Why is it obligatory to cool down the container of a sample to measure its mass in a lab? We were doing an experiment about hydrated crystals and more precisely how to determine $n$ in $\ce{CuSO4.nH2O}$. After we heated the crucible we were to cool it down using a Desiccator. Then this question came to my mind: Why must we cool it down? By doing researches in the Net I were only able to come up with these suggestions, but none seem very applicable for me:
Using Einstein's formula:
His famous equation:$$E = mc^2$$
This means that matter is able to be converted to energy. Thus, more energy should mean more mass because $C$ is constant. However, I argue that since $C$ is a very large number, normal and observable-in-normal-labs changes in energy aren't able to modify the net change of the mass dramatically (to a degree that the balance is able to detect)
A gut feeling, no backup research:
Anyone sane, and with regards to the value the balance has in the lab, would give slightest chances that the hot container would harm the surface of the balance and therefore, damage it. But chances are that I'm wrong too.
So, why must we cool the crucible in order to measure the mass of the sample in it using a balance?

A container that is not at ambient temperature will generate air currents around it.  If you place such a non-ambient container in a balance, air currents will develop around the container as it heats or cools to ambient.  These air currents will cause the balance to read incorrectly.


Calculating rate constant of intersystem crossing from Jablonski diagram I am to calculate the rate constant for $T_1 \to S_0$ intersystem crossing, with reference to the Jablonski diagram attached below. I know this should not be difficult, but for some reason I cannot put the number together to get the correct result.

I have that

$$
\Phi_\text{ISC} = \frac{k_\text{ISC}}{\sum_i k_i}
$$

where the denominator represents a sum of all decay pathways for the $T_1$ state. However, how can this sum be calculated if we do not know the rate constants to begin with?

The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:

$$\begin{align}
\tau(T_1) = \frac{1}{\sum_i k_i(T_1)} &= \pu{6 \times 10^-3 s} \\
\sum_i k_i(T_1) &= \frac{1}{\pu{6 \times 10^-3 s}} \\
&= \pu{167 s^-1}
\end{align}$$

The two lines going from the $T_1 \to S_0$ state represent intersystem crossing (ISC) and phosphorescence. Squiggly lines represent nonradiative transitions, of which ISC is one, so the quantum yield for ISC is 10%. (Straight lines represent radiative transitions, in this case phosphorescence.)

Therefore:

$$\begin{align}
k_\text{ISC} &= \Phi_\text{ISC} \cdot \sum_i k_i(T_1) \\
&= 0.1 \cdot \pu{167 s^-1} \\
&= \pu{16.7 s^-1}
\end{align}$$


Why does mesitylene absorb at a longer wavelength than benzene (UV-Vis)? According to UV-Vis spectroscopy, mesitylene (1,3,5-trimethylbenzene) absorbs at $\lambda_\text{max} = 210\ \mathrm{nm}$ while benzene absorbs at the slightly shorter wavelength of $\lambda_\text{max} = 204\ \mathrm{nm}$.

Why is that? One would think that $\pi$-ring's hyper-conjugation with the $\sigma^*$ orbitals of the methyl hydrogens would help stabilize the system thereby increasing the gap between the HOMO and LUMO (in turn decreasing absorption wavelength).

An opposite effect is observed when nitrobenzene ($\lambda_\text{max} = 252\ \mathrm{nm}$) is compared to 2-nitrotoluene ($\lambda_\text{max} = 250\ \mathrm{nm}$) and 2-isopropylnitrobenzene ($\lambda_\text{max} = 247\ \mathrm{nm}$). In these cases, the addition of larger and larger alkyl groups seem to be decreasing the absorbed wavelength (therefore increasing gap between HOMO and LUMO).

Is that because of sterics? Or because the the strongly electron withdrawing nitro group (conjugation) is being "counteracted" by the alkyl electron donating groups and therefore electrons are more concentrated in the ring rather than being spread though the molecule?

EDIT:

My question in more general terms: I understand HOW more conjugation results in a decrease in absorption frequency (due to molecular orbitals overlapping constructively vs destructively therefore decreasing gap between HOMO and LUMO). What I don't understand is HOW pumping electron density into a conjugated system (or withdrawing) effects the HOMO and LUMO. An explanation from a MO perspective would be greatly appreciated.

The general observation is that alkyl substitution intensifies and shifts the benzene spectrum to longer wavelengths. The effect is similar to that of groups with lone pairs such as $\ce{NH2}$ and O-alkyl. So I don't think that opposite effects are observed as you suggest. Some data are given below.

The 0-0 band transition, from  v=0 vibrational level to v' =0 in the first excited state of benzene is at 262 nm (due to symmetry this is a forbidden transition so its position has to be calculated. The maximum in the S0-S1 transition in benzene spectrum is at 245nm, and S0-S2 at 203 nm).  Toluene does not have a forbidden 0-0 transition, and has a 0-0 band at approx 266 nm (with maxima at 261 nm and S0-S2 at 207 nm), 0-zylene 0-0 is at approx 273 nm, so the substituents do stabilise the excited state very slightly. This has been attributed to 'hyperconjugation', i.e. some charge being donated by the methyl group to the ring, or the alkyl groups behaves as it it had a lone pair. (The idea of hyperconjugation is still hotly contested). Toluene shows a slight increase in intensity as the symmetry is broken $\ce{D_{6h} to C_{2v}}$ but otherwise the effect on the electronic spectrum is slight. Aniline has a relatively broad and intense  absorption maximum (not 0-0) at 280 nm, nitrobenzene max very broad spectrum with a maximum at approx 268nm, but with a long shoulder in absorption stretching out to almost 400 nm, which is thought to cover the n-$\pi^*$ transition.

With very broad and structureless spectra (almost all the molecules you mention except benzene and toluene) the maximum intensity does usually always correspond to the 0-0 energy of the transition, due to Franck-Condon factors (overlap of wavefunctions which control intensity of transitions) and the fact that extra groups add considerable vibrational intensity to transitions and so broaden them. Thus comparing absorption peaks is not a reliable way of comparing state energies. (Experimentally the only ways are to measure vapour  phase absorption spectra or measure in a solid matrix  at 4k if possible.) Secondly spectra are stabilised differently in different solvents, even relatively non-polar ones and this can lead to spectral shifts.

The general sort of handwaving explanation, that is nevertheless useful,  is to suppose that the molecule has energy levels that can be described like those by a  particle on a ring. The bigger the ring the lower the energy levels become and so the lower the energy of the electronic transition. Thus incorporating a substituent that involves conjugating the electron over more atoms, increases the ring size very slightly and so  produces a longer wavelength transition.


Last electron of a transition element The electronic configuration of nickel is

Ni: [Ar] $3d^84s^2$

Here, while writing the configuration, we fill the 3d after 4s.Hence the 28th electron enters the d orbital.

But

$\ce{Ni+}$: [Ar] $3d^84s^1$

Here we take out the last electron (28th) from 4s.
Are there any rules or conventions regarding this or am I conceptually misguided.

When you ionize an element, all the orbitals go down in energy but orbitals with higher angular momentum drop more than those with lower angular momentum. When you start ionizing Ni, the $d$-orbitals drop much more than the $s$-orbitals and so, in the ions, the highest energy orbitals are the $4s$.

Relatedly, I believe (and I just ran a quick calculation to confirm) your electron configuration for $\ce{Ni+}$ is incorrect and should be [Ar]$3d^9$ for the same reason.


What is the radius of the atoms? In a recipient there is an unknown monoatomic gas that occupies a volume of $\pu{230 cm3}$, at $\pu{300 K}$ and $\pu{1 atm}$. It is known that the atoms occupy a volume of $2\times10^{-4}$. Find the radius of the atoms.

I used the ideal gas law to calculate the number of moles, then I calculated the number of atoms, then divided the volume occupied by the atoms by the number of atoms to find the individual volume of each atom, then applied the formula for the volume of a sphere to find the radius, the problem is that I found 92x10^-10 and the correct answer is 1,25x10^-10, can anybody spot the mistake? Thank you.

Here are the maths I did:

I haven't dived into your math, but I think there are 3 things that are confusing:

intermediate calculations, even though you weren't asked to do any;
missing units and inconsistent notations (e.g. base 10 is not always shown);
ambiguous phrase "atoms occupy a volume of $2\times 10^{-4}$."

I always find it cleaner and more productive to derive the formula first, and only then plug in the numbers minding the proper units. As for the volume issue, I tend to believe that $2\times 10^{-4}$ means the fraction of the volume occupied by gas atoms and it is not the absolute value expressed in cubic centimeters. In fact, in this case you get the correct answer and you don't need to use gas volume ($\pu{230 cm3}$) at all.

Check this out. Radius of an atom $r_\mathrm{a}$ is indeed can be found from the volume of an atom $V_\mathrm{a}$

$$r_\mathrm{a} = \sqrt[3]{\frac{3V_\mathrm{a}}{4π}}.$$

At the same time

$$V_\mathrm{a} = \frac{V_\mathrm{atot}}{N}$$

where $V_\mathrm{atot}$ – volume of all gas atoms depicted with spheres; $N$ – number of the atoms that can be found via Avogadro's number $N_\mathrm{A}$ and amount $n$:

$$N = nN_\mathrm{A}.$$

Finally, you are allowed to use ideal gas law to determine the amount $n$:

$$n=\frac{pV}{RT}$$

where $p$ – pressure; $V$ – volume of the gas (macro scale); $R$ – the molar gas constant; $T$ – temperature. Now, lets's put all the things together assuming that  $V_\mathrm{atot}/V = 2\cdot 10^{-4}$:

$$
\begin{align}
r_\mathrm{a} &= \sqrt[3]{\frac{3V_\mathrm{atot}RT}{4πpVN_\mathrm{A}}}\\
  &= \sqrt[3]{\frac{3\cdot2\cdot10^{-4}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{300 K}}{4\cdot 3.14\cdot\pu{101325 Pa}\cdot\pu{6.022e23 mol-1}}}\\
  &= \pu{1.25e-10 m}
\end{align}
$$

Now that's the value that actually makes sense:)


What is the criteria for determining structure of a molecule? Does the shape of the molecule depends on symmetry?

Why is the shape of dinitrogen trioxide ($\ce{N2O3}$):

(oxidation number of nitrogen atoms: +4, +2)

Instead of my proposal, which is much more symmetrical in terms of shape and oxidation number (of nitrogen atoms: +3, +3)

As seen from the case of $\ce{N2O3}$, which is unsymmetrical, then molecule of pyrophosphorus acid $\ce{(H4P2O5)}$ should be unsymmetrical too.

But the actual structure is symmetrical:

Instead of my second proposal

Another interesting thing is: the central atom of the two previously mentioned molecules has the tendency to have have highest oxidation state. That is: +5 in case of phosphorus atom of pyrophosphorus acid and +4 in case of nitrogen atom of dinitrogen trioxide.

Is my above mentioned conjecture is valid?

No, not symmetry per se but generally the lowest energy arrangement of  the atoms.  What structure this is is not always obvious by just combining atoms according to the valency rules; if it were, experimental (analytical) methods would not be needed. This is particularly true of 'organic' molecules where one formula can give many similar structures, e.g. isomers, but also of the small molecules you list.

The rules we use are the result of numerous experiments over the last 150 years or so reacting one species with another and trying to make a consistent set of rules to explain things. But ultimately structural studies are needed, the definitive method is x-ray diffraction from a crystal of the material but of course this is a not always possible. In this case infra-red, Raman and NMR spectroscopies have been used as well as mass spectrometry.  These methods do not give complete structures directly (bond lengths and angles as does x-ray diffraction) but allow confirmation of one structure (as topology) over another, for example $\ce{N2O}$ is  NNO rather than  NON. More latterly computational methods are being used to predict structure but ultimately these will have to be backed up by experiments.


Entropy is a measure of unavailable energy? I came across a statement in my text book which said that entropy is a measure of unavailable energy.

What does this statement signify?

When the first law was first formulated it was assumed that the most efficient utilisation of a chemical reaction to produce work was to use all the heat  to produce work. The second law was not fully understood and thus at maximum efficiency $-\Delta H$ was expected to represent the maximum amount of work. Many experiments failed to confirm this. We now know that it is the free energy which measures the maximum capacity to do work, $\Delta G=\Delta H-T\Delta S$. Clearly $\Delta G$ and $\Delta H$ are equal only if there is no entropy change between the beginning and end of an isothermal reaction. If this is not the case the work obtainable in a reversible process can be either greater than or less than the heat of reaction.

From the first law the external work performed must be equal to the loss in energy of the system, unless some heat is taken from or given to the surroundings. This is exactly the point first clearly seen by Gibbs. In a reversible isothermal reaction $T\Delta S$ is the heat absorbed from the surroundings and if this is positive the work done will be greater than the heat of reaction.

So 'available/unavailable energy' seems to depend on the particular case. Hopefully some thermo experts will give you a more detailed argument.

edit: It occurs to me that perhaps the textbook was referring to the fact that one can 'convert all the work into heat but not all the heat into work'. In other words not all the random thermal motions of molecules can be made to do work.


Why is maximum buffer capacity for some diprotic buffers not when pH = pKa? According to Rajkovic et al. [1]:

$$β = \frac{[\ce{H+}]}{K_\mathrm{w}} + 2.303\cdot\frac{[\ce{H+}]\cdot K_\mathrm{a}\cdot c}{[\ce{H+}] + K_\mathrm{a}}$$
where $K_\mathrm{a}$ is the dissociation constant of the acid; $c$ is the concentration of the acid in all forms; and $K_\mathrm{w}$ is the ionization constant of water $(1·10^{–14})$. It can be seen that the buffer capacity of the acid is greatest when $[\ce{H+}]$ equals $K_\mathrm{a}$ (or when the $\mathrm{pH}$ is equal to the $\mathrm{p}K_\mathrm{a}$ of the acid). This relationship can be used for mixtures of monoprotic acids and many diprotic acids (by considering them to be made up of two monoprotic acids). Unfortunately, this is not true for most of the diprotic acids found in wines because the second dissociation is not completely independent of the first.

Is it true that for some diprotic acids maximum buffer capacity is not reached when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$? If so, why is it true? Also, then when is buffer capacity maximum? Can it only be determined experimentally?
Or did I just misinterpret the article? Is it saying that this equation of buffer capacity is invalid for some diprotic acids instead of the fact that buffer capacity is max when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$?
References

Rajkovic, M.; Novakovic, I.; Petrovic, A. Determination of Titratable Acidity in White Wine. Journal of Agricultural Sciences, Belgrade 2007, 52 (2), 169–184. https://doi.org/10.2298/JAS0702169R.

Any aqueous solution increases its buffer capacity toward $\mathrm{pH} = 0$ or $\mathrm{pH} = 14$, as more and more acid or basis is needed to change $\mathrm{pH}$ by a given interval. It happens even without presence of any buffering substance, aside of strong acids or bases.
If an acid, monoprotic or diprotic, is added, a peak(s) of maximum buffer capacity is added. But if $\mathrm pK_\mathrm a$ is small, the peak of maximum buffer capacity does not lay on a steady baseline of the capacity function. It lays on a sloped baseline. If the maximum absolute peak slope is smaller than the absolute baseline slope, the peak does not manifests itself  as a peak(=maximum capacity), but rather as just as a shoulder on the slope, without a maximum.
This phenomena is not specific for chemistry, but rather for mathematics and mathematical analysis. If you have a sinusoid function on a large enough slope, its peaks would disappear at some slope value, becoming just a shoulders.
The similar phenomena occurs in case of multiple $\mathrm{p}K_\mathrm{a}$ in the solution, of the same or of different compounds. On the background of non steady capacity function, a particular peak may not have a shape of a peak.
Additionally, the presence of close peak neighbors affects position of the peak maximum.
If 2 peaks of the similar height are close enough each other, their maxima shifts towards each other, and with enough closeness they merge into a single peak.


Plastics for food packaging There are many types of plastics that can be used for food packaging, for example:

Polyethylene terephthalate (PET)
Polyethylene (PE)
Polyvinyl chloride (PVC)
Polypropylene (PP)

What are the differences between them?
Which one is most suitable for packaging pastries, for example, biscuits?

This question is a little bit broad, but thankfully, there are many resources available.

In answer to part of your question

  Which one is most suitable for packaging pastries, for example, biscuits?

The Natural Resources Defense Council webpage Food Storage Containers, they state that the safest plastics are PET, HDPE, LDPE, and PP.

For comparing the various types of plastic, the Plasticology 101 website compares the types of plastic according to:

Temperature tolerances
Clarity
Chemical, scratch and impact resistance
Rigidity
Food contact acceptability
Recyclable material

They suggest that PET, HDPE, PVC, LDPE, PP and PS are suitable for food, but temperature tolerances would be a major factor in what food they store.  Another chemical comparison that needs to be considered for the storage of food is information about the $\ce{O2}$, $\ce{N2}$, $\ce{CO2}$ gas and moisture transmission rates are included in Evergreen's document The barrier performance of common plastic film.

Specifically for biscuits, and any other baked goods, according to sevral websites, it depends on what the ingredients are - things to consider are the dairy content, whether the plastic will react with some of the ingredients etc.


What happens when (+) 2-iodobutane is treated with NaI in acetone? After treating the compound with NaI, we should follow SN2 mechanism and the product will follow an optical inversion, giving us (-)2-iodobutane, but the answer is (±)2-iodobutane.

After going through various articles, I am getting reasons, such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.

And there is the same question where we are treating (R)-2-bromobutane with NaI in acetone, but the answer is (S)-2-iodobutane.

What could be the reason?

I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.

There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction


What is the difference between an industrial and a laboratory production method? I'm not sure if this is the correct SE in order to ask this, if that is true just said me that in comments and I will delete my question, there is no need of downvote or ban.

I was answering a question of World Building SE about how to make an atmosphere and I found that nitrous oxide can be used to make one. But then, I started reading how can I produce nitrous oxide (to post that information also) and I found with this on Wikipedia.

It talks about an industrial method and a laboratory method. Which is the difference?

I have the theory that a laboratory method can't be used at big scale. Is that true? But, How much is "big"? Maybe for a fabric it's too little the laboratory production (a few kilograms per minute per square meter of laboratoy???) but for me that is a lot, 1kg per minute is arround a cubic meter of air per minute (60 minutes per hour, 24 hours per day...)

P.S: Sorry for the bad tag. I don't have no idea of what to write.

This answer will be separate in few parts. In the first one I will just make a review of the scope between lab an industrial scale. In the other parts, you’ll find some things more deeply explained and exemplified. In any case, if something isn’t clear or the English grammar is wrong, please ask me a question and point it (the English mistake) out.

Part I - Short Overview

A laboratory method can perfectly be adapted at industrial scale however there are many things to take in consideration.

Let say you are working in a lab to perform the synthesis of a molecule. In the basics we will assume you will need a reactor (round-bottom flask), and a distillation column to separate your compound from the others (solvent, remaining reactants, possible side products).

If you do all that stuff at a laboratory scale, even if your reaction is super exothermic even involving very hazardous chemicals, you'll always have the possibility to do it safely or at least having the possibility to stop the reaction by cooling it down rapidly.

I would mention at this point that the ratio between the volume and the surface of an object isn't constant. For example, if you multiply by 10 the length of the sides of a cube, its volume rises by a factor 1000 whereas its surface by a factor 100. You can find more interesting information here.

So, if you make your reaction at laboratory scale, let say under few litre, even if it might be dangerous, you’ll control easily and readily the temperature inside your flask or small reactor.  So, at industrial scale, let say over 1 cube meter (in general I would say 5 to 50 cube meters is the average, but some reactors are bigger than 1000 cube meter), you must take in consideration safety and security details as well as the sizing of your operation. I recommend you to read here a really nice overview about most of the kinds of the reactors used in chemical plants with many details related to them.

Now I bet you start to understand the big difference. But let me go a bit deeper on that part. If you want to size a chemical operation, you must first take in consideration the hazards of all of your chemicals as well as their fundamental properties (ebullition temperature, state at a given temperature (gas, liquid, solid, supercritical fluid ,  …), the possible interactions between to compounds or more.

After that you’ll need to know about the kinetics of your reaction, is it fast or slow?, you’ll also need to know about the thermodynamics of all the chemicals to make a prediction for safety (see the next part) and to complete you’ll have to take in consideration if you have a system with only liquids, with gas and liquids or even harder with gas, liquid and solid (hydrogenation (using hydrogen gas) of a liquid organic molecule (let say an oil) catalysed by a given catalysor (a solid) (see the third part).

For years now engineers and researchers, developed methods to compute thermodynamic and kinetic properties as well as many different methods for sizing many different basic or very complex operation (distillation, crystallization, multiphase reactors, drying, precipitation, …) (and I’ll give when I’ll have time as many details as I can in a fourth part about that point). So now it exits a lot of extrapolations and models made from the experiments because of the too complex existing situations to make nice calculus involving mathematics and physics learned at school. This article gives you an idea of all the possibilities of having nightmare while sizing an operation at industrial scale.

Part II - Predicting the safety of a reaction

The Semenov diagrams

So, to give some concrete applications of what I said before, I will show you a very basic thing that safety engineers will do to know more about the danger of the reaction. They will use a Semenov-diagram (or plot).
The first idea is to consider your reaction having a zero-order kinetics, i.e. the kinetics doesn’t involve the concentration of the reactants. It’s a big assumption but it gives you really a good idea of how dangerous it can be. The second idea is, as your reactor will be quite big, to consider it as adiabatic. It’s not true at all of course because you wouldn’t be able to heat or cool it, but it allows you to know what could happen in the worst case.
Doing an energy balance on your system you’ll get:
$$Q_{acc} = k_0 \cdot C_0 \cdot \exp\left(\frac{-E_a}{RT}\right)\cdot \left(-\Delta_r H\right)-UA\cdot (T_r-T_c)$$
Where $Q$ represent the amount of heat generated by the reaction, $k_0$ is the pre-exponential factor of the Arrhenius’s law, $C$ is the concentration, $E_a$ is the activation energy, $R$ is the constant of ideal gases, $U$ is the global exchange coefficient of your reactor and $A$ the area of heat exchange.
Now if you plot this equation which tells you the amount of energy liberate by the whole process at a given temperature with the plot only $UA\cdot (T_r-T_c)$, you’ll get something like that,

reference

If you consider the blue dotted line crossing the red plot, the blue curve tells you about how much heat you can evacuate and the red one tells you about what the whole process liberates. The point A is the stable point as even if you go under in temperature your reaction will still stay in control, however the B point is the point from which you have instability. It’s clear to understand than between these two points you can control the temperature of your reactor.

Keep in mind the heat you can evacuate is a linear function of the temperature $UA\cdot (T_r-T_c)$ and the heat produced by the whole operation is exponential with the temperature as shown by Arrhenius.

Predict the dangerous nature of a reaction

Another thing you can do easily is to predict the kind of danger of your reaction (not dangerous, a little dangerous, dangerous, extremely dangerous) using the worst case (adiabatic reactor).

Then you have rises of the temperature equal to the amount of heat produced by the reaction. Then you have,
$$C_p \cdot \Delta T_{adiab} = [reactants]_0 \cdot V \cdot (- \Delta_r H)$$
Where $C_p$ is the heat capacity of your mixture, $[A]_0$ means the concentration of A at the beginning (you want to know what happens if all of it reacts), and V is the volume of the mixture.

You can then predict the variation of temperature knowing all the needed parameters. I would say if the rising of temperature is below 50 °C your reaction is quite safe; if you find something more than 100 °C you’ll have to find either another solution (a different reactor, …) or add a lot of security systems (emergency open valves, a possibility to almost instantly flush your reactor content in a inhibitor mixture under the reactor, …).

Part III - What if I have a multiphase system?

Well this is the case most of the time. It’s not common to have a system in which you only have soluble fluids (except for gases). If you are performing a treatment of your affluents or your effluents (because you can’t use row compounds and you can’t throw away your potential toxic, corrosive, … mixture in the nature) you will often have an exchange between a gas and a liquid. If you are doing a reaction involving a catalysor then you will probably have three phases if it’s a hydrogenation or a fluidized-bed reactor for a cracking operation or something else. You’ll have only one phase for vapocracking, polymerization (not in emulsion and relatives) and urea production for example.

Now in these cases you have to take in consideration of possibility of having an exchange of chemicals (the ones you want at least) between the two phases or three phases. If you have a liquid/gas system, the situation is common and even if a bit hard sometime you can get good approximation and correlation using the two films theory see that I wasn’t really confident with it some time ago.

If you use a catalysor, you have to predict the possibility of the species to go inside it so that they can react otherwise nothing will happen. Then you have to understand how your catalyst works and how to avoid the possibility of blocking it or poisoning it [ref] as in most of the cases catalysors are like sponge solids. I found two articles pdf1 and pdf2 if you want to got deeper on how complex it is.

So I hope this very long answer help you more than you need. I will also use it as a reference for some more questions in the future. You may say that all these parameters can also be taken in consideration even at a lab scale. I will anwser yes in the sense that you will never have all the data you want and you'll have to do experiment to extrapolate back to the industrial scale. But I will moderate that if you are only doing stuff in a lab scale it doesn't cost the same as when you are involving tons of chemicals and it's way more secure so even if you are wrong in something you can afford it much more easily.

  I will point out that in most of the cases the chemical plants are
  sized as the lab plants a company may have meaning that the lab plants
  are made by extrapolating down the data (size of the reactors, ratios
  of length especially, ...) so that when you do your experiment in the
  lab you can then extrapolate back using your own models and then be
  more close to the reality with really tiny variations.


Does aluminium dissolve in soda glass? Specifically, if a piece of aluminized glass is heated to the melting point of the glass will the coating dissolve into it or otherwise react with it?

Since soda[-lime] glass is specified, then Al will react with the alkali forming an aluminate, e.g. $\ce{Na2Al2O4}$. This may or may not dissolve in the molten glass, depending on concentration of the aluminate(s), temperature and cooling rate. If it precipitates, the glass would have a "frosted" appearance.


How to make graphite from wood/charcoal Would be very grateful if someone had the answer or can point me in a direction. I've searched a lot but didn't find it yet. It could benefit others as well.

I do not have a high tech lab, but I can invest, would prefer methods that don't require above \$100 or \$300, but am open to it. I can build a furnace and weld.

Edit: As I don't have 50 rep points to reply, I will answer here to comments the first comment as to what use I need it for;
It is for water Nanofiltration of 1 micron or lower ,and secondly batteries making.
--- Second comment; thanx alot for the document, I will check it and tell you about it.

There was a method patented in 1896 whose claims would seem to meet your requirements.  Even with your restricted budget, used laboratory equipment companies can probably provide you what you need.

An excerpt from the patent states:

  "My invention relates to the production of graphite from any of the
  cheap and common forms of carbonaceous materials, such as mineral
  coal, coke, charcoal, gas-carbon, and carbids and the like; and it has
  for its object to provide a process or method of cheaply and
  practically producing pure graphite..."

The reagents used are common, cheap and readily available.  The patent states that:

  "[the mixture is] composed of powdered coke, sand, salt, and sawdust,
  the coke forming about fifty per cent., by weight, of the mixture, and
  these materials are thoroughly mixed and introduced into the electric
  furnace..."

The key issue for you will be designing the high-temperature inductively heated furnace, though the patent describes this thoroughly, including diagrams.

The availability of used equipment to fashion the device as stated is of course uncertain, and you will have to be creative with what is available.  I would start my search with LabX, with whom I have no affiliation whatsoever.  I hope this does "point you in the right direction", and Good luck!


Rusting: alternatives to water? I have several questions about the rusting of metals (take iron as a specific case if you wish) that I have not found clear answers to:

1) Does water act as a catalyst in prototypical rusting reactions, or is it actually consumed? (keeping in mind that $H_2$ gas produced may be oxidized back to water by ambient $O_2$.)

2) Is dry rusting possible with just $O_2$ and metal? Am I correct that it is thermodynamically favored but kinetically discouraged?

3) Is it realistic to rust a metal using oxygen and a dry polar solvent? If so, what solvents would be suitable?

Rust is by definition the transformation of iron metal to iron oxide (and iron hydroxide as an intermediate). You cannot "rust" other metals, although you can form oxides and degrade them etc. It's just not called rust when you convert, e.g. $\ce{Cr}$ to $\ce{CrO3}$. So there is only one rusting process.

1) Water is not a catalyst in the reaction that forms rust. A catalyst by definition is not consumed during a reaction (its bonds are not broken). Water can be and is the medium that carries $\ce{O2}$ to the iron (otherwise $\ce{O2}$ from air can react with the metal), and it is consumed in the process by dissociation. The fact that it is reproduced does not make it a catalyst. First iron atoms are oxidized by $\ce{O2}$ as in:

(a) $$\ce{Fe + O2_{(aq)} \rightarrow Fe^{+2}_{(aq)} + 2e-}$$

Then the free electrons are quickly absorbed by free protons from water by:

(b) $$\ce{2e- + 2H+_{(aq)} \rightarrow H2}$$

It is highly unlikely for $\ce{H2}$ to react with $\ce{O2}$ to form water -- that is a combustion reaction, so high heat or pressure would have to be added. Rather, water is reproduced by a different reaction (d), below.

2) "Dry" rusting is not possible because the process requires reactions with water's ionization products $\ce{H+}_{(aq)}$ and $\ce{OH-}_{(aq)}$ (notice the aqueous states of all ions). The process continues by production of iron hydroxides:

(c) $$\ce{Fe^{+2}_{(aq)} + 2OH-_{(aq)} <=> Fe(OH)2_{(s)}} $$
(d) $$\ce{Fe^{+2}_{(aq)} + 4H+_{(aq)} + O_2_{(aq)} <=> 4Fe^{+3}_{(aq)} + 2H2O_{(l)} } $$
$$\ce{Fe^{+3}_{(aq)} + 3OH- <=> Fe(OH)3_{(s)}} $$

This is why more acidic media produces rust faster; it pushes (c) and (d) to the right because (b) consumes free electrons and renders more $\ce{Fe^{+2}}$ for further reaction.

Finally the iron oxides "dry" to form the actual rust compound, which is a hydrated oxide: $\ce{Fe2O3\cdot nH2O }$.

3) You can see by now that water is the required solvent because the reactions of interest all involve $\ce{H+}$ and $\ce{OH-}$.

One could say there exists a "rust triangle" where "time, oxygen, and water" are required as an analog to the "fire triangle": heat, oxygen, and fuel.


Dissociation of zinc hydride How can we determine if zinc hydride is a ionic compound? If it is a ionic compound then why doesn't it dissociate when put in water?

Zinc(II) hydride is not at all an ionic compound rather the elements are arranged in a one-dimensional network bonded covalently.  From wikipedia article of zinc(II) hydride:

  New evidence suggests that in zinc(II) hydride, elements form a
  one-dimensional network (polymer), being connected by covalent bonds.(...)Solid zinc(II) hydride is the irreversible
  autopolymerisation product of the molecular form, and the molecular
  form cannot be isolated in concentration.

  Molecular zinc hydride in the gas phase was found to be linear with a
  Zn-H bond length of 153.5 pm. Quantum chemical calculations predict
  the molecular form to exist in a doubly hydrogen-bridged, dimeric
  groundstate, with little or no formational energy barrier.The dimer
  can be described as di-μ-hydrido-bis(hydridozinc), according to IUPAC
  additive nomenclature

As it is not an ionic compound, it does not dissociate in aqueous solution, rather it gets hydrolyzed slowly. If you want to decompose into component element, just heat it at 90℃.

  Zinc(II) hydride slowly decomposes to metallic zinc and hydrogen at
  room temperature, with decomposition becoming rapid if it is heated
  above 90°C.

  $$\ce{ZnH2 → H2 + Zn(0)}$$

  It is readily oxidised and is sensitive to both air and moisture;
  being hydrolysed slowly by water .


Why Gibbs energy change of a process can be negative at constant temperature and pressure? This question has been asked before, but the answers are not clear and satisfactory.
If the pressure and the temperature are constant then G would be always zero. How would then G have a value other than 0?

G is a function of T and P, but also, in a mixture, the number of moles of each of the species in the mixture.  If these change because of a chemical reaction, G also changes (even at constant T and P).


If NO3- were to get oxidised to some compound in the way SO4 2- gets oxidised to S2O8 2-, what would be that compound? In electrochemistry, we learn of the sulphate ion getting oxidised to peroxydisulphate ion. If this was to occur in the case of the nitrate ion which has a larger discharge potential, what would be formed?

First, let’s take a look at both sulfate and nitrate. They both show similar characteristics when you compare the oxidation states: remember that nitrogen is in group 15 so it has five valence electrons while sulfur is in group 16 so it has six — therefore, the highest reasonable oxidation states for these elements should be $\mathrm{+V}$ and $\mathrm{+VI}$, respectively.

$$\begin{align}&\ce{\overset{$\mathrm{+V}$}{N}\overset{$\mathrm{-II}$}{O3}^-}&&\ce{\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O4}^2-}\end{align}$$

When oxidising sulfate $\ce{SO4^2-}$ to peroxidisulfate $\ce{S2O8^2-}$, sulfur cannot be oxidised any further; the only atom that can still be oxidised is oxygen. Hence, in an initial step, we generate $\ce{SO4^{.-}}$ with three oxygens of $\mathrm{-II}$ and a fourth of $\mathrm{-I}$. Two of these then dimerise forming a peroxo-bond.

$$\ce{^-\overset{$\mathrm{-II}$}{O3}\overset{$\mathrm{+VI}$}{S}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O3}^-}$$

Looking at nitrate, the only obvious theoretical would be to do the same, i.e. remove one oxygen electron to intermediately generate $\ce{NO3^.}$ and then have this dimerise to $\ce{O2N-O-O-NO2}$ or $\ce{N2O6}$. Interestingly, this would be a neutral compound unlike $\ce{S2O8^2-}$ which retains its charge. I don’t know whether it has been observed or not, but I would predict its formation from what I know.


Predicting pH of an acidic salt of an weak acid and weak base Suppose we want to calculate pH of an $\ce{NH4HCO3}$ solution of known concentration.
Now, $\ce{NH4+}$ will be hydrolised to give $\ce{H+}$ ions.
Some $\ce{HCO3-}$ ions will be converted to (i) $\ce{H2CO3}$ and $\ce{OH-}$
and some will give (ii) $\ce{H+}$ and $\ce{CO3^2-}$

\begin{align}
\ce{HCO3- + H2O &<=> H2CO3 + OH− }& 1/K_\mathrm{a1}\\
\ce{HCO3- + H2O &<=> CO3^2- + H3O+ }& K_\mathrm{a2}\\
\ce{NH4+ + H2O &<=> NH3 (aq) + H3O+ }& 1/K_\mathrm{b}\\
\end{align}

I am not sure how to take account of both $\ce{H3O+}$ and $\ce{OH-}$ in the calculation of pH.
Will the third equilibium costant be $1/K_\mathrm{b}$?

The main reactions which take place are

1)Ionization of $\ce{HA-}$: $\ce{HA- -> H+ +A^2-}$

2)Hydrolysis of $\ce{HA-}$: $\ce{HA- + H+ -> H2A}$

3)Hydrolysis of cation: $\ce{B+ + H2O -> BOH + H+}$

for reaction $1$, $$k_{a2}=\frac{[H^+][A^{2-}]}{[HA^-]}$$

for reaction $2$, $$k_{a1}=\frac{[HA^-][H^+]}{[H_2A]}$$

for reaction $3$, $$k_h=\frac{k_w}{k_b}=\frac{[BOH][H^+]}{[B^+]}$$

At any moment, $$\ce{[H^+]=[BOH] +[A^{2-}]-[H2A]}$$
$$[H^+]=\frac{[NH^{4+}].k_w}{[H^+]k_b}+\frac{k_{a2}.[HCO_3^-]}{[H^+]}-\frac{[HCO_3^-][H^+]}{k_{a1}}$$
Rearranging the terms to get $\ce{[H+]}$,
$$[H^+]=\sqrt{\frac{\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]}{1+\frac{[HCO_3^-]}{k_{a1}}}}$$
$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{k_{a1}+[HCO_3^-]}}$$

Now we will have to make 2 assumptions. First lets assume that $k_{a1}<<[HCO_3^-]$ so that $k_{a1}+[HCO_3^-]=[HCO_3^-]$

$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{[HCO_3^-]}}$$

If you assume that neither hydrolysis nor the dissociation goes too far, you can assume the concentration of bicarbonate ion to be equal to the initial concentration (say C). Since Ammonium bicarbonate dissociates to give two products in 1:1 ratio, concentration of ammonium produced is equal to the concentration of bicarbonate ion produced.

  I saw this formula for calculation pH of such salts in my textbook but
  no derivation was given:
  $$[H^+]=\sqrt{k_{a1}\cdot\big(\frac{k_w}{k_b}+k_{a2}\big)}$$ The above formula can
  be obtained if you consider the last assumption. (Clarification
  needed)


Monochlorination of 2-chloropentane and possible enantiomer products I found a question to determine the number of possible chiral products of monochlorination of 2-chloropentane.So, first the products are (with possible number of enantiomers in the parentheses:

2,2-dichloropentane(no chiral product),

1,2-dichloropentane(4 or 2, I am not sure is there a plane of symmetry here)

2,3-dichloropentane(2 or 4)

2,4-dichloropentane(obviously none for a plane of symmetry)

2,5-dichloropentane(2,4?)

But my surprisingly the answer says it's only 5. That's even lower than my lower estimate. I am not sure how there are 5 enantiomers.

2-chloropentane is optically active compound hence in some cases we will get only diastereomers. For example 1,2-dichloropentane is single product because it is formed by attack at carbon no. 1 hence no change in chirality of carbon no. 2.  So product formed by attack at carbon 1 (one chiral product) carbon3 (two chiral product i.e. diastereomers) carbon 4 (one chiral product, i.e. one chiral and one meso) carbon5 (one chiral product)


Is this rearrangement in carbocation possible? Generally, in the following type of reactions, positive charge on $\alpha$-Carbon causes ring expansion, but in the following example positive charge is formed on $\beta$-Carbon and the cation is cyclo-propyl methyl.

So is the above rearrangement even practically possible?

yes,  the rearrangement in carbocation is possible depending upon the condition of reaction and the stability of the new carbocation formed
 eg- the general order of stabilty for carbocation is 3degree >2degree > 1degree
NOTE- the stability of a carbocation depends on aromaticity resonance hyperconjucation inductive effect and sometime ring expansion also occurs


What is the correct Lewis structure of diazomethane? I tried to google and watch YouTube videos on how to write Lewis structures properly, but I can't seem to understand how to do this correctly.

My teacher, when class was in session, gave me a set of instructions to complete Lewis structures:

Sum Valence electrons
Add electrons to outer atoms, while obeying octet or duet rule.
Then add electrons to center atom, obey octet rule, unless they're third row element or below.
Do formal charges for each atom valance charge - (unbonded electrons + 1/2(bounded electrons))
Once formal charges have been determined, minimize formal charges by using unbound electrons to make double or triple bonds
Redo formal charges to check if formal charges are as low as possible, for central atom.

So I attempted to do this for the diazomethane, but my answer (circled in green) does not match the book's answer (in red).

What did I do wrong?

After further considerations, I realized that my structures didn't obey the octet rule. So then would this revised structure be also correct?

The image above depicts three mesomeric structures of diazomethane ($\ce{CH2N2}$). When you count the valence electrons for every atom, you will find that only the left and middle structure satisfy the octet / duet (for hydrogen) rule, i.e. that every $\ce{C}$ and $\ce{N}$ atom has 8 valence electrons, and every $\ce{H}$ has two valence electrons. The structure on the right side does not obey the octet / duet rule because the positively-charged nitrogen only has an electron sextet. The same is true for the two structures you have drawn with an $\ce{N-N}$ single bond. In both cases, at least one nitrogen has only 6 valence electrons.


Does ammonium react with water in an ammonia solution? In the following equilibrium:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

Is it possible for the ammonium to simultaneously establish another reaction with water?

$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$

If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle. I surmise that this system will eventually stabilize yielding hydroxide ions (with a negligible concentration of hydronium ions), hence the basic character of ammonia. Are these valid conclusions? Thanks in advance.

Is it possible for the ammonium to simultaneously establish another reaction with water?

It might help to add these two reactions to see what the net effect is:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
$$+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ $$
$$\ce{2H2O <=> H3O+ + OH-}$$

So yes, this is the autodissociation of water, and this reaction goes on at the same time. The three equilibrium constants are linked (K1 x K2 = Kw), so there is never a situation where two of these reactions are at equilibrium and the third one isn't, causing an endless change in concentrations. In fact, for reactions in solution, there is always a state where all of the reactions are at equilibrium, and if all reactions proceed, they will all reach equilibrium eventually.

You can take any reaction that includes hydronium (or hydroxide) ions and formulate another reaction that includes hydroxide (or hydronium) ions instead by combining it with the autodissociation reaction.

  If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle.

There is no cycle, unless you call reactants making products and products making reactants at the same time a cycle. The latter is a feature of all reactions that are at equilibrium. If you label a hydrogen, it will forever cycle between being part of ammonia, ammonium, hydroxide, water, and hydronium. There are some more complicated systems of reactions where the path toward equilibrium involves oscillations of concentrations, but in most cases, concentrations approach the equilibrium concentration in a monotonic way.


How to prepare a solution of accurate concentration? How does one accurately dose powders? I would like to know what tools are used and how.

For example, how would one apply 1 gram of a certain powder to a solution in a beaker? Specifically, how do you make sure all of your measured powder (or at least with a very small error margin), gets into the solution, without too much of the powder sticking to the spoon or other utensil.

I figured one might dip a spoon loaded with the powder in the solution, but then the solution might stick to the spoon.

This is how I did it in my analytical practicals, works for primary standards with a decent accuracy, works for everything else with a larger error. (You'd want to check concentrations with a primary standard then.)

Determine the mass of the empty volumetric flask
Prepare approximate mass of the substance you want the solution of in a beaker or watch glass or similar
Transfer the prepared substance into the volumetric flask
Determine the mass of the filled flask; the difference will give you the mass of the substance
Fill the flask with the solvent.
Given the precision of the instruments used, you can calculate the concentration quite accurately. If necessary check the concentration with another method.


Behavior of Natural Abundance and Mass Spec Ratios The presence of sulfur in a mass spec is 96% and 4%. I believe this to be because the natural occurrence of sulfur isotopes are within these proportions.
However, chlorine behaves in a 66%/33% ratio in mass spec, but the most abundance isotopes of chlorine are 75%/25%. Why is this so?

You are probably mixing natural abundance (NA) and relative abundance (RA).
In mass spectrometry RA is a more valuable parameter as it can be directly obtained as the $y$-coordinate of a plotted mass spectra: the most abundant ion (isotope) corresponds to the base peak, which is always $100\%.$
In other words, RA reflect isotope ratio, not NA.
For the isotopes of the elements RAs can easily be derived from NAs via normalization; however, the problem of finding RAs of the various isotopic molecular species is a bit less trivial [1].
The following table contains compiled data for NAs [2, p. 1-12] and RAs [3, p. 89] for both elements you've mentioned:
$$
\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lllrlrlr}
\hline
Z & \text{Isotope} & & &\text{Mass}/\pu{u} & \text{NA}&\hspace{-1em}/\% & \text{RA}&\hspace{-1em}/\%\\
\hline
16 & \ce{^{32}S} & \ce{[E]}      & \d{31}{9720711744(14)} & \d{94}{99(26)} & \d{100}{000} &\hspace{-1em} \\
   & \ce{^{33}S} & \ce{[E + 1]}  & \d{32}{9714589098(15)} & \d{0}{75(2)}   & \d{0}{789} \\
   & \ce{^{34}S} & \ce{[E + 2]}  & \d{33}{96786700(5)}    & \d{4}{25(24)}  & \d{4}{433} \\
\hline
17 & \ce{^{35}Cl} & \ce{[E]}     & \d{34}{96885268(4)} & \d{75}{76(10)}    & \d{100}{000} \\
   & \ce{^{37}Cl} & \ce{[E + 2]} & \d{36}{96590260(6)} & \d{24}{24(10)}    & \d{32}{399} \\
\hline
\end{array}
$$
References

Margrave, J. L.; Polansky, R. B. Relative Abundance Calculations for Isotopic Molecular Species. J. Chem. Educ. 1962, 39 (7), 335. DOI: 10.1021/ed039p335.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.


Change in concentration when acids are diluted I came across this formula to calculate the resultant [H]+ concentration when two weak acids are mixed (Both acids of equal volumes are mixed together).

The only problem i have with this is that, we are completely ignoring the fact that the concentration of both the acids are going to be halved after they are mixed together since, the volume is doubling. Is there anything i'm missing? The only way, this could be correct is if the concentrations that are given to me in the question are those after the acids have been mixed.

EDIT: This is the question i was talking about

This is the solution i was given

Let suppose $c, c_1, c_2$ are concentrations of weak acids, and $K_\mathrm{a}, K_\mathrm{a1}, K_\mathrm{a2}$ are their respective acidity(dissociation) constants.

The equations $\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$ for a weak acid, respectively for 2 acids $\ce{[H+]}=\sqrt{K_\mathrm{a1} \cdot c_1 + K_\mathrm{a2} \cdot c_2}$

are approximately valid, if $c, (c_1, c_2) \gg K_\mathrm{a}(, K_\mathrm{a1}, K_\mathrm{a2}) \gg 10^-7 = \sqrt{K_\mathrm{w} }$, what also implies $\ce{[H+]} \gg 10^{-7}$.

Then we can approximate all $\ce{H+}$ comes from the acid dissociation, but OTOH, all acid is in non-dissociated form, leading to $$K_\mathrm{a} = \frac{\ce{[H+][A-]} }{\ce{[HA]}} \simeq \frac{{[H+]}^2 }{ c}$$ and $$\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$$

If, OTOH, $c \ll 10^{-7}=\sqrt{K_\mathrm{w}}$  and $c \ll K_\mathrm{a}$ we can approximate all acid is dissociated.

Then $$K_\mathrm{w} =  \ce{[H+][OH-]}$$

$$\ce{[H+]} - \ce{[OH-]} = c_1 + c_2$$

$$\ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}} = c_1 + c_2$$

$${\ce{[H+]}}^2 - (c_1 + c_2) \cdot \ce{[H+]} - K_\mathrm{w} = 0 $$

$$\ce{[H+]} = \frac{c_1 + c_2 \pm \sqrt {{(c_1 + c_2)}^2 + 4 \cdot K_\mathrm{w} }}{2}  $$

$$\ce{[H+]} = \frac{\pu{2.1e-9} \pm \sqrt {{(\pu{2.1e-9})}^2 + \pu{4e-14} }}{2} = \pu{1.0105e-7 mol/L }$$

$$\mathrm{pH} = 6.995$$

It is obvious, that if with full dissociation $\mathrm{pH}$ does not reach even the value $6.99$, the result of partial dissociation cannot be $6.7$


Radiation in space and its effects on chemical reactions? I know that one current area of research is ways to protect astronauts from ionizing radiation when they venture out of the atmosphere of Earth, but would that same ionizing radiation be a cause of concern when performing chemical reactions in space or in the atmosphere of planets with little atmosphere?

I'm aware that radiation like ultraviolet, which is present in sunlight here on Earth, is used in organic chemistry for the homolytic fission of halides, but I'm not sure how much higher energy radiation might affect other reactions.

It depends....

First let's define the "problem space".

Chemical reactions are typically driven by a few eV (electron volts). "Ionizing radiation" can be keV or MeV which is vastly excessive. In fact that amount of energy is so massive that it would "rip up" the chemical bonding in any material. Only solid metals would be somewhat stable since the 3D lattice locks the atoms into place. $\ce{NaCl}$ for example could decompose to Na metal and $\ce{Cl_2}$ gas.

Any sort of chemical reaction in space has to have a container of some sort which would mitigate the penetration of the ionizing radiation somewhat. But it surely won't stop all of the ionizing radiation. There are nuclear experiments that are conducted a mile underground to try to remove cosmic particles.

The primary source of ionizing radiation in our solar system is the sun. So more distance from the sun is your friend.

"Performing chemical reactions in space" sort of implies that a human is doing the reactions. A biological entity is going to be more effected than any vat of chemicals. So in an environment which humans or plants could safely operate, then performing chemical reactions won't be a problem.


Why is there such a violent reaction when NaCl is formed? In school I saw how $\ce{NaCl}$ was formed and I noticed that there was a violent reaction. Can someone tell me why there is such a violent reaction? The $\ce{NaCl}$ was formed by reacting sodium metal and chlorine gas.

Sodium's single unpaired electron is partially occupying a much higher energy orbital than the unfilled orbital on the chlorine which will be occupied when the two chemicals are mixed. The excess electronic energy generated will rapidly become thermal energy giving a "violent" but rarely explosive reaction (as there is no gas generation only thermal expansion).


Why does the concentration of the acid equals that of the base at the mid-point of titration? Consider this graph of the titration of a weak acid and a strong base:

The buffering region is strongest at the midpoint of titration because there is equal concentrations of acetic acid and its conjugate base.

  My question is why at the midpoint of titration does the concentration of the acid equal the concentration of its conjugate base? Also, why does the $\mathrm{p}K_{a}$ of the acid = the $\mathrm{pH}$ at this point?

At the midpoint of titration, the concentration of acid is just approximately equal to its conjugated base, as there must be considered the acid dissociation and ( in less extend ) salt hydrolysis.

If we take the definition(simplified using concentrations):

$$K_\mathrm{a}=\frac {\ce{[H+][A-]}}{\ce{[HA]}}$$

we can see that by definition:

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

$\mathrm{p}K_\mathrm{a}$ is $\mathrm{pH}$ when both items of the conjugate pair have the same concentration.

$\mathrm{pH}$ near titration midpoint is approximately equal

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {n_\mathrm{base}}{
n_\mathrm{acid, init} - n_\mathrm{base}
}}$$

This function has minimal slope = maximal buffer capacity at $$n_\mathrm{acid, init} = 2 \cdot n_\mathrm{base}$$


Why is in the reaction of cyclic unsaturated aldehydes with hydrazine and hydrogen peroxide the aldehyde preserved? Question
  What is the product of this reaction?

  The answer given:

At first, I thought this was a typical Wolff-Kishner reduction, but then I saw the reagents didn't match up (no $\ce{OH-}$).

How is this getting formed? Shouldn't the $\ce{CHO}$ group be reduced? What am I missing here?

The reaction of $\ce{H2O2}$ with hydrazine forms diazene, $\ce{HN=NH}$.  This reacts in a syn fashion to deliver both hydrogens to the same side of the alkene.


Preparation of KHSO4 This is theoretical scheme for preparing $\ce{KHSO4}$:
$$\ce{K2CO3 + 2H2SO4 -> 2KHSO4 + CO2 +H2O}$$
using 1 eq $\ce{K2CO3}$ and 2 eq of $\ce{H2SO4}$ in approx 10 eq of $\ce{H2O}$ one gets $\ce{K2SO4}$ instead. How is it possible that $\ce{H2SO4}$ completely dissociated despite $\ce{HSO4-}$ is very weak acid in water? Does $\ce{K+}$ ion somehow influence the acidity of that anion? You have to use double the amount of sulfuric acid (4 eq) to get desired product.
Also when using 1 eq $\ce{K2CO3}$ and 1 eq of $\ce{H2SO4}$ what product would you expect? Mixture of $\ce{KHCO3}$ and $\ce{KHSO4}$?

The pKa2 of sulfuric acid is 1.99 (Table of pka).  This is definitely less acidic than pKa1, but calling this a 'weak' acid may be a distraction.

The relevant question is: does $\ce{KHCO3}$ readily deprotonate $\ce{KHSO4}$?  The pKa1 of carbonic acid is 6.35 (same source), making it a weaker acid than  $\ce{KHSO4}$.  The difference is more than 4 orders of magnitude, so yes, I expect that $\ce{KHSO4}$ will more strongly favor depronotation than carbonic acid, and hence $\ce{K2SO4}$ is the product I expect.

To get at this quantitatively, consider the sub reaction:

$$\ce{KHCO3 + KHSO4 <=> K2SO4 + H2CO3 }$$

Which leads to the expression:
$$\ce{K_{eq} = \frac{[K2SO4][H2CO3]}{[KHCO3][KHSO4]} }$$
which can be rewritten as
$$\ce{K_{eq} = \frac{[H^+][K2SO4][H2CO3]}{[H^+][KHCO3][KHSO4]} }$$
Which can be rewritten as
$$\ce{K_{eq} = \frac{[H2CO3]}{[H^+][KHCO3]}\cdot\frac{[H^+][K2SO4]}{[KHSO4]}}$$
which can be written as
$$\ce{K_{eq} = \frac{$K$_{a2,\ce{KHSO4}}}{$K$_{a1,\ce{H2CO3}}} }$$
Taking the log of both sides
$$\ce{pK_{eq} = p$K$_{a2,\ce{KHSO4}} - p$K$_{a1,\ce{H2CO3}} = } -4.36$$

Which is $$\ce{K_{eq} \approx 23,000}$$ heavily favoring products, and hence, $\ce{K2SO4}$ is the dominant product


What is "gold number"? I was reading about gold number and a book defined gold number as following:

  It is the minimum mass in milligrams of a stabilizing agent which is added to the 10 mL of red gold sol to protect it against coagulation caused by 1 mL of 10% (by mass) NaCl.

Is this the definition? Because I am not able to understand above definition.
Is NaCl used as the standard?

A red gold sol is a colloidal suspension of gold nanoparticles with average size of particles less than 100 nm. The suspension is quite stable — the nanoparticles don't aggregate because of the existence of an electrical double layer on the surface of particles which causes electrostatic repulsion between the particles.

If you add a sufficient amount of an electrolyte (e.g., NaCl) to the colloid, the ions will disrupt the electric double layer which will cause aggregation (coagulation) of gold particles. A stabilizing agent (e.g., citric acid) is a compound that is able to adsorb on the surface of gold nanoparticles and protect them from aggregation (it acts like a shield), even in a high electrolyte concentration.


Martensitic transformation in ceramics What is the specification of Martensite (displacive) transformation in ceramics comparing to the metals? I just know about Martensite transformation in metals. For example, these transformations are thermal and diffusionless. Also, the kinetic and morphology are dominant by strain energy.

Take a look at Kelly, P. M., Francis Rose, L.R., The martensitic transformation in ceramics — its role in transformation toughening, Progress in Materials Science, 2002, Vol.47(5):463–557, doi:10.1016/S0079-6425(00)00005-0.

Basically, a transformation is martensitic if it occurs from a collective shear motion of the atoms where each atomic displacement is less than the distance between atoms.  There is no diffusion (atoms moving to other lattice sites). The definition is the same as in a metal, but in a ceramic you have multiple atom types in the crystal, that's all.


Acidic strength of para substituted benzoic acids Comparing the acidic strength, the nitro-substituted group is definitely more acidic. This can be reasoned by fact that a higher delta positive charge which develops on the H due to the strong -M nature of the nitro group.
Another way to check the acidity is to check the stability of the conjugate base.

The conjugate base would be the carboxylate ion which being a -M group creates a lesser electron density at the ortho and para positions. This positive charge is however stabilised by the methoxy group, which is a +M group, and destabilised by the nitro group. So from this viewpoint, the methoxy derivative should have been more acidic.

Why is this not so? What is the fault in my reasoning?

You are somewhat on the right track. However, consider that acidity depends on the relative stabilization of the acid vs. the conjugate base, since the expression for $\mathrm{p}K_\mathrm{a}$ is as follows: $$\mathrm{p}K_\mathrm{a} = -\log\left(\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}\right)$$

Consider this figure for a visualization:

Electron-withdrawing groups like p-$\ce{NO2}$ (nitro group in the para position) increase acidity by making the carboxylate ion stable relative to the protonated form. However, electron-donating groups like p-methoxy decrease acidity relative to standard benzoic acid by destabilizing the (already negatively charged) carboxylate anion!

TL; DR: An electron-withdrawing group like p-$\ce{NO2}$ will better stabilize the negatively charged $\ce{COO-}$ group than an electron-donating group like p-$\ce{OCH3}$.

I found a neat picture from UC Davis that summarizes what I have said:


Why are group 1 elements so low in density? I was studying the s-block elements and found that they extremely low in density. Lithium is said to be the least dense solid in the entire periodic table and their Cohesive Energies are also very low. Could anyone please explain in detail the reason for this?

Because it is energetically easier for an atom (other than hydrogen) with only one outer electron to lose that electron, than to gain seven more, in order to have an inert gas configuration, that electron has a high probability of wandering off. In bulk samples, that causes metallic conduction. Having that lone, loosely-attached electron increases the effective atomic radius.

Hydrogen is an interesting special case. At the ionization potential is more than twice that of lithium, for example, and $\ce{H2}$ has a covalent, rather than metallic, bond. The gas is diatomic, with very little attraction between molecules, so its density is low. However, at sufficient pressure, it is believed that hydrogen would behave as a metal.


Effect of dilution on the degree of dissociation of an electrolyte My book says that on dilution, the degree of dissociation of an electrolyte increases. Consider the reaction:

$$\ce{A(aq) <=> C(aq) + D(aq)}$$

Let $\ce{A}$ be the electrolyte that splits into its ions $\ce{C}$ and $\ce{D}$.
According to my book, on dilution, the concentration of $\ce{C}$ and $\ce{D}$ will decrease. Hence according to the Le Chatelier's principle, the reaction must proceed in forward direction.

Won't the concentration of $\ce{A}$ also decrease? In that sense, shouldn't the reaction proceed both in the forward and backward direction?

Please do let me know where am I going wrong.

If our dummy dissociation degree is $$\alpha=\frac{[\text{C}]}{[\text{A}]+[\text{C}]}=\frac{[\text{D}]}{[\text{A}]+[\text{D}]}$$ and the total concentration of dissociated and undissociated species is $$x=[\text{A}]+[\text{D}]=[\text{A}]+[\text{C}]$$ the equilibrium constant, on which le Chatelier's principle is (well, should be) based is $$K=\frac{[\text{C}][\text{D}]}{[\text{A}]}=\frac{\alpha^2x}{1-\alpha}$$ Now, we'd like to dilute (to decrease) the total concentration $x$. Since the equilibrium constant cannot change, $\alpha$ must increase to keep the constant constant (sorry). And all the concentrations decrease upon dilution, which you can easily prove by checking whether derivatives $\frac{\partial[\text{A}]}{\partial x}$, $\frac{\partial[\text{C}]}{\partial x}$ and $\frac{\partial[\text{D}]}{\partial x}$ are positive.

I really hate them easy "intuitive" principles. In my opinion, resorting to the discussion of the equilibrium constants gives you a clearer picture.

And yes, the reaction proceeds in both forward and reverse directions, if only because it is an equilibrium process.


Can endergonic reactions be driven by heat? I am having trouble conceptualizing endergonic reactions and their difference from endothermic reactions. An endergonic reaction is one that has a positive Gibbs free energy change, such that energy is taken in for the reaction to take place; in biological systems the very exergonic breakdown of ATP is a major source. The expression for Gibbs free energy (for a reaction at constant temperature and pressure) is given by:
$$\Delta G = \Delta H - T\Delta S$$
It is known to me that some endothermic reactions can in fact be exergonic (at certain temperature thresholds) as the entropic increase will be able to counteract the enthalpy change. However, if the reaction is endothermic and associated with a decrease in entropy, the reaction is endergonic at all temperatures. I have a hard time understanding this; isn't heat energy a source of energy as well? If we raise the temperature high enough, surely there would be enough energy to drive the reaction. Intuitively, I imagine that heat energy can't flow into the system due to a thermodynamic law, but I can't seem to figure it out. Does it have to do with entropy or anything related to transfer of heat?

I think you can look at this entirely from the perspective of entropy (forget about the Gibbs free energy for a moment). You can divide the universe into two parts, the system and its surroundings.
When a change happens in your system, and some heat is exchanged with the surroundings then the change in entropy of the surroundings is: $$\Delta S_\mathrm{surr}=-\frac{\Delta H}{T}\tag{1}$$
(minus sign because if $\Delta H$ is positive, that means heat is taken in from the surroundings so the entropy of the surroundings must go down).
The change in entropy of the system itself is: ${\Delta S_\mathrm{sys}}$
(which you are referring to as $\Delta S$ in your question).
Now the total entropy change of the universe is:
$$\Delta S_\mathrm{univ}=\Delta S_\mathrm{surr}+ \Delta S_\mathrm{sys}=-\frac{\Delta H}{T}+\Delta S.\tag{2}$$
If you have a reaction where $\Delta H$ is positive, then the first term (i.e. $\Delta S_\mathrm{surr}$) is negative. If the entropy change in the system is also negative (i.e. $\Delta S_\mathrm{sys} < 0)$ then the total entropy change of the universe becomes negative. Of course, the universe is an isolated system, so the second law forbids any change that reduces the entropy of the universe.
Now what happens when you increase $T$? Well… nothing. Assuming that the signs of enthalpy and entropy change of the system stay negative regardless of rising temperature, the total entropy change is going to stay negative. All you do by raising $T$ is decrease the magnitude of the first term.
If you look at the third equation, and multiply both sides by $T$ and change signs, you will basically get the expression for Gibbs free energy. In the end, the sign of Gibbs free energy basically tells you if the total entropy of the universe is going down or not when you do that change in the system.
So even if you raise the temperature, an endergonic reaction will not go, and heat won't flow into the system because that would mean reducing the total entropy of universe.

As an aside, I should mention here that all of this applies to one system. In real world, there are many things you can do to drive an endergonic reaction. For example you can couple it with another reaction/change that makes more entropy than it consumes, and that can drive the endergonic reaction. You can supply electricity (think of it as the entropy that was produced during the generation of electricity feeding the endergonic reaction).
Lastly, even if a reaction has $\Delta_\mathrm{r} G < 0$ that does not mean that you won't get any product. It just means that the equilibrium lies to the side of the reactants. The values of $\Delta_\mathrm{r} G,$ $\Delta_\mathrm{r} H,$ $\Delta_\mathrm{r} S$ that we write consider only the reaction happening, it does not include the entropy of mixing that comes from mixing of the products and reactants. So, even though the reaction is disfavoured, you might get a tiny amount of the product.


Can beta D-glucose be present hydrolysate after hydrolysis of cellulose and how can it be converted into alpha D-glucose? When we perform the hydrolysis of cellulose by acid, we already knew that it can convert cellulose into glucose. As we know, cellulose is composed of beta glucose units linked by beta 1,4-glucosidic bonds. So I would like to know, resulting glucose after hydrolysis is beta or alpha glucose? If beta glucose is present, how can it be converted to alpha?

In aqueous solution, $\alpha$-D-glucose and $\beta$-D-glucose equilibrate. So glucose derived from aqueous acid hydrolysis of cellulose will eventually become an equilibrium mixture of α and β.

A 2006 paper goes into great detail on the kinetics and thermodynamics of the interconversion (in water).  They say:

  At 30 °C, the rate constants for ring closing toward α-pyranose and β-pyranose are 0.044 s−1 and 0.072 s−1, respectively.

Keep in mind those figures don't include the effect of acid catalysis, and are incomplete without rate constants for the formation of the ring-open aldehyde from the two different pyranoses.  But they should give you a rough sense of how quickly the process occurs.  A later figure in the same paper shows that it takes roughly 12,000 seconds or 3.3 hrs for nearly complete (~90%) relaxation to equilibrium.

Also see chemistry.stackexchange.com/a/7311/13184 for more information.


What current and voltage are best for electrolysis rust removal? I weld carbon steels all the time, and sometimes I get parts from the scrapyard. Unfortunately, these parts are generally oxidized, so I've been assembling an electrolysis tank for it. This tank removes the oxidation in about a day. I'm getting the feeling that I can make it go faster if I step up the current or voltage. What combinations of currents or voltages would make the fastest method of removal?

NOTE: I'm familiar with the risks of doing electrolysis, including hydroxy production, and the restriction on stainless steel electrodes.

In order to answer the question we also need how large a part you are electrolyzing.

The current density in $mA/cm^2$  is the key number. For electrolysis on stainless steel, I would say you need to stay less than about $ 1~mA/cm^2$ to stay safe.

The voltage is hard to gauge. In any two electrode system like yours, the electrochemical potential is only one part of the total potential. The total potential includes contributions from the resistance of the solution, resistance of the contacts, overpotential of the counter reaction, etc... The current should be a decent gauge of what you are seeing.


If magnesium reacts with oxygen to produce magnesium oxide only on the application of heat, then why isn't it categorised as an endothermic reaction? Why isn't the energy that is used to overcome the activation energy of magnesium metal atoms to react with oxygen, not taken into account?

Why isn't the energy that is used to overcome the activation energy of magnesium metal atoms to react with oxygen, not taken into account?

Typically, the activation energy is much higher than the difference in energy between reactants and products. The activation energy gets used when you go from reactants to activated complex, and more or less becomes available again when you go from activated complex to product. As Waylander stated in the comments, exo- or endothermic is based on the difference in energy between reactants and products, not the size of the activation energy (otherwise, all reactions would be endothermic).

  If magnesium reacts with oxygen to produce magnesium oxide only on the application of heat, then why isn't it categorised as an endothermic reaction?

Just because you are heating something to get it to react does not mean there is a net flow of heat from the surrounding into the reaction. It would be more accurate to say that you are increasing the temperature to allow the reaction to proceed.

  [from the comments]: You just need to heat up a 1/4 inch and get that burning. Then the burning itself supplies enough heat to keep the ribbon burning.

Again, this is about the temperature, not about exothermic or endothermic. There are multiple reasons why a reaction would happen at high temperatures but not at low temperatures:

Change in equilibrium constant: For example, water vapor forms dew at lower temperature, and dew evaporates again at higher temperature. The activation energy is sufficiently low that evaporation and condensation take place at either temperature, but the net direction changes.
Reaction rates: Some reactions have a high activation energy and are too slow at low temperature, so you have to raise the temperature to observe them. This does not mean they are endothermic (check out the thermite reaction).
Allowing the reactants to come together: Magnesium is covered in a layer of magnesium oxide, preventing oxygen in air to react with the magnesium. At higher temperature, magnesium metal melts and comes in contact with air.
A slow step in a multi-step chain reaction: In radical reactions, the initial formation of a radical has the highest activation energy, whereas reaction of a radical with other reactants to form another radical (chain propagation) has a lower activation energy. In this case, the first step would require higher temperature (like a spark) or a catalyst (radical starter) and then the chain reaction can proceed at low temperature.

For all but the first, it does not matter whether the reaction is endothermic or exothermic - higher temperature helps or is required to observe a reaction. You can have a endothermic reaction even at lower temperatures. The heat transfer can happen even under low temperature conditions because the molecules have thermal energy nonetheless.

Using day-to-day language, that sounds weird because we associate heat with temperatures higher than body temperature, so we might doubt that there is heat in cold water. However, reactions that cool down the water even more demonstrate that the water can transfer heat to these processes (e.g. cold water can melt ice).


Is there a single or double bond between S and O in propanethial S-oxide? I am doing some research on propanethial S-oxide, a chemical produced by onions. I am trying to find out whether they have a single or double bond between the sulfur and oxygen atom. I have seen both from different sources, simply search it up on Google Images. Is this because there is resonance going on? Also is this considered a sulfinyl functional group (I am guessing no because it cannot be on the end of a chain)?

In addition, is the C double-bond S part in this molecule considered a thial functional group? A Google search indicates that the atom bonded to the sulfur must be a hydrogen for it to be a thial. If not, what type of functional group (if any) exists between the C, S, or O?

The $\ce{S-O}$ bond in sulfoxides is best considered as a single covalent bond with positive charge on $\ce{S}$ and negative on oxygen. The oxygen is significantly nucleophilic (see for example the Swern oxidation mechanism 1) and the sulfur will stabilise an anion on the alpha carbon.
$\ce{C=S}$ has significant enol content and is better considered $\ce{HS-C=CR}$. The $\ce{S}$ is significantly nucleophilic and is easily oxidised to the ene-disulfide $\ce{RC=C-S-S-C=CR}$.


Assigning the 13C NMR spectrum of bromobenzene and 4-bromobenzophenone I've been trying to assign the peaks of 4-bromobenzophenone (I have the peaks, just need to assign them to the correct carbons), but my assignments and what it should be don't seem to agree. As part of this, I tried (and failed) to correctly assign the NMR of bromobenzene.

Bromobenzene's spectrum:
https://www.chemicalbook.com/SpectrumEN_108-86-1_13CNMR.htm

Bromobenzene, my labelling of the carbon environments: https://imgur.com/a/NiVUd

Here's what I've thought so far for bromobenzene:

The halogen atom is a deactivating substituent for the cyclic ring. The two effects of the halogen are the inductive and conjugative effect.

1) The inductive effect falls away with distance. Therefore, carbon environment 2 experiences a stronger inductive effect than carbon environment 3, and likewise for carbon environment 3 compared to 4. Therefore, due to the inductive effect only, carbon 2 has a more positive, and so higher shift than carbon 3, which has a higher one than 4 i.e. 2>3>4.

2) The conjugative effect acts mainly on carbon 2 and 4 by resonance arrows (as bromobenzene is ortho/para directing). Therefore, carbon environment 3 has a more positive environment, and therefore higher shift, than 2 and 4 i.e. 3>2=4

However, when it comes to carbon 1, it is directly attached to the bromine. The inductive effect is strongest for the carbon 1, therefore, and the conjugative effect isn't the same strength here as for carbons 2/4 and so carbon 1 should be the highest shift?

According to my rationalisations, carbon 1 should have the highest shift, followed by 2/3 (order is hard to determine) and then 4. This is not seen in the spectra.

Where am I going wrong then? I don't see how you can (qualitatively) predict that:

1) carbon environment 1 actually has the lowest shift

2) carbon environment 2 has a higher shift than carbon environment 3 i.e. the increased inductive effect outweighs the increased conjugative effect for carbon environment 2.

Bromine is an example of what is sometimes described as the 'heavy atom effect', where chemical shift follows an opposite trend than would be expected due to electronegativity. These types of atoms (Br to a lesser extent than for I and Te) have a large electron cloud distribution. The electron cloud has a significant effect on shielding of the ipso carbon, effectively immersing the carbon within a cloud of electrons. This contributes to the diamagnetic shielding around the carbon nucleus and results in an upfield shift. As an aside, because of their size, Br and I often have significant γ effects also, which are usually steric in origin.

For calculating 13C shifts at C1, the trend between halogen-substituted benzene series follows what is expected based on electronegativities. So, qualitatively, one might expect shifts to be in the order:

F  >  Cl  >  Br  >  I  >  H

However, you just need to be aware of the heavy atom effect for Br and I, and that these will result in an upfield shift for the ispo. Hence, one can expect, qualitatively, that the progression of shifts would look like:

F  >  Cl  >  H  >  Br  >  I

This heavy atom effect is also well demonstrated in the substituted methane series:

    CH4     CH3X    CH2X2   CHX3    CX4
Cl  -2.5    25.6    53.5    77.4    96.3
Br  -2.5    9.5     21.7    12.4    -29.4
I   -2.5    -24.0   -53.8   -139.7  -292.2

So, you can qualitatively predict shifts for C1, as long as you are aware of this contribution for Br. Predicting the order for C2/C3 is not so easy. Qualitatively, it is a reasonable assumption to say that the effect at C3 is negligible, that is, without any other knowledge, δ shift will be ~0. So you just need to look at the β carbon, or C2. Again, a reasonable guestimate might predict an upfield shift for those substituents with strong electronegativities. And, this holds for many cases. However, there are many exceptions to this, and the case becomes very complicated as soon as you start introducing multiple substituents. In this case, one should always assign peaks using 2D methods rather than just predictions.


Finding region of electromagnetic spectrum with wavenumber A line with wave number $\pu{1.028\times10^{-2} nm^{-1}}$ is emitted in the spectrum of atomic hydrogen. In what region of the electromagnetic spectrum does this line occur?

Do we find this by referring to the range of wavelength for UV, IR, etc., or is there a method for knowing which series (Lyman, Balmer, Paschen, Bracket, Pfund) this wavelength belongs to?

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


Is the phenoxide ion aromatic? Is the phenoxide ion aromatic? I read that technically it isn't aromatic because it violates some arcane rule, but I can't find any references in the literature.

I don't see why it couldn't be aromatic, since it's

flat/planar
has a system of conjugated p-orbitals.
has 4n+2 pi electrons: 2 from the oxygen lone pair that is delocalized and n = 0.
is cyclic.

I agree absolutely with Max, and while I was still running the calculations, he already provided the answer.
The goldbook provides a nice definition for aromatic compounds:

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Therefore not all aromatic compounds have to strictly full-fill the Hückel (4n + 2) rule, but all that obey that rule are aromatic. Phenols certainly behave like benzene and are therefore aromatic, please note that aromaticity is an experimentally observed feature.

You can see the aromaticity from the $\pi$ type orbitals, that are in both cases quite similar and extensively delocalised. The calculations were done at the DF-BP86/def2-SVP level of theory.

Benzene exhibits a very high symmetry $D_\mathrm{6h}$, therefore the last two of the shown orbitals are degenerate.

The restriction is not present in the phenolate, as the symmetry is only $C_\mathrm{2v}$.

Also due to the missing symmetry constraint the neutral phenolic radical, $\ce{H5C6O.}$, and the phenolic cation, $\ce{[H5C6O]^{\oplus}}$ would be aromatic.


Disubstituted cyclic alcohols - looking for IUPAC rule Given a simple cyclic alcohol:

We can name it cyclopentanol and using cyclopentan-1-ol wouldn't be correct. When more than one alcohol group is added:

we must use numbers and name it as cyclopentane‐1,1‐diol. However, what about situation like below?

In my understanding such compound should be named 1‐methylcyclopentan‐1‐ol and that's how I always named them. However I got into an argument with a person claiming the name of 1‐methylcyclopentan‐1‐ol is invalid and 1‐methylcyclopentanol must be used instead.

Does anyone maybe have a better google-fu and can locate a rule which specifies how such situation should be handled according to Gold Book or Blue Book?

For preferred IUPAC names, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), explicitly stipulates when locants are omitted. Most of these rules apply to the omission of the locant ‘1’.

In particular:

  P-14.3.4.2 The locant ‘1’ is omitted:

  (…)

  (c) in monosubstituted homogeneous monocyclic rings;

  (…)

One example for this rule given in the Blue Book is cyclohexanethiol. Likewise, cyclopentanol is the preferred IUPAC name for the first compound that is given in the question.

This rule, however, is not applicable to the other two compounds that are given in the question since they are not monosubstituted.

Furthermore, we can find a general rule that is relevant in this case:

  P-14.3.3 Citation of locants

  In preferred IUPAC names, if any locants are essential for defining the structure of the parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. For example, the omission of the locant ‘1’ in 2-chloroethanol, while permissible in general usage, is not allowed in preferred IUPAC names, thus the name 2-chloroethan-1-ol is the PIN. (…)

Therefore, the preferred IUPAC names are cyclopentane-1,1-diol and 1-methylcyclopentan-1-ol, respectively. The omission of the locant ‘1’ is not permissible for the preferred IUPAC name.

Nevertheless, the name ‘1-methylcyclopentanol’ describes the correct compound and may be used in general nomenclature.


Determine the order of the reaction, and the value of the rate constant. Predict the concentration of AB at 25s I've never posted a question about chemistry before so I'm not really sure how this community works but here is the question.

I do have the answer to this question but I have no idea on going about solving it.

I do know:

that second order is equal to all of the exponents added up
Rate is equal to $k[A]^{p1} [B]^{p2} $
Rate is also equal to $([A_2]-[A_1])/([t_2]-[t_1])$

And I believe $M$ stands for molarity.

I have little to no idea how I am to solve this other than those 4 things.

Here are the answers the book gives me:

Second Order
$k = 2.25\times 10^{-2}$
$[AB]$ at $25s = 0.619 M$

If n is the order of this reaction, then the rate equation is given by
$$-\frac{d[AB]}{dt}=k[AB]^n$$
If we take the log of both sides of this equation, we obtain:
$$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$
This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$.  So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk.  So we could determine both the reaction order and the rate constant all in one shot from this single graph.

But how would we reduce the given data to plot such a graph?  Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval.  If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as:
$$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$
The average concentration over this same time interval would be:
$$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$
So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table.  The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.


Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not? Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not? And is there a reason why the products of reaction 2) are written separately, instead of "2OH"?

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

2) $\ce{O^{2-} + H2O -> OH- + OH-}$

Correct, all reactions are reversible, just to different extents. Think about the combustion reaction in your vehicle. Is that reaction reversible? Sure, in that nothing is 100% irreversible. The reversing of the combustion just might take a while. But it's not reversible to any significant extent in the short and medium-term; i.e. you're not going to be re-capturing your combusted gasoline and be putting that back together as usable fuel for your vehicle.

Also, about the reaction of oxide ion with water - that reaction goes practically 100% because oxide ion is a very strong base. The strongest base supported in high concentration by a medium is the conjugate base of the medium. (Leveling effect). For water, that would be hydroxide ion (remember that conjugate bases lose a proton). Oxide ion is the conjugate base of hydroxide ion. We would expect oxide ion to be an even stronger base than hydroxide ion because of oxide ion's even higher negative charge density - we're talking about a negative two charge here (versus negative one for hydroxide) and we're missing a hydrogen proton (which only further decreases the size of the oxide ion and further increases its negative charge density). If you think about this Coulombically, then of course oxide ion is going to be a stronger base than hydroxide ion - opposite charges attract (recall that water has two partial positive charges/dipoles stabilized by its hydrogen atoms). F=kq1q2/r^2 and one of the q's here is massive.

Also about how the products are written - it doesn't matter how the products are written. Either way you are creating two HO- (hydroxide ions). Write hydroxide with the negative next to the O (so either -OH or HO- are acceptable) because that's what you're doing - heterolytic clevage of the O-H bond in H-O-H, which (in this case) creates a negative formal charge that is mostly stabilized by the oxygen atom (NOT the hydrogen atom). Plus it's the hydroxide ion not the oxyhydride ion (whatever the hell that is).

On second thought, when writing -OH, make sure the "-" is clearly a superscript, lest people think you're talking about a hydroxyl group.


What is thermal plasma in-flight oxidation technique? From Vijay, M.; Ramachandran, K.; Ananthapadmanabhan, P. V.; Nalini, B.; Pillai, B. C.; Bondioli, F.; Manivannan, A.; Narendhirakannan, R. T. Current Applied Physics 2013, 13 (3), 510–516. DOI 10.1016/j.cap.2012.09.014:

  Since large quantity of energy is consumed to vaporize the solvent in liquid precursors, a novel method that involves thermal plasma in-flight oxidation of $\ce{TiH2}$ has been developed to synthesize nanocrystalline $\ce{TiO2}$."

What does in-flight mean in this context? Obviously it doesn't have anything to do with actual flying, but everywhere I look it says that it is used for something that occurrs or is provided during an aircraft flight.

The article is about photocatalytic inactivation of bacteria by reactive plasma nanocrystalline $\ce{TiO2}$ powder. It suggests a new method for the physical preparation of $\ce{TiO2}$ for a novel reactor.

Original paper where this method (and, I guess, the term itself) has been introduced was written in Japanese in 1982 [1]. In 1997 there was a great overview published in English by Toyonobu Yoshida, author of the original paper, with a chapter (2.3) devoted to this particular method:

  Processes used to manufacture materials by injecting a solid, liquid, or gas into a thermal plasma to use its high enthalpy and reactivity are referred to as in-flight plasma processes (IFP processes).

Due to the nature of the method which utilizes plasma jet with reactants generated via various torch-types, there is actually a flow of chemicals reacting while they "fly", of course, in the plasma area.

Various plasma torches provide various times-of-flight, temperatures and velocities. For example, the direct current jet torch has the following characteristics:

  An argon plasma of about $\pu{80 kW}$ that has been generated in air reportedly produces a temperature of about $\pu{11600 K}$ and a velocity of about $\pu{410 m/s}$... The reaction time for reactants injected near the anode at high temperatures is about $\pu{400 \mu s}$ and the gaseous raw materials are
  mainly used for the synthesis of fine powders... The reaction time of the reactants is estimated to be $\pu{1 ms}$.

Bibliography

Yoshida, T.; Akashi, K. Tetsu-to-Hagane 1982, 68 (10), 1498–1502. DOI 10.2355/tetsutohagane1955.68.10_1498.
Ultra-fine particles: exploratory science and technology; Hayashi, C., Uyeda, R., Tasaki, A., Eds.; Materials science and process technology series; Noyes Publications: Westwood, N.J, 1997. ISBN 978-0-8155-1404-6.


Reasoning retention factor of benzoic acid In my organic chemistry lab, we ran a TLC plate on silica to determine the identity of an unknown compound. The two possibilities were benzoic acid and acetanilide. I know that the general rule is, the more polar the compound is, the more it absorbs onto the silica and slower and lower it would travel (corresponding to a lower $\mathrm{R_f}$ value). However, one question on my lab report asks:

  In practice, benzoic acid tends to have larger $\mathrm{R_f}$ values than acetanilide. Attempt to find an explanation why this is. And note that spots from more concentrated benzoic acid leas to larger $\mathrm{R_f}$ values.

My professor said the answer has something to do with the $\ce{-OH}$ group on benzoic acid.

The answer is the dimer formation of the benzoic acid. This is why higher concentrations lead to more extended dimer formation, thus higher $\mathrm{R_f}$ values as the carboxyl group "gets shielded."

One thing of note is that $\mathrm{R_f}$ values are not directly dependent on the polarity. They also depend for example on the size of the molecule (Oswald radius). And the relation between properties of the molecules (as stated for example in the form of molecular descriptors like $\log P$) and the resulting $\mathrm{R_f}$ is very complicated indeed. This is also illustrated by the fact that there are predictors for IR spectra, NMR spectra, solubility and lots of molecular descriptors but I have never heard of a good $\mathrm{R_f}$ value predictor.

This is why I would always be careful to talk about it like "the polarity" of the compound, especially as "the polarity" is not a clearly defined quantity.


Which gems can be prepared on a small scale basis? I am interested to know the following:

Which gems can be prepared at home (without sophisticated instruments)?
What will be the quality (appearance) of the product for a certain method?

I want these information just for hobby-scale purposes(no academic/commercial production). In addition to common lab equipment, I have $\pu{1000 K}$ of temperature and few atmospheres of pressure at disposal.

Most likely the only gems you will be able to make at home are plastic gems.
Wikipedia sheds some light on the production of some synthetic gems, Corundum and Cubic Zirconia, both of them not possible at home (or at least my home).


Different force constant in ground vs excited state in spectroscopy The force constant of iodine in the first excited state is $k=41.78$ N/m, and in the ground state it is $170$ N/m.

I know that the formula for calculating $k$ is:

$$ k= (\frac{ΔE_{vib} *2π}{h})^2 µ$$

What causes this difference?

Is it that the energy difference for the ground state is greater then that of the excited state and so the force constant is also greater?

At the risk of providing a somewhat circular answer, in the harmonic approximation of an interatomic potential, the force constant is a measure of the force of attraction between the atoms (in a classical description they would obey Hooke's Law $|\vec{F}|=-k(x-x_\circ$)). The answer to the question is therefore that the attractive interatomic force in the electronically excited state of $\ce{I2}$ is weaker than in the ground state.

The following figure from Ref. 1 illustrates the potentials in the ground and excited states:

The potential in the excited (upper) state is broader. In the Morse functional approximation, the potential has the form $$V(r)=D_e\left(1-e^{-\beta(r-r_\circ)}\right)^2$$

where the width of the potential is proportional to $\beta$ and $$\beta = \left(\frac{k}{2D_e}\right)^{1/2}$$

A smaller force constant results in a smaller value of $\beta$ and a broader potential well. This fits nicely with the results noted in the OP and the above figure (the equations are also outlined in the accompanying text). The energy accompanying a vibrational transition is proportional to the square root of the force constant and therefore inversely proportional to the width of the potential well. This is analogous to the behavior observed for a QM particle in a box, wherein the spacing between energy levels narrows as the box size increases. Working backward from vibrational transition energies leads to the conclusion that smaller transition energies are associated with greater oscillatory amplitudes (broader potentials) and smaller harmonic force constant estimates.

References

Bayram, S.B. and Freamat, M.V., A Spectral Analysis of Laser Induced Fluorescence of Iodine. arXiv eprint 1507.02600, 2015.