Is there any specific formula to find periodic table periods and valence electrons I am developing a program, that uses a lot of stuff related to the periodic table. As of now, I am making a part of it the determines the period and family. I know that both of those have to do with how many valence electrons each element has. So I want to know if there are formulas to find the amount of valence electrons, family, and periods?

The number of electrons in each orbital is double an odd number (the Pauli exclusion princilel states electrons can be in the same energy only if of opposite spin, so there can be two otherwise identical electron energy states [which can be separated in an EM field]).

s (sharp spectral lines): 2 (twice 1)
p (principal spectral lines): 6 (twice 3)
d (diffuse spectral line): 10 (twice 5)
f (fine, or "fundamental", spectral lines): 14 (twice 7)

Using that information, and the fact that the first row has only s electrons, second row s & p, usw., you should be able to calculate the first few rows easily. However, rules become much more complex for transition metals and heavier elements, where an outer shell may be added before all inner orbitals are filled.

Primary valence for the first few rows is simply the number of electrons needed to either complete the current shell (e.g. oxygen is two e- short), or revert to the previous complete, noble gas, configuration (e.g. Na minus one e- is neon). Again, there are complications, such as transition metals with multiple valences, fluorine reacting with some noble gases, usw.


Storage of methane in mineral oil or other liquid hydrocarbons Is it practical (or even possible) to store gaseous alkanes such as methane or ethane in liquid non-polar solvents, such as mineral oil or vegetable oil, and how much more (or less) methane can be stored for a given pressure with solvent vs. no solvent?

After some research, I found that butane is sometimes dissolved in gasoline to make it evaporate more easily in cold weather. So it's definitely possible. However, I cannot find solubility figures (hence this not being an answer).

"Practical" can mean different things to different people, so I am not going to comment on that.

Instead I am going to address the question, of solubility of methane in hydrocarbons, without any thought about "practicality" :)

This graph is taken from the book: Fuel and Lubricants Handbook (a simple google search lead me to it, and the relevant section appears in the preview :))

The data is reported at $50\ \mathrm{^\circ C}$ and $ 5000 \ \mathrm{psig}$ (pounds per square inch gauge, ungodly non-SI units. ca. $3.4 \times 10^7\ \mathrm{Pa}$ according to an online conversion)

Also, the following table from this publication:

I did not read it in detail, but the data was collected in the vicinity of 1 atm and between $25$ and $200\ \mathrm{^\circ C}$

FYI: Of possible interest could be this paper.


Calcium carbonate and hydrogen phosphate producing calcium phosphate, carbon dioxide, and water How does the balanced reaction look for these compounds? I found that phosphoric acid instead of hydrogen phosphate creates a much easier problem, but the teacher said hydrogen phosphate and I can't seem to balance it.

Most likely that your teacher meant to say phosphoric acid ($\ce{H3PO4}$) rather than the hydrogen phosphate anion ($\ce{HPO4^{2-}}$) so lets go with that.

$$\ce{ 3CaCO3 + 2H3PO4 -> Ca3(PO4)2 + 3CO2 + 3H2O}$$

In reality the carbon dioxide and water would exist in an equilibrium state with carbonic acid, $\ce{H2CO3}$:

$$\ce{ H2CO3 (aq) <=> H2O (l) + CO2(aq)}$$


Calculating singlet-triplet gap and open-shell singlet character I am trying to calculate the singlet-triplet energy gap ($S_0-T_1$) of a conjugated system (18 atoms). For the singlet $S_0$ state, assumed to be closed shell, the Gaussian options were

opt=tight b3lyp/6-311g(d,p) integral(ultrafinegrid)
...
0 1
...

and for the energy of the triplet $T_1$ state

opt=tight rob3lyp/6-311g(d,p) integral(ultrafinegrid)
...
0 3
...

Does the energy difference of those two calculations have any physical significance? (i.e. should have I used  ...td-ub3lyp...0 3 ... instead?)
I believe a system I am currently working on may have some open-shell character for the singlet ground state so I performed the following with the optimized geometry I got from a TD calculation of a triplet state:

#p ub3lyp/6-311+g(d,p) scf=(tight) guess=(mix,always) gfinput gfprint iop(6/7=3)
...
0 1
...

For the stability check, do I have to use the guess=(mix,always) on they keyword route as well?

There is a related discussion on which method to use for a singlet triplet guess, which I encourage you to also read:
U- or RO-method for Singlet-Triplet Gap?

This post is going to be a little longer, so I'll present the key points upfront. It is reasonable to calculate (and/or optimise) the wave function with a restricted method, then test it for instabilities. The same applies to to triplet (etc.) wave functions, where I would also try to start with a restricted open shell approach. To answer 1, it is a reasonable first approach for a singlet-triplet gap. If you are more interested in the electronic excitations a time-dependent treatment is necessary.

As a general recommendation I would advise to split the calculations up and read the respective wave functions from the checkpoint file to perform additional calculations and checks, so that you always have a fallback.
Hence the answer to 2, once you have created and converged a solution to the wave function you can read it in again. Do not generate a new initial guess.

You should also always keep an eye on spin contamination, because this is essentially telling you if the method you are using is suitable in the first place. When it comes to broken symmetry solutions, you should always consider to run multi-reference calculations, too.

(Note that guess(always) applies only to geometry optimisations; see below.)

I'm going to exercise a couple of calculations on dioxygen, because it's a nice study case. I will be using the BP86 functional with the def2-SVP basis set and density fitting. I have done the calculations with Gaussian 09 rev. D.01.
It is somewhat an expansion on the example in Exploring Chemistry with Electronic Structure Methods, 3rd Edition (p.71).

I am going to present the full input of the calculations and at the end the absolute energy to compare to, in case you'd like to run the calculations yourself. Please note that I have not run any geometry optimisations and assumed a equilibrium bond distance of 120 pm. I think that is a reasonable choice since this is just an example and the optimisations can be considered as additional exercise.

I labelled the the used methods RHF, ROHF, and UHF even though I am using density functional approximations.

Because the main focus is the use of the stable keyword, we start with a wave function that is already know to be not stable. The RHF solution of dioxygen, filename bp86svp.rhf.sing.gjf:

%chk=bp86svp.rhf.sing.chk
%nproc=2
%mem=10000MB
#p RBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)

title card unused

0 1
O  0.6 0.0 0.0
O -0.6 0.0 0.0

This input uses a few keywords, that are not strictly necessary to complete the calculations. For example the route section #p requests detailed output. DenFit is already implied with the choice of W06 as an additional basis set. I like to write it in nevertheless, to exactly know what I am doing. The same goes for symmetry(loose); occasionally I like to turn symmetry of, so I know this time I really wanted it. The line gfinput gfoldprint iop(6/7=3) is included so that molden (❤) can read (and display) the orbitals. There is no shame in always requesting population analyses.

In this very case even scf(MaxCycle=9999) will not lead to convergence; with scf(xqc) we are allowing the algorithm to switch to quadratic convergence if conventionally the convergence cannot be achieved. Note that this may converge to a different solution than originally attempted.

For publication type calculations I would also include int(ultrafine) and scf(tight); but I forgot it here.

From here I went on to calculate the stability of the wave function, i.e. bp86svp.rhf.sing.stab.gjf:

%chk=bp86svp.rhf.sing.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
scf(xqc)
stable

title card unused

0 1

Note that in this case the SCF will converge within the conventional cycles and hence the energy is different. You can additionally check this by reading in the checkpoint file and perform a new SCF cycle.

However, in the stability analysis you will find the following output:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:      Triplet-?Sym  Eigenvalue=-0.0560505  =2.000
       8 ->  9         0.70645

 Excited state symmetry could not be determined.
 Eigenvector   2:      Singlet-?Sym  Eigenvalue= 0.0372059  =0.000
       8 ->  9         0.70689

 Excited state symmetry could not be determined.
 Eigenvector   3:      Triplet-?Sym  Eigenvalue= 0.1233031  =2.000
       7 ->  9         0.70493

 Excited state symmetry could not be determined.
 Eigenvector   4:      Triplet-?Sym  Eigenvalue= 0.2035093  =2.000
       5 ->  9         0.70373

 Excited state symmetry could not be determined.
 Eigenvector   5:      Singlet-?Sym  Eigenvalue= 0.2902800  =0.000
       6 ->  9         0.70685

 Excited state symmetry could not be determined.
 Eigenvector   6:      Singlet-?Sym  Eigenvalue= 0.3353297  =0.000
       5 ->  9         0.70254
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    6 LETran=     118.
 The wavefunction has an RHF -> UHF instability.

As expected we see a couple of instabilities. At this point we can run another calculation with stable=opt to find such a wave function. For reasons I don't quite understand why it finds the the broken symmetry solution instead of the triplet solution.

For exercise purposes we'll manually generate the broken symmetry solution; i.e. bp86svp.uhf.sing.gjf:

%chk=bp86svp.uhf.sing.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
scf(xqc)
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
geom=check guess=(read,mix)

title card unused

0 1

The keyword guess(mix) generates a UHF guess by mixing the HOMO with the LUMO. You only need to do this once. Note that the keyword guess(always) only applies to optimisations and will perform a new initial guess at every point along the optimisation. I personally have no idea why this option exists or what it is used for.

Now we can go ahead and test for stability; i.e. bp86svp.uhf.sing.stab:

%chk=bp86svp.uhf.sing.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.uhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
scf(xqc)
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
geom=check guess=read
stable

title card unused

0 1

Note that we do not need to mix the orbitals in the wave function again. The stability section will now read like this:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:  1.012-?Sym  Eigenvalue=-0.0000016  =0.006
      8A ->  9A        0.70640
      8B ->  9B       -0.70640

 Excited state symmetry could not be determined.
 Eigenvector   2:  1.006-?Sym  Eigenvalue= 0.0930085  =0.003
      8A ->  9A        0.70688
      8B ->  9B        0.70688

 Excited state symmetry could not be determined.
 Eigenvector   3:  2.245-?Sym  Eigenvalue= 0.2031898  =1.010
      7A ->  9A       -0.70381
      7B ->  9B        0.70381
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    3 LETran=      64.
 The wavefunction is stable under the perturbations considered.

We already know that the triplet state should be more stable. As I hinted, I like to start with a ROHF calculation. In principle these calculations are always instable with respect to an UHF solution. However, usually this leads to faster results, especially if you have already a converged RHF solution. In this case we could readily work from the broken symmetry solution and directly calculate the triplet from there. Choose whatever is more convenient for you.
For this illustrative purpose I chose to generate a new initial guess; i.e. bp86svp.rohf.trip.com:

%chk=bp86svp.rohf.trip.chk
%nproc=2
%mem=10000MB
#p ROBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo

title card unused

0 3
O  0.6 0.0 0.0
O -0.6 0.0 0.0

And get the UHF triplet from there; i.e. bp86svp.uhf.trip.gjf:

%chk=bp86svp.uhf.trip.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rohf.trip.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)
geom=check guess=read

title card unused

0 3

And then obviously check for instabilities; i.e. bp86svp.uhf.trip.stab.gjf:

%chk=bp86svp.uhf.trip.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.uhf.trip.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)
geom=check guess=read
stable

title card unused

0 3

We will find the following stability section:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:  3.003-?Sym  Eigenvalue= 0.2313384  =2.004
      6B ->  8B        0.70567
      7B ->  9B        0.70567

 Excited state symmetry could not be determined.
 Eigenvector   2:  3.003-?Sym  Eigenvalue= 0.2313396  =2.004
      6B ->  9B       -0.70568
      7B ->  8B        0.70568

 LASSW sees vector with all elements less than  0.100000000000E-03
 LASSW sees vector with all elements less than  0.100000000000E-03
 Excited state symmetry could not be determined.
 Eigenvector   3:  3.002-?Sym  Eigenvalue= 0.2704717  =2.003
      6B ->  8B        0.70711
      7B ->  9B       -0.70711
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    3 LETran=      64.
 The wavefunction is stable under the perturbations considered.

A time-dependent calculation will look for excitations based on the reference wave function. Therefore the solution will not be different. If you don't need to perform this calculation because you are interested in for example the UV/Vis spectrum, then there is no reason to do it.

I guess we have run the necessary scenarios now, so here are the results:

Command file              Functional            Energy / Hartree ( cycles )
bp86svp.rhf.sing          E(RB-P86)       =       -150.155033509 (      3 )
bp86svp.rhf.sing.maxc     No energy statement found.
bp86svp.rhf.sing.scf      E(RB-P86)       =       -150.154053425 (     16 )
bp86svp.rhf.sing.stab     E(RB-P86)       =       -150.154053425 (     16 )
bp86svp.rhf.sing.stab.opt E(UB-P86)       =       -150.201668045 (      4 )
bp86svp.rohf.trip         E(ROB-P86)      =       -150.213780181 (      9 )
bp86svp.uhf.quin.stable   E(UB-P86)       =       -149.651924155 (      8 )
bp86svp.uhf.sing          E(UB-P86)       =       -150.201668044 (      8 )
bp86svp.uhf.sing.stab     E(UB-P86)       =       -150.201668044 (      1 )
bp86svp.uhf.trip          E(UB-P86)       =       -150.215687576 (      8 )
bp86svp.uhf.trip.stab     E(UB-P86)       =       -150.215687574 (      1 )
bp86svp.uhf.trip.td       E(UB-P86)       =       -150.215687575 (      9 )

Lastly and maybe most important: with DFA it is never enough to just use one method. For more on this: DFT Functional Selection Criteria. Winter is coming!


ammonium tartrate + lead tartrate=? From Googlebooks First Outlines of a Dictionary of Solubilities of Chemical Substances  By Frank Humphreys Storer:

  I find that lead tartrate is dissolved by tartrate of ammonia.

What is the final product?

The addition of ammonium tartrate turns lead tartrate into a water-soluble tartrato complex:
\[\ce{Pb(tart) + tart^{2-} <=>> [Pb(tart)2]^2-}\]


Why are PTFE and FEP named "ethylene" when they don't have double bonds? PTFE (Polytetrafluoroethylene) and FEP (Fluorinated ethylene propylene) have the following chemical formulas:

It's been years since I have taken organic chemistry, but I do remember alkenes have double bonds and alkanes have single bonds.  Neither of these compounds have double bonds.
Why do these compound names contain "-ene" in them, instead of "Polytetrafluoroethane" or something?

Polymers follow a slightly different nomenclature system than you are used to in organic chemistry. Instead of naming the polymer after the monomers as they are, we name them after the monomers as they originally were. In other words, a polymer made from ethylene monomers is called polyethylene. Since PTFE and FEP are made from fluorinated ethylene and propylene, we just put "poly" in front of the monomer names. This is despite the fact that after the reaction, the bond order is reduced by one.


Why does Nitrogen Oxide NO exist? The nitrogen oxide NO exists and has an uneven number of valence electrons. This is "impossible". Why does it exist ? Why is it not dimerized ?

It is possible due to the fact in NO oxygen has an electron configuration of a noble gas which makes it less energetic. And N shares 2 electrons and it gets close to a noble gas electron configuration.


Does increasing electronegativity imply more covalent character? If there is a larger electronegativity difference between two atoms, shouldn't the compound be more ionic? In an explanation of the diagonal relationship, it states that "on moving across the period, the elements become progressively more covalent" yet their electronegativity increases. What is even meant by an element being covalent? I thought only compounds/molecules could be covalent.

The main force responsible for holding together covalent bonds  and ionic bonds is different. Ionic bonds are held together through electrostatics. A transfer of electrons causes one atom to be positively charged and another one to be negatively charged. In other words, electrons are given away or taken away by atoms. Covalent bonds are held together due to the delocalization of electrons, i.e. electrons are shared by the atoms. For more detail look at What makes the difference between ionic and covalent bonds? (physics.se)

There is no meaning to the statement that an element is covalent. The reason is that the nature of a bond (metallic, covalent, ionic) is determined by the elements participating in the bond. Sodium may form a metallic bond with itself, but it will form an ionic bond with halogens. Halogens will form covalent bonds with themselves but will form ionic bonds with alkali metals. Furthermore, the conditions can determine the nature of a bond. For example, hydrogen is metallic under high temperature and pressure.
Just a side note, compounds can be ionic. I recommend that you read this thread explaining the difference between compound, element, mixture, etc: What is the definition of of 'compound', 'mixture', 'element' and 'molecule'?

Tldr: Wikipedia is not being exact. Saying that an element is covalent is inaccurate because the terms covalent and ionic refer to bonds not elements.


How do I write the deterministic rate equation for bimolecular reaction with similar particles? I was wondering how to write down the deterministic rate equation for a bimolecular reaction with similar particles.

e.g.

$$
\ce{A ->[k_+] 2B}
$$

and

$$
\ce{2B ->[k_-] A}
$$

Now the rate equations for the above reactions are:

$$
\mathrm{\frac{\delta A}{\delta t}= - k_{+} A + k_{-} B^2}
$$

$$
\mathrm{\frac{\delta B}{\delta t}= 2 k_{+} A - 2 k_{-} B^2}
$$

Now I have included a factor of 2 in the second equation to conserve $\mathrm{A + \frac{B}{2}}$.

But I don't understand the reasoning for this factor of 2 in the second equation.

There is really only one independent equation here because

$\frac{dB}{dt} = -2\frac{dA}{dt}$

The "second equation" in the OP is just -2 times the first equation.

Every molecule of A destroyed creates 2 molecules of B.

Every 2 molecules of B destroyed creates 1 molecule of A.


In a galvanic cell, why does an electrode with a higher potential "pull" the electrons from the electrode with a lower potential? I'm doing Grade 12 Chemistry and I'm unsure of this, taking the hydrogen electorde as having a potential of 0 volts.

I'll use the copper-zinc cell as an example. Just let me know if I've got the full logic right:

Zinc has higher electronegativity, so due to the equilibrium $\ce{Zn <=> Zn^{2+} + 2e-}$, we have more charge building up in the Zinc electrode than in the Copper electrode. (Given both half-cells are of same concentration and in same temp)
We have a potential difference between the two electrodes. This means some charge from the Zinc electrode will move to the Copper electrode in such a way that both electrodes will momentarily have equal charges.
Since the charge in the Copper electrode has gone up, we have added electrons to the equilibrium reaction $\ce{Cu^{2+} + 2e- <=> Cu}$. We therefore create more copper.

Is this why an electrode with a higher potential will "pull" the electrons from the electrode with a lower potential?

There are not many problems where electronegativity helps you find the answer.  It is popular for some reason amongst newer chem students but it shouldn't be.

Here's what happens in the Daniell cell:

Metallic bonds in zinc are broken and zinc atoms each lose 2 electrons.  Both of these processes are endothermic.

Zinc ions are hydrated. This is exothermic.

Copper ions are dehydrated. This is endothermic.

Copper ions gain 2 electrons each and metallic bonds are formed. Both of these processes are exothermic.

Each of the above processes also involves an entropy change.

Only when all of these factors are considered can the direction of the electrochemical reaction be determined. Or, you can look at a table of standard reduction potentials.

By the way, silver's electronegativity is 1.93 but the reduction potential for $\ce{Ag+}$ is $\pu{0.8 V}$ while $\ce{Cu^2+}$ is $\pu{0.34 V}$.

In terms of charge, electrons flow from anode to cathode. There is no charge built up because of the salt bridge.


Origin of broad CO stretch in infrared spectrum of tricobalt cluster I'm quite confident that the peaks around $\pu{2000 cm^{-1}}$ are $\ce{CO}$ stretches. But I cannot find any reference on that massive broad stretch following the $\ce{CO}$ stretches. Does it have anything to do with high symmetry?

I found an reliable article (http://www.ilpi.com/organomet/carbonyl.html) describing wide spectrum of metal carbonyls. Since Chemistry SE won't open that website, I have attached scheme with a range of carbonyl stretching frequencies in the article for your benefits:

The IR stretching of carbonyl can be varied based on the structure. According to the website:

  In the IR, typical stretching frequencies are: (i) Uncoordinated or "free" $\ce{CO}$: $\pu{2143 cm^{-1}}$; (ii) Terminal $\ce{M-CO}$: $\pu{2125}$ to $\pu{1850cm^{-1}}$; (iii) Doubly bridging ($\mu_2$): $\pu{1850}$ to $\pu{1750 cm^{-1}}$; (iv) Triply bridging ($\mu_3$): $\pu{1675}$ to $\pu{1600 cm^{-1}}$; and (v) Semibridging: somewhere between terminal and $\mu_2$.

In literature, most tricobalt clusters were discussed included with other metal cations. For example: (a) Rational synthesis of tricobalt-molybdenum and -tungsten butterfly clusters with alkyne ligands in Journal of Organometallic Chemistry, 2002, 659(1-2), 142-150 (https://doi.org/10.1016/S0022-328X(02)01720-5) and (b) Clusters as Ligands. 5. Tricobalt Cluster Alkoxycarboxylates of Titanium and Zirconium Exhibiting Novel Structures and Properties in Organometallics, 1997, 16(24), 5289–5301 (https://pubs.acs.org/doi/abs/10.1021/om970837b).

You can also read: A. S. Goldman, K. Krogh-Jespersen, Why Do Cationic Carbon Monoxide Complexes Have High $\ce{C-O}$ Stretching Force Constants and Short $\ce{C-O}$ Bonds? Electrostatic Effects, Not $\sigma$-Bonding, J. Am. Chem. Soc., 1996, 118, 12159-12166 (https://cdn-pubs.acs.org/doi/10.1021/ja960876z).

Since you have completely described preparation procedure of your tricobalt cluster, I assume the broasness of carbonyl absorptions are due to different structural aspects of carbonyl on $\ce{Co}$ and other metals in the cluster. It is also possibility that you have a mixture of clusters. Note that range of $\nu_\mathrm{CO}$ could be $\pu{2100}$ to $\pu{1300 cm^{-1}}$, based on stereo arrangement (see above references).


Rate law and rate constant How do we obtain the expressions for integrated rate laws ?!? What are the limits we integrate it under and with respect to what do we integrate it?

Remember first that the integrated rate laws that we are familiar with for zeroth, first, and second order reactions are based on systems involving a single reactant in their rate determining step (or those that have been manipulated to give very large excesses of one reactant such that its concentration remains essentially fixed during the reaction process.)

With that in mind, there are only two variables involved, the concentration of our reactant of interest, $[A]$, and time, $t$.  We can't directly integrate the differential rate law; it's a differential equation and must be solved as such.

For an example, consider the first order differential rate law:
$rate = k[A]$

The meaning of  $rate$ here is the rate of change of the concentration of reactant A with respect to time.  Since we want the rate to be positive, we use the opposite of that quantity.

$-\frac{d[A]}{dt} = k[A]$

This is a differential equation, specifically one that is separable (and thus fairly easy to solve.)  Multiply both sides by $-dt$ and we get:

$\frac{d[A]}{[A]} = -k\ dt$

$\int{\frac{1}{[A]}d[A]} = \int{-k\ dt}$

Integrating both sides and applying initial condition to find the constant gives the integrated first order rate law:

$ln[A]=-kt+ln[A_0]$

The other integrated rate laws are found in similar ways.


Does this reaction require catalyst/solvent? Planning on synthesising 2-formylphenyl benzoate by reacting benzoyl chloride with salicylaldehyde. Since both Benzoyl chloride and salicylaldehyde is liquid, will it require a catalyst/ solvent? Or do I just add them both in a conical flask and let a magnetic stirrer do the job?

Just three points why omitting a solvent might not be a good idea without experimental evidence that support the contrary:

There is a reason many organic syntheses are run with reagents in a concentration in range of 0.1 to 1 mol/L; to moderate both rate (kinetics) and heat (thermodynamics) of a reaction.  The omission of a solvent for an already exothermic reaction may lead to a "runaway", where the heat suddenly generated transforms your products into a tar.  The issue regarding the heat balance often becomes important when scaling the reaction to larger batches in process chemistry.
Secondly, your product target equally is an aldehyde.  At least potentially there already is a problem (chemoselectivity) discerning one of the starting materials (salicylic aldehyde) from your product (2-formylphenyl benzoate) in respect to benzoyl chloride.  In case of a highly concentrated (neat) reaction lacking any solvent, if the reaction of the acyl chloride with the aldehyde is exothermic, you may reach a temperature of your reaction mixture so elevated that both reaction pathways became accessible instead of only one (and other not envisioned here, too).
According to the information about who sells this pricy commercial product at PubChem and how -- per gramm, instead of per millilitre -- it appears safe to assume that the product of your reaction is a solid.  Having a solvent dissolving both your starting materials as well as your product equally may offer you a reaction more complete, because the progress of the reaction is not hampered by an interface of a liquid phase (your starting materials) and a solid (your product).  It equally may lead you to a product more clean, since these grains won't contain (this much) of starting materials (occlusion).


Shouldn't this compound be called 2-methyl-5-(1,2-dimethylpropyl)nonane? The book says that the name should have dimethylpropyl methyl nonane in that order. Shouldn't the methyl group come first since methyl comes before methylpropyl (in the alphabetic order)? Our prof told us to ignore prefixes such as mono, di, tri when putting groups in alphabetic order.

‘5-(1,2-Dimethylpropyl)-2-methylnonane’ is not the preferred IUPAC name. The preferred name according to current IUPAC recommendations is 2-methyl-5-(3-methylbutan-2-yl)nonane.
If this is a homework exercise, however, it might be based on obsolete IUPAC recommendations from 1979 or 1993.
You are partly correct; simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. The multiplicative prefixes are inserted later and do not alter the alphabetical order. For example, ‘1,2-dibromo-’ is considered to begin with ‘b’.
However, the name of a compound substituent is considered to begin with the first letter of its complete name. For example, ‘1,2-dibromobutyl-’ is considered to begin with ‘d’.
Likewise, ‘1,2-dimethylpropyl-’ is considered to begin with ‘d’. Therefore, it is arranged before ‘methyl’.


Solubility of compound in acid I have a substance that is insoluble in water but soluble in hydrochloric. If dissolve this substance in the acid, would neutralizing the solution with sodium bicarbonate then cause the substance to separate from solution?

Yes, you've described a property of amines. When protonated by an acid (e.g. $\ce{HCl}$), an amine is converted into into a water soluble salt. This is actually a thing to do in pharmaceuticals to increase the amount of a drug that can be dissolved in water and be taken up by the body. The next time you have the chance take a close look at the chemical names on a drug label and theres a good chance you might see "hydrochloride."

Just as something can be protonated, it can be deprotonated. For the hydrochloride salt of an amine, $\ce{NaHCO3}$ should be basic enough to deprotonate it, and cause it to precipitate out of solution, as it would no longer be a water-soluble salt.


Cis Orientation of dienophiles orientation of groups maintained in Diels Alder When I was reading about Diels-Alder reactions, I was confused about the cis and trans rules for dienophiles, as shown below.

The text told me that "a cis dienophile will generate a ring with cis substitution, and trans dienophiles will generate a ring with trans substitution."

I'm a bit confused.

Why is this the case?

To answer you question, imagine (or: build a molecular model) that the newly formed six-membered ring may be simplified as a flat disk.  Then

either you start with a (trans)-configurated diene, as in the first example drawn by you.  Since the Diels-Alder simultaneously forms the new $\sigma$-bonds on the expense of $\pi$-bonds and moves the location of the remaining double bond, the substitutents R are either both below, or both above this reference plane.  Relatively to each other, these substitutents R in the product conserved their (cis)-relationship from the dienophile.
or, if your dienophile is (trans)-configurated, yielding a product were one of the substitutents is above, and the other is below this reference plane; corresponding to the second example drawn by you where are R conserved their (trans)-relationship to each other.

On paper, the observed conservation of sterochemical information is both the synchronicity of the events (pushing the pairs of electrons, absence of [charged] intermediates as e.g. in nucleophilic substitutions), as well as the principle of conserving the symmetry of molecular orbitals envolved here.


Energy of electrons in the same subshell So I have this quote from wikia (it's similar to my lecture pdf file) but I don't fully understand the statement

  Although it is commonly stated that all the electrons in a shell have the same energy, this is an approximation. However, the electrons in one subshell do have exactly the same level of energy,[6] with later subshells having more energy per electron than earlier ones. This effect is great enough that the energy ranges associated with shells can overlap

In this statement, does it mean for example:

1) "Each" electron in the subshell 1s and "each" electron in subshell 2s have the same amount of kinetic energy.

Or

2) It means :"All electrons 1s have the amount of Kinetic Energy & All electrons 2s have the same amount of Kinetic Energy. However each electron in 1s doesn't have the same amount of Kinetic Energy compared to one electron in 2s" ?

It's important to remember (or to learn, if it hasn't been explained before) that the only atomic systems for which we can resolve the Schrödiger equation exactly are what we call hydrogenoid atoms - atoms with only one electron, such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$...

As soon as you add a second electron, atomic orbitals change and they aren't exactly the same as for the hydrogen atom (this is a form of the three-body problem that also comes up in other fields of physics; if you remember your Newtonian mechanics, that's why you cannot solve analytically a three-body gravitational problem). The source of this change is electron-electron interaction.

However, electron-electron interaction, in general, introduces only minor distortions in atomic orbitals - for the most part, the orbitals of poly-electronic atoms are very very similar to hydrogenoid atoms in terms of shape, size, etc, which is why we use the same labels for them as for hydrogenoid atoms ($\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p}$, etc).

There is one property of hydrogenoid atoms that is very importantly changed by electron-electron interactions, though: in hydrogenoid atoms, the energy of an orbital is only a function of its main quantum number, $\mathrm{n}$, which means that all the orbitals of the same shell will have the same energy. In a single-electron atom, an electron in a $\mathrm{2s}$ orbital will have exactly the same energy as an electron in a $\mathrm{2p}$ orbital, because they only interact with the nucleus through a Coulombic interaction - which has spherical symmetry (it's the same in all the directions of space; it only changes with distance to the nucleus).

If there are other electrons in the same atom, occupying different orbitals, this breaks down the energy equivalence of same-shell orbitals - because, for instance, an electron in a $\mathrm{1s}$ orbital interacts differently with an electron in a $\mathrm{2s}$ orbital or in a $\mathrm{2p}$ orbital - due to symmetry, overlap, and a number of factors (these interactions are very complex, can't be solved analytically, and to some extent aren't even completely understood). The energy corrections are generally small, but the orbitals belonging to the same shell are no longer equal in energy - that's why we can order subshells, and we know that $\mathrm{2s}$ orbitals fill up before $\mathrm{2p}$ orbitals.


What exactly is a potential energy surface? I first learned about potential energy surface (PES) when I was studying Born-Oppenheimer approximation. In the approximation, the PES is essentially electronic PES which is defined as follows:

Given the Hamiltonian of a molecule
$$\hat H = \hat T_{\text{nucleus}} + \hat T_{\text{electrons}}+\hat U_{\text{electron-nucleus}}+\hat U_{\text{electron-electron}}+\hat U_{\text{nucleus-nucleus}}$$
Under Born-Oppenheimer approximation, $\hat T_{\text{nucleus}} = 0$. So the equation about to be solved is:
$$(\hat T_{\text{electrons}}+\hat U_{\text{electron-nucleus}}+\hat U_{\text{electron-electron}}+\hat U_{\text{nucleus-nucleus}}) \psi_{\text{electron}} = E(\{ R_{\text{nucleus}}\}) \psi_{\text{electron}}$$
Clearly, $E(\{ R_{\text{nucleus}}\})$ gives a PES, and this PES is the relation:
$$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{ground state electronic energy}}$$

However, when I learned about molecular dynamics (also in wikipedia), a PES is defined to be the relation
$$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{energy of the system}}$$
How do we find this map in practice? The only way I can think of is:

We solve the electronic eigenequation for several $\{ R_{\text{nucleus}} \}$.
For each calculated $E(R_{\text{nucleus}})$ of electrons, we solve the equation
$$(\hat T_{\text{nucleus}} + E(R_{\text{nucleus}})) \phi_{\text{nucleus}} = E_{\text{nucleus}}\phi_{\text{nucleus}}$$
to find the energy of the nucleus. (I have never learned how to solve an eigenequation of a nucleus in quantum chemistry)

Is this the way that we find the 'generalized' PES? When we say PES, which is the PES that we are referring to?

The PES $V({R})$ (where $R$ are the nuclear coordinates, electronic coordinates are usually denoted by $r$) is the electronic energy of any electronic state. So there is one PES for the ground state and many more all the excited states.

  Clearly, $E(\{ R_{\text{nucleus}}\})$ gives a PES, and this PES is the relation:
  $$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{ground state electronic energy}}$$
  However, when I learned about molecular dynamics (also in [wikipedia][1]), a PES is defined to be the relation
  $$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{energy of the system}}$$

"Energy of the system" is here referring to the "electronic system", so those two relations are the same.

  How do we find this map in practice?

Your first step is correct, we need to solve for the eigenvalues of the electronic Hamiltonian, for many different parameters $R$. Each electronic eigenvalue is one point on the PES. So in the end we get a discretized representation of the (actually continuous) PES $V(R)$.
The second step is to solve the nuclear Hamiltonian
\begin{equation}
  \hat H_{\mathrm{nuc}} = \hat T_{\mathrm{nuc}} + V(R)
\end{equation}
There is only one such nuclear Hamiltonian for each PES.

So we need to do many times step 1, but only once step 2 (for each electronic state of interest).

How to solve the nuclear Hamiltonian?

This depends on what $V(R)$ looks like. In the simple case of a diatomic molecule, where $R$ becomes just the 1-dimensional internuclear distance, we may approximate it with a Morse potential or harmonic potential. We can fit the parameters of such analytic expressions to our discretized points from step 1 and then use the existing general solutions for those quantum systems. This will give us the vibrational eigenvalues and eigenfunctions.

Another option is to solve the nuclear Hamiltonian on a numeric grid over $R$. The grid is either defined by the points chosen in step 1, or we can interpolate to a new grid. How to solve this is a topic on its own, so will just give one literature reference here: The book "Introduction to Quantum Mechanics - A time-dependent Perspective" by David Tannor covers many things related to your questions (although it can be challenging to read it).


Radical monobromination of cycloalkane I'm trying to figure out the isomeric products of the radical monobromination of bicyclo[2.2.1] heptane. I'm not sure at which site the bromination would happen and what makes one site better than the other (how can one determine this without given the reaction enthalpy ?). I was also given the hint that there are stereoisomers but I can't think of one.

The radical monobromination of bicyclo[2.2.1] heptane could give $\ce{1}$ ,$\ce{2}$ and $\ce{3}$.

However , $\ce{2}$ and $\ce{3}$ , are unstable .On observing $\ce{2}$ , planar radical at bridgehead is unstable due to angular strain introduced by planarity of radical at bridgehead .

Similarly in $\ce{3}$ , due to angular strain introduced in to the neighboring bonds,this  radical also is unstable.

The situation leads to product from ${1}$ , that gives  stereoisomers as shown below.


What does Brightness/ intensity of spectral lines depend on? My book states that it depends upon "The number of photons of same frequency or wavelength ABSORBED or EMITTED"
From what I understood, if many hydrogen atoms within the discharge tube EMIT the same wavelength, then, on passing the radiation through a prism, these waves will overlap on the photographic plate (Because all of them will get deflected to the same extent by the prism). Hence increasing the intensity. Is my understanding correct?
Secondly, how can the number of photons of the same wavelength ABSORBED create any difference in the intensity of spectral lines? As far as I know, absorption only results in black lines on the bright background corresponding to the absorbed wavelengths.

The answer "The number of photons of same frequency or wavelength ABSORBED or EMITTED" from your book is a general way of saying when spectral lines are brighter (or darker). Your understanding of this general mechanism is (I think) mostly correct.

From what I understand, the differences in intensity of different spectral lines, however, is related to two (possibly more) things:

1) If a photon hits an atom (or molecule) and that photon has the exact energy needed to excite that atom does not necessarily means that the photon WILL be absorbed. There is an absorption probability associated with each energy transition. For example, one transition might have a 80% probability of absorption, while another has 0% (because the transition is forbidden).

2) Another factor that influences intensity is related to what state the system is in. For example, if two transitions, lets call them 0->2 and 1->2 both have a 100% probability of adsorption, but all the atoms/molecules are in state 0, then the 1->2 transition will not happen. That is, it will be dark/low intensity.

I hope this is helpful.


NMR sweeping magnetic field I am currently doing a experiment in the lab which is using a continuous wave NMR, by sweeping the magnetic field, and I am having trouble understanding how the theory works for this type of experiment.

So I currently understand if a proton is not placed in a magnetic field it will spin on it axis of rotation due to the magnetic moment.

My first question is on what causes this magnetic moment, from what I have found so far that the proton can be modelled a sphere with infinitely  many of  tiny circular current loops which run along the axis, but what I can understand is that for a dipole moment to be formed a current going in a circular loop generate the dipole moment.

  So dose the dipole cause the proton to spin which generates the magnetic moment and if so what causes the tiny dipoles along axis? It seem to me there a rotation being caused by a rotation.

So moving on when the proton is placed in the a uniform magnetic field it will either align it self with parallel or anti parallel. It would not align it self fully due to the rotation of the proton so the proton will actually precess around the axis in which is in the direction to the static magnetic field. Now when the proton is in a state parallel to the static magnetic field it is in a low energy state and when anti parallel it is in a high energy state so the energy difference is given by:

$$\Delta E=hf=g\mu _nB_0$$

Now applying this to a sample of high hydrogen 1 like glycerine that has been place in CW NMR. When the sampled is placed in the static filed according to the Boltzmann distribution there will be more protons parallel to the field that non-parallel so now in cw as I understand a constant RF signal is being applied and a varying magnetic field perpendicular to the static field.

So my second question is

  Dose the varying magnetic field vary the energy levels until resonate occurs?

I believe it dose as this would explain the nmr of glycerine that was take in the lab, as show below.

Now the real trouble I am having with these graphs is that one it looks to me that there is a dampening effect occurring, which I believe to be the relaxation time of the sample, but what I cant visualise is how the resonance is occurring within the sample it self in terms of the physics with respect to all the protons precessing.

  Are the individual proton within the sample at different depths become out of phase with each other and the varying magnetic field is decreasing over time, and why dose each resonance peak have a negative voltage associated with it? How dose this negative voltage tell me about what is occurring with in the sample?

If anyone could with any of the theory behind CW NMR and how the signals being receive can inform me on what occurring with in the sample it would, it would be appreciated. Thanks in advance.

What causes this magnetic moment?
  So does the dipole cause the proton to spin which generates the magnetic moment and if so what causes the tiny dipoles along axis? It seem to me there a rotation being caused by a rotation.

Spin is a property predicted when relativity theory is applied to electromagnetic theory, leading to the Dirac equation. It is an intrinsic property of subatomic particles, like charge. It explains magnetic phenomena such as NMR. It is a quantum property of NMR active nuclei (when measured it takes on discrete values). For spin-${\frac{1}{2}}$  nuclei, it is related by a multiplicative constant to the magnetic dipole moment of the nucleus. Thanks to spin, nuclei have an intrinsic and fixed magnetic moment in an external magnetic field. One component of this moment adopts a discrete orientation with respect to the applied field. Depending on the orientation, two states of different energy, usually called $\alpha$ and $\beta$, are possible for a spin-${\frac{1}{2}}$ nucleus.
We can define the resonance frequency in the presence of the external field using the equation you provide. For a spin-${\frac{1}{2}}$ nucleus it is proportional to the difference in energy between the $\alpha$ and $\beta$ states. It is the frequency at which the nuclear moment nutates about the applied field.

  Does the varying magnetic field vary the energy levels until resonance occurs?

Yes. In the CW experiment, you measure the electrical signal induced by the sample in a pick-up coil while simultaneously irradiating the sample with a second orthogonal RF coil, and sweeping an applied magnetic field. The sweep alters the resonance frequency of the spins. When this frequency equals the irradiation frequency spins are excited and transition between the two states - this is resonance, the points at which the oscillating red curve (the field sweep) crosses the origin. The spins now oscillate together coherently back and forth between $\alpha$ and $\beta$ states, yielding a fluctuating magnetic field which induces an oscillating current in the pickup coil. As the field sweeps past the resonance point, the spins begin to lose energy - to relax. Relaxation causes the signal to eventually decay to the baseline.

  Are the individual proton within the sample at different depths become out of phase with each other and the varying magnetic field is decreasing over time?

The NMR signal is an average, reflecting the behavior of a large ensemble of spins, acting coherently (nearly identically) during an experiment. The total signal is a sum over all excited spins in the sample. The applied field is as uniform across the sample as possible, so that all spins behave similarly. But in a way the spins act independently. Some oscillate faster or slower than the mean. The differences add up over time, leading to dephasing of the signal.

There is usually no spatial information in such an experiment unless special tricks are used.

  and why dose each resonance peak have a negative voltage associated with it? How dose this negative voltage tell me about what is occurring with in the sample?

This question is not clear. The spins exist in states whose coherent oscillations induce a sinusoidal magnetic field in the RF pickup coil. The electric signal induced in the pickup coil is an AC voltage, alternating between negative and positive values until it decays completely.


How do I name this alkane with a branch that has bromine? I identified the orange portion as the longest carbon chain which is the first step.

Next, I can't call it 3-ethylhexane as the bromine is a branch from a methyl branch.

This is the problem I am facing. Since the methyl branch is on the 3rd carbon, and bromine branch is in the "branched" 2nd.

So, is it called as 3-(2-bromomethylhexane)? Or have I got it wrong?

3-(bromomethyl)hexane

You can't have a 2-position on a methyl group, there is only a 1-position and we omit redundant 1-enumerations of position.


Problems with precipitation of transition metal solution I am trying to make a solution of "trace elements"--primarily transition metals-- for use in biological media; bacteria require elements like Cobalt, Molybdenum, iron, etc. in trace amounts. I am not a chemist, but my understanding is that these elements tend to form hydroxides in physiological pH ranges (slightly above 7). Therefore the authors of the protocol I am using employ EDTA as a complexing agent in 1:1 stoichiometric proportion with the metals in solution. The problem is that EDTA solubility (and complexing capability) and transition metal stability occur at opposite sides of the pH scale, and finding a balance to make both types of reagents happy is apparently really tricky; I've tried this 4 different times with no success. I am able to get all materials into solution, but any disturbance of my flask, (e.g. further stirring or shaking) inevitably causes precipitation.

Here is my recipe and protocol. I'll describe my observations after that. My question is:

  Any thought what the precipitate is? How can I avoid it?

I have scant chemistry experience, so any insight you can provide would be useful.

Recipe

190 mM $\ce{EDTA}$ $\ce{Na4.2H2O}$ (equimolar to subsequent di- and trivalent cations)
29 mM $\ce{FeCl3.6H2O}$
80 mM $\ce{CaCO3}$
6 mM $\ce{MnCl2.4H2O}$
0.5 mM $\ce{CuSO4.5H2O}$
0.5 mM $\ce{CoCl2.6H2O}$
5 mM $\ce{ZnO}$
2 mM $\ce{H3BO3}$
66 mM $\ce{MgCl2.6H2O}$
5 mM $\ce{NaMoO4.2H2O}$

Protocol (making 1 L)

Dissolve EDTA in $\ce{750 mL}$ of water. Bring pH down to about 5.8 with $\ce{HCl}$.
Add $\ce{FeCl3}$ and watch pH; don't let it drop below 4.8.
Add $\ce{CaCO3}$. Allow to dissolve (i.e. $\ce{CO2}$ to evaporate) by maintaining pH at 4.8 with $\ce{HCl}$ and occasionally shaking vigorously (not sure why the recipe doesn't just call for something more soluble like $\ce{CaCl2}$.)
Add remaining elements in order given in the recipe, watching the pH. It should drop to about 4.2 without any addition of acid or base.
At this point solution appears clear. Bring volume up to $\ce{1L}$.
Autoclave solution (heat under pressure at $123^{\circ}\ce{C}$ for 15 minutes) ---this is required for biological sterilization, and seems to improve solubility for a time.

Observations

Solution remains transparent throughout preparation, stirring constantly throughout at room temperature. Color is orangish-green, due to iron and copper. Once all ingredients have been added, additional stirring before autoclaving leads to gradual (over 15 minutes) accumulation of a white powdery precipitate. I thought this was EDTA coming out, so I added $\ce{NaOH}$, but this only caused the iron to precipitate, forming a brown precipitate. Letting the solution sit also seems to lead to white precipitation, although less rapidly. When I autoclaved the solution, it remained transparent (color had turned to deep rust-orange) even after sitting overnight, but as soon as I disturbed it by transferring from an Ehrlenmeyer flask to a glass bottle it precipitated again. The pH of the precipitated solution seems stable at 4.2, what it was after all ingredients had been added. I have also tried this preparation adding EDTA after the boric acid and keeping pH below 6 with $\ce{HCl}$, but this caused even more rapid precipitation. I have also tried EDTA (fully protonated) after the boric acid and adding NaOH to deprotonate, but the results were even worse.

Here's an idea.  Add the cations as their acetates or nitrates.  Control pH with acetate buffer. Fe(III) is a bad player.  "Ferric acetate," $\ce{[Fe3O(OAc)6(H2O)3]OAc}$, is water-soluble. You might sparge the solution with argon before sterilizing (sealed vessel) to prevent oxidation, and/or add ascorbate (or erythorbate, often biologically inactive) as anti-oxidant - if that is the problem.

Lactate chelates, and may give greater solubilities.  Fe(II) lactate is E585 via the European Food Safety Authority.

(Note that my singular contribution to biochemistry was telling a lab its phosphatase assay bombed because...they used phosphate buffer.  I can't say they were grateful for acquiring the knowledge.)


Does the addition of acid change alkalinity? I'm attempting to speed up the dissolution of calcium carbonate by the addition of hydrochloric acid. However, I'm using the calcium carbonate as a source of carbonate for a water sample. But would the addition of acid counteractively lower the alkalinity in the process, therefore lowering the carbonate concentration?

Calcium carbonate has a solubility of $\pu{13 mg/L}$ at room temperature. Adding acid to this solution will dissolve the calcium carbonate, by destroying the carbonate via the equilibria
$$\ce{CO3^2- <=>[+ H+][- H+] HCO3- <=>[+ H+][- H+] H2CO3 -> H2O + CO2,}$$
with the carbon dioxide having a tendency to escape the solution.


Wave-particle duality in electron Electrons show wave-particle duality

Particle (atom) containing a wave (electron) on Quora

Atoms contain electrons. We consider atoms as particles but electrons which are inside the atom as waves (50% chances). When we talk about orbits and orbitals, outer electrons, inner electrons, I sub-consciously assume that we are talking
about the particle nature of electron. But this model is true only half of the time. The possibility of the electron being a wave at any time is also 50% . Particle nature and Wave nature are two different things because scientists came up with wave-particle duality. So we cannot talk about outer electrons, inner electrons, degenerate orbitals in the case of wave nature of electrons.Even if we can say inner part of the wave(close to nucleus) and outer part of the wave(away from the nucleus) ,what we are going to do in that case to solve energy, wavelength, position of the electron (in wave form) in an atom (a particle)?

Interesting question and this is a complicated topic. Actually it is not true that the wave and particle models are each valid 50 percent of the time. I think it is better to say that the wave model is always valid, but under some conditions the wave gets compressed down to a small area, so it looks like a particle. I hope that makes sense, please comment if it doesn't?

I am not a quantum chemist, but I think the normal conditions for electrons in atoms and molecules are not ones that make them act like particle. So therefore we should always think of them as waves. Luckily we can still talk about the "inner" and "outer" electrons while they are waves. It is not "inner" and "outer" exact position, it is an area occupied by the wave, the "inner" one is nearer to the nucleus, the "outer" one if further away. For atoms these are called atomic orbitals - if you look them up on Wikipedia it might be very interesting.

If you want to solve the energy, wavelength and position, it's quite complicated. I would recommend studying quantum mechanics. It will take a long time to get there, although it is very interesting. There might be a quicker but less detailed course just for chemists to do these calculation too. I hope that's helpful.


What's the reaction between H2SO4 and NaH2PO4? I'm guessing it's going to be

$$\ce{H2SO4 + 2 NaH2PO4 -> 2 H3PO4 + Na2SO4}$$

Could anyone please let me know what the actual reaction is?

Thank you

Actually you have two reactions, since sulfuric acid can give off two protons:

$\ce{H2SO4 + NaH2PO4 -> NaHSO4 + H3PO4}$

$\ce{NaHSO4 + NaH2PO4 -> Na2SO4 + H3PO4}$

The $\ce{NaHSO4}$ is called sodium bisulfate.  Bisulfate is the thing you form first when sulfuric acid acts as an acid, then you get the sulfate ion if you have a strong enough base to react with the almost but not quite strongly acidic bisulfate ion.  In this case, actually, you can look up some acid dissociation constants and discover that phosphoric acid has almost exactly the same amount of strength as the bisulfate ion acting as the "second stage" of sulfuric acid.  So the second reaction is really reversible:

$\ce{NaHSO4 + NaH2PO4 <=> Na2SO4 + H3PO4}$

For more on what makes acid-base reactions reversible, see here.


pH Changes in primary cells Primary Cells - pH changes during discharge

Alkaline Cell:

Manganese Dioxide Zinc Alkaline Cell

Anode (-)
$\ce {Zn(s) + 2OH^{−}(aq) → ZnO(s) + H2O(l) + 2e^{−}}$

Cathode (+)
$\ce {2MnO2(s) + H2O(l) + 2e^{−} → Mn2O3(s) + 2OH^{−}(aq)}$

Overall
$\ce {Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s)}$

$\ce {OH-}$ ions are consumed at the anode
$\ce {OH-}$ ions are formed at the cathode
$\ce {same[OH^{−}]}$
pH - no change

Is this correct? Do pH changes occur during the discharge process?

No, the reaction does not produce hydroxide ions. But the solution is already basic (probably between pH 12 and 14).


If you are given ΔH for this reaction which occurs at constant pressure, what additional information you need to determine ΔE? Reaction: $\ce{2Al + 3Cl2 → 2AlCl3}$

Answer: We must have either temperature, or the values of P and ΔV in order to calculate ΔE from ΔΗ.

Since ΔH is given, we can use the formula ΔE=ΔH-PΔV to calculate change in internal energy, and only pressure and ΔV are required to calculate ΔΕ.
So my question is, why do we need temperature in this question?

Because:

1) your question assumes that the volumes of the solids ($\ce{Al}$ and $\ce{AlCl3}$) are negligible compared to the volume of the $\ce{Cl2}$ gas. This means that $\Delta V \approx V(\ce{Cl2})$.

2) your question assumes that chlorine gas behaves ideally. This means that $V(\ce{Cl2}) = \frac{nRT}{p}$.

In the end, it means you only need two of the three variables ($p$, $T$, $V$) in order to obtain all the information you want.

I would however be very careful about using molar quantities, in particular $V_m$ (since $p$ and $T$ are intensive properties), if the question is asking you to find $\Delta H$ and $\Delta E$ in terms of $\text{kJ mol}^{-1}$.


How should I go about picking a functional for DFT calculations? I've been looking to use time-dependent density functional theory, but reading the literature I keep getting overwhelmed by how many different functionals there are.  How should I go about selecting which functional to use?

For Structures and Energies: Answer Hazy, Try Again Later
People can talk on this subject for literally days.
It mostly comes down to computational expense vs accuracy - if you have a larger system, you can't afford the fancier functionals. The absolute highest accuracy state of the art functional-wise would probably be those using the Random-Phase Approximation (e.g.), and also the use of more complex range separation techniques (such as the Coulomb-Attenuated Method mentioned below).
Having said that, a lot of people just use B3LYP for everything. B3LYP is parametrised for low-Z main-group chemistry, and so does reasonably well there. It's a hybrid functional (i.e. it includes the exact Hartree-Fock form exchange operator), so it's not the cheapest. I tend to use the series of PBE, PBE0, TPSS - their only parameters are physical constants, so they should never give surprisingly bad results.
A couple of basic guidelines, though (which should be taken as "as far as I know"s):

If you have a delocalised system, you want exact exchange (so, a hybrid functional like PBE0, B3LYP, M06-2X, etc). Without this, most functionals will over-delocalise the orbitals.
If you have a system in which dispersion forces are important, worry. Then use something like $\omega$B97X-D which includes empirical dispersion corrections. Some packages can also apply Grimme's empirical dispersion corrections to any functional.
If you have a periodic system, you probably want to use screened exchange rather than full exact exchange - something like HSE (which tails off the exact exchange past a certain range) rather than PBE0 should be faster while not significantly less accurate.

For TDDFT: Probably CAM-B3LYP
Fortunately, this is a bit easier.
Firstly, you want something that incorporates exact exchange (a hybrid functional), which quickly rules out many functionals. Pure GGAs and LDAs give qualitatively wrong answers for states involving charge transfer. Secondly, functionals that include long-range exchange corrections, such as CAM-B3LYP, also give better results.
Also, look out for triplet instabilities, which can be a problem with the introduction of HF exchange.
(David Tozer is my go-to author for TDDFT testing, though I would warn that despite the above answer, I've never used TDDFT in "production".)
(Now, bring on the functional flamewar.)


How to measure quantities at standard state when this state is a hypothetical one? Consider the reaction

$$\ce{A(g) + H2O(l) <=> B(aq) + C(g)}$$

If all gases and solutions were ideal, the standard state would be pure water, the solute B at a concentration of 1 M, the gases A and C each at a pressure of 1 bar, and the system at a pressure of 1 bar.

For non-ideal gases and solutions, it would be the hypothetical state where all the activities are 1 (i.e. equal to the activity of an ideal gas at 1 bar, activity of a solvent with infinitely diluted solutes, and equal to the activity of a solute behaving ideally at a concentration of 1 M).

This hypothetical state can be quite different from a real state. How can the pressure be 1 bar while two gases are at a partial pressure of 1 bar each? What if the solute B is barely soluble, how do you get an activity of 1? How can you have pure water when it is acting as a solvent?

So how to you measure quantities like $\Delta_\mathrm{r} G^\circ$ when the standard state is a hypothetical one?

The standard states are limiting states. In these limits you can write the chemical potential of the component as

$\mu = \mu^\circ+RT\log(x)$

where $x$ is a measure of the concentration of solute (for Henry's law, where the standard state is a hypothetical state related to the behavior as $x \rightarrow  0$) or solvent (for Raoult's law, where the standard state is approached as $x \rightarrow  1$). In other words, in the limiting states the activity goes to 1. The meaning of $\mu^\circ$ depends on the limiting law (hypothetical with properties of an infinitely dilute solution or pure solvent).

A standard state can be hypothetical (as may happen when Henry's law is invoked), but it can also be real (as in Raoult's law, for a pure solvent). It is a state in which the sample behaves ideally as defined by the above equation, in the limit of infinite dilution or of a pure solvent. This limiting behavior provides a convenient and predictable reference, like setting the potential energy to zero when particles are separated by an infinite distance. In the case of Raoult's law, the standard state is the (not particularly mysterious) pure substance. In the case of Henry's law, the reference state is found by extrapolating behaviour at infinite dilution to the standard concentration (1 molal or 1 M, conventionally). This may be a hypothetical state.

Using reference standard states (at 1 bar of course) can simplify combining or comparing data for different substances or conditions. Of course when considering real solutions, activities may diverge from concentrations and activity coefficients deviate from unity.

For a mixture of substances, for each component you can describe individual thermodynamic properties relative to the ideal standard states. By ideal it is also meant that individual solutes do not interact.


Mechanism of burning of hydrogen So I was curious regarding a simple chemical equation :

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

You add heat to the system (i.e fire) and you get water vapor.

I'm curious of understanding the mechanism, which I'm sure at this point has been well studied.

Suppose I keep track of the location of each one of these atoms, and the energy nearby - my anticipation is that the reaction occurs when the atoms get close enough and there is sufficient energy present to conduct the reaction. Is there any way to summarize this critical point of water creation mathematically in this framework (i.e the combination of atoms being close enough and sufficient energy will make this reaction take place) - or is there a superior way?

Does anyone have any insight/publications that can help me understand the steps of water synthesis via reaction mechanism?

Martin - マーチン has given excellent answer here. I'll give here an supporting review article for your benefit, which also gives some mathematical overview:

Louis S. Kassel, “The Mechanism of the Combustion of Hydrogen,” Chem. Rev. 1937, 21(2), 331–345 (DOI: 10.1021/cr60069a010).

Overall, mechanism can be summerized as follows:
\begin{align}
\ce{H2 + \text{heat} &-> 2 H^.}\\
\ce{H^. + O2 &-> HO-O^.}\\
\ce{HO-O^. + H2 &-> HO-OH + H^.}\\
\ce{H^. + HO-OH &-> H2O + HO^.}\\
\ce{HO^. + H2 &-> H2O + H^.}
\end{align}

The propagation would continue with having $\ce{H^.}$ at the end and the presence of excess $\ce{O2}$.


How is it possible to increase mercury's specific weight? Is it possible to change mercury's specific weight to make it heavier? By mixing it with anything? But not by changing its temperature.

Specific weight is a property of the element. It is product of the atomic mass and the volume. You can't easily alter either.

You could mix an element with another element of a different density and the density of the mixture will change. But you won't be changing the specific mass of the elements, just creating a mixture with a different density.

You can't readily change the mass of an element. But an element might be made up of multiple isotopes (mercury has 4 significant ones with masses ranging from 199 to 202 atomic mass units). If you enrich the mercury for one of those isotopes, you can change the mass, very slightly (for other elements this can be important; fissile isotopes of uranium are concentrated by doing it when making nuclear fuels).

In practice there is nothing practical you can do to alter the density (specific mass) of an element though you can create mixtures with different densities (but then you still haven't altered the properties of the element itself).


How can a substance be diluted to nothing? If I start with a small amount of a substance, and I dilute that substance with billions of gallons of water, what is the scientific explanation for there being no (actually zero) remaining molecules of that original substance?

This question is related to the practice of homeopathic dilutions. I was reading some articles which stated that homeopathic dilutions don't work because at most homeopathic levels of dilution, there are very few or no molecules or the original substance.

What I'm confused about is how there can no longer be any molecules of the original substance (or even fewer molecules) of an original substance when diluted.

If there are X number of molecules in a solution, and Y liters of water are poured into the solution for dilution, don't the X number of molecules remain? How can a molecule just "disappear"? How can even one molecule just "disappear"? It makes more sense if these molecules are undergoing a chemical reaction, but if two kind of molecules (the substance, and water) are not undergoing any chemical reaction, then how can one molecule of either kind simply vanish?

You correctly point out that the number of molecules in a solution is finite and constant, however the volumetric concentration (that is, how many molecules per litre) changes upon dilution. If, for instance, you take one liter of a 1 mol/L solution of ethanol in water ($\approx{6.02\times{10}^{23}}$ ethanol molecules per liter) and add 9 liters of distilled water, you now have a 1 mole of ethanol in 10 liters, giving a concentration of 0.1 mol/L. If you were to draw out 1 liter of this and count the number of ethanol molecules in that liter, you would find approximately $\approx{6.02\times{10}^{22}}$ - a tenfold reduction. The rest of the ethanol is retained in the other 9 liters of solution.

Equivalently, if you were to measure out 100 mL of 1 mol/L ethanol solution and then make it up to 1 liter by adding distilled water, you would also get a 0.1 mol/L solution. This is the kind of thing taken to absurd extremes in homoeopathy, where tiny quantities of solution are repeatedly diluted, with some shaking. Obviously, they don't literally add billions of liters of water to their solutions, but rather pipette out a tiny amount of solution and discard the remainder, adding water to make up the desired volume. This process is repeated many times.

The finite number of molecules present in a solution leads to the concept of the dilution limit, which is where you have carried out sufficient dilutions that statistically speaking, your chances of finding even a single molecule of the dissolved ingredient become vanishingly small and that by extrapolation, you would have to drink approximately a zillion liters of the preparation to get a single molecule of it.

Where did all those other molecules go? They didn't simply disappear - they were poured down the drain during the dilution process, which curiously is the same place money goes when spent on homoeopathy.

Bonus: homoeopathic plutonium! (courtesy of Theodore Gray's periodic table)


What is the mechanism for electrospray ionization? I've been asked to

  Descibe the ionization mechanism used in electrospray ionization

I know that this is
a desorption ionization source, which uses a high electrical field as the ionizing agent. The best I've been able to put together is that there is a generation of $\ce{[M + _zH]^2+}$ ions. Surfing the web has just led me to articles saying the mechanism is still being debated. What's the mechanism, if there is a definitive one?

How electrospray works has been hotly debated for decades.  Two competing theories have been put forward.

Initial stages: theories agree on droplet collapse due to Rayleigh instability

The two theories agree on the initial stages of droplet formation.  An aerosol is formed by the extrusion of low ("nano", or 25 to 100 nL / min) to high (~1 mL / min) flows of eluent through a capillary charged to ~3000 V.  The volatility of the solvent used in electrospray techniques ensure that the aerosol droplets evaporate.  Charged spheres of liquid experience a self-repulsion due to electrostatic effects.  This repulsion is counterbalanced by surface tension.  As Lord Rayleigh noted in 1887, as long as $T > \frac{Q^2}{16 \pi a_0^3}$, where $Q$ is the droplet charge, $T$ is the surface tension, and $a_0$ is the droplet radius, then the droplet is stable.  But as heat evaporates the droplet, $a_0$ shrinks, while $Q$ stays constant and $T$ nearly so.  Thus the droplet eventually becomes unstable, exploding into a mist of finer droplets, each of which has a smaller charge. What happens after that has been the subject of debate.

The two theories for what happens after droplet collapse

ION EVAPORATION.  This theory holds that as droplets shrink, eventually the field strength at the droplet surface is high enough to force remaining solvated ions into the gas phase, an effect called "field desorption".
CHARGE RESIDUE.  This model holds that charged analytes remain solvated in droplets through many cycles of droplet evaporation / fission.  Eventually, droplets contain on average only one analyte molecule.  Remaining solvent molecules evaporate, leaving the ion in the gas phase.

Current evidence suggests that most small molecules ionize primarily by ion evaporation, but that for large molecules such as proteins, charge residue mechanisms may be important.  In the ion evaporation mechanism, molecules or ions must translocate to the surface of a droplet before they can become gas phase ions.  This  means that molecules which preferentially segregate toward the surface of liquid droplets will ionize more easily.  Thus, the ion evaporation mechanism, but not the charge residue mechanism, predicts that surfactants will have higher ionization efficiencies than non-surface active agents.

An extension of the ion evaporation model

EQUILIBRIUM PARTITIONING MODEL.  This theory divides droplets into an electrically neutral core, where ion pairing maintains local electroneutrality, and a charged shell, from which small molecule ions can move into the gas phase by ion evaporation.  The amount of "excess" charge in the shell available for ionizing neutral molecules from the core is determined by electrospray flow parameters, in particular by the ratio of electrospray current to flow rate, or $[Q] = \frac{4.2}{F} \sqrt{\frac{K \gamma }{V_f \kappa_e}}$, where $F$ is Faraday's constant, $K$ is the conductivity of the medium, $\gamma$ is surface tension, $\kappa_e$ is the dielectric constant,  and $V_f$ is the volumetric flow rate. Conditions where excess charge is limiting could cause severe ion suppression.  For more discussion, see this paper.

Conclusion:

This answer is very long which should be a clue that this topic is complex.  Usually, saying that things are complex is a sign that people don't understand something very well, and that is the case here.  But in general, perhaps the closest to a consensus in the community seems to be that small molecules ionize primarily by ion evaporation, but that for large molecules such as proteins, charge residue mechanisms may be important.


Number of total stereoisomers of a piperazine derivative How do I find out the total number of stereoisomers of 3,6-dimethylpiperazine-2,5-dione?

As far as I can see, the compound has two chiral carbons. Would the nitrogens make a difference? They seem chiral as well (three different groups and a lone pair).

My book says there are 3 total stereoisomers, but I expected more. It has considered only two chiral carbons, and nothing has been said about the nitrogens. Why?

I believe this is a sophomore organic chemistry text book question, so I won't elaborate possible stereoisomers involving lone pairs on nitrogen atoms. Nonetheless, 3,6-dimethylpiperazine-2,5-dione has 2 chiral carbon centers so that it should theoretically have $2^2 = 4$ stereoisomers. If you look at the molecule, it has center of symmetry (i):

Therefore, two of the isomers are identical and we call it meso-isomer. Thus, total isomers of this molecule are three: ($3R,6R$)-, ($3S,6S$)-, and  ($3R,6S$)- (($3S,6R$)- is essentially ($3R,6S$)-).


Le Châtelier's Principle and heat Consider the following reaction at equilibrium.
$$\ce{A->B}, \Delta H < 0 $$

Suppose I increase the temperature. Now, quite a few people would invoke Le Châtelier's Principle and say that since "heat" is a product of this reaction, the equilibrium should shift backwards. This is clearly wrong because "heat" is not a species you can have in a reaction. You can't incorporate it into the reaction quotient.

If you take
$$\Delta G = \Delta H - T\Delta S$$
for $\Delta S < 0$, you could make a claim of the equilibrium shifting backwards at higher $T$. This doesn't have anything to do with the sign of $\Delta H$ however.

Is there an actual theoretical basis for the following claim that does not invoke a principle that does not apply:

  For an exothermic reaction at equilibrium, increasing the temperature
  will cause the equilibrium to shift towards reactants.

Before to show you that what you said is false (I'm sorry ^^) be sure you understand the constant of a reaction depends on the temperature.

Let the same reaction you want $$\mathrm{A \rightleftharpoons B}$$

With a constant $\mathrm{K^{\circ}}$. Imagine you heat your system a little with $\mathrm{d}T>0$, then by Van't Hoff's law we get:

$$\frac{\mathrm{d}\ln(K^{\circ})}{\mathrm{d}T}=\frac{\Delta_rH^{\circ}}{RT^2}$$
Where $R$ is the perfect gas constant which is positive. Then $RT^2>0$.
 So $\mathrm{d}\ln(K^{\circ})$ as the same sign as $\Delta_rH^{\circ}\cdot \mathrm{d}T$.

So if you have an endothermic reaction $\Delta_rH^{\circ}>0$ because $\mathrm{d}T>0$ then, $$\mathrm{d}\ln(K^{\circ})>0$$

Then $K^{\circ}$ will increase with the temperature. If you take $\mathrm{d}T<0$ for an endothermic reaction then the constant of the reaction will decrease with the temperature.

I let you do the same final reasoning for an exothermic reaction. So the "Le Châtelier principle" is still true.

Note: If you know the chemical affinity, you can do the same proof, just a bit longer!


Does coffee have the ability to dissolve less sugar into it than water? Like water can absorb such and such an amount of sugar, can coffee absorb less as it already has coffee in it.

I will agree with Ivan Neretin. That being said, just because other components are dissolved in solution does not mean that solubility of other components will be affected. (i.e. if you dissolve sodium chloride in water, the solubility of sucrose may or may not be affected). If you have some spare sugar, a scale, coffee, and a little extra time, I'd recommend playing with it a little bit. Look at the literature solubility of sucrose in water, make a known volume of coffee, mass a fair amount of sugar, and see how much you can get to dissolve. Do a few unit conversions and compare the value you get to literature. Could be a good weekend experiment.


Combustion properties of Dimethyl Ether I have a C++ code which calculates the combustion thermodynamic variables (temperature, density) and the mass fractions of Dimethyl Ether for different equivalence ratio. I would like to check my results but I couldn't find anything on google. Most of the papers about DME that I've found show the laminar flame speed.

Please refer me to some paper or book if you know any. A graph showing the adiabatic flame temperature vs equivalence ratio would be ideal.

Please refer me to some paper or book if you know any. A graph showing the adiabatic flame temperature vs equivalence ratio would be ideal.

In 2014 it appears that Glaude, Sirjean, Fournet et al. published a paper that gives exactly the data you are looking for.  See Figure 4 from their paper, which I have also included below.

Possibly relevant text from the paper:

  the
  maximum flame temperature is obtained for slightly rich
  mixtures, at an equivalence ratio around 1.05-1.1, because of
  the lower impact of dilution by nitrogen when the amount of
  air decreases. At the strict stoichiometric conditions, the rate
  of the endothermal molecular dissociations is also at its
  maximum which tends to depress the temperature.

Also this next bit:

  The results show that alcohols have flame temperatures
  comparable to those of conventional fuels, contrary to DME
  that develops significantly higher temperatures. This is
  consistent with the respective values of the heating values of
  these fuels and thermochemical equilibrium data.

  I couldn't find anything on google.

I found the paper by Googling dme adiabatic flame temperature.


Why does the Second Law of Thermodynamics hold true? The Second Law of Thermodynamics states that the state of entropy of
  the entire universe, as an isolated system, will always increase over
  time. The second law also states that the changes in the entropy in
  the universe can never be negative.

According to that, $\Delta S$ of the universe should always be positive for a spontaneous reaction to occur. But what is the reason for that?

In Advanced Chemistry, Philip Matthews states:

  Pg-287  This result cannot be proved; rather it is a statement of
  experiment

While there is true reasons for every point in chemistry, I feel there should be a real reason for it rather than experimental statement. But I searched it on Google but can't to find an answer. So, can you help me to give the reason?

According to second law of Thermodynamics,

During spontaneous process:

$$ ∆S_{total} > 0 \ or ∆S_{total} = 0 $$

So total entropy is positive.

But $ ∆S_{system} \ or \ ∆S_{surroundings} $ may be less than 0.

Which $ \implies $ any one of $ ∆S_{system} \ *OR* \ ∆S_{surr} $ may be less than 0.

Now, according to Big Bang theory, the Universe, soon after existence , is expanding more and more and so - its entropy is increasing.

 Why entropy should be positive to occur a spontaneous reaction?

An example first:
• Heat flow is unidirectional. That is, from higher to lower temperature.
Similar to it, all Physical or Chemical Process will tend to proceed spontaneously in one direction only(unless altered by some external agency).
Hence,
Spontaneity is the potential to proceed without assistance of external agency.

Now, Entropy means randomness or disordered behaviour.
And there is always a tendency for the system's energy to become more disordered. An example would be diffusion.
Where gases diffuse and become more chaotic. And during the expansion of our universe, its entropy increases.

So, we interpret that entropy for a spontaneous process increases, till it reaches maximum. And our universe having a positive $∆U$ and gaining entropy in the direction of spontaneity.

$∆S=0$ at equilibrium.

If $∆S$ is negative that would mean that reaction or process is going opposite to the spontaneous direction, and which is not possible in case of our universe.


In what sense is the Inductive effect a permanent effect? Almost everywhere I checked, the Inductive effect is described as a permanent effect with almost no mention of what that actually means.

What exactly is permanent about the effect?
If the effect is permanent, does that mean that even if the attached electron
withdrawing/donating group is removed, the induced effect still continues to
exist?

The effect is permanent because the group that does the 'induction' is always there in the molecule. For example, if the induction is caused by a t-butyl group, that group is held by a strong carbon to carbon ($\ce{C-C}$) bond that will not easily be broken. One could also consider the inductive effect of the Fluorine held by a very strong $\ce{C-F}$ bond.

A temporary effect might be caused by protontation of an oxygen atom where the protonation can catalyse a reaction. The protonation is reversible. The bond between the proton and oxygen would be much weaker, as $\ce{C=O^+-H}$ will be in equilibrium with $\ce{C=O  +  H^+}$.

Removal of the group that does the induction will remove the inductive effect.


Non stationary effects of molecular orbitals We have been learning a bit about molecular orbitals in class.

However, we have always considered things under a quasistatic approximation.

What are some effects that can only be explained by not using the quasistatic approximation ?

First off, let's define our terms. The quasistatic approximation:

  More generally, the quasistatic approximation can be understood through the idea that the sources in the problem change sufficiently slowly that the system can be taken to be in equilibrium at all times.

So the question boils down to "are there ever any quantum mechanical behaviors that are non-equilibrium?" Or. more precisely, are there fast changes where the system is not really in equilibrium.

Well, as noted in the comments, there's the Born-Oppenheimer approximation which, at least in my opinion, reflects a quasi-static view of electronic structure. That is, the nuclei move slowly relative to the electrons, so despite vibrations, etc. the electrons can be considered in equilibrium with nuclear positions.

Obviously the Born-Oppenheimer approximation fails.

The poster suggested the question was about other non-static behavior. In that case, you fall into time-dependent quantum mechanics, e.g. time-dependent Hartree Fock (i.e., RPA) or TD-DFT.

  Linear-response TDDFT can be used if the external perturbation is small in the sense that it does not completely destroy the ground-state structure of the system. In this case one can analyze the linear response of the system. This is a great advantage as, to first order, the variation of the system will depend only on the ground-state wave-function.

Are there cases where linear response breaks and the variation in the system doesn't depend only on the ground-state wave-function. Yes. Consider the effects of large perturbations, like large electric fields, nonlinear optical effects (two photo absorption), etc.

As another fairly easy example, consider photo-ionization. That is, remove an electron from a system. In the quasi-static approximation, you'd have something like Koopmans' theorem with frozen ground-state orbitals. In reality, there's orbital relaxation even though the ionization process is very fast.


Is it possible for monobasic and dibasic potassium phosphate to complete break down into phosphates? Here is the picture of monobasic potassium phosphate:

And here is the picture of dibasic potassium phosphate:

Is it possible for the phosphorous-containing anions to further break down into actual phosphates and floating protons?

At basic conditions over pH 10, those phosphorus-containing anions can get further deprotonated, as can easily be seen on the corresponding Wikipedia article on phosphates.

Phosphoric acid $\rightarrow$ Dihydrogen phosphate $\rightarrow$ Hydrogen phosphate $\rightarrow$ Phosphate


What exactly is basicity? In my textbook, it’s given that the trend of the basicity of the elements of group 15 is $$\ce{NH3 > PH3 > AsH3 > SbH3 \ge BiH3}$$
Also, it’s said that the reducing character decreases down the group. (Which is the ability to lose $\ce{e-}$)
It’s also give (on a different page) that $\ce{H3PO3}$ and $\ce{H3PO4}$ are di- and tri-basic respectively.

Phosphoric acid (presumably, a Brønsted-Lowry base) has a basicity of 3. This would imply that basicity is a property of acids. However, Ammonia (which is a Lewis base) has the highest basicity. Further, quickly Google search of the definition of basicity states: “Basicity is the number of hydrogen atoms replaceable by a base in a particular acid.”

My question is, how does Ammonia (a Lewis base) have a basicity (i.e. acid character)... shouldn’t it be a weak base? Or is my understanding of the definition of basicity wrong?

The issue is that basicity has a couple of meanings and you have to be able to tell from the context what it means.

The first is logical, and is used to describe how basic a base is. That is the trend that you mentioned about the Group 15 hydrides. (In my experience, this is by far the more common usage of the word.)

The second is somewhat confusing, and it is used to describe how many protons of an acid can be removed by neutralisation with base. For example, $\ce{HCl}$ has one acidic proton, so it is monobasic; $\ce{H2SO4}$ has two acidic protons, so it is dibasic.


Mechanism of ring formation of an epoxide under acidic aqueous conditions I am trying to figure out the mechanism of this reaction:

I think the $\ce{H+}$ will first attack the oxygen of the epoxide and another $\ce{OH-}$ will be added to the most substituted carbocation, which is the secondary carbon of the epoxide. This will form a glycol, but don't know where to go from there.

What is the mechanism that leads to that product?

This is a compilation from the comments (some of them deleted by now) showing the thought process in finding the reaction product and eliminating some of the steps that might seem plausible. None of the insight is mine.

Formation of a cation

  I think the H+ will first attack the oxygen of the epoxide

Yes, that is the first step.

No attack by a nucleophilic oxygen at this point

  [...] and another OH− will be added to the most substituted carbocation, which is the secondary carbon of the epoxide. This will form a glycol,but don't know where to go from there.

This does not happen.

  H3O+ means H+ in water so OH- is available.

The solution is acidic, so there is very little hydroxide available.

  H20 will act as a nucleophile and attack the carbocation, this will result in a H+ being released from +(H2O)

This does not happen.

  Why is the H2O not attacking before the formation of 5 member ring?

Compared to hydroxide, water is not a strong nucleophile. ANBENZENE comments: "[...] the molecule will always try to generate a stable carbocation [...]".

Rearrangement to make a more stable carbocation

As Waylander states in a comment: "Have the double bond electrons attack the protonated epoxide to give the most stable cation. Remember, forming [5] rings is favorable".

This is an intramolecular reaction, going from a carbocation on a secondary carbon to one on a tertiary carbon. Also, a five-membered ring is formed and a three-membered ring was opened.

Final step: water reacts with the carbocation

Finally, the oxygen on water does attack a carbocation, resulting in the second hydroxyl group found in the product.


How photoluminescence (PL) spectroscopy works Photoluminescence (PL) spectroscopy is used to find the energy levels in a compound but how does PL spectroscopy works? After exciting a material  with a particular wavelength whether PL machine looks for photons emitted with in a range of wavelenght?

The below image was taken from here.

To get this spectra the $\ce{CeO2}$ was excited at 345 nm and got peaks at 490 and 520 nm(approx) ie 2.53 and 2.38 eV respectively and its band gap is 3 eV. Whats happening here?

The explanation that follows is valid for compound dissolved in liquid solution where PL is understood either as fluorescence or phosphorescence.

PL is measured with a fluorimeter. The main optical components of this instrument are: a lamp that provides the light used to excite the sample, two monochromators (M), one for the excitation (Mex) and the other one for the emission (Mem) and a detector, usually a photomultiplier.

To measure PL the Mex is fixed at the excitation wavelength (in your case 345 nm) and the Mem does the scan, i.e. it let pass the luminescence (a narrow band) in a fixed wavelength range. This is then measured with the photomultiplier. At the end you obtain a spectrum, i.e PL vs wavelength.

From the spectrum that you show this range goes from 200 nm to 1000 nm. Usually the range starts after the excitation, this is to avoid to record the excitation itself that is not desired. Because in the spectrum that you show there is not the excitation band i suppose they use a filter (cut-off) to cut it out.


Difference required for compound to be a phenol vs benzene compound When naming organic compounds, how should one decide whether a compound should be named as a benzene compound, or as a phenol?

For example, the compound 3,4-dihydroxybenzoic acid has an -OH group, so why is it not named as a phenol derivative? I feel like I'm missing something here, but none of the numerous sources I've looked at have provided a clear-cut rule to differentiate with.

In organic nomenclature, when you have a compound with multiple functional groups, you follow the priority rules that have been nicely summed up on Wikipedia:

  When compounds contain more than one functional group, the order of precedence determines which groups are named with prefix or suffix forms. The highest-precedence group takes the suffix, with all others taking the prefix form. However, double and triple bonds only take suffix form (-en and -yn) and are used with other suffixes.

These rules apply to benzene derivatives as well as aliphatic and other cyclic compounds.

Continuing,

  Prefixed substituents are ordered alphabetically (excluding any modifiers such as di-, tri-, etc.), e.g. chlorofluoromethane, not fluorochloromethane. If there are multiple functional groups of the same type, either prefixed or suffixed, the position numbers are ordered numerically (thus ethane-1,2-diol, not ethane-2,1-diol.) The N position indicator for amines and amides comes before "1", e.g. $\ce{CH3CH(CH3)CH2NH(CH3)}$ is N,2-dimethylpropanamine

Here's a shorter (and quite handy) version of the Wikipedia table (shamelessly copied from masterorganicchemistry.com, I've taken a liking to their color-scheme :-))

So in the example you've posted:

You give the carbon atom that has $\ce{-COOH}$ attached to it the first locant (because $\ce{-COOH}$ takes precedence over $\ce{-OH}$, according to the priority rules), and you continue numbering clockwise so the the two $\ce{-OH}$ groups take locants 3 and 4 respectively (Lowest set of locants rule).

That's how you end up with 3,4-dihydroxybenzoic acid :-)


Why can we not use water instead of blank control solution in enzyme catalysis with Peroxidase experiment? I have to do a lab about enzyme catalysis with peroxidase in my chemistry class. We are to use the blank control solution in the experiment. My question is, why can we not use water instead? I did some research about this and someone says something like "It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy." Is this the exact reason why we do not use water in this experiment?

It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy.

For water, the absorption coefficients $\alpha\ [\mathrm{m^{-1}}]$ at $\lambda$ = 680, 580, 535, 480 and 380 nm are 0.47, 0.09, 0.045, 0.013 and 0.011.

(Source: R. M. Pope, E. S. Fry, Applied Optics, 1997, 36, 8710-8723)

That is not a lot.

On the other hand, you never know about impurities in your buffers or other components (apart from your probe).

Therefore, you usually do not measure against "just the solvent", but against the solution of "everything except the probe" to record the complete background.


Products of radical bromination and mechanism My organic chemistry final is next Tuesday, and I have worked through the practice final. The last question is the following: Give and name all the compounds resulting from the radical monobromination 2,3-dimethylpentane. Indicate the major and minor products.

I drew:

My prof says that I am missing my R/S. I can clearly see my chiral centres, but how do I determine the orientation of the H? Red is the major product, blue is the minor product, and magenta is the very minor product.

The second question is: write the mechanism of this reaction identifying the initiation, propagation, and termination steps for the formation of one of the products. This is what I drew:

My prof says that I am missing a propagation step with another bromine-bromine molecule.

EDIT:

My prof says that I am missing my R/S. I can clearly see my chiral centres, but how do I determine the orientation of the H? Red is the major product, blue is the minor product, and magenta is the very minor product.

You don't determine the orientation of the H you just acknowledge the formation of both the R and S stereo isomers in your products in 50/50 proportions.

  My prof says that I am missing a propagation step with another bromine-bromine molecule. I would greatly appreciate your input!

All you need to show is the organic radical reacting with diatomic bromine to form you brominated molecule and another bromine radical.


Taking volume contraction into account when mixing water with ethanol To calculate the amount of purified water and the amount of ethanol (96% Vol.) which is needed for a desired/variable amount of a mixture - given a desired/variable percentage of ethanol - I use the following formula:

$$W = \left(\frac{T * P}{ 100}\right) \times \frac{1}{0.96}$$

and

$$E = T -  W$$

Where $T$ is the desired Total amount of product (in liters); $P$ is the desired Percentage of ethanol in the final product (in %Vol.); $W$ is the calculated amount of Water (in liters) and $E$ is the calculated amount of Ethanol (in liters containing 96 %Vol ethanol).

For example - using this formula - when I need $\pu{40 liters}$ product containing 50% vol. ethanol, the calculated amounts of water and ethanol for the mixture become $19.1667 : 20.833$.

The problem in the above calculation is that because of volume contraction the desired total amount of the final product will be less as given in the formula. To me, the problem became even more confusing after I learned that volume contraction depends on the mixing ratio of both components in a nonlinear way.

What changes should I made to the formula to take volume contraction into account?

Measuring out volumes can sometimes be easier than dealing with mass, but in this case, converting the calculations to mass-centric can make the process easier. (I'll use your symbols, but they will now be in mass units (kg). If you desire percentages of ethanol by volume, you still need to convert to mass units because of the volume contraction!))

First, calculate the total mass  in kg of the volume of solution needed; divide the volume by the density of the specific ethanol-water solution. You can look up this density in a handbook or Google it. One simple table is at https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html, although many tables are available, even to tenths of a percent.

Second, calculate the mass of 96% ethanol needed:  = (∗/100)×1/0.96. The ratio / will be higher than , the % desired, because the ethanol is only 96%. Convert this to volume in liters, if you desire, by dividing by 0.80138, the density of 96% ethanol.

Third, calculate the amount of pure water needed:  =  − .   / will be less than 100 -   because of the 4% water already in the ethanol. But the total mass will give you the volume required, and the % ethanol will be correct - BY MASS.

Commercial isopropanol solutions are frequently sold as 70% by volume; this calculates out to 64.7% by weight. Your procedure, if you desire to make variable % solutions BY VOLUME, will need to do a calculation before you even do the first step: convert the % ethanol by volume to % ethanol by mass.