The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$

The additional pressure compresses the liquid and thereby increases its chemical potential, resulting in an increased vapor pressure. However, since $V_{m,~l}$ is relatively small,  the effect on the vapor pressure is also small except at very high pressures. The effect of air at atmospheric pressure and $\pu{298 K}$ on the vapour pressure of water ($\pu{3.2 kPa}$ in the absence of inert gas) is negligible, an increase of $\pu{2.3 Pa}$.

So $P_A \neq P_B$, but they become nearly identical when $PV_{m,l}\ll RT$.


Can rate constant for forward reaction be equal to rate constant for backward reaction?

When forward and backward reaction have the same rate, $K_c$ is equal to $1$ in the case of a reaction where the total number of reactants is equal to the total number of products in the chemical equation describing the change. If the total of these stoichiometric coefficients is not the same before and after the horizontal arrow, $K_c$ may be equal to $1$, but must have a dimension, like mol/L or L/mol, or any combination of these units.

As the forward and backward rections have different activation energies, their rates change differently with increasing the temperature. So the equilibrium constant $K_c$ can be equal $1$ at only one temperature.


Oxidation means increase in Oxidation state or loss in electrons.

Example: $\ce{Fe^2+ -> Fe^3+ + e-}$

Oxidizing agent is a substance that oxidizes other substances and reduces itself.

In your example, $\ce{F2}$ will act as oxidizing agent as in the reactant side its oxidation state is $0$ but in the product side its oxidation state is $-1$, which means that its oxidation state in decreased, i.e. $\ce{F2}$ is reduced.


Associated occupation numbers of frozen natural orbitals (FNOs) from MP2 method

Virtual or unoccupied refers to the Hartree-Fock solution - the virtual orbitals are thus unoccupied in Hartree-Fock, i.e. have occupation number 0, whereas the occupied orbitals have occupation number 2 (spin-traced occupation in restricted HF) or 1 (for a given spin in unrestricted HF). The first-order MP2 correction to the wave-function will have different occupation numbers - the occupied orbitals will become slightly less occupied, whereas the virtual orbitals become slightly occupied by the same total amount as a result of electron correlation. This doesn't change the nomenclature: occupied and virtual orbital still refers to the HF reference wave-function.


How to draw Fischer projections for carbon chain having more than two carbons?

The first things to note when drawing Fischer projections are the general rules for drawing them. The first general rule concerns the choices of ‘top’ and ‘bottom’ while the second concerns how each cross is interpreted.

In a Fischer projection, the carbon atom with the highest oxidation state should be closer to the top than the bottom.
In cases where the highest oxidation state is in the centre or where multiple carbons have the same highest oxidation state, the second-highest applies analogously and so forth.

In a Fischer projection, the vertical bonds are always considered to point behind the paper (or screen) plane (hash) while horizontal bonds are considered to point in front (wedge).
This is exemplified in figure 1.

Figure 1: How stereochemical information is encoded in Fischer projections.
You compound presents a problem according to these rules as there is no most oxidised carbon. Rather, all three carbons 2, 3 and 4 are in the oxidation state $\pm 0$ while the two terminal ones are both $\mathrm{-III}$. To ease this limitation, I am going to discuss the Fischer projection with the related molecule shown in figure 2, which has one terminal carbon oxidised to a carboxylic acid breaking the tie.

Figure 2: L-5-deoxyribonic acid; a replacement molecule with the same stereochemistry on the central hydroxy groups.
For this molecule, it is clear that the acid should be put on top. Going down each bond, we need to rotate them until we have the longest chain in the vertical positions and pointing backwards (cf figure 1). I’ll let you use a modelling kit, pen and paper or your 3D viewing to perform the actual transformations; the final product you should end up with is the following in figure 3.

Figure 3: Fischer projection of the molecule in figure 2.
There should be little debate on the carboxylic acid; and whether the carbonyl group or the hydroxy group of it is pointing to the left does not matter; we can rotate around the $\ce{C{1}-C{2}}$ bond. To see the correct orientation of $\ce{C{2}}$ and $\ce{C{4}}$, you can take the skeletal formula from figure 2 and look from above; for $\ce{C{3}}$ you should be looking from below. We find that all hydroxy groups are pointing in the same direction; here: to the left. This means my molecule can be seen as a derivative of ribose; and since the lowest asymmetric carbon’s hydroxy group is pointing to the left, it derives from L-ribose.
Back to your molecule. Not only is the designation ambiguous (see above on the highest oxidation state) but it is also $C_\mathrm{s}$ symmetric, i.e. achiral. I have drawn your molecule along with its plane of symmetry (dashed) in figure 4.

Figure 4: The original molecule in question; 1,5-dideoxyribitol or (2R,3s,4S)-pentane-2,3,4-triol.
We need to arbitrarily choose a ‘top’ carbon here. For reasons of consistency, let’s choose the one on the left that was an acid in the former structure (figure 2). This means, we have a very easy job of drawing the Fischer projection as we already did it. It is shown in figure 5.

Figure 5: Fischer projection of the triol shown in figure 4.
We may be tempted to assign an L-configuration immediately but remember we realised that the molecule should have a plane of symmetry. Well, it does in the Fischer projection, too; the plane is horizontal and encloses the $\ce{HO-C-H}$ bonds of the central carbon atom. We can also turn around our piece of paper or PC screen and realise that we can thereby match it perfectly with its mirror image, shown in figure 5 to the right of the dashed mirror plane. Thereby we have twice shown that your molecule is achiral, that both a hydroxy group pointing to the left and to the right on the bottommost asymmetric carbon is possible and that therefore assigning a D/L stereodescriptor is meaningless.


Why does ice crack when a drink is poured on it?

I believe the ice cracked due to residual strains from freezing. Since ice freezes from the outside inward and it expands as it freezes, that as the inner water freezes, it imparts a tensile force on the surrounding ice (like the opposite effect of tempering glass).  As the warm liquid removes ice, the cross sectional area under tension decreases while the tensile force remains the same, causing an increase in stress. Once the stress reaches a certain point, the ice fails to counter the tension mechanically, causing the cracks.


How do molecules know how to behave when a disturbance occurs in an equilibrium?

The molecules don't know, they aren't alive in any way for that matter!

Here's an easy way to see it (note that I'm avoiding a mathematical approach to give an intuitive reason):

Consider a hypothetical reaction where $\ce{A}$ and $\ce{B}$ combine and gives $\ce{C}$:

$$\ce{A + B <=> C}$$

Let's say initially, there's an equilibrium. The rate of combination of $\ce{A}$ with $\ce{B}$ is the same as the rate of dissociation of $\ce{C}$. Let's say we added an excess of $\ce{B}$. If there's lots of $\ce{B}$ around, molecules of $\ce{A}$ have more chances of bumping into molecules of $\ce{B}$. This pushes the equilibrium forward, until there's enough $\ce{C}$ to counteract the amount of $\ce{A}$ and $\ce{B}$ present. A similar argument can be given if $\ce{A}$ was in excess than in $\ce{B}$.

You could now think about what happens if we removed some amount of $\ce{C}$. In such a case, the rate dissociation of the limited amount of $\ce{C}$ in the vessel would not balance with the rate of combination of $\ce{A}$ and $\ce{B}$, and thus equilibrium would again flow towards the products.

Now, lets see what happens if you add an inert gas $\ce{D}$ to that mixture. In such a case, $\ce{C}$ will continue to dissociate in it's usual rate, but $\ce{A}$ and $\ce{B}$ cannot combine with each other as fast because they may bump into $\ce{D}$ instead of each other. Thus equilibrium shifts towards the reactants.

I hope this makes sense. Leave a comment if you're not understanding any part of this.


So, you only need a descaler/softener if you have what is known as hard water, which is basically just water with a lot of minerals such as calcium and magnesium. You can do a test that shows how many GPGs (grains per gallon) your water has. According to http://home.howstuffworks.com/question99.htm,

  If your water tests at 1 GPG (17.1 mg/L) or less, then you have soft water. Water around 1-3.5 GPG (17.1-60 mg/L) occupies a gray zone between soft and slightly hard water and 3.5-7 GPG (60-120 mg/L) is moderately hard. Hard water is around 7-10.5 GPG (120 - 180 mg/L), and very hard water is above that.

So if it does turn out that you have hard water, then you look into descalers and softeners. The reason for this is that the calcium and magnesium come out of the water as scale, which builds up on the insides of pipes, water heaters, etc., to the point where they can become completely clogged.

A descaler is used to remove this buildup. They do this because they are acids and release hydrogen ions when dissolved in water. These ions, in turn, eat away at the scale by pulling at them with the charge. A softener is used to clean the water as it comes, disposing of the minerals safely. Water softeners are mechanical appliances that are put into your home's water system. They exchange the minerals for something else, usually sodium, in what's called an ion exchange. A diagram of a water softener is shown below:

This website provides more information about the exact workings of a water softener.

Hope this helps!


How to calculate the charge distribution at the ortho/para or meta sites of a substituted benzene ring

Summary: The method used does not handle π-conjugation

The partial charges in Avogadro (and many other software programs) are assigned by default by the Gasteiger-Marsili scheme†

Importantly, while the Gastieger charges are fairly good for atomic partial charges of organic-ish molecules, they're unlikely to reproduce effects in conjugated systems very well.

  A method is presented for the rapid calculation of atomic charges in σ-bonded and nonconjugated π-systems. Atoms are characterized by their orbital electronegativities. In the calculation only the connectivities of the atoms are considered. Thus only the topology of a molecule is of importance.

My suggestion is to use some type of quantum chemical calculation, e.g.:

NWChem
OpenMOPAC
ORCA
(etc.)

Even semiempirical quantum chemical methods (e.g. PM7) will allow you to assign charges based on the quantum electrostatic potential around the molecule. They'll be much more accurate at the trends you want.

† "Iterative Partial Equalization of Orbital Electronegativity - A Rapid Access to Atomic Charges," Tetrahedron, Vol. 36, pp. 3219-3228, 1980.


What chemicals are present in burnt toast?

First of all, one should have look a look on the ingredients of your toast. It mainly consist of starch, proteins (2% maybe), fat (a few %), water and some inorganic compounds such as NaCl.
Well Salt normally won´t interact with anything below 1400°C so no need to look at this.
Water just evaporates, i think we must not discuss that here.
Fat: that's more interesting because fat is known to degenerate into different compounds when heated.

But the definitely most interesting reactions will be present when one heats carbon hydrates together with protein: so called Maillard Reaction(s) those kind of reactions are still issues of ongoing research although this type of reactions have been discovered long ago. As a good starter on the various compound which can occur during Maillard Reaction you might read Mottram or Monnier. Especially acrylamide could be of interest when looking a toxicity. I don't have a certain paper in mind but i know that there has been research about toxicity of acrylamide in fish and other animal. So it's likely that there was some research about plants.

When you look at starch and protein separately you will encounter heat degeneration as well as caramelisation.


What is the magnitude of the current in an electrolyte?

By Kirchoff's Current Law ("what goes in must come out") the current flowing in total in all the branches of a circuit must be equal at all points. Hence if A amperes are flowing in the wire, as no parallel branches are present we conclude that A amperes must also be flowing in the electrolyte.


Factors that influence the kinetics of an irreversible exothermic reaction

I assume by kinetics you mean the rate of the reaction (i.e. how fast it happens), in which case the factors you mention are fine.

Regarding the temperature: the rate constant and therefore the reaction rate at a given concentration always increases with temperature as you can see in the Arrhenius equation. The fact the the reaction is exothermic means that it releases heat which, in the case where no adequate temperature control is applied (such as an ice bath), will increase the temperature of the solution and therefore the rate of the reaction releasing even more heat and so on, resulting in a runaway exotherm. This is the reason why exothermic reactions must be well temperature controlled especially when performed in large scale.

If your reaction was reversible, then heating it would not only accelerate the reaction (both forwards and backwards at equilibrium) but also shift the equilibrium towards the reactants.


Relation between Gibbs energy and net reaction rate

Thank you for putting together an excellent, well-specified, thoroughly documented question.  The answer, as you suspected, largely comes down to what "small" and "large" mean.

  I have come across some papers in biochemistry that seem to claim that a reaction can be "near equilibrium" even though the flux through the reaction (net reaction rate) is large[....] How can this be?

Consider the equilibrium approximation of Michaelis-Menten kinetics:

$$  \ce{E + S <=>[k_f][k_r] E\cdot S ->[k_3] E + P}$$

When $k_r \gg k_3$, the enzyme-substrate complex $\ce{E \cdot S}$ is assumed to be in equilibrium with $\ce{E + S}$, i.e. free enzyme and free substrate.  Making this equilibrium assumption eventually provides the rate law for overall flux through the reaction:

$$\frac{dP}{dt} = k_3 E_0\frac{S}{S+\frac{k_r}{k_f}}$$

Note that the flux is not always zero even though we have assumed that $\ce{E\cdot S}$ is in equilibrium with $\ce{E + S}$!  This is one hint -- hopefully a familiar example if you have studied biochemistry -- that assuming (quasi-) equilibrium doesn't mean zero fluxes.

  In my understanding, the net rate of a reaction depends on the Gibbs energy $\Delta G$ so that, near equilibrium where $\Delta G \approx 0$, there is no net flux through the reaction.

This isn't quite right.  First, you are right that the net rate of a reaction depends on the Gibbs energy $\Delta G$.  But near equilibrium where $\Delta G \approx 0$, it would be better to say "there is little net flux through the reaction".  Then the question is little relative to what?  The answer is that "little" is relative to the exchange flux.  So mathematically,

$$\nu_{net} = \nu_f - \nu_r = k_f(E)( S) - k_r(E\cdot S) $$
$$\nu_{exchange} = \nu_r = k_r (E \cdot S) $$

Reactions that are "approximately" at equilibrium have $\nu_{exchange} \gg \nu_{net}$.  Thus, "little" is relative to the exchange flux, and not necessarily the rates of other reactions in a metabolic network.

  ... "Small changes in $Q$ are accordingly adequate to tip the thermodynamically favored flux direction."

Remember that $\Delta G = \Delta G^\circ + RT \ln Q$.  The question of whether a reaction is reversible is asking about the sign of $\Delta G$.  The authors' statement that small changes in $Q$ can change the sign of $\Delta G$ only make sense if $\Delta G^\circ \approx 0$.  From the context of their remark, I think they are aware of this.

  But if this is true, the reaction direction seems unpredictable and the system should be difficult to control?

I wouldn't say "unpredictable" but I think you're right to say "difficult to control".  Another way of saying it would be that those reactions in the metabolic network that are approximately at equilbrium are not likely to be control points.  This 2006 paper examines that topic in more detail.

  Is this all just a question of what "small" or "large" $\Delta G$ means?

Yes!

  Can a reaction really be considered "near equilibrium" if the net reaction rate is large?

It depends what you mean by "large".

If you mean "large" relative to some other flux in the network, then yes, a reaction can be near equilibrium even though its flux is "large".
If you mean "large" relative to the exchange/equilibration flux, then no, a reaction near equilibrium has a net flux that is very small compared to the exchange flux.


Perhaps not a complete answer, but this might put you on the right track. It looks as though a lot of the system we have today grew out of recommendations made by August Wilhelm von Hofmann in a paper titled On the Action of Trichloride of Phosphorus on the Salts of the Aromatic Monamines

He actually suggested using the Latin numerals as a prefix for the alkanes with n>3. So pentane was quintane, hexane was sextane, etc. Over time, the Latin prefixes fell out of favor and were replaced with the Greek one's except in the case of nonane (and quadrane which became butane rather than tetrane). A Yale website, which led me to this article and helped me answer a previous nomenclature question about the alkanes (Etymology of saturation degrees (-ane, -ene, -yne) in aliphatic compounds), posits that it might just be that enneane just doesn't sound as good. Plus the trend from the polygons probably reinforced keeping it around.

So not a full answer, but at least it shows that nonane wasn't chosen to break the trend, but remained in spite of new naming convention.

The author of the Yale page does mention that their source was The Name Game by Alex Nickon and Ernest F. Silversmith, so perhaps there is more information there.


This is a really quick and dirty method since I don't know much about the calculations involved in crystal field theory. This method is based on some general observations I've made.

First of all, if your complex metal ion is not in group 8 ($\ce{Ni}$, $\ce{Pd}$, $\ce{Pt}$) or it doesn't have a $d^8$ configuration (like $\ce{Rh(I)}$ or $\ce{Au(III)}$), it is usually tetrahedral (Assuming you already know that the coordination number is 4).

If your metal ion is in group 8 or has a $d^8$ configuration, look at the crystal field splitting diagram. Square planar complexes have a four tiered diagram (i.e. four different sets of orbitals with different energies).

If it has a two tiered crystal field splitting diagram then it is tetrahedral.

But this assumes you have the crystal field splitting diagram of the complex.

As to how you obtain these diagrams (the calculations involved), I don't know exactly how it's done for specific molecules.

Like I mentioned before, this is just a very basic way to distinguish between the two geometries. I'm sure the calculations involved to obtain these diagrams would be really complicated but would make more sense in some cases if you're looking for the actual reasons behind why some complexes are square planar while others are tetrahedral.

EDIT:

Predicting the structure of a complex depends on how much info you're given. If you only have the number of ligands in the complex or the molecular formula of the complex, then you can predict the structure based on this list here.

Then again, steric factors and electronic factors are also important in determining the structure and your prediction would be more accurate if you know what ligand(s) you're dealing with (for steric effects) and the electronic configuration and oxidation state of the central metal ion.

I don't know much about how exactly these factors influence the structure but in general, you only have to be worried about these if:

the ligands are bulky or have some other special features
certain special electronic configurations (I've read something about certain electronic configurations causing distortions in the structure especially in copper(II) complexes)


Can someone tell the me the data bank which has all the STO-3G basis sets for many electron atoms?

A wide range of Gaussian basis sets for all elements can be found at the Basis Set Exchange. Alternatively, you can try to extract that data from the basis set library of a quantum chemistry software package.
The functional form of a primitive cartesian gaussian with $l=0$ is $\exp(-\alpha r^2)$, independent of the $n$ quantum number.


Major product of the reaction of m-hydroxyphenol in boiling aq NaHCO3

The major product (proportions dependent on conditions) is option 2 (beta-resorcylic acid) according to these papers$\ce{^{[1]} ^{[2]}}$. The relatively low proportion of option 1 (gamma-resorcylic acid) is attributed to steric hindrance$\ce{^{[3]}}$.

References

The carboxylation of resorcinol and the separation of β-and γ-resorcylic acid by ion-exchange chromatography by
D. K. Hale,  Audrey R. Hawdon, J. Idris Jones  and  D. I. Packham. DOI: https://pubs.rsc.org/en/content/articlelanding/1952/JR/JR9520003503#!divCitation
The chemistry of resorcinol carboxylation and its possible application to the CO2 removal from exhaust gases by Vincenzo Barbarossa, Francesco Barzagli, Fabrizio Mani, Sarah LaicGiuseppina Vangaa
DOI:https://doi.org/10.1016/j.jcou.2015.04.004 (picture source)
https://www.studocu.com/en-us/document/university-of-north-carolina-at-charlotte/organic-chemistry-2/assignments/lab-report-4-carboxylation-of-resorcinol/2141490/view


A Pourbaix diagram is a special kind of phase diagram. As in a phase diagram, the lines of all kinds represent phase boundaries. Each line represents the conditions in which two phases are in dynamic equilibrium. Without information on what the phases enclosed by lines look like, a phase diagram is pretty meaningless. Below is the phase diagram of water. Imagine how hard it would be to interpret without the labels. We still need to go look up what is meant by Solid IX - is it ice IX or ice 9?

Here is the iron-carbon (steel) phase diagram. Again, the phases are labeled.

In a Pourbaix diagram, the axes are pH and applied electric potential $(E_0)$ in volts instead of temperature and pressure (pure substance) or temperature, pressure (sometimes omitted), and composition for mixtures. The lines represent boundaries, but the "phases" are now the predominant iron-containing ion in solution. These diagrams seem to be primarily for aqueous systems, and so there is never an absence of $\ce{H+}$ ions. They may not be needed in the redox equation, but they are in solution. Otherwise, how would we have different pH along the horizontal axis?

Below is the Pourbaix diagram for aqueous solutions of iron. Notice that not all lines are horizontal or vertical. Each line represents a set of conditions ($E_0$ and $\text{pH}$) where two ionic species are in dynamic equilibrium.

For example, at $E_0=1.0\ \text{V}$ and $\text{pH}=2$, the predominating species is $\ce{Fe^{3+}(aq)}$. If we increase the $\text{pH}$ to $5$, the predominating iron species becomes $\ce{Fe2O3}\cdot\ce{nH2O}(s)$, rust. The reaction might be (we increase the pH by added $\ce{OH-}$):

$$\ce{2Fe^{3+}(aq) +6OH- (aq) -> Fe2O3(s) + 3H2O(l)}$$

Note that this reaction is redox neutral - all elements remain in the same oxidation state. Vertical lines represent redox neutral transitions/reactions.

If we start at the same place $E_0=1.0\ \text{V}$ and $\text{pH}=2$, and change $E_0$ from $1.0 \ \text{V}$ to $0.4 \ \text{V}$, the predominating species becomes $\ce{Fe^{2+}(aq)}$. When we lower $E_0$, it is like adding electrons to the system (it is really encouraging electrons to flow through the system in a specific way), we get this reaction, which is pH-neutral.

$$\ce{Fe^{3+}(aq) + e- -> Fe^{2+}(aq)}$$

As you noted, there is no $\ce{H+(aq)}$ in this reaction. Horizontal lines are pH-neutral transitions/reactions.

A diagonal or curved line is neither pH or redox neutral. For example, let's go from $E_0=0\ \text{V}$ and $\text{pH}=6 \ (\ce{Fe^{2+}(aq)}$ to $E_0=0\ \text{V}$ and $\text{pH}=10 \ (\ce{Fe2O3\cdot nH2O(s)}$ crossing a diagonal line. The reaction now involves a change in oxidation state and in pH. Since we are traveling horizontally, the electrons lost by $\ce{Fe^{2+}}$ must go somewhere.

$$\ce{2Fe^{2+}(aq) + 4HO- (aq) -> Fe2O3(s) + 2H2(g) + H2O(l)}$$


When and how in anaerobic digestion is water dissociated?

This is not a trivial question to answer, as many diagrams are often simplified and do not state the full chemical reactions that occur, such as

from Aerobic and Anaerobic Digestion and
Types of Decomposition education webpage from the Mountain Empire Community College.

However, according to this image (below) from the Ecocorp page - Metabolic Chain of the Anaerobic Process for the Utilization of Organic Wastes, water is a reactant in the acetogenesis
 stage and is removed as a product in the methanogenesis stage.

(Sample reactions are included in the image)

(Just in case the image becomes a dead link), the chemical processes involving water from the diagram are:

acetogenesis

$$\ce{CH3CH2COOH + 2H2O -> CH3COOH + CO2 + 3H2}$$

methanogensis

$$\ce{4H2 + CO2 -> CH4 + 2H2O}$$


What happened to mesoporous materials for drug delivery?

I think that research of mesoporous materials for several uses is alive and well.  The link referenced within your question is for an article written in 2008.  Wikipedia has a good, though brief, article on its current research status without respect to medical use in particular:

  A procedure for producing mesoporous materials (silica) was patented
  around 1970, and methods based on the Stöber process from 1968 were
  still in use in 2015. It went almost unnoticed[8] and was reproduced
  in 1997. Mesoporous silica nanoparticles (MSNs) were independently
  synthesized in 1990 by researchers in Japan. They were later produced
  also at Mobil Corporation laboratories and named Mobil Crystalline
  Materials, or MCM-41.

This is an abstract available from PubMed published in 2007:

  Research on mesoporous materials for biomedical purposes has
  experienced an outstanding increase during recent years. Since 2001,
  when MCM-41 was first proposed as drug-delivery system, silica-based
  materials, such as SBA-15 or MCM-48, and some metal-organic frameworks
  have been discussed as drug carriers and controlled-release systems.
  Mesoporous materials are intended for both systemic-delivery systems
  and implantable local-delivery devices. The latter application
  provides very promising possibilities in the field of bone-tissue
  repair because of the excellent behavior of these materials as
  bioceramics. This Minireview deals with the advances in this field by
  the control of the textural parameters, surface functionalization, and
  the synthesis of sophisticated stimuli-response systems.

One possible competitor for sustained drug delivery is transdermal drug delivery.  It seems that this method, which has been used for centuries, is well established and thoroughly approved by agencies like the FDA in the US.  In the US anyway, approval of new medicine or medical devices happens very slowly, and mesoporous materials seem to still be more in the research phase and may not have the level of agency approval that competing systems have attained.


Can I use harvested plants to gain electric power?

A company in the USA called Voltree Power has gone one step further to harvest electricity generated directly from live trees and other plants!

It seems trees are able to provide somewhere between 20 and a few hundred millivolts as well as some current, derived from a difference in pH between the trees and the soil, maintained by the trees metabolic processes.

The science behind this technology was published in 2008 by researches at MIT.

The company claims to have demonstrated the technology by developing the first wireless sensor network powered exclusively by trees to detect forest fires! They are also developing the technology to monitor humidity and temperatures in the forest using electrical power harvested from the trees themselves and transmit the data using low-power wiremess mesh technology.

Another collaborative group from University of Michigan, MIT & UC Berkeley published their research in 2009 into micro-power electricity scavenged from evaporative flows within the microchannels in trees.


Why do liquids freeze into a polycrystal solid when they freeze fast?

Embryo crystals spontaneously and independently form in a supercooled liquid. A crystal successfully growing in one location doesn't prevent others from growing elsewhere. As the temperature decreases, the rate of embryo formation and chance of successful growth increase, so at a low temperature you're likely to get multiple successful crystals.


An irritant is a chemical, which is not corrosive, but which causes a reversible inflammatory effect on living tissue by chemical action at the site of contact. There are specific lab tests to determine the ranking of an irritant, but something which is rated as very hazardous is likely to cause a very severe reaction upon exposure.

A permeator is a chemical which can pass through the outer protective dermal layers and into the body, and can either expose the body to toxic effects of that chemical OR act as a carrier for other toxic/hazardous chemicals.

Other common classes of hazardous chemicals include corrosive and sensitizer.

o-Toluic acid is a strong irritant, but moderate permeator. Depending on your level of experience in handling chemicals, simply getting the advice to 'wear gloves' may leave you a little underprepared. You should take this to mean that you should not allow any skin contact; full length sleeves (lab coat), eye protection and THE CORRECT TYPE of gloves. Using the incorrect type of gloves is worse than using no gloves at all, as it gives you a false sense of security. Select the correct type of gloves using a suitable selection tool or guide, such as this one from Ansell.

As a strong irritant, you should also avoid breathing any vapours or fumes, so it means handling this chemical in a suitably ventilated area.

AS with any risk assessment, it also means that before you handle the chemical you have an understanding of the response procedures in the event of things going wrong. That is, from your MSDS, you need to know Section 4:First Aid Measures prior to any procedure you undertake. It is no good having to find and read the MSDS after you have splashed yourself with this stuff.


Is HDPE affected by long term exposure to methanol?

HDPE should be fairly stable to MeOH, but it crazes in the presence of UV, when subject to oxidization or to environmental stress.

Since only some of the containers failed, were those subjected to more UV (e.g. placed by windows or at the front of a shelf in a well-lit location), placed near oxidizers, such as bleaches or were they subject to higher temperature, which would have caused the methanol liquid and vapor to expand, increasing pressure and stressing the container?

My experience has also been that apparently identical polyethylene containers have different failure rates, perhaps due to subtle differences during synthesis or manufacture.

Given polyethylene's tendency to craze, it might be best to use another type of container for long-term storage.

Note, though, that ~50% MeOH/water solution used for windshield antifreeze is usually shipped in polyethylene containers. Though the water lowers the vapor pressure, perhaps the UV susceptibility is addressed with additives.


2,2,6,6-Tetramethylpiperidine - Brønsted Basicity

I don't see much hinderance[note that they all have been optimised to stable configurations]
Blue: nitrogen,White: Hydrogen,Grey: Carbon

Side View:

Top View:


Why quaternary nitrogen but not tertiary oxygen?

This has to do with the electronegativity of the species and how good of a base (or how good an acid the conjugate acids) are.

$\hspace{9ex}$

$\hspace{7ex}$

[source]

To put this in perspective, nitric acid ($\ce{HNO3}$) has a $\mathrm{p}K_\mathrm{a}$ of $-1.3$. Yes, $\ce{HNO3}$, a strong acid, is more stable than $\ce{H3O+}$! The central oxygen in $\ce{H3O+}$ is too electronegative to effectively share these electrons and form a stable bond. In $\ce{NH4+}$, its lone pairs are far more available due to nitrogen's lower electronegativity.

How common each species is only has to do with the number of acids stronger than $\ce{H3O+}$ or $\ce{NH4+}$. As you can imagine, there are going to be a lot more acids stronger than $\ce{NH4+}$, which starts out around the middle of the $\mathrm{p}K_\mathrm{a}$ spectrum, than compared to $\ce{H3O+}$, which is near the end.


Does the electrochemical series reverse in electrolysis?

No, only the first is. The oxidising or reducing property of a species is tied to the species, not to the process in which the species is used. Therefore regardless of whether you are running a galvanic cell or an electrolytic cell, $\ce{F2}$ will always be a strong oxidant and $\ce{Li}$ a strong reductant.

The only difference about the two types of cells is that in the galvanic cell, you are letting the spontaneous cell reaction happen, which produces energy. In the electrolytic cell, you are using energy to force the non-spontaneous cell reaction to happen. The spontaneous cell reaction refers to the direction of the reaction with $\Delta G < 0$ (or equivalently, $E > 0$).

By this, I mean: you can use a galvanic cell to carry out the reaction $\ce{2 Li + F2 -> 2Li+ + 2F-}$. Since $\ce{Li}$ is a strong reductant and $\ce{F2}$ is a strong oxidant, this reaction will generate a lot of energy.

You can use an electrolytic cell to reverse that reaction (although you'd have to use molten and not aqueous $\ce{LiF}$). But since $\ce{Li+}$ is a weak oxidant and $\ce{F-}$ a weak reductant, you're going to need lots of energy to carry that out. Which makes perfect sense!


As a chemical engineer, my perspective on this differs from those of the other responders.  I regard a process occurring at constant temperature and pressure as one in which

The system is held in contact with a constant temperature over the portion of its interface with the surrounding through which heat transfer is occurring, not necessarily the initial temperature of the system, throughout the process

The external pressure (or more precisely, the external compressive stress) at the portion of the system interface with its surroundings at which displacement work is being done is held constant (not necessarily at the initial pressure of the system) throughout the process.


Do we prefer the substituent or the longest chain in IUPAC nomenclature?

The related question and answer that were already linked in this question also cover this case. If the other answer is too long and confusing, we can limit it to the rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) that are relevant in this case:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-45.0 INTRODUCTION
  Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. A parent structure is defined (P-15.1) as a parent hydride, for example, benzene, a functionalized parent hydride, for example cyclohexanol, or a functional parent compound, for example, acetic acid. Application of the criteria in this section will generate the preferred IUPAC name.

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

Therefore, for a correct name, only the longest chain is relevant for the choice of a principal chain in this case. Only if there is still a choice, substituent groups that are expressed as prefixes (e.g. ‘bromo-’) are taken into account for the preferred name according to P-45.2.1 and P-45.2.2.

Thus, the correct name for the compound that is given in the question is indeed 3-(bromomethyl)-3-methylpentane and not 1-bromo-2-ethyl-2-methylbutane since the pentane chain has the greater number of skeletal atoms.


Reaction of ethene with ammonia and sodium

No, because for ethene, $\mathrm{p}K_\text{a} = 44$. You won't be able to abstract even one proton.

Maybe, you were thinking in ethyne ($\ce{HC#CH})$? In that case, you are right! The bis-alkylation would work with sodium amide as a base!

But is the simultaneous abstraction of both protons the most plausible reaction then?


Why does electron velocity increase with the increase of atomic
  number?

Newton's second law tells us that for an electron moving in a stable circular orbit around a nucleus the centripetal force pushing out must equal the electrostatic attraction pulling in, or

$${\frac{m_e~ v^2}{r}} = {\frac{Z~k_e ~e^2}{r^2}}~~(1)$$

where $m_e$ is the electron's mass, e is the electron charge, $k_e$ is Coulomb's constant and $Z$ is the atom's atomic number.  Rearranging equation (1) yields

$$v = \sqrt{ Z~k_e~ e^2 \over m_\mathrm{e} r}~~(2) $$

From equation (2) we can see that the electron velocity increases as the atomic number, $Z$, increases.

Edit: Response to OP's comment

  I am expecting some quantum mechanical explanation

As the electron velocity given by equation (2) becomes comparable to the speed of light, the mass of the electron increases due to relativistic effects.

$$m_{rel}=\frac{m_{e}}{\sqrt{1-(v_e/c)^2}}$$

As the electron mass increases, the s (and somewhat for p) orbital radius contracts.  So relativistic effects affect the electron mass and the orbital radius.  In terms of electron velocity, these 2 relativistic effects offset one another and the electron velocity is not affected; equation (2) is still valid.


How will the periodic table change due to redefinition of the kilogram (May 2019)?

Molar masses

The molar mass of the carbon-12 isotope used to be 12 g/mol exactly (by definition). Now, it is experimentally determined, with the current value $\pu{11.9999999958(36) g mol-1}$ (CODATA 2018). So the molar mass of carbon-12 changed. The standard atomic weights (relative atomic masses) shown in the periodic table are not affected by this. However, atomic weights (dimension one) are no longer exactly equal to molar mass divided by 1 g/mol. Instead, they differ by a tiny bit (you would need 9 or 10 significant digits to see the difference).

No visible changes

Here is an animated gif of a portion of the periodic table at ptable.com over the past four years (Jan 23rd 2016, Jan 10th 2017,Jan 6th 2018, May 2nd 2019). Nothing much has changed near carbon:

The standard atomic weights change a bit each year, as better estimates of the isotope compositions of elements with multiple stable isotopes are made. Also, the publishers of this specific periodic table decided to truncate atomic weights to 5 significant figures in 2018. The biggest change occurred when elements in period 7 were discovered and named.

The last image is from May 2nd 2019, before the change of the kilogram (and mole) definition. On May 29th 2019, the data shown at ptable.com is identical.


What element or compound could be used to make the ocean more basic?

The ocean is becoming more acidic due to the dissolution of excess $\ce{CO2}$ from anthropogenic activities to form carbonic acid, $\ce{H2CO3}$, which even though mildly acidic is decreasing the pH of the ocean as you rightly point out.

So if we're trying to reduce the effect of this $\ce{CO2}$ on the ocean, we might consider removing it via some sink. One of these theoretical methods could be to react a base, such as $\ce{CaCO3}$,  with this carbonic acid in a conventional neutralisation. In symbols:

$$\ce{H2CO3 + CaCO3 -> Ca(HCO_3)2}$$

This could work, and we could derive everything we wanted to know about how much base to use from information such as the amount of residual $\ce{CO2}$ left over from emissions after photosynthesis has been taken into account (and other sinks), and then assuming that all other $\ce{CO2}$ will be taken up by the sea. From here we can calculate the amount of $\ce{H2CO3}$ present in seawater from the equilibrium constant between $\ce{CO2}$ and $\ce{H2CO3}$.

But we won't do that!

Notwithstanding these nonsense assumptions which could grossly overpredict/underpredict the amount of base required, neutralisation reactions don't deal with the excess carbonate effectively. What you really want is a permanent sink (a sink so stable its contents are practically constant through time) at the bottom of the sea. See the long-term carbon cycle for more information on this.

So your question becomes: how do we get the $\ce{CO2}$ dissolved in the ocean to the deep-ocean sediments? And how do we do it 'permanently'?

An interesting paper came out in the Journal of Global Biogeochemical Cycles in 1990 trying to address the question about why phytoplankton in the Antarctic don't use the myriad nutrients in the Southern Ocean to photosynthesise. One might expect that it could be temperature or sunlight which limit photosynthesis in the Antarctic, but this actually isn't the case. The limiting factor in the Antarctic ocean is actually a lack of iron.

It has since be proposed to increase the amount of iron in the ocean (simply by adding it ourselves) which would promote the photosynthesis of phytoplankton in such locations. When these organisms die, they sink and take their biomass with them which contributes to the ocean sediments.

Both neutralisation and iron supplementation may work to offset ocean acidification, however we really don't know the full consequences of these bioremediation efforts—there are simply too many variables involved.


Essentially, 'how good of an electrolyte' something is depends on how many particles the substance dissociates into. Let us discuss acidity. Now, from experimental evidence, it may seem that 'stronger' acids and mixtures, such as HCl, $\ce{H2SO4}$, Muriatic, are all good electrolytes. These are not because of their relative acidities. You will notice that in my last sentence, I put 'stronger' in quotes. This is because when talking about acids, 'stronger' has a different meaning than the one you are used to. A strong acid is one that dissociates virtually 100%. In other words, when you put a strong acid in water, all of it turns into ions, and you have none of the original acid left over. Similarly, a weak acid is one that does not completely dissociate into ions. By the definition of an electrolyte, stronger acids are better electrolytes (more ions in solution). This leads to our next point, ions in solution. As mentioned by Zhe above, the 'Van't Hoff Factor' is important. Essentially, it is just a measure of how many particles a substance dissociates into, for example NaCl has a VHF of 2, whereas $\ce{CaCl2}$ has a VHF of 3. The higher a substances VHF, the better of an electrolyte it is, the reasoning being that with an increase in number, there is an increase in dissolved particles, and therefore an increase in the state of being an electrolyte. Now, going back to acids for a moment, you may be wondering which acids are strong. There are 6, and they are HCl, HBr, HI, $\ce{HClO4},\ \ce{H2SO4},\ \ce{HNO3}$. Weak acids are everything else. These strong acids will have the best electrolytic results if you use them as such, much better than lemon juice (citric acid) or vinegar (acetic acid). In your example, you used muriatic acid (HCl) as one of your ion-producers, which explains the rapid voltage change, occurring from the dissociation of a strong acid.

$$\text{In essence, the answer to your question is:}$$ $${\text{How many particles a substance dissociates into determines how good of an electrolyte it is.}}$$


Obviously, there are three carbons in propene to which a phenyl group might be attached. Let's have a look and name the compounds:

1 with the phenyl group at C-2 of propene is 2-phenylpropene, also known as $\alpha$-methylstyrene.

2 with the phenyl group at C-1 of propene is 1-phenylpropene. To be exact, the compound drawn is (E)-1-phenylpropene. E is for entgegen, that is on opposite sides of the double bond.

3 is 3-phenylpropene, also known as allylbenzene.


Potassium permanganate oxidation of sugars

Each of your molecules also contains at least one primary alcohol appendage.  Primary alcohols are readily oxidized by potassium permanganate to the corresponding carboxylic acid (see this link for more information).


What does my textbook mean by saying this about petroleum?

Petroleum is a mixture of many different compounds.
Some of those compounds are gas.  Some are liquid.  Some are solid.
Many of them are hydrocarbons, and there are many different hydrocarbons included in the mix (e.g. different numbers on the C and H).
Petroleum is generally refined into more specific products before use, at a refinery.
Another example to consider might be "mud" - that's a mixture of some liquids, maybe some dissolved gases, and a bunch of different solids, usually including clay, maybe a bit of gravel, and some organic materials formed from decaying plants and animals.
If you take mud and compress it underground for thousands of years, then you'll have petroleum.

The second sentence seems clear enough.


How can I balance the redox reaction of sulfur dioxide with oxygen?

WELL I think you put something erroneous in your reduction formula.

You will never get SO3 out of just O2 and water. Instead of SO3

The reduction formla will be 4e- + O2 + 4H+ --> 2 H2O

Multiply oxidation formula by 2 and add it to new reduction formula and you got your answer


Why doesn't the carbon dioxide produced by a fire put out said fire?

The fuel for the chemical reaction is the CH4 (in this case) and the oxygen.  Assuming the CH4 isn't going to run out any time soon, the chemical reaction will continue as long as there is enough oxygen accessible. The force behind fire extinguishers can replace the oxygen available to the fire in its immediate area with CO2, and without one of the reactants (oxygen) available, the reaction won't continue.

The CO2 produced by the fire in an open area isn't going to replace the oxygen of the surrounding area.  However, in a closed system (for example, if you put a lid on top of a burning candle), then the air inside of the candle will become deficient in oxygen and rich in CO2 after a certain amount of time and the flame will go out, even without blowing on it or forcing it out.


What chemical characteristics give types of steel their properties?

(Source: Roy Beardmore, http://www.roymech.co.uk/ (defunct, via the Internet Archive)

As carbon content increases, the ability to resist a sudden impact decreases, as measured by the Charpy Impact Test.

As carbon content increases, ultimate tensile strength and Brinell Hardness increases.

As far as a phase diagram:

increasing $\ce{}$ (cementite) content makes the steel more brittle and hard.

Near 0.76% carbon, the steel will be pearlite which is ductile (good for wires).  Below 0.76% carbon, there will be increased alpha-iron (ferrite). Above 0.76% there can be Ledeburite.  Austenite only exists at high temperature.  Depending upon the cooling rate of Austenite, Bainite (slow cooling) or Martensite (rapid cooling) can form.


Grignard reagent is a hard base. The HSAB theory states that soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and form stronger bonds with hard bases, all other factors being equal (Wikipedia). Thus, Grignard reagent (a hard base) would preferably attack the harder position when it is available and in offer, like in the situation with its reaction with ketone carbonyl carbon even with the presence of $\alpha$-$\ce{H}$s. Keep in mind that the Grignard reagent probably react with $\alpha$-$\ce{H}$s to certain extent, but when the carbonyl group is readily accessible, that reaction proceeds to much faster rate. Yes, Grignard reagents do react with ethyne to give hydrocarbons, but it is a different story if the molecule contains a carbonyl group.


To find the number of degrees of freedom of a system, ask yourself the question, "in how many ways can I describe the motion of this system?"
Let us begin with monoatomic gaseous molecules. Visualise them as minute spheres.
Image source: http://vle.du.ac.in/mod/book/print.php?id=13264&chapterid=29093
Do you think there are any other ways to describe its motion? No, there aren't, but you might argue that it can also rotate about its own axis, so why do we exclude that? If that's the case, refer to this answer (tl;dr: The moment of inertia of an atom about its own axis is negligible due to the extremely small size and can be safely ignored.)
Now, let's switch to diatomic gases:

Source: https://physics.stackexchange.com/a/210955/168273
The diatomic molecule too is free to translate in any three coordinates of space. Also, it can rotate about the two axes perpendicular to the axis joining the two constituent atoms.
Again if you are wondering about the rotation about the axis joining the two atoms, apply the moment of inertia argument here too.
Alternative definition: The number of terms appearing in the energy equation of the molecule is called its degree of freedom.
$$E_\text{monoatomic}= \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2
+ \frac{1}{2}mv_z^2$$
$$E_\text{diatomic}= \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2
+ \frac{1}{2}mv_z^2 + \frac{1}{2}I_x\omega_x^2+\frac{1}{2}I_y\omega_y^2$$
where $I$ denotes the moment of inertia about an axis and $\omega$ the angular velocity about the same axis.
Do you see how this agrees with our qualitative analysis? There are $5$ terms in the energy equation of the diatomic molecule and $3$ in the case of the monoatomic one.
But there's a problem, at higher temperatures, under the provision of sufficient amount of energy the molecules can vibrate too, leading to an increment in the number of terms in our energy equation.
From Wikipedia:

The number of vibrational degrees of freedom, or vibrational modes, of a molecule is determined by examining the number of unique ways the atoms within the molecule may move relative to one another, such as in bond stretches or bends. This can be determined mathematically using the $3N-6=n_\text{DOF}$ rule, where $N$  is the number of atoms in the molecule and $n_\text{DOF}$ is the number of vibrational degrees of freedom. Note that for linear molecules, the rule instead becomes $3N-5=n_\text{DOF}$, meaning a linear polyatomic made up of $N$ atoms will have one less vibrational degree of freedom than a non-linear polyatomic with $N$ atoms.

Why are degrees of freedom important in thermodynamics?
According to the equipartition theorem an ideal gas distributes its energy equally in all degrees of freedom.  The average energy of a molecule in a gas associated with each degree of freedom is $kT/2$ where $k$ is the Boltzmann constant and T the temperature.
Task: Use the equipartition theorem to derive the concise internal energy equations for a diatomic and monoatomic gas molecule at low temperatures (i.e. ignore the vibrational degrees of freedom.)
Besides, knowing the degrees of freedom of the gas under investigation, we can find out its approximate molar specific heat capacities $C_p$ and $C_V$ which we often use in our thermodynamics calculations.


Is the amount of cubic inches in a cubic foot exact up to infinite significant figures?

Dealing with imperial measures in science is problematic. The main reason for that is that you have (at least) two different definitions (fixed values) for a lot of them, one based on the laws of the United States and one based on the laws of the United Kingdom. While the differences are quite insignificant for everyday use (as in baking a cake), in the science world these small deviations can cause further inaccuracies. On Wikipedia you can find more, but for the argument, let's look at the values for the yard:
United States    0.914401829 m
United Kingdom   0.9143993   m
International    0.9144      m

In principle the definition of cubic inch to cubic foot is exact, while the conversion to the SI system may carry an error.
Unit        Abbreviation    Definition  SI equivalent (rounded 4sf)
Cubic inch  cu in or in³                16.38 cm³
Cubic foot  cu ft or ft³    1728 cu in  0.02831 m³
Cubic yard  cu yd or yd³    27 cu ft    0.7646 m³

Most of the time this shouldn't affect your measurements more than the use of significant figures, which is used as stating an error implicitly. That means it is in itself inaccurate. In any scientific capacity you would state the error explicitly.
When a value is stated as $1.00$ it has three significant figures it actually means that the measurement falls within the interval $0.99$ to $1.01$, or $1.00\pm0.01$.
Let's have a quick example. You measure the volume of water in a beaker to be $0.35~\pu{in^3}$. The beaker doesn't specify how accurate the measurement is, so we simply assume $\pm0.01~\pu{in^3}$. You would state that value as $0.35\pm0.01~\pu{in^3}$, or $0.34-0.36~\pu{in^3}$, or simply $0.35~\pu{in^3}$ with implied error. On the other hand, when we know significant figures are used, we would interpret a value of $0.35~\pu{in^3}$ as $0.35\pm0.01~\pu{in^3}$.
When you convert this to the SI system, your value doesn't become more or less accurate. You cannot use significant figures to express the error in one system in the same way you would do it in the other, because the implicit error is different.
The exact conversion (of the international yard) is
$$\frac{0.9144~\pu{m}}{1~\pu{yd}}=\frac{0.9144~\pu{m}}{3~\pu{ft}}=\frac{0.9144~\pu{m}}{36~\pu{in}}\implies1~\pu{in^3}=\left(\frac{0.9144}{36}\right)^3~\pu{m^3}.$$
The actual trust of the value is $\pm0.01~\pu{in^3}=\pm0.16387064~\pu{cm^3}$. When you convert your value $0.35~\pu{in^3}=5.74~\pu{cm^3}$ (keeping the same number of significant figures since the conversion is exact), you would imply an error of only $0.01~\pu{cm^3}$, which is far less than the actual trust. Instead your measured value falls somewhere between $5.58-5.90~\pu{cm^3}$. So the correct way of implicitly stating the error would be $6~\pu{cm^3}$.
TL;DR
The conversions are exact, the system of significant figures is not. In doubt, state error margins explicitly.


Do catalysts increase the rate of exothermic reaction more in a reversible reaction?

Look at the form of the Arrhenius rate equation (this is fine since they mostly all have this form):

$$k = Ae^{-\frac{E_{a}}{RT}}$$

If you make the substitution, $E_{a}->E_{a}+\Delta E$, then you have:

$$k = Ae^{-\frac{E_{a} + \Delta E}{RT}} = Ae^{-\frac{E_{a}}{RT}}e^{-\frac{\Delta E}{RT}}$$

This means that you're just scaling the rate constant $k$ by another constant. This scaling applies to both the forward and reverse rate constants. If the concentrations were such that they were in equilibrium before, they're still in equilibrium now.

You'd have to prove this in general for complex reactions with complicated rate laws, but for simple reactions, this should suffice to help you see that the equilibrium is unaffected.


Can any atom donate more than 1 lone pair as a coordinate bond to the same metal?

There is a class of compounds, called carbodiphosphoranes, which formally has a divalent carbon (0) central atom. Employed as a ligand it acts as a sigma and pi donor. This capability is also one of the major features to distinguish these compounds and derivatives from carbenes, which are sigma donors and pi acceptors. The following is reproduced from Thi Ai Nhung Nguyen and Gernot Frenking, Chem. Eur. J., 2012, 18, 12733-1748:

Similarly carbodicarbenes (substitute the phosphine with a carbene), also known as bent allenes, are also double donors.

This post shall only serve as an example, I am quite certain, there are more compound with this kind of bonding motive out there. Nicolau mentioned $\ce{OsO3N^{-}}$ and other metal nitrido complexes, which he will hopefully extend in an answer.


Why do Krypton and Xenon have high electronegativity?

As you point out, krypton ($\ce{Kr}$) and xenon ($\ce{Xe}$) are members of the Noble gas family.  They are generally unreactive (noble) because all of their occupied orbitals are filled with electrons - they really don't want to gain or lose an electron.  However, in the 1960's researchers found that $\ce{Kr}$ and $\ce{Xe}$ did react with extremely electronegative elements such as fluorine and oxygen to form new molecules.  In the reaction process, $\ce{Kr}$ and $\ce{Xe}$ basically gave up an electron to the very electronegative elements they reacted with.

$$\ce{Xe + 3F2 → XeF6}~~(1)$$
$$\ce{XeF6 + 3H2O → XeO3 + 6HF}~~(2)$$

The most common definition of electronegativity is based on Pauling's work and is given by the equation

$${\chi_{\rm A} - \chi_{\rm B} = ({\rm eV})^{-1/2} \sqrt{E_{\rm d}({\rm AB}) - [E_{\rm d}({\rm AA}) + E_{\rm d}({\rm BB})]/2}}$$

where ${\chi_{\mathrm{A}}}$ and ${\chi_{\mathrm{B}}}$ are the electronegativities of atoms $\ce{A}$ and $\ce{B}$ and ${E_{d}(AB)}$ represents the bond dissociation energy of molecule $\ce{A-B}$.

Using $\ce{HBr}$ as an example, if we know the bond dissociation energies of $\ce{HBr,~ H2}$ and $\ce{Br2}$, then we can calculate the difference in electronegativity between $\ce{H}$ and $\ce{Br}$.  From this, if we know the electronegativity of hydrogen, then we can determine the electronegativity of bromine.

In essence, for any element that can react and form a molecule, we can calculate its electronegativity.  Since Kr and Xe do react to form molecules, we can use the above methodology to calculate its electronegativity.

Note: In the case of atoms like Kr and Xe that do not form a diatomic species like $\ce{Xe2}$, we need an extra step to find their electronegativity.  Since there is no $\ce{E_{d}(Xe-Xe)}$, we must use Pauling's equation twice, once with reaction (1) and once with reaction (2).  We now have 2 equations and two unknowns (${\chi_{\mathrm{Xe}}}$ and ${E_{d}(Xe-Xe)}$) so we can solve for ${\chi_{\mathrm{Xe}}}$, the electronegativity of $\ce{Xe}$.


Molecular mechanics force fields for metals?

The modelling of metals using a 'simple' potential is relatively difficult. I have a suggestion for a force field, though it does not fit exactly all your criteria, but I would like to offer it anyway.

Adri van Duin has developed a force field called ReaxFF. This force field is non-proprietary and is readily available in the open-source package LAMMPS. This force field gives pretty good results and can be applied to any kind of elements, it can describe chemical reactions and is also valid for hetero-atomic interactions. That being said, the biggest challenge when using the ReaxFF force field is that you have to fit / train it to your system. This requires, for instance, a database of ab initio calculations that are representative for your system.

If you cannot find any force field, than this might be a interesting alternative to developing a force field completely from scratch.


What does a change in temperature do to the concentration of gases at equilibrium?

You are correct in saying that the increase in temperature causes the equilibrium to shift to the left, in the direction of the endothermic reaction. As the reaction continues in the endothermic direction, the concentrations of $\ce{SO3}$ will decrease in favour of the evolution of more $\ce{SO2}$ and $\ce{O2}$, due to the additional heat available.


A substance that is immisiscible with water and that can be made cloudy

It smells but I think chloroxylenol 48 mg in 250 ml ethanol mixed with water would be what your looking for, I'm not sure what it's called where you are but it's a common antiseptic sold everywhere, in Australia the most common brand is Dettol.

Also , if something is completely immiscible with water it will form two phases , both relatively clear, eg water/xylene.

If you mix the solution it will go cloudy but then seperate, the longer it takes to do this depends on how much it was mixed, an emulsion can be formed but that's not what your looking for.

You want something that's only very slightly or sparingly soluble but doesn't seperate into a different phases once it's mixed.

Dettol came straight into my mind but I'm sure there's many more. Arrowroot or cornflour mixed with water would produce a milky/cloudy solution.


A Book on Preparation and Reactions of organic compounds

For example, why do we chose Nickel as the catalyst other than anything else to prepare ethane by catalytic hydrogenation of ethene?

Palladium is expensive ($690~\$\mathrm{US}/\mathrm {oz.tr}; 1~\mathrm{oz.tr = 31.1 g}$).
Nickel is cheap ($11~\$\mathrm{US}/\mathrm {kg}$).

Catalytic hydrogenation of ethene is the model reaction performed to understand the mechanism of this type of processes, but I'd be astonished if it had any technical relevance.

There is enough ethane available from the processing of natural gas by distillation (demethanizer, deethanizer).
Ethene is the more interesting starting material for polymerization, oxidation, etc.

  And why this reaction has to take place at 300 °C and not at a lower or greater temperature?

Does it really? In earlier mechanistic studies, such as  Review of Physical Chemistry of Japan, 1937, 11, 153-165, or Discuss. Faraday Soc., 1950, 8, 152-159, the hydrogenation of ethene on a nickel catalyst was performed at much longer temperatures (-78 to 0 °C).

  preparation and reactions of alkyl halides, alcohols, phenols, ethers, aldehydes, ketones, carboxylic acids and esters

For lab-scale reactions, any decent introductory book on Organic Chemistry  will do.

If you're looking for books that focus on experimental procedures (cook books), there are two which I can recommend (Note that have only used the original German versions):

Organikum – Organic Chemistry Basic Practical Course
Reactions and Syntheses: In the Organic Chemistry Laboratory


Is atmospheric pressure acting only on the contents or also on the container?

According to the way this question is worded, adding 1 atmosphere would be incorrect. The reason is that they specify an empty container - if you have a rigid container and it is truly empty (high vacuum), then the absolute pressure would be zero, and the gauge pressure would be -1 atm (see this wikipedia article for reference to these terms).

Once the $\ce{CO2}$ sublimates, the final absolute pressure would be given by $PV=nRT$ - there is no need to add atmospheric pressure. To find the gauge pressure, you would actually subtract one atmosphere.

It is possible that the question wants you to assume that "empty" means "filled with air at 1 atm," but that is a guess and I would argue that it's not safe to assume that based on the way the question is written.


Why are full and half filled orbitals the most stable?

In short, the so-called Fermi correlation which is a consequence of the Pauli principle and applies only to electrons of the like spin keeps such electrons further apart comparing to the case of electrons of unlike spin. Naively one might think that a larger average distance between electrons of like spin reduces electron-electron repulsion energy which is the reason for the stability of high spin states. And while it is true that for each and every pair of electrons electron-electron repulsion energy is decreased as the average distance increases, it was shown that in the whole electron-electron repulsion energy for the molecular system in a high spin state is actually greater that in a lower spin spate.

So, the modern explanation of the lower energy of high spin states is that a larger average distance between electrons also effectively leads to less screening of the nucleus, i.e. each electron experiences a larger nuclear effective charge. As a result the unpaired electrons can approach the nucleus more closely and the electron-nuclear attraction is increased. And this increase in the electron-nuclear attraction energy (which is negative) overcomes the above mentioned increase in the electron-electron repulsion energy (which is positive) leading to a greater stability of high spin states.

For details, see Boyd, R. J. Nature 1984, 310, 480-481, or any modern textbook on quantum chemistry.


The major product of the following reaction

When you form the allene, the methylene hydrogen has enhanced acidity because deprotonating it leads to a carbanion in which the negative charge is delocalized:

$\ce{CH3-CH=C=CH2} - \ce{H^+ -> CH3-CH=C=CH^- <-> CH3-C^-H-C#CH}$

The proton then reattaches to the opposite end of the allene chain to form a more stable product.  In effect this is a (strong) base catalyzed tautomerization.


Isopropyl alcohol to clean Nylon plastic

Your nylon plastic should be thoroughly resistant to concentrated isopropyl alcohol.

According to this encyclopedia.com entry, nearly all of the nylon produced (in the US anyway) is one of two types: type 66 and type 64. Each of these is stated in the article to be insoluble in water and most organic solvents, but soluble in strong acids.

As some anecdotal evidence of nylon's resistance to common solvents, this Quora discussion referenced a frustrated biker's need to find a suitable solvent for removing some nylon that had melted onto a motorbike exhaust pipe. There was really no good consumer-type of solvent listed (and nothing remotely chemically similar to isopropanol), and it actually included a link to a patent for a compound specially designed for dissolving nylon-based materials, just because they are so chemically resistant.

That all said, it wouldn't be a bad idea to do a "spot test" of the isopropyl alcohol on a small, inconspicuous as possible piece of the plastic. There's always an outside chance that it's not exactly what you think it is or that for some reason your particular nylon-based plastic is not as immune to a solvent like isopropanol as the above arguments imply that it should be.


Specificity is the term used to define the selectivity of enzymes
for their substrates. The selective qualities of an enzyme are collectively recognized as its specificity.

Other texts have synonymised enzyme selectivity with substrate specificity:

The non-covalent forces through which substrates and other molecules bind to enzymes are similar in character to the forces
that dictate the conformations of the proteins themselves : Both involve van der Waals, electrostatic, hydrogen
bonding, and hydrophobic interactions.
Molecules that differ in shape or functional group distribution
from the substrate cannot productively bind to the
enzyme; that is, they cannot form enzyme–substrate complexes
that lead to the formation of products.

The extraordinary ability of an enzyme to catalyse only one particular reaction is a quality known as specificity.

Specificity means an enzyme acts only on a specific substance,
its substrate, invariably transforming it into a specific product. That is, an enzyme
binds only certain compounds, and then, only a specific reaction ensues.
Some enzymes show absolute specificity, catalyzing the transformation of only one
specific substrate to yield a unique product. Other enzymes carry out a particular
reaction but act on a class of compounds. For example, hexokinase (ATP-hexose-6-
phosphotransferase) will carry out the ATP-dependent phosphorylation of a number
of hexoses at the 6-position, including glucose.

  Pioneering enzyme specificity studies at the turn of the 20th century
  by the great organic chemist Emil Fischer led to the notion of an
  enzyme resembling a “lock” and its particular substrate the “key.”
  The specificity of an enzyme (the lock) for its substrate (the key) arises from their geometrically complementary shapes. This was later
  modified to a more accurate induced fit hypothesis given the high
  flexibility, conformationally dynamic properties of enzyme molecules.

References

Voet and Voet Biochemistry 4th ed: Introduction to Enzymes
Biochemistry Grisham & Garret How Can Enzymes Be So Specific?


How do bonds form without violating the second law of thermodynamics

Note that 2 isolated hydrogen atoms will not form a stable hydrogen molecule, unless they are able to dissipate the energy, released by the bond formation. It can be done e.g.  passing mechanical energy to neighbour molecules, or emitting a photon.

In a macro scale, a system can decrease its entropy, if its Gibbs energy decreases, as it means the dissipated heat causes the bigger entropy increase of the surrounding, than is the entropy decrease of the system, so the total entropy increases.

$$\Delta G = \Delta H - T \cdot \Delta S$$

If $\Delta G_\mathrm{sys} \lt 0$ then $\Delta S_\mathrm{tot}=\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr} \gt 0 $

as $\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr}=\Delta S_\mathrm{sys} - \frac{(\Delta G_\mathrm{sys} + T \cdot \Delta S_\mathrm{sys})}{T}=-\frac {\Delta G_\mathrm{sys}}{T}$


Why is tautomers not possible in benzoquinone?

You could sort of devise a tautomer by transferring an $\alpha$ hydrogen from carbon to oxygen.  But if you draw the resulting structure you find a pair of cumulated double bonds.  That tends to be unstable, and since the cumulated double-bonded carbons favor a 180° bond angle putting them into a relatively small ring makes them a lot more unstable.  Your proposed tautomer is way out there in energy.

Tautomerization of a ketone works better when you have an $\alpha$ hydrogen on a saturated carbon atom.


As an acidic compound, phenol started to degrade the protective layers of your skin, and partially diffused into deeper layers.  This, and because of the moderate solubility in water are the reason the scent of phenol was still recognisable after some time.  While phenol is better soluble in ethanol and methanol, one refrains from these solvents to clean the skin because these actually would help phenol to reach even deeper tissues.
 Your writing that it was recognisable even after 3-4 days, even after extensive rinsing with water and soap, hints to an extended exposure of the skin towards pure phenol, which however should be avoided.

Not visible to human eye, but thanks to antiseptic power, it killed some germs, too.  Earlier in times, it was known and used as carbolic acid.


Why would there be differences in UV/Vis spectrum?

$\ce{H2O}$ does not absorb the radiation almost at all, so it will not interfere in the absorbance of the analyte we are analysing, thats why we use water almost always as a solvent in instrumental methods of analysis.Therefore $\ce{MeCN}$ wil absorb some of the radiation so then of course the absorbance of the analyte will also change ( decrease ) and the spectrum will be different ( the peaks of the analyte will be lower ).


Need for enthalpy of vaporization in calculating bond dissociation enthalpy

Well, I think that you have to look up the actual way bond dissociation enthalpy is defined in your textbook/lecture notes, since it is a not quite standard term. IUPAC defines only bond-dissociation energy but not enthalpy.

bond-dissociation energy, $D$
The enthalpy (per mole) required to break a given bond of some
specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

Many resources on the Internet claim that the IUPAC definition of the bond-dissociation energy refers to $\pu{0 K}$ (see e.g. Wikipedia, UC Davis ChemWiki) and that it differs from bond dissociation enthalpy, which refers to the enthalpy change at room temperature ($\ce{298K}$).
Wikipedia

The bond-dissociation energy is sometime also called the
bond-dissociation enthalpy (or bond enthalpy), but these terms are not
strictly equivalent, as they refer to the above reaction enthalpy at
standard conditions, and differ from $D_0$ by about $\pu{1.5 kcal mol-1} \pu{(6
kJ mol-1)}$ in the case of a bond to hydrogen in a large organic molecule.

UC Davis ChemWiki

Officially, the IUPAC definition of bond dissociation energy refers to
the energy change that occurs at $\pu{0 K}$, and the symbol is $D_0$. However,
it is commonly referred to as BDE, the bond dissociation energy, and
it is generally used, albeit imprecisely, interchangeably with the
bond dissociation enthalpy, which generally refers to the enthalpy
change at room temperature ($\pu{298K}$). Although there are technically
differences between BDEs at $\pu{0 K}$ and $\pu{298 K}$, those difference are not
large and generally do not affect interpretations of chemical
processes.

If we follow this definitions, then yeah, at standard conditions $\ce{HBr}$ is liquid so we have to add up $\pu{12 kJ mol-1}$ for its enthalpy of vaporization, and (almost) get the desired $\pu{364 kJ mol-1}$. The small discrepancy is probably due to slightly wrong value of enthalpy of vaporization which I just quickly found somewhere.


How to determine mobility when two mechanisms exist simultaneously

I needed to add the inverses of the given mobilities and take the inverse of that to get my final result.

$1/\mu = 1/\mu_L + 1/\mu_I$


Can a first degree carbocation form if a methyl shift is possible?

The cation is produced when a methyl group migrate from C-2 to C-1 as the C-Br bond is broken. The simultaneous migration of alkyl group and departure of leaving group to form a tertiary cation  is therefore faster and more favourable than formation of primary cation due to simple loss of leaving group.

The products observed are those that result from further reaction of the so formed tertiary cation. The mechanism mentioned in the pic you've attached is incorrect, 1° carbocation intermediate is not formed at all.

Several comments correctly mention that formation of primary carbocation is not favourable. So, it is important to note that methyl shift and departure of leaving group in this case are simultaneous processes - primary carbocation is not formed at all!

P.S.
Also, I'd like to quote some evidence I found for a very similar reaction of 1-bromo 2,2-dimethyl propane:

Above is the solvolysis reaction of 1-bromo 2,2-dimethyl propane.

Source: Reactive Intermediates in Organic Chemistry: Structure, Mechanism, and Reactions - By Maya Shankar Singh (Wiley Publishers)


Electrolysis of brine solution gives sodium hydroxide , chlorine and hydrogen gas but catner kellener cell it forms amalgam

I like what Maurice said "When nobody understands a scientific phenomena, we give it a name. Here the name is overpotential. Here the hydrogen is said to have a big "over potential" on mercury cathode. That is a bright and remarkable way to hide our ignorance."
This is little on the extreme side but people have spent their life on studying the hydrogen evolution reaction. It is a kinetic issue. It was one of the favorite topics of electrochemists. As one can see in the figure, each electrode evolves hydrogen at a different potential in water. Using mercury in the Castner-Kellner cells, allows sufficient negative voltage that can reduce sodium ions even in the presence of water. Additionally, mercury can form amalgams with many metals. Sodium is not an exception. The moment sodium is reduced at the mercury surface, it is dissolved in the mercury pool, which is periodically tapped off.
Figure taken from: An Electrochemistry Experiment: Hydrogen Evolution Reaction on Different Electrodes in the Journal of Chemical Education.


Why does a certain number of moles of gas occupy a set volume?

The reason the gas doesn't expand to infinity is because it is in a vessel.

Or, to be more precise, the ideal gas equation defines the relationship between temperature, pressure and volume for a fixed amount of gas.

What is normally done to understand the relationship is to put the gas into a vessel of fixed volume and measure the pressure it exerts at a given temperature. The pressure changes with temperature according to the formula (real gases may deviate slightly).

Alternatively, we can use a more complex apparatus that allows us to measure the volume occupied by the gas as we vary the temperature while maintaining a constant pressure (say atmospheric pressure). Same equation, different constraint.

Your misunderstanding is that you have not thought through the conditions under which you are applying the rules. The thought experiment you are doing is releasing the gas in the vacuum of space where there is no external pressure. In those circumstances the gas will expand to infinity but this doesn't violate the ideal gas equation as there is no constraining volume (you could say volume is infinite).

When applying the rules you have three variables (you usually keep the amount of gas, the number of moles, fixed) and you have to constrain two of them. If you don't, are an infinite range of possible combinations of the variables (e.g. constrain temperature and there is still nothing that tells you what the volume and pressure are as they can vary together over any values that satisfy the equation).


How do I get the molarity of solution 1 given volume 1, molarity 2, and volume 2?

The reaction that occurs: H+ + OH- --> H2O

When you multiply the volume of HCl by the molarity of HCl you get the total amount of moles of H+ that reacted (mol/L x L = mol).
This means that the same amount of OH- was needed for the reaction (1:1).
Volume of NaOH is known and thus yields the answer.


Computer killed by insecticide; what compounds are likely?

Ingredients

So we could probably guess what's in the insecticide based on what's used in some of their other products.

Black Flag Flying Insect Killer

Tetramethrin - Sounds like the "active ingredient" here. By "active ingredient", I mean poison.

D-phenothrin - Another poison. This product sounds like it kills more than one insect.

Petroleum Distilates - Likely a solvent. Unspecified

Liquefied Petroleum Gas - Propane or butane. An aerosol propellant. Was your product from a bucket or can? If a can, then it probably has this stuff.

So what happened?

My guess is that one of the products created a short circuit

  A short circuit is an abnormal connection between two nodes of an electric circuit intended to be at different voltages. This results in an excessive electric current limited only by the Thévenin equivalent resistance of the rest of the network and potentially causes circuit damage, overheating, fire or explosion.

Typically, when something like a computer experiences a short circuit, it will stop functioning, "die", once the irreversible damage is done. As specified in the above text, circuit damage is likely done, and the circuits will need to be rewired. If there was no damage, then removing any remaining substances through a solvent, followed by thoroughly drying the computer, would remove the short circuit and allow it to function normally. It's possible that if the short circuit was created by a solvent, that the solvent has already been removed by evaporation, due to common organic solvents' melting points.


Is it possible to functionalize silica gel as Activated Carbon can be? How?

Yes, silica gel can be modified to make its adsorption more specific. E.G., epichlorohydrin dimethyl amine copolymer (Epi-DMA) can be used to make silica gel effective for adsorbing anionic dyes.


What is the joke that is shown in this picture?

As commented by @long, the elements shown are not completely randomised. The elements in all the rows, except in the 3rd row ($\ce{Na, K, Ca}$), are arranged in same order as they are given in periodic table.

After studying them on periodic table for 15 minutes, it seemed to have no meaning. There are no periodic trends. $\ce{H, Li, Na, K, Rb}$ belongs to IA group, $\ce{Be, Ca, Sr}$ belongs to IIA group. And they are arranged in a random way, having no meaning.

The picture is clearly a joke showing that when scientist have eye check-up, they have to read elements instead of alphabets.


Reversibility of an acid's reaction with water depending on its strength

$$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$

This reaction owes its ability to go to completion to the great stability the ions it forms when dissolved in water ($\ce{H+}$ and $\ce{Cl-}$). This ionization is caused by stronger attractive forces between the water and the atoms in a molecule (or ions in a salt) than between the atoms within the molecule have themselves, an is an endothermic process. In the dissociation of salts like $\ce{KNO3}$, the solution becomes notably cooler. When an acid dissociates, however, the hydration of $\ce{H+}$ to $\ce{H3O+}$ releases a great deal of heat.

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$

This reaction also results in the formation of $\ce{H+}$ ions, though I wouldn't say that $\ce{CH3COO-}$ is less stable because it is larger than the $\ce{Cl-}$ ion. Consider that $\mathrm{p}K_\mathrm{a}(\ce{HCl})=-7$, $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=4.75$, and $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=0.75$. In case you are unfamiliar, the $\mathrm{p}K_\mathrm{a}$ is a way of expressing how much an acid dissociates in water, and a drop in $\mathrm{p}K_\mathrm{a}$ by $1$ means an acid releases $10$ times as many $\ce{H+}$ ions. $\ce{HCl}$ releases $10^{11.75}$ times as many $\ce{H+}$ ions as $\ce{CH3COOH}$, but when acetic acid's $\ce{C-H}$ bonds are swapped out for $\ce{C-F}$ bonds to make $\ce{CF3COOH}$ (an even larger molecule), its acidity increases by $10^4$. This shows that ion stability has more to do with bond strength and the inductive effect than ion size.

Both of these reactions are reversible. This just means that the forward reaction is occurring as well as the backward reaction (when occurring at the same rate this is known as equilibrium), but the first reaction lies more towards the formation of the products due to its greater stability. All reactions proceed in the backward direction, unless they are irreversible. Though the dissociation of $\ce{HCl}$ is often considered irreversible due to how shifted its equilibrium is, it is not truly irrversible because it has the potential to reform a molecule of $\ce{HCl}$. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, which does not happen in this case.


Why does a reduced compound become a better electron donor?

Simple electrostatic repulsion can be used to explain this. For example, if you have $\ce{O^2-}$, there are 8 protons in the nucleus, but 10 electrons, so there is a net negative charge of 2. Like charges repel, pushing the electrons away.

BTW, compounds rarely "gain an electron", the reactive species are usually radicals.


How to explain the discrepancy between the calculated absorption wavelength within the particle in the box theory and the observed wavelength?

Luckily I found a source containing your molecule as an example.

You need your equation for the energy difference between the HOMO and the LUMO, which is (nearly) defined as you wrote:
$$\Delta E = \frac{(N+1)h^2}{8 m_e L^2}$$

As you want a wavelength, you need to convert the energy difference, which yields then:
$$\lambda = \frac{8m_ecL^2}{h(N+1)}$$

Now they state, that the length of the box can be guessed by the amount of carbons (p) between both nitrogens by using this formula:
$$L = (p+3+\alpha)l$$
... with $l$ being the length of typical benzene-C-C-bonds, thus being 0.139 nm and $\alpha$ is an empirical correction factor for polarizable end groups.

The amount of pi-electrons (N) that you need to inlude can be calculated by a similar equation:
$$N = p+3$$
... where the additional "+3" come from both nitrogens (would be actually +4, but it's +3 because one nitrogen atom had to form an ion).

This leads to their equation 4:
$$\lambda = 63.7\frac{(p+3+\alpha)^2}{p+4}$$
... where all the physical constants have been combined to give 63.7.

As you have three carbon atoms between both nitrogens, you come up with 328 nm and when you include their "correction factor for polarizable end groups" of $\alpha = 0.675$, you come up with 405 nm, which is more than close to the experimental result.

When I use your box length and use 12 electrons (3 from the "left" benzene ring, 4 from the two double bonds to the "right" nitrogen" and 1 from the lone pair of this nitrogen), I come up with about 356 nm, which is not that far away as it sounds. In eV this is an error of 20% (the 405 nm are wrong by about 5%), whereas your 200 nm would have been an error of 120%!


Is NADPH sensitive and unstable in UV light?

NADPH isn't destroyed by absorbing 340 nm light; if it were, then spectrophotometry wouldn't be a good technique to monitor its concentration.  The primary stability concern with NADPH and NADH is acid-catalyzed decomposition.  A great 1986 article summarizing this phenomenon is available here.

Some highlights from the article:

At 30 °C, the pseudo-first order rate constant for NADPH degradation in aqueous solution goes from 0.5 min$^{-1}$ at pH ~3 to $10^{-5}$ min$^{-1}$ at pH 10.  At pH 7 the rate constant is $10^{-3}$ min$^{-1}$.
The rate is elevated slightly by higher concentrations of phosphate buffer, about 10x higher for 1 molar (!) phosphate vs. extrapolation to infinite-dilution  phosphate.

A secondary pathway for NAD(P)H decomposition is oxidation by atmospheric air, but it is usually pretty minor compared to acid degradation.

So I don't think there will be a big problem with light destroying your NADPH.  You might be able to re-use it, but it depends how long you let it sit in solution, and at what temperature.


Is 3,5-diethyltoluene an accepted IUPAC name?

Names like toluene are "retained names".

In chemistry, a retained name is a name for a chemical compound that is recommended for use by a system of chemical nomenclature (for example, IUPAC nomenclature), but that is not fully systematic. - Wikipedia
‘Retained names’ are traditional or common well-established names that may be used either as preferred IUPAC name or as an allowed alternatives in general nomenclature, for example, naphthalene, benzoic acid and pyridine. IUPAC's Blue book,  P-12.3

Later in the same draft, we read

Toluene, xylene and mesitylene are specific parent hydrides that are composed of two components, one cyclic and the other acyclic and saturated. These names are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not substitutable; toluene is substitutable under certain conditions for general nomenclature. P-22.1.3
Furthermore, anisole and toluene cannot be substituted by other
methoxy or methyl groups, respectively. P-46.2.3.2

Conclusion:

Names like toluene, anisole and xylene are retained names. They are preserved for historical reasons; however, their unsubstituted forms are PINs (Preferred IUPAC names).
The name 3,5-diethyl toluene isn't acceptable according to the systematic IUPAC nomenclature.
If you want to name the compound below, it's called 1,3-diethyl-5-methylbenzene.

Source


Do hybrid orbitals exist in unbonded molecules? What would they look like?

Hybrid orbitals do not exist. Individual atoms have electronic configurations which can be explained by considering atomic orbitals. Molecules have electronic configurations which can be explained by considering molecular orbitals. Hybrid orbitals are just one mathematical way of arriving at molecular orbitals from combining atomic orbitals. They are a pure mathematical fiction and have no physical meaning (this does not mean they aren't useful).

Some schools or textbooks teach that in order for a reaction to occur energy is required to 'promote' electrons into hybrid orbitals to create some sort of excited atom prior to bonding. This does not happen in real life and is not a useful way of looking at bonding. What does happen is that as atoms approach each other, their atomic orbitals start to overlap significantly, giving rise to a new set of molecular orbitals (in quantum mechanics you are changing the potential in which the electrons reside and so the allowed energy states change). If the electrons in the new molecular orbitals are lower in energy than in the atomic orbitals then the reaction will proceed.

As for the electron distribution of atoms - it is spherically symmetric. The two $\ce{2p}$ electrons will reside in different orbitals because this does not require pairing of their spins (which is a higher energy state). However, it is meaningless to ask which $\ce{p}$ orbital they are in because all three orbitals are degenerate and so are indistinguishable.


ChemDraw: how to change the default aromatic ring style for drawing from SMILES

From pp. 91–92 ChemDraw 20.0 User Guide, PDF version shipped with the ChemOffice suite:

Toggle Aromatic Display
An aromatic structure can be displayed either as an alternating double and single bond or as an aromatic circle. Toggle Aromatic Display option lets you change an aromatic structure from Alternating double/single bond to Aromatic circle
and vice versa. To do so:

Select the structure.
Go to Structure → Toggle Aromatic Display.

When opening a file of non native chemical format or pasting as SMILES, you can now define the way aromatic structures are displayed.
To do so:

Go to File → Preferences. The Preferences Dialog appears.

Click Open/Save tab.
Click the Normalize Aromatic Display checkbox and select either Alternating Double/Single Bond or Aromatic Cirlces as desired.
Click Apply and then click OK. All aromatic structures in the file loaded or the SMILES string pasted will be of the defined representation.


Is the carbon in a carbonyl functional group electropositive?

Electronegativity and electropositivity are usually used in a relative sense, X is more electronegative than Y.  The carbonyl carbon in acetone is more electropositive than the central carbon atom in isobutane, and less electropositive than the methyl carbocation carbon.

  because he defined electronegativity earlier as the ability to
  stabilize negative charge

Weak definition, wouldn't you say; that's why you won that argument.  Atoms can be electronegative in the abstract, charge doesn't need to be involved.  The statement, "fluorine is more electronegative than carbon according to the Pauling Electronegativty scale" is accurate and understandable, yet no one is describing a situation where charge stabilization is involved.

  because electrons cannot stabilize other electrons.

I understand what you mean, but you should know by now that you can't make such sweeping statements.  Don't the electrons in one of the double bonds in 1,3-butadiene interact with and stabilize the electrons in the other double bond?

  Can we describe the electron deficient carbon as electronegative

Again, I understand what you mean, but... Yes, it is more electronegative than the carbon in an alkane, but not as electronegative as $\ce{F+}$.

  Is electronegativity the ability to stabilize negative charge or the
  ability to attract electrons within a covalent framework? I guess the
  two go hand in hand - i.e. you wouldn't attract electrons if you
  weren't able to stabilize them

Yes, the two go hand in hand.  More electronegative atoms are better able to stabilize negative charge than less electronegative atoms.  It's all about lowering the overall system energy.

  As a side question: can we calculate EN for ions?

I'm sure there is a way.  For example, using studying the $\ce{^{13}C}$-nmr spectrum of various carbocations leads to a relationship of (somewhere around) 180 ppm shift/ electron.  So you can estimate the fractional positive charge on any carbon atom, the carbonyl carbon in acetone for example.  Once you've estimated the amount of charge on an atom, I suspect there is a way to relate it to EN.


If correctly prepared, the pH should have been the same. Logically, consider if the pH would be different if measured in 5 separate 100 ml samples poured from the 500 ml preparation.

Why might the pH differ in your tests? Some possibilities:

One container was not completely rinsed after cleaning and had some (often alkaline) detergent clinging to it.
The preparation absorbs $\ce{CO2}$ from the air, so the layer on top is more acidic.
Error in measurement.

Unless you have a buffered system, pH may vary abruptly with a slight change in ingredients.


The main issue is that the Al needs to remove its hydride. With a carboxylic acid and/or an aldehyde, it can stick its hydride onto the carbonyl carbon without issue. But the carbon bonded to the alcohol cannot take on a hydride.


Do I understand why heat capacity is called 'heat *capacity*'?

Expanding on Nicolau Saker Neto's comments above, the heat capacity dictates the (reversible) heat flux required to increase the system's temperature by one unit.

Definitions of the heat capacity ($C_V$ and $C_p$)

Another way to look at it is that the heat capacity dictates the response in internal energy ($U$)/enthalpy ($H$) of the system to temperature, as described below. Note that the heat capacity at constant volume ($C_V$) is not the same as the heat capacity at constant pressure ($C_p$) - and that both may be defined in several ways, the most common being:

$$ C_V = \left(\frac{\partial U}{\partial T}\right)_{V,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{V,\mathbf{n}} $$

$$ C_p = \left(\frac{\partial H}{\partial T}\right)_{p,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{p,\mathbf{n}} $$

Contributions to the heat capacity

In should be added that while translation - the kinetic energy of molecules in motion - are an important contribution to the heat capacity of a gas, both vibrations and rotations also contribute for all but the simplest molecules. For high temperatures (approx. $T \geq 3000\ \mathrm{K}) $, the excitation of outer electrons will also contribute to the heat capacity. For even higher temperatures, the substance in question will behave as plasma.

In short

There are many complicating factors to the precise understanding of the origin of contributions to the heat capacity - both at the high end and low end of the temperature range. In all cases, however, the heat capacity will give you the answer to the question:

  "How much energy do I have to add to this system at the given
  conditions in order to increase the temperature with one unit?"

Further reading on Wikipedia is recommended for anyone interested.


pH Testing Without Cabbage/Disposal of Ferric Chloride

You can add powdered calcium carbonate (limestone) until the liquid component was clear upon settling or until the addition of calcium carbonate did not produce additional bubbling. Calcium carbonate is ideal because it is cheap, neutralizes acids to a non-corrosive pH with no need to measure, and is not itself caustic.


How is the crystal direction important for etching silicon surfaces?

Certain types of wet etching agents have differential reactivity on different crystal surfaces. Consider that the Si-Si distance and coordination environment will be quite different.

Let's start with an isotropic etch - it just eats silicon. Well, if you started with a pinprick (i.e., just a point and not a gap in the mask) you'd etch out a hemisphere. Normally, we etch through gaps in the mask, so you'll get rounded sides.

With an anisotropic wet etch you can get flat, angular side-walls defined by crystal planes.

  For instance, potassium hydroxide (KOH) displays an etch rate selectivity 400 times higher in <100> crystal directions than in <111> directions.

OK, so let's consider that etch through a gap in a mask into a <100> surface. As we go down 400 atomic layers of Si, we'll clearly have some etching along the side-walls (e.g., an "undercut"). But also consider that depending on the size of the feature for the etch, you'll go down into the <100> but not completely uniformly. Different wet anisotropic etches will have different selectivity rates between <100> and <111> and thus you get different aspect ratios.

Incidentally, dry etching (e.g., RIE, plasma) can give more vertical side-walls.


In acidic solution, what does ClO₃⁻ reduce into?

The first equation is correct because $\ce{ClO3-}$ is reduced to $\ce{Cl-}$ and oxygen maintains its -II oxidation state in both educts and products. As chlorine changes its oxidation state from +V in chlorate to -I in chloride, $6 \ \mathrm{e^-}$ are required for this reduction.

In your second equation, chlorate is reduced to chloride as well; however, only $2 \ \mathrm{e^-}$ are explicitly supplied. The remaining electrons would have to come from the concurrent oxidation of oxygen from -II in $\ce{ClO3-}$ to 0 in $\ce{O2}$, and this is unlikely to happen, as oxygen is more electronegative than chlorine.


The stability of the cupric and cuprous ions

Hund's rules, which are where we get the idea that half filled and completely filled orbitals are more stable than partially filled orbitals, are specifically about the ground states of atoms. Hund's rules do not apply to ions in solution which are complexed with ligands (like water).

What you fail to consider is that $\ce{Cu+}$ is larger than $\ce{Cu^{2+}}$ but with only half the charge, so its charge density is much lower. As a result, so its enthalpy of hydration for $\ce{Cu+}$ is much less, meaning it is much less stabilized by the formation of the metal-ligand bonds. As a result, in aqueous conditions $\ce{Cu+}$ tends to disproportionate to $\ce{Cu}$ and $\ce{Cu^{2+}}$.