Does storing apples alongside potatoes keep potatoes from sprouting?

Ethylene has been used in storing potatoes for a while, see this article. In it, they address the puzzle of a ripening agent acting to prevent sprouting:

  Sprout suppressant properties   of   apple   ‘vapours’   were   first   reported   in   1932   and   subsequently,  ethylene  and  apple  ‘vapours’  in  1933  (Elmer  and  Huelin,  respectively).    This  was  preceeded,  however,  by  work  demonstrating  a  ‘dormancy  breaking’  effect  of  ethylene  (Rosa,  1925).    These  apparently  conflicting  responses  to  ethylene  were  resolved  with  the  publication of work showing both effects, but under different ethylene exposure conditions.

The mechanism of the different effects are largely unknown. The cited article (by the British Potato Council) gives some references. Potatoes treated with ethylene do have slightly different properties (they mention the color after deep frying), and there is ongoing research to get an overview of all the differences.

  [OP:] Chemically speaking, do we actually know the exact reason and the necessary conditions to make this work?

We don't know the exact reason, but there are companies who figured out how to make this work in a warehouse under controlled conditions. I think on your shelf would be more difficult because you have no way of monitoring the ethylene concentration, so the apples might influence the potatoes in etiher way.


Comparing Basic strengths of different species

Comparing acidic strengths is fine - I remove an H+ and compare the stability of the anion coz my teacher told me strong acid has a stable anion and explained me the reason from equilibrium point of view.
  When it comes to comparing basic strengths what do we do ?

You follow the same general idea - the stronger the base, the weaker its conjugate acid. Consider a generic base such as $\ce{B-}$ reacting with water, the acid in this case.

$\ce{B- + H2O <=> HB + HO-}$

The conjugate acid of $\ce{B-}$ is $\ce{HB}$. And the stronger that $\ce{HB}$ is as an acid, the more likely it is to protonate hyroxide ion and favor the left side of the equilibrium, thereby making the base weak.

  Can I add an H+ to the species and compare the resulting stability by resonance ,hyperconjugation and inductive effect and then can I conclude the more basic one will have a stabler cation?

Yep, that's it.


When will acid-base reactions occur before substitution reaction

Acid-base reactions (like the one shown, involving proton transfer between electonegative atoms) are almost always extremely fast - for all practical purposes instantaneous (diffusion-controlled), and therefore much faster than substitution reactions. In the example given, the bromocyclohexanol is of comparable acidity to tert-butanol (maybe even more acidic), so we would expect instant proton transfer, and the observed result.

The initial proton transfer might not be so favorable if you used a much weaker base, so you could get a different result - or no reaction.


What is the formula for migration rate of DNA in agarose gel electrophoresis?

I guess my answer is very old, but here it is for newcomers searching the internet for this formula:

$$D = a - b \log(m)$$

where

$D$: Distance traveled for the restricted fragment.
$a$ and $b$: Constants depend on the electrophoresis experiment.
$\log(m)$: The logarithm of the molecular weight ($m$)of the fragment.

Knowing $a,b$ and the distance traveled in your gel, you can easily determine the molecular weight of your fragment and its size by the following formula:

$$m = e^ \left(\frac{a-D}{b}\right)$$


Collision rates between two molecules in a liquid

If you assume that the reaction is diffusion controlled  then this is the fastest that a bimolecular solution phase reaction can be and the second order rate constant is $k=8000RT/(3\eta)$ (dm$^3$/mol/s) where $\eta$ is the solution viscosity and assuming that the molecules are approximately the same size and even if they are not its still a good estimation. In mobile solvents (hexane, water etc)  $k\approx 10^{10} \to 10^{11}$ dm$^3$/mol/s.

If the species a, b are ions (charge $Z_{a,b}$) then the rate constant is multiplied by $\delta/(e^{\delta}-1)$ where  $\displaystyle \delta = \frac{Z_aZ_bq^2}{(4\pi\epsilon_0)\epsilon k_BTd_{ab}}$ where $d_{ab}$ is the centre to centre molecules size, $q$ electronic charge, $k_B$ the Boltzmann constant, $\epsilon$ the solvent dielectric constant and $\epsilon_0$ the permittivity of free space.

[If the species are not the same size radius $r$ then $\displaystyle k= \frac{2000RT}{\eta}\frac{(r_a+r_b)^2}{r_ar_b}$.  ]


Least likely product of a hydrocarbon cracking reaction?

There are two ways to look at this question.  One is simple carbon atom counts.  Another is the saturation of the products.

Carbon atom counts.  The question doesn't specify which of the 24894 isomers of heptadecane is to be cracked.   Suppose it is "normal" or n-heptadecane.  Cracking usually begins with the homolytic cleavage of a carbon-carbon bond.  To make a $\ce{C16}$ product from n-heptadecane by cleaving carbon-carbon bonds, there are only two potential bonds to cleave, one at one end of the molecule, another at the other end.  In either case, to obtain a $\ce{C16}$ product, only a single cleavage would be allowed, because further reactions would break more C-C bonds and change the product.  Cracking is a high temperature reaction that is entropically driven.  The system is increasing the disorder.  Imagine rolling four 6-sided dice, and the total comes up 17.  Then you take away either one or two of the dice (whether its one or two is random).  What is the probability that the remaining dice sum to 16, compared to say 8 or 4?

Saturation.  Choices B and C are unsaturated hydrocarbons.  Choices A and D are saturated hydrocarbons.  Both saturated and unsaturated compounds can be formed by cracking, but most modern cracking processes tend to focus on making unsaturated hydrocarbons.

Thus, primarily because of simple statistical unlikelihood, but also because of saturation considerations, D is the correct answer.

Wikipedia's article on cracking is very helpful.  See here.$%edit$


Concentration uncertainty due to volume measurement with cylinder

1. Let's assume that you're talking about random error. If you specify the error in the volume with $\pm$, this is probably the case---systematic error in this case would have a single sign; for example, if the scale marks on the side of cylinder were shifted by $\rm 0.5\ cm^3$ down from what their true values should have been, you'd have a systematic error of $-0.5\ \rm cm^3$ (the negative sign tells you that your readings are $0.5\ \rm cm^3$ lower than they should have been).

You can say that the error in a sum will be less than or equal to the sum of errors. It's better to say that the uncertainty in a sum is the square root of the sum of squared uncertainties:

$$\delta(V_1+V_2) = \sqrt{\delta(V_1)^2 + \delta(V_2)^2} = \sqrt{\rm(0.5\ cm^3)^2 + (0.5\ cm^3)^2} = 0.70_7\ \rm cm^3$$

For a product or quotient, you can say that the relative error will be less than or equal to the sum of the relative errors in the measurements. It's better to say that the relative error in a product or quotient is the square root of the sum of squared relative errors:

$$\begin{array}{rcl}
\delta(C_{dilute}) &=& \delta\left(\frac{C_{stock}V_1}{V_1+V_2}\right)\\
\frac{\delta(C_{dilute})}{C_{dilute}} &=& \sqrt{\left(\frac{\delta(C_{stock})}{C_{stock}}\right)^2 + \left(\frac{\delta(V_{1})}{V_{1}}\right)^2 + \left(\frac{\delta(V_{1}+V_2)}{V_{1}+V_2}\right)^2
}
\end{array}$$

...can you take it from there?

2. Let's assume you have systematic error, that is, you actually meant something like "The cylinder always reads $\rm 0.5\ cm^3$ too high". You will need a definite sign on a systematic error, not a $\pm$.

There are several ways to proceed here. Let's take the easiest route, which is to estimate the systematic error in the concentration as the difference between the concentration and its perturbed value (the value with all of the volumes having the systematic error added in). If your systematic error in volume was $\delta V = +0.5\ \rm cm^3$, you can estimate the systematic error in the concentration as

$$\begin{array}{rcl}
 \delta{C_{dilute}} &\approx& C_{dilute}^{perturbed} - C_{dilute} \\
&\approx& \frac{C_{stock}(V_1+\delta(V)}{(V_1+\delta(V))(V_2+\delta(V))} - C_{dilute} \end{array}$$

The sign on this error will be important; it tells you whether the systematic error in volume makes the concentration higher or lower than its "true" value.

Note: The above calculation is just an estimate of the error. You can do a better job using derivatives here. I'll show you that if you want to see it.


Lye (Sodium Hydroxide) is sometimes added to pool water to bring about increase in pH, without making significant increases to total alkalinity. It will bring about quite rapid changes in pH. I assume the intention is to bring the overall pH back within a sensible range of 7.4-7.6.

Adding high concentrations of sodium hydroxide (liquid lye is about 50% in water) might degrade the vinyl over long periods, but so too will high levels of acid, bleach, exposure to sun, and just about every other chemical used in pools. Vinyl liner can be damaged by low pH (below 7.0), but there will be other symptoms of high pH (above 7.8) (cloudiness, calcification, precipitation) long before damage to the vinyl liner kicks in, I think.

I think the nutty neighbour may be not too far from the mark to say "only use a little bit". All things in moderation.


There are many factors and the ideal way to understand this would be with the quantum mechanical description of atoms, which gets too complicated for answering such questions. But in general, following are few simple-to-understand face that helps in understanding the SRP values-

Electronegativity and Electron gain enthalpy: Higher the electronegativity, greater is the tendency to accept electrons, be reduced and hence greater (more positive) is the Reduction potential. For example, reduction potential of fluorine is greater than chlorine and that is greater than bromine. As we move down the halogen group, electronegativity decreases.
Ionization potential: This is a consideration for metals, which usually loose electrons and hence electronegativity is not the right quantity to consider. The lesser the IP, easier is the oxidation, and more negative (lower) is the SRP. In gas phase, the SRP of cesium is the most negative (in the group) and it becomes less negative as we move upwards.
Hydration: This is an important fact to consider since most electrochemical reactions occur in aqueous phase and hence the commonly cited SRP values corresponds to the aqueous phase. Greater the hydration of the ion formed after oxidation/reduction, greater is the tendency to lose/gain electrons and hence lower/higher is the SRP values. Lithium ion has very high hydration enthalpy and hence lithium has a strong tendency to lose an electron to form an ion which can be excellently solvated. This lends it one of the most negative SRP in the periodic table. The same holds true for fluorine, which is excellently solvated in the ionic form by water and hence has a high SRP.
Size, EC etc: Several factors like ionic size, it's electronic configuration, the inner orbital structure and the nuclear force of attraction determines how the incoming electron will be received and consequently the SRP values. Higher atomic size, higher inner electron repulsion, almost-stable configuration(just 1 electron more) and lower nuclear force, all favour lower SRP values.

All of these consideration, in varying proportions can be applied to the transition elements, but they act differently and to different extent in different compounds. Hence, predicting the SRP order without experimental data is very difficult and not trustworthy.


Do melt flow rate and density tell you everything about pure HDPE?

Yes, as a rule the melt flow rate and density do tell you what the other material properties will be. This is why HDPE is often specified by those two properties.


Silver phosphate is soluble in aqueous ammonia. AgBr is only slightly soluble in ammonia. AgI is not soluble in ammonia.


The information you provided ("two moles of NOCl react for every three moles of products which react") confirms that the reaction occurs as depicted, but is insufficient to establish that the system is at equilibrium. The important missing information is not the relative proportions of these three species in the mixture, but whether those proportions are changing (system is not at equilibrium) or staying the same (system is at equilibrium).


Holding your hands in this way merely proves that your hands are mirror images. If you take any object (chiral or not) and hold it up to a mirror, you can always align common features.

Imagine instead placing one of your hands inside the other. You may be able to align the overall thumbs and fingers, but they will be facing opposite directions, thus are not completely superimposable.

If you'd like further confirmation that your hands are chiral, try shaking someone's right hand with your left. Or try putting a left hand glove on your right. The interaction of mirror image forms (enantiomers) of a chiral object (like your hands) is different with the mirror image forms of another chiral object (like a glove or someone else's hand).


In simple molecules like what you've drawn above, see if the hydrogens are on the same side of the double bond or not.  If they are on the same side, then the compound is cis; if the hydrogens are on opposite sides, then the compound is trans.  In more complex cases, prioritize the four substituents on the double just as you've been doing with "R\S" nomenclature.  If the two highest priority groups are on the same side then the compound is cis, if they are on opposite sides, then the compound is trans.  The terms "cis" and "trans" are still used, but especially in the more complex cases they have been replaced with the first letters of the German words "entgegen" (E, opposite) and "zusammen" (Z, together).  So cis-2-butene could also be called Z-2-butene.


There's no technical reason why you couldn't make a powdered dishwashing detergent and I suspect that someone probably does sell it, just as there are liquid, tablet, and powdered automatic dishwasher detergents and laundry detergents. It does seem more convenient to use a liquid detergent for washing dishes in the sink. A liquid doesn't need time to dissolve, is easy to dispense without a scoop and if you're only washing a couple of dishes, you can simply squirt a small amount directly on a dish or use one of those brushes with a detergent reservoir.

I haven't seen any dishwashing detergent that had a full ingredients list, but the one I have says "contains anionic surfactants and enzymes". Anionic surfactant typically implies sodium dodecyl sulfate, which is itself a solid, usually found as a powder. The enzymes, while they could be freeze-dried, could retain more activity if stored in solution. Likewise, the fragrances often added to these detergents may be more suited to a liquid form.


How does the "kreosan" Coca-Cola rocket work?

Here's a paper written about propane forming small ice crystals when sprayed into a cloud of water vapor: James R. Hicks and Gabor Vali, J. Appl. Meteor., 1973, 12, 1025–1034.

I'm no expert, but I'm guessing the propane forms small ice crystals as it passes through the Coca-Cola, and when flipped, the Coca-Cola warms up the liquid propane, which causes it to expand as it travels upwards. This provides an immense surface of nucleating sites, similar to a Mentos reaction, but with millions of microscopic Mentos (this means larger surface area). The pressure waves caused by the violent nucleation further drives the reaction, as the liquid propane expands further and freezes more ice crystals and so on. This all happens quite fast, and with one easy way out, everything shoots out the mouth of the bottle.

Of course this was written by a Materials Engineering undergrad, so don't take this explanation too seriously.


Why plastic sheets contract when heated?

When plastic sheets are produced, they are rapidly cooled to keep the polymer chains oriented in a way that makes the sheets nice and flat. This is a relatively high-strain orientation since it is associated with the energy level of the molecules at the casting temperature.

Once the plastic is heated above its glass transition temperature, the polymer chains are no longer locked in that high strain orientation. They relax to a low energy orientation- curled and bending in a way that shrinks the bulk material.

As for the precise mechanism, I'm not exactly sure. I would guess that the shrunken conformation is entropically favorable because there are more arbitrary bends. This would decrease the Gibbs free energy, making it a more stable shape.

Alternatively, it could be that hydrogen bonding between chain elements makes the folded shape more enthalpically favorable.


ChEMBL database provides API access: https://www.ebi.ac.uk/chembl/ws
properties can be retrieved from the 'molecule' endpoint, for example aspirin:

https://www.ebi.ac.uk/chembl/api/data/molecule/CHEMBL25.json

Disclaimer: I'm the author of this API.


Diastereomers, enantiomers, and meso compounds of molecules with lots of chiral centres

Could we generalise that flipping an even number of chiral centres always gives either an enantiomer or the same molecule if it's a meso compound, and flipping an odd number of chiral centres always gives a diastereomer?

No. The generalization is slightly different: Comparing chiral centers in two stereoisomers with more than two chiral centres, if

all chiral centres are the same, it is the same molecule.
some but not all are switched, it is a diastereomer.
all are switched, it is an enantiomer (or meso compound).

Open-chain carbohydrates are a good example (compare D-glucose, L-glucose, D-galactose, D-talose etc.). For this class of molecules, there is one more term, epimer (when exactly one chiral centre is switched).

This generalization excludes the more exotic cases of stereoisomers, see e.g. Why are allenes chiral? and Why atropisomers are called conformers?


How could I easily quantify the amount of EGCG in a sample of green tea?

To analyze specifically this component, you will have to use HPLC.
In case you can use a more generic approach, you can analyze the total phenolic content with the Folin-Ciocalteu method.

The Folin-Ciocalteu method is based on a reaction with polyphenols that gives a blue color.
This color can then be measured in a UV-Vis spectrophotometer and the concentration can be established.

This method has been used to quantify polyphenols in many plants, as for example here.


Why is maltose a reducing sugar but not sucrose, even though they're both disaccharides?

The difference in stability between maltose and fructose boils down to the different structural elements their aldehyde/ketone groups have been turned into during the formation of the disaccharide.

In maltose, you combine two glucose units using the 1-hydroxy group of one and the 4-hydroxy group of the other. The 1-hydroxy group has been created in a process called acetal formation by the attack of what used to be the 5-hydroxy group and is now the ring oxygen. This is shown in equation $(1)$, the explicit carbon atom being carbon-1; the aldehyde.

$$\ce{R'-OH + RCHO <=>> R'-OH^+-CHRO^- <=>> R'-O-CHR-OH}\tag{1}$$

Notice that there are two oxygens bound to that carbon after the reaction; if we wanted, we could label it $\ce{RCH(OR)(OH)}$; this structure is called hemiacetal. If we use that hemiacetalic hydroxy group to bond to a second glucose unit,[1] we have replaced the hydrogen with an alkyl residue. Like the difference between alcohols ($\ce{R-OH}$) and ethers ($\ce{R-OR'}$), this group has a different reactitivy and is termed full acetal or O,O-acetal.

Hemiacetals are rather easy to break down as two mechanisms are available: you can either initially protonate the other oxygen or you can use a base to facilitate proton removal of the hydroxy group — ideally perhaps while offering the other oxygen a proton by another base. Full acetals, however, are not as easy to break down; the require protonation, implying a sufficiently strong acid.

Coming back to maltose, remember we have two glucose subunits. We used one to generate a full acetal but the other one (on the right-hand side of your scheme) is still hemiacetalic and can reconvert back to an aldehyde. This is why you preserve the reducing abilities; remember that equation $(1)$ is fully reversible and according to Le Châtelier removing aldehyde (e.g. by oxidation) will enhance the reverse reaction.

In sucrose, on the other hand, carbon 1 of glucose and carbon 2 of fructose are connected. These are the two carbon atoms that form hemiacetals in isolated glucose/fructose. Because of this, there is no hemiacetal group $\ce{RCH(OR)OH)}$ in sucrose — only full acetals — and no way it can easily revert back to an aldehyde or a ketone. Therefore, it has lost its reducing ability.

This is the key information of your quotation by the way. The anomeric carbon is the one that will be a hemiacetal. If two anomeric carbons are connected, they have formed a full acetal. Therefore, identifying the anomeric carbons and seeing if they still have a free hydroxy group (i.e., if they are hemiacetals) allows quick determination whether a disaccharide is reducing or not.

Note:

[1]: While formally correct; the actual reaction mechanism involves the loss of this hemiacetalic hydroxy group.


Why is the temperature graph sigmoidal in simple distillation of components with very different bps?

You must understand... in an experiment, the sigmoidal shape will never have infinite slope... This would be a discontinuity... the sigmoidal shape (rightmost) and the "curve" (middle image) are essentially representing identical phenomena... there is a transition from one pure component to another. The only difference is how abrupt the change is... and this is determined by the endpoints/boiling points for each component. The more abrupt the change from one to another pure component, the steeper the slope of the change. I hope this helps. I can speak to this further, if necessary.


Why are some chemicals packaged in incompatable containers?

The shrinkage is called panelling and bottles that are otherwise incompatible are made compatible by a process called fluorination. In fluorination the HDPE is exposed to fluoride gas with precise control in time, temperature and pressure. This exposure converts the outer layers of polyethylene to a highly fluorinated polymer similar to teflon.

$$\ce{-(C2H2)_n-~+ 4 F2~~ \longrightarrow ~~-(C2F4)_n-~ + 4HF}$$

This fluoropolymer coating greatly reduces diffusion of the solvent into the plastic and thus the solvent is unable to soften the plastic, preventing paneling.


What are unit cells, and how do you determine the number of atoms in one of them?

The unit cell is the most primitive 3D structure that contains the arrangement of atoms which form the crystal of a compound. The unit cell is probably most meaningful in the study of minerals, so it is more of an inorganic analytical technique rather than an organic analytical technique. There are 7 3D crystal lattice systems which form 14 primitive 3D forms called the Bravais lattices. The 14 Bravais lattices can be further subdivided into 32 point groups.


Has it been observed that the dihydrogen molecule dissociates into ions?

Surely1. At the end of the day, it is just the matter of supplying enough energy: the homolysis requires approximately 4.5 eV per hydrogen molecule, while for heterolysis more than 17 eV per molecule must be provided.

1) G. Busca, A. Vaccari, Heterolytic dissociation of hydrogen on high-temperature methanol synthesis catalysts, Journal of Catalysis, 108(2), 1987, 491-494, http://dx.doi.org/10.1016/0021-9517(87)90198-9.


Preparation of acetate buffer from sodium acetate and hydrochloric acid

To answer this question, we have to assume that the given buffer solution is an ideal solution of which the total volume of it always equals to the summed volumes of sodium ethanoate ($\ce{NaOAc}$) and $\ce{HCl}$ solutions when they are added together. It should also be assumed that $\mathrm{p}K_\mathrm{a}$ of ethanoic acid (acetic acid; $\ce{HOAc}$) is given ($\mathrm{p}K_\mathrm{a} = 4.76$).

When added together, $\ce{HCl}$ (a strong acid) and $\ce{NaOAc}$ (weekly basic salt) react completely to give $\ce{HOAc}$ and $\ce{NaCl}$, according to the following equation:
$$\ce{NaOAc + HCl -> HOAc + NaCl} \tag{1}$$

Suppose you added $x$ volume of $\ce{NaOAc}$ and $y$ volume of $\ce{HCl}$ to make the $\pu{1.0L}$ of buffer ($\mathrm{pH} = 4.43$). Thus if units of $x$ and $y$ are $\pu{L}$, then:
$$x + y = 1 \tag {2}$$

And, after acid-base reaction, according to the chemical equation given above and equation $(2)$, remaining amounts of $\ce{NaOAc}$ in $\pu{mol}$ is $(0.3x-0.2y)=\left(0.3x-0.2(1-x)\right)= (0.5x-0.2)$ and amounts of $\ce{HOAc}$ formed in $\pu{mol}$ is $0.2y= 0.2(1-x)=(0.2-0.2x)$. Thus, since total volume of buffer is $\pu{1.0 L}$, respective concentrations of $\ce{NaOAc}$ and $\ce{HOAc}$ in the solution are as follows:
$$\ce{[NaOAc]} = (0.5x-0.2)/1.0=(0.5x-0.2) \text{ and } \ce{[HOAc]} = (0.2-0.2x)/1.0 = 0.2-0.2x$$

Now you can substitute these values to Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[NaOAc]}}{\ce{[HOAc]}}$$

$$4.43 = 4.76 + \log \frac{(0.5x-0.2)}{(0.2-0.2x)}$$
Thus,
$$ \log \frac{(0.5x-0.2)}{(0.2-0.2x)}= 4.43 - 4.76= -0.33$$
$$ \frac{(0.5x-0.2)}{(0.2-0.2x)}= 10^{-0.33} = 0.468$$
Once you solved for $x$, you would get, $x=0.495$ and $y=1.0-0.495=0.505$ in $\pu{L}$, respectively (or $495$ and $\pu{505 mL}$). Thus, I'd say the given answers for your question is incorrect.


Both mechanisms will probably be present but $S_N2$ will dominate because the enolate anion is a strong nucleophile. $E2$ will not occur because there are no $\beta$ hydrogens available on the benzyl chloride. Despite the relative stability of the benzyl cation, especially in protic solvents, the strong nature of the enolate nucleophile and the lack of an $E2$ pathway means that $S_N2$ will probably dominate.

For a comparison of alkyl halide reactivities with different nucleophiles, see this link.


The colorimetric method using rosaniline and formaldehyde is more sensitive and it is well documented.

A somewhat safer version of the test (avoiding use of highly dangerous $\ce{HgCl2}$) is documented at MDPI - Open Access Publishing:

  "A concentrated pararosaniline stock solution (3.088 × 10 −3 mol/L) was made by dissolving pararosaniline hydrochloride in a 10% aqueous ethanol solution. The concentrated stock solution was used to make... [a] working solution contain[ing] pararosaniline hydrochloride, hydrochloric acid (HCl) and formaldehyde in the molar ratio of 1:1255:125 respectively. For example, a working solution with PRA concentration of 4.68 × 10 −4 mol/L in 25 mL contained 3,790 μL of concentrated PRA stock (3.088 × 10 −3 mol/L), 1,439 μL of 32% HCl, 189 μL of 36% formaldehyde solution and 25 mL with water."

Pararosaniline is available from Sigma-Aldrich, though it seems a bit pricey at ~$US82/25 g; perhaps your school would obtain some for use by yourself and by other students, as you would need perhaps a gram for your experiments.


Would there be need for periodic DFT if we are able to get all the forcefields of all the elements in the periodic table in ReaxFF?

Quoting from the article The ReaxFF reactive force-field: development, applications and future directions

"Such empirical methods, including reactive force-field (ReaxFF),1 trade accuracy for lower computational expense, making it possible to reach simulation scales that are orders of magnitude beyond what is tractable for QM."

So yes, if you want accuracy, especially for something which the approximations made in generating a ReaxFF force field are not good, you are going to have to get out the Quantum Mechanics.


If electrons could have $m_s = -1/2, 0,$ and $1/2$, then the entire structure of the Periodic Table would be different. The s-block would have 3 elements instead of the 2 that we see now, because each s orbital can accommodate 3 electrons, and similarly the p-block would have 9 elements instead of 6.

It is not possible to say, for example, "sodium would be in so-and-so position in the new Periodic Table" because in such a universe, the element sodium would not even exist the way it does in our universe. The new Periodic Table would not have an element with the electronic configuration $\mathrm{(1s)^2(2s)^2(2p)^6(3s)^1}$. What I am basically saying is, there is no one-to-one correspondence between the elements in our universe and the elements in such a hypothetical universe (mathematically speaking, you can, but it would not make any chemical sense). Nevertheless, you could say that sodium is defined to be the element with 11 protons and 11 electrons. In that case, it would have a configuration of $\mathrm{(1s)^3(2s)^3(2p)^5}$ and it would be the fifth element in the Period 2 p-block.

Actually, there is no evidence that the aufbau principle should still hold true in such a universe: the large repulsion that arises from putting 3 electrons into the same orbital might well make filling the 2s orbital before the 1s orbital more favourable. So, our hypothetical "sodium" might have a weird electronic configuration of $\mathrm{(1s)^2(2s)^2(2p)^7}$. The Periodic Table is useful partly because it groups elements with similar electron configurations - which implies similar chemical properties - together. However, if these elements do not obey the aufbau principle, then there is no guarantee that the elements in the same group should have similar chemical properties, and therefore no guarantee that a Periodic Table in such a hypothetical universe would even be useful.

I should end off by saying that the laws of quantum mechanics stipulate that the different values of $m_s$ must increase in steps of $1$. Therefore, it is not actually possible that $m_s$ could ever take on values of $-1/2, 0$, and $1/2$, since these increase in steps of $1/2$. You might say: "Ah! But we could say, what happens if $m_s = -1, 0, 1$." Well, in that case, you have much more to worry about than just the orbital occupancy. Without going into too much detail, the Pauli exclusion principle would no longer apply to electrons, and you would have more problems than just "one orbital can fit 3 electrons".


Can a liquid turn to a solid by applying pressure?

It is possible to change a liquid into a solid, or in general change the phase a substance finds itself in by just altering the pressure.

You might want to refer to some graphs known as Phase diagrams that portray the effect of temperature and pressure on the physical state of a substance.

(source: purdue.edu)

In this example if you're to draw a straight line from the point B to D, it'll have a subtle positive slope. So there's a possibility to transform a liquid into a solid without making any changes to the temperature conditions.

You might have heard of solid $\ce{CO2}$ or dry ice. I might be wrong but a part of me says carbon dioxide is pressurised and refrigerated for it to form.

An interesting thing to note here is that in case of a gas turning into a liquid, there's a temperature known as Critical Temperature ($T_c$) above which the gas cannot be liquified no matter how much pressure is applied.


Two places where you can find predicted values of LogP (the oil:water partitioning coefficient) are chemicalize.org and Wolfram Alpha.  Both provide a predicted value for most of the compounds you are interested in (chemicalize provides a complete set, assuming you mean betacarotene).
Examples
This link provides a direct link to the output of the cholesterol page from Chemicalize.org and shows the logP value on the left hand side (7.11)
To get the logp from Wolfram alpha you must search for logp  and then chose the predicted link that appears when no results are returned.  This link should get you to the value obtained from Wolfram Alpha (7.02).


Can potassium nitrate and sulfuric acid be used to nitrate organics?

A simple, but dangerous path to nitrating a metal oxide (say MgO) is to treat with NO2 gas (caution: toxic and corrosive), as to quote Wikipedia on Nitrogen dioxide:

NO2 is used to generate anhydrous metal nitrates from the oxides:[10]
$\ce{MO + 3 NO2 -> M(NO3)2 + NO}$

where heating mix of Cu, KNO3 and H2SO4 will generate NO2.
A safer path to nitrating, starts with the introduction of the powerful  radicals, like the hydroxyl radical (•OH). The latter can transform commonly available nitrate salts  (like KNO3) to the nitrate radical (•NO3) especially in the presence of an acid catalyst per the reaction:
$\ce{•OH + NO3- <=> OH- + •NO3}$
A supporting source reference, Nitrate radicals and biogenic volatile organic compounds: oxidation, mechanisms, and organic aerosol. Per the paper's abstract:

Oxidation of biogenic volatile organic compounds (BVOC) by the nitrate radical (NO3) represents one of the important interactions between anthropogenic emissions related to combustion and natural emissions from the biosphere.

Also, see The Reactivity of the Nitrate Radical (•NO3) in Aqueous and Organic Solutions, by
Stephen P. Mezyk, Thomas D. Cullen, Kimberly A. Rickman and Bruce J. Mincher, per the abstract to quote:

ABSTRACT
Rate constants for the nitrate (•NO3) radical reaction with alcohols, alkanes, alkenes, and several aromatic compounds were measured in aqueous and tert‐butanol solution for comparison to aqueous and acetonitrile values from the literature. The measured trends provide insight into the reactions of the •NO3 radical in various media. The reaction with alcohols primarily consists of hydrogen‐atom abstraction from the alpha‐hydroxy position and is faster in solvents of lower polarity where the diffusivity of the radical is greater. Alkenes react faster than alkanes, and their rate constants are also faster in nonpolar solution. The situation is reversed for the nitrate radical reaction with the aromatic compounds, where the rate constants in tert‐butanol are slower. This is attributed to the need to solvate the NO3− anion and corresponding tropylium cation produced by the •NO3 radical electron transfer reaction. A linear correlation was found between measured rate constants in water and acetonitrile, which can be used to estimate aqueous nitrate radical rate constants for compounds having low water solubility.

With respect to the hydroxyl radical, the UV photolysis of a photocatalyst like TiO2 in aqueous KNO3, could be a path to the powerful, but very transient, hydroxyl and nitrate radicals.


@Tyberius: I think you're neglecting Berry Holmes' comment above. Dilution is not always inconsequential to the pH, especially when concentrations get quite low. As my analytical chemistry teacher once put it, if you squeeze a lemon into a lake...

@notiarogers: Tyberius already showed you that you miscalculated the formal concentrations of $\ce{HA}$ and $\ce{A^-}$. Don't skip any steps: $50 \space mL$ of $0.1 \space M$ solution contain $50 \cdot 0.1 = 5 \space mmol$ of substance; the final volume is $100 \space mL$, therefore the final concentration is $5 \space mmol / 100 \space mL = 0.05 M$.

The formula used above (for the calculation of the pH of an aqueous buffer solution made from a weak acid + its salt with a strong base) is only an approximation. Its main assumption is that the concentrations of acid $\ce{HA}$ and anion $\ce{A^-}$ are equal to the nominal ones (0.05 M, as explained above). In reality, when you dissolve $\ce{HA}$ in water, some of it already dissociates (forming some $\ce{A^-}$), and when you dissolve $\ce{NaA}$ in water, it dissociates to $\ce{Na^+}$ and $\ce{A^-}$, and some $\ce{A^-}$ already reverts to $\ce{HA}$ (which, BTW, is why the former solution is slightly acidic and the latter is slightly basic). So things are quite a bit more complicated than it seems: even your initial solutions don't contain 'only' $\ce{HA}$ or 'only' $\ce{A^-}$.

Now, it may be that whoever is asking you to do this calculation wants you to stick to such approximation, and in fairness the final concentration after dilution is probably still large enough, unless you want a very precise result.

In case they want you to do a more complicated calculation, you can write an equation for each equilibrium, one for each mass balance and one for the charge balance. Your equation is the dissociation equilibrium of $\ce{HA}$: $$\ce{K_A=\frac {[H^+] \cdot [A^-]}{[HA]}}$$
You can add the dissociation equilibrium of water: $$\ce{K_w=[H^+] \cdot [OH^-]}$$The mass balance of all acid-related species is:$$\ce{[HA] + [A^-] = C_{HA} + C_{NaA}}$$ where $\ce{C_HA}$ is the nominal concentration of $\ce{HA}$, i.e. 0.05 M as calculated above, and $\ce{C_{NaA}}$ is the nominal concentration of $\ce{NaA}$, i.e. also 0.05 M in this case.

The charge balance is: $$\ce{[H^+] + [Na^+] = [OH^-] + [A^-]}$$
And as $\ce{[NaA]}$ is a strong electrolyte, we know that: $$\ce{[Na^+] = C_{NaA}}$$

If you put all these equations together and eliminate all unknowns except $\ce{[H^+]}$, you get a 3rd degree equation in $\ce{[H^+]}$ and known constants alone, which you can solve (probably best numerically) to find the pH.

The exact solution would look very different from the approximate one you were using, which was essentially:$$\ce{[H^+] = K_A \cdot \frac{C_{HA}}{C_{NaA}}}$$

I can tell you that the pH after your 10-fold dilution is only slightly higher than 4.76 (it's 4.763). You would have to dilute 100-fold to start seeing some difference (4.79); and frankly that looks to me like a minor difference for most practical applications.

Note: we are neglecting the difference between activity and concentration; but I guess we have seen enough maths for one day...


Migratory aptitude in pinacol pinacolon rearrangement

Groups with higher electron density have a greater tendency to migrate. So, o-methoxy phenyl should be the answer but due to steric factors p-methoxy phenyl can always be an equally likely answer.  It can only be found out accurately by experimentation and considering other reaction conditions.


Does deuterated water slow down the overall metabolism of a cell?

Yes it would, by a few percent. It may or may not be a goal worth pursuing, but there is more to it.

Different reactions would be slowed down to a different extent. Tiny as they are, these discrepancies suffice to disrupt the delicate biochemical machinery of the living cell. No life form more advanced than bacteria is OK with that. Deuterated water in high concentrations is unpleasant for plants and downright lethal for animals, including us.

So much for your slowed aging.


Significant Figures in almost-scientific notation

Most of the answers to your questions can be found in this wikipedia article. You are right if you say:

  I've also learned that for a number to be in scientific notation, it
  must have a stem between 1 and 10.

This is the convention in scientific notation. The number format of $0.40 \times 10^3$ is called exponential notation (at least it was called that way in a book about the programming language Fortran I'm reading at the moment): Here the number has a stem between 0 and 1 followed by a power of 10. This is quite common in computer science.
As to the significant digits: This wikipedia article (especially the paragraph "Scientific notation") contains the information you need. According to its information I would say that the number $0.40 \times 10^3$ has $2$ significant digits.


What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers?

Here are the electron configurations of the relevant elements:

Source: https://chemistrytheperiodictable.files.wordpress.com/2017/11/electronic-config-lanthanides.png?w=474&h=399

The first $f$ electrons get filled in the 6th period of the table of elements. First, the $6s$ get filled, then (for lanthanum) the $5d$ gets filled for the first time, and then the $4f$ subshell. As you go across the lanthanides, the $5d$ is sometimes filled with one electron and sometimes empty.

  What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers: n=4,ml=+3

The answer depend on how you interpret "last electron". Going from Thulium to Ytterbium, you could argue the last electron is going into $4f$, whereas from Ytterbium to Lutetium, the last electron is going to $5d$. On the other hand, the $5d$ is filled before $5f$ going from Barium to Lanthanum. Comparing Lanthanum to Lutetium, all 14 electrons seem to be going into the $4f$ subshell, supporting the notion that the answer to the question is 71.

In the end, it is not the best question - it does not have a clear answer, and you don't really learn anything from the answer other than that the energies of $4f$ and $5d$ are close to each other.


How to calculate Ka for hydronium and Kb for hydroxide?

To an extent, you are right; the calculation of the $K_\mathrm{a}(\ce{H2O})$ does include $[\ce{H2O}]$.

Consider the following equation:
$$\ce{H2O(l) +H2O(l) ->H3O+(aq) + OH- (aq)}$$

The equilibrium for this equation is:

$$K_\mathrm{a}(\ce{H2O})=\frac{[\ce{H3O+}]\cdot [\ce{OH-}]}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Since $K_\mathrm{w}=[\ce{H3O+}]\cdot [\ce{OH-}]=1.01 \times 10^{-14}$ at $25^\circ \mathrm{C}^{[1]}$, we can make such a substitution in our equation:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Now here you will have to bear with me. In a usual $K_\mathrm{a}$ calculation, $[\ce{H2O}]$ is disregarded, as it remains mostly constant ($55.5084\ \mathrm{M}$), except when working with extremely concentrated solutions. In this equation, there are two $[\ce{H2O}]$ terms, and only one of them is disregarded (as a solvent). The other one can be considered concentration of the acid. Therefore:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]}=\frac{1.01 \times 10^{-14}}{55.5084}=1.819544 \times 10^{-16}$$

Thus we have calculated the $K_\mathrm{a}$ for water, or the $K_\mathrm{b}$ for $\ce{H3O+}$. The relation between these is:

$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$

So:

$$K_\mathrm{a}(\ce{H3O+})=\frac{1.01 \times 10^{-14}}{1.819544 \times 10^{-16}}=55.5084$$

and:
$$-\log(K_\mathrm{a}(\ce{H3O+}))=\mathrm{p}K_\mathrm{a}(\ce{H3O+})=-\log(55.5084)=-1.7404$$

The calculation for the $K_\mathrm{b}(\ce{OH-})$ is exactly the same for the calculation of $K_\mathrm{a}(\ce{H3O+})$ (pretty much just replace all the $\ce{H3O+}$ with $\ce{OH-}$) and yields the same result.

[1] Starkey, R.; Norman, J.; Hintze, M. Who Knows the Ka Values of Water and the Hydronium Ion? J. Chem. Educ. Journal of Chemical Education. 1986, 63, 473.


Finding region of electromagnetic spectrum with wavenumber

Wavenumbers are a unit of energy. Given that $1\mathrm{eV}$=$8100\mathrm{cm^{-1}}$, and the relations $E=h\nu$ and $c=\lambda\nu$, where $h$ is Planck’s constant, $c$ is the speed of light, and frequency $\nu$ and wavelength $\lambda$, you should be able to solve this with some simple algebra. Once you calculate the wavelength, try to find where it falls on the electromagnetic spectrum$^{[1]}$.

 $^{[1]}$ EM Spectrum - cyberphysics


Anhydrous magnesium sulfate to replace epsom salt?

Yes, they become the same, with some rare exceptions. There is usually difference in the dissolving kinetics, but the final result is the same.

Sometimes there are different thermal effects, Hydrates have sometimes tendency to cool the solution, while anhydrates to warm it. It is caused by released hydration energy of the anhydrate.The typical example is calcium chloride.

The obvious thing is to recalculate the needed amount for the contained water.


Why is lithium the most reducing alkali metal, and not caesium?

The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated.

$$\begin{array}{cc}
\hline
\ce{M} & E^\circ(\ce{M+}/\ce{M}) \\
\hline
\ce{Li} & -3.045 \\
\ce{Na} & -2.714 \\
\ce{K} & -2.925 \\
\ce{Rb} & -2.925 \\
\ce{Cs} & -2.923 \\
\hline
\end{array}$$
Source: Chemistry of the Elements 2nd ed., Greenwood & Earnshaw, p 75

However, a full description of the middle three metals is beyond the scope of this question. I just thought it was worth pointing out that the trend is not really straightforward.

The $\ce{M+}/\ce{M}$ standard reduction potential is related to $\Delta_\mathrm{r}G^\circ$ for the reaction

$$\ce{M(s) -> M+(aq) + e-}$$

by the equation

$$E^\circ = \frac{\Delta_\mathrm{r}G^\circ + K}{F}$$

where $K$ is the absolute standard Gibbs free energy for the reaction

$$\ce{H+ + e- -> 1/2 H2}$$

and is a constant (which means we do not need to care about its actual value). Assuming that $\Delta_\mathrm{r} S^\circ$ is approximately independent of the identity of the metal $\ce{M}$, then the variations in $\Delta_\mathrm{r}H^\circ$ will determine the variations in $\Delta_\mathrm{r}G^\circ$ and hence $E^\circ$. We can construct an energy cycle to assess how $\Delta_\mathrm{r}H^\circ$ will vary with the identity of $\ce{M}$. The standard state symbol will be dropped from now on.

$$\require{AMScd}
\begin{CD}
\ce{M (s)} @>{\large \Delta_\mathrm{r}H}>> \ce{M+(aq) + e-} \\
@V{\large\Delta_\mathrm{atom}H(\ce{M})}VV @AA{\large\Delta_\mathrm{hyd}H(\ce{M+})}A \\
\ce{M (g)} @>>{\large IE_1(\ce{M})}> \ce{M+ (g) + e-}
\end{CD}$$

We can see, as described in Prajjawal's answer, that there are three factors that contribute to $\Delta_\mathrm{r}H$:

$$\Delta_\mathrm{r}H = \Delta_\mathrm{atom}H + IE_1 + \Delta_\mathrm{hyd}H$$

(the atomisation enthalpy being the same as the sublimation enthalpy). You are right in saying that there is a decrease in $IE_1$ going from $\ce{Li}$ to $\ce{Cs}$. If taken alone, this would mean that $E(\ce{M+}/\ce{M})$ would decrease going from $\ce{Li}$ to $\ce{Cs}$, which would mean that $\ce{Cs}$ is a better reducing agent than $\ce{Li}$.

However, looking at the very first table, this is clearly not true. So, some numbers will be needed. All values are in $\mathrm{kJ~mol^{-1}}$.

$$\begin{array}{ccccc}
\hline
\ce{M} & \Delta_\mathrm{atom}H & IE_1 & \Delta_\mathrm{hyd}H & \text{Sum} \\
\hline
\ce{Li} & 161 & 520 & \mathbf{-520} & 161 \\
\ce{Cs} & 79 & 376 & \mathbf{-264} & 211 \\
\hline
\end{array}$$
Source: Inorganic Chemistry 6th ed., Shriver et al., p 160

This is, in fact, an extremely crude analysis. However, it hopefully does serve to show in a more quantitative way why $E(\ce{Cs+}/\ce{Cs}) > E(\ce{Li+}/\ce{Li})$: it's because of the extremely exothermic hydration enthalpy of the small $\ce{Li+}$ ion.

Just as a comparison, the ionic radii of $\ce{Li+}$ and $\ce{Cs+}$ ($\mathrm{CN} = 6$) are $76$ and $167~\mathrm{pm}$ respectively (Greenwood & Earnshaw, p 75).


This answer is based on the educated guess that the green excitation Raman spectrum was acquired with a dispersive Raman instrument*.

  But why is there general nonzero background in intensity? What caused it?

For any real-life instrumentation, you should never assume to measure 0 signal for 0 analyte concentration (see also Does a calibration curve of absorbance vs concentration need to pass exactly through the origin?) nor expect the baseline to be at zero unless corrected for, even if there really were no Raman scattering outside those two bands, the instrument and/or sample preparation procedures (electronics, optics, masking, cross-sensitivity, ...) will often lead to some non-zero signal.

The CCD in the spectrograph has a so-called dark current. So the raw signal you observe is dark current plus photon signal. Dark current is influenced by settings of the CCD (you trade off sensitivity and dark current) and by temperature.

The constant part of the dark current shouldn't trouble you too much: you can measure and then subtract it (which is in fact what spectrometer software does where you see the baseline at zero). However, as subtracting spectra sums their noise** (error propagation). As Raman measurements typcially suffer from low signal-to-noise-ratio, you may want to avoid this, so Raman spectrometer software usually either outputs raw spectra or offers you the choice whether you want to subtract the dark spectra.

For SNR of Raman spectra, you'll have to take into account shot noise (Photon noise): the noise is directly proportional to the signal intensity, and that includes optical background such as fluorescence or elastically scattered or reflected light as well as your Raman signal. (And then come the other sources of noise such as flicker noise and the various sources of noise related to your application).

For a thorough discussion of the noise properties of Raman spectrometers, there's a whole chaper of McCreery: Raman Spectroscopy for Chemical Analysis, ISBN: 978-0-471-23187-5 on the topic
(for FT instruments, see Zhao & McCreery: Multichannel FT-Raman Spectroscopy: Noise Analysis and
Performance Assessment, ApplSpec, 1687 - 1697, 1997.)

About the 785 measurement in comparison:

The green excitation y-axis looks like raw counts (the data itself would reveal more: raw counts are usually integer numbers). In contrast the red spectrum's y-axis goes from 0 to slightly above 1.0. Which means that some transformations have been done (which is impossible to say from the data alone - you'll have to talk to the manufacturer what they do).

   water absorption

Absorbance of in the range below 1100 nm is < 0.2/cm. So yes, you loose about 1/3 of the intensity through 5 mm water and back - but you can avoid that by focusing not as deep into the water.
What hurts your signal more is that the quantum efficiency down there is pretty awful for many CCDs (around 10 %, there's a reason why their spectral range ends in the mid of the -CH stretching vibrations).

* A rather safe guess, as FT instruments for such short wavelength AFAIK exist at the moment only as "home-built" instruments developed in some research facilities - whoever uses one of those is going to be fully aware that they are using something special.

** more precisely: the variances of independent random variables are additive


Increasing available area of a solid will increase effective concentration?

The concentration of solid in itself is always the same, like m3wolf said in the comments (unless the solid is not pure, which is not relevant anyway, because grinding wouldn't change that either). So no, you can't change the effective concentration this way.

If anything, you will have higher surface area and hence higher overall rate of reaction (or dissolution, if the reaction happens in the solution and not on the surface, and thus has to be preceded by dissolution). In this manner, you will get the same yield, but in less time.

Secondary effects can eventually change your yield in an unpredictable way, as likely down as up.


Here is how I would do it.

Part I: Blue

First, ethanol 1 is oxidised to acetaldehyde 2, which undergoes nucleophilic addition with $\ce{KCN}$ and $\ce{HCN}$ to form the hydroxynitrile 3, and finally dehydrated and hydrolysed at the same time with sulfuric acid to give acrylic acid 4.

Part II: Red

First, ethanol 1 is reacted to form the Grignard reagent 5, then combines with the aldehyde 2 to form the alcohol 6. It is dehydrated to form butene 7, dihydroxylated to form the diol 8, and dehydrated again to get the diene 9.

Part III: Black

In the final part, 4 and 9 participate in a Diels-Alder reaction with the Lewis acid $\ce{TiCl4}$ to give 10. Selenium dioxide gives the diol 11 and subsequent dehydration gives us benzoic acid, 12.

Although it may be longer, it does not involve things such as high reaction temperatures and therefore may be more feasible in the context of a laboratory.

EDIT: Here is another attempt:

Ethanol 1 is oxidised to acetaldehyde 2, and subsequently with excess formaldehyde and acidic conditions forms pentaerythritol 3, and is oxidised again to form the quadruple aldehyde 4. Strong heating causes 3 to turn into the diol 5. The Grignard reagent 6 is formed and oxidation and reduction of the alkene forms the reactant 7.

Now, both 4 and 7 react to form the ring product 8, and with strong heating thus produces benzoic acid 9.


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.


You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.

The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.

So you should follow these rules:

When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).

In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.


Understanding the use of Le Chatelier's principle in the industrial production of ammonia

Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?

The rate constant would be high, but that does not mean that the net forward rate is high. If you start without product, the initial rate would be high, but would drop to zero very quickly as equilibrium is reached. The equilibrium concentration of product would be very low at high temperature, as you said, so the yield would be very low.

If you had a very cheap and highly efficient process to remove product directly from the reaction vessel, you could run the reaction at high temperature. However, the product removal is slow and expensive (cool down the reaction mixture, remove ammonia, and reheat the remaining reactant to feed back into the reactor). Because this is costly, the reaction is run at a compromise temperature where at equilibrium, about 15% of the reaction mixture is product.


How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed?

The volume can change in the course of the reaction. For example, if you add $\ce{Zn}$ to $\ce{HCl}$ the volume of the system changes. In the reaction $\ce{2H2 + O2 -> 2H2O}$ three volumes of gases turn into 1 (or even zero, if you allow water to condensate).

While mass remains constant, volume can change in chemical reactions.

Behavior of gases (under low pressure) is described by ideal gas theory. In gas a molecule is a small particle that has a large free space around it. The amount of free space is independent from of the molecular mass. lightest molecule $\ce{H2}$ occupies as much space as a heavy gas $\ce{SF6}$. As a result, density of $c\ce{H2}$ is much smaller than density of air (this is how helium baloons fly) which in turn is smaller than density of $\ce{SF6}$.

If two molecules form a dimer the density of the gas doubles, and the volume decreases by a factor of two (if the pressure is kept constant.

From 12 carbons and 24 hydrogens you can make 6 molecules of ethane ($\ce{C2H4}$) or 4 molecules of cyclopropane ($\ce{C3H8}$) or three molecules of cyclobutane ($\ce{C4H8}$). The amount of atoms is the same, the weight is conserved, but the volume and density are not.


What is the hybridisation in dimolybdenum?

The electronic configuration of $\ce{Mo2}$ is given in Bursten, B.E.; Cotton, F.A.; Hall, M.B. Dimolybdenum: nature of the sextuple bond. J. Am. Chem. Soc. 1980, 102 (20), 6348–6349:

  $$\ldots\mathrm{
9\sigma_g^{1.88}
5\pi_u^{3.78}
2\delta_g^{3.42}
10\sigma_g^{1.92}
9\sigma_u^{0.08}
2\delta_u^{0.58}
5\pi_g^{0.22}
10\sigma_u^{0.12}
}$$

Which can be simplified as:

$$\ldots\mathrm{
(9\sigma_g)^2
(5\pi_u)^4
(2\delta_g)^4
(10\sigma_g)^2
(9\sigma_u)^0
(2\delta_u)^0
(5\pi_g)^0
(10\sigma_u)^0
}$$

$9\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{4d_{z^2}+4d_{z^2}}$.
$5\pi_{\mathrm{u}}$ is doubly degenerate; it is $\mathrm{4d_{xz}+4d_{xz}}$ and $\mathrm{4d_{yz}+4d_{yz}}$.
$2\delta_{\mathrm{g}}$ is doubly degenerate; it is $\mathrm{4d_{xy}+d_{xy}}$ and $\mathrm{d_{x^2-y^2}+d_{x^2-y^2}}$.
$10\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{5s+5s}$.

Now, let us have a look at the beautiful 4d orbitals, kindly provided by our user Philipp:

From here, you can see that no hybridization is necessary, because the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals can all form bonds within themselves.


Euler's Theorem and Extensive properties

It really is that simple. Stating that a thermodynamic system observes Euler's Theorem can be considered axiomatic if the geometry of the system is Cartesian: it reflects how extensive variables of the system scale with size. In a homogeneous system with a Cartesian geometry all extensive properties scale identically and so any extensive property can be written as a homogeneous function of degree 1 of the remaining properties.


Why is the change in enthalpy computation for a change of state done like this?

You're confused because the equation is incorrect.  It should read $$\Delta H=\Delta U+\Delta (PV)$$If you insist on writing it the way you did, then P should be $\frac{P_1+P_2}{2}$ and V should be $\frac{V_1+V_2}{2}$.  That will give the same result as the first equation I wrote.


It is a quirk in how temperature would be measured using quantum mechanics. In general thermodynamics is about "bulk" properties and ignores quantum mechanics. Resolving these sort of quirks is why it would be so wonderful to have a theory of everything that would tie all the forces together.


Would ionic conduction be halted by shielding at the electrodes?

Without electrolysis, there would be just an extremely short spike of a temporary initial current, until capacitors formed by electrodes charge.
There is no significant accumulation of ions, as even a tiniest such charge separation would cause strong electrostatic forces acting against it.


Is isobutyl chloride or ethyl chloride more reactive in an SN1 reaction?

The inductive effect of the methyl groups should help stabilize the carbocation formed from 1.
The answer may be a bit clearer if you add equilibrium to your thinking.

The initial dissociation of the 1 to form carbocation A is the rate determining step.  However the rearrangement of A to B should be relatively quick, and will drive the initial dissociation forward by Le Châtelier's Principle.

In contrast, 2 has less stabilization by the inductive effect, and there is no possibility of a rearrangement.


Compounds with aluminum and nitrogen atoms directly bonded to each other as part of an aromatic ring?

Again no.  The aluminum-nitrogen bond is much stronger as a sigma bond than as a pi bond.

Borazine, with boron rather than aluminum in the six-membered ring, is known as an aromatic ring and even called "inorganic benzene".  Like its benzene counterpart, the borazine ring can be fused to give polymeric structures ultimately leading to hexagonal boron nitride.  But with aluminum nitride, where pi overlap is weaker and the bonding more polar, such a layered phase is not known, this nitride most commonly having a wurtzite structure.  In effect, the Al-N pi bonds are broken up by cross-linking with other nitrogen atoms even where the pi bonded structure is ostensibly aromatic.

Take a cue from this answer.  You are on firmest ground with aromaticity when the elements involved are a subset of boron, carbon, nitrogen, oxygen, sulfur.


Why can Bi not achieve its 4+ oxidation state?

Actually bismuth(IV) is known, and in not so exotic conditions at that. In fact both even oxidation states +2 and +4 appear when acidic bismuth(III) solutions are irradiated:

Bi(IV) has been observed; see A. I. Aleksandrov, I. E. Makarov (1987). "Formation of Bi(II) and Bi(IV) in aqueous hydrochloric solutions of Bi(III)". Bulletin of the Academy of Sciences of the USSR, Division of Chemical Science. 36 (2): 217–220. doi:10.1007/BF00959349.

From the reference cited above:

The formation of bismuth(II) and bismuth(IV) under the action of ionizing radiation on hydrochloric acid solutions of bismuth(III) has been shown by low-temperature and pulse radiolysis. The kinetics of bismuth(IV) disappearance at 300°K have a complex nature and would seem to be caused both by the disproportionation of bismuth (IV) and the reaction of bismuth(IV) with bismuth(II), which leads to bismuth(III).

The properties of bismuth (IV), which is formed from the reaction of Cl2 with bismuth (III) agree with previously established relationships on the variation in optical transition energies and the unpaired electron density at the s atomic orbital of the ion for the series of isoelectronic ions mercury (I), thallium (II), lead (III), and bismuth (IV) in the $^2S_{1/2}$ state.


According to the Molecular Orbital Theory, what makes a compound aromatic?

There's 4 criteria for aromaticity:
1. The compound must be cyclic. As in a ring structure of some kind.
2. The compound must be planar. All the atoms making up the ring must be on the same plane. The hybridization of bonds causes this which is why benzene is planar and cyclohexane is not.
3. The compound must be conjugated. Therefore it must alternate between single bonds, double bonds, lone pairs, or empty orbitals. For example, two double bonds in a row mean it is not conjugated. But a single bond connected to an atom with an empty orbital still satisfies the rules of conjugation. Another example, benzene alternates single bonds and double bonds, so it is conjugated.
4. Must satisfy Huckel's rule. So the number of pi electrons must fit in the equation 4n+2, with n being a whole number or 0. So 6=4n+2 works but 4=4n+2 would mean n=1/2 so that does not work.


Why doesn't every pure enantiomer racemise?

Because the activation energy for the enantiomeric H-Atom to jump to the other side of the carbon atom's sp3 tetraeder (by whatever mechanism) is so high that it is just not available.

You're right, statistically, every energy is available (with very bad odds). But very likely other processes, and more destructive ones, have a lower activation energy. Your compound will likely degrade and rot away long before it racemises. But even that, for alanine at room temperature, does not happen on any timescale that you can observe.


Sulfur hexafluoride creation process and risks/safety

The preparation of sulfur hexafluoride from elemental sulfur and elemental fluorine is not suitable for amateurs or beginners; it is definitely not a home experiment. Fluorine is corrosive and very toxic. It causes severe skin burns and eye damage and it is fatal if inhaled. Furthermore, it is an oxidiser and reacts violently with all combustible materials; thus, it may cause or intensify fire.
Self-made sulfur hexafluoride is not suitable for inhalation or similar experiments since it may contain various toxic impurities.

In the laboratory scale, a nickel boat is filled with sulfur and placed in a reaction tube. The tube is connected to a cold trap that is cooled with liquid nitrogen. The apparatus should be dry and ungreased. A stream of fluorine is then introduced.

The reaction
$$\ce{S + 3F2 -> SF6}$$
is strongly exothermic (standard molar enthalpy of formation at $298.15\ \mathrm K$: $\Delta_\mathrm fH^\circ=-1220.5\ \mathrm{kJ/mol}$[1]). In the fluorine stream, sulfur burns with a blue flame.

Sulfur hexafluoride is collected in the cold trap. The product contains various impurities, such as $\ce{F2}$, $\ce{HF}$, $\ce{SOF2}$, $\ce{SO2F2}$, $\ce{SF2}$, $\ce{S2F2}$, $\ce{SF4}$, and $\ce{S2F10}$. Most of the impurities are easily hydrolysable; therefore the gas is passed through a washing bottle filled with $\ce{KOH}$ solution. However, $\ce{S2F10}$ has to be removed with activated charcoal or decomposed by pyrolysis at $400\ \mathrm{^\circ C}$ according to
$$\ce{S2F10 -> SF4 + SF6}$$

[1] “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


Order of reactivity of carbonyl compounds towards nucleophilic addition

Carbocation doesn't come into the picture at all, when we are discussing about nucleophilic addition reactions. Steric hindrance will also not play a role because the compounds are planar (attack is going to take place on $sp^2$ hybridised carbon). The nucleophile is going to attack only from the top and bottom* as shown:

The only factors that come into picture in the options are the electron withdrawing or releasing nature of the groups attached to the carbonyl carbon. If the group is $e^-$ withdrawing in nature, the deficiency on the carbon atom increases, making it susceptible to nucleophilic attack.

So, the correct order is : B > A > C > E > D

P.S.:* Since the attack is from top and bottom, both of which are equally likely, the resulting product is a racemic mixture (if the carbon on which attack happens is chiral after the attack).


What is the term used as correction for attractive forces in the Van der Waals Equation?

Nothing states that the correction is itself a force, only that the correction is for attractive forces. There is thus no need for the correction to have the dimensions of a force.


Difficult to say. After a fair amount of Googling, I was unable to find any reliable sources on the compound you ask about, $\ce{H3Si-SiH2OH}$. From the Wikipedia article on silanols, it appears that the simple silanols are not stable, but no reference is provided.

It stands to reason that if $\ce{H3SiOH}$ is not stable, there is a good chance that $\ce{H3Si-SiH2OH}$ is also not very stable. Disiloxane $\ce{H3Si-O-SiH3}$ is much more stable, and silanols likely convert into siloxanes.

Good knowledge about the health hazards of a substance requires thorough investigation, which is not possible when the substance is not stable.

Alkyl silanols, like trimethylsilanol are stable and used for a variety of applications. Acutley, trimethylsilanol(pdf) is a volatile liquid that irritates skin, eyes, and mucus membranes (e.g. the lungs). The language on the MSDS is "may be harmful if..." This language suggests that you want to avoid contact with it but that some exposure will not cause serious harm. Note the msds for ethanol(pdf) has much scarier language.


What happens during oxidative and hydrolytic rancidity of fats?

I am currently finishing my Ph.D. in rancidity, and therefore I know quite a bit about it. The picture is not a fair representation of what happens, since free fatty acids should be present, though you just show the triacylglycerols. Which rancidity happens first will depend on the matrix and on the storage condition of the oil. Usually, but not always, oxidative rancidity happens first.


Does the rate constant depend on pressure?

According to hard-sphere collision theory of gas-phase reactions, the rate constant ($k$) for the elementary bimolecular reaction $\ce{B + C -> products}$ is
$$ k = N_\text{A} \pi (r_\text{B} + r_\text{C})^2 \left[ \frac{8RT}{\pi}\left( \frac{1}{M_\text{B}} + \frac{1}{M_\text{C}}\right)\right]^{1/2}e^{-E_\text{thr}/RT}$$
where $N_\text{A}$ is the Avogadro constant, $r_\text{B}$ and $r_\text{C}$ are the radii of the hard-spheres, $R$ is the ideal gas constant, $M_\text{B}$ and $M_\text{C}$ are the molar masses of B and C, and $E_\text{thr}$ is the threshold energy on a per-mole basis. Therefore, no pressure dependence.

For nonideal systems, the rate law for the same elementary reaction can be written as
$$ r = -\frac{d[\text{B}]}{dt} = -\frac{d[\text{C}]}{dt} = k_r [\text{B}][\text{C}]$$
The Brønsted–Bjerrum equation relates the apparent rate constant, $k_r$ with the infinite-dilution rate constant, $k_r^\infty$, and the activity coefficients of the envolved species:
$$  k_r = \left( \frac{\gamma_\text{B}\gamma_\text{C}}{\gamma^\ddagger}\right)k_r^\infty$$
where $\gamma_\text{B}$, $\gamma_\text{C}$, and $\gamma^\ddagger$ are the concentration-scale activity coefficients of B, C, and the transition state respectively. These quantities depend on temperature, pressure, and on the concentration of all species on the same phase. For this reason, the rate constant of nonideal elementary reactions can have a dependence on the pressure.


Can acid concentration be determined from Ka values in a titration?

The Henderson-Hasselbalch equation will help:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} + \log {[\ce{A-}]\over [\ce{HA}]}$$

Once the $\mathrm{p}K_{\mathrm a}$ is known, the equation will tell you the ratio of acetate to acetic acid (ethanoic acid). The ratio changes at various $\mathrm{pH}$ values.

The $\mathrm{p}K_{\mathrm a}$ is 4.76, so for a $\mathrm{pH}$ below 3.75 the dominant species in solution will be acetic acid; at pH 5.76, the dominant species in solution is the acetate ion. If you want to make a solution at a particular $\mathrm{pH}$ value, use the equation to give you the ratio of acetate to acetic acid needed.


What does the intramolecular aldol condensation of 6-oxoheptanal form?

Intramolecular-Aldol reactions involves heat, as you have mentioned, and the base used (NaOH) isn't much affected by steric factors. Therefore, the preferred product should be thermodynamically controlled product. Moreover, as you commented about the formed carbanion being less stable in case of 2° carbon, actually the carbanion is converted to an enolate then the reaction proceeds further. So, the 5-membered cyclic ring should be the major product.


How could I cause metal to rust rapidly?

Here are factors that would speed up rusting:

Presence of water: make sure that the iron is wet.
Presence of oxygen: make sure the iron have access to air (the dissolved oxygen in water also works).
Presence of metal below iron in the metal reactivity series (see the picture below): usually copper is used. Tie copper to iron. Make sure that they are in contact. The larger the surface area of contact the better.
Temperature: make the iron as warm as possible.
Presence of salt: explained here. In short, salt increases the conductivity of water which makes the iron rusts faster, since rusting involves electron transfer.
Acidic environment: since rusting involves hydrogen ions ($\ce{H+}$), an acidic environment would increase the concentration of hydrogen ions, making the iron rust faster. However, this would dissolve the rust.
Presence of hydrogen peroxide: hydrogen peroxide ($\ce{H2O2}$) is an oxidizing agent which would not dissolve the rust:
$$\ce{3H2O2 + 2Fe -> 3H2O + Fe2O3}$$

Metal reactivity series:

(source: bbc.co.uk)
Equation of rusting:

$\ce{Fe -> Fe^3+ + 3e-}$
$\ce{O2 + 4H+ + 4e- -> 2H2O}$
$\ce{2Fe^3+ + 3H2O -> Fe2O3 + 6H+}$

More equations can be found here.
Overall equation:
$$\ce{4Fe + 3O2 ->[H+,H2O] 2Fe2O3}$$


What is the reason behind electron transition in the flame test

The electron releases its energy because it can. It happens because it has a non-zero probability to do so. If it would never happen it would have zero probability.

Whether the electron just got excited prior to emitting some light does not matter. Particles do not have a memory. Instead such processes simply happen based on whether the requirements are met (e.q. absorbing energy is only possible if energy is available) and on probabilistic rules (quantum mechanics).

Consider some particles in thermal equilibrium: Even the though the temperature does not change on a macroscopic level, the particles still constantly exchange heat with each other. This is called dynamic equilibrium.

The difference to the flame test is, that the particles exchange energy not only by heat, but by light as well.

Exchange of heat is basically a collision of two atoms, so this is a 2-particle process: its rate dependents on the likelihood of those 2 particles "hitting" each other.

Emission of light is a 1-particle process: it can just happen spontaneously. However, absorbing light is a 2-particle process: a photon needs to intercept an atom in order for the atom to absorb its energy.

Heat is the source of energy, and it allows for redistribution between all particles. But for the light, the emission happens much more often than absorption, because 1-particle processes are much more likely to happen than 2-particle processes. Therefore we see the overall emission of light until the heat source is gone.


Computing the commutator of the Fock and density matrices in AO basis (Hartree-Fock theory)

Start from the AO Hartree-Fock equation and its adjoint
$$\mathbf{F}^{AO}\mathbf{T}=\mathbf{S}\mathbf{T}\epsilon \text{ and } \mathbf{T^\dagger}\mathbf{F}^{AO}=\epsilon\mathbf{T^\dagger}\mathbf{S}$$

where $\mathbf{T}$ is an $N\times n$ matrix that is essentially the occupied block of $\mathbf{C}$ ($n$ is occupied, $N$ is total orbitals). We use this $\mathbf{T}$ matrix because it has the convenient property that

$$\mathbf{T}\mathbf{T}^\dagger=\mathbf{D}^{AO}$$

Now, we can multiply the HF equation by $\mathbf{T^\dagger}\mathbf{S}$ on the right and multiply its adjoint by $\mathbf{S}\mathbf{T}$ on the left, which gives

$$\mathbf{F}^{AO}\mathbf{T}\mathbf{T^\dagger}\mathbf{S}=\mathbf{S}\mathbf{T}\epsilon\mathbf{T^\dagger}\mathbf{S}$$

$$\mathbf{S}\mathbf{T}\mathbf{T^\dagger}\mathbf{F}^{AO}=\mathbf{S}\mathbf{T}\epsilon\mathbf{T^\dagger}\mathbf{S}$$

Subtracting the first equation from the second and yields the desired commutator relationship.

$$\mathbf{F}^{AO}\mathbf{D}^{AO}\mathbf{S}-\mathbf{S}\mathbf{D}^{AO}\mathbf{F}^{AO}=\mathbf{0}$$

The notation I use here is based on a similar derivation given in Chapter 6 of McWeeny's Methods of Molecular Quantum Mechanics, 2nd edition.


What type of reactions are used in this synthesis?

Most reactions can be classified in a number of different ways, just like many compounds can be classified in a number of different groups. For each transformation, your question only allows for two choices, so you either work out which one works or you just need to identify the wrong choice.

The first reaction is to be classified as either a substitution or an addition. If we look at the structures of reactant and product, we realise that all the reactant’s atoms are present in the product — but an additional $\ce{HBr}$ have been added.

That should give you more than enough clues that it has to be an addition.
The second reaction is to be classified as either an addition or a substitution. Again, just by checking the atoms present in reactant and product, we realise that we lost the bromine but gained a nitrile group.

‘Loss and gain’ is definitely not an addition but a substitution.
The third reaction is to be classified as either a hydrolysis or a reduction. It does get a bit more tricky here. To identify a reduction, you would need to do one of the following:

allocate oxidation states and determine whether they have changed
check if formally an oxygen has been removed
check if formally $\ce{H2}$ has been added.

Identifying a hydrolysis is easier. It is a reaction with water that can occur in one of two ways:

the original molecule is split into two fragments, one of which contains a hydrogen, the other a hydroxy group
the molecule has formally added $\ce{H2O}$

Just by counting carbon atoms, we realise that the molecule has not fragmented, and by identifying atom symbols we can see that water has not formally been added. Thus, it cannot be a hydrolysis, it must be a reduction. In fact, two of the three reduction identification conditions are met, but I’ll leave the proof up to you.

Note that reduction is not the only way in which you can classify reaction 3. You could also call it:

hydrogenation
hydrogen addition
reductive nitrile cleavage
saturation
and probably others.


From the Gibbs-Helmholtz equation we usually write
 $$\left (\frac{\partial \ln(X)}{\partial T}\right )_p = \frac {\Delta H^0_{fus}}{RT^2}$$
where X is the molec fraction of solute. Integrating this equation gives
$$ \ln(X)=\frac{\Delta H^0_{fus}}{R}\left(\frac{1}{T_{fus}} - \frac{1}{T}  \right)    $$
thus a plot of log mole fraction vs reciprocal temperature should produce the enthalpy. The eutectic point is the point where the melting curves meet with the solid phase, its about 0.4 to 0.5 on your figure. On the left should be the melting curve of liquid with solid biphenyl as you plot T vs naphthalene mole fraction, and on the right, liquid plus solid naphthalene. Solid biphenyl + naphthalene should be observed below $\approx 30 $ C at all mole fraction and liquid above the blue dots.


Mercury is an element with poor sensitivity when analyzed by traditional flame atomic absorption. This happens because the element is extremely volatile (remember mercury is a liquid at room temperatures and has significant vapor pressure at room temperature). Cold vapor is a widely used method, that's EPA approved, and shows a significant increase in sensitivity (nearly four fold).
Source

There's many different devices out there that can be used to do elemental analysis, but from a personal experience, if you're working with dry solid material (ex. soil), then X-ray fluorescence (XRF) could be worth looking at. It's relatively cheap and does multi-element analysis. With mercury however, you're looking at a LOD of about 5 ppm, which is... okay. It depends on your experiment.


Why do SN1 reactions prefer tertiary carbons?

The intermediate in an SN1 reaction is a carbocation. The stability of that intermediate determines how favorable that particular reaction is, so a more stable intermediate means a more favorable reaction.

Carbocations are quite unstable on their own, but the inductive effect of nearby C-C bonds provide some electron density to balance out the positive charge, making the carbocation intermediate more stable. Tertiary carbons have the largest number of adjacent C-C bonds, the largest inductive effect, the most stable carbocation intermediate, and are thus favored in SN1.


Does water vapor in the air oxidize copper?

Although $\ce{H_2O}$ contains oxygen, it is there in the most reduced state $\ce{O^{2-}}$ and therefore cannot act as oxidizer. It is the oxygen in the air $\ce{O_{2}}$, which is the main and strongest oxidizing agent.

On the other hand, presence of water as a solvent can speed up some reactions. For a short overview of copper corrosion, see e.g. Why does copper oxidize. As the corrosion products form compact (or even protecting) layer, the corrosion itself is not a problem for copper, as compared to e.g. steel corrosion.


In terms of a simple bonding concept the answer of ron is correct, however due to symmetry constraints it is not possible to distinguish the two $\pi$ systems. The reason for this is, that these $\pi$ orbitals are degenerated and perpendicular to each other.

Carbon dioxide

The molecule is very symmetric and one of only a few with the point group $D_{\infty h}$. That also imposes constraints on how the spatial distribution looks like.

Along the molecular axis ($\ce{O-C-O}$) is a rotational axis of infinite fold, $C_\infty$, as the molecule is linear. Perpendicular to that axis there is a horizontal mirror plane $\sigma_h$ passing through the centre of mass, i.e. the carbon. Additionally there is an infinite number of two-fold symmetry axis $C_2$ lying in that plane, which is the criterium for $D_{xh}$ point group. In line with the main axis there is also an infinite number of vertical mirror planes $\sigma_v$ (which is implicit due to the horizontal mirror plane).

For the sake of convenience we can define an upper and a lower half of the molecule (taking the cartesian $z$ axis as the main symmetry axis). These two halfs are indistinguishable, hence their orbitals must reflect that.

Having a look at the $\pi$-orbitals, clearly shows that (enlarge):

So your first drawing was not too wrong after all. The only thing you forgot is, that there is an identical orbital perpendicular to that.

$\ce{CO2}$ is also a nice example to see the boundaries of the hybridisation concept. In this example, due to symmetry, all atoms have to be $\ce{sp}$ hybridised. And that matches all the $\sigma$ orbitals (enlarge).

Allene

The allene on the other hand is not linear and has a much lower symmetry. There is the main symmetry $C_2$ axis (for convenience) along the $\ce{C-C-C}$ axis, accompanied be two $\sigma_v$ mirror planes. There are also two diagonal symmetry planes $\sigma_d$ right between them. Perpendicular to the main symmetry axis we have two more $C_2$ axis (also perpendicular to each other). The point group is therefore $D_{2d}$ and has a lot less symmetry constraints.

However, the orbitals are actually (surprisingly) quite similar though. We can define $\sigma$ and $\pi$ orbitals accordingly. The biggest distinction is still in the $\pi$-orbitals (enlarge).

The Lack of symmetry and therefore less constraints make is possible, that the orbitals combine in a different manner. The top and the lower half of the molecule does not have to be the same anymore, hence the orbitals are more localised to one bond in particular. A notably thing is, that hydrogens take part in $\pi$ bonding (Hyperconjugation).

The sigma orbitals are quite the same as in $\ce{CO2}$ (enlarge).

However, the traditional bonding picture of localised orbitals is quite valid here. The central carbon is still being $\ce{sp}$ hybridised and the moieties are $\ce{sp^2}$ hybridised.

Ketene is a combination of the two above, with even lower symmetry ($C_{2v}$). Therefore the conventional bonding picture should hold.

For Oxygen it is usually always good to check, if hybridisation is a good concept to describe the bond. When bonded to more than one partner, the answer is usually yes. If less, then there is a good chance, that hyperconjugative effects outweigh the information gained from the hybridisation concept.

Additional fun note

One could argue that the carbon suboxide ($\ce{C3O2}$) is not too different, but here again you will be surprised, as it has a very shallow bending potential. (crystal structure, vibrations)


In general, the reaction of metals with water could produce either the oxide or the hydroxide.  Roughly speaking, most metals are capable thermodynamically of displacing hydrogen from water to form the hydroxide, but in many cases you need such a high temperature to run the reaction that the hydroxide is not stable and thus we go over to the oxide.

To show the variety of possible reaction routes let's look at a few cases:

Iron

Iron does not react with water alone under ambient conditions; a simple reaction between iron and water requires steam at a high enough temperature to destabilize any hydroxides.  You therefore get only oxides, generally $\ce{FeO}$ or $\ce{Fe3O4}$.

But if you add the supplemental oxidizing power of air, then iron can react with both the water and the air to generate rust -- a mixture of hydroxide and oxide compounds, or even  oxide-hydroxides like $\ce{FeO(OH)}$ with iron(III).  The hydroxides become possible precisely because the combination of air and water lowers the reaction temperature to make it so.  See Wikipedia for a fuller explanation of the many reactions involved in the rusting of iron.

Magnesium and Calcium

Magnesium is something of an intermediate case between metals that clearly react with water alone at ambient temperatures and those that do not.  Magnesium reacts slowly at ambient temperature, possibly generating enough hydrogen bubbles to temporarily float pieces of the light metal in the water.  Reaction is faster in boiling water.  In both of these cases the hydroxide $\ce{Mg(OH)2}$ is a stable product; but this decomposes at $350 \, \pu{^\circ C}$   From that temperature upwards, including burning the magnesium in steam, you will get $\ce{MgO}$.

Calcium more clearly reacts with water under ambient conditions to give $\ce{Ca(OH)2}$, but this also transitions to forming $\ce{CaO}$ as the hydroxide decomposes at $580 \, \pu{^\circ C}$.

Alkali metals

In general, alkali metals react so rapidly with water and the hydroxides are so stable that at least for all practical purposes, we expect exclusively the hydroxides.


Why is a bridgehead atom bearing a negative or a positive charge unstable?

The hybridisation of a bridgehead carbon when its not charged is sp3. Now, if you remove a hydrogen atom, its hybridisation becomes sp2.

Normally, in sp2 hybridisation, the 3 hybridised orbitals lie in a single plane for minimised repulsions. But the bridgehead carbon with positive charge would not have hybridised orbitals in a single plane. This increases repulsion and makes the compound unstable.

Here is a link - Bredt's Rule - that explains the same thing, but instead of a positively charged bridgehead, they discuss a double bond on the bridgehead atom.


What is the role of dichloromethane in the ozonolysis / Criegee oxidation?

Sometimes the solvent choice is very important to enhance or decrease reactivity of either of the reactants.

Dichloromethane is a widely used organic solvent. Chemists sometimes say tongue-in-cheek that it can dissolve almost anything organic since it is rather mid-range in polarity. Other than that, it is also non-nucleophilic, non-electrophilic and generally does not take part in many reactions. (There are certain exceptions, however!) This means that it is a standard solvent choice for many reactions and the ozonolysis is no exception. Most reactions require a solvent to allow all reactants to get into close proximity of each other for the reaction to occur at all — not many reactions can also be conducted in solid state.

The standard mechanism for ozonolysis — the Criegee mechanism — is given for solvents such as dichloromethane, hexanes or any others that do not participate in the reaction in any way (that are inert). However, when using methanol or other alcohols the mechanism must be modified as also highlighted by Criegee in the original mechanism publication. I have included the mechanism for ozonolysis both in presence and in absence of methanol in the scheme below.

The resulting peroxy-methyl-hemiacetal is much more stable than the carbonyl oxide and thus various side reactions, e.g. formation of Staudinger ozonides with remote carbonyl groups are inhibited. Thus, in the case of the ozonolysis, adding methanol decreases the reactivity of the ozone in solution. Conduction the reaction in dichloromethane should be considered the standard case — unless one wants to say that it enhanced ozone’s reactivity.

These details can also be found on organic-chemistry.org and this page from Michigan State.


Hot and cold air interaction in a sealed environment

In order to work out this problem, several parameter values are required to define the initial state, preferably

amount $n_1$, volume $V_1$, or mass $m_1$ of the hot air in the cylinder
initial temperature $T_1$ of the hot air
initial pressure $p_1$ of the hot air
amount $n_2$, volume $V_2$, or mass $m_2$ of the injected cold air
initial temperature $T_2$ of the cold air
initial pressure $p_2$ of the cold air

Unfortunately, only the initial temperature of the hot air is defined as $T_1 = 1200\ ^\circ\mathrm F$, which is $T_1 \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$ in proper units (the use of the unit degree Fahrenheit is deprecated). This leaves us with five unknown variables, which is certainly too many for a vivid analytical description.

Nevertheless, we may investigate this problem by guessing reasonable parameter values as follows. By way of example, we arbitrarily assume that equal masses of hot and cold air $(m_1 = m_2 = 1\ \mathrm{kg})$ are mixed. The assumed temperature of the cold air sample is equal to room temperature, i.e. $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$, whereas the temperature of the hot air sample is given as $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$. The assumed pressures of the air samples is equal to normal atmospheric pressure, i.e. $p_1 = p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$. These conditions define the initial state and also allow to determine other interesting properties using tabulated reference values for air, e.g. density $\rho$, Volume $V$, specific internal energy $u$, and internal energy $U$.

1. Hot air sample

temperature $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$
pressure $p_1 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_1 = 1\ \mathrm{kg}$
density $\rho_1 = 0.3827\ \mathrm{kg/m^3}$
volume $V_1 = \frac{m_1}{\rho_1} = 2.613\ \mathrm{m^3}$
specific internal energy $u_1 = 819.4\ \mathrm{kJ/kg}$
internal energy $U_1 = u_1 \cdot m_1 = 819.4\ \mathrm{kJ}$

2. Cold air sample

temperature $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$
pressure $p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_2 = 1\ \mathrm{kg}$
density $\rho_2 = 1.2046\ \mathrm{kg/m^3}$
volume $V_2 = \frac{m_2}{\rho_2} = 0.830\ \mathrm{m^3}$
specific internal energy $u_2 = 335.3\ \mathrm{kJ/kg}$
internal energy $U_2 = u_2 \cdot m_2 = 335.3\ \mathrm{kJ}$

3. Mixture at constant volume

The hot air sample is confined by a piston in a cylinder. The cold air sample is added, thus allowing the two air samples to mix. The total mass $m_3$ is given by
$$\begin{align}
m_3 &= m_1 + m_2 \\
&= 1\ \mathrm{kg} + 1\ \mathrm{kg} \\
&= 2\ \mathrm{kg}
\end{align}$$

For simplicity’s sake, we ignore any work $W$ performed by or on the system (e.g. the work that is required to inject the cold air into the cylinder). Furthermore, we assume that the process is adiabatic, i.e. no heat $Q$ is transferred into or out of the system. Therefore, in accordance with the first law of thermodynamics, the total internal energy $U_3$ is given by
$$\begin{align}
U_3 &= U_1 + U_2 \\
&= 819.4\ \mathrm{kJ} + 335.3\ \mathrm{kJ} \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and the new specific internal energy $u_3$ is
$$\begin{align}
u_3 &= \frac{U_3}{m_3} \\
&= \frac{1154.7\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$

At constant volume $V$, i.e.
$$\begin{align}
V_3 &= V_1 \\
&= 2.613\ \mathrm{m^3}
\end{align}$$
the new density is
$$\begin{align}
\rho_3 &= \frac{m_3}{V_3} \\
&= \frac{2\ \mathrm{kg}}{2.613\ \mathrm{m^3}} \\
&= 0.7654\ \mathrm{kg/m^3}
\end{align}$$

The new temperature $T_3$ and pressure $p_3$ can be determined from the density $\rho_3$ and the specific enthalpy $h_3$ using tabulated reference values:
$$\begin{align}
T_3 &= 658\ ^\circ\mathrm F \approx 348\ ^\circ\mathrm C \approx 621\ \mathrm K \\
p_3 &= 1.37\ \mathrm{bar} = 0.137\ \mathrm{MPa}
\end{align}$$

4. Mixture at constant pressure (ignoring work)

Allowing the system to expand to normal atmospheric pressure by moving the piston, the new pressure $p_4$ is
$$p_4 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$

We know that the mass $m$ of the air in the closed system does not change, i.e.
$$\begin{align}
m_4 &= m_3 \\
&= 2\ \mathrm{kg}
\end{align}$$
and we still (incorrectly) assume that no work $W$ is performed by or on the system and that no heat $Q$ is transferred into or out of the system. Therefore,
$$\begin{align}
U_4 &= U_3 \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and thus
$$\begin{align}
u_4 &= u_3 \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$

The new density $\rho_4$ can be determined from the pressure $p_4$ and the specific internal energy $u_4$ using tabulated reference values:
$$\rho_4 = 0.5681\ \mathrm{kg/m^3}$$

Thus, the new volume $V_4$ is
$$\begin{align}
V_4 &= \frac{m_4}{\rho_4} \\
&= \frac{2\ \mathrm{kg}}{0.5681\ \mathrm{kg/m^3}} \\
&= 3.521\ \mathrm{m^3}
\end{align}$$

The volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_4 - V_3 \\
&= 3.521\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.908\ \mathrm{m^3}
\end{align}$$

The corresponding pressure-volume work $W$ can be calculated assuming an external pressure of $p_\mathrm{ex} = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$.
$$\begin{align}
W &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.908\ \mathrm{m^3} \\
&= 92.0\ \mathrm{kJ}
\end{align}$$

The amount of energy transferred to the surroundings as work $W$ significantly reduces the internal energy $U$ of the system. We ignored this fact when we assumed that $W_4 = 0$ and $U_4 = U_3$. We can eliminate this error by calculating a new state as follows.

5. Mixture at constant pressure (including work)

If the internal energy is reduced (due to energy transfer to the surroundings as work) to
$$U_5 = 1087.8\ \mathrm{kJ}$$
and thus the specific internal energy is reduced to
$$\begin{align}
u_5 &= \frac{U_5}{m_5} \\
&= \frac{1087.8\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 543.9\ \mathrm{kJ/kg}
\end{align}$$
the new density $\rho_5$ after expansion to normal atmospheric pressure
$$p_5 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$
can be determined from the pressure $p_5$ and the specific internal energy $u_4$ using tabulated reference values as
$$\rho_5 = 0.6111\ \mathrm{kg/m^3}$$

Thus, the new volume $V_5$ is
$$\begin{align}
V_5 &= \frac{m_5}{\rho_5} \\
&= \frac{2\ \mathrm{kg}}{0.6111\ \mathrm{kg/m^3}} \\
&= 3.273\ \mathrm{m^3}
\end{align}$$
the volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_5 - V_3 \\
&= 3.273\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.660\ \mathrm{m^3}
\end{align}$$
and the corresponding pressure-volume work $W_5$ is
$$\begin{align}
W_5 &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.660\ \mathrm{m^3} \\
&= 66.9\ \mathrm{kJ}
\end{align}$$

We find that
$$\begin{align}
U_5 &= U_3 - W_5 \\
1087.8\ \mathrm{kJ} &= 1154.7\ \mathrm{kJ} - 66.9\ \mathrm{kJ} \\
\end{align}$$
This result is in agreement with the first law of thermodynamics, which states that the change in the internal energy $U$ of a closed system is equal to the amount of heat $Q$ supplied to the system (which is $Q = 0$ in this case), minus the amount of work $W$ done by the system on its surroundings.

$$\small
\begin{array}{llllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Unit} & \text{State 1} & \text{State 2} & \text{State 3} & \text{State 4} & \text{State 5} \\
\hline
\text{Temperature} & T & ^\circ\mathrm F & 1200 & 68 & 658 & 658 & 580 \\
& & ^\circ\mathrm C & 649 & 20 & 348 & 348 & 304 \\
& & \mathrm K & 922 & 293 & 621 & 621 & 577 \\
\text{Pressure} & p & \mathrm{bar} & 1.01325 & 1.01325 & 1.37 & 1.01325 & 1.01325 \\
& & \mathrm{MPa} & 0.101325 & 0.101325 & 0.137 & 0.101325 & 0.101325 \\
\text{Mass} & m & \mathrm{kg} & 1 & 1 & 2 & 2 & 2 \\
\text{Density} & \rho & \mathrm{kg/m^3} & 0.3827 & 1.2046 & 0.7654 & 0.5681 & 0.6111 \\
\text{Volume} & V & \mathrm{m^3} & 2.613 & 0.830 & 2.613 & 3.521 & 3.273 \\
\text{Specific internal energy} & u & \mathrm{kJ/kg} & 819.4 & 335.3 & 577.3 & 577.3 & 543.9 \\
\text{Internal energy} & U & \mathrm{kJ} & 819.4 & 335.3 & 1154.7 & 1154.7 & 1087.8 \\
\text{Work} & W & \mathrm{kJ} & \text{n/a} & \text{n/a} & 0 & 0 & 66.9 \\
\hline
\end{array}$$


Can we oxidize aliphatic alkane to produce carboxylic acid?

It's tricky, because the bond enthalpies of sp3 C-H bonds are high (so they are hard to break) and because alkanes typically have a large number of CH bonds so selectively oxidizing any one of them can be challenging. Generally speaking the more electron-rich CH bonds (i.e. tertiary positions) are oxidized more easily that less-substituted, less electron rich ones.

TFDO will do it, see here: https://en.wikipedia.org/wiki/Oxidation_with_dioxiranes and there are a bunch of other methods (just try googling "aliphatic CH oxidation").

It's quite a hot area of research, since if you can control it then in theory you can turn feedstocks from crude oil into more useful functionalized molecules.