.
Or you could start with solvent molecules near the solid-liquid interface that already have a higher kinetic energy than the average become part of the solid, thus lowering the average kinetic energy of the remaining solute molecules.
Or you could say that the water molecules solvating the ions use parts of their kinetic energy in the desolvation process, directly lowering the temperature of the solvent.
Because all the species have a distribution of kinetic energies around the average, all these scenarios are correct.

As a result, the temperature of the remaining solid will drop a little.

I don't think the bulk of the solid will have a lower average kinetic energy, so I don't think the temperature of the solid will drop faster than that of the solution. I would think that there is a temperature gradient between the solid-liquid interface on the one hand, and the bulk solid and liquid on the other hand. This is maintained as long as there is a net transfer from solution to solid. Once equilibrium is reached, of course, this will include thermal equilibrium.

this energy will be sufficient to overcome the 4 kJ/mol barrier

"Barrier" sounds like an activation energy, something you can step over. Here, I would rather talk of "energetic cost" or "step up", something that you don't get back. The solvation energy is higher than the lattice energy, so there is a cost for taking the ions out of solution and adding them to the ionic solid.


Why is SnCl2 produced when tin reacts with hydrochloric acid? $$\ce{Sn + 2HCl -> SnCl2 + H2}$$
Our teacher told us that it's because $\ce{Sn^2+}$ is more stable, but I can't find that in any textbook.

Why is $\ce{Sn^2+}$ in the chloride product over any other oxidation state?

First, let's consider possibility, that indeed some cation is more stable than others. Let's look at ionization energies of tin (in eV)

\begin{array} {|r|r|}
\hline
1 &7.34 \\
\hline
2 &14.63 \\
\hline
3 &30.50 \\
\hline
4 &40.73 \\
\hline
5 &72.28 \\
\hline
\end{array}

As you can see, the ionization energies increase monotonically, meaning that each subsequent step to higher charged cation is progressively harder. This means, that if only cation stability (in vacuum) was considered, first all available tin should produce tin (I) compound and only then move further. This is, obviously, not the case.

Still, if you are lazy, it is perfectly OK to go with notion of 'stable' valence states, that are quite easy to memorize. Standard electrode potentials are sometimes quite puzzling from uninvolved point of view and it is easier to use the tables without much thoughts about them.

Still, if we want a reasonably acceptable explanation. We can see, that difference between 2nd and 1st ionization energies (~7 eV) and between 3rd and 4th (~10) is lower than between 2nd and 3rd (~15). This suggests that (II) state should oxidize disproportionally harder. Since cations are stabilized by ion-dipole or ion-ion interactions that strengthens linearly with charge, this suggests extra stability of tin (II) ionic compounds comparing to (I) and (III).

The oxidation state (IV) of tin is a completely different matter. Cations of charge $3+$ and higher very rarely truly bear said charge. Typically, they are involved in Lewis acid-base interactions reducing the charge considerably. Tin (IV) compounds have Tin bound covalently. In this case stability of compound is guarded by 8 (18 for d-elements) electron rule.

Mind you, explanation above are not very convincing from quantum mechanical POV, but it is good enough for most cases
.


Why is the equilibrium constant defined that way intuitively? Suppose we have chemical equation: $$\ce{ aA + bB  <=>   cC + dD }$$

then equilibrium constant is defined: $$K=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$

but why don't we define it as: $$K=\frac{cd[C][D]}{ab[A][B]}$$

or even as: $$K=\frac{c[C]+d[D]}{a[A]+b[B]}$$

As adding concentrations is additive and not multiplicative. And because: $$\ce{ aA + bB <=> cC + dD }$$
is the same as: $$\ce{ cC + dD <=> aA + bB }$$
then we should have: $$K=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$
and specifically: $$\frac{[A]^a[B]^b}{[C]^c[D]^d}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
but trying some examples shows that this isn't the case.

It can be understood in simple terms with just few intuitive keys (of course it is not very rigorous).

Take in mind that $K$ is a measure of how displaced to reactive or equilibrium the reaction is. So it should be greater when products concentration is relatively high (that is why they go into the numerator) and small when reactive concentrations are relatively high.
In elemental reactions (technical term) is like a balance between the speed of the reaction in the direction of products and the reverse reaction.
A reaction only occurs when its reactives are in the same region of the space. So, if you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. This is the main point: How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^n$.
The reaction speed in that direction is proportional to the number of encounters.

That is why you need to make products. Of course equilibrium constant $could$ be defined in other ways, for example, suppose that we define them as $K_{new}=ln(K_{old})$. In such case for the logarithm properties you'll have other expressions that includes sums. But there is not advantages of defining $K$ in that way.


What is the difference between these wave functions? My chemistry textbook introduces the wave function as

$$\psi(x)= A \sin\left(\frac{2\pi x}{\lambda}\right)$$ Therefore, the Schrödinger Equation is:
$$\frac{d^2\psi(x)}{dx^2} = -~\left(\frac{2\pi}{\lambda}\right)^{2}\psi(x)$$
and by multiplying each side by $\frac{-h^2}{8\pi^2m}$ you get:
$$-~\frac{h^2}{8\pi^2m} \frac{d^2\psi(x)}{dx^2} = \frac{p^2}{2m}\psi(x) = (T)(\psi(x))$$ where $T$ is kinetic energy.

Now, I don't fully understand that last bit, and I am willing to see an explanation on it, but my main question is that, when the book transitions from discussing the Schrödinger Equation in the context of a "particle-in-a-box" to the context of a one-electron atom, it seemingly changes the wave function to
$$\psi_{(n,\ell,m)}(r,\theta,\phi) = R_{n,\ell}(r) Y_{\ell,m}(\theta,\phi)$$
where n = principle quantum number, $\ell$ = angular momentum number, m = magnetic quantum number, r = radius (distance between nucleus and electron), R = the "radial part" of the equation, Y = the "angular part" of the equation, and $\theta$ and $\phi$ are some angles that I still don't know how to get.
I guess my question is did the wave function $\psi$ change? Why is suddenly so different and why do r,$\theta$ and $\psi$ suddenly come into the picture? Is it because you are going from the one-dimensional, Cartesian "x" version of the wave function to the three-dimensional, polar coordinate version of it?

The first function you have there

$\psi(x)=A\sin(\frac{2\pi x}{\lambda})$,

is very similar to the function of a particle in a monodimensional box.

This function is a very helpful example in order to understand how does the quantum mechanics works.

The other function is

$\psi_{(n,m,l)}(r,\theta,\phi)=R_{n,l}(r) Y_{m,l}(\theta,\phi)$.

This function represents how an electron moves in a hydrogen atom and the full expression is quite different to the first function. Just see the representations:

Ok. The functions do not mean the same concept. So... Why do we use "$\psi$" in both?

In math we use $f(x)$ to speak about any function. In quantum mechanics we use $\psi(x)$ for the same: represent a function easily.

Yeah, but... Why do i have to study the first function if it's just a lie and not the full story?

Remember it's only a very good example. However, the particle in a box function can be used to determine the energy of an electron in a conjugated system as beta-carotene and it's energy for the first excited level. If you subtract the second energy to the first one it gives you the energy for a photon in the visible region: the carrot's orange color.

This means that the particle in a box is a good example if you start to study quantum chemistry and it even works as a good approximation for some systems.

Ok. Too much information but I want to know more: How can I get the second function from the first?

Well, that's... Impossible...

To obtain the functions we use the Hamiltonian operator, which contains the kinetic energy operator and the potential energy operator.

For the particle in a box, we only use the kinetic energy operator because we don't have any potential into the box. This leads to the first function you have.

For the hydrogen atom we must use both operators (all the Hamiltonian) which means to include a function for the potential and we must work in 3D. The potential function can be obtained using the angular momentum. The spherical coordinates makes easier to work in 3D. This leads to radium and angular functions.

Summarizing, we have a different wave function for every different situation. This situation is what determines the function and it's conditions:

The particle in a box → Quantization.
The free particle → No quantization.
The electron in a hydrogen atom → Quantization + quantum numbers + rotation +... → Atomic orbitals.

Ok, seems legit, but it's hard to understand

I recommend you the books "Physical Chemistry" from Atkins, Levine or Engel to understand better the ideas and see the complete development.

In case you hate textbooks you can visit hyperphysics web page for a general looking.


Disproportionation of hydrogen peroxide Hydrogen peroxide is decomposed as follows:
$$\ce{2H2O2 -> 2H2O +O2}$$
This is a disproportionation redox reaction of $\ce{H2O2}$ involving the 2 half reactions
$$\ce{H2O2 -> O2 + 2H^+ + 2e-}$$
$$\ce{H2O2 + 2e- + 2H^+ -> 2H2O}$$
But I noticed semantically that it can also be the sum of the two following half reactions:
$$\ce{2H2O2 -> 2O2 + 4H^+ + 4e-}$$
$$\ce{O2 + 4H^+ + 4e- -> 2H2O}$$
Both reactions are feasible according to the following:

$$\begin{array}{ccc}
\text{Oxidised species} & \text{Reduced species} & E^\circ (\mathrm{V}) \\
\hline
\ce{H2O2} & \ce{H2O} & 1.763 \\
\ce{O2} & \ce{H2O} & 1.23 \\
\ce{O2} & \ce{H2O2} & 0.695 \\
\end{array}$$

Can it happen in the second way?
If yes, can you explain the mechanism in simple terms?

In principle you could find a lot of equations, all leading to the same result. But these are only (half-)equations. Most of the times the reaction mechanism is much more complicated than just these reaction equations.

The decomposition of hydrogenperoxide is still an active field of research in various media with various catalysts.

Unfortunately with the massive overflow of the above, I was only able to find a quite old PhD thesis by Trice Walter Haas (Iowa State University, 1960)

Here is a small line-up of reaction equations that will play a role in the decomposition, at least to that part that I can imagine it. And I am just assuming that there are radicals involved, because there usually are.
\begin{align}
\ce{H2O2 &<=> H+ + HO2- }\\
\ce{HO2- + H2O2 &<=> HO2 + HO + OH-}\\
\ce{HO2 + OH &<=> H2O + O2}\\
\end{align}


Internal Energy vs Enthalpy I struggle to see the difference between internal energy and enthalpy, or at least the need for differentiation between the two. I understand that enthalpy, supposedly unlike internal energy, accounts for the expansion work required for the system to exist, but it seems that that would only be necessarily stated if one had failed to account for the negative contribution to work in the first place.

Could enthalpy, as $$H = U + pV$$ and thus, $$H = q + w + pV$$ not be written as equally validly as, $$H = q + w_1 + w_2$$ i.e. $$H = U \ ?$$

It would seem that this definition of enthalpy implies that internal energy accounts only for work done to the system, disregarding work done by it.

Trying to figure out a good intuitive physical interpretation of enthalpy has caused thermodynamics student over the last couple of centuries to waste huge amount of their valuable time (in my judgment).  The really fundamental physical quantity is internal energy U, not enthalpy, and its physical interpretation is simple and straightforward.  I regard enthalpy as just a convenient function to work with in solving many types of physical problems.  If the function enthalpy had never been invented, we would still be able to solve these problems, but the results would not be as concise.


In rdkit, how to preserve stereochemistry in a SMARTS subgraph? In rdkit, when doing a MCS search for molecules bearing a chirality center, (how) is it possible to preserve the stereochemical information when exporting the subgraph to a SMARTS string?

Consider the following three molecules:

>>> mol_ccw = Chem.MolFromSmiles('C1=C[C@H](Cl)CCC1')
>>> mol_cw  = Chem.MolFromSmiles('C1=C[C@@H](Cl)CCC1')
>>> mol_lin = Chem.MolFromSmiles('C=C[C@H](Cl)CCC')

Doing a chirality-sensitive subgraph search leads to the somewhat expected result:

>>> rdFMCS.FindMCS([mol_ccw, mol_cw], matchChiralTag=True).smartsString
'[#6](=[#6])-[#6]-[#6]-[#6]-[#6]-[#17]'
>>> rdFMCS.FindMCS([mol_ccw, mol_lin], matchChiralTag=True).smartsString
'[#6]=[#6]-[#6](-[#17])-[#6]-[#6]-[#6]'
>>> rdFMCS.FindMCS([mol_cw, mol_lin], matchChiralTag=True).smartsString
'[#6]-[#6]-[#6]-[#6]-[#6]'

The subgraph over mol_cw and mol_lin includes the stereocenter. But unfortunately, the chirality information is not stored in the SMARTS string, and using [#6]=[#6]-[#6](-[#17])-[#6]-[#6]-[#6] for a chirality-sensitive substructure match leads to the expected result of the pattern made from that SMARTS string matching all three molecules:

>>> patt = Chem.MolFromSmarts('[#6]=[#6]-[#6](-[#17])-[#6]-[#6]-[#6]')
>>> len({mol_cw, mol_ccw, mol_lin}.GetSubstructMatches(patt, useChirality=True))
1

Manually inserting a lazy @H or a &*&H1 at the stereocenter leads -of course- to the desired result:

>>> patt = Chem.MolFromSmarts('[#6]=[#6]-[#6@H](-[#17])-[#6]-[#6]-[#6]')
>>> len({mol_ccw, mol_lin}.GetSubstructMatches(patt, useChirality=True))
1
>>> len(mol_cw.GetSubstructMatches(patt, useChirality=True))
0

Now, when matching mol_ccw and mol_lin, how do I get the stereochemistry-aware SMARTS string [#6]=[#6]-[#6&*&H1](-[#17])-[#6]-[#6]-[#6] as the substructure in the first place?

For anyone interested in this topic: I just learned that this is a bug that should already be filed on GitHub, but apparently the developers themselves do not find it anymore - probably that's why I didn't find it. Nonetheless, the bug is known and will be fixed in the next release.


What is this glassware between flask and condenser to collect liquid? I've been trying to identify this piece of glassware for ages now. I've been combing the internet and can't seem find anything similar. It looks like some kind of adapter/collecter/trap that sits on the pot and the condenser goes on top.

This glassware is used to separate /remove the light phase from a biphasic (immiscible) liquid mixture.  This can be used primarily to enhance liquid-liquid extraction from the light phase to the denser phase. As a mixed distillate collects,  the denser phase will eventually siphon back through the side-arm to the flask below and the light phase remains in the collector.


Freeze distillation: Does efficiency depend on solubility? I was surprised to learn that freeze distillation has the same inefficiencies as vapor distillation: Namely, that it is impossible to get two miscible fluids to fully separate in one "pass" of subjecting the mix to a temperature between the melting point of its constituents.

I can understand why this is the case in vapor distillation, because vaporization of a liquid still occurs below its boiling point.

But AFAIK the same is not generally true of freezing: assuming it has a nucleus, a single liquid will either crystallize completely at a given temperature or it won't at all.

My guess is that the inefficiency of freeze distilling has something to do with the chemistry of miscible fluids: I.e., whatever phenomenon allows them to mix also drags molecules of the lower-melting chemical out of the mix above its melting point when the higher-melting chemical crystallizes.  If so then the efficiency of freeze distilling should have some relationship to a measurable quantity (like a miscibility coefficient?).  What theories and models of miscible fluids are applicable to the question of efficiency of freeze distillation?

(If there is no attraction between the two fluid chemicals then presumably "single-pass freeze distillation" could be perfect.  It would probably require agitation to help the more volatile of the two stay out of the crystals forming from the first chemical to freeze.  For example, agitation during freezing is how ice is formed without visible bubbles from dissolved gasses.  But if "there is no attraction" then presumably the fluid chemicals are not miscible, right?)

Solids have solubility, just as liquids do. This is easily seen in metal alloys, where different amounts of different compounds can cause different crystal structure, low-level phase separation, and large-scale phase separation (if the solid solution is not stable). The changing of these nanostructures of the material can even occur with a material fully in the solid state, such as tempering of steel.

This relates to your question in that liquids freezing will behave very similarly with respect to phase composition as vapors condensing. The composition of the solid will be proportional to the composition of the liquid and affected by differing freezing points, just like you would expect from a liquid-vapor Txy diagram.

There are some differences, however. Molecules in solids can, as I said above, move around within the solid. However, this diffusion is substantially slower in solids than in liquids, and in non-metal solids, slow enough to be largely insignificant within the process of freezing. This means that the freezing process will create regions of differing composition within the solid. Further and related to this phenomena, the speed of freezing will impact the concentration of the solid, as faster freezing will push it even further from equilibrium.

The slow diffusion in solids also means that once trapped in the crystal structure, compounds with higher freezing points will not as easily exit the solid as they might from a liquid. This means that melting can behave very differently from vaporization, and solids might become a slush in the process of melting.

An explanation as to why this all occurs could go as follows: solubility is a factor of the strength of the interaction between different types of molecules. Those interactions must be stronger than interactions within the individual types to avoid phase separation. The process of freezing tends to not fundamentally shift the strength of these interactions. It is possible for materials partially soluble in liquid to be mostly insoluble in solid, but pairs of miscible liquids have such intense cross-compound interactions that freezing typically can only impact concentration of the separate phases.

The really important chemistry in this case takes place at the phase boundary between the solid and liquid, which behaves almost identically to that between a liquid and a gas. Molecules with weaker interactions with the solid phase (weaker interactions with other molecules in that phase) will tend to leave the solid phase more often - but there is an equilibrium where molecules will enter their less-favored phase. Essentially, everything you would have learned about liquids and vapors applies here, except diffusion is too slow to allow good mixing within the more condensed phase.

Of course, there are, in specific cases, more complicated explanations. One is of interstitial compounds, where a second compound might occupy spaces within a solid. While this happens to a degree in all solids, interstitials that are small enough to fit fully in a crystal structure without deforming it might exhibit unusual solid solution behavior, such as a compound with a lower freezing point showing a greater affinity for solid phase than the other compound.

To answer the title question, yes, separation depends on solubility. Unlike vapor distillation, though, it can also depend on temperature difference (thus speed) and mixing.


Help identifying glassware I am helping clean out my lab and there is some glassware that I can't identify. Attached are the photos:

Thank you!

Ok. Finally found some time to look up references.

The first piece of glassware is a bubbler. See:
https://www.chemglass.com/product_view.asp?pnr=AF-0513

You use this to help keep positive pressure in your inert atmosphere system. You need to fill it with mineral oil or mercury and you can vent your sealed system though this. This allows you to keep piping in say nitrogen or argon and keep positive pressure without blowing anything up.

The second piece of glassware is a fritted filter used for inert atmosphere chemistry. See:
https://www.chemglass.com/product_view.asp?pnr=AF-0540

The side arm allows you to equilize the pressure between the two sides (otherwise the liquid may not flow from top to bottom). There are ground glass joints on both ends so that you can create a sealed system free from water or oxygen.


Ozone resonance structure: differing opinions I've always thought that the two-form resonance structure of ozone was the scientific concensus (from Wikipedia]:

Then today I found two other opinions. A three-form structure:

And a four-form structure:

Is there a consenus? Are the two and three-form representations just simplifications of the four-form structure? Please include references.
EDIT: The second resonance structure on the four-form image above is drawn incorrectly, as pointed out by @ron. The upper oxygen atom should have a lone pair and a positive charge, just like the second structure of the two-form.

Those are all valid resonance structures for ozone (except the second from the left in the "four-form structure" row, that one has two negative charges, maybe it's a typo).  In fact you can draw even more structures like this one, where this is no single bond between the middle and terminal-right oxygens.

All of these structures contribute to the "real" structure of ozone - however, some resonance structures contribute more and some resonance structures contribute less.  Here is a link to a nice set of rules (see page 2, the "Resonance Contributor Preference Rule" section) that describes how to weight the contribution of various resonance structures.  The most important rule is that each atom in the resonance structure should have an octet of electrons.  Only the structures labeled I and II in your "three-form structure" row meet this requirement, so they are the two most important resonance structures for ozone - but all of the others do contribute, just to a lesser extent.


Why is this structure impossible? I'm asked the following:

  Explain why the structure $\ce{H-C=C-H}$ doesn't make sense chemically.

Not only am I just looking for a check up of my answer, but am I making any sense in what I'm saying here? Am I mixing up or misunderstanding concepts here?

My answer was:

The hydrogen and carbon atom form a covalent bond. Hydrogen has only 1 electron in its valence shell, so it 'needs' 1 of carbon's electrons to complete it. Now there are 3 electrons left to take in the carbon atom. But this isn't enough to make a double bond with the other carbon atom: we're one electron short.

Does this sound about right?

Your analysis was a good start, but then it dropped. You need to consider four electrons on each carbon atom. Each of the two carbons can then form a single bond to one hydrogen atom using one of its electrons. Each of the carbon atoms then uses two of the remaining electrons to form the double bond.

Thus, each of the carbon atoms has one extraneous electrons; the structure would be a diradical structure. It should better be represented e.g. like this: $\ce{H-C^.=C^.-H}$.

It is possible for these two radicals to be alongside each other — see for example the structurally strongly related dioxygen molecule which is a triplet (diradical) in ground state and not adequately described in simple Lewis structure formalisms. However, it is highly unlikely for two carbons without additional constraints to form a diradical like that; forming a triple bond ($\ce{H-C#C-H}$) would be much better energetically (for carbon!).


Why does copper react only with nitric acid? When we are entering the chapter for acid and base, one of the reactions is about reacting metal with acid to form a salt and hydrogen gas. My teacher also informed us that a metal which is at a lower position than hydrogen in the electrochemical series, such as copper, cannot react with acid. It is because copper cannot displace the hydrogen atom from the acid.

However, I have seen several videos from YouTube which show copper can react with nitric acid as follows:

$$\ce{Cu + 4HNO_3 -> Cu(NO_3)_2 + 2H_2O + 2NO_2}$$

My question does why nitric acid react with copper, unlike the other acids? And are there any other acids which can react with copper?

Copper is an unreactive metal and doesn’t react in normal circumstances with dilute acids. However, it does react with nitric acid. Why is this?

Nitric acid is an oxidizing agent and the reaction is not the usual acid + metal reaction. The products are oxides of nitrogen instead of hydrogen. The actual nitrogen oxide formed depends on the concentration and temperature of the acid.

There are actually two equations for the reaction of copper with nitric acid. It depends on whether the nitric acid is concentrated or not. If it is concentrated and in excess then the ratio is 1:4 copper to nitric acid. If it is dilute then the ratio is 3:8.

$$
\begin{align}
\ce{Cu + 4HNO3 &–> Cu(NO3)2 + 2NO2 + 2H2O}\\
\ce{3Cu + 8HNO3 &–> 3Cu(NO3)2 + 2NO + 4H2O}
\end{align}
$$

Copper also reacts with hot conc. $\ce{H2SO4}$ but, here $\ce{H2}$ gas is released as follows :

$$
\begin{align}
\ce{Cu(s) + H2SO4(aq,conc) &-> Cu^2+(aq) + SO4^2-(aq) + H2(g)}\\
\ce{Cu(s) + H2SO4(aq, dil) &-> no ~reaction}\\
\ce{Cu(s) + HCl &-> no ~reaction}
\end{align}$$


The acidity of boric acid The “real way” boric acid behaves as an acid in water is first by acting as a Lewis acid (bottom reaction, in black).  In fairness to the undergrads, I’d bet >25% of graduate students around the country would not be able to correctly explain the acidity of boric acid.

How are we supposed to know that boric acid heterolytically cleaves water as to leave a proton behind?

Is this something that is supposed to be immediately obvious? Or do we just have to read about experiments and then realize that this is how boric acid behaves as an acid - as exclusively a Lewis acid?

Other structurally analogous acids such as sulfuric acid also have a electrophilic center atom because of the electron-withdrawing oxygens, but these acids do not behave the way that boric acid does; these acids simply donate protons - they don't chop up water molecules to create protons.

Is a hint that $\ce{B(OH)_3}$ behaves this way found in the fact that boron can form either three or four bonds with elements? Should this clue us off to the possibility of $\ce{B(OH)_4^-}$?

Is another hint found in the fact that boric acid is trigonal planar with an unused p-orbital? Sulfuric acid on the other hand doesn't have any empty orbitals (well, it might have empty d-orbitals which may or may not be accessible).

I think your suppositions and observations are basically correct. On the basis of general principles, it is certainly possible to imagine that such a mechanism might be at play. If one considers the structure of boric acid, it should be readily apparent that the boron atom could be quite electrophilic and hence act as a Lewis acid: it possesses a formally vacant $p$-orbital; it is bound to highly electronegative oxygen atoms, which withdraw electron density and create a partial positive $(\delta^+)$ charge on it (though this effect may be mitigated by the oxygen lone pairs, which can donate electron density into the aforementioned $p$-orbital); and finally, it has a planar geometry that makes it quite sterically accessible to approaching nucleophiles. Such a mechanism is also lent plausibility by the well-known propensity of various other boron compounds for forming Lewis adducts and being highly susceptible to hydrolysis (where applicable).

All of that having been said, I don't think it's fair to claim that such a mechanism is "obviously" correct for the reasons given above. Such arguments are strictly qualitative, and while they're very suggestive of the possibility, they are in no sense definitive. The only way to even approach something vaguely approximating certainty is to perform experiments (perhaps in concert with theoretical computations), and even there considerable difficulty exists in reaching firm conclusions. The confirmed presence of $\ce{B(OH)4-}$ under one particular set of conditions is not sufficient proof that the mechanism involving boron as a Lewis acid is always operative, nor does it rule out other pathways. (Note that I'm not necessarily disputing that mechanism, but only pointing out that extrapolating too widely from specific experimental results has certain liabilities.) As ron points out in his answer, the issue apparently isn't settled.

Edit: I would also add that there are arguments to be made against such a mechanism. There's likely some degree of $\pi$-bonding between the oxygen lone pairs and the boron, which would be disrupted by the pyramidalization of the molecule upon formation of the intermediate Lewis adduct resulting from nucleophilic attack by $\ce{H2O}$. The associated energy penalty might be relevant to the reaction kinetics under some conditions. My instinct is that this is probably not significant, but it may be a factor worth considering. Of course, this is all speculative on my part, though I think it further demonstrates that qualitative reasoning in chemistry is rarely clear-cut, and there's an awful risk of falling prey to fuzzy thinking. Hence, rigorous experimentation and calculation are vital.


Does dissolving something in water change the specific heat of the solution? I’m a little confused on this topic. If I have a solid that I dissolve in water and that undergoes an exothermic reaction, can I assume that the specific heat $c$ of the solution instantaneously after the dissolution occurs is equal to $\pu{4.18 J g-1 °C-1}$? Does dissolving something in water change the specific heat of the solution? Or can I just say that the specific heat of the solution (solid plus water) equals the specific heat of water?

My first instinct was to think that yes, dissolving a solid in water does change the specific heat of the solution. If this is true, then I thought I might determine the overall specific heat of the solution by taking the specific heat of water and multiplying it by the percentage of the solution that is water, then adding that to the specific heat of the solid times the percent of the solution that is ions from that solid.

I’ve overthought this quite a bit, and I’ve tried to find an explanation on the internet. I’m pretty sure I’m thinking about this incorrectly, and I apologize if this is a bad question. Any clarity you could provide would be much appreciated.

If you were adding two liquids ideally, that is, assuming that the mixing occurs without any change in enthalpy - which occurs for chemically similar substances, then yes, the heat capacity of the mixture is a weighed average of the heat capacity of the components ($A$, $B$):

$$c_p = x_Ac_{pA}+(1-x_A)c_{pB}$$

However, when we are talking about dissolving solids, the situation is a little different. That is because infinite-dilution properties of a solute that derives from a solid source do not mimic the properties of the (crystalline) solid. For example, table salt ($\ce{NaCl}$) has a certain density as a powder, but this value has nothing to do with the density it acquires in solution once the ions dissociate and live freely in solution. The same goes for heat capacity, which is also a thermodynamic property.

So the answer is yes, the heat capacity of the solution does change by adding a solute and, in fact, it depends on the amount that is added. Only experimentally can you measure the exact function of concentration that defines the heat capacity of a particular solution. From a molecular perspective, it will be highly dependent on how the phenomenon of solvation develops.

As always, you can always look up the scientific literature to get the data of the thermodynamic functions of certain solutions including heat capacity.


Phosphoric acid spill cleanup While working with some rust converter (MSDS, ~20% phosphoric acid by weight), a sequence of events that would make Rube Goldberg proud led to me spilling about 500mL of it all over the place. Yes I was wearing gloves and an acid gas respirator.
I'm not really well versed in chemical cleanup procedures and I needed to act fast so I kind of winged it by:

Wiping the majority with a bath towel.
There happened to be a box of baking soda right there so I dumped a bunch of it on the remaining spill.
It fizzed for a bit then I wiped that up.
I shop-vac'd the rest of the baking soda.

My question is just: Did I handle that OK? And, if not, how should I be prepared if that happens again? (I've been working with the chemical a lot lately.)
The reason I ask is, when I took off my respirator, it still smells like rust converter.
Also, it just happened so I won't be able to see any effects for a few hours but I'm hoping I don't start seeing rust spots on the metals it contacted, or dried crystals, or tear up any of the finish on the wood it contacted. But I'm hoping I got it.
Also, there are cats here and I don't really want residue to be left laying around.
So I'm wondering if that was a decent reaction, or if it was dangerous (I dunno what phosphoric acid + baking soda produces, although the space was well ventilated); and if it was effective, or if there is a better way.
Context is home shop, primarily woodworking; not a lab. So I don't have those nice metal lab tables or anything; lots of cracks, crevices, porous surfaces (concrete and wood), and stuff. And while I normally work with this stuff on a tray for small splash containment, this was a rather... violent spill (heh).

The product you were using is very likely a "rust converter". Used to convert rust to a stable products before painting . I have used it several times , usually I just rinse the  surface , let it dry and paint . It is also used for industrial maintenance coating of poorly prepared surfaces ( those not sand blasted). That is ;it is commonly used and when I worked with coatings I never heard of a problem. Baking soda will neutralize it . I once inspected super phos storage tanks : it starts as 107 % phosphoric acid used to blend liquid fertilizers. When I stepped out of the tank and walked across the limestone gravel ,each step of the rubber boots left some jelly-like acid which foamed on the gravel. The people that worked at the facility thought nothing of it; from that I conclude neutralized phos acid it not a problem . Sorry this is all anecdotal ,but I have never needed to research the situation.


What can be used to thionate a monosubstituted urea? I'm looking to make a thiourea from a monosubstituted urea. I haven't found any references that use Lawesson's Reagent, which I have been using for the thionation of di and trisubstituted ureas. Is there something that would prevent Lawesson's Reagent from working on a monosubstituted urea?

I believe Lawesson's Reagent could do this. It is useful to thionate a large number of carbonyl compounds including amides, so I think it can do for urea as well. I don't see any effect on a monosubstituted urea that would negatively affect the reaction.


What is the basis for naming the p-orbitals with reference to the Cartesian coordinate system as x, y, and z? There are 3 p-orbitals per subshell, which are $p_x$, $p_y$ and $p_z$. As well as there are 5 d-orbitals, which are $d_{xy}$, $d_{yz}$, $d_{zx}$, $d_{x^2-y^2}$, and $d_{z^2}$.

My Question: What is the basis for choosing the axes as $x$, $y$, and $z$? Can we rearrange the axes, suppose lets replace $x$ with $y$, $y$ with $z$, and $z$ with $x$. Then we will have the d-orbitals to be $d_{yz}$, $d_{zx}$, $d_{xy}$, also you would get $d_{y^2-z^2}$, and $d_{x^2}$, orbitals. Why do we choose that particularly?

The choice of axes is arbitrary in this case (isolated atom with no external influence).
You could choose about any set of co-ordinate axes, and that wouldn't really affect the behavior of any of the electrons there ;)
However, in the presence of external influences, such as a magnetic field, it's customary to take the positive axis in the direction of the field as the positive z axis. Though ultimately, the choice of notation is left to you.

EDIT: Since you seem to have an issue here, I'll mention this as well:
Quoting from chemguide.co.uk
A)  $\mathrm{d_{xy}}$,  $\mathrm{d_{yz}}$,  $\mathrm{d_{zx}}$

The names tell you that these orbitals lie in the x-y plane, the x-z plane and the y-z plane respectively.

Each orbital has four lobes. Notice that each of the lobes is pointing between two of the axes - not along them.

B) $\mathrm{d_{x^2 - y^2}}$, $\mathrm{d_{z^2}}$

Although these two orbitals look totally different, what they have in common is that their lobes point along the various axes. That's different from the first three where the lobes pointed in between the axes
The $\mathrm{d_{x^2 - y^2}}$ orbital looks exactly like the first group - apart, of course, from the fact that the lobes are pointing along the x and y axes, not between them.

The $\mathrm{d_{z^2}}$ orbital looks like a p orbital wearing a collar! The main lobes point along the z axis.


Why does sodium sulfate have an unusual solubility-temperature curve? For many salts there are solubility curves as a function of temperature that are smooth (don't have any kinks). Sodium sulfate, however, has a kink in the solubility-T curve at 30 degrees as shown below:

What happens when the temperature is about 30 degree?

"Sodium sulfate" is a mixture of hydrated phases in shifting equilibrium given temperature and concentration.


What causes Copper Rose? We made a solution of Tetrakis Copper(I) hexafluorophosphate and the usual color is yellowish. As it oxidizes to Cu(II), it should turn blue.

However, our current mixture has an off color that is pink. Looking up alternative copper colors, there will be copper rose. What would cause that color?

I assume you mean tetrakis(acetonitrile)copper(I) hexafluorophosphate. This compound contains a tetrahedral complex of $\ce{Cu+}$ with 4 $\ce{CH3CN}$ ligands which stabilize the copper(I) center. They are, however, easily exchanged in the presence of other potential ligands, and with water, disproportionation to $\ce{Cu^2+}$ and elemental $\ce{Cu}$ will occur.

$$\ce{2[Cu(CH3CN)4]^+ + 6H2O ->~ [Cu(H2O)6]^2+ + Cu\downarrow + 8 CH3CN }$$

The pale blue color of the hexaaquacopper(II) complex might not be visible at low concentrations, and the pink color that you have observed arises from fine particles of the elemental $\ce{Cu}$ precipitate.


Difference between vapour and gas This is the question i have been asking from months to my friends,teachers,and everyone but i haven't got a satisfactory answer yet(in fact i haven't got any answer).These two terms (vapour and gas) are confusing me from a long time and i can't get the difference.I shall be very thankful if someone of you give me a detailed explaination for the question.I m sorry if this question or a similar question have already been asked on the site and i shall be thankful if you provide me the link of the question so that i can get my answer there.

Finally the question is "What is the difference between vapour and gas".

I think vapor have a temperature near to the condensation point so it can become liquid easily but gas have much higher temperature than it's condensation point hence we need a process to make it liquid for example if steam is at around 100°c then it can be termed as water vapor and if you consider nitrogen it liquifies at -198°c or near to that hence it is referred as gas.


Do processes involving synthesis need heat? I came across a textbook that stated a combination process that required heat. There was no explanation whether it was a prerequisite for a combination process to have heat or it was just an example just that it was grouped under "heat process chemical changes".

Was doing a test paper that asked about 2 differences between thermal decomposition and combination processes and one of the difference was that "combination processes do not need heat but thermal decomposition processes do".

Would like to ask which interpretation is correct. From preliminary searches on Google it seems that the consensus is that any reaction that simply involves two separate substances combining in a reaction to become a single product is considered to be a combination reaction regardless of whether there is heat involved or not.

Would like to know what is the correct interpretation.

Definitely. Although combination processes may be exothermic, they do have activational energy. As an example, consider the reaction:

$$\ce{H2 + I2 -> 2HI}$$

While the formation of two $\ce{H-I}$ bonds is quite energy releasing, you still have to go through breaking a $\ce{H-H}$ and a $\ce{I-I}$ bond. You need to supply heat to give thermal agitation to allow the molecules with sufficient energy to break the bonds. A typical reaction like this one would have this energy-reaction coordinate graph:

The reaction proceeds into a high-energy transition state, and then comes down to the low energy products. To reach the transition state however, you must supply heat.

Edit: As @porphyrin mentioned, there are some reactions that have essentially zero activational energy. Combination of radicals may be an example:

$$\ce{Cl. + Cl. -> Cl2}$$

This one has no activational energy because there is no unstable transition state involved. Another popular example is the basic neutralization reaction:

$$\ce{H+ + OH- <=> H2O}$$

Here again, there isn't any highly unstable transition state.


Is every compound that has a H-bond donating group soluble in water? I am wondering if there is any general rule of thumb

Definitely not.
For instance, palmitic acid with a pK$_\mathrm{a}$ of 4.75 has a water solubility of only 0.04 mg/L.

A general rule of thumb is that every oxygen can solubilize four carbon atoms but there are many exceptions.
I addressed a similar question here.

Nitrogen also makes the compound more soluble, but there it depends very much on the pH.


Electrophilic addition on cyclopropane using sunlight I encountered a weird question in an exam. The statement says that instead of radical substitution, cyclopropane undergoes electrophilic addition reaction in sunlight.
I know that all saturated compounds give substitution reactions in sunlight but the answer key states that electrophilic addition takes place.
I couldn't find any source to prove this. Any help with an explanation and a verified source would be appreciated.

It should say "in addition to" rather than "instead of".
Cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal ~109.5° when carbon makes four single bonds. In the absence of light, when in contact with $Cl_2$ or $Br_2$ the ring will break and addition occur.
In sunlight, both the expected substitution reactions and the addition reaction will occur.
Can't find the reference I thought I remembered, but here's a site or two discussing it:
https://www.chemguide.co.uk/organicprops/alkanes/halogenation.html
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.07%3A_Reactions_of_Alkanes


Why does copper not react with nitrogen? Please describe why copper only reacts with oxygen and not nitrogen in the air. Isn't nitrogen a negative ion, and copper positive, so wouldn't that mean that copper and nitrogen would want to form bonds?

Sketching with broad strokes: the first key point is that oxygen and nitrogen are major components of the atmosphere in the form of $\ce{O2}$ and $\ce{N2}$ molecules, and not as ions. $\ce{N2}$ is a very stable compound because the two nitrogen atoms in its molecules are united by a particularly strong triple bond. $\ce{O2}$ is overall more reactive than $\ce{N2}$, largely as a reflex of the bond between the atoms not being as strong. Specifically concerning oxidation reactions (of which those you suggest are typical examples), oxygen atoms having more affinity for electrons than nitrogen atoms contributes towards making $\ce{O2}$ being a better oxidiser than $\ce{N2}$.


Why basicity of oxides increases down a group? Take group $13$ oxides as an example. The elements in this group are $\ce{B,Al,Ga,In,Tl}$.
Boron oxide $\ce{B2O3}$ is acidic and $\ce{Al2O3}$ and $\ce{GaO3}$ are amphoteric while $\ce{In2O3}$ and $\ce{Tl2O3}$ are basic in nature.
What is the reason for this?

As suggested in comments, major factors play a part

Atomic size

As the atomic size of the element increases, its electron affinity decreases consequently. It turns out that when moving vertically in the periodic table, the size of the atom outdoes its electronegativity with regard to basicity. The atomic radius of gallium, for example is approximately twice that of boron, so in a gallium ion, the charge is spread out over a significantly larger volume.

Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.

Electron affinity

The ability of an atom “to accept an electron” decreases down the group therefore increases the tendency of the elements to be good Lewis bases (recall a Lewis base is an electron-pair donor). Thus the decrease in electron affinity down the group means the elements readily donate an electron rather than accept them.

Electronegativity

The more electronegative an atom, the better it is able to bear a negative charge. Down the group, the electronegativity of the elements decrease down the group.  B is very electronegative and doesn’t readily release electrons but instead holds on to it, additionally is able to stabilise a negative charge, and consequently has good acidic properties unlike proceeding elements down the group.


Stereochemistry of product in Grignard-Gilman addition reaction I would like to know whether the stereochemistry given in the following reaction is correct or not?

Which attack is more likely? axial or equatorial? I think I should say pseudo axial or pseudo equatorial....since there is a double bond in the ring.

Am I missing anything here...?

Ok. I am adding what I got from other forum, though I do not follow it properly.

I have supplied a diagram that is more legible than the small one you have provided. The cis stereochemistry of the product is a function of axial, but not equatorial attack of the reagent. The stereochemistry is established after the addition occurs. Structures 1 and 2 illustrate the "chair-chair" interconversions of the cyclohexene ring. As the C-C bond is being formed by axial addition in 1, the carbon bearing the red p-orbital will rehybridize from sp2 to sp3 during the transition state (TS) for bond formation. This transition state (TS) will lead to enolate 3 through a low energy, chair-like pathway.
Alternatively, TS 2 would proceed through a higher energy, twist boat-like TS. Both routes occur through axial attack with 1 ---> 3 prevailing over 2 ---> 4. Enolate 3 is in equilibrium with enolate 5, which is unfavored by what is known as an allylic A1,3 interaction.1 The critical step is the kinetic protonation of enolate 3 on workup. Protonation of enolate 3 on carbon from the underside of the double bond (remote from the phenyl group) again occurs through a chair-like TS. Protonation on the top of the double bond would occur through a twist boat-like TS. This pathway is higher in energy and would have provided the trans isomer 9. Structures 6 and 7 are the two chair conformers of the cis-isomer. If one uses to vigorous a set of conditions upon workup, enolization of 6 and 7 can lead reversibly to enol 8 and eventually to the thermodynamic isomer 9.

1) Johnson, F., Malhotra, S. K., J. Am. Chem. Soc., 1965, 87, 5492.


Point Groups and MO Theory I am a little confused about how point groups give information about MO of polyatomic molecules.

Is it true that all molecules of a given point group such as, $\mathrm{C_{2v}}$, will have the same MO diagram, the only thing that will change is the number of electrons?

Is it true that all molecules of a given point group such as, $C_{2v}$, will have the same MO diagram, the only thing that will change is the number of electrons?

No. There are so many different possibilities of how molecules with $C_{2v}$ may look, that they cannot possibly all have the same energy level diagram. Take water and acetone for mild examples.

What you can do, however, is estimate how different molecular orbitals will look based on the atomic orbitals you put into the estimation. Let’s take formaldehyde as it is a nice and simple example.

We have two hydrogen atoms, one carbon and one oxygen. Carbon and oxygen both lie on the $C_2$ axis so we can consider them separately, but we should use group orbitals for the hydrogens (I am defining the $yz$ plane to be the plane of the molecule). Luckily, they only have s orbitals, so it doesn’t get too complicated. Check out how the s-orbitals transform with the different symmetry operations to find their irreducible representations: $\mathrm{A_1 + B_2}$. Carbon and oxygen each have an s orbital and three p orbitals which will transform in the same way: the s orbital as $\mathrm{A_1}$ (naturally), the p orbitals transform as $\mathrm{A_1 + B_1 + B_2}$ (for $\mathrm{p}_z, \mathrm{p}_x$ and $\mathrm{p}_y$, respectively).

Thus, we have five atomic (or group) orbitals of the $\mathrm{A_1}$ type which will combine to form most of the σ bonds, three of the $\mathrm{B_2}$ which will also combine to σ bonds and two of the $\mathrm{B_1}$ type which will combine to form π bonds. Had we chosen water, we would only have had six orbitals altogether transforming as $\mathrm{3 A_1 + B_1 + 2 B_2}$ — a clear distinction showing that the diagrams cannot always be the same.

We now have to estimate the energy levels the orbitals initially have. It is a good guess to put the highest-energy atomic orbital at the respective atom’s ionisation energy; then you would just have to esimate a lower level for the s orbitals of carbon and oxygen. With these levels in hand, you can educatedly guess the different energy levels of the resulting molecular orbitals and educatedly guess what the orbital would look like: The resulting MO must transform in the same way as the AOs used to build it. So any $\mathrm{A_1}$ orbital will be symmetric to any symmetry operation while $\mathrm{B_2}$ will always have a nodal plane in the $\ce{C=O}$ bond axis, etc. I’m sure you can now come up with a well-educated guess of the formaldehyde orbitals; you can check out what they actually look like on the homepage of Professor Zipse. (You can ignore the accompanying German text. The higher-level antibonding MOs LUMO+2 and LUMO+3 that are required by theory are not shown; one must be $\mathrm{B_2}$, the other $\mathrm{A_1}$.)


How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination? How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination?

I think that three distinct products may result namely
a) 1,1-Dichloro-2-methylpropane
b) 1,2-Dichloro-2-methylpropane
c) 1,3-Dichloro-2-methylpropane

But the book says only two distinct products may form. Can anyone explain how?

The following three products wil be formed:

Your book may be giving the answer to the number of monochloro products formed, as there are only two distinct monochloro products formed on chlorination of isobutane.


Do Neon Lights Work Do To Stokes Shift? I was wondering whether or not the reason some noble gases, like neon, fluoresce at particular frequencies of the visible spectrum is because of a Stokes effect? I mean, do the neon dimers get stretched out dramatically at a more LUMO state, resonating with high energy electricity due to the high stability of noble gases, and then contract back to a more HOMO state, releasing red light, which has a much lower frequency? Noble gases, like neon, have a high quantum yield, correct?

Ed has already explained the main ideas. Stokes shift is associated with the concept when you excite a substance with light (or electromagnetic radiation). In neon discharge tubes you are using electrical energy to excite the neon atoms. All the emission you see is from neon atoms. Also recall, that molecular emission has band like structure. Neon light spectrum is nowhere close to this band structure.

Buy a small pocket spectroscope or build your own and see how the spectra look like. The black and white picture has not done justice to the actual extremely beautiful spectrum of Swan bands (observed due to diatomic carbon species).

  a more LUMO state, resonating with high energy electricity due to the high stability of noble gases, and then contract back to a more HOMO state, releasing red light, which has a much lower frequency?

This has no scientific meaning, because molecules do not resonate with electricity or electrical energy. I do not know the exact mechanism how electrical energy or even heat is "transformed" into electronic energy, but electrical discharge is quite efficient at exciting atoms or molecules in the gas phase. I am sure it is quite complex, which involves ionization, collisions, recombination etc.


Can ethyl alcohol fumes be suppressed by a household air purifier/filter? Can ethyl alcohol fumes be suppressed by a household air purifier/filter? In general, how are the fumes suppressed? How is it done?

Can somebody direct me to some information regarding this? I've been searching but do not have the strongest background in chemistry (I'm a mechanical engineering student). I've done some research on filtering media as well as ethyl alcohol itself.

No home air purifier removes alcohol vapor from the air. There are a number of ways it can be eliminated, though, depending on quantity, purity and other factors.

Bakeries have to comply with clean air standards for VOC (volatile organic compounds), primarily ethanol, which can cause smog. One way the alcohol is removed is by catalytic oxidation, similar to the way an automobile catalytic converter works. Even hot copper is sufficient to oxidize alcohol.

Another way to remove ethanol is by condensation, which allows the alcohol to be saved for resale, or by washing (scrubbing) the oven gases with water.

Sadly, though this regulation of ethanol emission is necessary, it also removes the sublime aroma of baking bread from the vicinity. sigh...


Why don't we use an alkyl halide directly in Gabriel Phthalimide Synthesis? In Gabriel Phthalimide Synthesis, why do we use $\ce{KOH}$ in first step rather than directly adding an alkyl halide?

It is used to make the conjugate base of the imide which is far more nucleophilic than the imide itself, but not too much to attack another molecule of the alkyl halide.


Can heteroatoms with lone pairs be chiral centres? If a compound has a carbon atom with four different groups covalently bonded to it, it is called asymmetric and enantiomers of the compound can exist.

But imagine if one has a different central atom, such as a nitrogen or a sulfur where one of the four "groups" is an electron pair (see examples below).

Would such a thing still behave like a "normal" asymmetric carbon? If not, how else does it behave, i.e. are there enantiomers? Is my assumption that those will still have $\mathrm{sp^3}$ hybrid orbitals wrong? I would also enjoy pointers to literature, I was unable to find any.

Generally, amine nitrogens will not behave like a normal asymmetric carbon. Simple amines are roughly $\mathrm{sp^3}$ hybridiized and the molecules you use as examples do have 4 (we include the lone pair of electrons as a substituent) different substituents around the central nitrogen atom. So in principle me might consider it asymmetric or chiral.  But simple amines can undergo a process called nitrogen inversion, which essentially coverts one enantiomer into the other.

(image source)

However, if one can find a way to slow down or eliminate the nitrogen inversion process, then chiral amines can be isolated. One way to achieve this is to incorporate the amine nitrogen into a 3-membered ring (an aziridine). Now to achieve the planar state necessary for nitrogen inversion requires the bond angle in the 3-membered ring to open from 60° to 120°, an impossible task. Aziridines containing a chiral nitrogen atom have been isolated and characterized.


How to calculate the kinetic order of an enzymatic reaction? This question is concerning the metabolism of ethanol by an alcohol dehydrogenase enzyme.

Usually people metabolize alcohol equivalent to "one beer per hour". One beer is said to contain 33cL. The ethanol content of a beer is 4.6%(V/V). If this approximation is correct for an average lean person of 75kg after drinking two beers, then what is the kinetic order of reaction of the enzymatic reaction metabolizing the alcohol?

The alcohol content in one beer is:
$(33 \cdot 0.046)cL = 1.518cL$

The ethanol content in two beers is then:
$(2 \cdot 1.518)cL = 3.036cL$

How do I link this to the weight of the person (the 75 kg)? And how to I determine of this follows a zero, first, or second-order reaction?

I know that a first order reaction is given by the following rate expression:
$-\frac{dA}{dt} = k \cdot A$

The second order reaction is given by the following rate expression:
$-\frac{dA}{dt} = k \cdot A^2$

The legal limit in some states of 0.08% BAC corresponds to about 17 mM ethanol in the blood. The $K_M$ value for human liver alcohol dehydrogenase is 0.5 mM according to this paper. So at this particular legal limit, the enzyme works at 97% of its maximal capacity. ($v = v_{max} \frac{[S]}{[S] + K_M}$).

Above or very much above the legal limit, the rate does not change much, making it approximately zero order in ethanol. Because the breakdown of alcohol does not get faster with more intake, it can take past the next morning to metabolize sufficient alcohol to reach values below the legal limit if someone drinks heavily at night.

For ethanol concentrations below the $K_M$, the reaction is approximately first order in ethanol, so it gets slower and slower as the ethanol concentration approaches very small concentration.

While the $K_M$ is the same for different individuals, the amount of enzyme is not (depends on the size and health of the liver, gender, and on genetics). That's where the body mass comes in.


Calculating molar solubility of calcium hydroxide from solubility product On the Wikipedia page for calcium hydroxide, the stated solubility product for $\ce{Ca(OH)2}$ is $K_\text{sp} = \pu{5.5 * 10^{-6} M^3}$. It doesn't stipulate the temperature, but I have a chemistry book that says the exact same value for $\pu{25 ^\circ C}$.

However, when I calculate the solubility of $\ce{Ca(OH)2}$ in water using this value ($\pu{5.5 * 10^-6 M^3}$), I get a result of $\ce{0.082 g/100ml}$.

However, according to solubility tables on Wikipedia, calcium hydroxide is only this soluble at around $\pu{85 ^\circ C}$. How can this apparent contradiction be explained?

A cursory search suggests the solubilities in g/100 mL on the wikipedia are closer to a consensus, for what it's worth:

solubility 0.2% ($0^oC$):
https://www.cdc.gov/niosh/npg/npgd0092.html
solubility 0.159 g/100 mL
(0.16% by weight) ($25^oC$):
https://www.mdpi.com/1996-1944/10/10/1219/pdf
the value you reported from the National Lime association.

These are reasonably consistent with the values in

https://en.wikipedia.org/wiki/Solubility_table#C

but not with the $K_{sp}=5.5 \times 10^{-6} M^3$ on the wiki page.

Skoog, West and Holler's Analytical Chemistry textbook (7th Edition) reports $K_{sp} = 6.5 \times 10^{-6} M^3$ at 25$^oC$. They provide a ref:

  most of the data were taken from A.E. Martell and R. M. Smith, Critical Stability Constants, Vol. 3-6.

Note the apparent inconsistency recurs (I called this discrepancy an "error" before but am no longer so sure):

http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
http://www5.csudh.edu/oliver/chemdata/data-ksp.htm

It reminds you that you should generally reference your data sources if you want to be helpful to fellow researchers.

The value for $K_{sp}$ in the wikipedia (which is not referenced as far as I could tell) and in a highly respected textbook which references its source is smaller by roughly an order of magnitude compared to values derived using standard equations from reported solubilities in a number of other sources noted above. This inconsistency may be because some of the reported data is for hydrates with higher Mw. Not sure how you could explain such a discrepancy otherwise. If the $K_{sp}$ were much smaller you could argue that the dissociation constant of water affected the solubility.

My current guess is that there is a reasonable explanation for the discrepancy (ie it has a chemical or physical explanation and is not due to human error). It is unfortunate that the solution is not more obvious given that $\ce{Ca(OH)2}$ is a recurrent example in problems (textbook calculations or experiments) teaching solubility to undergraduates.

See also:

Solubility of calcium hydroxide determined by titration


Why is ethoxyethane the preferred IUPAC name for diethyl ether rather than 1,1'-oxydiethane? Diethyl ether has a few systematic names. Its preferred IUPAC name is generated using substitutive nomenclature: ethoxyethane. But multiplicative nomenclature could be used, in the symmetric spirit of its common name, to generate the name 1,1'-oxydiethane. The Nomenclature of Organic Chemistry 2013 (blue book) makes it clear that multiplicative nomenclature is preferred to substitutive. So why is the preferred name substitutive?

The book says these about multiplicative nomenclature:

P-51.1.5: “Multiplicative nomenclature (P-15.3, P-51.3), as a subset of substitutive nomenclature, is preferred to simple substitutive nomenclature when the criteria for its use are met…”
P-51.3: “Substitutive nomenclature is used when the conditions for constructing multiplicative names are not met.”
P-51.3.1: “Multiplicative nomenclature is preferred to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical parent structures other than alkanes when
“(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical and
“(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
“(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.”

The deciding criterion is in the preamble to P-51.3.1, not in the numbered criteria: “other than alkanes.” The identical parent structure in diethyl ether is a straight-chain alkane, so multiplicative nomenclature does not apply.


Mechanism for the conversion of an unsaturated alkyl halide to an alcohol Question:

Predict the mechanism for the following reaction:

Attempt:
I thought it will proceed through SN1 mechanism as $\ce{H_2O}$ is involved (solvolysis). So I proceeded with the formation of a primary carbocation, which then rearranges via 1,2 methyl shift. Then the nucleophilic attack of $\ce{OH^{-}}$ leads to the following product, which is different form the required one.
Can any further steps lead to the required product or am I wrong from the start (most probably)?

For what it's worth, this reaction does not appear in the literature. So, whether it actually forms the product or not is anybody's guess.

However, if I had to suggest a mechanism... I would note first that an SN1 reaction on a primary centre, as you have proposed, is difficult. I think it more likely that the alkene assists in the expulsion of the chloride.

The resulting cyclopropylmethyl carbocation is stabilised and is electrophilic at multiple sites. I have only drawn one resonance form, but there are a few more. Attack of water at the tertiary carbon leads to the desired product after deprotonation.


Why can't Al3+ be reduced to solid, pure aluminum? There are efforts underway to develop ionic liquids for the purpose of reducing bauxite using exotic intermediates such as aluminum fluoride. However, the aqueous $\ce{Al^3+}$ ion is tantalizingly close to pure aluminum. What prevents it from being reduced?

It is aluminium’s standard electrode potential that prevents it from easily being reduced. For the half-cell $\ce{Al^3+ + 3e- <=> Al(s)}$, the standard potential is $E^0 = -1.66~\mathrm{V}$. For comparison a few other values:

$\ce{Fe2O3 + 3 H2O + 2e- <=> 2 Fe(OH)2(s) + 2 OH-}$ and $\ce{Fe(OH)2 + 2 e- <=> Fe(s) + 2 OH-}$ have potentials of $E^0 = -0.86~\mathrm{V}$ and $-0,89~\mathrm{V}$, respectively;
$\ce{Zn^2+ + 2 e- <=> Zn(s)}$ has a potential of $E^0 = -0.76~\mathrm{V}$;
$\ce{Cr^3+ + 3 e- <=> Cr(s)}$ has a potential of $E^0 = -0.74~\mathrm{V}$;
$\ce{Co^2+ + 2 e- <=> Co(s)}$ has a potential of $E^0 = -0.28~\mathrm{V}$;
$\ce{Cu^2+ + 2 e- <=> Cu(s)}$ has a potential of $E^0 = +0.34~\mathrm{V}$.

(All values taken from Wikipedia)

The higher the standard potentials are, the easier the ion can be reduced. Hydrogen, of course, has a potential of $E^0 = 0~\mathrm{V}$ at pH 0 by definition. So in acidic media, hydrogen is reduced first. So let’s lower the pH to impede the formation of hydrogen. According to the Nernst equation:

$$E = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\mathrm{Ox}}{a_\mathrm{red}}$$

Where $z$ is the number of electrons transferred in the half-cell, and $a$ is the activity of the reduced/oxidised species. $a_\mathrm{red}$ is $1$ by definition and $z = 2$ for the hydrogen half-cell. So we need to solve the following simplified equation to determine the acid concentration that will allow reduction of aluminium:

$$-1.66~\mathrm{V} = 0~\mathrm{V} + \frac{0.059~\mathrm{V}}{2} \lg \frac{[\ce{H+}]}{1}\\
\frac{0.059}{2} \mathrm{pH} = 1.66 \\
0.059 \times \mathrm{pH} = 3.32 \\
\mathrm{pH} = 3.32 / 0.059 = 56$$

Whoops, so we would need a pH of $56$ to allow the reduction of aluminium in the presence of water. This calculation was highly theoretical, because:

A pH of 56 is impossible in anything that once could have been water. Wikipedia gives the $\mathrm{p}K_\mathrm{b}(\ce{O^2-}) \approx -38$; because $\mathrm{p}K_\mathrm{a}(\ce{HA}) + \mathrm{p}K_\mathrm{b}(\ce{A-}) = 14$ in water that must mean that $\mathrm{p}K_\mathrm{a}(\ce{OH-}) \approx 52$ — thus we would be dealing with an ‘oxide solution’ rather than water.
Although it is nice that both equations (pH and Nernst) use acitivities, we cannot really say anything about the activities at this large a concentration range.
The redox system of both species would be totally different in these environments.

Note that you can reduce most air-stable metals in thermite reactions from their oxides thereby creating aluminium oxide. An example reaction would be:

$$\ce{Fe2O3 + 2Al -> 2Fe + Al2O3}$$

That alone should show the unwillingness of aluminium to be reduced in aquaeous solutions; think about the heat that the thermite reactions generate.


Why do we not using concentrated HCl and zinc reaction to prepare hydrogen? First apologize for this kind silly question.One of my student ask me this question but I was unable to get a perfect answer from internet.Generally dilute HCl and impure zinc are using for hydrogen production in laboratory.My question is why  do reaction stoped after some little time reaction if we use pure zinc and concentrated HCl?
Is it for coverup zinc by zinc sulphate?
It will be thankful for any kind help.

If metallic zinc $\ce{Zn}$ is in contact with concentrated $\ce{HCl}$ solution, $\ce{H2}$ is produced, but the reaction is exothermic  : the solution gets hot. As $\ce{HCl}$ is not so highly soluble in hot water, a fraction of the dissolved  $\ce{HCl}$ will be vaporized. So the gas produced will be a mixture of $\ce{H2}$ and $\ce{HCl}$, which is not wanted.
Anyhow, the reaction between pure $\ce{Zn}$ and pure concentrated $\ce{HCl}$ is not really rapid. And, after some time, when the solution contains a certain amount of $\ce{Zn^{2+}}$, the reaction rate decreases in such a manner that the reaction looks finished. It is rather surprising, and even deceiving.
However, if some metallic impurities are present in solution, like $\ce{Co^{2+}}$ or $\ce{Cu^{2+}}$, the zinc metal reduces these ions according to $$\ce{Zn + Co^{2+} -> Co  +  Zn^{2+}}$$or $$\ce{Zn + Cu^{2+} -> Cu  +  Zn^{2+}}$$ As a result, some small amounts of metallic cobalt or copper are deposited on the zinc metal. The contact of the two metals $\ce{Zn + Co}$ or $\ce{Cu}$ produces a galvanic cell, and the reaction rate increases a lot. $\ce{Zn}$ gets dissolved much quicker, producing electrons that go onto the copper or cobalt spots, where they reduce $\ce{H+}$ to produce a lot of $\ce{H2}$.
So the best way of producing H2 by the reaction $\ce{Zn + HCl}$ is to use not too concentrated solutions of $\ce{HCl}$ solution, with small amounts of copper or cobalt salts added into the acidic solution.
The same phenomena of galvanic cell happens if  $\ce{HCl}$ is replaced by $\ce{H2SO4}$


Why can some ions have different valence? Like $\ce{Fe}$, it can be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$, same with $\ce{Ni}$ and $\ce{Co}$, what determines whether the ionic compound will be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$?

An empty, half-filled or fully-filled electron shell is especially stable.  $\ce{Fe}$ metal is [Ar] 3d6 4s2.  $\ce{Fe^{2+}}$ emptiess the 4s-shell.  $\ce{Fe^{3+}}$ give that plus a half-filled 3d-shell.  Vanadium II, III, IV, V.  Now try chloride, hypochlorite, chlorite, chlorate, and perchlorate.

The most stable oxidation state depends on the medium and added ligands, if any.  Cu(I) is a Lewis soft acid. It is stabilized by soft bases like pyridine, $\ce{MeCN}$, olefins, $\ce{I^{-}}$.  $\ce{Cu^{2+}}$ is a hard Lewis acid.  It is stabilized by hard bases like ammonia, water, $\ce{Cl^{-}}$.  Dissolve $\ce{CuCl2}$ in pyridine or $\ce{MeCN}$ and you have a powerful oxidizing agent. Dissolve a $\ce{Cu^{1+}}$ salt in water and it disproportionates into $\ce{Cu}$ metal and $\ce{Cu^{2+}}$.  Put $\ce{Cu}$ metal and $\ce{Cu^{2+}}$ into $\ce{MeCN}$ and you get $\ce{Cu^{1+}}$.


Rate Constant Units and Eyring Equation Rearranging the Eyring equation leads to the following:

$$\Delta^\ddagger S^\circ = R \ln{\frac{k \times h}{{k_\text{B}}{T}}}+\frac{\Delta^\ddagger H^\circ}{T}$$

where $k$ is the rate constant, $h$ is the Planck constant, $R$ is the universal gas constant, $k_\text{B}$ is the Boltzmann constant, $T$ is temperature, $H$ is enthalpy, and $S$ is entropy. The entropy term should ultimately be in units of $\mathrm{\frac{J}{mol\ K}}$. The right term of the equation will clearly yield the correct units. For the left term, the universal gas constant can be expressed in $\mathrm{\frac{J}{mol\ K}}$, which means that the log term should be dimensionless. For the bottom part of the fraction, the units evaluate to $\mathrm{J}$. For the top part, the Planck constant has units of $\mathrm{J\cdot s}$. To get the top part of the fraction to evaluate to $\mathrm{J}$, the rate constant would have to be in units of $\mathrm{s^{-1}}$. For first order reactions, this is indeed the case. For reactions of higher order, this will never work out to the correct units.

This makes me think that the Eyring equation is simply using the unitless values of these variables. This still leads to some ambiguity about which value of the rate constant to use. Take for instance a second-order reaction in which the rate constant is in units of $\mathrm{M^{-1}\ s^{-1}}$. If we rearrange the units to SI units ($\mathrm{m^3\ mol^{-1}\ s^{-1}}$), the value of the rate constant changes. So is the rate constant $k$ in the Eyring equation simply the value of the rate constant when that rate constant is expressed in certain units? And if so, are the concentration units always expressed in molarity? I kept missing a homework problem because I was converting $k$ from molarity to SI units, which makes me believe this is indeed the case.

Using Transition State Theory (TST) one has to be aware of at least some of the many flaws of it. And may also often lead to wrong assumptions. However, it is a wonderful tool to describe basic kinetics in a theoretical manner.

The most important/ fundamental assumptions are

separation of electron and nuclear movement (comparable to Born-Oppenheimer approximation)
(all possible) transition states (TS) are in equilibrium with the reactants  $$\ce{A + B <=> [AB]^{\ddagger}}\tag1$$
combined Boltzmann statistic for reactants and TS
All molecules that reach the transition state from the reaction coordinate of the reactants will leave the TS to the reaction coordinate of the product(s) - and vice versa
\begin{align}
\ce{A + B <=> &[AB]^{\ddagger}->P}\tag{2a}\\
\ce{P <=> &[AB]^{\ddagger}-> A + B}\tag{2b}
\end{align}
at the TS the transition vector can be separated from all other motions and hence may be treated classical as translation

In other words TST reduces a very complicated multidimensional problem to a one-dimensional translation. That implies that all treated molecules must be connected on a potential energy hypersurface with one vector. (Borrowing from the amazing wikipedia)

Hence reactions that are described with TST are of first order by default. One would have to make various adjustments to treat higher orders (and I would assume that not matching units would be of minor concern).


Is a galvanic cell only called that if copper and zinc are used? I know other metals can be used to get the same effect but I'm interested in the name. Is it only called a galvanic cell if copper and zinc are used? Or is the name "voltaic cell" used only when copper and zinc are used? I'm confused because I've heard them being used interchangeably.

Voltaic cell and galvanic cell are both synonyms for the same thing: An electrochemical cell, which produces an external electric current. So the two terms can be used interchangeably.

Check out also a site from Georgia state University, where it says:

  Electrochemical cells which generate an electric current are called voltaic cells or galvanic cells, and common batteries consist of one or more such cells.

Addendum: Nevertheless there might be some galvanic cells, like the Daniell Cell, which have a defined electrode material and defined electrolyte concentrations to supply a specific voltage.

For the Daniell cell these would be:

$$\begin{array}{lc}
\text{Electrode material:} & \ce{Zn|Zn^{2+}||Cu^{2+}|Cu}\\
\text{Electrolyte concentrations:} & c(\ce{Zn^{2+}}) = c(\ce{Cu^{2+}}) = 1\ \mathrm{mol\ l^{-1}}\\
\text{Voltage:} & U(25\ \mathrm{°C}) = 1.10\ \mathrm{V}
\end{array}$$


Oxidation number vs charge? From what I understand:

*oxidation numbers: are the "act" of electrons moving

while

*charge: is what the atom "wants" to do with its electrons.

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Can anyone please clear up the distinction between the two and when its appropriate to use one, the other, or both at the same time?

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Is it that charge is only used for ions?

If an atom loses electrons, it develops a positive charge and becomes a cation. Likewise, if an atoms gains electrons, it develops a negative charge and becomes an anion.

"Charges only EXIST on IONS."

Covalent bonding rose the oxidation number.
We are aware that in a Covalent Bond sharing of electron takes place.
The more electronegative atom will take more bonding electrons and it will get a negative oxidation number (It depicts Anion).
Similarly less electronegative atom mimics a Cation.

In other words you can say Oxidation Numbers are fictional ionic charges