Is the specific heat capacity of liquid iron constant? Does liquid iron have a constant specific heat capacity?  Everywhere I've looked (save one sourceless online reference to $c_p = 611\ \mathrm{J\ kg^{–1}\ K^{–1}}$ here), I find no value listed for liquid iron.  Is this because the heat capacity is temperature-dependent?  Or is this related to a measurement problem?  Or does an accepted constant value exist, but I'm just a poor researcher?

Okay, let's answer bit by bit:

Measurement of heat capacity of liquids: this isn't particularly hard to do, and in fact it's much easier to obtain $C_p$ experimentally than $C_V$… however, once you know one, the other can be determined from it and some other thermodynamic properties. Different types of calorimeters are used for liquids, but the best-known is probably the Callender and Barnes apparatus.
“Does an accepted value exist?” and “Is the specific heat of liquid iron temperature-dependent?” — Well, I found a good paper which includes measurements and comparison with previously published values: Thermophysical Properties of Liquid Iron, M. Beutl, G. Pottlacher and H. Jäger, Intl. J. Thermophys., 1994, 15, 1323.

which confirms that considering that heat capacity is roughly constant in the 1800–4000 K range. The authors thus extract the constant pressure specific heat capacity as 825 J/kg/K, apparently in reasonable agreement with previous work that they cite:

“Or am I a poor researcher?” — Well, the above paper is the first result return by Google for a search for "specific heat" and "liquid iron", so I'll let you be the judge of that ;-)


How many signals would be seen in total in the 13C NMR spectrum This is NanoBalletDancer.
How many different carbon environments are there in NanoBalletDancer – i.e. how
many signals would be seen in total in the 13C NMR spectrum?

I counted 8 signals from benzene ring carbons
4 from methyl group carbons
4 from triple bond carbons

What is the total amount of signals? Which did I miss?

3 different methyl signals: (1, 16) (40, 41) (27, 28, 29, 33, 34, 35)

4 different methylene signals: (17) (2, 15) (3,14) (37, 39)

1 aliphatic CH signal: (36)

2 different aliphatic quaternary C signals: (26, 32) (38)

4 different triple bond C signals: (24, 30) (25, 31) (4, 13) (5, 12)

8 different aromatic carbons: (7) (10) (6, 8) (9, 11) (18) (21) (19, 23) (20, 22)

22 different signals

Just looked at a model, methyls 40 and 41 are different; one is on the same side of the 6-membered ring as the benzene ring, the other is on the same side of the 6-membered ring as the hydrogen at C-36

23 different signals!


How to connect structures in Avogadro at certain angles? I am working with this carbon nano-structure in Avogadro and the junction between my two nano-structures I try to join never seems to be aligned well. It always seems to be off kilter, which is most likely due to how I move the junction into position (manually dragging).

I posted a picture of my issue above. Where the two white arrows are is what is bothering me. I would like those to be as vertical as possible but I do not know of ways to do this aside from manually adjusting the atoms/bonds, which I assume would not get them to align exactly.

UPDATE:

As per khaverim's advice, I followed his first instructions, using MMFF94 instead of UFF (forgot to change it and my structure is large so it takes a while to optmize, will try with UFF later), and got the following structure:

I am pretty sure they are all equal angled, at least they look to be so, so thanks.

Did you try a (Universal Force-Field) geometry optimization? I would:

Select all atoms
Deselect (hold Ctrl + click) the 4 (6? 8?) carbons of interest
Click Extensions -> Molecular Mechanics -> Ignore Selection
Click Extensions -> Optimize Geometry

Alternatively, you could optimize a few more atoms around/in the bridge while the others are constrained. The G.O. algorithm is pretty robust and handles many systems quite well. A simple example of hand-drawn cyclohexane:

I got a result like this by making 2 nanotubes with chiral index (4,4), drawing 6 bonds and optimizing the bridge:

If this is ineffective, I would resort to:

Change the 4 carbons to different elements e.g.

$\ce{O        H}$

$\ce{N        S}$

(like the arrangement you have in the picture, so you can identify them later)

Click File -> Save As... and save the file as an .xyz
Calculate the O-N and H-S distance by $r_{ij} = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$ from the coordinates in the .xyz file.
Ideally, Avogadro saved the .xyz coordinates aligned in z (which is default when you draw molecules without rotating the initial view).
(Even if not,) re-write the coordinates of the atoms aligned in x and z, e.g. $\ce{O}$ at $(x, y, z)$; $\ce{N}$ at $(x, y-r, z)$.
Or (yikes) write a quick program to generate possible systems with the fixed $r$ you calculated, name them system-000.xyz, system-001.xyz, system-002.xyz, etc., and view the "trajectory" frames until you find one pleasing. (Viewing can be made much easier with VMD, e.g. vmd system-*.xyz).

Avoiding Avogadro: If you use Molden and load your system (as an .xyz, .car, .pdb etc.), you can use the ZMAT-editor to manually change bond angles and dihedrals:

Finally, if all else fails, it may be time to consider a full quantum-mechanical treatment of the system (if you're familiar with High Performance Computing) with a free (for academia!) program like Orca (where one can constrain certain atoms during the optimization). An example input would be:

! hf 6-31g* opt  # or copt, for cartesian optimization

%pal
nprocs 128   # num. processors
end

%scf
maxiter 2000 # max self-consistent field iterations
end

%geom
Constraints # atoms to keep fixed starting at index 0
{C 61 C}
{C 62 C}
{C 290 C}
{C 291 C}
{C 292 C}
end
end

*xyz 0 1  # charge=0; mult=2S+1 = 1 (closed shell)
  C      -1.306743      4.170005      1.012072
  C      -0.042414      3.718873      0.686759
  C       0.210647      2.367809      0.785537
  C      -2.267754      3.271952      1.435683
  C      -2.920909      0.889649      1.948837
  C      -4.153277      1.187877      2.498723
  C      -4.989183      0.155126      2.896328
  C      -4.571713     -1.152339      2.735382
  C      -3.332221     -1.380404      2.172864
....... etc.
*


Silicone water repellent help Is there a chemical or solution that can penetrate silicone water repellents in fabric? Need to break down or penetrate the silicone water repellent in selected areas.

"Silicone water repellent" is barely helpful information. They are a category of materials which are used in an enormous variety of industrial, commercial and retail applications. Some products react and bind to the surface they are coating, others are simply dried onto the surface. Some cross-link, most don't. A typical consumer fabric silicone water repellent will be "penetrated" by solvents such as acetone, methy ethyl ketone (2-butanone) or toluene; mineral spirits might work. There's no way for us to determine whether these solvents can also damage the fabric or clothing item they are poured onto. Another possibility is to use a fairly harsh (alkaline) detergent, but again I can't say whether the fabric will be damaged. Many of the retail products gradually wear away, diminishing repellency slightly after each wash/dry cycle.


Potential energy versus reaction coordinate (graph of Gibbs free energy) Does the potential energy scale in the above mentioned graph measure the Gibbs free energy of the system?

As the system always tries to achieve a minimum of free energy and chemical reactions are taking place can be considered a non-mechanical work (I'm not sure of it, as the system isn't in a state of chemical equilibrium if a reaction is taking place), potential energy should represent Gibbs free energy.

If not, how should I understand the potential energy of a reaction mixture?

There is a long explanation for that, however in short they are not the same thing. Free energy relates to thermodynamic systems with many particles and is governed by Boltzmann's distribution, a sum of all possible energy configurations, where F is the free energy of the system.

$Z = e^{- \frac{F}{kT}} = \sum_i e^{- \frac{E_i}{kT}}$

Potential energy is non-statistical and well-defined for one particle. Thermodynamic free energy (one type being Gibbs) is a subset of the total potential energy. Even though they may seem similar, they should not be confused and should not be interchanged.


Origin of Azide's Name Today, I learned that  $\ce{N_3^-}$ is referred to as azide. I was wondering why $\ce{N_3^-}$ is referred to as azide. I checked Wikipedia for an answer, but I didn't find one. Does anyone know why this ion is assigned this name?

Nitrogen was originally called 'azote' by Lavoisier. This name persists in many nitrogen containing species such as azide, hydrazine, diazonium etc.

The '-ide' ending is the standard ending for anions of a single element such as carbide, oxide, sulfide etc.


Suggest a method, other than spectrophotometry, for determining the reaction rate For the hydrolysis of $\ce{\mathit{trans\hyphen}[Co(en)2Cl2]Cl}$ (and cis) we measured reaction rates via spectrophotometry. Plots of change in absorbance were used to determine the rate law and the rate for the reaction.

The lab report asks:

  Suggest another method, other than spectrophotometry, for determining the reaction rate measured in this experiment.

Knowing little about measuring rates, I said one could measure the enthalpy and entropy of the reaction at a given temperature, and use those to calculate the rate constant. This is based off of the equation

$$\ln\frac{k}{T}=-\frac{\Delta H_\text{act}}{RT}+\ln\frac{k_\mathrm{B}}{h}+\frac{\Delta S_\text{act}}{R}$$

but, my professor said thermodynamics couldn’t be used to characterize kinetics. Also, reflecting on my answer, I didn’t really specify a technique.

What’s another technique that could be applied here?

Spectroscopy is often the best method, since it allows for (almost) continuous monitoring of the concentration.

If that is not wanted, ask yourself:

What is released in the hydrolysis of the complex?

Chloride would be a good guess.

Can I take samples and perform a wet chemical analysis to determine the concentration of new species?

Looking at the graph in your other related question, the answer is yes. The hydrolysis is slow enough.

Which methods can I possibly use to determine the concentration $\ce{Cl-}$ in solution?

An argentometric titration, based on the low solubility of $\ce{AgCl}$ is a good idea. What about the Mohr titration, using $\ce{K2CrO4}$ as an indicator?


Can Cellulose Molecules be Synthesized from Glucose? I am reading that cellulose may be broken down into glucose. This is usually done with some sort of solvent.

Is there any way to 'rebuild' these cellulose molecules from glucose?

Cellulose may be broken into smaller pieces (though not necessarily into separate glucose molecules) and regenerated into cellophane with the viscose process.

For the home lab, rather than using the poisonous and extremely flammable $\ce{CS2}$ in the viscose process, the somewhat safer cuprammonium process is a better choice. Do be careful -- don't wind up with blue stains on skin or clothing, it shows poor technique. See this video for a demonstration.


Glass joint stuck in flask I am doing distillation using system of glassware including a flask and a glass joint. Recently I've faced a problem: the glass joint stuck in the flask. I tried to remove it but it broke and hurted my finger. I want to ask how to prevent this from happening again: can I use a paper masking tape or continue using grease? I am thinking to try the adhesive tape do you think it deserves trying?

You have multiple options.  All of them assume that the distillation setup is empty, at ambient pressure.

Of course, like mentioned in the comment by @Waylander, put a little of (silicon) grease on the glass joints prior to the distillation, just enough that they become transparent.  There are multiple kinds of grease, suitable for different temperatures / pressures during the application.
If grease is not an option, there are sleeves made of PFTE / teflon to be put on the joints.  They are available in different diameters, and forms, e.g.

(source, source)

Especially for vacuum distillations, but not limited to this: the best remedy for this issue is to dismount the distillation right after the recovery of the product fraction(s), while the parts (e.g., column, head) are still considerably warm.  Otherwise, once cooled to rt, the setup then often will stuck much better than ever intended.  Depending on size of the distillation and remaining temperature of these pieces, wearing gloves (to protect your hands against the heat) may be helpful.  Then, it is better to allow the warm pieces to cool to rt while being put e.g. on cork rings.
«Forgotton» and already stuck glass joints may be detached from each other if you immerse them in the basket of a ultrasound bath.  Addition of some of detergent to the water bath helps while occasionally gently pulling the pieces apart.
If you have the dexterity (some develop this over time / by experience), wear both gloves to protect your hands against heat and goggles: you may heat the circumference of the outer wall of the glass joint with a heat blower or even a heat gun / pocket burner.  The aim here is that the outer part of the joint dilates quicker than the inner part.  It is the most risky variant because uneven heating may shatter your distillation into pieces; especially if the parts joined together are made of different types of glass with different expansion coefficients.  It this is not for you, just ask your local glass blower for assistance.


What makes bond break apart from heat? What makes bonds break apart from heat?

I am wondering if anyone could explain the reasons why new bonds are formed in the absence of energy. I have found that they form in order to lower the overall energy, presumably because electrons move closer to the nuclei via bonds.

But in a reaction like $\ce{2 K (s) + 2 H2O (l) -> 2 K+ (aq) + 2 OH- (aq) + H2 (g)}$, I wonder what makes the bonds break? I just don't understand why $\ce{2 K}$ would suddenly split to form new bonds unless I'm missing some sort of energy being put into the system. Maybe potassium is able to attack $\ce{H2O}$ since it is a polar molecule?

But, I truly can't find a reaction mechanism that explains these phenomena.

You have a two part question so I'll answer both of them:

  Question - What makes bonds break apart from heat?

"If you start to heat the system, two things can happen:

The atoms gain enough energy to move into higher, excited states in the potential well. Eventually the additional energy may be high enough to make the total energy positive. The nuclei can escape the potential well - the bond is broken.
The electrons that participated in forming the bond, i.e. those on orbitals overlapping both nuclei, get excited into different, higher energy orbitals. This makes the potential well shallower, and the nuclei can escape it easier - the bond becomes weaker and easier to break. Eventually the potential well can even become so shallow, that no bound state can exist." - Source

  Question - Why do alkali metals ($\ce{Li, Na, K ...}$) break bonds in water ($\ce{H_2O}$)?

"All alkali metals melt as a part of the reaction with water. Water molecules ionise the bare metallic surface of the liquid metal, leaving a positively charged metal surface and negatively charged water ions. The attraction between the charged metal and water ions will rapidly increase the surface area, causing an exponential increase of ionisation. When the repulsive forces within the liquid metal surface exceeds the forces of the surface tension, it vigorously explodes."- Source

I hope that my answers were useful to you.


Is the 'central Tin atom' in crystalline tin oxide really offset like this? Is the position of the central Tin atom as shown in the first figure ($\ce{SnO2}$) drawn offset towards the oxygen to the right for a reason? The oxygen also seems to reciprocate and move towards the tin. Usually the rutile structure is shown as symmetric with the metal the center. I'm showing a GIF alternating images of $\ce{SnO2}$ and $\ce{RuO2}$, with complete image source information following.

note: I'm interested to know about the locations of the atoms within a unit cell as determined by crystallographic information - not one's impression of the drawing. I'd like to know if the offset shown in the drawing is there to represent an actual asymmetry, or if it's just an error.

Figure 1. Tin Dioxide (Tin IV oxide), also found as Cassiterite, image from Rutile unit cell 3D balls.

Figure 2. Ruthenium(IV) oxide, image from Ruthenium(IV) oxide unit cell 3D vdW.

Figure 3. Rutile structure, image from Rutile structure

 Fig. 1

 Fig. 2

 Fig. 3

From http://www.crystallography.net/cod/1000062.html one can see that Sn sits in the middle, without any displacement, and mimics the classic Rutile-type structure:

I checked original Wikimedia image, and according to metadata

  CIF retrieved from The American Mineralogist Crystal Structure Database

is no longer available. I also failed to discover this source:

  X-ray crystallographic data from:
  R. W. G. Wyckoff (1963) Second edition. Interscience Publishers, New York, New York. Crystal Structures 1, 239-444

Ideally, if you have *.cif file, it might be worth checking occupancies for the atoms you are interested in. Probably there is a 50-50 disordered Sn atom about center of unit cell, and you only observe a single site. As long as this is not the case, I would rely on the available COD data from 1971 instead.


Is there a way to calculate bond energies using computational chemistry? I'm currently trying to see if a particular set of ligands will react with a metal ion to form a complex. While I'm waiting for the wet lab results to arrive, my adviser suggested we look into computational chemistry to see if we can model the proposed structures theoretically.

Neither of us (or anyone in the department for that matter) have any experience in computational chemistry. I got my hands on a copy of Avogadro and GAMESS-US and have optimized the geometries of the complexes we think might arise. Is there a way to determine if the complexes would form? Thanks.

Is there a way to calculate bond energies using GAMESS?

Yes. You would normally model the bound and unbound states

$$\ce{ML_{n - 1} + L -> ML_{n}}$$

The definition of bond energy is based on the enthalpy of reaction in the gas phase (thanks DSVA!), so you would calculate it as follows:

$$\Delta H^\circ = H_{\ce{ML_{n}}} - (H_{\ce{ML_{n - 1}}} + H_{\ce{L}})\text{.}$$

While not strictly bond energy by definition, a value for the reaction Gibb's free energy might also be useful in some contexts:

$$\Delta G^\circ = G_{\ce{ML_{n}}} - (G_{\ce{ML_{n - 1}}} + G_{\ce{L}})\text{.}$$

This takes entropy into account.
For the calculation of Gibb's free energies, most codes calculate the molecular partition function from translational, rotational and vibrational contributions and get enthalpy and entropy from it.
Due to the vibrational contributions, you'll need to diagonalize the Hessian matrix of the system. This will give you an association (minus dissociation) energy value.
(You might take solvation into account through explicit/implicit solvation, use decent level of theory, validate/benchmark, etc.)
There are a lot of approximations and considerations in the process. More can be found here (some equations) and here (more practical).

Observe that

$$K = e^{- \frac{\Delta G^\circ}{R T}}$$

is the equilibrium constant associated with the process above, which is called the stability constant, formation constant or binding constant.
This is widely used to "measure the strength of the interaction between the reagents that come together to form a complex" (source).
Bear in mind that $\Delta G^\circ$ goes inside an exponential, i.e., small errors in $\Delta G^\circ$ become (exponentially) large errors in $K$, so $\Delta G^\circ$ might be more useful in a computational context than $K$ itself.

  Is there a way to determine if the complexes would form?

This is complicated.

Your reaction might be too slow though, i.e., actual formation might depend on transition states. Furthermore, there might be faster processes in competition that you might miss.


SMART string for alkyl group excluding alcohols I'm quite new to the usage of SMARTS strings. I would like to find all alkyl groups (but no $\ce{CH3})$ in a component, here e.g. cyclohexanol (C1CCC(CC1)O).
My SMARTS string is: [CX4;H0,H1,H2]
However, this matches every carbon. I would like to write my smart string, so that C:3 is not matched, because it is bound to something else than a carbon (here: oxygen). How is this possible with SMARTS?

How to find carbon

with four connections
not -CH3
and only connected to carbon

As in your SMILES there are only implicit hydrogen, you do not need to count Hs.
Count the explicit bonds and exclude the carbon with one explicit bond to avoid -CH3.
[CX4;D2,D3,D4]
To find the carbons that are connected to others than carbon
C[!C]
Now you can use recursive SMARTS, and set the second pattern to NOT.
[CX4;D2,D3,D4;!$(C[!C])]
Now only carbons would be found that match both patterns.
CX4   C with total connections of 4
D<n>  number of explicit bonds
$()   recursive SMARTS
!     NOT
;     AND
,     OR
In cyclohexanol it will find

and here is a dummy molecule for testing it.
CCC(F)C=CCC1CCC(O)C(C(C)(C)C)C1


How to detect nM levels of Copper(I) Oxide in blood? I'm trying to design an experiment to detect nM quantities of Copper(I) Oxide in fish blood. Are there any specific tests for Copper(I) Oxide that can be used to detect trace quantities(~nM)? Any non-destructive tests such as using optical methods would be preferable.For purposes of designing the experiment, let's assume the particles are nano-sized and dispersed in blood.

The main analytical problem in your case of detecting Cu(I) oxide in blood is that you do not know the fate of the particles once they enter blood stream. As I said earlier, Cu(I) is thermodynamically not stable in any aqueous system. It oxidizes to Cu(II), which is water soluble. The non-destructive technique to detect species is X-ray photoelectron spectroscopy, but its detection limit is very poor. It may take a day to collect weak signals from such a low concentration of Cu(I) oxide. The sample has to be bone-dry as well! This is a species specific analysis, i.e. one can distinguish oxidation states of metals by deconvolution.

A practical suggestion is that you do a control test, i.e. measure copper levels before treatment in blood. As a result, all copper originates from your copper (I) oxide dispersion. You would not know the specific species of copper but you will get the total content. For that one can use common techniques such as graphite furnace atomic absorption analysis (only a few microliter of sample is required).


How do I comment on the Gibbs energy values standard and reaction in different situations? I have the following graphs:

Taking into account the following graphs, how do I comment on the Gibbs energy values
standard and Gibbs energy of reaction in different situations (a, b and c), and how do I relate the values of the latter to the spontaneity of the reaction? I'm really lost here.

Here is a labeled and annotated version of your graph (from OpenStax Chemistry).


Mechanism of Friedel-Crafts acylation with succinic anhydride Here is my proposed mechanism, something tells me that the intramolecular deprotonation happens after, and that this is not concerted. Is my mechanism correct?

The reactivity of acid chlorides and acid anhydrides are actually quite similar. It may be helpful for you to think of an acid anhydride like this:

When we do, the acylation reaction is almost identical to that of an acid chloride, where an acylium cation $\ce{R-C#O+}$ is generated:

The subsequent proton exchange restores aromaticity:


How to change CO₂ to a less polluting gas? I want to change $\ce{CO2}$ coming out of the tailpipe of a car to some other compound that is considerably less harmful to the environment with a filter of some sort.

What are some ways that this is possible?

$\ce{CO2}$ to carbon and oxygen?

It makes no difference where you grab the $\ce{CO2}$.  Don't burden the car.

Plant mammoth sterile forests with gene-gineered high lignin wood for building.  Or, deep bury the trunks in the Sahara Desert of the Empty Quarter.  RuBisCO is the slowest, least efficient enzyme on Earth, the first carbon fixation step in photosynthesis.  Gene-gineer a faster, more efficient (e.g, not poisoned by oxygen) RuBisCO for the forests.

http://www.cea-jaea-collaboration.net/literature/cea/ICCAP%20Nice%202007%20NaCO2finalpaper.pdf

A reactor for your tailpipe.


How to calculate atoms of *each* element in a compound I am trying to do the following exercise in a chemistry workbook. I have searched over internet but I couldn't find or understand it.

How many atoms of each element are found in:

5 moles $\ce{H2SO4}$

Now the only confusing thing for me is the 5 moles.
For example lets take $\ce{H2}$ element

so n=N/Na --> n × Na = N (number of atoms)
2 × 1gr/mol H = 2 moles H x 5 moles in the compound = 10 moles
10 moles $\ce{H2}$ × 6.02 × $10^{23}$ = 60.2 × $10^{23}$ atoms

Or do I do like this?

5 moles $\ce{H2}$ × 6.02 × $10^{23}$ = 30.1 × $10^{23}$ atoms

Think of the mole as a SI prefix, like 'kilo' or 'mega'. If you have 5k $\ce{H2SO4}$ molecules, you will have 10k hydrogen atoms, not 5k. Just as 'kilo' means '1000', a mole means $6.02 \cdot 10^{23}$ of something.

Therefore, the first approach gives the correct answer, but I don't understand what you mean by 1gr/mol. Grams are out of the picture here.


How to know whether s-p mixing will happen in heteronuclear molecules? How do I know whether s-p mixing would occur in some diatomic heteronuclear molecules like CN, CO, NO etc?

I don't think you can know, a priori, whether or not s-p mixing occurs without first having some experimental data.  I'll use water, a triatomic molecule, as my case in point.

For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H $\ce{sp^3}$ sigma bonds and two equivalent lone pairs, also in $\ce{sp^3}$ orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.

With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:

2 $\ce{sp^3}$ O-H sigma bonds
one lone pair in a p orbital
and the second lone pair in an $\ce{sp}$ orbital

If chemists knew how to predict s-p mixing, it wouldn't have taken them over 50 years to understand how they mix in water.


Why is KI oxidized by H2SO4, whereas NaBr isn't oxidised? My textbook says that alcohols react with sodium bromide and sulphuric acid to form the corresponding alkyl bromides. At the same time they also mention that sulphuric acid mustn't be used with potassium iodide ($\ce{KI}$), because $\ce{HI}$ formed would be oxidized to $\ce{I2}$. Won't $\ce{HBr}$ formed from $\ce{NaBr}$/$\ce{H2SO4}$ oxidise to $\ce{Br2}$?

Iodide ions are much more powerful reducing agents than bromide ions:

$$\ce{2I- ->I2 + 2e-} \hspace{1cm}E^{o}=-0.54\ \mathrm{V}$$

$$\ce{2Br- ->Br2 + 2e-} \hspace{1cm}E^{o}=-1.0873\ \mathrm{V}$$

Iodide ions are so powerful, in fact, that they are capable of reducing sulfur from a +6 oxidation state in $\ce{H2SO4}$ all the way to -2 in $\ce{H2S}$ via the following half reaction:

$$\ce{HSO4- +9H+ +8e- -> H2S + 4H2O} \hspace{1cm}E^{o}=+0.80\ \mathrm{V}$$

The full reaction being being:

$$\ce{HSO4- +9H+ +8I- -> H2S + 4I2 +4H2O}\hspace{1cm}E^{o}=+0.26\ \mathrm{V}$$

A quick glance at the reduction potentials tells you that the analogous bromide reaction has a negative reduction potential and is therefore non-spontaneous.

[Source]


Does aldol condensation give different products in different conditions? I know the mechanism of aldol condensation and general method to determine products in crossed aldol reactions, or reactions between dissimilar molecules. But the following statement in my class notes baffles me:

  In presence of a base (alkaline medium) the $\alpha$ hydrogen of the lower aldehyde is more acidic and hence is removed in the first step, while in presence of an acid, the $\alpha$ hydrogen of the higher aldehyde is more acidic and hence migrates.

According to this, on reaction of propanal with ethanal, basic medium yields 3-hydroxy pentanal while acidic medium yields 3-hydroxy 2-methyl butanal.

Why is this statement true (if it is)? And can aldol condensation proceed in an acidic medium, since the presence of the base is necessary for the removal of the slightly acidic $\alpha$-hydrogen to start the reaction?

I believe your conception of the need for deprotonation under acidic conditions is misconceived.  Whenever you are under acidic conditions, your mechanisms must necessarily start with protonation.

The key concept here is the enol tautomerization.  The enol is more favored than for propanal as compared to ethanal because it results in more substituted alkene.  Thus there is likely a higher concentration of this nucleophile in the reaction mixture.  The mechanism of tautomerization is difficult to elucidate, but presumably starts with protonation of the carbonyl in acidic conditions.  Then, likely another carbonyl deprotonates the the recently protonated carbonyl at its alpha position, forming the enol.  This transformation has produced a nucleophile (enol -- 1-propene-1-ol ) and also a strong electrophile (a protonated carbonyl -- protonated ethanal).  The reaction occurs, leading to a protonated aldol product (proton on the carbonyl) which then is deprotonated by the next carbonyl and the process continues.

As far as the base catalyzed case, ethanal is more acidic because it does not have the extra methyl group donating electron density disfavoring formation of a negative charge.  Hence ethanal becomes the nucleophile and propanal becomes the electrophile.

In each case, these are only major products.  I don't expect the energy differences are large enough to give great selectivity.


What is the OH in alcohol? What is the $\ce{OH}$ functional group in alcohol (ethanol)? My teacher told me it isn't the same as the $\ce{OH-}$ ion or hydroxide.
But it does donate one electron to the respective carbon chain and it's very tempting to say it's hydroxide as it fits perfectly into the role. What exactly is it?

You could call ethyl alcohol "ethyl hydroxide" analogously to "sodium hydroxide". But it is a poor analogy, since (unlike NaOH) ethyl alcohol is not an ionic compound, does not dissociate in water to form ethyl cation and hydroxide anion, and does not raise the pH when dissolved in water.


When finding hybridisation of atoms, why are double bonds and triple bonds only considered once? To figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom.   (Chem.Libretexts)

Why does this rule hold? Why are double and triple bonds counted only once?

Wouldn't a double bond have the equivalent number of electrons as 2 lone pairs?

And if orbitals are all about electrons, and there are three times as many electrons in a triple bond, shouldn't it count three times?

The hybridised orbitals are almost always involved in $\sigma$-bonding, and in housing the lone pair(s) of electrons. This helps us determine the geometry of the molecule.

Consider acetylene, $\ce{C_2H_2}$.

The carbon is $\mathrm{sp}$-hybridised. So, it has 2 $\sigma$-bonds - one to a hydrogen, one to another carbon. All this is in the third shell. So, a total of 4 orbitals (1 $\mathrm{s}$, 3 $\mathrm{p}$) undergo hybridisation and form 2 $\mathrm{sp}$-orbitals and 2 $\mathrm{p}$-orbitals. The $\pi$-bonding occurs due to these unhybridised $\mathrm{p}$-orbitals.

Hope this clears your doubt!

PS: The only exception to the first statement that I know is https://en.wikipedia.org/wiki/Aryne.


Why doesn't LiHCO3 exist in solid state? I have read that $\ce{LiHCO3}$ does not exist in solid state. Is it because of the size of anion-cation concept or is it something else? So why is it only for lithium and not for sodium too?

Claiming that a particular chemical compound "doesn't exist" is difficult to prove. The only convincing proof would be to show that neither LiHCO3 nor any of its hydrates can be isolated from (aqueous) solution. Essentially, you would have to show that using any of an almost infinite number of methods, that it couldn't be prepared at any temperature or pressure - not very likely to have ever been done, imho. So, I'll ask you for your source for such a claim, or better yet, I'll ask you to check and verify that your source - which you accept as being accurate - has made stringent efforts that demonstrate the claim as well as having published that work in peer reviewed literature. As an aside, there are plenty of compounds which require "waters of hydration" in their crystal structure. If it really doesn't exist, then the reason is simple: kinetically the reaction 2LiHCO3 → Li2CO3 + H2O + CO2 is spontaneous and energetically favored. It just seems to me that if I placed Li2CO3 in a pressure cell with water and CO2 at a billion gigapascals of pressure, I could make the bicarbonate salt, but perhaps some other structure is more likely...and anyway, generally what they mean by "doesn't exist" is that it "isn't known to exist at standard temperature and pressure".


How to calculate pH of a solution with two dominant equilibria? Say you have acid X for which $K_a(X)$ = 1E-7 and acid Y for which $K_a(Y)$ = 0.8E-7. In separate solutions, it would be easy to calculate $[H^+]$ because there would be only one dominant equilibrium. But in solution together, both contribute significantly to $[H^+]$, and as a result, both equilibria must be driven to the left by Le Chatelier's Principle. How do you calculate the pH of such a solution? Since both contribute to $[H^+]$, the resulting $[H^+]$ is not additive, but rather a bit less than that because of suppression.

Assume X and Y are 0.1M and monoprotic.

The first step is to write down the variables of interest, and the equations that relate them.  If there exactly as many (non-redundant) equations as variables, then the system might be solvable.

In this case, the variables are:

$\ce{[H+]}$
$\ce{[X^{-}]}$
$\ce{[Y^{-}]}$
$\ce{[HX]}$
$\ce{[HY]}$
$\ce{[OH^{-}]}$

The equations are:

water dissociation equilibrium: $K_w=\ce{[H+][OH^{-}]}$
HX dissociation equilibrium $K_{a,x}=\frac{\ce{[X^{-}][H+]}}{\ce{[HX]}}$
HY dissociation equilbrium $K_{a,y}=\frac{\ce{[Y^{-}][H+]}}{\ce{[HY]}}$
charge balance: $\ce{[H+]}=\ce{[OH^{-}] + [X^{-}] + [Y^{-}]}$
X mass balance: $\ce{[X^{-}] + [HX]}=\text{constant}=0.1\text{ molar}$
Y mass balance: $\ce{[Y^{-}] + [HY]}=\text{constant}=0.1\text{ molar}$

So, "all" you have to do is solve those equations for the six unknown variables.  That isn't easy to do with pen and paper, but it can be done.  Or you could use a computer to numerically solve the equations since you know all the numbers $K_w$, $K_{a,x}$, $K_{a,y}$, and the two different 0.1 molar total concentrations.

Once you solve the equations for $\ce{[H+]}$, the pH is just calculated from $pH = -\log_{10}\ce{[H+]}$.


When was the amu renamed the dalton? I taught chemistry for 30 years and never encountered the unit until I saw it on a juice bottle!

The atomic mass unit (amu) was not renamed to dalton (Da). These are different, albeit related, units.

The atomic mass unit (amu) referred to the relative atomic mass of oxygen, which was taken as 16. However, physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. This unit became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$.

The unit dalton (Da) and the unified atomic mass unit (u) are non-SI units that are accepted for use with the SI. Actually, ‘dalton’ and ‘unified atomic mass unit’ are alternative names for the same unit, equal to 1/12 times the mass of a free carbon-12 atom, at rest and in its ground state, i.e.

$$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$

Therefore, using the old definition of the atomic mass unit (amu) in physics

$$1\ \mathrm{Da} = 1\ \mathrm{u} \approx 1.0003\ \mathrm{amu}$$

and using the old definition of the atomic mass unit (amu) in chemistry

$$1\ \mathrm{Da} = 1\ \mathrm{u} \approx 1.00006\ \mathrm{amu}$$


Comparing the strength of metallic bonds Why is the boiling point of iron higher than magnesium?

There are two main reasons:

Metallic bonding is more prevalent for transition metal elements because there are more electrons available for delocalization (from s and d orbitals), while Magnesium has only s electrons
Fe has a higher atomic mass than Mg --> van der Waals forces are stronger.


IUPAC nomenclature of coordination complexes Where can I get the latest rules on IUPAC nomenclature of coordination complexes?
I have J D Lee book which is also assumed to  have outdated rules.

I think IUPAC's Red Book in the 5th edition by 2005, is the most recent reference of this topic, IR-9 (pages 142 onward) about coordination compounds.  Corrections of errors in the print are published here.


How to choose a UV lamp for photoisomerization of 4,4'-azobenzene dicarboxylic acid? I have prepared a thin film containing 4,4'-azobenzene dicarboxylic acid and I want to study the photoresponsive behaviour of them. For the trans-cis isomeration , I used a 365 nm UV LED with irradiation intensity varied in the range of 5-50 mW/cm². Not seeing any isomeration I have the doubt if I am using the right lamp.  Articles which have prepared thin films containing 4,4'-azobenzene dicarboxylic acid have used other types of lamps such as 200 W Xenon-doped mercury lamp(using a 355 nm cut off filter), 500 W high pressure mercury lamp or 400 W high pressure mercury lamp. I also took the UV-Visible spectroscopy of my films and noticed that the absorption of trans azobenzene-4,4'-dicarboxylic acid is centered at 330 nm.

Am I using the right lamp?

I'd probably try a more powerful lamp. Plain low-pressure mercury tubes have their emission maximum at 254 nm, which isn't what you want here. But these tubes also come in various coated variants with emission maxima around 300 or 365 nm.

A reasonable choice would be a Rayonet RPR-3000A or a Luzchem LZC-UVB lamp in a proper housing.


But what are anti-bonding pi-orbital? In search for an intuitive explanation Imagine that you want to explain to an undergraduate why they have to to shade the pi-orbitals in a symmetrical way, i.e. dark on top (+), white on bottom (-) for two neighbouring pi-orbitals because they are bonding (or anti bonding), but then the student asks: why?!.. Can anyone  explain why this is without knowing about quantum mechanics and diving into molecular orbital theory? Intuitively, it makes no sense that an electron should 'care' why if they are 'on top' or 'on the bottom' of the nucleus.
I have a hard time both intuitively understanding, but also explaining why we care about bonding and anti-bonding orbitals. I know that I should draw them, and that one is stabilising while the other is destabilising. However, for a person with limited chemistry knowledge my reasons for why is simply:  'because it is so'. This is not very satisfying. I think this has something to do with the phase and wave equations (..etc) - but again,..this seems to go into heavy chemistry (..or?).
Question: Does someone have a nice analogy, or way of explaining why this is - and what anti-bonding orbitals are without any very high-level chemistry concepts?
I think there are others struggling with this concepts, as I found a few questions here on stack exchange that failed to give a satisfying answer. This might be because the concept is extremely complexes and not possible to understand without quantum mechanics... but thats why I want to ask the question here in hope of an intuitive explanation.
The linked questions:

Anti-bonding molecular orbitals
Why do higher orbitals have more energy?
Why is the antibonding pi bond higher in energy than the normal pi bond?

First, remember that the + and - on the lobes of an electronic wavefunction do not ever represent or suggest electronic charge. An electron is always negative; a proton (or nucleus) is always positive. In order to imbed this firmly in your mind without + or - signs, imagine one lobe of a p-orbital flashing red while the other flashes blue. In this picture, an s-orbital would flash either red OR blue.
Plus and minus signs are just the way mathematicians describe the orbitals: they don't use red and blue pens, they use + and -. Now take the orbitals and move them about. Wavefunctions have a wave nature; waves have a frequency; so two orbitals of the same kind on similar atoms will have the same frequency. Two hydrogen s-orbitals flashing red at the same frequency look like one object, that is, they seem to be bonded into a hydrogen molecule (if they are close enough). Their waveforms are constructive.
But if one flashes red while the other flashes blue, the picture is not so attractive - the individual atoms are not in sync, not bonded. If I now switch the metaphor to a musical (audio) frequency rather than a visual wavelength, the idea of wave addition or neutralization images better (see picture).

To clarify, the red and blue orbitals don't totally collapse each other and disappear - just where they might "touch". So each orbital, red and blue, still contains one (negative) electron, but there is no continuity of electron travel from and around one nucleus to the other, because there is a gap, a node, a location where there is no electron density, where they might come in contact. This represents two  atoms that have been dissociated from their bonded state as a molecule. They could have been bonded together using two red-flashing orbitals, but got excited by some energy input that brought one electron up to a blue energy which is now out of sync with the red orbital, and finds it repulsive. A $pi$ bond between two carbon atoms would show the two upper lobes of each atomic orbital to have the same color and the two lower lobes of each atomic orbital to have the other color: both upper and lower regions would be bonding. A $pi$ antibond would have a red and a blue in the upper region and a blue and a red in the lower region: repulsive!
So we usually show the direction of bonding or repulsion in bonding or antibonding molecular orbital by using + and - signs on the atomic (and molecular) orbitals, saving red and blue colors for PowerPoint presentations.


Electrolysis of water - why not pure water? In the electrolysis of water, $\ce{H2O}$ molecules are being reduced at the cathode (forming $\ce{H2}$ and $\ce{OH-}$) and oxidized at the anode (forming $\ce{O2}$ and $\ce{H_3O+}$).

It is said that:

  The electrolysis of water usually involves dilute, or moderately concentrated, salt solutions in order to reduce the power loss driving the current through the solution, but the presence of salt is not a requirement for electrolysis.

What is the reason for "power loss"? How does the presence of foreign ions change that?

The electrolysis seems possible to happen in pure water alone (as the quote also says).

The terrible conductivity of water can explain this.

Pure water is a really bad electric medium, since it has a low tendency to ionize by itself (we know the low concentrations of ions involved by looking at the Kw).

This means that the voltage needed to get a current flow from anode to cathode would be really high, with consequences both in term of safety and energetic waste.

A salt dissolved provides ions, which make the diluted solution a good electric medium, and allow a current to easily flow between anode and cathode.


Density Functional Definition Is it wrong to say that Density Functional means that Electron Density is a function of the orbitals (wave function) of all electrons in 3 dimensions, if so, why ?

In general, a functional $F$ is a mapping from an arbitrary set $\mathcal{X}$ of functions to the set of complex numbers $\mathbb{C}$ or the set of real numbers $\mathbb{R}$:
$$
F : \mathcal{X} \mapsto \mathbb{R}.
$$
or
$$
F : \mathcal{X} \mapsto \mathbb{C}.
$$

For example, if you consider $\mathcal{X}$ as the set of polynomials with real coefficients, you can define a functional $F$ as
$$
F[f] = \int_0^1f(x)\,dx
$$
i.e. your functional $F$ takes a polynomial function $f\in\mathcal{X}$ (for example $f(x)=3x+1/2$) as an argument and returns a scalar (2 for $f(x)=3x+1/2$, as you can easily verify).

A density functional is simply a functional $F[f]$ where the argument $f$ is the electron density $\rho(\vec{r})$ (i.e. a density functional is a functional of the electron density). For example Hohenberg and Kohn showed that the energy $\epsilon$ of a quantum system is a functional of the density
$$
\epsilon=E[\rho]
$$
This means that when you plug the electron density of your system $\rho(\vec{r})$ into the energy functional $E[\rho]$ you get a number $\epsilon$, which is the energy of your system. The whole energy functional is not known explicitly, but some of its components are known. For example for the external potential energy we have
$$
V[\rho] = \int v(\vec{r})\rho(\vec{r})d\vec{r}
$$
and for the Coulomb interaction between electrons we have
$$
J[\rho] = \frac{1}{2}\iint \frac{\rho(\vec{r})\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d\vec{r}d\vec{r}'
$$
which are clearly functionals of the electron density.


Does boiling water really treat it to make it safer than filtered/chlorinated water? It doesn't seem like it would, but some folks I know are adamant about boiling their water before ingesting it and the online articles I've found on this seem very subjective.

What do you mean by "safer"? If the water is pure to start, there's no point in boiling it. For mildly contaminated water, chlorination is inexpensive, but can form some toxic halogenated carbon compounds. Boiling is very energy intensive, and may require use of fossil fuels (unless you use a solar water heater). Both eliminate most microorganisms, though there are a few, mostly nonpathogenic ones, such as Deinococcus radioduran and others found at undersea volcanic vents, that can survive radiation, heat and other chemicals. Bacteria have even been found living in the common biocide, povidone-iodine! See this source for more information on biocides.

Better alternatives might be ultraviolet sterilization or ozonation.


Magnetic nature of tetraamminedichlorocobalt(III) chloride I know how to predict magnetic nature when the compound contains only strong field or only weak field ligands. But in $\ce{[Co(NH3)4Cl2]Cl},$ $\ce{NH3}$ is a strong ligand whereas $\ce{Cl-}$ is a weak ligand.

So, how do we know if the ligands are causing the electrons to pair up or not? I understand VBT and CFT to predict the hybridisation and magnetic nature of compounds.

The ammine ligand is not really a strong-field ligand. It is not a strong π acceptor which is characteristic of strong-field ligands such as the nitrosyl cation or carbonyl. Instead, it is a weak π donor. Quite unlike other strong-field ligands such as cyanide, there are a great number of high-spin complexes with ammine ligands. Ammine (and aqua) ligands are better termed intermediate field. They can stabilise low-spin complexes but they typically also support high-spin environments.

Understanding when ligands cause the electrons to pair is very non-trivial but essentially guesswork. This is exemplified by some complexes being isolated in both the high-spin and low-spin states as is the case in $\ce{[Fe(ptz)6]^2+}$ ($\ce{ptz}$ being 1-propyl-1H-tetrazol-κN4). Furthermore, the metal has even more influence on the question with higher oxidation states typically stabilising low-spin configurations while lower oxidation states often favour high-spin complexes. The only clear cases are 4d and 5d transition metals which can be safely assumed low-spin unless stated otherwise.

With that said, cobalt(III) complexes feature a combination of a high (-ish) oxidation state of the central metal combined with a $\mathrm d^6$ environment which generally favours low-spin a bit more as it allows for maximum electronic stabilisation. Therefore, cobalt(III) complexes can be guessed as low-spin if they are not entirely made up of weak-field ligands. Ultimately, however, the experiment or calculation will have to answer.


The delocalised nature of hyperconjugation I have always thought I understood hyperconjugation well until recently, in my post on Regioselectivity of bromination of alkenes, I began to question my understanding of this concept. Hyperconjugation is essentially an MO concept which can be generalised to include any sort of intramolcular orbital interaction between one set of occupied molecular or atomic orbitals with another such set of unoccupied orbitals, belonging to adjacent atoms or groups of atoms in a molecule. A most excellent introduction to this concept can be found here. This is not the IUPAC definition, but anyway, the emphasis of my question is not on the definition so there is no point being pedantic here. This term is most commonly evoked to explain the mechanism by which alkyl groups "donate" electron density to stabilise carbocations.

I am aware that there have already been multiple posts, with good answers, on this concept:

What is hyperconjugation?
Stabilisation of carbocations through hyperconjugation
Inductive effect and hyperconjugation - one elephant, different parts?
Hyperconjugation involving other bonds

In my post which discusses the nucleophilic attack on the cyclic bromonium ion intermediate, I was enlightened by Martin through one of his comments:

  I am certain there is hyperconjugation, as there already was hyperconjugation in a substituted alkene. There probably is always something to which we refer to as hyperconjugation. You are thinking in localised orbitals, which is only one viewpoint...

Just like he said, my conception of hyperconjugation had always been how Ron frequently describes it as - a localised phenomena. Specifically, in the context of stabilising carbocations, I see it as an interaction between a $\sigma$ bonding MO of the $\ce {C-H}$ or $\ce {C-C}$ bond with the empty p orbital of the carbocation. However, this comment made me think about the delocalised nature of these interactions. MO theory is supposed to be based heavily on delocalised interactions, yet we are considering rather localised interactions here. Well, it is not as localised as a 2-centre 2-electron bond, but still it is a rather localised picture. I would not like to contend this. But... I would just like to pose the following related questions below:

If hyperconjugation should be more seen as a more delocalised phenomena, would distance and geometry play such important factors? Please take note of my examples below and pay attention to them as you attempt a response at my question.

For example, if I have a tert-butyl group bonded to a carbocationic centre, would there also be (significant enough) hyperconjugative interactions between the $\ce {C-H}$ $\sigma$ bonding MOs with the empty p orbital of the carbon atom. Note that by looking at the more localised picture, such as that given by Ron, we would likely only expect the $\ce {C-C}$ bonds to be significant hyperconjugative donors. Additionally, specifically, in my case of the bromonium ion, would there be significant hyperconjugative interactions between the bonding MOs of the $\ce {C-C}$ and $\ce {C-H}$ bonds of the adjacent alkyl groups with the low-energy $\ce {C-Br}$ $\sigma$* LUMO?

Any help is much appreciated!

I'll try to answer your question through the exemples you give us.
Where are the electrons ?
If you look at the hybridized orbitals making the C-H and the C-C bonds, you can see that the more the bond is polarized, the worse electron donor you have. It's quite understandable since if the bond is polarized, the electron density will be closer to the heteroatom but further to the carbocation that needs stabilization.

Usually the donor ability goes in this order increasingly:
C-F < C-O < C-N < C-H < C-C
Hyperconjugation from C-C to carbocation
Following the "rule" stated before, the stabilization of a carbocation will be best with an adjacent C-C bond. And this is in accordence with what we learn in school since the more alkyl groups nearby, the more stable is the carbacation. Not only because of the increased overall electron density (donor inductive effect) but also because of the hyperconjugation. This is also the base of some reactivity (1,2-C-C bond migration)
Hyperconjugation in the bromonium case
We can easely extend the thing we just saw for the exemple of a bromonium intermediate. In fact it is important to dissociate the cases of the bromonium and the transition state for the bromonium openning.
The bromonium has two bonds with a carbon atom. One could expect the opening reaction to occur in competition between these two sites. We can see that the C-Br bond (Scheme 2, left) can be aligned with a C-C bond, so if we take a look at the molecular orbitals, we will see that this filled sigma C-C bond can interact with the empty C-Br sigma*. Thus creating a destabilization of the C-Br bond (filling a anti-bonding molecular orbital destabilizes the bond) trough an hyperconjugation intercation. But if we look at the second C-Br bond (right) we see that there are no such interaction since there are no filled molecular orbital which can interact with the sigma* of the C-Br bond.

In term of reactivity, the destabilization of the left-hand C-Br bond will favor the breaking of this  bond instead of the other one.


Removing ozone from a gas flow I have a potential problem in that I need to reliably remove ozone from a stream of ionized gases, mostly air. The flow rate is quite low, less than 400 sccm. I was thinking of something catalytic, maybe platinum mesh? There may well be a related need to remove any $\ce{NO_x}$ created. I am open to suggestions — it's mainly a health and safety issue if it vents into the air.

Ozonolysis is a synthetic procedure deploying diluted ozone as a reagent.  This reaction may be relevant in the context of your question, because i) during the reaction, the gas typically is bubbled across a solution (and you have some exhaust to deal with), and ii) even more importantly, after the reaction, the organic products are to be isolated safely, too.

Here, I would like to suggest to use what is otherwise described as reductive workup.  Ozone, which may dissolve in solvents, hereby is quenched (deactivated) by a material which in turn is oxidized by ozone.  Because ozone is a powerful oxidizer, the deactivation is performed while keeping the concentration of ozone entering the quench solution low, and simultaneously keeping the temperature of this solution low, too.

The provision of the flux of 400 mL/min only partly describes your problem.  It neither states the concentration of ozone in this input, nor the regime of this input -- it may be in batches, it could be a contiously working line.  With these uncertainties in hand, bubbling the ozone-containing gas input into an aqueous solution of sodium hydrogensulfite (with "i", $\ce{NaHSO3}$) seems both simple to setup, manage, and scale.  The reaction basically will yield sodium sulfate ($\ce{Na2SO4}$).

As an example, Ragan et al. published about an Ozonolysis at large industrial scale I hereby refer to.  For about 19 h, their recation was running at low temperature with a constant flow of 15-20 L/min, 4-5 wt% $\ce{O3}$ in air bubbled across their reaction vessel.  After completion of reaction, their workup reported was pouring the reaction solution slowly (over 30 min) to a prechilled quench solution of 3.0 kg $\ce{NaHSO3}$ in 6 L water at 0 degC.  Only then  they allowed the mixture to warm slowly to room temperature.

Depending on the conditions in your example, maybe you do not need to cool your sulfite solution as much, yet monitoring the temperature of the stirred quench solution is recommended.  With the parameters missing in your question, the minimal quantity of bisulfite you need might be estimated.  Equally, there are KI/starch indicator stripes, which may be used to check if the concentration of oxidizing ozone still is significant; if present, they will change from colourless to dark blue (ref, example).

Literature reference: Ragan et al., Safe Execution of a Large-Scale Ozonolysis: Preparation of the Bisulfite Adduct of 2-Hydroxyindan-2-carboxaldehyde and Its Utility in a Reductive
Amination, Organic Process Research & Development 2003, 7, 155−160 (DOI: 10.1021/op0202235).


How can you explain the energy change in this phenomenon? Tagged is the diagram of the apparatus used in this thought experiment. Pure water is taken in one side of the beaker and a solution of common salt in water in the other. After a very long time, the pure water should get transferred to the NaCl solution side because according to Raoult's law, the NaCl solution should have lesser vapor pressure than pure water, thus causing the water-vapour equilibrium in the pure water side to move forward, resulting in the condensation of the vapour in the other side (the solution side) . But how has the potential energy (=mgh) of the pure water increased during the process?

Initially suppose we have a mixture with mole fractions $x_1$ and $x_2$.   Let $x_1$ be the solute and $x_2$ the  water and assume for simplicity that  deviations from  Raoult’s law behaviour are insignificant.  In this case the heat of mixing is zero and entropy alone drives the mixing process.

In this case the free energy of mixing per mole at constant temperature is
$\Delta G =RTx_1\ln(x_1)+RTx_2\ln(x_2)$
and in as there are two components $x_2=1-x_1$ (and changing $x_1 \rightarrow x$  for simplicity ) gives
$$\Delta G =RTx\ln(x)+RT(1-x)\ln(1-x)$$
This energy is negative over the whole range, except at $x=0,1$ where it is zero. The minimum value at $x=1/2$ is $-RT\ln(2)$

After the pure water is added in the left hand compartment and equilibrium is established again, the initial mole fraction can only drop to half its value by mixing equal volumes of water which is a mixing energy of
$$\Delta G\left(\frac{x}{2}\right) = RT\frac{x}{2}\ln\left(\frac{x}{2}\right)+RT(1-\frac{x}{2})\ln\left(1-\frac{x}{2}\right)$$

Again this quantity is negative over the whole range, thus the solution will continue to be diluted if nothing prevents this.

Dilution of the mixture could, in principle, be prevented by the gravitational energy $mgh$ where m is the (molar) mass and h the height the liquid has to be raised to. To prevent further dilution
$mgh  - \Delta G(x_f)  \gt 0 $
where $x_f$ is the final mole fraction. If $mgh \rightarrow 0$ then $x_f=x/2$.

Using values for water gives $mgh = 0.176h$ Joule for height h whereas thermal energy $RT=2494$ J.  If, for example the height is $h=9800 $ m then $mgh \approx RTln(2)$ which is the minimum value of $\Delta G$ then no dilution can take place as $\Delta G \gt 0 $ for all x.

At smaller heights whether or not  dilution occurs depends on the mole fraction of the initial mixture but at a height of $1000$ m as long as the initial mole fraction is greater than $\approx 0.035$ then complete dilution will occur. Below $1000 $ m the final mole fraction is effectively half the initial value, $x/2$.

Of course the heights involved in effecting any change are quite ridiculous, but this is just because gravity is a weak force; you may not find it so if you ever go hill running :)

The figure shows different values of $\Delta G/RT$ per mole.

(a) shows the normal mixing free energy $\Delta G/RT$ vs. mole fraction solute x (b) the change upon no dilution and a potential of $1000$ m .(c) maximal dilution (doubling volume) at $1000$ m, thus dilution can occur for all initial values of molec fraction above $\approx 0.035$.(d) $\Delta G/RT$ plus $mgh = RT\ln(2)$ for water $h \approx 9800$ m. No transfer of water can occur in this case as $\Delta G \gt 0$


Functional Isomer of Formic Acid is? I have read like ''Functional Isomers of Carboxilic Acids are Esters'', then my doubt is; What about Formic Acid?

Note the image of acetic acid (top) and the image of the carboxylic structural isomer methyl formate (middle).  As you can see the bonding location of the proton and the methyl group switch between the structural isomers.  Note the image of formic acid (bottom).  If you switch a hydrogen and a hydrogen you still get formic acid.  Yes, structural isomers of carboxylic acids make esters, but formic acid does not have a structural isomer, so the rule does not apply.

 Acetic Acid
 Methyl Formate
 Formic Acid


crystallinity of seed crystals Does the crystallinity of seed crystals affect the final crystal structure? and is it different depending on the crystallized material?

In my field, protein crystallography, we often use seeding to improve our crystallization experiments. Typically seeding speeds up crystallization by introducing well defined nucleation cores for crystal formation.

It can often be observed that seeding influences the crystal formation, as crystals grown from seeded crystallization conditions often have better diffraction. This is very important as all our crystals must be formed in their native solvent enviroment (water, lipids/detergents, ions etc).

Sometimes the act of seeding can influence the crystallographic symmetry, it is not uncommon for seeding to cause crystallographic symmetry to turn into non crystallographic symmetry. Personally I have had P12(1)1 crystals generate P1 crystals due to small changes in the asymmetric unit. However my case were also rather extreme as seeding cut my crystallization time from ~3-6 months to 3-6 days.

Also note that crystal seeds, does not necessarily have to be crystalline in nature. Small glass fragments, dust particles and hairs, can also induce crystal growth (I have personally seen examples of all three on a project I have worked on). This probably occurs when the protein being crystallized is highly stable, but cannot form strong inter-molecular interactions in the crystal packing. Then having a foreign surface to interact with may help keep the protein molecules close together and allow weaker indirect interactions (such as hydrogen bonds via water molecules) to form and stabilize the crystal structure.

So in my field, seeding can have an enormous impact on crystallization experiments.


Synthesing tetrahydropyran from ethylene oxide I thought that the first step must be using a Grignard reagent to add carbons, but I need a hydroxyl group on the carbon on the other side to make it a diol and produce the product given, but nothing seems to be working.
How do I proceed?

Prepare the tetrahydropyran ether of commercially available 3-bromopropanol using dihydropyran, catalytic PTSA in DCM.

Make the Grignard of the THP-protected bromopropanol, react this with ethylene oxide. Work up and isolate to give the mono-THP-pentane-1,5-diol.

Make the tosylate or mesylate of the mono-THP-pentane-1,5-diol.

Remove the THP group using cat. PTSA in MeOH

Form the alkoxide using LiHMDS in Et2O and cyclisation will occur to give the pyran. Isolate by careful distillation.


Enantioexcess for a non-chiral molecule? I am trying to understand this quote in a section on "Biotransformations of Organic Molecules" from a handbook I am reading:

[...] Euglena gracilis could convert acetophenone to 2-phenylethanol; however, its enantioexcess is very poor (10%) (Takahashi, 1994).
Ref.: Takahashi, H., 1994. Biotransformation of terpenoids and aromatic compounds by some microorganisms. PhD thesis. Tokushima Bunri University, pp. 1–115.

What does "enantioexcess" mean in this context? Neither of those molecules are chiral, right?  I cannot spot any asymmetric C atoms. Nor any higher order symmetries.
I looked up Wikipedia entries on both molecules and neither mention any d / l isomers.
What am I missing?

The Ph.D. thesis that is cited is probably near-impossible to obtain and may very well be in Japanese, which certainly doesn't help.

2-Phenylethanol is not a chiral compound, so as Ivan pointed out, the concept of enantiomeric excess (ee) makes no sense. The fact that the direct reduction product of acetophenone (e.g. via hydride transfer) is 1-phenylethanol strongly suggests that this is nothing more than a typo. However, I thought that it might be fun to go to great lengths to verify the applicability of Occam's razor to this situation.

A SciFinder search for "acetophenone Euglena gracilis" did throw up one Chinese patent, CN 103757061. There is a translation on Google Patents. The patent describes the reduction of various prochiral ketones with various microphytes (also known as microalgae) to form chiral alcohols, with near-perfect enantioselectivity.

I managed to figure out the technical terms in the patent. E. gracilis in Chinese appears to be 纤细裸藻 (xian1 xi4 luo3 zao3). Acetophenone is 苯乙酮 (ben3 yi3 tong2). The patent does not, in fact, describe the reduction of acetophenone by E. gracilis, and the text in Table 2 is so small that I cannot quite make out what compounds they tested (they seem to be aliphatic). Perhaps they simply did not include this combination in the patent if the enantioselectivity they obtained was very low.

However, there are experiments with different species that lead to the reduction of acetophenone to "S-苯乙醇" (ben3 yi3 chun2). This translates to (S)-phenylethanol, and leaves open the possibility of it being either 2-phenylethanol or 1-phenylethanol. Since only the 1-phenyl isomer is chiral, I assume the reduction product is (S)-1-phenylethanol.

This is supported by Figure 1 in the patent. I have scribbled some loose translations. In hindsight I think "Photosystem" should be "Light-harvesting complex". The text translates to "Light-harvesting system" which is a bit of a mix and match of the two, so I wasn't sure what to go for.

Having $\ce{R_1 = Me}$ and $\ce{R_2 = Ph}$ would mean acetophenone is reduced to 1-phenylethanol.

In conclusion, I would therefore assume with confidence that the handbook contains a typo and the authors meant 1-phenylethanol.

(On a side note, the Chinese also got >95% ee for all their acetophenone reductions, go figure...)


How to convert an oxirane to an alkanol with one additional carbon atom? This is a question given in our exam.
In the reaction,

Find the reagents X and Y required for steps(ii) and (iii) are _______ and ________ respectively.
I have tried in this way but I have stuck in second step. The given compound reacts with PhCO3H to form an epoxide.
Am i correct? Then please help me continue the reaction.
The answer given is CH3MgBr and H3O+.

You're right: The reaction of the alkene with perbenzoic acid forms an oxirane.

My suggestion for the next step is:

Draw the epoxide.
Draw the final product.
Map the atoms: Count and assign the carbon atoms. Do you spot any difference?

Supposed that the oxygen atom of $\ce{OH}$ group in the final alkanol is the same atom introduced by the action of perbenzoic acid, the oxirane was apparently opened.

Which general ways to open an oxirane exist?

Try to think in general synthons first.

One of the two steps in the sequence might just be a neutralization, quenching or workup step.

UPDATE

With a bit of drawing, atom counting and mapping, you will have realized  (by now) that:

The final alkanol has one carbon atom more than the oxirane intermediate.
The additional methyl group was added to the less substituted side of the oxirane.

Now, try to think in synthons, such as "$\ce{CH3+}$" and "$\ce{CH3-}$", rather than in reagents.

Where would a "$\ce{CH3+}$", thrown at the oxirane, add? Probabaly at the centre of high(est) electron density. This is the oxygen atom, and thus the wrong position. Apparently, "$\ce{CH3+}$" is not your synthon.

What about "$\ce{CH3-}$"? It's a strong nucleophile that would preferably attack centres of low(er) electron density, such as carbon atoms with electron withdrawing substituents. You have two of these, namely the two carbon atoms that constituted the $\ce{C=C}$ double bond and which are now in your oxirane. If "$\ce{CH3-}$" attacks one of these carbon atoms, the oxirane is cleaved in such a way that one $\ce{C-O}$ bond is broken and the oxygen atom gets the negative charge. Note that sterical hendrance comes into play here too.

In summary,"$\ce{CH3-}$" seems to be the right synthon. Now, you have to translate the synthon, "$\ce{CH3-}$", into a reagent that you can either find in the lab or make in situ. In order to get the negative charge on the carbon atom, it might be a good idea to attach an electropositive to it.

Consequently, a Grignard reagent, $\ce{CH3MgBr}$ is a good choice.

As I mentioned initially, while trying not to spoil the fun of solving, the last step is just an aqueous, acidic workup to release the alkanol ;-)


Solubility of potassium nitrate and potassium chloride Why more water is needed to dissolve potassium chloride than potassium nitrate of equal amount? Is it to do with the reaction that occurs when the compounds dissolve or the properties of the compounds?

This is for an experiment on the effect of temperature on solubility in various chemicals.

Recently, I answered a similar type of question. The reasoning for that question is also applicable for this answer also.

Electrostatic force between potassium cation and chloride ion is way stronger than that with nitrate ion. So, more energy is needed to break the  bond between potassium chloride than potassium nitrate. Hence, lattice enthalpy is more in case of potassium chloride and thus more water is needed to dissociate the bonds. So, potassium chloride is less soluble in water.

You can compare the solubilities of potassium nitrate and potassium chloride by observing the graph taken from this site:

You can observe that solubility of both potassium chloride and potassium nitrate is same at approximately 23℃ but then as temperature is increased, solubility of potassium chloride increases steadily but solubility of potassium nitrate increases drastically.

So solubility order:

Potassium nitrate > potassium chloride


What are Mulliken atomic charges, and how are they computed? I was reading about the self-consistent extended Hückel method (SC-EHT), and stumbled upon various formulae for the diagonal elements of the Hamiltonian matrix. I noticed that many formulae depend on either the gross atomic population or Mulliken atomic charges. I know the formula for the Mulliken population analysis, but what is the Mulliken atomic charge and how can I evaluate it?

Thanks!

Mulliken atomic charges can be defined as[1]:

$$q_A = Z_A-\sum_{\mu\in A}\left( \mathbf{P\cdot S} \right)_{\mu\mu} \tag{Szabo 3.196}$$

Here I have used the same notation as in Szabo[1], with $\mathbf{P}$ being the density matrix and $\mathbf{S}$ beign the overlap matrix. $Z$ is the nuclear charge, and using the greek letter $\mu$ as indicies indicates that we are working in atomic orbital basis (not in in molecular orbital basis).

The sum runs over $\mu\in A$ meaning, that we only consider atomic orbitals that are centered on the $A$th atom.
we can therefore note that Mulliken charges are only defined when we used atomic-centered basis functions (which we most often do).

[1] : Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Attila Szabo and Neil S. Ostlund


Why is $1s-1s$ orbital overlapping stronger than $2p_x-2p_x$ orbital overlapping? Assume the x-axis to be the internuclear axis. Hence, both $1s-1s$ and $2p_x-2p_x$ will form $\sigma$ bonds.

Also consider the fact that $p$ orbitals have more directional characteristics.

Given the above facts, shouldn't the $2p_x-2p_x$ overlapping be more effective than $1s-1s$ overlapping? Why is $1s-1s$ orbital overlapping stronger than $2p_x-2p_x$ orbital overlapping?

Your question is hinged on a false premise. On a deeper level, the premise makes no sense at all and can't even be classified as true or false. If a book says otherwise, then maybe it contains some very specific piece of context that would turn the situation upside down and make everything fall in place. Or maybe the book just says nonsense; such things happen, after all.

To begin with, orbital overlap depends on the internuclear distance more than slightly. Now, $1s$ orbitals are a good deal smaller than $2p$ (not like "you vs Dwayne Johnson", but more like "you vs an elephant"). Naturally, at a distance where $2p_x$ orbitals are in their best overlap, the $1s$ orbitals barely feel each other. Likewise, if the distance is good for $1s$, it is way too short for $2p_x$ and makes them feel pretty uncomfortable. So you see that considering one of these overlaps to be universally stronger than the other is plain untrue.

There is more to it, however. See, the $1s$ and $2p$ orbitals of the same atom are pretty far apart in energy. They never mix. In other words, they can never be important at the same time.

If both your atoms are $\rm H$, then their $1s$ orbitals overlap and make a good strong covalent bond, while the $2p$ orbitals are empty and hence effectively don't exist.
On the other hand, if both your atoms are (say) $\rm O$, then their $2p_x$ participate in bonding (together with other $2p$'s and $2s$), while the $1s$ orbitals are pretty much kept apart. What's more important, both $1s$'s are fully occupied, hence whatever little interaction they have will end up producing a bonding and an antibonding orbital which are also both occupied, thus negating the effect.

At this point you might be wondering what if one of your atoms is $\rm H$ and another $\rm O$, but that's a story for another day.

So it goes.


What is the difference between "molecular mass", "average atomic mass" and "molar mass"? I don't understand the difference between "molecular mass" and "average atomic mass". They seem like the same thing to me. Is it that average atomic mass is just the weighted average of the "weights"/masses of the isotopes whereas the molecular mass is the average of "weights"/masses of the average atomic masses of each element in the molecule.

Also, what's the difference between molar mass and molecular mass? Is it just that molar mass is expressed in Daltons and molecular mass is expressed in g/mol?

The hardest part about chemistry is keeping track of which people are using which terms, and which terms are outdated. Is this one of these "oh, we use this term now," type of thing?

Atomic mass refers to the average mass of an atom. This has dimensions of mass, so you can express this in terms of daltons, grams, kilograms, pounds (if you really wanted to), or any other unit of mass. Anyway, as you said, this is an average of the masses of the isotopes, weighted by their relative abundance. For example, the atomic mass of $\ce{O}$ is $15.9994~\mathrm{u}$. $\mathrm{u}$ is short for unified atomic mass unit and 1 u is equivalent to $1.661 \times 10^{-24}~\mathrm{g}$. It is exactly the same as the dalton, but from what I've seen, the term dalton is used more when discussing polymers, biomolecules, or mass spectra.
Molecular mass refers to the average mass of a molecule. Again, this has dimensions of mass. It's just the sum of the atomic masses of the atoms in a molecule. For example, the molecular mass of $\ce{O2}$ is $2(15.9994~\mathrm{u}) = 31.9988~\mathrm{u}$. You don't need to calculate the relative isotopic abundance or anything for this because it's already accounted for in the atomic masses that you are using.
The term molar mass refers to the mass per mole of substance - the name kind of implies this. This substance can be anything - an element like $\ce{O}$, or a molecule like $\ce{O2}$. The molar mass has units of $\mathrm{g~mol^{-1}}$, but numerically it is equivalent to the two above. So the molar mass of $\ce{O}$ is $15.9994~\mathrm{g~mol^{-1}}$ and the molar mass of $\ce{O2}$ is $31.9988~\mathrm{g~mol^{-1}}$.
Sometimes you may come across the terms relative atomic mass ($A_\mathrm{r}$) or relative molecular mass ($M_\mathrm{r}$). These are defined as the ratio of the average mass of one particle (an atom or a molecule) to one-twelfth of the mass of a carbon-12 atom. By definition, the carbon-12 atom has a weight of exactly $12~\mathrm{u}$. This is probably clearer with an example. Let's talk about the relative atomic mass of hydrogen, which has an atomic mass of $1.008~\mathrm{u}$:
$$A_\mathrm{r}(\ce{H}) = \frac{1.008~\mathrm{u}}{\frac{1}{12} \times 12~\mathrm{u}} = 1.008$$
Note that this is a ratio of masses and as such it is dimensionless (it has no units attached to it). But, by definition, the denominator is always equal to $1~\mathrm{u}$ so the relative atomic/molecular mass is always numerically equal to the atomic/molecular mass - the only difference is the lack of units. For example, the relative atomic mass of $\ce{O}$ is 15.9994. The relative molecular mass of $\ce{O2}$ is 31.9988.
So in the end, everything is numerically the same - if you use the appropriate units - $\mathrm{u}$ and $\mathrm{g~mol^{-1}}$. There's nothing stopping you from using units of $\mathrm{oz~mmol^{-1}}$, it just won't be numerically equivalent anymore. Which quantity you use (mass/molar mass/relative mass) depends on what you are trying to calculate - dimensional analysis of your equation comes in very handy here.

Summary:

Atomic/molecular mass: units of mass
Molar mass: units of mass per amount
Relative atomic/molecular mass: no units

A small (and unessential) note about the definition of the $\text{u}$. It's defined by the $\ce{^{12}C}$ atom, which is defined to have a mass of exactly $12 \text{ u}$. Now the mole is also defined by the $\ce{^{12}C}$ atom: $12 \text{ g}$ of $\ce{^{12}C}$ is defined to contain exactly $1 \text{ mol}$ of $\ce{^{12}C}$. And we know that one mole of $\ce{^{12}C}$ contains $6.022 \times 10^{23}$ atoms - we call this number the Avogadro constant. That means that $12 \text{ u}$ must be exactly equal to $(12 \text{ g})/(6.022 \times 10^{23})$, and therefore,
$$1 \text{ u} = \frac{1 \text{ g}}{6.022 \times 10^{23}} = 1.661 \times 10^{-24} \text{ g.}$$


Why can anthracene, but not phenanthrene, take part in Diels–Alder reactions? Both anthracene and phenanthrene seem to be structurally similar, with three aromatic rings. The middle rings in particular seem to be similarly activated towards cycloadditions.

Have a look at the reaction mechanism of the Diels-Alder Reaction, e.g. at Wikipedia.
We have talked about this reaction before on this site, and concluded that a sufficient explanation is only possible with the help of molecular orbitals: How accurate is this polar mechanism for the Diels-Alder reaction?
Borrowing from Wikipedia,[1] here is the most important part of the mechanism, the HOMO of the diene and the dienophile:

Now the most important part of this picture is the symmetry of the HOMO/LUMO of the diene, since this is the subject of your question. We need either a HOMO or a LUMO of these symmetries and then could match a suitable dienophile.

In anthracene we find the appropriate symmetry of the HOMO:

The reason for the regioselectivity has also been discussed before on this site: Diels Alder with Anthracene and Maleic anhydride.
We could also tune our reaction to be of inverse demand, as the LUMO has also appropriate symmetry:

In phenanthracene we do not have the appropriate orbital symmetry in the HOMO:

The LUMO is also not suitable for an inverse Diels-Alder reaction.

Reference and Notes

"FMO of Diels-Alder reaction" by Chrito23 - Own work. Licensed under CC BY-SA 3.0 via Commons.
Calculations on the DF-BP86/def2-SVP level of theory employing Gaussian09 rev. D. Contour values for the orbitals are at 0.05 e/rBohr. Occupied orbitals are depicted in blue and orange and unoccupied orbitals are depicted in yellow and red. Phases are chosen at random.


What is spin spin coupling? I am studying nmr and the concept of multiplicity is very hard to undersand. Specifically I wanted to know where this arises form. Can anyone explain the concept of spin spin coupling in detailed and in simple terms?

As it stands, your question is very general. It is unlikely that a single answer here will describe spin coupling and multiplicity in a simple and detailed manner any better than any publicly available reference source. There are hundreds available. I suggest you read one or more of the following extremely useful introductory NMR references, and ask more specific questions that you do not understand from them.

Reich (http://www.chem.wisc.edu/areas/reich/chem605/)
Hornak (http://www.cis.rit.edu/htbooks/nmr/)
Libretexts (https://chem.libretexts.org/Core/Organic_Chemistry/Spectroscopy/Nuclear_Magnetic_Resonance_Spectroscopy
and
https://chem.libretexts.org/LibreTexts/Purdue/Purdue%3A_Chem_26200%3A_Organic_Chemistry_II_(Wenthold)/Chapter_12._Nuclear_Magnetic_Resonance)


Find the uncertainty in position The question says:

A proton is accelerated to one tenth of the velocity of light. If it's velocity can be measured with a precision $\pm 1\%$. What must be its uncertainty in position?

Therefore,
$v=0.1 \times c =3\cdot 10^7\ \pu{m/s}\\
 \Delta(v)=\frac{1}{100}\\
 m= 1.6\cdot10^{-27}\ \pu{kg}$
Then I directly substituted these values into the formula:
$$\Delta(v)\cdot\Delta(\text{position})\cdot \text{mass}=\frac h{4\pi}$$
To get the uncertainty in position, however, the answer I got was approx. $3.5 \cdot 10^{-6}\ \pu{m}$ which is way too different from the correct answer:

$0.5\cdot10^{-13}\ \pu{m}.$

Can anyone please explain me how to solve this question?

The question tells you
\begin{aligned}
&&v&=3\cdot10^7\:\mathrm{m/s}\pm1\%\\
&&\Delta v &= |(v+1\%v) - (v-1\%v)|\\
\implies&&\Delta v &= \frac{2}{100}v\\
\therefore&&\Delta v&= 2\cdot3\cdot10^5\:\mathrm{m/s}
\end{aligned}

Hence
\begin{aligned}
&&\Delta x\cdot\Delta p &\geq \frac{\hbar}{2}\\
\implies&&\Delta x\cdot\Delta (v\cdot m) &\geq \frac{\hbar}{2}\\
\implies&&\Delta x&\geq \frac{\hbar}{2\cdot\Delta v\cdot m }\\
\therefore&&\Delta x &\gtrapprox\frac{1.05\cdot10^{−34}\:\mathrm{J\cdot s}}{2\cdot6\cdot10^5\:\mathrm{m/s}\cdot1.6\cdot10^{−27}\:\mathrm{kg}}\\
&&\Delta x&\gtrapprox5.5\cdot10^{-14}\:\mathrm{m}\\
\end{aligned}


Does Copper reacts with Sodium Sulfate solution? Okay I am trying to make a capacitor in which current collector is copper foil and electrolyte is sodium sulfate. Would this work? Do these two react with each other or not?

If you're trying to make an electrolytic capacitor, you've picked inappropriate materials.

An electrolytic capacitor works because a material such as aluminum or tantalum develops a tightly-adherent insulating oxide layer when immersed in certain solutions and "formed". Copper does not have an adherent insulating oxide; instead, cuprous oxide is a semiconductor and $\ce{Cu2O}$ can be used to make a rectifier.

You can make your own aluminum electrolytic capacitor with common household chemicals.


Simple way to understand why chromophores often have double bonds? Chromophores — atoms or groups of atoms within a molecule that absorb some visible wavelengths better than others — result in compounds that have color. For example, see this answer about azo dyes, and this mention of bleaching stains by oxidation of double bonds.

Without going too deep into quantum mechanics and/or molecular orbital theory (MO), is there a way to understand in a qualitative way why chromophores are often related to the presence of double carbon or double nitrogen bonds?

No. Without going into quantum mechanics and molecular orbital theory, there is no way to understand why chromophores typically have double bonds.

Everything related to colour is related to the excitation of electrons from one orbital to another and therefore any explanation that does not rely on orbitals in at least some way is doomed to fail.

If you decide to accept a little MO theory then you can be helped. Double bonds are formed by an underlying σ bond (or single bond) and a π bond. Each bond is formed by two atomic orbitals (one from each atom) that combine to give a bonding and an antibonding orbital. The energy difference between the bonding and antibonding orbital corresponds to the excitation energy: a photon that carries exactly that energy will be absorbed and the electron excited to the higher level. The greater the overlap between the atomic orbitals, the higher the energy difference between bonding and antibonding.

Now in π orbitals, the atomic p orbitals are oriented perpendicular to the bond axis, so any overlap is doomed to be much smaller than the corresponding single bond’s σ overlap. Therefore, the energy difference from the bonding to the antibonding π orbital is not far into the ultraviolet region; typical values would correspond to a wavelength of $150$ to $200~\mathrm{nm}$. Extending the π network will move those orbitals closer together energetically (again, the complete answer requires much more MO theory) which is why benzene has an energy difference corresponding to a wavelength of $250~\mathrm{nm}$. Thenceforth, it is not a long way until you reach visual wavelengths ($400{-} 700~\mathrm{nm}$).


Which one is the stronger base? I'm studying Wolff-Kishner reduction and I don't understand why after this step

the leaving group is $\ce{OH-}$ instead of $\ce{^-HN-NH2}$. I guess it's because hydroxide ion is less strong than the other although it would be useful to know if i can calculate pkb of diazanide (namely $\ce{^-NH-NH2}$) starting from data on hydrazine. Also I'm considering pkb of $\ce{OH-}$ equal to $0$ because it's acting as a base.

Mirko: The mechanism you present for the early stages of the Wolff-Kishner Reduction includes hydroxide necessary for the decomposition of the yet to be formed hydrazone 6 of the ketone 1 (or aldehyde). One may preform the hydrazone and then do the actual reduction in the presence of hydroxide in a high boiling hydroxylic solvent (Huang-Minlon variation1).

The mechanism for hydrazone formation is a reversible reaction and is often conducted by removal of water to shift the equilibrium to the right.2 Realizing that most commercial hydrazine contains 15% water, it can act as the agent for proton exchange in adduct 3. One need only use the electron pair in structure 4 to eject hydroxide contrary to the elimination proposed in the last step of your mechanism. There is no need to be concerned about the anion of hydrazine as a leaving group. Hydroxide only has to deprotonate the acid 5.

1) Huang-Minlon, J. Am. Chem. Soc., 1946, 68, 2487.
2) Hydrazones and oximes are not as susceptible to hydrolysis as are imines owing to the resonance stabilization provided by the second heteroatom.


Do bonding and anti-bonding orbitals exist for ionic bonds? Is there any concept of bonding and anti-bonding orbitals for ionic bonds, as there are in case of covalent bonds?

OR is it possible to break an ionic bond by pushing electrons into the antibonding orbitals?

I know that this question may sound stupid to some, but please excuse my ignorance.

Bonding and antibonding orbitals in a homonuclear diatomic molecule  A-B should mix due to polarization to give rise to pair of atomic orbitals with delocalized tails  for an ionic heteronuclear diatomic molecule. Now suppose A more electronegative and is the rightful anion and B the cation. The tailed atomic orbital of A is the bonding orbital(ie. A:- B°+) and the one belonging to B is the antibonding (ie. A°+ B:-). See original paper on Coulson Fischer theory.


Why are fluoride ions (and ions in general) reactive? Please bear with me—-I have a pretty meager background in chemistry but am trying to learn some!

My question is about ion reactivity—-specifically of the reactivity of $\ce{F-}$.
I often hear that the main reason the “weak” acid $\ce{HF}$ is so dangerous is because of the fluoride ion rather than the $\ce{H+}$. The impression I’ve gotten is that fluoride is the real hit man in this situation, with the $\ce{H+}$ only really serving to corrode barriers to entry.

A search for “fluoride ion reactive” yields the following sentence: (source)

  Unlike other hydrogen halide acids, the anion (fluoride ion) is quite reactive, and can form fairly insoluble salts with alkaline earth metals such as calcium and magnesium.

However, I think I’m fundamentally failing to grasp something. Why is this insoluble salt effect specific to fluoride? It’s obvious why fluorine is reactive, but why is the octet-satisfied fluoride reactive as well? And how does fluoride reactivity compare with the other halides?

I have also heard people say that fluoride is a bad nucleophile, which makes sense given how tightly it holds on to its electrons. So I can’t quite wrap my brain around why fluoride would want to react with anything at all. Wouldn’t that just force it to share its extra electron with another atom? (Is there such a thing as “electronegativity” but for ions rather than atoms?)

After thinking about it, I realized my confusion extends to all ions in general. Why would “fully dressed” ions with full octets do anything but sit around on the couch eating Cheetoes?

There’s definitely something stupidly fundamental that I’m missing — hopefully someone can point me in the right direction!

Fluoride activity, or more properly danger, comes from the fact that formation of some salts of fluoride is energetically highly favorable for aforementioned metals.

Calcium for instance, rather than in calcium carbonate, is more stable in $\ce{CaF2}$ form. Then, consequently, it will etch your bones eventually if there is a dermal contact with $\ce{HF}$. Other acids like sulfuric acid make a superficial effect and with a copious amount of water you mostly get rid of it. $\ce{HF},$ on the contrary, is a weak acid and slowly seeps through your epidermis to your bone.

$\ce{HF}$ is also reactive with glass, through silicon. It may also etch the beaker or any other container through time so you don't want to store $\ce{HF}$ for a long time in glass containers, actually same with bases to a lesser degree.

$\ce{F}$ can even react with noble gases, you should just look at the last product and compare the overall stabilities of the current form and the final form. Fluoride simply more stable in $\ce{CaF2}$ solid than in solution.


Acidity order checking for α-furoic acid and acetic acid α-Furoic acid's conjugate base is aromatic and hence is more stable. Then why acetic acid is more acidic than α-furoic acid?

The conjugate base of 2-furoic acid is aromatic, but so is the acid.
'

The aromaticity does not contribute to the stability of the conjugate base because the negative charge from the carboxylate anion cannot delocalize into the ring. The major resonance contributors for the carboxylate anion are:

However, because of the oxygen atom, furan is an electron-rich arene, and one position that has higher than expected electron density is the 2-position:

Furan is therefor electron-donating, and electron-donating groups destabilize anions. Consider the following resonance contributor of 2-furoic acid and the close proximity of the formal negative charges:


Only d orbital electrons for spin only magnetic moment During my chemistry lessons and self studies, I have come upon many instances where I observed that during calculations of spin only magnetic moment of transition elements only the d orbital electrons are to be counted.  To gain a deeper insight into these instances of "calculations", I would like to expand my question as follow around the same point (i.e. spin only magnetic moment):

Is spin only magnetic moment calculated only for ions? If not, what would be it for $\ce{Cr^0}$: I mean would the $\mathrm{4s^1}$ electron be taken into consideration? If yes, then why?
Are there some other instances apart from $\ce{Cr^0}$ which supports your views?

Any unpaired electron is taken into consideration while trying to quantify spin only magnetic moment.
valence electrons can be unpaired (not always), and inner shell electrons are paired so the latter don’t contribute (since they are fully filled). That’s why usually only the outer shell electrons are considered.
and no it’s not calculated for ions only. You can use it for individual atoms too!
In your example $\ce{Cr^0}$, you can certainly consider that one unpaired electron in the s orbital,

so giving a total of 6 unpaired electron it’s magnetic moment would be $\sqrt{48}$

By using the spin-only formula $M =\sqrt{n(n+2)}$  where  $n$ is the number of unpaired electron and $M$ has the units of magnetic moment.

Caution this formula is applicable for d block elements only.

Another example would be niobium, molybdenum it also has one unpaired electron in the 5s orbital and certainly we consider that electron while quantifying  spin only magnetic moment.

Furthermore, the spin only magnetic moment of molybdenum is very close to that of chromium.

I am telling very close but not same because theoretically calculated magnetic moment don’t usually coincide with experimentally calculated they differ ever so slightly

Sources  -

NCERT Chemistry for class 12 (pg no 221)
Descriptive Inorganic Chemistry by James E. House and Kathleen E. House by Elsevier. 2nd edition (pg no 329)
Shriver and Atkins Inorganic Chemistry, 5th edition (pg no 478)


How is volume a State function? My question as stated above, is how Volume a state function? As described by the  definition of a state function, that it is independent of the path adopted to bring about a change in it, how can volume be one, wouldn't volume change differently if we brought about a change in it by say increasing/decreasing temperature rather than say pressure?

Thanks

PS:Please explain in simple terms, as i am beginner in thermodynamics.

Thanks again.

As described by the definition of a state function, that it is
  independent of the path adopted to bring about a change in it [...]

This is not the definition of a state function, since it is missing the crucial first part of it. Here is the definition: a state function is a property of a system that depends only on the current equilibrium state of the system, not on the way in which the system acquired that state.

The point is that something is a state function if you can tell its value by knowing just the equilibrium state of a system and nothing else. As Chester Miller said in his comment: tell me the temperature and pressure (i.e. specify the state of a system), and I'll tell you the volume. It is irrelevant how the system acquired that equilibrium state, the volume is determined by just the state.

  Wouldn't volume change differently if we brought about a change in it
  by say increasing/decreasing temperature rather than say pressure?

In general, of course, when changing the system differently, its volume will also change differently. But, if by doing different changes you eventually will reach the same equilibrium state, all the state functions (and not just volume) will have the same values.


What (stored-as-solids) acids/liquids can clean corroded copper? I'm looking for a liquid that can remove copper corrosion (oxides and basic carbonates - the black and green stuff) off copper piping.
The corrosion will be located on the inside of the pipe, in hard-to-reach areas, so ideally I should be able to just pour the liquid in, leave it for a while, and just flush it out with water. It will be practically impossible to scrub the corroded areas.
Additionally, the piping is brazed or soldered together with an unknown filler metal.
I've read around, and it seems people commonly use acetic acid (vinegar) or hydrochloric acid for this. Is this correct?
However, I'm interested in if any acids (or other liquids) that can be stored as solids can be used for this instead. Mainly because it's easier to store and transfer solids than liquids. (As far as I'm aware, acetic and hydrochloric acid are only readily available in liquid forms)
Some common solid acids that seem to be effective on rust (iron oxide) that I can buy off Ebay include citric acid, oxalic acid, and sulfamic acid. But will these also be effective on copper corrosion, and keep the byproducts in solution?
Additionally, will I need to use any corrosion inhibitors, to prevent the acid for eating away from the raw metal itself? If so, what should I use, and what acids are they compatible with? I'd like to avoid anything that would leave phosphates/silicates/etc coated on the metal.
TL;DR: What (stored-as-solids) acids/liquids can clean corroded copper, that will leave the byproducts in solution? Do I need to use any corrosion inhibitors?
Note: I do not have access to chemical suppliers, so only things commonly available on Ebay and the like please. Additionally, I do not have fume hood, so no toxic fumes too.

Per a source to generally clean copper metal surface, to quote:

A few hard rainstorms may clean the surfaces sufficiently to start operations. However, it is always advisable, particularly if the coloring is to be done immediately after installation, to go over the surface with a commercial chemical metal cleaner. Cleaners of the trisodium phosphate type should be satisfactory. Avoid cleaners which leave a coat of oxide on the copper surface.

However, assuming one has already a basic copper carbonate, the application of an acid, like HCl or more friendly acetic acid, is likely suitable, especially with the addition say NH4Cl or NaCl to increase the activity coefficient of the acid (so less strong acid may be employed).
However, a review of the cited source notes that these reagents are the very agents suggested to create a patina coating (which can be characterized as electrochemical based corrosion) occurring in the presence of air/oxygen.
So, do limit air/oxygen exposure by immersing the copper surface to be cleaned in say a HCl/NH4Cl mix.
Also, employing the concept of a sacrificial anode, experiment with performing the latter cleaning in the presence of a more anodic metal (say a small piece of Aluminum metal in acetic acid and NaCl), especially when the requirement of limiting air exposure could be difficult to implement. This may prove to be actually beneficial in avoiding surface
oxidation.