Supposed low temperature of calcium carbide synthesis I came across an interesting reaction on allreactions.com under silicon tab:
$$\ce{SiO2 + 5 C(coke) + CaO ->[\pu{800–1000 °C}] Si + CaC2 + 3 CO}$$
Unfortunately the website doesn't provide sources and I wanted to read further about this reaction, particularly about the reaction time. If anyone here could point me to another source I'll be very grateful.

Liu et al. in 2012 [1] showed experimental data that lead me to believe this to be inaccurate. Two relevant informations from their findings:

$\ce{SiO2}$ and $\ce{Al2O3}$ do not react with coke at temperatures lower than $\pu{1460 °C}$, but they do react with $\ce{CaO}$ to form $\ce{Ca3SiO5}$ and $\ce{Ca2Al2O5}$, respectively […]
Note that the behaviors of $\ce{SiO2}$ and $\ce{Al2O3}$ observed in this work do not agree with some literature, which reported that both $\ce{SiO2}$ and  $\ce{Al2O3}$ reacted with coke at temperatures of $\pu{∼ 1200 °C}$.

Also, in Fundamentals of Metallurgical Processes: International Series on Materials Science and Technology [2, p. 174] it is written that calcium carbide is not stable in the presence of silicon and undergoes the reaction:

$$\ce{2 SiO2 + 2 C + CaC2 <=> Ca + 2 Si + 4 CO}$$

References

Li, G.; Liu, Q.; Liu, Z. $\ce{CaC2}$ Production from Pulverized Coke and $\ce{CaO}$ at Low Temperatures—Reaction Mechanisms. Ind. Eng. Chem. Res. 2012, 51 (33), 10742–10747. DOI: 10.1021/ie300671w.
Coudurier, L.; Hopkins, D. W.; Wilkomirsky, I. Fundamentals of Metallurgical Processes, 2nd ed.; Hopkins, D. W., Ed.; International series on materials science and technology; Pergamon Press: Oxford [Oxfordshire]; New York, 1985; Vol. 27. ISBN: 978-0-08-032536-1, eBook ISBN: 9781483162409, DOI: 10.1016/C2013-0-03773-5.


What are the properties of alloys? You spend a lot of time in chemistry talking about molecular amd ionic compounds, but I'm interested in what you can do with metals only. What is the difference between the properties (physical & chemical) of pure metals and those of alloys? Is it possible to predict the properties of alloys before making them? Can you make alloys with mercury?

What seems to be a set of simple questions leads one into the field of metallurgy, computational thermodynamics, mechanical properties, and goodness knows what else.

For example, steel has very different mechanical properties from pure iron.  There are about a bazillion different kinds of steel, each with different impurities mixed in to give specific mechanical properties, optimizing strength, fatigue life, corrosion resistance, etc. Much of the development has been through insight gained by experience.  More recently, the CALPHAD approach is giving more fundamental insight through thermodynamic modeling, but predicting the mechanical properties $\it a\ priori$ is still difficult to impossible.

As for mercury, there are a large number of binary phases of Hg with other elements.  These are readily found in binary alloy references, or online at the ASM Alloy Phase Diagram Database (assuming your library has access).  ASM has info on >60 Hg binary diagrams, and a number of ternary systems as well.


How to recast Henderson–Hasselbalch when fitting data on a chemical shift vs. pH graph for NMR I've just been getting into organic and been reading about H-H related to $\mathrm{NMR}$. I've been seeing articles state how recasting of H-H is necessary and want to know how equations like below are derived.

$$σ_\mathrm{peak} = σ_\ce{HA} + \frac{Δσ}{1 + 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}\tag{1}$$
$$y = y_\mathrm{min} + \frac{y_\mathrm{max} - y_\mathrm{min}}{1 + 10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}}\tag{2}$$

Eq. 1: https://www.nature.com/articles/srep43748
"In the equation, $\sigma_\mathrm{peak}$ is the measured $\mathrm{NMR}$ chemical shift of the peak of interest, $\sigma_\ce{HA}$ is the chemical shift of the protonated form and $\Delta\sigma$ is the difference between $\sigma_\ce{HA}$ and the chemical shift of the deprotonated form of $\ce{HA}$ ($\sigma_\ce{A}$)."
Eq. 2: https://pubs.acs.org/doi/full/10.1021/ed3000028?mobileUi=0
"$y$ is the experimentally measured parameter (chemical shift, in this case), and $(y_{max} - y)$ and $(y - y_{min})$ are proportional to the concentrations of the deprotonated and protonated forms of the acid, respectively."

The derivation of these expressions begins from the premise that the observed value of the NMR parameter (chemical shift) is an average over values for two states weighted by their mole fractions: $$\sigma = \sigma_1 \chi_1 + \sigma_2 \chi_2= \sigma_1 \chi_1 + \sigma_2 (1-\chi_1)$$ This is a reasonable assumption in NMR experiments if there are two states (protonated and deprotonated) with distinct shifts and exchange between the two states is fast compared to the chemical shift difference such that the observable signal is a single time average. This allows us to write that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma_2)} = \chi_1 $$ Similarly we can write that $$\frac{(\sigma-\sigma_1)}{(\sigma_2-\sigma_1)} = \chi_2 $$ such that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{\chi_1}{\chi_2} $$ or, in terms of concentrations, that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{c_1}{c_2} $$ Introducing this last expression into the Henderson-Hasselbalch equation $$pK_a=pH+\log_{10}\left(\frac{c_{HA}}{c_{A}}\right)$$ gives upon rearrangement
$$10^{(pK_a-pH)}=\left(\frac{c_{HA}}{c_A}\right)=\frac{(\sigma-\sigma_A)}{(\sigma_{HA}-\sigma)}$$ which can be further rearranged into $$1+10^{(pK_a-pH)}=\frac{(\sigma_{HA}-\sigma_A)}{(\sigma_{HA}-\sigma)}=\frac{\Delta \sigma}{(\sigma-\sigma_{HA})}$$ and then further into the expressions in the articles (note the sign convention for $\Delta \sigma$).


Epsom salts - heavy metal contamination myth? This recipe for home-made handwash (liquid soap) consists of bar soap, water, Epsom salts, and coconut oil.

The author specifically warns against using 'artificial' Epsom salts, preferring the naturally occurring type, because the 'artificial' type is supposedly contaminated by heavy metals incl. arsenic.

  Note – Be sure to get natural salts as some are man made and a full of
  unwanted chemicals. A lot of Epsom salts are made from synthetic
  materials, they are manufactured in factories from neutralized
  Sulfuric Acid, which is very high in heavy metals and arsenic. The
  REAL Epsom salt is naturally occurring mineral deposit in spring.

According to Wikipedia, Epsom salts are made by reacting sulfuric acid with magnesium oxide or magnesium carbonate.

I find it difficult to believe that 'artificial' Epsom salts would have appreciable amounts of heavy metal contamination, as neither sulfuric acid nor magnesium oxide are known to be frequently contaminated with heavy metals. A quick search on Google Scholar gives no results on the topic of heavy metal contamination in Epsom salts.

The only way I am aware that heavy metals might enter the process is if the sulfuric acid was produced from smelter stack gases, which do have appreciable heavy metal content. This is done at Mount Isa Mines, where $\ce{SO2}$ from the copper smelter stack gas is scrubbed out to make sulfuric acid, but the entire output of this plant goes straight to fertilizer production.

Is the above claim about heavy metal contamination of Epsom salts credible, or just an urban myth?

Here in the US, they sell different grades of Epsom salt, from USP down to plant food.  I'd presume that depending upon what the product will ultimately be used for, the manufacturer may use MgO with different purity levels - the lower the purity, the cheaper the product.  Even the "bath salts" grade contain trace amounts of arsenic and selenium (see the first link below).  Among the "medicinal" offerings, some are labeled "for external use" and others are labeled "for external and internal use."  I suspect that anything with the "for external and internal use" label would be more than safe enough to use in handwash.

Here are some links you might find helpful:

http://k.b5z.net/i/u/2182313/f/Tech%20Data%20Sheets/Tech_Data_Sheet-bath_salt-epsom_USP-3.pdf

http://www.drugs.com/otc/104235/pepsom-flying-p-epsom-salt.html


Why does my professor use both the upper, and the lower, case literal of a letter to symbolize the same concept? In one of our slides, the professor has written

  $ \Delta E_{sys} = -\Delta e_{sur}$

Where $E_{sys}$ symbolizes the energy of a given system, and $e_{sur}$ symbolizes the energy of the surroundings of that system.

On a few occasions, she's used both the upper, and the lower, case literal of a single letter as symbols of a single concept.

Is there a notational convention that requires the use of an upper case literal of a letter in some circumstances and the use of its counterpart in other circumstances?

Of course, there is a convention. In physics $E$ usually stands for energy, $e$ for elementary charge. You professor is inconsistent, to say the least. In my opinion using lower case $e$ for anything rather than for elementary charge is wrong by itself, and using it for energy is simply ridiculous. Besides, I don't see any reason for doing so.

P.S. During my university years I had a professor who randomly picked letters for physical quantities every lecture. :D Today I think of it with a smile, but at that time I hated both his lectures and him personally.


Keeping Li-Ion battery on charger vs. plug the charger in and out regularly I have read on numerous occasions a set of recommendations to keep a Li-Ion battery happy: avoid discharging the device at all costs, don't charge it up completely, keep it as much as possible between 40% and 80% and don't charge it too quickly.

I'm wondering, however, which one of those two workflows are least harmful to a Li-Ion battery:

Keeping the device on (slow) charger most of the time the whole day while using the device, the battery level remaining at 100% most of the time, and dropping sometimes to 80%.
Unplug the (slow) charger regularly during the day, and then plug it again a hour or two later, the battery level constantly fluctuating between 50% and 100%.

I'm asking this question because while it is quite unpractical for me to keep the battery level of my phone between 40% and 80%, it is perfectly possible for me to either keep the phone on charger nearly all the time when I'm at home or at work, or to switch regularly between on battery and on charger modes.

One thing worse then overtemperature is overtemperature at charging.

One thing worse then overtemperature at charging is overtemperature at charging with charging >70-9x%.

That speeds up the side reactions of eventual metallic lithium deposits with dialkylcarbonate at high charge.

  The reaction of either DMC or DEC with lithium was initiated at 180 K, a temperature much lower than their bulk melting temperatures, producing lithium methyl carbonate, methyllithium and lithium ethyl carbonate, and ethyllithium, respectively. At temperatures greater than 270−300 K, the lithium alkyl carbonates start to decompose with $\ce{Li2O}$, elemental carbon, and alkyllithium as products on the surface.

Those reactions decrease the cell capacity and increase the cell internal resistance.

Things go worse in summer.

The rule number one is never heavily use the phone ( e.g. at gaming ), while charging it at high percentages.

If it must be charged, then when it is just topping,
or better at low charge percentage
or the best not being charged at all.

Charging scheme is also a question of the cell static aging, that applies more to laptops than to short life cycle of mobile phones.

Another thing is that cells prefer rather  more of shallow cycles than less of deep cycles. That may be unfortunately counteracted by durability of charging connectors.

Fully charged cell is aging several times faster than about half charged one. As at full charge, lithium ions may form a metallic lithium deposit instead of lithium graphite intercalate, which very slowly reacts with dialkylcarbonate solvent.

I frequently charge my phone from my powerbank to avoid immobility of the phone during charging, what gives me the option to charge the phone whenever I want. I do not charge the phone overnight nor keeping  it long time being fully charged.  But the truth is, phone life is typically 2-3 years max, what does  not have much chance to show consequences of not optimal cell maintenance.

My Lenovo T400 laptop from 2009 with 40-80% battery maintenance had its battery ( in late 2017) in better condition than T430 from 2013. (70-75% versus 60% capacity). The latter running RHEL Linux was fully charging for several years, until I found the way to achieve 40-80% mode as well. Both were running mostly with AC charger attached on the office table.


Understanding the Jahn–Teller elongation and compression for octahedral complexes (stabilisation/destabilisation of d orbitals) I am currently learning about Jahn–Teller effect. For elongation I wondered why the d orbitals with $z$ component are stabilised when the metal–ligand bonds are getting longer? I thought a longer bond would mean a less efficient overlap of the d orbital on the transition metal with the orbital on the ligand.
And if yes, shouldn't it be raised so the difference in energy between the d orbital and the ligand’s orbital be greater? I thought the ligands’ orbitals would be lower in energy each time.
Where am I going wrong with my thinking?

Before you consider J-T distortion, take a look at the d orbitals of an octahedral complex. Note that the orbitals that are on the axes where the ligands are are higher in energy. This is because these "d orbitals" are actually anti-bonding orbitals comprised primarily of the metal d orbitals. Because they are anti-bonding, they get higher in energy as the bonding interaction between the ligand and metal gets stronger. Thus, a longer distance between the ligand and metal (ie weaker bonding interaction) results in less destabilization.
Another way to think about it is to consider the interaction of a ligand lone pair (ie negatively charged electron density) with a filled d orbital on the metal (also negatively charged electron density). The closer they are together, the more unfavorable the interaction is.


What is the maximum number of water molecules that has been recorded till date to present as water of crystallization in a compound? Context: I was casually looking through some old questions when I found this question. The OP wanted a compound that has all the hydrates ranging from monohydrate to decahydrate. I found chromium(III) sulfate [$\ce{Cr2(SO4)3.$x$(H2O)}$ where $x=0,\ldots,18$] to be suitable. But, the actual formula of the compound is $\ce{[Cr(H2O)6]2(SO4)3·6H2O}$, so 6 of the 18 molecules are present as water of crystallization. So, I wondered, what is the maximum number of water molecules that can present in a compound as water of crystallization?
I can agree to what @BuckThorn said in his answer. It might be protein crystals. Wikipedia also supports the fact:

Compared to inorganic salts, proteins crystallize with large amounts
of water in the crystal lattice. A water content of 50% is not
uncommon for proteins.

Now, the question is what protein has the maximum number of water molecules as water of crystallization?

I did some cursory searches and looked for inorganic hydrates (see edit history: 36-hydrate is maximum I could find). I am also looking for such compounds which is in inorganic domain.

It might be a protein crystal. Proteins are large and the interstitial space is filled to a large extent with water. The volume of a protein crystal occupied by the solvent can be 50% or so. However the fraction of waters observed in crystals (relatively fixed) is often low, on the order of 10%.
For instance, from a report on the crystal structure of crambin (chosen arbitrarily, see Ref. 1):

The volume of solvent (32%) is low but not atypical for proteins [values of 30-78% solvent are found.] Crambin is small (Mr 4720), and most residues are solvent accessible. Although crystals were grown from 60% ethanol, only 7%(vol/vol) ethanol has been located. Of the 85 water molecules expected (based on the density and the presence of 2 ethanol molecules), the model includes 64 [...]

It's not clear whether the 85 expected or the 64 found water molecules would be considered crambin's water of crystallization (presumably all count, even if not observed).
As another example, the model crystal structure of BPTI contains 63 waters (observable waters, see PDB structure 5PTI).
While I was looking for the protein crystal structure with greatest number of waters in its model, Karsten Theis suggested as an example the pyruvate-bound crystal structures of ADP–glucose pyrophosphorylase from the bacterium Agrobacterium tumefaciens (pdb 5W5T, see the file for a description of the solvent model), which has 3752 waters in the crystallographic model (59.12% of the volume).
Many waters may have irregular positions, and some may form clathrate-like solvation structures (as in gas hydrates, as andselisk comments), and some can be placed at regular locations.
References

M. M. Teeter. Water structure of a hydrophobic protein at atomic resolution: Pentagon rings of water molecules in crystals of crambin. Proc Natl Acad Sci U S A. 1984 Oct; 81(19): 6014–6018. doi: 10.1073/pnas.81.19.6014


Why do Rydberg atoms have electrons that can transition above the first ionization energy? Why do Rydberg atoms have electrons that can transition above the first ionization energy?
This can only happen in a multi-electron system but I don't see how the Rydberg formula would even work for n = infinity + 1. I thought it might have to do with the second ionization energy but the definition of that is just another electron being ionized not anything above the first ionization energy.

A useful concept in Rydberg atoms is the notion of a channel. This concept is borrowed from scattering theory and describes all solutions of the Schrodinger equation for the scattering of the Rydberg electron of specified angular momentum with the ionic core in a certain energetic state. The bound Rydberg states correspond to scattering at negative energy and the scattering at positive energy corresponds to the ion-electron continuum of the channel.
If the ion has multiple electrons, it can have several excited states. Each of these excited states can form a channel. The energy of a Rydberg electron is given by
$E=E_\text{IP} + E' - \frac{hcR}{(n-\delta_\ell)^2}$
Where $E_\text{IP}$ is the first ionization energy, $E'$ is the excitation energy of the ionic core, $R$ is the mass-corrected Rydberg constant, $n$ is the principal quantum number and $\delta_\ell$ is the quantum defect for a Rydberg electron with orbital angular momentum $\ell$.
Each excited state of the ionic core thus gives a new Rydberg series with each own set of principal quantum numbers. So, Rydberg states do exist above the first ionization energy, but their principal quantum number is finite; it just belongs to a different channel.

Image source


Do SALC-AOs really belong to their symmetry species? I'm working through a molecular symmetry textbook and something keeps nagging at me. If I derive the SALC-AOs for NH3 (using the projection operator method), I'll get

A1:  $  \frac{1}{\sqrt{3}}(\phi_1+\phi_2+\phi_3)$

Doubly-degenerate E:

$  \frac{1}{\sqrt{6}}(2\phi_1-\phi_2-\phi_3)$

$  \frac{1}{\sqrt{2}}(\phi_2-\phi_3)$

where the $\phi$s are the H 1s orbitals, ie.:

(This was the best image I could find, but in the second E orbital the black s-orbital should be enlarged)

I'm having trouble seeing how these SALC-AOs would be a member of their symmetry species. The A1 SALC-AO is identical under each operation, so it makes sense that it would belong to A1. But if I want to confirm that the two E SALC-AOs belong to E, how would I do that?

My intuition is that I should be able to apply each C3v operation to the SALC-AO and get back the E row of the character table (Below). But if you apply a C3 rotation to the e2 orbital, you get the black orbital taking the place of the white orbital, the white taking the place of the node, and the node taking the place of the black orbital. This doesn't seem like it could be expressed as a number in a character table, since the SALC-AO isn't being taken to itself or -itself.

Am I thinking about this all wrong? How should I understand the SALC-AO as a whole? Any guidance is appreciated, the textbook seems evasive on this.

To show that your two functions do represent the $\mathrm{E}$ irreducible representation, we can approach the problem algebraically.

We start by representing the operators of the $C_{3v}$ point group in the basis of the AOs. The identity element is easy in any basis:
$$E=\pmatrix{1 &0 &0\\0 &1 &0\\0 &0 &1\\}$$

We can see that applying this operation to $e_1$ and $e_2$ will just return these functions, so $\chi(E)=2$. We can also represent this operation in the basis of $a_1$, $e_1$, and $e_2$, which will be important for explaining the remaining operations.

For the $C_3$ rotation, we can notice that this just has the effect of moving each AO to the next site (i.e. $\phi_1\to\phi_2$, $\phi_2\to\phi_3$, and $\phi_3\to\phi_1$), so we can write it in matrix form as
$$C_3=\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}$$

Lets see what happens when we apply this to $e_1$:
$$C_3e_1=\frac{1}{\sqrt{6}}\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}\pmatrix{2\\-1\\-1}=\frac{1}{\sqrt{6}}\pmatrix{-1\\-1\\2}=e_1'$$

Writing $e_1'$ in the basis of the original functions, we obtain:
$$e_1'=0a_1+\frac{-1}{2}e_1+\frac{-\sqrt{3}}{4}e_2$$

Doing the same with $e_2$:
$$e_2'=0a_1+\frac{-\sqrt{3}}{4}e_1+\frac{-1}{2}e_2$$

We can then take the trace of this matrix (or, to simplify, the 2x2 submatrix spanned by $e_1$ and $e_2$) to get the character of $C_3$:
$$\chi(C_3)=\mathrm{Tr}\pmatrix{\frac{-1}{2} & \frac{-\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4} & \frac{-1}{2}}=-1$$

You can go through the same sort of exercise with $\sigma_v$ (determine form in AO basis, apply to each function, write resulting functions in original function basis, take the trace to get the character).


How to measure the electrical conductivity of a substance? Suppose you are given these substances:

$\ce{NaOH(aq)}$

$\ce{HCl(aq)  }$

$\ce{C6H12O6(aq)  }$

$\ce{NH3(l)}$

Are there rules to figure out how conductive each compound is? Is it based on intermolecular forces or polarity?

A lot of work was done in the 19th and 20th century on determining the conductivity of aqueous solutions. It was a huge field called conductometry (check Wikipedia). You experimentally measure the resistance of a solution by placing the desired liquid between two inert electrodes and pass high frequency alternating current to prevent electrolysis. In solutions, electrical current is carried by ions. If your solution has ions or if it is ionizable, it will conduct electricity. Now you can determine which solution has ions based on ionization, e.g. you can write $\ce{HCl (aq) -> H+ + Cl-}$ and can predict that it must be a conductive solution. In fact, the aqueous $\ce{H+}$ has the highest conductivity known in water. $\ce{OH-}$ is next.


The increasing order of reactivity of substituted benz0yl chlorides towards nucleophilic substitution Arrange the following compounds in the order of their increasing reactivity towards nucleophilic substitution:

The answer to this question is $[3<2<1<4]$. We have to compare the positive charge density on the carbons to which leaving group is connected. In $(1)$, the phenyl ring shows negative inductive effect but my teacher did not mention anything for that. In $(3)$ and $(4)$ the connected groups ($\ce{OCH3}$ and $\ce{NO2}$) show +M and -M respectively. But, they also show -I effect. Why are we ignoring the inductive effect?

Electronic effects govern in the order
Resonance/Mesomeric > Hyperconjugation > inductive
+M or +R dominate over inductive. That's because +M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons. So therefore +M effect has priority over +I.
Therefore the order of reactivity towards nucleophilic substitution is 4>1>2>3


How to remove a stuck stopper from a flask? How do I remove a stopper that has been stuck in the neck of a 100 mL round bottom flask since I forgot to wipe the neck of the flask. A salt was formed in the 100 mL flask and the stopper is just stuck like cement. Is there anyway to remove it without breaking the flask, breaking the stopper, or heating the flask (since I don't want to interfere with the product)?

Is it a hollow glas stopper?

Sometimes, it helps to hold the flask at the body and hit the corner of the lab bench with the stopper. If that doesn't help, heat the neck of the flask carefully with a heat gun.


Why does the electron configuration for some elements not follow the diagonal rule? I'm doing a high-school assignment, and I came across a question that I didn't quite understand.

  Explain how the electron configurations for the following elements do not follow the diagonal rule:

  Gold
  Curium
  Thorium
  Molybdenum
  Palladium

From what I understand, all of these elements do follow the diagonal rule. How exactly do they not follow this rule?

The reason of these exceptions (from the diagonal rule) is that some elements are more stable with fewer electrons in some subshells and more electrons in others:

Electronic configuration of molybdenum: $[\ce{Kr}] \ce{4d^{5} 5s^1}$, instead of  $[\ce{Kr}] \ce{5s^24d^{4}}$ according to the diagonal rule, because a half-full $\ce{4d}$ subshell and a half full $\ce{5s}$ subshell are more stable than $\ce{4d}$ filled with four electrons and a full $\ce{5s}$ subshell.
Electronic configuration of Gold: $[\ce{Xe}] \ce{4f^14 5d^{10} 6s^1}$, instead of  $[\ce{Xe}] \ce{4f^14 5d^{9} 6s^2}$ according to the diagonal rule, because a full $\ce{5d}$ and half full $\ce{6s}$ subshell is more stable than $\ce{5d}$ filled with 9 electrons and a full $\ce{6s}$ subshell.
Electronic configuration of Palladium: $[\ce{Kr}] \ce{4d^{10}}$, instead of  $[\ce{Kr}]\ce{ 5s^24d^{8}}$ according to the diagonal rule, because a full $\ce{4d}$ orbital is more stable than $\ce{4d}$ filled with eight electrons.
Electronic configuration of Curium: $[\ce{Rn}] \ce{7s^2 5f^7 6d^{1}}$, instead of  $[\ce{Rn}] \ce{7s^2 5f^8} $ according to the diagonal rule, because a half full $\ce{5f}$ orbital is more stable than $\ce{5f}$ filled with eight electrons.


How do we know if a compound is amphoteric or not? My teacher told me if it reacts with an acid and a base, then the compound is amphoteric.

However, I am not satisfied. Is there any other way of telling? For example we are give a compound say BeO, how can we say if it's amphoteric or not?

Sorry to tell you but your teacher's opinion is not right, just because the word "react" is not enough. Usually you label oxides as "amphoteric". You may say that if an oxide reacts with an acid and forms a salt or if an oxide reacts with a base forms a salt then you can say that it is an amphoteric oxide. Take the example of aluminum oxide, when dissolved in HCl it will form aluminium chloride, but we can also dissolve it in NaOH, where it forms sodium aluminate.

However, we can broaden this definition further (good for advanced classes). One formal definition is "A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: H2SO4 is an acid when studied in water, but becomes amphoteric in superacids."

See: https://goldbook.iupac.org/html/A/A00306.html


what is the acceptable percentage for food grade ethanol?Is 95% ethanol plus 5% water consider as food grade ethanol? Is 95% ethanol plus 5% water consider as food grade ethanol?

The grade of ethanol does not depend on the water and ethanol percentages, but rather on what else is in there. If your mixture is truly 95-5 ethanol-water, with nothing else in it, than that would be safe to consume (accounting for the health hazards of consuming too much ethanol of course).

However, there might be traces of e.g. methanol or benzene in the ethanol, which are both highly toxic to humans and therefore cannot be present in food grade ethanol mixtures. According to this website the FCC grade and above are food grade, in particular due to some heavy metal specifications, so apparently there might also be trace heavy metals in the ethanol.


Reduction of copper oxide by hydrogen in the lab to copper In the reduction of copper oxide by hydrogen in the lab to copper, why does hydrogen need to flow through the tube even after the reduction is completed and the burner is turned off until the tube is cool?

As the system cools, the gases will contract. Unless a positive flow of hydrogen (or some other inert gas) is maintained during the cooling process, this will result in room air being pulled into your tube. The oxygen present in air would then oxidize some of $\ce{Cu}$ back to $\ce{CuO}$, so long as some regions within the tube remained hot enough for this reaction to take place (at least $\pu{300^oC}$ or so).


Cyclohexene reduction reaction with hydrogen gas and ethanol I have a following reaction (see below). I found the section of reduction and oxidation states in my book but there is nothing about cyclohexenes. I know that the double bond will be reduced in this reaction but what I have no idea what happens to those substituents. Why do they move around and what are their orientations?

Note: This is a homework question, so I will not give a complete answer.

$\ce{H2/Pd}$ is a reducing agent, and it acts so by adding hydrogen to the double bonds. Alkenes are converted to alkanes. This proceeds via a concerted pathway, so rearrangements due to carbonation shifts do not occur. Now I think you can proceed from here.

As far as those substituents are concerned, think carefully on what would happen.


How to name an alkyne with cyclic substituent? My attempt: 1-(1-methylcyclopentyl)-3-methylpent-1-yne. Looks so messy though.

As far as I am concerned, your attempt was incorrect.
You must use a parent chain with the highest number of carbon molecules. If I am not mistaken, if a cyclic carbon chain has the same number of carbon molecules as a straight carbon chain, the cyclic carbon chain should be the parent chain.
In addition to what I said above, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), rings/cyclic carbon chains are always selected as the parent chain.

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

Thus, the name should be: 1-methyl-1-(3-methylpent-1-yn-1-yl)cyclopentane.
Note that an alkyl substituent that is also an alkyne (i.e. 3-methylpent-1-yn-1-yl in your question above) is called an alkynyl group. It is also worth noting that rings/cyclic carbon chains are always selected as the parent chain for the preferred IUPAC name as I mentioned previously.


Does the electrolysis of water produce chlorine gas? http://orbitingfrog.com/2014/11/02/electrolysis-of-water-with-pencils-and-a-9v-battery/

This article seems to indicate that "a small amount of Chlorine may be produced as you [perform electrolysis on water]". Is this true? I thought it was just H2 and O2 produced.

It's the tablespoon of salt he added, giving the chlorine when electrolysis is done. I have done electrolysis with salt-free water (though a bit acidic to get the water to electrolyze faster) and chlorine was nowhere to be found.

Electrolysis of saltwater produces hydrogen and chlorine; and leaves sodium hydroxide behind.


Why is copper(II) sulfate added drop by drop during the Biuret test? The Biuret test is a test for proteins.  The procedure is to add a strong base, such as NaOH, to dissolve the protein.  After that, copper(II) sulfate is added drop by drop into this until a (violet) colour change is observed.  I have a few questions.

How does NaOH dissolve the protein exactly?
What is the significance of adding copper(II) sulfate drop by drop?
How would the reaction vary if copper (II) sulfate was added first and NaOH added subsequently, drop by drop?

The Biuret Test is a chemical assay that detects the presence of proteins in a sample. The test is not named after any famous scientist, but after an urea dimer called biuret ($\ce{H2NC(O)NHC(O)NH}$). However, funny thing is biuret is not a part of the test at all. The test is called Biuret Test because biuret gives a positive result to the test (See the image below).

The test relies on the characteristic color change (from blue to deep purple or violet; see the image), which confirms the presence of proteins. Biuret isn't a protein, but it gives a positive result to the test, because it has two amide bonds.

Your question (1): How does $\ce{NaOH}$ dissolve the protein exactly?

Proteins has a lot of amide (peptide) bonds, terminal amine and carboxylic acid groups, and along the $\alpha$-helix, a lot of amino acid side chains including amine groups (e.g., lysine), carboxylc acid groups (e.g., aspartic acid), thiol groups (e.g., cystine), etc. All of them can be reacted with $\ce{NaOH}$ to become ionic, which make it more soluble in water. Specifically, recall most of zwitterionic form of amino acids resist to dissolve in water. Regardless, $\ce{NaOH}$ is essential for the test, not only to make protein more soluble, but also to extract $\ce{H}$ from amide $\ce{N-H}$ bond to make $\ce{N}$ negative (Ref.1).

Your question (2): What is the significance of adding copper(II) sulfate drop by drop?

When you perform the Biuret Test in qualitative manner, you need to be careful to see color change clearly, I'd say picture perfectly. Suppose, you have very dilute protein sample. Adding the blue solution into a colorless solution is tricky, because if you are not slow enough, you won't realize there is a color change happens due to both (before and after) solutions been colored but intensities are low (see the near pipette tip of the picture). If you are doing the quantitative test, it doesn't matter how fast you add the solution. After color development, you use UV-Vis-Spectrometer to measure color intensity: at appropriate wavelength to purple color ($\pu{540 nm}$) where absorbance is zero for the $\ce{CuSO4}$ blue color, thus interference is minimal.

Your question (3): How would the reaction vary if copper(II) sulfate was added first and $\ce{NaOH}$ added subsequently, drop by drop?

When you search for the Biuret Test, you'd find out that biuret solution is already alkaline with $\ce{NaOH}$ (see the description of solution in the picture). Thus, I'd say, the order does not matter. Yet, if you want to do a crude test with aqueous $\ce{CuSO4}$ solution and $\ce{NaOH}$ solution, you should be careful, because initial $\ce{Cu(OH)2}$ precipitation might happens. If interested, you may read Ref.2 to see the effects of all solutions on the test, which gives very comprehensive review of the test.

References:

H. Sigel, R. B. Martin, “Coordinating properties of the amide bond. Stability and structure of metal ion complexes of peptides and related ligands,” Chem. Rev. 1982, 82(4), 385–426 (DOI: 10.1021/cr00050a003).
A. G. Gornall, C. J. Bardawill, M. M. David, “Determination of Serum Proteins by Means of the Biuret Reaction,” Journal of Biological Chemistry 1949, 177, 751-766 (http://www.jbc.org/content/177/2/751.short).


Enthalpy - unit confusion I think I have trouble interpreting reaction enthalpy in these kind of situations:

I am given that $\Delta _rH^0 = 25,9$kJ/mole in the reaction:
$$\ce{1/2 H2 + 1/2 I2 \rightarrow HI}$$
which I interpret as: if this reaction takes place where 1 mole HI forms, the the produced heat is equal to 25,9kJ.

Now suppose we double the amounts, so that we have the reaction
$$\ce{H2 + I2 \rightarrow 2HI}$$
then clearly the difference in enthalpy should be doubled and be equal to 51,8kJ. Then why do chemists write that  $\Delta _rH^0 = 51,8$kJ/mole and not $\Delta _rH^0 = 25,9$kJ/mole? Because I would think that we can calculate the enthalpy change like this: $2mole\times25,9$kJ/mole.

There is cleary some fundemantel thing I'm missing here or misinterpreting, so I would like some help.

25.9 kJ/mol refers to one mole of $\ce{HI}$ produced while 51.8 kJ/mol corresponds to one mole of $\ce{H2}$ consumed. It is good practice to give the full equation, especially when dealing with reaction enthalpies/energies $\Delta_rX$. For formation quantities $\Delta_fX$, I would always assume one mole of product.


Why is dp/dv zero for a real gas in critical conditions? I have read this fact in various websites but cannot understand the reason. What is an intuitive explanation of this?

This is merely a shard of a fact which does not make much sense in and by itself. After all, in systems with gas/liquid equilibrium there is nothing really special about $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. On the contrary, this is pretty typical. See all those points where the blue lines (isotherms) are horizontal? They make up a whole huge area on the diagram.

(source)

Indeed, if you have a liquid in equilibrium with its saturated vapor, then the pressure does not depend on volume (in an isothermal processes, that is). As you compress it a little more, some gas turns to liquid; as you release it, some liquid evaporates. The pressure still remains the same - namely, the saturated vapor pressure at given temperature. Things start changing only when you pull the piston all the way out or push it all the way in.

Now, the two-phase area, huge as it is, does not comprise all diagram. If you heat the gas past the critical temperature, you'll get an isotherm of a different sort (it is not shown on the picture, so just imagine one, positioned above the rest of them). It will be all smooth and monotonic, without those horizontal parts.

Well, then there must be a temperature where this behavior changes, and that's what we call the critical temperature. Sure enough, it produces a very special isotherm: there is only one point on it where $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. (That's what we call the critical point, BTW.) Moreover, since it is an inflection point, it also has $\left(\dfrac{\partial^2\mathfrak p}{\partial V^2}\right)_T=0$. Now that's the fact one should remember about the critical point.

The peculiar behavior of pretty much all physical properties around the critical point gives rise to various weird phenomena like critical opalescence, but that's another story.

So it goes.


Sign of enthalpy for exothermic and endothermic reactions In Cambridge Chemistry Coursebook [1, p. 94] it’s written that

A rise in temperature is given a positive sign. So the value of $\Delta H$ is negative for an exothermic reaction. A fall in temperature is given a negative sign. So the value of $\Delta H$ is positive for an endothermic reaction.

Why is the sign of enthalpy for exothermic reaction negative? Doesn’t exothermic reaction rise the temperature of surrounding?
Why is the sign of enthalpy for endothermic reaction positive? Doesn’t endothermic reaction take in energy and cool down the surrounding?
Reference

Ryan, L.; Norris, R. Cambridge International AS and A Level Chemistry Coursebook, 2nd ed.; Cambridge International Examinations; Cambridge University Press: Cambridge, 2014. ISBN 978-1-107-63845-7.

Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give that heat to surroundings. If you do so, than since our system lost heat its enthalpy decreases (first law of thermodynamics) and because of that enthalpy of exothermic reaction is taken as negative. The opposite reasoning holds for endothermic reactions. You need to bring heat from surroundings into system to keep it at the same temperature and that heat increases enthalpy of the system.


Is this the lowest chair conformer of Beta D-galactose I drew the conformer (below) and I'm wondering if this is the most stable chair conformation.

Three of the bonds are equatorial so it looks like the most stable to me.

The chair conformation that you have drawn (4C1) is likely to be the most stable one, as it minimizes the number of heavy axial groups.

To determine the chair conformation of a hexose, it is generally easiest to draw it and compare it with β-D-glucose, where all heavy groups are equatorial and the conformation is 4C1. If the number of heavy axial groups becomes smaller when the conformation is changed to 1C4 (all equatorial groups in 4C1 become axial and vice versa), then it is likely that the conformation is 1C4.


Surface energy for different crystallographic planes For a given crystal structure, say body centred cubic, on what factors does surface energy of each plane depend?
A basic approach considers the atomic density on each crystallographic plane, the number of bonds broken if the surface is cleaved along that plane and combines these two.

Is this a realistic and reasonable approach if I want to compare the surface energies of two crystallographic planes, say (100) and (111)? What are the other factors, if any, that should be included in the comparison?

Edit : Background : My problem related to the wettability of a crystal surface by a liquid. The contact angle that the liquid makes with the crystal surface is determined by the interfacial energies between solid-liquid, liquid-vapour and solid-vapour. Keeping all other factors, I only change the nature of the solid crystal surface, by presenting a different crystallographic plane. I would like to theoretically estimate which crystallographic plane would have the highest energy, and which would have the lowest, so that a prediction can be made about the extent of wettability of the surface.

Thank you!

You can think about this problem from the definitions of surface energy.

The surface energy quantifies the disruption of inter-molecular bonds that occurs when a surface is created

The composition or structure of a surface depends on how the bulk crystal has been cut.
The electronic structure (or number of dangling bonds) depends on surface composition
The reactivity of a surface depends on the electronic structure.

Imagine a battenberg cake:

The cake has been cut horizontally and the surface is comprised of tasty red and yellow squares.
If however we had cut the cake vertically straight down the middle, you would only see a red line next to a yellow line (Cut 1).
If you cut vertically and diagonally across the cake, you might only see either a purely red or purely yellow surface. In the case of the example image, you would only see a red surface (Cut 2).

I hope this example shows that (by relating the coloured squares to atoms), different cuts result in different surfaces with different properties.

When calculating the surface energy of a system:

You need to calculate the bulk crystal energy.
You need to calculate the energy of the surface slab. Make sure your model is converged with respect to slab thickness. The central layers should be representative of the bulk crystal. Look at magnetic change, charge difference, degree of ionic relaxation.
You then calculate the difference between the slab energy and the bulk crystal energy.

Note: In periodic calculations, a surface slab is constructed by cutting the bulk in a specified plane and introducing a vacuum gap. This vacuum gaps should be sufficient that the surface does not interact with its periodic mirror images. Around 15 Angstroms.

There are a number of codes that can help in the construction of surface slabs (METADISE, Materials Studio, Vesta)

If you wish to calculate the energy of a surface (and assuming you are using a periodic DFT code), I would highly recommend the paper by Wenhao Sun and Gerbrand Ceder (https://www.sciencedirect.com/science/article/pii/S003960281300160X)

Note: be careful of magnetic influences.


How are related the electrons magnetic dipole moments in atomic orbitals? I wondering, has somebody ever try to take in account the electrons magnetic dipole moment for the shapes of the probability distribution in atom? Does this magnetic dipole moment play any role in the inner bounds of molecules?

In my paper, mentioned above, I have had some doubts about the role of the electrons dipole moment distribution in chromium atom which is quite symmetrical if I use my octahedral model instead of s- and p-orbitals.

Now a have read on Wikipedia about the magnetic characteristics of pure chromium: "Chromium is remarkable for its magnetic properties: it is the only elemental solid which shows antiferromagnetic ordering at room temperature (and below). Above 38 °C, it transforms into a paramagnetic state." This matches perfect into my model abot octahedral structure of neon and argon shell electrons.


Ketone prefix organic chemistry nomenclature Is there a difference between the prefixes keto- and oxo-?

The prefix ‘keto’ doesn’t exist in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). Already in the 1979 Recommendations, the prefix ‘oxo’ is used to indicate $\ce{=O}$ when the group cannot be cited as suffix.

Nevertheless, the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995) still remembers the use of the prefix ‘keto’:

  To indicate a double-bonded oxygen that is part of a ketonic structure, the term keto is sometimes used as a prefix, but such use has been abandoned by IUPAC for naming specific compounds. A traditional use of keto is for indicating oxidation of $\ce{CHOH}$ to $\ce{C=O}$ in a parent compound that contains $\ce{OH}$ groups, such as carbohydrates. E.g. 3-ketoglucose.


Is it possible to oxidize ether to ester? Is it possible to synthesize an ester from an ether in a single step? For example if I had an ether $\ce{(R-CH2-OR')}$, can I oxidize it so it will be an ester $\ce{(R-CO-OR')}$.

Yes, it is possible.  One reagent to do so is acidic bromate salts.  This paper discusses a number of optimizations and characterizations of the bromate-powered ether oxidation reaction.

$$\ce{R-CH2-O-CH2-R + BrO3- ->[H+][H2O; {r.t.}] R-CO-O-CH2-R + Br^-}$$

Hydrolysis of the formed ester leads to an undesired byproduct: the free acid $\ce{RCOOH}$.  In addition to that byproduct, overoxidation of the substrate can also happen, leading to anhydrides or even $\ce{CO2}$.

There may well be plenty of other oxidants and catalysts than can achieve the same reaction, but I thought this one was notable because it works in water at room temperature.  Yield depends strongly on reaction conditions and the vagaries of the particular substrates you use, though.


Suggestions for 200-600 nm range transmitting polymers/materials I am a student of physics who is doing research related to Cherenkov measurements. For that purpose, I need suggestions for some materials that have the following properties:

$80\%$ transparency in the range of $200$-$600$ $\mathrm{nm}$
Low scintillation light produced from electrons
Relatively lower density
A high refractive index

Some materials in this regard that I have tested are PMMA, TPX, NAS, LURAN, TOPAX, and Plexiglass. I want to test more materials to see if something does better; what polymers or other materials might I try to achieve these properties?

No plastic of any practically useful thickness$^\dagger$ has $80\%$ transparency down to $200~\mathrm{nm}$. Period.

Below is a UV transmission chart for various materials from a marketing brochure published by BrandTech, a manufacturer of disposable UV/Vis cuvettes (click image to enlarge):

The chart only goes down to $210~\mathrm{nm}$; extrapolating the trend, not even quartz is $80\%$ transparent at $200~\mathrm{nm}$, but it's probably a decent choice for your purposes as long as it sufficiently meets your other criteria.

Beyond that, fluorite $\left(\ce{CaF2}\right)$ is the only specific material I know of with better UV transparency; it appears to have better than $90\%$ transparency down to around $200~\mathrm{nm}$ (figure is from ThorLabs online catalog):

A chart at that ThorLabs link indicates two additional candidate materials: $\ce{MgF2}$ and sapphire. As with quartz, the choice among these will boil down to how well they meet your other criteria.

$^\dagger$ As a caveat to the above: If a very thin plastic film ($\leq 100~\mu \mathrm{m}$ or so) is acceptable, a specialty material might give sufficient transmittance. You would have to convince the manufacturer of such a material to make it for you, however, which would probably not be trivial, or cheap.


Finding moles of different compounds using balancing $$\ce{K2Cr2O7 + $x$\,H2SO4 + $y$\,SO2 -> Cr2(SO4)3 + K2SO4 + $z$\,H2O}$$

For $\ce{Cr}$ change in oxidation number is: $+12$ becomes $+6$

For $\ce{S}$ change in oxidation number is: $+4y$ becomes $+6$

For $\ce{H}$ change in oxidation number is: $+2$ becomes $+2z$

since $\ce{Cr}$ decreases by 6, $\ce{S}$ and $\ce{H}$ have to increase by 6, so

$$(4y+2)-(6+2z)=6$$

$$2y-z=5$$

Also using the principle of atom conservation (POAC) for $\ce{H}$ atoms from $\ce{H2SO4}$ to $\ce{H2O}$, we get:-

$$2x=2z$$

$$x=z$$

I can't find any more equations to solve this.

Never having heard of the Principle of Atom Conservation (POAC) before I'm curious.

Looking up the concept, I understand it simply to mean that atoms can't be created or destroyed in a chemical reaction. (So we're ignoring radioactive decay for example.) I start blindly with a variable for each reactant and product, and will then look for linear relationships in the variables.

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $d$\,Cr2(SO4)3 + $e$\,K2SO4 + $f$\,H2O}$$

So with $a$ $\ce{K2Cr2O7}$ molecules, I must get $a$ $\ce{Cr2(SO4)3}$ molecules to balance $\ce{Cr}$, and $a$ $\ce{K2SO4}$ molecules to balance $\ce{K}$. So:

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $f$\,H2O}$$

Now $\ce{H}$ is only in $\ce{H2SO4}$ on the left and $\ce{H2O}$ on the right so $f=b$.

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $b$\,H2O}$$

To balance $\ce{S}$:

  $$b + c = 4a$$

To balance $\ce{O}$:

  $$7a + 4b + 2c = 16a + b$$ which reduces to $$3b +2c = 9a$$

At this point just balancing atoms leaves you with 2 linear equations and 3 unknowns. So there are not enough equations to explicitly solve the problem using linear equations. However relationships can be established.

\begin{array}{cccc}\\
3b& +2c & = & 9a \\
-2(b& +c & = & 4a)\\ \hline
 & b & = &a \\
\end{array}

$\therefore c = 3a = 3b$

The final step in this problem is to realize that an integer solution (a Diophantine problem in mathematics) is desired. So $a=b=1$ and thus $c=3$. So the final chemical equation is:

$$\ce{K2Cr2O7 + H2SO4 + 3SO2 ->Cr2(SO4)3 + K2SO4 + H2O}$$

PS - Many thanks to user Mathew Mahindaratne for suggesting a cleaner solution in the comments.


How dangerous is taking a bath in Coca Cola? What can be the effects for the body to be immersed in Coca Cola for a while? A friend commented me that the coke is quite acidic (pH 2.5 if I recall properly), and we were guessing what could be the effects for the body after a time of immersion.

I read 20 Practical Uses for Coca Cola – Proof That Coke Does Not Belong In The Human Body and I see some things like:

  E338 – Orthophosphoric Acid. This can cause irritation of the skin and
  eyes.

But I am guessing: what are the problems of a long contact of coke with the body - not drinking, but being in touch with it.

Cola and soft drink products are generally in the pH=2.5-3.5 range.
This is the same pH range as vinegar, lemon juice and most fruit juices. Note apple itself is around 3.0-3.5, so it is not about some magical trick of food industry to kill us all.

Does Cola dissolve your body traceless? No.
Do you want to soak your body in it for hours? Would you soak your body in vinegar for hours? Or even water...


Is the notion of orbitals different in theoretical chemistry? I am just an average chemistry student that is currently exposed to theoretical chemistry. And I am seeing very puzzling things, such as 1p, 2f, 3g orbitals. In my attempt to reconcile this, I was thinking perhaps these are just so-called additional basis functions that are included in the calculations to better approximate the electron behaviour. Would that be correct?

Additionally, there is this idea of single-electron wavefunctions and N-electron wavefunctions, but no double-electron wavefunctions. Does this mean that there are only 1-electron orbitals in theories like DFT and HF, and no 2-electron orbitals?

This is rather weird because both these theories in theoretical chemistry and the the theory that we have learnt in high school (i.e. the theory of atomic structure that most students would know) are based on the same foundations of quantum mechanics. So how can they be so different? Could someone help to reconcile these differences?

Unfortunately, it only gets more complicated the deeper you dig. There is some explanation here: What exactly is an orbital?, but you should bear in mind that electronic structure theory is something that the average undergraduate student only barely touches. A strong background in QM is IMO mandatory to understand what some of these things mean. I'll see how far I can get with some of these queries, though, and some of it will be necessarily simplified.

The traditional "orbitals" that introductory chemistry teaches resemble what theoreticians would call spatial orbitals. These are distinct from spin orbitals, i.e. one-electron wavefunctions which describe both the spatial and spin orientations of the electron. The spatial orbitals only describe the spatial component.
For every spatial orbital, there are two possible spin orbitals, because you can have one electron with up spin and one with down spin. The simplest possible way to represent an "orbital" being filled with two electrons is by taking the product of the two spin orbitals, i.e. the product of two one-electron wavefunctions. So, the "two-electron orbitals" from introductory chemistry still have a mathematical form in theoretical chemistry, although it does depend on exactly what theory you are looking at (there are cases where the notion of a two-electron spatial orbital breaks down).
The reason why you don't see "two-electron wavefunctions" being discussed is because they have no real meaning in electronic structure theory. The N-electron wavefunction is useful because that is the way we choose to describe the entire molecule, and we can use it to extract information such as the energy of the system. The one-electron wavefunctions are useful because (1) they are used to construct the N-electron wavefunction (most simply by direct multiplication, or more properly using Slater determinants), and (2) we have developed methods to obtain the best possible forms of these. Two-electron wavefunctions are somewhere in the middle: they're not useful for describing the molecule as a whole, but at the same time there's no mathematical motivation for using them.

As for basis sets, it's a different ball game: we are no longer talking about what orbitals theoretically are ("theoretical chemistry"), but rather what is the best way to model them for a computer to get accurate results ("computational chemistry"). AOs themselves have no intrinsic physical meaning in a molecule, they merely serve as useful building blocks for MOs. Also, we don't even use the actual AOs as basis functions: we use things that resemble AOs as basis functions, to increase computational efficiency. So it is not too much of a stretch to add "unphysical" functions such as 1p, 2d, ... all for the purposes of getting better and/or faster results.
Also, orbitals in DFT have an entirely different meaning from orbitals in HF, but that's a story I'm not qualified to tell.

The problem with chemistry is that these theories cannot be taught to young kids - they are far too complicated and require far too much existing knowledge. And so we resort to teaching simple things, such as

"Orbitals are trajectories of electrons" - no.
"Orbitals are a region in which you have a 90% (or x%) chance of finding the electron" - no. Orbitals may be visually depicted by an isosurface which encompasses 90% of the electron density, but the orbital is not that region - it is a wavefunction.

For most people (cough organic chemists cough), simplified models work just fine. However, unfortunately, it can sometimes be a constant process of finding out that what you have been taught before is... wrong, or at best, heavily simplified. To understand exactly what justifies the simplification, though, you need to delve into the theory: and as I mentioned at the start, it is a very long journey. There are some good books in the Resources post, but again, a good foundation in QM is needed.


In a copper wire, how many electrons does each atom give away? As an ion, copper can give off 1, 2, 3 or 4 electrons. But it has 1 s electron in the last shell and 10 d electrons. So as a metal, how many of those are delocalised and free to move around, and how many are staying with the atom?

This is a number which can be measured via the Hall effect. This reference give the Hall coefficient as $-5.4\times10^{-11}\,\mathrm{m^3/C}$ for a number density of charge carriers as
$$n_\mathrm e=\frac1{\left(-5.4\times10^{-11}\,\mathrm{m^3/C}\right)\left(-1.602\times10^{-19}\,\mathrm C\right)}=1.16\times10^{29}/\mathrm m^3$$
The number density of copper ions is
$$n_\ce{Cu}=8920\,\frac{\mathrm{kg}}{\mathrm{m^3}}\times\frac{1000\,\mathrm g}{\mathrm{kg}}\times\frac{1\,\mathrm{mol}}{63.546\,\mathrm g}\times\frac{6.022\times10^{23}}{\mathrm{mol}}=8.45\times10^{28}/\mathrm m^3$$
So that works out to about $1.37$ charge carriers per ion.


Why do alkenes with more surface contact have greater London forces? Why is it that alkenes with greater surface contact have greater London forces? I thought greater London forces were dependent on the size of the molecules, or the number of electrons, rather than the surface contact.

You are correct that London forces are dependent on the size of a molecule, but think about why this is. One reason is that a larger molecule tends to have a greater surface area than a smaller molecule, and thus there is more surface area available to interact via London (and other) forces.

Also as stated in this Wikipedia article:

  London forces become stronger as the atom in question becomes larger, and to a smaller degree for large molecules. This is due to the increased polarizability of molecules with larger, more dispersed electron clouds.


Relating change in internal energy and to that in enthalpy The question is

  For which exothermic reaction is $\Delta\,E$ more negative than
  $\Delta\,H$?

  (A) $\ce{Br2(l) <=> Br2(g)}$

  (B) $\ce{2C(s) + O2(g) r 2CO(g)}$

  (C) $\ce{H2(g) + F2(g) -> 2HF(g)}$

  (D) $\ce{2SO2(g) + O2(g) r 2SO2(g)}$

r is supposed to be reaction? Either a one headed or equilibrium arrows. The "correct" answer is supposed to be D.

However, using the formula $H = U + PV$ to $\Delta H = \Delta U + \Delta PV$, I found that $\Delta U$ would be less if $\Delta V$ was positive.

Thus, more gaseous products should be produced. As a result, I chose B.

Note, A is not exothermic, so would not be considered.

Could anyone confirm my answer or offer an explanation for my answer being wrong?

You are right. B is the correct choice.

$$\Delta H = \Delta E + \Delta PV$$

Since $\Delta E $ is more negative than $\Delta H$, $$\Delta H > \Delta E$$.
Hence $$\Delta PV>0$$

$\Delta E \space $has to be negative. So option A is ruled out.

Among the other choices, B is the only one where number of gaseous moles are increasing from left to right, leading to a positive value of $\Delta V$ and hence $\Delta PV$.


which starting material on heating with concentrated sulfuric acid generate cyclic ester I have just learned esterification of normal reactions with aliphatic compounds, but this one is a cyclic compound. Is it different? There are both a hydroxyl group and a carboxyl group in choice B and D, but why the correct answer is B, not D?

The correct answer is glycolic acid, B.

It forms the cyclic di-ester glycolide (1,4-dioxane-2,5-dione).


What provides the energy allowing membrane proteins to facilitate diffusion across the membrane? We know that facilitated diffusion requires no energy.  But the transport proteins, when bind to a molecule from ECF (extra-cellular face) , the proteins rotate and deliver the molecule into ICF (intra-cellular face). Now my question is that--- Which force provides with the torque to rotate and how it works with out energy?

[...]how it works with out energy!

The transport proteins can change conformation in a stochastic manner. Even if the conformation opening to the extra-cellular side is a bit different in energy than the one opening to the intra-cellular side, there is no net change in energy/entropy/Gibbs energy in a complete cycle. The only thing that changes is that the cargo molecules are on the other side, a process for which the change in Gibbs energy is negative.

This argument is the same for any catalyst. If the catalyst is recovered in the same state as it was at the beginning of the reaction (and it has to be, otherwise it would not be a catalyst), the catalyst does not change the thermodynamic quantities of reactants or products. As a consequence, it speeds up the reaction, but does not influence the equilibrium constant.

If the catalyst has a high activation energy to do what it needs to do, then it is not an efficient catalyst. After all, the role of the catalyst is to lower the activation energy. So the transport proteins must be able to change from one to the other conformation both with cargo and without cargo in a facile manner. Otherwise, they could not go through the cycle of 1) picking up cargo 2) switching conformation 3) releasing cargo 4) switching conformation again.

  Which force provides with the torque to rotate [...]

Random forces associated with thermal energy are responsible for the conformational change. The binding constant for cargo has to be moderate so the the cargo dissociates easily. One the other hand, the binding constant for non-cargo has to be so low that nothing other than cargo binds.


Why do only unsaturated hydrocarbons undergo addition reactions? Why do only unsaturated hydrocarbons undergo addition reactions? For example, ethene or ethyne will undergo an addition reaction with chlorine, whereas ethane will not.

Furthermore, why does benzene not undergo addition reactions, even though it is not saturated?

To what would you add given a saturated hydrocarbon? You must substitute, abstract, or displace.

Benzene is aromatic — different pattern of reactivity.  It photochlorinates.  It UV reacts with maleic anhydride, then adds another mole thermally.  Palladium catalysis will swap appropriate substituents.


Why do alkyl chlorides have lower boiling points than the corresponding alkyl bromides and iodides? To my knowledge, the $\ce{C-Cl}$ bond is stronger than the $\ce{C-Br}$ and the $\ce{C-I}$ bond.

I always thought that the stronger a bond, the harder it is to break so the boiling point would be higher.

If that is the case, why does, among $\ce{RCl}$, $\ce{RBr}$ and $\ce{RI}$ (R = alkyl), $\ce{RI}$ have the lowest boiling point?

Your knowledge of the bond strengths is correct, but your understanding of boiling is flawed.

As orthocresol has mentioned in his comment, boiling is not about breaking the carbon-halogen bonds in your tetrahalomethanes.

Boiling does not mean decomposition!

You only transfer your material from the liquid to the gaseous state. Here, the molecules are still intact. But upon heating to the the boiling point, you have overcome the interactions between the molecules that existed before and held the whole bunch of molecules together as a lquid.

So, why is there the observed trend in the boiling points? Did you have a look at the molecular weights? ;)

What about the boiling points of alkyl halides, $\ce{R-Hal}$?

Let's have a look at the data for haloethanes, $\ce{C2H5-Hal}$:

$$\begin{array}{cr}
\mathrm{Hal} &\mathrm{bp~[°C]}\\
\hline
\ce{F} & -37 \\
\ce{Cl} & 13 \\
\ce{Br} & 38 \\
\ce{I} & 71 \\
\end{array}$$

Again, the trend is similar: Fluoroethane has the lowest, iodoethane has the highest boiling point.

Apparently, compounds with a higher molecular mass have a boiling point.

As a rule of thumb, the heavier the molecule, the more kinetical energy is needed to transfer it to the vapour phase. However, please note that this is a simplified model which might break when strong interactions between molecules, such as hydrogen bonds are involved!


Validity of van der Waals equation Van der Waals  had stated that at high pressure, pressure exerted by ideal gas on the walls of container is more than real gas.

However in this graph, one can clearly see that, the at high pressure volume of real gas tends to be more than volume of ideal gas. Hence real gas should be exerting more pressure on the walls of the container, as it occupies more volume. So why does this not hold true?

It's because in ideal gases there are no intermolecular forces whereas in a real gas the force with which it collides is reduced due to intermolecular forces.Now for the volume occupying part ideal gas particles are point sizes but in a real gas the molecules do occupy some volume.That's why some corrections were made to volume and pressure in the ideal gas equation.
Note that the above were assumptions of the Kinetic theory of gases.


Depression of freezing point Recently, I learnt that by adding a non-volatile solute to a solvent, the  vapour pressure of the solution is decreased and consequently its boiling point.

Facts and definitions

freezing point: "The temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase."

So as the solution cools down, the vapour pressure decreases and eventually equals the vapour pressure of the solid.

My argument as to why depression in freezing point shouldn’t take place

consider this argument:the atmospheric pressure is constant regardless of the temperature of the solution.

As we cool down a solution, we are basically removing the kinetic energy of the solution's particles  so due to less kinetic energy and due to atmospheric pressure the Van der Waal's forces come to play, hence turning the solution into a solid state.

the pressure exerted on the liquid by the atmosphere is the same whether you add a solute or not, so no depression of freezing point should take place.

Reality check

I know that this argument is not true, as experiments proves the colligative properties are true.

Where and why my "intuitive" argument is failing?

You argument is largely based on enthalpy, whereas freezing point depression and boiling point elevation is (in the situation when the solute only weakly interacts with the solvent) mainly an entropic phenomenon.

Because of the increased entropy by adding a solute (i.e. there are more configurations by having two types of molecules instead of one), the solution will liquify at a lower temperature as compared to the pure solvent and with a similar reasoning, the boiling point will increase with respect to the pure solvent.


Electrolysis: Water to Hydrogen gas? I recently conducted an experiment where I electrolyzed a solution of sodium chloride. I was using a copper coin as the anode, and I observed a pale blue precipitate in the cup when I was done with the experiment. I then deduced the following:

The chloride ions get oxidized at the anode:

$$\ce{2Cl- -> Cl2 + 2e-}$$

The chlorine gas then reacts with the copper anode to form copper(II) chloride which goes into solution as it is completely soluble.

At the cathode, the hydrogen in the water gets reduced to form hydrogen gas:
$$\ce{2H2O + 2e- -> 2H2 + 2OH-}$$

I know the above half reactions exist from the list of standard reduction potentials.

Next, the copper(II) ions react with the hydroxide ions being formed to produce copper(II) hydroxide, which is both insoluble and pale blue.

However, this source mentions that the hydronium ions in the water get reduced at the cathode:

$$\ce{2H+ + 2e- -> H2}$$

Is this correct? In my understanding, the concentration of hydronium ions ($\pu{1e-7 M}$) would be much too low, and where would the hydroxide ions come from to form copper(II) hydroxide?

In answer to your last statement about the formation of Hydrogen. The actual amount of hydronium ions doesn't matter because it is at equilibrium ($\ce{H2O <=> OH- + H+}$) and so once you use an $\ce{H+}$ you'll simply have more $\ce{H2O}$ dissociate into $\ce{OH-}$ and $\ce{H+}$ (Le Chatlier states) so if you have battery current running through the system you should experience a healthy, visible stream of hydrogen bubbles. I built a similar thing myself back in high school.

The hydroxide ions will come from the water also. As mentioned above, the $\ce{H2O}$ will dissociate into $\ce{H+}$ and $\ce{OH-}$ and you are using both. The $\ce{H2O}$ will keep dissociating and try to keep the $\ce{H+}$ and $\ce{OH-}$ concentrations at $10^{-7} \, \small\text{M}$  as you keep using them.

Feel free to question my knowledge.


Is there a way to tell what the coordination of a complex ion will be without knowing the ligand? Do complex ions tend to form a certain number of coordinate bonds, or does it completely depend on the ligand molecule?

There is no way you can predict this a priori. For example, cobalt(III) forms an octahedral hexaaquacomplex $\ce{[Co(H2O)6]^3+}$. If you add chloride to the solution, you will form a tetrahedral tetrachloridocomplex $\ce{[CoCl4]-}$. This was a pure ligand exchange reaction as shown in the equation below.

$$\ce{[Co(H2O)6]^3+ + 4 Cl- <=>> [CoCl4]- + 6 H2O}$$


Finding the pH among HF and KOH Okay I ran into this question in homework.

A 100.0 ml sample of 0.20M HF is titrated with 0.10 M KOH. Determine the pH of the solution after adding 200 ml of KOH. The Ka of HF is 3.5 x 10-4?

Oh yeah its multiple choice

3.46
10.54
8.14
9.62
7.00

I've found answers to similar questions.Here

But I come into a different problem with this question. The HF and the KOH cancel out each other for they have the same amount of moles

I assume what ever we're looking at on the other side will have 0.02 moles and is an acid

HF -> 0.1L * 0.2M = 0.02 mol
KOH -> 0.2L * 0.1M = 0.02 mol
HF-KOH = 0

I beleive this means I can't use the hasselbach equation so I did this:

(M of ???) = 0.02 mol / 0.3 L = 0.667*10^(-2)
3.5*10^(-4) = x^(2) / (0.667*10^(-2))
sqrt(3.5*10^(-4)*(0.667*10^(-2))) = x

so there is another way of solving it

I have had some feed back on this reaction: "You assume the first reaction goes to completion when the Ka is not that high"

If it doesn't go to completion what do I do?!

I really just need to see how this works out and know why things are happening

From the comments:

  [OP] Okay Thank you all I finally got it. so K is just thrown out the window. and we are left with the equation (F- + H2O <-> FH + OH-) So i take the Ka to make Kb by (Kw/Ka) and my concentration of F- and OH- was (0.02/0.3) which I plugged into sqrt((Kb)*(M of OH- or F-)) took the -log of that got the pOH and then converted it back to pH (14 - pH) One final thing why did I K out the window does that mean its neutral. Oh and the answers 8.14

Paraphrased with paragraphs:

A strong base ($\ce{KOH}$) reacts with a weak acid ($\ce{HF}$) at stoichiometric ratio:

$$\ce{KOH + HF -> H2O + F- + K+}$$

The major species is fluoride, a weak base with pKb = 14 - pKa, where pKa is that of hydrofluoric acid. The potassium ion is a spectator. To find the pH, use your favorite strategy for a pure weak base. The result is pH = 8.14.

As MaxW pointed out in the comments, this relies on getting the stoichiometric ratio just right. If $\ce{KOH}$ is even in slight excess (let's say one tenth of a millimole per liter too much), the pH will be strongly basic (pH 10). If you err in the other direction, you will have an $\ce{HF, F-}$ buffer with an acidic pH.


What is the structure of Cl₂O₆? I couldn't find which of these is correct and even someone I asked said there's no general covincement on it, but I'm unsure whether he's correct?

I reproduced the second structure a little wrong maybe, but arrangement and bonds are correct(I think there should be no charge on oxygen).

Dichlorine hexoxide has different structures depending upon whether it is in the gas or solid phase. In the gas phase, the molecule adopts structure (a) (reference 1, reference 2).  It is a relatively unstable molecule having a half-life measured in minutes.

In the solid phase it exists as the dark-red ionic material as shown in (b) and is often referred to as chloryl perchlorate.  Both forms are very strong oxidizing agents and react explosively with organic compunds.


What would happen to a Water/Lard Mix under pressure and high heat? If water is mixed with lard oil and heated (creating some super-critical water), how does this affect the volatility of the mixture in comparison if it is a pure mix of either water, or lard oil?

What happens when a mixture of lard oil and water is compressed and then heated to over 800 degrees Fahrenheit almost instantaneously (up to 1200 degrees Fahrenheit)? If it is in a well sealed copper container, would it be volatile enough to burst through the container?

Note: I'm no scientist or chemist, so use of laymen's terms, if possible, would be appreciated

I'm not sure what aspect you are asking about exactly… but all supercritical liquids are completely miscible with each other, so you would get one supercritical phase with the mixture of the two… supposing the oil doesn't decompose.

However, the smoke point of lard is 182°C while the critical temperature of water is 374°C, so what most probably happens is that your lard will burn, breaking down to glycerol, acrolein and free fatty acids.


What happens when Hinsberg derivative is added to acidic solution? The Hinsberg test of $\ce{C5H14N2}$ compound produces a solid that is insoluble in 10% aq. $\ce{NaOH}.$ This solid derivative dissolves in 10% aq. $\ce{H2SO4}.$ Which of the following would best fit these facts?

I think b or d is the answer.
The reason being it's insoluble in NaOH. That implies it should not contain 1 degree amine. Can anyone clarify any further?

" compound produces a solid" indicates that a reaction has occured.(hence no 3° amine group(according to this)). Further this is insoluble in base so there is a 2° amine group.

"This solid dissolves in acid" directly implies that there is a 3° amine group.

Hence there is both 2° amine group as well as 3° amine group in the compound.

Hence option b is the correct answer.

References:
1. Solomon and Fryle Hinsberg Test


Naming functional groups in a molecule Did I name the functional groups that are present in the following molecules correctly?  All I was given is the picture below. I have drawn my attempt in the picture.

Edit1: response to k--

Edit2: response to Mithoron

First molecule:
The ester group is correct - you have a carbonyl (C=O) carbon bonded to an oxygen and both sides are bonded to an R group (any carbon/hydrogen group).

Phenol is incorrect - for a phenol group, the hydroxy/OH must be directly bonded to the ring, like so:

So instead, what you've labelled comprises of two functional groups: the ring is an arene group, and the OH is a hydroxy/alchohol group.

You haven't labelled the two branch alkyl group on the left of the molecule - this is called an isopropyl group (it has three carbons, so it's basically a propyl group with something attached to the middle carbon). This group might not always be considered a functional group though (maybe ask your lecturer?). Or it can just be called an alkane group.

Second molecule:
The amide group is pretty much correct, except it should only include at most the atoms bonded directly to the nitrogen and carbonyl carbon.

The alkene should only include the double bond (while the whole chain could be considered an alkene, the alkene functional group only contains the double bond).

One that you haven't included is the ether group, R-O-R.

To learn these I would suggest just using a table of common functional groups until you start to remember them. I think the Wikipedia page is good because you can look at your molecule and see what unusual atoms you have (e.g. nitrogen, oxygen) and see if any of the groups under that heading look like they're in your molecule, except it might be missing a few e.g. phenol.


Is argon preferred over nitrogen for maintaining an inert atmosphere? I recently started a project in a new lab-group over Summer that does medicinal chemistry. Unlike in the undergraduate labs, they use argon instead of nitrogen for their inert atmosphere work.  When I asked around, the lab members said that it was 'just because they'd always has argon', though a few people said that they thought argon was better, without explanation.

As far as I am aware, both gases are unreactive and can be dried, therefore it makes no sense to me why argon would be preferred given how cheap and abundant nitrogen is. There are some air sensitive steps in my route but generally nothing too unstable

Is there any real difference between the two that would make one better, or is it, just as my lab said, that it's just what they have and they've never bothered to change it?

Speaking purely from an organic point of view, there isn't really a huge difference, labs tend to stick to one or the other, largely depending upon the availability (some universities use the boil off from the liquid nitrogen cylinders to furnish labs with nitrogen gas, whilst others have no such facilities meaning labs must buy cylinders of either nitrogen or argon).

For serious organic groups (organometallics, total synthesis) doing a lot of air-sensitive chemistry, there can be some advantage to argon due to its higher density [argon (1.714 g/L) nitrogen (1.251 g/L)], meaning it tends to blanket the reaction, and if you need to quickly open the flask to put a reagent in then the argon doesn't escape so quickly.

For inorganic chemists, there are situations where nitrogen will react with certain chemicals, and in these cases argon must be used.


How to pronounce G°? To denote standard state like enthalpy of formation in standard state, or change in Gibbs free energy in standard state, we use a degree-like symbol above the $G$ as in $\Delta G^0$. How should I refer to this superscript in spoken language?

G-zero and G-prime.

G-zero has also in my presence been referred to as "G-ought", "G-nought", "standard-G" or "G at standard state". I am particular towards "G-zero" myself. (Mostly because of a witty limerick I heard during my studies. Zero rhymes with hero.)


Developing Computational Chemistry Software I want to learn to develop novel methods in computational chemistry (e.g. Grimme's empirical dispersion, DFTB, new DFT functionals, etc.) What is the ideal way to learn this by oneself?

I have read books on the Hartree-Fock theory and I understand it to some degree. However, there seems to always be a large amount of details which are left unsaid and brushed under the carpet. I am looking for quite practical ressources which would teach me how to actually translate the theory into code. Is there a way to learn this other than just reading books until it sinks in?

How do I get into method development?

As was said in the comments, one would join a group that works in this field and learn from them. From up close, there is some derivation from previous results and some trial and error involved. And yes, a PhD-sized investment is typically required.

How do I get into method implementation?

It depends on how deep you wish to go. There are open-source projects to look at and contribute to. There are publicly available exercises on implementing HF and beyond. I would not advise to implement DFT from scratch because the numerical integration grid is both critical and difficult. Same is true for the integrals, use a library for them.

How do I learn to use Program X?

It is helpful to learn one program suite well and learn others later. The reason is that many concepts are similar between the programs. Learning one program comes down to locating good worked examples, reading the manual, attending a workshop if possible. One may for instance attempt to reproduce data from a DFT benchmark, since the supporting material usually has a detailed listing, possibly down to absolute energies.


What is deuterium-depleted water actually used for? I noted, flipping through the paperback Alfa Aesar catalogue today, that they sell deuterium-depleted water. Under the usage category, they list 'NMR'. I'm trying to work out what specific use this water has. Unfortunately, on the basis of very limited clinical evidence of improved cancer outcomes I've been bombarded mostly with companies recommending deuterium depleted water for drinking, which trips my snake oil sensor and is something I doubt Alfa has in mind (at US$1930 per liter, and a US RDI of 3.7 L/day for an adult male, that would seem to be a comically expensive drinking habit.)

I suspect biophysics experiments are a small part of the market for this water, but that there's some other NMR-related use I can't think of, possibly calibration related. What is it?

When you perform 1H-NMR spectroscopy in solution, there are many cases where you want to use deuterated solvent, so that signals coming from solvent hydrogen nuclei don’t interfere with the signal from your target molecule. Hence the frequent use of deuterated water, deuterated acetone, deuterated methanol, and deuterated chloroform. Aprotic solvents are also in common use, including carbon tetrachloride.

The situation is reversed when you want to perform 2D-NMR: you then need deuterium-free solvents, so their signal doesn’t mess with the molecule you're interested in. Hence, you will be interested in buying deuterium-depleted water.

(If you're wondering about real use cases for solution deuterium NMR, you can browse the literature for articles, there are plenty. One recent exemple is this one.)


Ashes to diamonds --- How? Several companies are offering a service to turn ashes of human remains into diamonds.
Does anyone know any details about the chemical aspects of the process?

I have no idea, but suspect that the ashes are composed mostly of phosphates and oxides. Where does the carbon hide? Any carbonate would have decomposed at the creation temperature. Or does the cremation chamber has such a high $\ce{CO2}$ partial pressure that $\ce{CO2}$ from burners is captured by the $\ce{CaO}$?

How is the carbon then extracted?

Also, diamond is normally made out of graphite, which agrees with a statement on the LifeGem website.

Let's assume they can squeeze out a little $\ce{CO2}$ by heating the ashes in an oxygen stream. Then they concentrate the $\ce{CO2}$ on a molecular sieve, use the Sabatier reaction to create $\ce{CH4}$ which they decompose on a hot metal wire. Then they dissolve that in acid and get a little bit of powder rich in $sp^2$ bonds and similar to graphite. Could one make diamonds out of that?

On the other hand, the page mentioned above states that they first extract the carbon and then turn it into graphite. How do you turn a few mg of fine carbon powder into graphite?

According to US published patent application 20030017932 "Cremeting; filtering carbon particles; sublimation; polishing"

  The preferred process for collection begins with the oven operator positioning the body in the oven so that the head and chest area are not positioned directly underneath the main burner. This can be accomplished by positioning the body to the left or right side of the main burner, or positioning the body so that the legs and feet are underneath the main burner rather than the head and torso. Positioning the body in this manner assures that carbon will remain in the body's head area. The carbon can then be gathered by hand, or by using a metal shovel or scoop, or the like.

So, in this technique (or proposed technique) the idea is to not let the normal cremation process occur, which would change much of the carbon to CO2.  It seems more like pyrolysis to charcoal.

They go on to say:

  Another alternative embodiment for carbon collection involves collecting carbon from pulverized cremated remains. These remains consist mostly of ash, but depending on how the cremation was performed, there may be traces of carbon particles mixed in with the ashes. The remains can be placed in a vacuum induction furnace. The furnace is heated to 2000 degrees centigrade in a vacuum ranging from 30 toir to 500 torr. Chlorine gas is injected into the furnace, and reacts with the impurities to form chlorides. The impurities leave the carbon in the form of chloride gases, and are filtered as they exit the furnace.

For more possible methods read this and other patent applications such as US20040031434.

  In another alternative embodiment of the invention, the remains can be cremated conventionally, mixed with additional carbon from another source, and purified as described above. It is contemplated that, using this technique, a gem containing at least some of the original carbon from the cremated remains can be prepared, even if the amount of carbon present in the remains alone is insufficient to make a gemstone of desired size or type.

There is no way to say what actually occurs at a particular company, unless someone has first hand knowledge, which I don't.


Tetragonal (z-in/z-out) in tetrahedral complexes? Is it possible to have tetragonal distortion in tetrahedral complexes? Because distortions in octahedral complexes take place along the C4 axis to reduce the symmetry to D4h, but there is no C4 operation in the Td point group, I was inclined to think there cannot be z-in/z-out as such?

Since $D_{h4}$ is not a subgroup of $T_d$, the z-in/z-out distortion you talk about does not give rise to a tetragonal distortion for a tetrahedral structure. It is possible to have tetragonal distortions in tetrahedral complexes, and has been observed in some cases (Bates and Chandler 1975) and (Sharnoff 1965). There is an illustration in Figure 1 of (Virot, Hayn, and Boukortt) which shows what turns out to be the easiest and most common tetragonal distortion for a tetrahedron, the one to the point group $D_{2d}$. So while z-in/z-out is associated with octohedral groups (and is usually the easiest way for them to have a tetragonal distortion), there are other distortions which occur in tetrahedrons.


Finding reactants in a azo-coupling reaction whose product is given Find the reactants of the following azo-coupling reaction:

My solution would look like this:

My reason for the above answer
The amino Group is more activating than the nitro group and therefore the diazonium salt had to be on the left ring.
Is this correct? If not, what is the correct explanation?
Edit: To get to the azo-coupling product shown above I should use a high temperature, or? So, the reaction has to be thermodynamically controlled because the azo-group is localized on the  carbon.

The comments summarized the reasons well: the diazotization reaction for the initial structure would produce a mixture of $\alpha$-azo and $\beta$-azo compounds, which would give a mixture of products. Also, diazonium salts couple with activated aromatic systems. The diazonium salt shown would not couple with nitrobenzene, which is highly deactivating and does not favour electrophilic aromatic substitution.
Also, having an electron withdrawing group on the diazonium salt favours the reaction. Thus, the correct solution would have the azo group on the smaller, less activated benzene ring rather than the larger, more activated naphthalene ring.


Hygroscopic material absorption level I live in a high humidity location and every winter, mold forms on one of my walls due to water sticking on it during the cold times. This also ruins the paint.

I found out that certain materials absorb water from the air, so I decided to make my own dehumidifier based on sea salt. The problem is alleviated slightly, but not fully - the salt becomes moist, but I think after a certain point it just can't absorb any more humidity.

So I was wondering how much salt I would need to bring the humidity in my room to a normal level.

I looked for things like "hygroscopic materials water absorption levels" on Google, but I did not have much luck.

What I am looking for is a formula like:

X kg of sea salt will absorb Y kg of water at Z% humidity

I have no background in chemistry, unfortunately, so I am not even sure if I am searching for this formula using the right terms.

I tried to do something similar when I had a humid car.

In chemistry, we often use $\ce{CaSO4}$ or anhydrite to serve as a desiccant. You could probably get cost-effective plaster for similar purposes. You could also get gypsum and heat it to generate plaster of Paris yourself.

As you're probably aware, people also use silica gel in products (and food) to minimize water content.

Either material should be much better as a desiccant than regular salt.

One key criteria is to increase the surface area - make sure the material is a fine powder and enclose it in something porous. I've seen people use various cloth materials, paper towels, etc.

As to how much you need, you can work out how much it will eventually absorb. If I take plaster of Paris to gypsum, each mole of $\ce{CaSO4}$ will take ~1.5 moles of water. So for every 136.14 g of anhydrite or ~145.15 g of plaster, you can absorb ~27 g of water. (I use ~ here, since I'm not convinced that regular plaster of Paris will be completely $\ce{CaSO4*0.5H2O}$.. some will likely already have extra water in it.)

The trick is that humidity will continually increase in the room if there's exchange of outside air.

Personally... I'd go with an electric dehumidifier.


How can one test if a ruby is real? My friend recently bought a ruby gem from a store and asked me how to test whether its an original one or a duplicate.

Are there any ways to test it using only domestic materials?

To test if it's ruby at home check it hardness. If it's harder than 7 even if it's not ruby it still something precious. :-) You can also try to find refraction index however for this you need some equipment. If you really want 100% prove, ask on a local chemistry faculty if they could "as a teaching example" make single crystal diffraction experiment. Simple run with 30 frames would be enough to find a cell what is enough to prove it's ruby.

Fitting your ruby on diffractometer would be challenge as a new diffractometers are working on crystals with size of 10--300 micrometers. To find if it natural or synthetic you need to go to specialize jeweler. They look for imperfections (natural stone are usually less perfect than synthetic).


Entropy of Irreversible Isothermal Compression If I want to find the entropy change of universe in isothermal irreversible compression, I use 2 approaches:

I use q/T. Which gives me positive entropy change for surroundings positive and negative entropy change for system. So, using clausius inequality to say that net entropy change is positive.
I find entropy change for irreversible isothermal expansion which comes out to be positive (again using clausius inequality) and say that the compression process is just the reverse of the expansion. So, entropy change being a state function must be negative(as it is just the reverse process).

What am I doing wrong here?

There is no reversible process that can take both the system and the surroundings to the same final states (simultaneously) that they reached in the irreversible process.  So, to get their changes in entropy for the irreversible process, we need to first separate the system from the surroundings and then subject each of them separately to new reversible processes (perhaps even involving a second and third "surroundings").

For the case of the original surroundings, we know that $Q_{irrev, surr}=-Q_{irrev, syst}$.  So, from the first law applied to the irreversible process $$\Delta U_{irrev, surr}=-Q_{irrev, syst}$$The alternate reversible process for the surroundings must take it between the same two states.  This means that the internal energy for the alternate reversible process must suffer the same change as for the irreversible process.  The only way this can happen is if the heat flow for the alternate reversible process is the same as for the irreversible process.  So, the entropy change for the surroundings in the irreversible process is given by $$\Delta S=-\frac{Q_{irrev, syst}}{T}$$

Here is an example of how this plays out for a specific irreversible isothermal compression or expansion.  The example works equally well for both expansion and compression.

The initial state of the system is specified by $P_1V_1=nRT$.  The irreversible process is specified by suddenly changing the external pressure on the piston from $P_1$ to $P_2$, and then allowing the gas to re-equilibrate irreversibly to its new equilibrium state with volume $V_2$ given by $V_2=P_1V_1/P_2$.  Since, during this irreversible process, the external pressure is constant, the work done by the gas on the surroundings is given by $$W=P_2(V_2-V_1)=nRT\left(1-\frac{V_1}{V_2}\right)$$This is also equal to the heat Q transferred from the surroundings to the system:  $$Q=nRT\left(1-\frac{V_1}{V_2}\right)$$So the entropy change of the system during this irreversible process is given by:  $$\Delta S_{syst}=nR\ln{(V_2/V_1)}$$And from the analysis I presented above, the change in entropy of the surroundings during this irreversible process is given by
$$\Delta S_{surr}=-nR\left(1-\frac{V_1}{V_2}\right)$$The overall change in entropy for the system plus surroundings for this irreversible process is thus given by
$$\Delta S=nR\left[\ln{(V_2/V_1)}-\left(1-\frac{V_1}{V_2}\right)\right]$$The term in brackets in this equation is positive for all values of the ratio of the volumes, whether the ratio greater than unity or less than unity.  Therefore, the Clausius inequality ($\Delta S>0$) is satisfied for both expansion and compression in this irreversible process.


Does this produce the anti-Markovnikov product because of the ROOR? I know that the anti-Markovnikov product is produced when there is a free radical present. Does the ROOR separate into two free radicals and that changes the reaction? But ROOR doesn't end up in the product at all right? Why? Is my thinking correct?

Yes, having a peroxide will change the regiochemistry to an anti-Markovnikov addition. The reason, as suggested, is because of a change in mechanism to a radical process (below). The homolytic decomposition of the peroxide initiates the radical chain, generating two hydroxyl radicals. The hydroxyl radical abstracts a hydrogen from HBr to give a bromine radical. This is where the cycle really begins, with bromine radical adding to the terminal end of the alkene, leaving a radical at the internal position. The stability of a tertiary radical relative to a primary radical drives this selectivity. The tertiary carbon radical abstracts a hydrogen from the next molecule of HBr, giving the product and a new bromine radical which reenters the radical chain.


How to draw chair-form for trans-decalin (with slight variation)? I understand how to draw chair form for trans-decalin (figure 1) for the usual structure, but I am having difficulty if taken as shown in figure 2 wherein the hydrogens at bridge-head carbons are in opposite orientation. Please help me with relevant chair form diagram.

Use of 3D models
3D version of Figure 1:

Viewed upside down (rotate about z by 180 degrees):

Now it looks like we are viewing from the bottom, so we will turn the molecule a bit around the x axis:

Use of 2D projections
If you want to use the 2D projections, consider whether hydrogens are pointing up or down on the bridge head. First, we will interpret the chair as viewing the top of the ring from above (left panel) rather than viewing the bottom of the ring from below (right panel) to avoid ambiguity.

No matter whether you use two fused chairs as drawn by the OP (Fig 1 bottom panel) or its mirror image, the front hydrogen will point down and the back hydrogen will point up. This works well for Fig 1.
In order to use the same chair building block but show the front hydrogen pointing up, you could rotate the flat model, and connect two chairs accordingly, as shown below. In effect, you choose the location of the two bridging position to match the stereochemistry you want.