What are the degrees of freedom that define the temperature of an ionic solid (such as sodium azide)? Degrees of freedom describe the different ways atoms move in a sample. For a pure ideal gas made of non-linear molecules, there are 3N degrees of freedom (N is the number of atoms in the molecule), 3N-6 of which are vibrational, 3 rotational and 3 translational. So per atom, we have three degrees of freedom.

How many degrees of freedom does solid sodium azide have (per formula unit) and what are these degrees of freedom classified as?

For a solid metal, I know there are three degrees of freedom per atom, and they are related to lattice vibrations (phonons). I also know that black body radiation of solids is used as a "mise en pratique" to define the SI unit Kelvin. That makes me think that solids, no matter whether metallic or ionic, share some feature related to kinetic energy when they are at the same temperature.

I am less concerned with the nomenclature and more with how these degrees of freedom change with temperature, giving rise to the same black body radiation at the same temperature, and being at thermal equilibrium with other samples of the same temperature through collisions at the interface between them.

Degrees of freedom describe the different ways atoms move in a sample.

The classical equations of motion for a system of N atoms describe the time evolution of 3N positions and 3N momentum coordinates for the atomic nuclei (ignoring electronic degrees of freedom). The total energy is a function of those 6N degrees of freedom. Sometimes some of those coordinates do not provide a significant way of storing energy so they can be ignored. For instance, in a dilute gas, translational momentum provides a means of storing energy, but the potential is essentially flat, so position does not.

  How many degrees of freedom does solid sodium azide have (per formula unit) and what are these degrees of freedom classified as?

Sodium azide ($\ce{NaN3}$) has $4\times6\times \text{N}_\text{A}$ degrees of freedom per mole of substance. Half are associated with momentum coordinates, the other half with position coordinates (ignoring once again electronic degrees of freedom). Positions can be expressed in terms of 3N-6 internal and 6 global coordinates, and internal coordinates expressed approximately as harmonic oscillator modes. Each oscillator mode is a linear combination of the 3N-6 spatial coordinates, the entire set being orthonormal. If the solid does not move we can ignore global coordinates.

  I am more [concerned] with how these degrees of freedom change with temperature, giving rise to the same black body radiation at the same temperature, and being at thermal equilibrium with other samples of the same temperature through collisions at the interface between them.

Ignoring electronic degrees of freedom (that is, focusing only on nuclear motions), each atom contributes 6N degrees of freedom. Unlike in a dilute gas, here both momentum and position provide a means of storing energy. The amount of energy that can be stored in  each degree of freedom is, following classical statistical mechanics (high T limit), equal to kT/2, for a total of 3RT for one mole. However, quantum mechanics limits occupation to certain states rather than a continuum, and this can reduce the amount of energy each degree of freedom can store (the heat capacity).

There are some concepts in your question that need to be disentangled:

1) Equipartition, a principle that states that energy is dispersed equally among all degrees of freedom. At low T quantization may determine to what extent different degrees of freedom are thermally available. Too cold and only a reduced number of states may be occupied, and there is no equipartition. This is well explained in the wikipedia. The implication for the example you present is that, when compared to a metallic solid with softer vibrational modes, a molecular solid, in which atoms form covalent bonds having large force constants, might be expected to have a lower heat capacity: degrees of freedom corresponding to high frequency vibrational modes that distort covalent bonds can accommodate less energy (on average) at lower temperatures. Such arguments justify some trends in the table below, showing the heat capacity of various solids on a per atom basis (computed from heat capacities from Ref. 1).  Sodium azide is a combination of our definitions of ionic solid and molecular solid. For solid sodium azide the heat capacity (on a per atom basis) is $\pu{19.2 J/molK}$, a value intermediate to that of metallic and molecular solids. Note also that $3\text{R} \approx \pu{24.9 J/mol K}$, the heat capacity of a solid (on a per atom basis) predicted by the partition theorem. This value excludes an estimate of an electronic contribution to the heat capacity.

2) Temperature, an equilibrium thermodynamic property: bodies that exchange energy as heat are not in thermal equilibrium, their temperatures differ. The concept of thermal equilibrium also applies to degrees of freedom. For instance, if you (somehow) freeze particular motions (translational motion in a gas, say) you lower the temperature associated with those particular modes. If the system is then allowed to return to equilibrium, heat flows from warmer into the (accessible) colder degrees of freedom.

3) Blackbody radiation, which is intuitively thought of as a "photon gas" in a cavity. The concept of thermal equilibrium can be applied to this "gas". You can consult the wikipedia for more information on these last aspects:

  Just as the Maxwell–Boltzmann distribution is the unique maximum entropy energy distribution for a gas of material particles at thermal equilibrium, so is Planck's distribution for a gas of photons.[8][9] By contrast to a material gas where the masses and number of particles play a role, the spectral radiance, pressure and energy density of a photon gas at thermal equilibrium are entirely determined by the temperature.

References

Engineering ToolBox, (2003). Specific Heat of Solids . [online] Available at: https://www.engineeringtoolbox.com/specific-heat-solids-d_154.html [Accessed 23/9/2019]. The source does not state the measurement temperature but it is presumably ~room T.


Adsorption on surface of adsorbent Why is energy released during the process of adsorption? How is this energy change observed and how is it calculated?

Under normal conditions, atoms will bond to one another and produce a molecule, only if the bond formation process is exothermic (e.g. it releases energy) - this translates into the products (the molecule) being thermodynamically more stable that the starting material (the atoms).  The same is true for the adsorption process.  A gas will adsorb on a surface only if this configuration is more stable than a gas and a free surface.

Typically when a gas is adsorbed on a surface, something akin to bond formation between the surface and the gas takes place and we wind up with a chemisorbed species.  There are many ways to study this process.  One simple way is to measure the amount of gas admitted to the system containing the surface to be studied and measuring the change in pressure over time.  When done in a calorimeter, the heat given of during the process can also be measured.  Combining these two measurements(change in moles of free gas and energy released) allows one to understand the basic energetics of the process. The "isoteric method" is one experimental procedure for accomplishing this in a laboratory.  It relies on measuring moles of gas vs. the pressure in the system (an isotherm) at various temperatures.  From these measurements the enthalpy of adsorption can be determined.

The study of the adsorption of a gas on a surface is complicated.  There is not a surface with a monolayer of gas adsorbed on it with free gas above this.  Rather there is a surface and bulk gas and in between the two is a multilayer structure of adsorbed gas where the gas concentration varies throughout.  Here is a link to a recent Cal Tech thesis on the characterization of hydrogen gas adsorption onto various surfaces; take a look at section 1.5 and the beginning of chapter 3 for a nice overview of the adsorption process and the interfaces involved. The rest of chapter 3 is a discussion of the thermodynamics and various experimental techniques used to study the process.  It sounds like you might also be interested in performing calculations on the adsorption process.  If so, section 5.8 delves into computational methods used on one of the surfaces studied.


What is "chemical pressure"? While reading some publications on solid superconductors, I encountered a term "chemical pressure" a few times, which is usually attributed to the changes of superconducting transition temperature $T_\mathrm{c}$ alongside with the hydrostatic, or physical, pressure. For the reference, consider this paper:
Chemical Pressure and Physical Pressure in $\ce{BaFe2(As_{1-x}P_{x})2}$ [1].
Whereas I can understand what the physical pressure is and how it's defined and estimated, I found no sources which would explicitly define either what chemical pressure is, or how it's measured (and what units are used) and/or analyzed.
All explanations are somewhat expansive and usually point to the effects of isomorphic substitution in the crystal lattice (ionic structure), or intermolecular repulsion between ligands (molecular structure). So, what chemical pressure is in the nutshell, how it's measured and expressed?
Reference

Klintberg, L. E.; Goh, S. K.; Kasahara, S.; Nakai, Y.; Ishida, K.; Sutherland, M.; Shibauchi, T.; Matsuda, Y.; Terashima, T. Journal of the Physical Society of Japan 2010, 79 (12), 123706 DOI: 10.1143/JPSJ.79.123706. (Open Access)

External and internal pressure

To study the effect of pressure on properties of a solid, is equivalently to learn how changes in volume transform physical parameters. For external pressure at constant temperature, this relationship manifests through compressibility $\kappa$.

$$\kappa =-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$$

An alteration in volume will modify the energy of a system. The relationship is internal pressure, something chemists first encounter when discussing real gases.

$$\pi_T=\left(\frac{\partial U}{\partial V}\right)_T$$

In solids, one might discover different notations. For example[1],

$$\alpha =\left(\frac{\partial W}{\partial V}\right)_T \sim \frac{\mathrm{d}W}{\mathrm{d}V} $$

where $W$ is potential energy, and $\alpha $ is called energy–volume correlation. Disregarding possible sorption, varying external pressure is the purest option of probing internal pressure.[2] Unfortunately, this might not always be feasible.[2][3] Most actinides, for instance, are highly radioactive, thus limiting their study to very specialised laboratories.[3] Many actinides are too scarce to perform necessary large scale measurements.[3] Here is where chemical pressure comes in.

Chemical pressure

Instead of external pressure or thermodynamic internal pressure, we now discuss chemical pressure. Much akin to external pressure, scientists take interest in how energy is dependent on volume. Chemical pressure comprises of substituting in or adding chemical entities to a solid network.[2] [4] [5] [6]

External pressure presses down on the solid. Volume is modified via a compressibility relationship.[1] A more direct or pure method.[2]
Chemical pressure is induced by changing cell volumes via the introducing of new chemical entities, such as atoms or compounds, in the lattice. An indirect, less pure approach.[2]

Obviously, chemical doping modifies a system in other ways besides volume alone.[2] [5] [7] By choosing an element with similar electronic configuration, these deviations from ideality are minimised.[2] [7] Another advantage of chemical pressure over extrernal pressure is that the former may be positive and negative.[2] [8] Positive pressure implies that pressure from chemical substitution is directed inwards, and adds to external pressure. Negative pressure pushes the solid apart, hence acting against external pressure.[2]

Immunisation against terminological confusion

Condensed matter science, like other fields of study, is entitled to its own nomenclature. For chemists specialising elsewhere, this might seem a superfluous obfuscation. This is especially true for cutting-edge chemistry where terminology is fresh and perhaps not entirely settled. For this reason, synonyms are widespread, and overlap with other terms may occur.

External pressure is also called physical pressure[4] to establish a physical pressure – chemical pressure dichotomy.
Chemical pressure is often enclosed between quotation marks. More confusingly, internal pressure is used as an alternative to chemical pressure.[5] Perhaps this is to emphasise an external pressure – internal pressure dichotomy. Or, it could be, that since chemical pressure is used as a tool to measure internal pressure, the concepts have blurred lines because the distinction seemed unnecessary.

Units and experiments

Internal chemical pressure will probably have concurring units with external physical pressure. Pascals ($\ce{Pa}$) when operating in the SI framework. Very few studies explicitly mention this fact[5] because more often the changes in some concrete parameter are discussed[1] [9]. Because of what chemical pressure constitutes, we do not apply a chemical pressure of $10\ \mathrm{kPa}$. We might, however, bring about (or apply) a chemical difference in pressure ... equivalent to several $\mathrm{kbar}$ of external pressure[5]. For solids where external pressure experiments are currently unthinkable, this equivalence cannot be discussed. But the goal is not to compare chemical pressure to external pressure anyway; as before, we wish to learn more about how volume affects (potential) energy.

As far as I am aware, chemical pressure cannot be measured directly. Chemical pressure is induced inside a solid material by substituting or introducing new chemical entities. Where would one put the barometer? If it is possible to carry out external pressure measurements alongside chemical doping, we may compare one to the other. To emphasise, the effects need not match. In these and other cases, volume changes in cells are estimated or measured, and their effect on the system as well.

/(short overview/list of experiments in progress, next week)/

In my opinion, this subsection deserves a question on its own.

TL; DR

/(in progress, next week)/

Additional reading

For an overview of highly-correlated electron systems in metallic solids, I recommend C. Marini. Pressure-induced metallization process in Strongly Correlated Electron Systems [Online]; G. Stefani, G. Altarelli, P. Postorino, Eds. Università degli Studi Roma TRE: Rome, Italy, 2009–2010. Archived link: bit.ly/2yPJcCo

References and bibliography

[1] G. Borelius. 'Internal Pressure in Solids and Liquids'. Physica Scripta, 1970, 1, 2-3, pp 141–147. DOI: 10.1088/0031-8949/1/2-3/011
[2] J-M. Fournier. 'Chemical pressure in actinide systems'. Physica B: Condensed Matter, 1993, 190, 1, pp 50–54. DOI: 10.1016/0921-4526(93)90441-8
[3] U. Benedict. 'Properties of Actinide Metals Under High Pressure'. Journal de Physique Colloques, 1984, 45, C8, pp C8-145–C8-148. DOI: 10.1051/jphyscol:1984826
[4] D. M. Friedman. New Research on YBCO Superconductors; Y. Itoh, Ed. Nova Science Publishers: 2008. Chapter 1: Recent Advances of NMR and NQR Studies of $\ce{YBa2Cu3O_{7-δ}}$ and $\ce{YBa2Cu4O8}$, pp 25–69 Subsection 6: 'Chemical Pressure, Physical Pressure, and Site Disorder', pp 59–59. ISBN: 978-1-60456-084-8
[5] A. Hauser, A. Nahid, S. Delahaye, A. Sadki, S. Schenker, R. Sieber, M. Zerara. 'Chemical Pressure'. CHIMIA International Journal for Chemistry, 2002, 56, 12, pp 685–689. DOI: 10.2533/000942902777679858
[6] R. S. Liu, C. H. Shen, T. S. Chan, R. Gundakaram, S. F. Hu, J. G. Lin, C. Y. Huang. 'Chemical Pressure Induced Phase Transition in Single and  Double Perovskites with Magnetoresistance Effect'. Tamkang Journal of Science and Engineering, 2002, 5, 1, pp 59–61. Archived link: bit.ly/2fvxrsv
[7] A. Huon, D. Lee, A. Herklotz, M. R. Fitzsimmons, H. N. Lee, S. J. May. 'Effect of chemical pressure on the electronic phase transition in $\ce{Ca_{1-x}Sr_xMn7O12}$ films'. APL Materials, 2017, 096105, pp 096105-1–096105-7. DOI: 10.1063/1.4994089
[8] Masatomo Uehara, Tsuyoshi Amano, Sachiko Takano, Tatsuya K[o]ri, Takahiro Yamazaki, Yoshihide Kimishima. 'Chemical pressure effect on the superconductor $\ce{MgCNi3}$'. Physica C, 2006, 440, 1–2, pp 6–9. DOI: 10.1016/j.physc.2006.03.017
[9] M.Tropeano, C.Fanciulli, F.Canepa, M.R.Cimberle, C.Ferdeghini, G.Lamura, A.Martinelli, M.Putti, M.Vignolo, A.Palenzona. 'Effect of the chemical pressure on superconductivity and SDW in undoped and $15$% $\ce{F}$ doped $\ce{La_{1-y}Y_yFeAsO}$ compounds'. Physical Review B, 2009, 79, 174523, pp 174523-1–174523-6. DOI: 10.1103/PhysRevB.79.174523


What is some advance reference books about alloy that gives a lot historical background and applies a lot of theoretical equation or formulas What is some advance reference books specifically about alloy that gives a lot historical background and applies a lot of theoretical equation or formulas with explanation?

I seem to find nothing really very interesting for myself to learn general theory of alloy while searching the google books because these books is too wordy in detail but too few formula and laboratory skill to support it.

The classic would be Porter and Easterling's "Phase Transformations in Metals and Alloys", apparently still available. This should give you a good basis for understanding the thermodynamics of binary (and higher) systems.

For a deeper look at how computational thermodynamics is done, like CALPHAD, the Lukas, Fries, and Sundman book "Computational Thermodynamics" might do, but that has zero historical background.

You might also try asking this in Physics using the materials-science tag.


What is the difference between an Adiabatic System and an Isolated System? In my school text book there is a passage that says:

The presence of reactants in thermos flask or any other closed insulated vessel is an example of an isolated system.

Then a few passages of text later it says:

we take a system containing some quantity of water in a thermos flask or in an insulated beaker. This would not allow exchange of heat between the system and surrounding through its boundary and we call this type of system as adiabatic.

Does that mean all adiabatic systems can be called isolated systems and vice versa?
This got me confused. Please provide some clarity on this.

There are three ways a system can interact with its surroundings:

heat flow across the system boundary

work flow across the system boundary

a flow of matter across the system boundary

An adiabatic system is one for which there is no heat flow into or out of the system.
A closed system is one for which there is no flow of matter into or out of the system.
An isolated system does not interact with its surroundings at all, so there is no heat flow into or out of the system, no work done on or by the system, and no flow of matter into or out of the system.
Thus all isolated systems are adiabatic, but not all adibatic systems are isolated.  To put it another way, being adiabatic is a necessary but not sufficient condition for being isolated.
Also, your textbook is not quite correct in its language when it says:  "This would not allow exchange of heat between the system and surrounding through its boundary and we call this type of system as adiabatic." [emphasis mine]   If you're a high school student using a high school textbook, this imprecision is understandable (they don't want to overly burden you with subtleties), so don't worry about it.
But, if you are interested in exploring the topic more deeply, read on:
Heat is not something that is exchanged between the system and surroundings.  Rather, it is thermal energy that is exchanged, and we call this exchange of thermal energy "heat".
More specifically: If systems actually contained "heat", then they could exchange heat with their surroundings.  But they don't.  Instead, what they contain (assuming $T > 0\,\mathrm{K}$) is thermal energy.  It is this thermal energy that can (if the system is not adiabatic) be exchanged with its surroundings.  And we call this exchange of thermal energy, which manifests itself only during a change in the system, and appears only at the system boundary, "heat".  I.e., heat is not a property of the system.  It is a property of the path through which a system undergoes change.
To give an analogy:  Thermal energy is like cars, and heat is like traffic.  Suppose cars are being moved from one parking lot to another along a connecting road.  The flow of cars from one parking lot to another is called traffic. The parking lots don't themselves contain traffic, nor are they exchanging traffic; they're exchanging cars.
This distinction between system properties (like internal energy) and path properties (heat and work) will become increasingly important if and when you continue your studies in thermodynamics.


If the gaseous molecules didn't have any kinetic energy would the surface of a filled balloon still be firm? The most of the ideal gas law equation is based on the fact that the gaseous molecules are moving around and cause collisions and thus have some pressure. If these molecules were to not have any kinetic energy, and thus consequently would not suffer any collisions would the surface of let's say a balloon, a filled balloon, which is firm only due to the pressure offered by the gas inside, still be firm? would the volume of the gad contribute to its firmness and give it shape? won't the volume be affected too

If ideal gas molecules had had no kinetic energy, there would have been no gas pressure. Inside nor outside the balloon. The outer gas would have been laying on the ground, the inner gas on the bottom of the balloon.

In such a case, the only remaining force, aside of gravity, would have been the elastic forces of the balloon, pulling the balloon rubber together to its original shape and volume.

For real gases compressed to volume about equal to tight arrangement of molecules, a sudden repulsive force and therefore pressure would have progressively appeared, due repulsion of electrons.


What could possibly cause a "melting" reaction between vinyl and plastic? This question was posted in Electronics. Something is really odd about it! What could possibly be causing this "melting" reaction between a common flexible covering (vinyl, apparently) on test lead clips and the plastic there?

  a couple of breadboards had been lying in a box with a rubbered alligator wire sandwiched between. They've been there for maybe a month untouched. No other chemical or electricity has been present and only ambient heat (hot summer but 30°C tops here in Sweden). I'm pretty shocked that the plastic of these two different styles of breadboards both got so messed up simply by touching this rubber:

This is a very common alligator clip and very common electronics breadboard, often used together. I'm wondering whether the asker is telling us everything.

It is probably a consequence of the plasticiser not the polymer

It is hard to be sure what has happened here without checking the details of which polymers are involved which the picture can't make clear. But we can make some intelligent guesses based on commonly used polymers.

First: the cable isn't made from natural rubber. Most likely it is made from plasticised PVC. This is a good and cheap cable material with good insulating properties and doesn't decay like rubber from interacting with atmospheric oxygen. This involves a lot of plasticiser, usually something like a phthalate. The cable might contain 10-40% of plasticiser by weight.

Depending on the exact polymer formulation and the specific plasticiser, the plasticiser can migrate into other materials if they are in contact for long periods of time. When I used to work in a lab, for example, the cheap PVC tuning we used in low pressure gas lines would sometimes accumulate a thick liquid in the lower parts of the tubing loops if used for years: this liquid was plasticiser that had migrated out of the PVC over years of use. This migration is worse when other volatiles are in contact with the polymer.

This may be what is happening in this case. In warm conditions the plasticiser in the PVC might migrate and damage other plastics. It isn't possible to tell from the picture what the breadboards are made from but ABS is most likely in a good quality breadboard, though acrylic, polystyrene or polycarbonate are all possible. All of these polymers can be affected by the migration of plasticiser from PVC as this study shows (though it depends on the specific plasticiser). Good choice of materials for the board and plasticiser can minimise this problem but cheap breadboards and cables are likely to have worse problems.

In short, this is the likely problem here.


How to convert Mn2O3 to Mn3O4? I was preparing $\ce{Mn3O4}$ but it was when characterized by XRD, a $\ce{Mn2O3}$ phase existed. Now how can I convert $\ce{Mn2O3}$ to $\ce{Mn3O4}$?

You may need a good lab oven.
Manganese(II,III) oxide on Wikipedia

Preparation
$\ce{Mn3O4}$ formed when any manganese oxide is heated in air above 1000 °C. Considerable research has centred on producing nanocrystalline $\ce{Mn3O4}$ and various syntheses that involve oxidation of $\ce{Mn^{II}}$ or reduction of $\ce{Mn^{VI}}.$


Salt used in Perkin Reaction The Perkin reaction gives an α,β-unsaturated aromatic acid by the aldol condensation of an aromatic aldehyde and an acid anhydride, in the presence of an alkali salt of the acid. Here is an a diagram showing what the reaction does

The anion of the salt acts as a base for abstracting the β-hydrogen in the acid anhydride. Later it attacks the carbonyl carbon of an intermediate but however leaves in the same step. Since the base is not ultimately involved in the product, is there any kinetic or thermodynamic reason as to why the salt of the acid of the acid anhydride is preferred over any other base?

The base, as you pointed out is nearly always the salt of the acid corresponding to the annhydride.

This is because the base which differs in structure will react with the given carbonyl compound and yield some other compound rather that abstracting a proton.

Eg.
Let's say you have tertbutyl salt there instead of what we are having right now.The tert butyl could have directly attacked on the Carbon having the double bonded oxygen as it has a partial positive charge due to polarization.Which will result in to eliminating the other symmetric half of the annhydride and replacing it by tertbutyl.This changed your basic reacting molecule.And hence the
 product you obtained is definitely not the right one. Whereas using the salt of the acid that formed the annhydride will not change the annhydride even if it reacts with the annhydride the way I stated above.

Using the salt of the acid they produced the annhydride does nothing but abstracts the proton or simply regenerate the original carbonyl compound/annhydride if it reacts rather than abstracting a proton.


What are these stains on my stainless steel pans? I've had enough of the homemaker advice of baking soda or barkeeper's friend products - their advice isn't working.  I need a chemist!

Please see the two photos below.  This pan isn't quite as clean as I could possibly get them if I scrubbed and scrubbed with a Brillo pad, but it's close - and as clean as I can get it would only last for about one use before it needed more intense scrubbing.

The brand of this pan is Cuisinart which is relevant only for the particular variant of stainless steel they might be known to use, but I don't think my experience with this brand is much different than I've had with, say, Calphalon stainless steel pans.

I refer to these blackened areas as stains because they have no significant dimension to them, as opposed to "baked-on-crud" which has a dimension and is more expected/accepted by me than the stains.  The stains are rougher than the shiny parts though, impacting use.

1) Why am I getting these stains?  Am I doing something wrong (heating technique or kinds of food chemicals that adhere to SS)?  Are these pans inferior and really I need to buy very high-end/expensive pans (for a different grade of stainless steel) for them to be indestructible?

2) With knowledge of the chemistry involved that leads to this visual effect on the stainless steel, is there an industrial chemical (like those in carburetor cleaner or something) that could clean this stuff off without damaging the pan and which I could completely remove (to avoid toxicity in my food)?

It could be that metals in your tap water are plating out;  i.e. the more active iron in stainless steel is being replaced by a less active metal such as copper. Manganese also tends to precipitate, leaving stain, and these, in particular, are dark colored, but can be removed.

Have your tap water tested, either at a nearby school or with a commercial kit(s) that include copper, manganese and lead tests. If there is a high concentration of anions, you might want to install an ion-exchange water softener.


Twisting stereoisomers with rings to determine R/S I know that if the smallest atomic number group is not in the back, we can twist the molecule so that it is in the back and then determine if it is R/S (Left image). But if the chiral center is attached to a ring, can we still simply twist the molecule (right image)? If so, how? Or is there another way to make the smallest atomic number group go to the back for molecules with rings?

Rotating structures is a recipe for disaster because errors in transcription will invariable occur. Use the best chirality tool you have---your hands. Leave the structure where it is! Use either hand to make a "thumbs up" gesture. Point your thumb in the direction of the lowest priority group. The remaining fingers should point in the direction of the highest to next to lowest priority group (OH>C2H5 >CH3). If not, you are using the wrong hand. Left hand = S. Right hand = R. Try to rotate structures 1-3 and see how it compares with the "hand method". Structures 1-3 are of the S-configuration. Cyclopentene 4 is Jon's structure with substituents. It is of the R-configuration (=CH>CH2>CH3), a conclusion reached without redrawing the structure.


Mechanism for the conversion of dimethyl cyclohexa-1,4-diene-1,2-dicarboxylate to dimethyl benzene-1,3-dicarboxylate using bromine Question

Convert the first compound into the second using only $\ce{Br2}$

I tried solving it by making a 3-ring with $\ce{Br+}$, and then making one of the oxygens break that ring. Than I used $\ce{Br-}$ to steal an $\ce{H}$ and get a lone pair on one of carbons. Lastly, I used the lone pair to create a double bond. Am I on the right track?

Firstly, I didn't get which oxygen you are talking about to break the ring. If you are considering the oxygens that are part of esters, that is not possible. Secondly, $\ce{Br^-}$ is the conjugate base of s very strong acid $\ce{HBr}$. So, $\ce{Br^-}$ is a very weak base. Therefore, it is very very difficult for that $\ce{Br^-}$ to abstract the hydrogen atom and act as a base.

Instead what you can simply do, you can first perform bromination i.e.  treating the substrate with $\ce{Br_2/ CHCl_3}$. You will first get a bromonium intermidiate, like this:

The intermidiate will form on the farther double bond as the $\pi$ -electron cloud of the other one is pulled by the electron withdrawing ester groups, and also to form that ring, there will be huge steric crowding. The other $\ce{Br^-}$ will act as nucleophile to break this ring, (normal mechanism) and you will get this product:

Now, the two $\ce{Br}$ are in anti position. So, if you now treat the compound with excess $\ce{EtO^-}$ in $\ce{EtOH}$ and heat it, anti-periplanar E2 elimination will occur, and finally you will get your desired product as the one mentioned above.

The elimination will occur from two different carbons as the bromines are in anti position. So, to follow anti elimination of $\ce{Br}$ with $\ce{H}$, the elimination has to be from two different non adjacent carbons.


Why is Buna-S called so? Styrene Butadiene synthetic rubber is also called Buna-S. It is the abbreviation of Butadiene (Bu) Sodium (Na) Styrene (S).
Why is Sodium included?

Sodium is included in the name because it was essential in the chain polymerization that forms the rubber.
Sodium served as the initiator for the anionic polymerization.


Can nitrous dioxide develop when copper and dilute nitric acid react? When doing an experiment in the laboratory with copper, dilute nitric acid and finally hydrochloric acid, the first reaction between copper and dilute nitric acid should end in nitric monoxide, cupric nitrate and water:
$$\ce{3 Cu(s) + 8 HNO3(aq) -> 3 Cu(NO3)2(aq) + 4 NO(g) + 2 H2O(l)},\tag{R1}$$
but somehow nitric dioxide, a reddish-brownish gas appears with a poignant smell, develops from this type of reaction:
$$\ce{Cu(s) + 4 HNO3(aq) -> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)}.\tag{R2}$$
There was 5 mL of nitric acid and as we added nitric acid we quickly covered it with a watch glass. Then we put it in a sand bath instead of heating it normally with a Bunsen burner. Unfortunately, I cannot as it was one of the experiments that we just wanted to know what was going to happen and write down our observations and conclusion.

There are two ways to interpret an observation nitrogen dioxide.

Reactions between a metal and nitric acid rarely proceed by a single path to generate a specific nitric acid reduction product; in all likelihood there will be a combination of multiple reactions and thus a mixture of products.  With copper you could get both $\ce{NO}$ and $\ce{NO2}$.  And with a more powerful reducing metal like magnesium even more products could get into the mix, including ammonium nitrate.

You may have conducted the reaction in the presence of air.  $\ce{NO}$ is readily oxidized by the air to make $\ce{NO2}$ "after the fact".  This is in fact part of the natural production of nitric acid in thunderstorms.  Further complicating this is the fact that nitrogen dioxide in turn reacts with water, forming nitric oxide and regenerating some of the nitric acid (this is also indicated in the above link).  To get the true product mix out of the nitric acid you needed to use an environment free of air and unintended water.


Why is ZnO reduced in this reaction? Consider the following Ellingham diagram.

At 673K, the Gibbs energy value of oxidation of Zn is more negative than the Gibbs energy value of oxidation of C (coke). So, Zn should be oxidised to ZnO. But in various sources like the textbooks of class 12 of CBSE (India), it is mentioned that ZnO is reduced to Zn and C (coke) is oxidised to CO at 673K in the reaction between ZnO and C. Looking at the Ellingham diagram, coke should not reduce ZnO to Zn at 673K.

So why is ZnO is reduced to Zn in this case?

This method works because one of the reaction products, $\ce{CO}$, is gaseous (at the stated reaction temperature) and is continually removed. Following Le Chatelier's Principle, the equilibrium:

$$\ce{ZnO(s) + C(s) <=> Zn(l) + CO(g)}$$

is pushed to the right and the reaction proceeds.

There are numerous of these reactions where Ellingham data don't tell the full story, notably also the reduction of magnesia ($\ce{MgO}$) by coke, is in accordance with the same principle.

One of the most interesting cases is a lab preparation of caesium metal that takes place according to:

$$\ce{CsCl(s) + Li(l) <=> Cs(g) + LiCl(s)}$$

The reaction is carried out under vacuum and at about $970\ \mathrm{K}$, conditions under which $\ce{Cs}$ is much more volatile than $\ce{Li}$. The former thus distils off and is caught in a condenser, pushing the reaction equilibrium to the right.


Why is the melting point of hydrogen iodide larger than that of hydrogen fluoride? Compare the melting points of hydrogen fluoride (HF) and hydrogen iodide (HI).

I know the following things:

$\ce{HF}$ melts at $189.6~\mathrm{K}$ and $\ce{HI}$ at $222.35~\mathrm{K}$.
There are a lot of factors that have to be considered while comparing melting point, i.e., lattice enthalpy, packing fraction, intermolecular forces.
Generally, compounds that show hydrogen bonding show it in the solid and liquid phases because in the gaseous phase the molecules have enough energy to break the electrostatic interactions of hydrogen bonding.
So, HF shows hydrogen bonding in solid state. In fact, it even shows hydrogen bonding in gaseous state, though its not relevant here.
As molecular mass increases, van der Waals interactions between molecules also increase. If we only consider this factor, then melting point of HI must be definitely higher.
The strength of hydrogen bonding of HF is so much that the boiling point of $\ce{HF}>\ce{HI}$ although boiling point of $\ce{SbH3}>\ce{NH3}$, i.e. the hydrogen bonding dominates the van der Waals interactions in HI molecules because of their high molecular mass.
It's the combined effect of all the factors that determine the melting point of a substance.
Lattice energy is inversely related to the internuclear distance, it is also inversely proportional to the size of the ions. So, I expect HF to have more lattice energy.

Every factor that I can think of is in favour of HF except molecular mass. But if molecular mass of iodine is the reason that melting point of HI is higher, than boiling point of HI must have also been higher. Then why is the melting point of HI higher than that of HF?

Why is this contradiction?

Do let me know if I am wrong in the above mentioned facts or if I have any gaps in my understanding of the above points.

The increase in dispersive attraction between HI molecules is greater than the loss of dipole interactions. Only in $\ce{H2O}$ do dipole forces dominate (Intermolecular and Surface Forces, Jacob N. Israelachvili). Contribution of dispersion to intermolecular bonding for various molecules:

$\ce{CH4-CH4}$ - 100%
$\ce{HI-HI}$ - 99%
$\ce{HBr-HBr}$ - 96%
$\ce{HCl-HI}$ - 96%
$\ce{H2O-CH4}$ - 87%
$\ce{HCl-HCl}$ - 86%
$\ce{CH3Cl-CH3Cl}$ - 68%
$\ce{NH3-NH3}$ - 57%
$\ce{H2O-H2O}$ - 24%

What makes the contribution of dispersion so great is that it's non-directional and always attractive. Think of atoms as sticky spheres.


Differentiating between Elementary and Non Elementary reaction Is it possible to tell just by looking(reaction's equation) that whether a given is reaction is elementary or non elementary ?

Reactions that are probably not elementary

Whenever there are more than three reactants, it is unlikely that the reaction is elementary. Tenary elementary reactions are already quite rare. If you have a reaction like

$$\ce{2C8H18 + 25 O2 -> 16 CO2 + 18 H2O}$$

it is impossible for 27 particles to come together at once, so you would expect some intermediates. Even for a reaction with just 3 reactant molecules such as

$$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$

it is highly unlikely that a single collision would be sufficient to break 6 bonds and make 6 new ones.

Conversely, a reaction with a single product such as

$$\ce{O + O2 -> O3}$$

is probably not bimolecular because there is no way of getting rid of the excess energy. Instead, the reaction proceeds like this:

$$\ce{O + O2 + X -> O3 + X}$$

where X is "a third molecule" that leaves the collision chemically unchanged but with higher kinetic energy (source: https://cimss.ssec.wisc.edu/wxwise/ozone/OZONE2.html).

Reactions where you can't tell

Bimolecular simple reactions could well be elementary, but often are not. The reaction

$$\ce{H2 + I2 -> 2 HI}$$

shows two distinct mechanisms, one with one elementary step as written (the iodine has to be vibrationally excited, favored by high temperature) and the other one with iodine first dissociating into iodine atoms (source: https://link.springer.com/chapter/10.1007/978-3-642-80299-7_12).

Is it possible to tell just by looking whether a given is reaction is elementary or non elementary?

No, but sometimes it is obvious that the reaction has to be multi-step. In cases where a mechanism is dependent not only on the net reaction, but also on reaction conditions, it becomes obvious that just knowing the net reaction is not sufficient.


Adding powdered Pb and Fe to a solution My book states that if we add powdered $\ce{Pb}$ (lead) and $\ce{Fe}$ (iron) to a solution containing $1\ \mathrm{M}$ each of $\ce{Pb^2+}$ and $\ce{Fe^2+}$ it would result in formation of more $\ce{Pb}$ and  $\ce{Fe^2+}$
What I think:

1) Since both $\ce{Pb}$ and $\ce{Fe}$ have negitive standard electrode poential both will have tendency to form $\ce{Pb^2+}$ and $\ce{Fe^2+}$, then as a result $\ce{Pb^2+}$ and $\ce{Fe^2+}$  will increase in solution which is against the result given by book.

2) Another thing which comes to my mind is that since $\ce{Pb}$ has greater standard reduction potential than $\ce{Fe}$ so only  added Fe will convert into $\ce{Fe^2+}$ And the electrons released by it will reduce pre-existing  $\ce{Pb^2+}$  to $\ce{Pb}$ and hence the result as given by book

So my question is what exactly will happen if we conduct the following experiment.

Your second thought is the right one. As for the first, it implies the reaction of both metals with $\ce{H+}$ which releases $\ce{H2}$; for that to happen, the solution must contain some significant amount of $\ce{H+}$, i.e., be acidic, which is not the case.


Is it possible to have same mass and volume for a solvent and its solution Is there any solute-solvent interaction possible, in such a way that a solvent with mass X grams and volume Y litres, and a solute with mass A grams and volume B litres, to form a solution that has exactly the same mass and volume as that of the solvent (X grams and Y litres)?

As Klaus suggests above, you can't add mass to another mass without it increasing.  That would defy the first law of thermodynamics that energy cannot be destroyed since mass and energy are directly linked.

However, much of this depends on what degree of precision you're measuring to.  If you're working with a cubic metre of water and only measuring to the first two significant figures, mass would be 1000kg and volume 1000l, and you add 40kg of sodium chloride to make a 4% saline solution, you would still only have 1000kg and 1000l because you're not counting the third significant figure.

It's all about context ;-)


Silver "polish" with aluminum foil left a hard yellow coat I cleaned my silver teapot with the aluminum foil-baking soda-hot water method.
Usually this works well on my silver plated flatware.
However,  this time it left a hard yellowish coating.  I was only able to remove it by very strenuous manual polishing with a silver polish paste and when desperate, I even used toothpaste. The trade off was that it could scratch very finely,  but that was preferable to the yellow.
Now I'm afraid to use this method on unusual pieces.
--> How can I prevent this from happening again?
Are there kinds of silver and metal combinations that produce this reaction?
Is there an easier way to remove it?
Thanks!

The source of the "hard yellowish coating" may be Silver carbonate (Ag2CO3), as per Wikipedia:

Silver carbonate is yellow but typical samples are grayish due to the presence of elemental silver.

I suspect the teapot in question is "electroplated Britannia metal" per Wikipedia with a nickel alloy as the base metal, as to quote:

After the development of electroplating with silver in 1846, Britannia metal was widely used as the base metal for silver-plated household goods and cutlery.[5] The abbreviation EPBM on such items denotes "electroplated Britannia metal". Britannia metal was generally used as a cheaper alternative to electroplated nickel silver (EPNS) which is more durable.

The Baking Soda will interact with any soluble Ag ions to form the unstable Silver bicarbonate which breaks down to Ag2CO3 on warming.
The question is what is the source of Ag+?
My speculation is that the Al foil was completely coated given the surface area of the teapot, and the nickel-based alloy is notoriously cathodic (hence, the use of Nickel alloys in coins). So now, the Silver itself can be subject to an anodic attack in the presence of H+ (from amphoteric NaHCO3) and O2 (from air exposure), resulting in Silver ions.
This was a somewhat unique situation, hence the successful use of the Al/NaHCO3 process on the Silver-plated flatware.


Can Iron(III) oxide oxidize elemental iron in a nitrogen environment under room temperature without any catalysis? Will there be electron transfer to allow this process to happen:
$\ce{2 Fe(III) + Fe(0) -> 3 Fe(II)}$

Whether this qualifies as "without catalysis" is a matter of perspective, but it certainly does happen under ambient conditions.  One mechanism of atmospheric rusting involves the $\ce{Fe(III)}$ bearing rust reacting with fresh iron to form $\ce{Fe3O4}$, which then oxidizes in air to form more rust.  See Section 1.6 here.

We can capture the $\ce{Fe3O4}$ stage by merely placing the rusty material in water with a little electrolyte, such as rinse water after an acid pickling process, where air cannot easily penetrate.  The rust, being converted to $\ce{Fe3O4}$, turns black.  This emphasizes the fact that the iron reacts with the $\ce{Fe(III)}$ in the rust without oxygen, once the rust has first formed.

I've actually seen the $\ce{Fe3O4}$ formation happen on steel products during rinsing after pickling, giving black stains or stains bearing a black component.  To prevent it, we had to design and maintain the rinsing (and drying) processes to prevent the rusting.


What is the meaning of the "dot" notation in chemical formulas? For example for 4 given elements A,B,C,D, what's the meaning of $\text{AB} \cdot x \text{CD}$?

By writing $\ce{AB.xCD}$ chemists mean that there are CDs are found in the crystalline framework of AB. The most common example of this is water trapped inside the crystal structure of ionic compounds. (See water of crystallization in wikipedia)
An example that's often taught is $\ce{CuSO4.5H2O}$. See that 5 that's a representative of $x$? It means that water can be bound only with intermolecular interactions, but it exists in the structure in a stoichiometric ratio.

Source
Also, according to this article

In summary, when a dot is used to break a formula into subunits, it may signify ignorance of how the subunits are structurally related, as in our inorganic example; or it may correspond to actual significant structural subunits, as in our organic example; or it may represent the combining ratios of the binary starting materials required for the synthesis of the compound, as in our phase diagram example.

As a result, for example in direct addition reactions, where there is ignorance about the structural relation of the compounds, we use a dot.


How does hyperconjugation explain the stability of alkenes? The more the number of alpha hydrogens, the higher is the stability of alkenes, because the more the number of hyperconjugative structures. Why is this so? Has it got something to do with the electrons being easily delocalised?

As always in chemistry one phenomenon (experimental observation) can be explained with different models, that live in different "physical model worlds", and more often than not these models are not consistent among themselves and might be even bare of a physical reality, especially if they are analyzed in rigorous way, nevertheless might be very useful tools, that take the justification of existence out of the fact that they work in many cases.

That said,  especially the concept of hyperconjugation itself can be employed at different levels. A very primitive model, but one that works astoundingly well, is to assign a positive inductive (electron releasing effect) to alkyl chains that is increasing with the degree of substitution of the ipso atom. That leads to the idea that this (however small) charge that is pushed to the sp$^2$ carbon atom stabilizes its slight positive polarization (since the majority of electrons in the double bond is above and below the plane of the atoms engaged attached to the doubly bound atoms. The physical reality of this model if often drawn into doubt, which is good to know, but still its simple and powerful and thus there is no reason to abandon it.

At a higher level of model, using a MO picture the idea is that the (empty, anti-bonding) $\sigma^*$ orbitals that come inevitably with the C-C and C-H $\sigma$ bonds can act as kind of "acceptors" for the $\pi$-electrons of the double bonds if in the right orientation. It is important to note that in this idea to actual overlap that mediates the interaction is thought to be established with the part of $\sigma^*$ orbital that is opposing the bound atom. That donor-acceptor interaction ultimately leads to a stabilization of the system. Again the physical reality is also here at least disputed, but however less than in the former case. But still both have in common that the Substituent acts as the donor and that this causes stabilization.

So yes in this higher level of description of the hyperconjugation the reason for the stability is the increasing number of possibilities to form stabilizing overlap.


Reaction quotient and Gibbs free energy at the start of a reaction According to the equation 33 above, shouldn't be all the reactions be spontaneous initially? Because the reaction quotient Q, is zero at the start of the reaction and logarithm of Q makes it a very large negative value and ultimately we'll have ∆G to be a large, negative value which makes the reaction spontaneous?

Your interpretation of the equation is correct. $\Delta G^o$ gives you thermodynamic favorability of a reaction under standard conditions (Q=1) and even reactions with positive values of $\Delta G^o$ (unfavored under standard conditions) can be driven to proceed if the concentrations of the reactants and products are made extreme enough.

The best way to think about the process is via a curve of G vs reaction extent $\xi$, this is related to Q and ranges from 0 (pure reactants, Q = 0, #1 on the graph) to 1 (pure products, Q = ∞, # 2 on the graph):

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/23.4%3A_Free_Energy_and_the_Gibbs_Function

Any reaction will have an equilibrium point on this curve, corresponding to a most stable composition (# 3 on the graph). The value of the equilibrium constant (K) tells us the x coordinate of the most stable position (i.e. how much reactant vs product we have in the equilibrium mixture).

No matter what the value of K, it will always be in the range $0<K<∞$. If we start the process with pure reactants, $Q = 0$ and we are to the left of the equilibrium point. The system will shift to the right and some of the reactants will convert into products.

This is what you mean by

  ultimately we'll have ∆G to be a large, negative value which makes the reaction spontaneous?

you can calculate a negative $\Delta G{_r}$ for the reaction when you have no (or infinitesimal amounts of) products.

Many texts are moving away from the term "spontaneous" for this situation, because the system may not shift to the right (reactants convert to products) if there is a kinetic barrier to prevent this, so the forward reaction may not actually be observed. The term "thermodynamically favored" is often used instead, meaning having a negative $\Delta G{_r}$ under these conditions.

You can find lots more about these concepts at the following posts:

If change in free energy (G) is positive, how do those reactions still occur?

Calculating ΔG at the extremes of reaction extent


Output of the Sugar & Sulfuric Acid Reaction? When a reaction of sugar and sulfuric acid occurs, some form of carbon is produced. Is this carbon bonded as graphite? If not, than what is it bonded as?

$\ce{C6H12O6 ->[H2SO4] 6C + 6 H2O }$

The product is not pure elemental carbon, but instead an amorphous solid that consists of mostly carbon atoms, with at least 5-10% of the atoms being other elements.  For example, one academic paper from 2001 found that the "carbon" formed by sulfuric acid catalyzed dehydration of sugar had a $\ce{C}:\ce{H}$ molar ratio of 14:1.

Such materials are sometimes called hydrogenated amorphous carbon.


Synthesis of Haloperidol with 4-chlorobutanoyl chloride and piperidin-4-one I searched for a easy synthesis on Haloperidol.

Step 1 and 2 are taken from https://www.slideshare.net/SagarJoshi2/antipsychotics-33785528 (Slide 23). I just tried to work out a way to prepare 4-(4-chlorophenyl)piperidin-4-ol via Grignard.

Does the 4-chloro group has any negative influence on the reaction.

The reaction mechanism for the step in question simply involves initial formation of a Grignard reagent followed by nucleophilic attack into the ketone.

When forming Grignard reagents from halides, there is a general trend that the rates of formation are in the order  I > Br >> Cl, making it possible to get selectivity where multiple halides exist.


Can the following molecules be called meso compounds (cyclic structures which usually exist in non-planar forms)? In the planar form the following molecules have one plane of symmetry.However in
  the chair-form (the most stable form) for the first molecule and the non-planar form for the second molecule there is no plane of
  symmetry.Will the molecules still be called meso?

BTW I do not think the molecules ever attain completely planar form.

Just for quick reference I'm adding the non-planar forms of the two molecules.

These molecules can never be planar. There are no internal symmetry planes and therefore they are not meso.


The last element's atomic number I was just thinking what can be the last atomic number that can exist within the range of permissible radioactivity limit and considering all other factors in quantum physics and chemical factors.

Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as the innermost electrons would have to be moving above the speed of light. This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the nucleus is not a point particle, then there seems to be no real issue with arbitrarily high atomic numbers, although unusual effects start happening above $Z \approx 173$. These results may be overturned by an even deeper analysis with current quantum electrodynamics theory, or a new theory altogether.

As far as we can tell, however, we will never get anywhere close to such atomic numbers. Very heavy elements are extremely unstable with respect to radioactive decay into lighter elements. Our current method of producing superheavy elements is based on accelerating a certain isotope of an relatively light element and hitting a target made of an isotope of a much heavier element. This process is extremely inefficient, and it takes many months to produce significant amounts of material. For the heaviest elements, it takes years to detect even a handful of atoms. The very short lifetime of the heaviest targets and the very low collision efficiency between projectile and target mean that it will be extremely difficult to go much further than the current 118 elements. It is possible that we may find somewhat more stable superheavy isotopes in the islands of stability around $Z=114$ and $Z=126$, but the predicted most stable isotopes (which even then are not expect to last more than a few minutes) have such a huge amount of neutrons in their nuclei that we have no idea how to produce them; we may be condemned to merely skirt the shores of the islands of stability, while never climbing them.

EDIT: Note that the best calculation presented above is based on quantum electrodynamics alone, i.e. only electromagnetic forces are taken into account. Obviously, to predict how nuclei will behave (and therefore how many protons you can stuff into a nucleus before it's impossible to go any further), one needs detailed knowledge of the strong and weak nuclear forces. Unfortunately, the mathematical description of nuclear forces is still an incredibly tough problem in physics today, so no one can hope to provide a rigorous answer from that angle.

There must be some limit, as the residual nuclear forces are very short-ranged. At some point there will be so many protons and neutrons in the nucleus (and the resulting nucleus will have become so large) that the diametrically opposite parts of the nucleus won't be able to "detect" each other, as they are too far away. Each additional proton or neutron produces a weaker stabilization via the strong nuclear force. Meanwhile, the electrical repulsion between protons has infinite range, so every additional proton will contribute repulsively just the same. This is why heavier elements need higher and higher neutron-to-proton ratios to remain stable.

Thus, at some atomic number, possibly not much higher than our current record of $Z=118$, the electrical repulsion of the protons will always win against the strong nuclear attractions of protons and neutrons, no matter the configuration of the nucleus. Hence, all sufficiently heavy atomic nuclei will suffer spontaneously fission almost immediately after coming into existence, or all the valid reaction pathways to reach an element will require events which are so fantastically unlikely that if even all the nucleons in the entire observable Universe were being collided with each other since the Big Bang in an attempt to synthesize the heaviest element possible, we would statistically expect some sufficiently heavy atom to not have been produced even once.


What are the border definitions in the Ramachandran plot? I've heard some people mention that this or that program isn't "up to date" with respect to the borders used in the Ramachandran plots to classify φ/ψ as being in the most favorable/acceptable/disallowed regions.

How are the accepted regions defined?  What is the current accepted paper that covers this?

I wish I could answer your question outright, but this is outside my area of expertise. The best I can give you is a place to start. I expect any recent book on proteomics would cover this topic. For example Introduction to Proteomics: Principles and Applications (Methods of Biochemical Analysis) by Mishra and Blobel, Wiley 2010.

I found some promising papers with a little searching with SciFinder. Neither I nor my institution subscribe to Protein Science, but the abstract looks promising. I have skimmed the article in PNAS. Hopefully one of these articles has your answer.

Zhou, et al, "Revisiting the Ramachandran plot from a new angle", Protein Science 2011 1166-1171.

Porter and Rose, "Redrawing the Ramachandran plot after inclusion of hydrogen-bonding constraints", PNAS, *2011, 108, 1, 109-112


Calculation of entropy for an isothermal irreversible expansion How do I calculate the entropy change for an irreversible isothermal expansion against $p_\text{ext}=p_0$?

For a reversible isothermal expansion we know $\delta S_\text{uni}=0$ so $\delta S_\text{sys}=-\delta S_\text{surr}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$

However I can't figure out why $\delta S_\text{sys}=nR\ln\frac{V_\mathrm f}{V_\mathrm i}$ remains for irreversible expansion.

I also found out an equation for irreversible expansion
$\delta S_\text{uni}=q_\text{rev}/T-q_\text{irrev}/T$
Is it correct? If yes? Can you explain how do we get it?

Edit

Is this correct ?

$\delta S_\text{uni}=\delta S_\text{sys}+\delta_\text{surr}$

$\delta S_\text{sys}=\frac{q_\text{rev}}{T}$

$\delta S_\text{surr}=-\frac{q_\text{irrev}}{T}=p_\text{ext}\delta V/T$

Here are the steps to determining the change in entropy for an irreversible process on a closed system:

Use the first law of thermodynamics to determine the final thermodynamic equilibrium state of the system for the irreversible path.
Totally forget about the irreversible path.  It is of no further use.  Focus only on the initial equilibrium state of the system and the final equilibrium state.
Devise a reversible path for the system that takes it from the initial equilibrium state to the final equilibrium state.  This reversible path does not have to bear any resemblance whatsoever to the real irreversible path, other than it must pass through the same initial and final end points.  (Entropy is a function of state).
Calculate the integral of dq/T for the reversible path that you have devised.  This will give you the change in entropy of the system for the irreversible path as well as for the reversible path.

For your problem, this procedure will give you the equation that you have written.  Note that, for the irreversible path, the system temperature matches the surroundings temperature (isothermal) only at the interface with the surroundings, but not throughout the interior of the system.  For an irreversible expansion, this is what we define as "isothermal."


Fluorochrome (fluorescein) emission decay when linked to macromolecule I measure the intensity and polarization of light emitted from a blood sample that contains excited fluorescein (experiment known as fluorescence polarization assay). The excitation is done, as usual, with vertically polarized light.
In case the sample has certain macromolecules of interest (an antibody, in this case), the fluorescein links to the much-larger macromolecules, yielding a higher polarization factor, giving a positive- or "sick"- result. Alternatively, if there is no antibody, the excited fluorescein randomly rotates and then emits light, and the polarization factor is smaller giving a negative-or "healthy"- result.
The question:
I find that the light intensity (in both vertical and perpendicular polarizations) decreases for the positive case. Why is that so? Could it be that a non-radiative process like FRET acts stronger for fluorescein linked to macromolecules than for free fluorescein?
EDIT 1: I have used two different intensity detection systems, PMTs and SiPMs, and got similar behaviours.

There should be no FRET for the free fluorescein unless you have specifically added an acceptor and in any case this would reduce the fluorescein emission.  When bound to the protein fluorphores are in a different environment (e.g. pH  change, hydrophobic pocket, tight binding, polar/non-polar nearby residues etc.)  and this can affect their fluorescence yield and lifetime up or down, ANS (anilino naphthalene sulphonate) is a good example.  The same type of thing is probably happening in your case, you know the motion is restricted as the polarisation is greater so there has to be some interaction. Fluorescein emits at a relatively long wavelength so aromatic protein residues cannot act via FRET as they absorb in the UV. You would need a chromophore (such as heam/chlorin/porphyrin) that absorbs in the green to red region to be a FRET acceptor from the fluorescein and so reduce its intensity.


Are bulkier molecules more likely to react? I have heard a theory from someone and just wanted to confirm if its true or not. Is it true that more bulkier an organic compound is, more likely it is to react? So does this mean that higher the molecular weight of an organic compound, more will be the probability of it to react?

The relative rates are going to depend not on molecular weight so much but the structure of the molecule.

The specific reaction is also very important.

Nucleophilic substitution reactions (some of the first reactions most students learn about in an organic chemistry course).

Let us compare 1-chlorobutane and 2-chloro-2-methylpropane, both $\ce{C4H9Cl}$. However, the 1-chlorobutane is a primary alkyl halide (not very bulky) and 2-chloro-2-methylpropane is a tertiary alkyl halide (bulky).

The bulkier compound reacts faster in a solvolysis reaction:

$$\ce{(CH3)3CCl ->[\ce{H2O}] (CH3)3COH}\ \ \mathrm{fast}$$
$$\ce{CH3CH2CH2CH2Cl ->[\ce{H2O}] CH3CH2CH2CH2OH}\ \ \mathrm{slow\approx no\ reaction}$$

The reason that the tertiary halide reacts faster in this reaction is that its mechanism involves the formation of a carbocation intermediate and the extra carbon groups at the tertiary center provide more inductive stabilization to that carbocaion than the single alkyl group attached to a primary position. In general, in reactions that generate carbocation intermediates, tertiary substrates will react faster.

However, in many reactions bulkier molecules have steric interactions that negatively effect the rate. In a nucleophilic displacement with $\ce{NaI}$ in acetone, 1-chlorobutane will react faster.

$$\ce{(CH3)2CCl ->[\ce{NaI}][\mathrm{acetone}](CH3)3CI}\ \ \mathrm{slow\approx no\ reaction}$$
$$\ce{CH3CH2CH2CH2Cl ->[\ce{NaI}][\mathrm{acetone}]CH3CH2CH2CH2I}\ \ \mathrm{fast}$$


Thermal decomposition of K2CO3 $\ce{K2CO3}$ doesn't boil. It thermally decomposes at $\pu{1200 ^\circ C}$. However, I find conflicting information online about this. Roughly what I have found say three inconsistent things:

It doesn't decompose
At $\pu{1200 ^\circ C}$, it decomposes into $\ce{K + O + CO}$
(Chemiday)
At $\pu{1200 ^\circ C}$ it decomposes into $\ce{K2O + CO2}$ (allreaction)

What actually happens when $\ce{K2CO3}$ thermally decomposes? Most sources agree with (3). But how can that be? Wouldn't it just decompose some more? $\ce{K2O}$ decomposes at $\pu{300 ^\circ C}$ into ($\ce{K2O2 + K}$) and $\ce{K2O2}$ decomposes back into ($\ce{K2O + O2}$) at $\pu{500 ^\circ C}$.
What is going on here? Is $\ce{K2O2}$ really decomposing back again to $\ce{K2O}$? If so, then at $\pu{500 ^\circ C}$ do both of these decompositions occur, cyclically pumping out $\ce{O2}$ and $\ce{K}$?
I read elsewhere that at temperatures over $\pu{500 ^\circ C}$, $\ce{K2O2}$ decomposes into ($\ce{2K + O2}$). This would resolve the conflict. But I find the source is unreliable. What is right?

This question serves as a prime example of how much fake and scientifically incorrect information, or more famously, fake news is there on the web. This is why ordinary web should not be anyone's first priority to search about scientific topics. There is a famous saying that "paper never refused an ink" which applies to the web as well in the sense that anyone can write anything. Use Google Scholar, which only searches research level articles (relatively more authentic).
With this type of search one can arrive at Thermal stability of potassium carbonate near its melting point. This paper rules out the possibilities of (1) and (2). The authors note that there is significant vapor pressure of carbon dioxide near the melting point, and suggest $\ce{K2O}$ formation. They also advise the readers to consult the following as the most comprehensive study on alkali carbonates. I will let you try Google Scholar for this one.
References

A. Reisman, Reactions of the Group VB pentoxides with alkali oxides and carbonates. IX. A DTA study of alkali metal carbonates, J. Am. Chem. Soc. 80 (1958) 3558-3561.


Does the amount of acid change when I add water to vinegar? I have a question that asks me to describe why adding distilled water to my vinegar sample would not affect my calculated Molarity (determining amount of acid in the vinegar be titration with a base).

At first thought, it wouldn't, because the same amount of acid is still there, we'll still have the same amount of base reacting. So you'll still get the same amount by measuring how much base was needed.

But that got me thinking. There is a lot of diluting acids with water, but isn't water amphiprotic? Why doesn't it react with the acid, or react with it enough to matter at least? What information am I missing here?

First off, your thinking about the titration volume of a dilute solution vs the concentrate from which it was made is correct.

Why we don't need to worry about water acting as an acid or a base in this context:  Consider the reaction that occurs between a weak acid, HA, and water (a very weak base.)
$\ce{HA + H2O <-> A- + H3O+}$

The products of the reaction are a weak base and a strong acid (hydronium, the strongest acid that can exist in aqueous solution.)  If I add a base to this mixture, it doesn't matter whether that base reacts with an undissociated HA molecule or with a hydronium ion, in either case it has removed a particle of one acidic species from the solution, leaving behind A-, the conjugate base of HA.  The total number of such events that has to occur before the titration is done is a fixed value, regardless of how many of the events involved hydronium or HA.


How to distinguish mass concentration and density? I am writing a proof where I have to mention both with density and of a material and mass concentration (mass/volume) of a material.

I am using the letter $\rho$ to represent density, and I didn't want to abuse the same symbol for concentration because it will make the whole proof much harder to follow. I would also prefer to avoid subscripts as the only distinguishable element.

What other symbol or syntax should I use to distinguish density from mass concentration of a material?

According to IUPAC's Green Book [1], both $γ$ and $ρ$ can be employed to denote mass concentration. Fragment of a table from the section 2.10 GENERAL CHEMISTRY [1, p. 48]:

  $$
\begin{array}{llll}
\hline
\text{Name} & \text{Symbol} & \text{Definition} & \text{SI unit} \\
\hline
\ldots \\
\text{mass concentration, (mass density)} & γ, ρ & γ_\ce{B} = m_\ce{B}/V & \pu{kg m-3} \\
\ldots \\
\hline
\end{array}
$$

  […]  In polymer science the word “concentration” and the symbol $c$ is normally used for mass concentration.

Also, from the SI Brochure, [2, p. 139]:

  $$
\begin{array}{lll}
\hline
\text{Derived quantity} & \text{Typical symbol} & \text{Derived unit expressed} \\
  & \text{of quantity} & \text{in terms of base units} \\
\hline
\ldots \\
\text{mass concentration} & γ, ρ & \pu{kg m-3} \\
\ldots \\
\hline
\end{array}
$$

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 9th ed.; Bureau international des poids et mesures: Sèvres, 2019.


Nitroxyl anion electron structure Is the $\ce{NO-}$ a radical species?  If so, can someone please explain why because $\ce{NO}$ has an odd number of electrons (11), so adding an additional electron would give 12 valence electrons and a formal charge of -1 on the nitrogen. Is my general chemistry failing me or?

The nitroxyl anion is isoelectronic to oxygen. It has pretty much the same orbitals as oxygen does and it also features the same number of electrons. The principal difference is the orbital shape: since oxygen and nitrogen are two different atoms with a different number of protons, there is no plane of symmetry bisecting the $\ce{N=O}$ bond axis (in group theory terms: the point group is $C_{\infty \mathrm v}$ rather than $D_{\infty \mathrm{h}}$). This difference means that bonding orbitals will be more oxygen-centred while antibonding ones will be more nitrogen-centred.

However, if you take a look at the molecular orbital scheme, the key orbitals are identical to those in oxygen. The highest occupied molecular orbitals are two symmetry-equivalent π* orbitals and the unoccupied σ* orbital is higher in energy. Since the two π* orbitals are symmetry equivalent, there is no way the molecule can pair the spins; the highest energy electrons will be parallel and occupy one of the π* orbitals each.

Considering that we have exactly the same situation as oxygen in the key orbitals, this also means that $\ce{NO-}$ features exactly the same states. Most notably, the ground state is a triplet $\ce{^3NO-}$ state, while a higher-energy singlet $\ce{^1NO-}$ state also exists — entirely analogously to the singlet $\ce{^1O2}$ and triplet $\ce{^3O2}$ states of dioxygen.

When calling the nitroxyl anion a radical, one should remember that it is, like the oxygen molecule, more accurately described as a diradical. Its properties are distinct from those of monoradicals such as $\ce{NO}$, which feature a doublet state.


What is the difference between statistical mechanics and quantum mechanics in terms of accuracy? It is my impression that everything relevant to 'quantum' is the truth and others is just a simplification. Is this the case in those two kinds of mechanics?

Quantum mechanics is about the physics of very small things, molecules and smaller.

Classical mechanics is about macroscopic things. Quantum mechanics covers the whole of classical mechanics as well, but in the macroscopic limit both become equivalent. For example discretized energy states become so close, that you can thing of them as a continuum of states.

Statistical mechanics can be employed to consider many, quantum or classical, like systems and how they evolve. One application is to get thermodynamic quantities from quantum mechanical energy states, by considering how these states are statistically populated.


How to interpret TD-DFT results for finding λmax and amplitude I am trying to understand how to interpret the TDDFT results ( I know there is another post for this but it does not answer my question ).   Here is part of the result from GAMESS:

I downloaded Chemissian and this is its result for this state:

I have two related questions:

1-How did it calculate this 1.41 from the amplitude of this state ?
2-Does it have anything to do with λmax ?

You are looking at the wrong numbers.

You should look for excitation  that has non-zero oscillator strength. In the text file "State # 1,2 .." and in Chemissian on the left, list with columns "# Energy Strength" The transitions with zero or very small oscillator strength has a very low intensity in the spectrum, so you don't see them.
In your case the relevant excitation are #2 (Energy: 6.691 eV, Strength: 0.027) and #4 (Energy: 8.353 eV, Strength: 0.042), and you can see them as small peaks on the graph down.

The "amplitude .. 1.41" has nothing to do with the oscillator strength. It tells you which orbital contributions are the most important ones for the given excitation.


Does my data show quadratic regression in a calibration curve? This data was taken for my science fair project, and I would like to know the relationship between spectrophotometer absorbance and cell density (taken manually) of algae after a toxicity test. The different levels of toxicity are IC 0 (no chemical), IC 25, IC 50, IC 100 (the most chemical). The IC value is basically the estimated about of the chemical needed to destroy a certain amount of the population. I took the manual density and absorbancy of four samples of different levels of toxicity.

IC value/Cell densities(cells/ml)/absorbancy:

IC 100/28000/0.408

IC 50/42600/0.302

IC 25/70000/0.01

IC 0/138000/0.79

Based off of these points, I would like to make a calibration curve. But when I do this, the highest r^2 is quadratic. Is this an acceptable calibration curve for finding other densities based off of known absorbances? Or would it have to be linear with a very low r^2?

As you have correctly assumed, the relationship between absorbance and absorbant concentration is linear (law of Lambert-Beer):
$$ A = \varepsilon \, l \times c$$

This should give you a straight line through (0,0) and your data.

Now, for a graphical analysis of data, let's not look at the numbers themselves but on what kind of a graph it gives us. So after 2 minutes with gnuplot I get something like this:

Now, there are two things that I note here:

The absorbance of the $\mathrm{IC_{25}}$ value is very very low (nearly 0) which indicates some wrong measurement. Maybe the cells weren't in the pathway of the light beam, only the medium?
You have four data points.

My advice is the following:

Double check your measurements (the absorbance and the cell density). It would be even better if you did several measurements for the different inhibitor concentrations and then calculate the mean (and standard deviation if you so desire).
Generate more data. It is very wonky to try and justify a linear regression through these 4 data points (although mathematically allowed). However, if all intermittent points are equally chaotically distributed along a straight line the linear relationship still holds (with a fairly high $R^2$). What this does to the interpretation of the data is a completely different story.


Is benzene a polymer of ethyne? In one of my classes, my teacher told me that benzene can be formed by cyclic polymerization of ethyne. But I think that benzene isn't a polymer because only 3 ethyne molecules are combining to form benzene whereas, thousands of monomers are combined to form a polymer in a polymerization reaction. So, after all, can benzene be called a polymer of ethyne?

The scientist who coined the term polymer, Jöns Jacob Berzelius, used the word to refer to different substances which had the same empirical formula. In this sense, benzene is a polymer of ethyne (acetylene), because benzene $\ce{C6H6}$ and ethyne $\ce{C2H2}$ have the same empirical formula, namely $\ce{CH}$. In other words, all substances with molecular formula $\ce{(CH)_x}$ were polymers of the smallest unit in the family (in this case, $\ce{C2H2}$, because $\ce{CH}$ molecules are only known as extremely fleeting species on Earth, though they are abundant in space).
However, that was back in the 1830s, when we had no clue about atomic and molecular structure (in 1808, one of the eminent chemists of the time, John Dalton, defended $\ce{HO}$ as the formula for water). The meaning of the term has evolved over time to accommodate our increase in knowledge. The most current IUPAC recommendations on polymer nomenclature (Compendium of Polymer Terminology and Nomenclature - IUPAC Recommendations 2008) states the definition of polymer as:

A polymer is a substance composed of macromolecules.

And, fortunately, they also define a macromolecule as:

Molecule of high relative molecular mass, the structure of which essentially comprises the multiple  repetition  of  units  derived,  actually  or  conceptually,  from  molecules  of  low  relative molecular mass.

Note  1:  In  many  cases,  especially  for  synthetic  polymers,  a  molecule  can  be  regarded  as  having  a  high  relative  molecular  mass  if  the  addition  or  removal  of  one  or  a  few  of  the  units has a negligible effect on the molecular properties. This statement fails in the case of certain properties of macromolecules which may be critically dependent on fine details of the molecular structure, e.g., the enzymatic properties of polypeptides.

Note  2:  If  a  part  or  the  whole  of  the  molecule  has  a  high  relative  molecular  mass  and  essentially  comprises  the  multiple  repetition  of  units  derived,  actually  or  conceptually,  from   molecules   of   low   relative   molecular   mass,   it   may   be   described   as   either   macromolecular or polymeric, or by polymer used adjectivally.

Note  3:  In  most  cases,  the  polymer  can  actually  be  made  by  direct  polymerization of  its  parent  monomer  but  in  other  cases,  e.g.,  poly(vinyl  alcohol),  the  description  ‘conceptual’  denotes that an indirect route is used because the nominal monomer does not exist.

As you can see, there is quite a bit of detailing to try to keep things consistent. Observe that the definition speaks of high relative molecular mass, and this point is expanded upon in Note 1. With these considerations, benzene is not a polymer - it is certainly a very small molecule, and adding or removing a portion of it (presumably cyclooctatetraene $\ce{C8H8}$ and cyclobutadiene $\ce{C4H4}$, respectively) would result in a substance with drastically different properties.
In some sense, Berzelius wasn't really wrong to call benzene a polymer, because for all he knew back then, benzene could well have the formula $\ce{C20000H20000}$ (actually, it was not widely believed that "large" molecules could exist at all, with some resistance to the idea existing as late as the 1920s-1930s). He did the best with what knowledge was available to him. But now we can do better, and it just is more convenient and self-consistent to not label benzene as a polymer.
This isn't to say that, however, that benzene and ethyne are completely chemically unrelated. There is in fact a close relationship, in that ethyne can efficiently undergo a [2+2+2] cycloaddition reaction to form benzene. This is not exploited industrially, however, as there are much cheaper (and safer) ways of making it.


What will the final product of this synthetic sequence be? What is the major product formed in the following reaction?

NBS should do allylic bromination but where will it attack in the larger ring or smaller one ?

Potassium carbonate is not going to give you hydrolysis of a tertiary bromide. It will however close the halohydrin to form an epoxide (1). This intermediate undergoes Lewis acid catalyzed rearrangement to form the spiro compound that is choice (D).

Note that 1 or the carbocation could undergo certain elimination reactions to form (B) and (C). These are probably minor products.


Is there a metal that melts and solidifies quickly? Is there any kind of metal that melts very quickly with very low heat or temperature, and at the same time, becomes hard very quickly?

The melting points of some eutectic mixtures are tabulated in an article by French.1 Some options for low-melting point metals/mixtures are:

Gallium-indium eutectic, mp 15.7 °C (Sigma-Aldrich catalogue)
Galinstan, mp –19 °C (Wikipedia article)
Various gallium-based Indalloy® alloys, mp between 7.6 and 29.78 °C (corporate page)
A general search for "fusible alloys" brings up several sources with many choices:

Wikipedia page
Rotometals sales page
Belmont sales page
Bolton Metal sales page

Reference

French, S. J. Melting Points of Eutectics. Ind. Eng. Chem. 1936, 28 (1), 111–113. DOI: 10.1021/ie50313a026.


Enthalpy change for a van der waal gas One mole of Van der Waal gas is compressed from $\pu {20 dm^3}$ to $\pu{10 dm^3}$ at $\pu{300 K}$. Given $a = \pu{3.60 dm^6 mol^-2}$ and $b = \pu{0.44 dm^3 mol^-1}$. Find $\Delta H$.

$\Delta H = -\mu C_p \Delta P + C_p \Delta T$

Since for VDW gas $ \displaystyle \mu C_p = {2a\over RT}- b$.

The process took place at $\pu {300 K}$ which means $\Delta T = 0$.

$\therefore  \Delta H = -\mu C_p \Delta P + C_p \Delta T = \left(-{2a \over RT} + b\right)\Delta P = \pu{0.4399 kJ/mol}$.

But the answer is $\pu {-30.5 J/mol}$

I think my mistake is taking $\Delta T = 0 $ but if that were the case what should I take as $\Delta T$, I don't even know the process.

Is the author's answer correct ?

The equation for $\Delta H$ for a van der Waals gas should be:
$$\Delta H=C_p^{IG}\Delta T-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+b\left(\frac{RT_2}{V_2-b}-\frac{RT_1}{V_1-b}\right)$$where $C_p^{IG}$ is the heat capacity of the gas in the Ideal Gas (IG) region (low pressures and high specific volumes).  For constant temperature ($T_1=T_2=T$), this equation reduces to:$$\Delta H=-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+bRT\left(\frac{1}{V_2-b}-\frac{1}{V_1-b}\right)$$

Explanation:
In the link https://physics.stackexchange.com/questions/321968/internal-work-done-by-a-gas/322170?noredirect=1#comment725750_322170, I showed that, for a van der Waals gas, the internal energy U is given by:  $$U(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}$$
where $U(T_{ref},\infty)$ refers to the internal energy in the ideal gas reference state $T=T_{ref}$ and $V\rightarrow \infty$ (i.e., a reference state in which the gas exhibits ideal gas behavior).

The enthalpy H of the gas is equal to U + PV:
$$H(T,V)=\left[U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}\right]+\left[-\frac{a}{V^2}+\frac{RT}{V-b}\right]V$$
Combining the terms on the right hand side gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTV}{V-b}$$
But, $$\frac{V}{V-b}=\left[1+\frac{b}{V-b}\right]$$
So, substituting into the equation for H gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}+R(T-T_{ref})+RT_{ref}$$
But, $$U(T_{ref},\infty)+RT_{ref}=H(T_{ref},V)$$and$$\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}+R(T-T_{ref})=\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}$$where $C_p^{IG}(T)$ is the ideal gas heat capacity at constant pressure.  So, finally, we obtain:$$H(T,V)=H(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}$$


Why and how is rust forming on moon? note: @Mithron's proposed duplicate Why can't rust form without water? does not have anything about the conditions on the Moon, so no, it's not a duplicate.

Several popular news articles mention that "rust" has been found on the Moon, and this is surprising because rust requires the presence of both oxygen and liquid water.

nasaspaceflight.com: Rust on the Moon. How is that possible without oxygen and liquid water? "But how can rust form far from water ice deposits on a barren oasis devoid of oxygen?"
Axios: Researchers find rust on the Moon "Scientists were surprised by the findings because rust requires oxygen and water to form on Earth."

When I open a plastic-wrapped cake or muffin there is often an "O-buster" packet with finely powdered iron. When I open the packet (chemistry is more interesting than cake) and watch the powder, indeed it rusts. Of course air has plenty of both oxygen and water, so this is no surprise.
Question: But why and how is rust forming on moon? Why does it need both oxygen and water there, and more importantly, how is it getting them?

UPDATE:
Today I got a chance to read the abstract the original paper which made news everywhere that moon has rust on the side which faces the Earth.
The authors propose that upper atmosphere oxygen from the Earth reaches the moon. In their words
"Oxygen delivered from Earth’s upper atmosphere could be the major oxidant that forms lunar hematite. Hematite at craters of different ages may have preserved the oxygen isotopes of Earth’s atmosphere in the past billions of years. Future oxygen isotope measurements can test our hypothesis and may help reveal the evolution of Earth’s atmosphere."
Widespread hematite at high latitudes of the Moon in Science Advances (open access

Let us split your query into three parts:
a) Will iron oxidize without water and form oxides?
Iron would happily burn in oxygen to form iron oxides. But we cannot call it iron rust. So iron can oxidize without moisture. In fact, iron powder may spontaneously burn in air.
b) Will iron form rust without water?
Rust is more of a semantic issue.  Iron rust may not have constant composition. As per the dictionary definition (OED) "

A red, orange, or yellowish-brown substance which forms progressively
as a flaking, permeable coating on the surface of iron and its alloys
as a result of oxidation, esp. through exposure to air and moisture.

Therefore, rust is an product of ambient environmental factors that affect iron at (approx.) typical room temperature. Rusting in sea water might different than rusting on a mountain. Acidic gases such as carbon dioxide, sulfur dioxide, nitrogen oxides certainly accelerate this process. Iron rust can be green and it can be common brown version. Since, chemically rust is a hydrated oxide of iron plus its hydroxide, water's presence is a must. More importantly water is a necessary ingredient which works as mediator of the electrochemical cell that forms between oxygen and iron surface. If you search Google Scholar you would find tons of articles on the electrochemical mechanism of iron rusting.
Coming to the more interesting question of rust on the moon:

nasaspaceflight.com: Rust on the Moon. How is that possible without
oxygen and liquid water? "But how can rust form far from water ice
deposits on a barren oasis devoid of oxygen?" Axios: Researchers find
rust on the Moon "Scientists were surprised by the findings because
rust requires oxygen and water to form on Earth."

Since you would know more astronomy than I do, wasn't moon a part of the Earth? I recall this from a documentary. I do remember seeing layers of "rust" in moutainous paths, so the oceans had a lot of iron which settled as a layer in mountains. If moon were a part of Earth, why is that surprising?
a) The question is what was the age of this rust?
Secondly, Mars is also "rusty", its surface is highly oxidizing with a huge amount of perchlorate in the soil (there is a long story how that was discovered by the folly of analytical chemists). Who knows what is the moon soil like. What about the shower of radiation on the moon. That may promote oxidation as well.


Is hydrogen peroxide safe for cleaning porcelain? From my understanding hydrogen peroxide only reacts with bacteria in a way that it should burn them more or less.

If that is true, can I then safely use hydrogen peroxide to clean my bathroom sink which are made of porcelain?

While hydrogen peroxide is not normally used for cleaning surfaces, there is nothing in normal glazed porcelain that it would damage. Porcelain is about as unreactive to most things as glass, which is unreactive enough to be used for the bottles that store hydrogen peroxide.

But you should probably be careful with it as it can be dangerous if the solution is a strong one. Normally available hydrogen peroxide should not be as strong as the solutions used in the lab, though, which is often a 30% solution in water and should only be handled if you know what you are doing (see the safety data sheet). Consumer hydrogen peroxide is usually a 3-6% solution and isn't any more dangerous that hypochlorite bleach (and is a safer way to bleach hair).

Standard household (hypochlorite) bleach is usually better for cleaning, though.


Planes of symmetry of a molecule Are we limited to use only xy, yz and zx planes to check for planes of symmetry of a molecule? Because I see a plane of symmetry in  the following molecule through the two shown H atoms or OH groups. I am aware of the 3D structure shown by this projection. And since this molecule is Chiral, and has enantiomeric mirror image, this existence of a plane of symmetry appears contradictory.

I don't see planes of symmetry through H atoms or OH groups. Actually, these planes are not present. It is easy to check this by looking at the position of the CH3 group after reflection in one of these planes: you find CH3 at a completely different position (none of the original positions is mapped and hence you can distinguish the starting configuration from the final one).


Arrhenius equation and rate constant I am a high-school student and I am currently taking a course on chemical kinetics. Towards, what seems, end of the course at highschool level, the Arrhenius equation came in. Now, I'm not sure if what I'm presented is just a general form of Arrhenius equation and there's more to it, or this is it. Well, anyways, one of the things I am being told to learn is the plot between graph of rate constant $k$ vs temperature $T$. $$k=A\mathrm e^{-E_\mathrm a/(RT)}$$
The graph is sort of looks like with increasing $T$, the value of $k$ is skyrocketing (kinda like $y=x^2$ in first quadrant) but if I'm not mistaken, the rate constant $k$ should equal $A$ as temperature tends to infinity. Now I believe there must be some other factor in the equation that is also a function of temperature and is causing the graph to be the way it is. Can someone help me understand the same?
(In case you didn't get, because I am unable to attach an image, the looks like $y=x^2$ instead of $x=y^2$, which basically interrupts my intuition of Temperature on $x$ axis ever approaching infinity. $k$ on $y$ and $T$ on $x$ axis.)

This is how the graph between the rate constant and temperature looks like:

This might not seem intuitive looking at the Arrhenius equation which says:
$$k=Ae^{\frac{-E_a}{RT}}$$
After using some mathematical prowess someone might say that as $T\rightarrow \infty, ~k\rightarrow A$ and so the graph should look something more like:

Which on first glance seems fair but doesn't match the aforementioned curve.
To understand what is causing the problem we must check everything that might change during a change in temperature.
Upon examination we can find that only two things might cause this discrepancy, $E_a$ and the pre-exponential "constant", $A$.
The activation energy of a system is also independent of the temperature and thus the only possible problem point left is the pre-exponential "constant". Turns out that this constant isn't a constant in true sense, which isn't that disturbing of an idea if you know what that constant signifies.
The pre-exponential constant shows how many effective collisions can take place in a reacting mixture. It is defined as,
$$A=pZ$$
Where,
$A$ is the pre-exponential constant.
$p$ is the steric factor, which takes into account the steric effects during an effective collision.
$Z$ is the collision frequency.
Now as it might seem intuitive, the collision frequency is related to temperature and increases with increase in temperature according to the following relation(using equations for collisions between similar molecules to prevent unnecessary complexity):
$$Z=4nr\sqrt{\frac{\pi RT}{M}}$$
Where,
$Z$ is collision frequency.
$n$ is molecules per unit volume.
$r$ is the radius of molecule.
$R$ is the universal gas constant.
$T$ is temperature.
$M$ is the molar mass of the gas.
Thus the collision frequency and corresponding the pre-exponential constant tend to infinity when $T\rightarrow \infty$, which matches with our observation of $k\rightarrow \infty$ when $T\rightarrow \infty$.
Reference:
The Effect of Temperature on Reaction Rates https://chem.libretexts.org/@go/page/24271 (accessed May 14, 2021).


Purify gas produced after reactions with sulfuric acid Say, I preheat 50ml of 98% $\ce{H2SO4}$ and proceed with adding 25ml of 85% $\ce{HCOOH}$.
The result of which is:

$$\ce{H2SO4 + HCOOH -> CO + H2O}$$

Sulfuric acid will dehydrate formic acid producing $\ce{CO}$, however, gas could still be contaminated with vapor. What is the correct way of removing these impurities?

Pass the gas mixture through Anhydrous Calcium Chloride or Silica Gel. Calcium Chloride is quite hygroscopic and absorbs water vapors from the gas mixture. Alternatively you can use Silica gel, which adsorbs water due to its nature as a good adsorbent.

Edit: To get rid of the acids present in the vapor, bubble the mixture through a concentrated base solution. Acids would react with the base, while $\ce{CO}$ being a neutral oxide, simply doesn't react and can be collected.


Why is '-ethane' in 'methane'? Is there a chemical or historical significance in the fact that 'ethane' is just 'methane' without the 'm'?

meth vs eth

[OP] Why is '-ethane' in 'methane'?

This is a coincidence.
Methyl is

ultimately from Greek methy "wine" + hylē "wood.

Source: https://www.etymonline.com.
The terminology was created by Dumas and Péligot in 1834 to distinguish wood alcohol from ethanol, and was published first in the French language:

Source: https://gallica.bnf.fr/ark:/12148/bpt6k6569005x/f15.item
Ethyl is from

Greek aithēr "upper air; bright, purer air; the sky" (opposed to aēr "the lower air"), from aithein "to burn, shine," from PIE *aidh- "to burn"

Source: https://www.etymonline.com.
Liebig had argued in 1834 that what now is known to be $\ce{-C2H5}$ is a "radical", i.e. a reoccuring group in molecules such as diethyl ether, ethanol and ethyl ethanoate, and gave it the name ethyl.

Source: https://babel.hathitrust.org/cgi/pt?id=uva.x002457887&view=1up&seq=30
Using the ending -yl for both the methyl and the ethyl group is a suggestion by Berzelius from 1835 (the correct atomic weight of carbon was not know yet, so the chemical formulas are off by a factor of two):

Source: https://books.google.com/books?id=1DM1AAAAcAAJ&pg=PA376#v=onepage&q&f=false, see Wikipedia entry on Ethyl group.
methane vs ethane
That these both end in -ane is not a coincidence - all alkanes end in -ane. As stated in M. Farooq's answer, this nomenclature was introduced by A. W. Hofmann in 1865.
Acknowledgements
Maurice's answer inspired me to find the first work by Dumas that led to the methyl terminology. I have edited my answer to incorporate information from the excellent answers by M. Farooq and Jan and the insightful comments. In trying to figure out the sequence of events in the 1830s, I made use of this account by Frederick E. Ziegler.
References
Dumas, J.-B, Mémoire sur l'esprit de bois et sur les divers composés éthérés qui en proviennent. lu à l'Académie des Sciences les 27 Octobre et 3 novembre 1834, Paris, 1834
Jean-Baptiste Dumas et Eugène Péligot, Mémoire sur l’Esprit de Bois et sur les divers Composés Ethérés qui en proviennent, Annales de chimie et de physique, 58 (1835) p. 5-74
Justus Liebig (1834) "Ueber die Constitution des Aethers und seiner Verbindungen" (On the composition of ethers and their compounds), Annalen der Pharmacie, 9 : 1–39
Jacob Berzelius, Årsberättelsen om framsteg i fysik och kemi [Annual report on progress in physics and chemistry] (Stockholm, Sweden: P.A. Norstedt & Söner, 1835), p. 376
A. W. Hofmann 1866, in Monatsbericht der Königl. Preuss. Akad. der Wissenschaften zu Berlin, 1865 653


Adiabatic expansion of an ideal gas A. For certain ideal gas Cp=8.58 cal/mol-deg. What will be the final temperature and volume when two moles of the gas at 20 degrees celsius and 15 atmosphere are allowed to expand adiabatically and reversibly to 5 atm pressure? Find W, ∆U, and ∆H.
B. Suppose that the expansion takes place adiabatically against a constant pressure of five atm, what will be the final volume and temperature of the given gas?
I already solved part A but I have no clue for part B. Especially on how to get the final temperature and pressure.

In the following answer I am using the Physics convention of work done. Where, $W= \int PdV$ and the first law of thermodynamics is as follows: $q = \Delta U + W$. If you want answers in the chemistry conventions than just reverse the sign of the work done. Internal energy change and heat change will be same in both conventions.
As you said you have done the first part, I will continue with the second.
First of all, $C_V=C_P - R = 8.58- 2 = 6.58$
As the process is adiabatic,
$$q=0\\
\Delta U=-W\\
nC_V(T_2-T_1)=-nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
T_2= 247.67 K = -25. 53 ^\circ C\\\\$$
$\\Now, \\\\$
$$W= P_{ext} \Delta V\\
W= nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
W= 599.2 cal\\\\$$
$$\Delta U= -W \\
\Delta U= -599. 2 cal\\\\$$
$$\Delta H= \Delta U + \Delta (PV) \\
\Delta H= \Delta U + \Delta (nRT)\\
\Delta H= \Delta U + nR\Delta T\\
\Delta H= (-599. 2 -181. 32) cal\\
\Delta H= -780. 52 cal\\\\$$