Vibrational state dependency of the reaction cross section

Vibrations are very important for reactions. In the particular example of $\ce{H2+(g) + He}$

vibrational levels 0-3 have an energy less than E0 (minimum energy for reaction), so additional transitional energy is needed to induce reaction. That’s why we see a threshold.
vibrational levels > 4 already have more than E0 energy so they react even with no additional energy

The greater the vibrational energy level of $\ce{H2+}$ the easier it will be for the substitution reaction, and this makes sense considering the transition state diagram

You may also be interested in the "harpoon" reaction which demonstrates how vibrations are important 1.
This reaction is known as a harpoon reaction because an electron is first transferred from M in $\ce{M...XR}$ or $\ce{[M...XR]^*}$ to $\ce{RX}$ and thereby forms the ionic transition state $\ce{M+...RX-}$. The negatively charged $\ce{RX-}$ is unstable and can dissociate, forming the products $\ce{M+X-}$ and $\ce{R}$.
These type of reactions are, in general, valuable for studying how rovibrational states affect the reaction probabilities because they can be triggered by light, and are small enough to model with extremely accurate theories.
In particular, the reaction probability has notable "resonances" spaced precisely 10.9 $cm^{-1}$ apart. These resonances are most likely due to low frequency bending mode of the T-shaped complex.

References:

Skowronek, S., Jiménez, J. B. & González Ureña, A. Resonances in the Ba...FCH3+hv→BaF+CH3 reaction probability. J. Chem. Phys. 111, 460–463 (1999).


Stoichiometry Chemistry Lab: Copper (II) Sulphate Pentahydrate and Iron Single Displacement Reaction?

Your results are actually very good.  The reason for thoroughly washing your copper is to remove any residual water-soluble un-reacted reagents.  If the copper were simply dried by sitting in room-air, the bulk of your extra 6% could still be water.  That's only 70 mg or 70 ul of water.  They way you produced your copper would likely result in a high-surface area product which can be tricky to dry.  Ideally, you would do multiple rinses in hot DI water then dry overnight at 105 C (really 70+C should do it).  If your yield goes down just a couple percent or so, you can still verify the excess water theory by repeating the drying.  If your yield goes down again then you are just having trouble getting all the water out.  Make sure you cool the copper, in a desiccator, before weighing.

If you don't have lab time for this here's what I'd try: take you copper home and put it an oven on low, about 200F or so, for several hours.  While still hot, place in the smallest reasonably air-tight glass jar you can find an get it back to the balance ASAP!  Good luck!  Oh, and NEVER touch the copper with your sweaty, oily fingers ;-)


The need for excess acid in dehydration of alcohols

Alcohols generally are very poor acids and have pka's in the range of 16-17.

Hence, if the medium is only slightly acidic, only some alcohols will protonate, after which the unprotonated oxygen (of ROH) will attack the generated carbocation, following which the H+ ion will leave making H+ act as a catalyst, resulting in the formation of ether.

But, if the solution is highly acidic, we would be generating the carbocation in full and there would be no nucleophile left to attack the carbocation.

We need to maintain the optimum pH as substitution reactions are generally favoured over elimination reactions especially in the case of primary and secondary alcohols.


There is a program called mapman that can convert between different formats. The .xplor format is a text-based one, the others are all binary, so that's the one you need. The program runs on linux and mac platforms.
Other crystallographic software suites (ccp4, etc) might have similar capabilities, but this one might be the easiest to install (standalone program).

I want to find the electron density around two atoms within a protein.

This is much easier with a molecular browser. You could just set the contour levels to the ones your are interested in, and take a look. If you need to do this programmatically for thousands of structures though, that will not help you.
It will be kind of tricky to find the right spot in the density, though, because they use fractional coordinates (makes generation of symmetry much easier) and your coordinates are in orthogonal coordinates. If this is a problem, just ask another question.
Here is the description of how data is organized in an .xplor map:

Following the map header, the density matrix is then written section-by-section with c moving slowest (in z-sections). Each section of the density map is preceded by a section number.
Thus, for the formatted map file each section of the density map is written using FORTRAN statements of the type
WRITE(OUNIT,'(I8)') KSECT
WRITE(OUNIT,'(6E12.5)') ((SECTON(I,J),I=1,ISECT),J=1,JSECT)
and the resulting map is written with six pixels on each line. The binary format is identical except the format statements are missing, so that each line that is written contains the entire length of map along the `fast' (a-axis) direction.


How does 1-bromo-2-fluorobenzene react with lithium amalgam?

Quoting from Page 1063 of Advanced Organic Chemistry FIFTH EDITION, Part B -
Reactions and Synthesis, FRANCIS A. CAREY, and RICHARD J. SUNDBERG:

  Benzyne can also be generated from o-dihaloaromatics. Reaction with lithium amalgam or magnesium results in the formation of transient organometallic compounds that decompose with elimination of lithium halide. o-Fluorobromobenzene is the usual starting material in this procedure.

As @orthocresol said in the comments that Benzyne undergoes Diels-Alder reaction with furan.

The first Diels–Alder reaction of benzyne with furan was performed before giving a 76% yield of the adduct, which was converted to 1-naphthol with acid${^1}$ ${^2}$.

From Diels-Alder adduct to 1-naphthol

In the absence of the furan trapping agent, biphenylene and triphenylene were obtained in yields of 24 and 3% respectively${^3}$.

References

${^1}$:https://www.researchgate.net/publication/273556121_The_Benzyne_Story

${^2}$ :Shashidher et al., Pharm Anal Acta 2011, 2:7
DOI: 10.4172/2153-2435.1000137

${^3}$ :https://www.researchgate.net/publication/273556121_The_Benzyne_Story


No, it is not toxic to touch solid lead.  Lead poisoning results from

ingestion: paints used to contain lead-based materials, and kids would sometimes eat the peeling paint leading to health problems.
inhalation: if you were grinding or polishing a block of lead and fine particles were generated, you might inhale them if you weren't using the proper protective equipment and health problems could develop.
dermal contact: if you were working with organic lead compounds that were easily absorbed through the skin, serious health problems would result if you weren't using the proper protective equipment.  Gasoline used to contain lead compounds, but they were removed from gasoline in order to prevent problems.

Being around or touching solid lead would not lead to ingestion, inhalation or dermal contact issues.

Here's a link to Wikipedia article on lead, see the section on "Health effects" if you'd like to read more or read about the biochemical mechanism of action.


Indeed, $\ce{XeF6}$ has a distorted tetrahedral structure in all three phases. Two of solid state crystal structure modifications ($\bf{A}$ and $\bf{B}$) are shown in following diagram (Ref.1):

The abstract of Ref.1 tells them all:

  According to single crystal X-ray diffraction, neutron powder diffraction, solid state MAS NMR data, and differential scanning calorimetry, $\ce{XeF6}$ exists in at least six different modifications. Three of them are formed at temperatures above room temperature, one exists at room temperature, while two have been found at low temperatures. In the high temperature modifications $\ce{XeF6}$ forms a non-symmetric tetramer, better described as a cyclic trimer with a weakly associated monomer. The normal temperature modification is the previously described cubic phase IV, having disordered tetrameric and hexameric units. The low temperature modifications are regular tetramers. Only in presence of $\ce{HF}$ symmetric dimers are formed.

The tetramer $\bf{C}$ (ball and stick version of $\bf{A}$) represent the general version of all modifications, a cyclic trimer with a weakly associated monomer. The symmetric dimer $\bf{D}$ is the structure of $\ce{XeF6}$ in the presence of $\ce{HF}$ (highlighted by $\color{green}{\text{green}}$ ovals). Nonetheless, all of these modifications display distorted octahedral structure for individual $\ce{XeF6}$ molecule as represented by ball & stick 3D structure in the central box. None of $\ce{F-Xe-F}$ angles are equal to $90^\circ$, which is the normal angle of central atom with two adjacent ligands in regular octahedron. As a consequence, $\ce{XeF6}$ has net permanent dipole moment since all six $\ce{Xe^\delta+-F^\delta-}$ bonds are polar.

For your question about ammonia, look at Ivan Neretin's comment and here where there is an accepted answer.

References:

Sevim Hoyer, Thomas Emmler, Konrad Seppelt, “The structure of xenon hexafluoride in the solid state,” Journal of Fluorine Chemistry 2006, 127(10), 1415-1422 (https://doi.org/10.1016/j.jfluchem.2006.04.014).


Can Aquatabs be used to clean swimming pool water?

Aquatabs can certainly be used as a disinfectant in a small swimming pool like you've described. The active ingredient has many uses, including drinking water treatment as you've pointed out, but it is also used as a disinfectant in swimming pools.

According to this Wikipedia article regarding the active ingredient of Aquatabs, sodium dichloroisocyanurate:

  As a disinfectant, it is used to sterilize drinking water, swimming pools, tableware and air, fight against infectious diseases as routine disinfection.

A study was done[1] comparing the effectiveness of sodium hypochlorite (common household bleach, also used for pool chlorination) with sodium dichloroisocyanurate. From the abstract:

  A comparison of commercial sodium hypochlorite (NaOCl) and sodium dichloroisocyanurate (NaDCC) products was made. Solutions of NaOCl and NaDCC containing the same levels of available chlorine (av.Cl) exhibited very similar bactericidal activities, despite significant differences in pH.

The ideal level of active chlorine for a swimming pool at neutral pH is about 0.5 to 1.5 mg of active chlorine per liter of pool water. So, if you aim for the middle of this range, you would want to add one Aquatab tablet for every 300 liters of water in your pool. Also, note that chlorine is less than half as effective at a pH of 8 as it is at a pH of 7, so I would pay some attention to the pH anyway.

D. Coates, J. Hosp. Infect. 1985, 6 (1),31-40. A comparison of sodium hypochlorite and sodium dichloroisocyanurate products. DOI: 10.1016/S0195-6701(85)80015-3


Difference between aldehydes and ketones

Ketones are to aldehydes as ethers are to alcohols. You could extend this to water. Is water formalcohol? Once you begin to explore the periodic table deeply enough, you will see that there is no distinctions really exist. Only trends. The same is true of organic structure. Organic chemistry is a continuum and the groups we create are constructs -- but they are not arbitrary. Especially in organic chemistry, we group things into these categories because they help us to make assumptions based on consistent trends in reactivity. Ketones and aldehydes have significantly different chemistry. There is only a minor difference between the chemistry of acetone and butanone, but there is a pretty big difference between the chemistry of acetone and acetaldehyde.

To make a long story short, aldehydes are categorized differently because their chemistry varies significantly enough that different sorts of reagents are needed for the same types of reactions.

Aldehydes are much more electrophilic. This is because they don't have the electron donating effect of an alkyl group. Aldehydes have much more acidic alpha-protons. Aldehydes are also more easily oxidized. Aldehydes pose problems different problems in synthesis. For Their non-acidic proton, for example can be hard to remove for substitution.

An interesting example of a useful tool for dealing with aldehyde protons is dithiane chemistry. You can use a dithiane protecting group to make the aldehyde proton acidic, substitute it, then put the oxygen back on in a deprotection step.


Don't worry too much about it, this educational exercise will be over when the exam is over. In real life nobody can predict the melting point or solubility in water just by looking at the structure of a given molecule. There are software like ACD, ChemDraw and SciFinder database which have predicted melting points and boiling points of most known compounds but those values are often off. Sometimes the values are correct. This is why only the experimental values of are considered reliable. Maybe in future, our understanding will become so much better that melting/boiling points and solubilities in water would be accurately predictable.
All your teacher wants is to remember some general trends and periodic properties, which should be in your mind (in other words memorize, just like 2+2 is always 4). Take the example of halogens (diatomic molecules). As you move down the group the melting point increases. Fluorine has the lowest melting point and iodine has the highest.
In the alkali metals, the melting point decreases down the group. All you have to keep in mind the melting point trends down in the groups as well as across the periodic table for simple molecules. Similarly, look for the trends in alkali halides, hydrogen halides, other hydrides of the oxygen group.
Read this article on Is there a trend in melting points on the periodic table? https://antoine.frostburg.edu/chem/senese/101/periodic/faq/melting-point-trend.shtml


Europium (II) is an anomaly because it is, well, "II".  Most metals in period 4 or higher, apart from the alkali and alkaline earth metals, favor higher oxidation states which lead to highly insoluble hydroxides or oxides.  Europium is different because the $+2$ ion is stabilized by having a spherically symmetric electron configuration $[\ce{Xe}]\mathrm{4f^7}$.

So we have a $+2$ ion similar in size and spherically symmetric electron configuration to the heavier alkaline earth metals, and $\ce{Eu(II)}$ compound tend to resemble their alkaline earth counterparts -- including the solubility of the hydroxide.


Can atoms delocalize over multiple equally probably bonding structures in certain cases?

If it happens at all, it exceedingly rare. Through my 10-years work in chemistry, I didn't once found a case where it may be relevant. However, what does and indeed affect chemistry is reversible rearrangements.

The simplest case is rotation of molecules around single bonds. Another one is nitrogen inversion. More complex cases also exist. Such things sometimes make interpretation of NMR quite difficult. NMR gives a peak for each non-equal nucleus in the molecule. When internal rearrangement is involved, it sometimes makes two atoms 'equal', i.e. exchanging places too fast for NMR to catch the difference and producing a single peak instead. This, however, is 'quasi-delocalisation', since there is no need in resorting to delocalisation to explain said phenomena.

Another quantum phenomena often observed is tunneling. It affects slightly effective reaction rates/barriers and vibrational spectra.

Anyway, de Broglie wave length for proton is thousand times smaller than for electrons, so quantum effects in nucleus movement usually overshadowed by quantum effects in electrons movement.


Why is Kc not affected by change in pressure?

Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration (seen in the rate laws), temperature, activation energy, and how often molecules collide with the correct orientation (together these last three determine the rate constant, $k$, as shown in the Arrhenius equation).

The equilibrium constant, $K_c$ is the ratio of the rate constants, so only variables that affect the rate constants can affect $K_c$. Pressure doesn't show in any of these relationships.

$$K_c=\frac{k_\text{forward}}{k_\text{reverse}}=\frac{[\text{products}]^y}{[\text{reactants}]^x}$$

$$\text{rate}_\text{forward}=k_\text{forward}[\text{reactants}]^x$$

$$\text{rate}_\text{reverse}=k_\text{reverse}[\text{products}]^y$$

$$k=A\mathrm e^{-\frac{E_\mathrm a}{RT}}$$

It may also be useful to think about different ways pressure can be changed.

The volume of the reaction can be changed. This will affect concentrations of reactants and products, and the reaction will likely have to shift left or right to reestablish equilibrium. If volume were decreased, the rates of the forward and reverse reaction will each increase due to higher concentration. If the exponents in the rate laws are different, the amount of increase for each will be different. Although the rates change, the rate constants do not, so $K_c$ doesn't change either.
The temperature can be changed. This will affect the rate constants of the forward and reverse reactions according to the Arrhenius equation. An increase in temperature will increase $k_\text{forward}$ and $k_\text{reverse}$, but by different amounts because their activation energies are rarely equal. This changes $K_c$, which is the ratio of the two rate constants.
More gas can be added without changing the volume of the container.

a. If the gas is one of the reactants or products, this would affect one of the concentrations, and the reaction will have to shift to reestablish equilibrium. Again, this changes one of the rates, but does not affect the rate constants, so $K_c$ is unaffected.

b. If the gas is inert and doesn't react with anything involved in the equilibrium, the concentration of each reactant is unchanged. Neither the rates nor the rate constants are affected, so the system should still be at equilibrium and $K_c$ is unaffected.


Does 2-chloro-1,3-dinitrobenzene show resonance?

2-chloro-1,3-dinitrobenzene is shown below:

I believe your question is referring to the lack of conjugation between the nitro groups and the aromatic ring. this is best shown in an X-ray crystal structure such as this one:

https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=722415

Due to the sterics between the nitro groups and the chloro substituent, the nitro groups cannot adopt a conformation planar with the ring. This prevents the $\pi$ system of the nitro group from achieving substantial overlap with the $\pi$ system of the ring and prevents full conjugation of the electrons. The sterics are better seen in a spacefilling view:

So, it is not the chloro group that is out of the plane, but rather the nitro groups.


Is it safe to mix methylated spirits with isopropyl alcohol?

Yes, as far chemical safety is the concern. No dangerous reactions nor dangerous phenomena will occur. The mixture, a solution to be more precise, will retain at least qualitatively the properties of the two components, eg will be a disinfectant, a flammable liquid, a solvent, and so on.
Flammable materials are better stored in separated containers in separated places.
For the case in question, going camping with two small bottles or one double sized bottle in the backpack won't make any real difference.


What is the IUPAC systematic name for methane?

The latest edition of the Blue Book, the 2013 Revision, clarifies this:

  Methane is a retained name (see P-12.3) that is preferred to the systematic name ‘carbane’, a name never recommended to replace methane, but used to derive the names ‘carbene’ and carbyne for the radicals $\ce{H2C^2-}$ and $\ce{HC^3-}$, respectively.

So it seems to be the case that the systematic IUPAC name is only there for consistency with the ions, and is not recommended (and not used) to describe methane. Similarly, "ethane", "propane", and "butane" are preferred to their systematic equivalents "dicarbane", "tricarbane", and "tetracarbane".


Terminal alkynes are weak acids, but they are still stronger ones than alkenes and alkanes. This actually means they would readily react with conjugate bases of weaker acids, giving them the proton. So yes, since ethyne is a stronger acid than ethane (and ethyl anion is a stronger base than ethynyl anion - the list for bases is reverse due to obvious reasons), they will exchange H and Li.

Below you can see a mechanism I would use to describe the problem in your question. C-Li bond has significant contribution of both covalent and ionic components, so I wrote it as both in equilibrium:


Formation of Calcite from Calcium carbonate

It's my understanding that the production of the calcite takes a decent amount of time as there are 2 stages being undertaken for the crystallization of the amorphous calcium carbonate (ACC). The first of these is relatively fast whilst the second stage is the one that you want to be looking at as it is roughly 10 times slower. This is the stage which looks at a dissolution and precipitation mechanism.


What exactly is a potential energy surface?

The PES $V({R})$ (where $R$ are the nuclear coordinates, electronic coordinates are usually denoted by $r$) is the electronic energy of any electronic state. So there is one PES for the ground state and many more all the excited states.

  Clearly, $E(\{ R_{\text{nucleus}}\})$ gives a PES, and this PES is the relation:
  $$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{ground state electronic energy}}$$
  However, when I learned about molecular dynamics (also in [wikipedia][1]), a PES is defined to be the relation
  $$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{energy of the system}}$$

"Energy of the system" is here referring to the "electronic system", so those two relations are the same.

  How do we find this map in practice?

Your first step is correct, we need to solve for the eigenvalues of the electronic Hamiltonian, for many different parameters $R$. Each electronic eigenvalue is one point on the PES. So in the end we get a discretized representation of the (actually continuous) PES $V(R)$.
The second step is to solve the nuclear Hamiltonian
\begin{equation}
  \hat H_{\mathrm{nuc}} = \hat T_{\mathrm{nuc}} + V(R)
\end{equation}
There is only one such nuclear Hamiltonian for each PES.

So we need to do many times step 1, but only once step 2 (for each electronic state of interest).

How to solve the nuclear Hamiltonian?

This depends on what $V(R)$ looks like. In the simple case of a diatomic molecule, where $R$ becomes just the 1-dimensional internuclear distance, we may approximate it with a Morse potential or harmonic potential. We can fit the parameters of such analytic expressions to our discretized points from step 1 and then use the existing general solutions for those quantum systems. This will give us the vibrational eigenvalues and eigenfunctions.

Another option is to solve the nuclear Hamiltonian on a numeric grid over $R$. The grid is either defined by the points chosen in step 1, or we can interpolate to a new grid. How to solve this is a topic on its own, so will just give one literature reference here: The book "Introduction to Quantum Mechanics - A time-dependent Perspective" by David Tannor covers many things related to your questions (although it can be challenging to read it).


If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.

Well … you get an average oxidation state. This calculation arguably implicitly assumes that all nitrogen atoms be equivalent. In some cases (e.g. hydrazine) they are and the result you get algebraicly is as you would expect from a Lewis depiction. In other cases (e.g. here) this is not the case as the nitrogens are not equivalent (only one is bound to oxygen). Thus, it is obvious that a simple algebraic approach should fail.
But what about the ‘real’ result? What about the resonance structures? Well, this is where things get really difficult. You essentially have two π systems orthogonal to each other and each occupied by four electrons which can manifest as a lone pair on either end and a multiple bond to the other atom. If you really wanted to play the game, you could include another resonance structure as shown below, where all π lone pairs are centred on the terminal nitrogen:
$$\ce{N#\overset{+}{N}-\overset{-}{O} <-> \overset{-}{N}=\overset{+}{N}=O <-> \overset{2-}{N}-\overset{+}{N}#\overset{+}{O}}$$
(It is clear that this third resonance structure contributes least to the overall picture as it has a greater charge separation and charges are separated opposite to what electronegativity would predict.)
The experimental structure shows that the $\ce{N-N}$ distance is slightly shorter than the $\ce{N-O}$ distance which one could use to assume that the $\ce{N-N}$ bond have a slightly larger bond order than the $\ce{N-O}$ bond. But ultimately, they are still very similar (the difference is merely $\pu{4pm}$) so equal bond orders might also be an option. Long story short: short of calculating the electronic distribution (i.e. solving the Schrödinger equation) you won’t be able to arrive at a definite answer for the ‘real’ oxidation states.
So what can you do on paper? Comparing the resonance structures, the leftmost as I have ordered them is slightly better than the central one as the formal charges are distributed according to the different electronegativities. Thus, I would be inclined to give it a slightly higher weight and – in a classroom setting – use it to determine oxidation states. That said, it is too much of an ambiguous example to be seriously used in any examination unless the goal of said examination is to develop the chain of arguments that formed this answer.


Is enthalpy of activation the same as activation energy? Is there a purely mathematical way to find that value?

(1) Enthalpy of activation is a component of activation energy. The other component is temperature * entropy of activation. Enthropy controls "heat difference" while entropy controls "measure of disorder difference".

$\Delta$G = $\Delta$H - T*$\Delta$S. This applies to activation energy too.

(2) DFT functionals rely on 4 parameters per atom make predictions. So, if you give 4*5=20 parameters, then you can in theory predict behaviour of any compound, build with only C, H, N, O, S atoms. This covers most of biochemistry, excluding nucleic acids (you would need additional 4 parameters to characterize P).

These 4 parameters are known for most of the atoms (excluding some elements, going after uranium). But those four parameters were experimentally measured by fitting behavior of modeled molecules to real molecules.

Should we count it as "purely theoretically predict?"

Also there are purely ab initio methods like Hartree-Fock with even less input required. You would still need some input. In math you can build many different plausible worlds. For example purely "newtonian" vs "newtonian + quantum mechanics" words. experiment is the only way to tell, which model is correct.


A Clemmensen reduction of a 1,3-diketone probably isn't the best idea. The reduction is likely to proceed via a ketyl radical.

In the case of a 1,3-diketone, formation of a dihydroxy cyclopropane, which undergoes opening/rearrangement cannot be excluded.

This reaction has for instance been observed for dimedone, where 2,4,4-trimethyl cyclopentanone was formed.

EDIT 1

Which other viable procedures exist?

If you insist on using the given starting material, the following might (!) work:

regioselective (asymmetric) catalytic hydrogenation of the acetophenone-type carbonyl using a 1,2-diamine-substituted Noyori catalyst
Barton-McCombie reaction of the resulting alcohol

If you are rather interested in the product, you might want to rethink the strategy:

The intended product is an arylbutanone, related to the raspberry ketone. A possible retrosynthesis would be:

If you get your hands on the necessary benzaldehyde, an aldol reaction with acetone, followed by catalytic hydrogenation might give the right product.

EDIT 2

5-tert.-butyl-2-hydroxybenzaldehyde is commerically available, but not really cheap.

However, 4-tert.-butylphenol is cheap as dirt. In case you need significant quantities of your final product, you might want to consider the synthesis of the benzaldehyde. The substitution pattern cries for a Reimer-Tiemann reaction.


Why is tetraamminecopper(II) a square planar and not a tetrahedral species?

It is very convenient to use crystal field theory to discuss this.

It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy.

The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2-y^2}$. However, if these two electrons occupy $d_{z^2}$, two corresponding opposite ligands become well shielded and leave. This is the common case of $d^8$ complexes, such as $\ce{Ni^{+2}}$, $\ce{Pd^{+2}}$, $\ce{Pt^{+2}}$, $\ce{Rh^{+1}}$, $\ce{Cu^{+3}}$ etc.

Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $\ce{Cu^{2+}}$ is large enough to bond four charged ligands if available, such as in $\ce{[CuCl4]^{2-}}$, but the ligands are bound weakly and easily dissociate.

However, $\ce{Cu^{+2}}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the $\ce{NH3}$ complex is written with only four molecules; the two other are so weakly bound.

This type of distortion is a case of Jahn–Teller distortion, very common in chemistry.


Relation between freezing point depression and vapor pressure

A freezing point is a temperature at which solid and liquid phases are in equilibrium with each other. In equilibrium the chemical potentials of each component in all phases present are equal. It means that if there are three phases in the system, solid, liquid and gas, then the chemical potential should be equal in all the three. As a consequence, the vapor pressure of a given component over the liquid phase should equal that over the solid phase. But if there are only solid and liquid phases, I see no need to involve gas. The basic derivation, which I was taught, of the freezing point depression of liquid solvent A (when the solute is insoluble in solid A) starts with $$\mu_\mathrm{A}(\mathrm{solid})=\mu_\mathrm{A}(\mathrm{solution})=\mu_\mathrm{A}(\mathrm{pure\,liquid})+RT\ln{x_\mathrm{A}}$$ - no gases here.

I might miss something here, but my guess is that your lab manual is just trying to be more obscure than it ought to be (no offence to its authors intended). While the expression for $\mu_\mathrm{A}(\mathrm{solution})$ itself can be derived from the Raoult's law, which requires gases, that's slightly different, although related, story. Provided that the expression for $\mu_\mathrm{A}(\mathrm{solution})$ is already known, I can see no sound reason for inviting the vapor pressures to explain cryoscopy.


Increase in Potential Energy When Bonds Are Broken

First, changing a solid into a liquid does not involve breaking bonds in the normal sense of the word, i.e. covalent bonds. Mostly intermolecular interactions are broken as heat is put in to melt something. These may be hydrogen bonds e.g. water, $\pi-\pi$ interaction e.g. benzene or dispersion as in cyclohexane and similar non polar but (slightly) polarisable molecules.

Starting with free atoms, when these are brought together and a molecule is formed then energy is released to the surroundings and a stable molecule is formed, say, H$_2$, or HCl etc. The figure shows the potential energy plotted vs atom separation. The energy is zero when the atoms are separated at infinity; in reality about 10 bond lengths is equivalent to infinity.

To dissociate the molecules, potential energy has to be added as shown by the dissociation energy. Any energy above this is released as kinetic energy of the atoms.

The heats of formation of different types of molecules have a wide range of values. Ignoring vibrational energy, which is small compared to bond enthalpy, this is in effect moving from the total minima in the potential wells of reactants (of the whole molecule not just a bond) to the minima in the products. This may involve absorption or release of heat to the surroundings and so may involve gain or release of potential energy overall. It just depends what the reaction is. (In calculating the Free energy changes entropy changes have to be considered also.)


Writing equilibrium expressions with aqueous species

I will say the latter form is more correct, and in fact an important way to consider equilibrium with solubility and the common ion effect.  You should not count any solids in equilibrium expressions though as they do not determine the final equilibrium.


What does the "N" in N-nitrosoamine mean/stand for?

The "N" means that the nitroso group ($\ce{-NO}$) is attached to the amine nitrogen.

So, for example, in N-nitrosomorpholine the nitroso group is attached to the nitrogen atom of the saturated, heterocyclic morpholine ring:

You could never have an "N-nitroso" group attached to a plain benzene ring, since there is no nitrogen atom in a benzene ring for the "N-" prefix to refer to.

In general, though, you would name aromatic nitroso compounds in the same way as you would those with any other substituent like 'chloro-' or 'nitro-'. For example, the following compound would is named nitrosobenzene:


Order of internal pressure resistance to expect from laboratory glassware?

Put the flask in a metal container, with a large (at least three times the volume of the flask, for 3 atm pressure) bladder or sylphon (metal bellows) in the supply to the inner flask. Pressurize the outer metal container with air or inert gas.

The absolute pressure in the reaction vessel should be the same as the outer vessel (unless the reactants outgas rapidly), but there would be no differential pressure difference, ergo, no additional stress on the glass.

Or just conduct the whole experiment in a hyperbaric chamber... not a wimpy 0.1 atm plastic bubble, but something like this 10 atm unit, with margin for error. ;-)


(a) Determine the symmetries (irreducible components) of the MOs responsible for the d-d metal-metal quadruple bonding in this molecule

I'll try to walk through how it works for $d_{xy}$. You will need to consider $d_{yz}$ and $d_{xz}$ as a pair later when you attempt those.

Obviously $\chi(E)=2$. Now for the $C_4$ rotation, just think about rotating one of the orbitals in the $xy$ plane (the same thing will happen to the other orbital, so we will then double the character). When we do this, the orbital remains in place, but it's positive and negative lobes invert. So it will contribute -1 and for both orbitals $\chi(C_4)=-2$. For $C_2$, the rotation again leaves the orbital in place and leaves the lobes in each location with the same sign so $\chi(C_2)=2$.

$\chi(C_2')=\chi(C_2'')=\chi(i)=\chi(S_4)=\chi(\sigma_h)=0$ since all of these operations involve a rotation/reflection around a parallel line/plane so they will move the orbitals.

$\sigma_v$ runs along the $\ce{Re-Cl}$ bonds. Along and on either side of this plane, the sign of the lobes is the same and so $\chi(\sigma_v)=2$. $\sigma_d$ runs between the $\ce{Re-Cl}$ bonds and so this plane inverts the sign and $\chi(\sigma_d)=-2$.

Plugging these into a calculator I get a $b_{1g}$ and a $b_{2u}$ irrep. I will leave it to you to figure out which of these is the bonding combination and which is antibonding.


Why does potassium permanganate turn colourless when reacted with sulfur dioxide?

This is a redox reaction. The permanganate ion (purple) is reduced to manganese(II) ion which is a very pale pink, or almost colourless.

$$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$

while sulfur dioxide is oxidized to the sulfate ion, which is colourless.

$$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}$$

So, the solution is decolourised. The overall equation is

$$\ce{2MnO4- + 5SO2 + 2H2O -> 2Mn^2+ + 5SO4^2- + 4 H+}$$


When two German chemists H.O. Wieland and A.O.R. Windaus were awarded Nobel prize in two consecutive years 1927 and 1928 for their research of bile acids and sterols, despite the enormous effort of structure determination, their alleged cholesterol structure was not yet correct at a time, but already had similar numbering:[1,2]

It's difficult to answer why this one exactly was chosen. But see, here you have counter-clockwise numbering in first three rings, however clockwise in the last one, and with methyl and ethyl intermission. I don't think that it's an important factor anyway. In fact, fused ring systems nomenclature usually numbers the ring system peripheral atoms continuously (through the peripheral carbons, if including the fusion carbons), and in the clockwise sense, when the skeleton is in preferred orientation. However, for natural compounds, especially terpenes, there is large variety of different parent structures with various numberings.

So, when the correct structure was determined in 30’s (I think I was R. Robinson; and later confirmed by X-ray analysis in mid 40’s), it adopted similar numbering scheme (changes were made in methyls / side chain numbering), and the first systematic nomenclature of steroids[3] (pre-IUPAC though) from early 50’s also uses it, without explanation, rather referring to tradition:

  STEROID NOMENCLATURE.

  The following proposed rules for steroid nomenclature are based on agreements reached at a Conference held at the CIBA Foundation, London, on May 30th to June 1st, 1950. It is hoped to submit them for approval to the International Union of Pure and Applied Chemistry.

  NUMBERING.

  Rule 1.—Steroids shall be numbered as in formula (I). If one or more of the carbon atoms shown in (I) is not present, the numbering of the remainder shall remain undisturbed.

  Remarks. The numbering in (I) is that generally accepted in steroid chemistry, although it does not conform to that given in the “Ring Index” for cyclopenta[a]phenanthrene (No. 2561). It has been the general, but not invariable, practice to assign the number 18 to the carbon atom of the methyl group attached to C(13), as in (I), because this methyl group is present in all natural steroids whereas the methyl group (involving C(19)) attached to C(10) is absent in members of the œstrogen series.

  (…)

…and it's used still (that's for the ring numbering. There are some changes – C-28 and C-29, not present in cholesterol, now have numbers 241 and 242).

If chemists were to decide for steroids numbering revision, an interesting option would be to choose numbering based on the biosynthesis facts. The numbering would be changed only in ring A (starting with today’s C-4, clockwise), and would be entirely continuous:

(or opposite direction)

… Based on squalene (farnesyl, or farnesene units joined tail to tail depicted in different colors, their isoprene units bold) cyclization (with a couple of methyl shifts, and also hydride shifts (not depicted)) after squalene epoxidation (bottom leftmost double bond) to lanosterol (30 carbons), which then losts three methyls via oxidation and decarboxylation, and double bond is shifted. (The correct topology of of squalene cyclization was first correctly proposed by R.B. Woodward, if I'm not mistaken, as an alternative to R. Robinson proposal[4]).

References:

Wieland Nobel lecture, PDF
Windaus Nobel lecture, PDF
Steroid nomenclature. Chem. Ind. June 23, 1951, SN 1-11.
Woodward, R. B.; Bloch, K. THE CYCLIZATION OF SQUALENE IN CHOLESTEROL SYNTHESIS. J. Am. Chem. Soc. 1953, 75 (8), 2023–2024.


The energy gap between a pi-conjugated system with (2 bonding and 1 anti-bonding orbital) and (1 bonding and 2 anti-bonding) orbitals

If you want to really understand why energy levels of conjugated $\pi$ systems are the way they are I suggest you take a good look at the Hückel theory as @ssavec suggested - this method is only an approximation but because of its simplicity it is really good for helping you to understand the underlying concepts.
Nevertheless, doing as you suggested and looking at the overall overlap between the $\text{p}$ orbitals in a given molecular orbital does indeed seem to work although I don't have a strict justification for that.

From Hückel theory it is known that the coefficients of the $\text{p}$ orbitals in a given molecular orbital can be calculated as

\begin{equation}
  c_{j,r} = \sqrt{\frac{2}{N + 1}} \sin\Bigl(\frac{r j \pi}{N + 1}\Bigr)
\end{equation}

giving the coefficient $c_{jr}$ for atom $j$ in molecular orbital $r$ of a conjugated system of $N$ atoms (so that $j$ and $r = 1, 2, 3, \dots , n$).

Let's apply this to butadiene.

The $\pi$ molecular orbitals of butadiene are $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, and $\Psi_{4}$.
They are built from a linear combination of 4 atomic $\text{p}$ orbitals ($\phi_{1}$, $\phi_{2}$, $\phi_{3}$, and $\phi_{4}$), so

\begin{equation}
  \Psi_{r} = c_{1,r} \phi_{1} + c_{2,r} \phi_{2} + c_{3,r} \phi_{3} + c_{4,r} \phi_{4}
\end{equation}

When you calculate the atomic orbital coefficients for each molecular orbital $r$ then you get to the following picture:

where the numbers above the different $\text{p}$ orbitals are the coefficients.

Doing the same for ethene you get the following:

Now, you need to compare the overall overlap in the HOMOs of ethene and butadiene.
Since it is already assumed in the Hückel theory only the overlap between neighboring $\text{p}$ orbitals is considered.
Furthermore, it is useful to set the overlap integral of two "normal" $\text{p}$ orbitals (both with coefficient $c = 1$) to 1 (its exact value doesn't matter because only the relative amounts of overlap in different MOs are important for this comparison), so that you can calculate the overall overlap in a molecular orbital simply by multiplying the coefficients of neighboring $\text{p}$ orbitals and summing over these products.
So, for the overlap in the ethene and butadiene HOMOs you get:

\begin{align}
  S_{\text{HOMO, ethene}} &= 0.707 \cdot 0.707 \\
  &= 0.500 \\
  \\
  S_{\text{HOMO, butadiene}} &= 0.600 \cdot 0.371 + 0.371 \cdot (-0.371) + (-0.371) \cdot (-0.600) \\
  &= 0.307
\end{align}

There is less overlap and thus less bonding in the HOMO of butadiene than in the HOMO of ethene which would imply that the HOMO of butadiene is higher in energy than that of ethene.

Now, what about the LUMOs:

\begin{align}
  S_{\text{LUMO, ethene}} &= 0.707 \cdot (-0.707) \\
  &= -0.500 \\
  \\
  S_{\text{LUMO, butadiene}} &= 0.600 \cdot (-0.371) + (-0.371) \cdot (-0.371) + (-0.371) \cdot 0.600 \\
  &= -0.307
\end{align}

There is more overlap and thus more bonding (or better to say less antibonding) in the LUMO of butadiene than in the LUMO of ethene which would imply that the LUMO of butadiene is lower in energy than that of ethene.

So, the HOMO of butadiene is higher in energy than that of ethene and the LUMO of butadiene is lower in energy than that of ethene.
Thus, the HOMO-LUMO gap in butadiene is smaller than that in ethene.


Does the natural abundance of radioactive elements change?

Yes, you are right. The abundances of (radioactive and stable) nuclides change due to radioactive decay.

Regarding this, most environmental radionuclides can be divided into four categories:

Cosmogenic radionuclides are constantly produced by cosmic irradiation of the atmosphere. Therefore, their average concentrations in the atmosphere stay roughly constant even if their half-lives are short.

Primordial radionuclides have lifetimes comparable to the age of the earth. These nuclides have been formed at the time of the formation of the earth, the solar system, or even before. Their abundance is slowly decreasing due to radioactive decay. Therefore, the radiation exposure was higher some billion years ago when life evolved on earth than today. Furthermore, due to different half-lives, the relative abundance of these nuclides (e.g. the natural $\ce{^{235}U/^{238}U}$ isotopic ratio) also changes.

Radionuclides that are part of the natural decay chains of $\ce{^{238}U}$, $\ce{^{235}U}$ and $\ce{^{232}Th}$ are constantly reproduced by the radioactive decay of their mother nuclides. Therefore, their abundance is only slowly decreasing with the long half-life of $\ce{^{238}U}$, $\ce{^{235}U}$, or $\ce{^{232}Th}$ even if their own half-lives are short (this situation is called secular equilibrium).

Anthropogenic radionuclides are produced by modern technology. Since the discovery of nuclear fission, the activity concentrations of various anthropogenic radionuclides in the environment were increasing until the limited nuclear-test-ban treaty banned atmospheric tests of nuclear weapons. After that, the activity concentrations of most anthropogenic radionuclides decreased again.

If you isolate and chemically purify a radioactive sample, you remove any other mother or daughter nuclides and thus you disturb any decay chain or radioactive equilibrium. Thereby you start a radioactive clock.

For example, a solid rock sample may contain $\ce{^{226}Ra}$ $\left(t_{1/2}=1600\ \mathrm{a}\right)$ and its daughter nuclide $\ce{^{222}Rn}$ $\left(t_{1/2}=3.83\ \mathrm{d}\right)$ in a secular equilibrium. If you dissolve the sample and release the $\ce{^{222}Rn}$ (radon is a noble gas), the released $\ce{^{222}Rn}$ decays with its half-life of 3.83 days. If you purify the $\ce{^{226}Ra}$ and put it in a closed container, initially the container does not contain any $\ce{^{222}Rn}$, but new $\ce{^{222}Rn}$ is produced by radioactive decay of $\ce{^{226}Ra}$.

A similar thing occurs when natural minerals solidify. Radionuclides become trapped and decay. As they decay, decay products accumulate in the closed mineral.

By measuring the amount of mother and daughter nuclide, it is possible to calculate how long this sample has existed. This is the basis for nuclear dating.

If you isolate a sample that contains multiple radionuclides of the same chemical element, you can observe changing isotopic ratios – as you have expected in your question.

Your example, freshly purified thorium from a natural sample, mainly contains six different thorium nuclides:

$\ce{^{234}Th}$ $\left(t_{1/2}=24.10\ \mathrm{d}\right)$
$\ce{^{232}Th}$ $\left(t_{1/2}=1.405\cdot10^{10}\ \mathrm{a}\right)$
$\ce{^{231}Th}$ $\left(t_{1/2}=25.5\ \mathrm{h}\right)$
$\ce{^{230}Th}$ $\left(t_{1/2}=7.54\cdot10^{4}\ \mathrm{a}\right)$
$\ce{^{228}Th}$ $\left(t_{1/2}=1.91\ \mathrm{a}\right)$
$\ce{^{227}Th}$ $\left(t_{1/2}=18.7\ \mathrm{d}\right)$

The individual half-lives differ significantly; thus, the different thorium nuclides decay at different rates. The isotopic ratios change accordingly.
However, the calculation is slightly complicated since $\ce{^{234}Th}$ indirectly decays to $\ce{^{230}Th}$, $\ce{^{231}Th}$ indirectly decays to $\ce{^{227}Th}$, and $\ce{^{232}Th}$ indirectly decays to $\ce{^{228}Th}$.


Why are moles used in the ideal gas law and not simply mass?

The overall notion is that gases exhibit "ideal" behavior so that the identity of the particular gas molecule doesn't matter. Thus regardless of what compound the gas molecule is, or if the gas is a mixture of different compounds, a mole of gas molecules at standard temperature and pressure (STP, 0 °C and $1.000\times10^5$ Pa) occupies a volume of 22.7 liters. So the equation for an ideal gas is

$$PV = nRT$$

In addition, there is the Maxwell–Boltzmann distribution which is for the speed of molecules. The gist is that regardless of the compound, gas molecules have the same kinetic energy spread. Since kinetic energy is given by $mv^2/2$, that means that lighter molecules have faster speeds than heavier molecules.

If you try to define the relationship by the molecular weight of the gas, then the value of the gas constant, $R$ would be different for different compounds and/or mixtures. Now in reality the ideal gas law works reasonably well, but there are empirical equations which fit the relationship between P, V, and T better. For those empirical equations the extra coefficients are fitted to the experimental data for particular gas compound under consideration.


Viscosity of branched vs. linear short alkanes

Viscosity is related to Van der Waals/London forces between molecules.  This in turn is related to packing and surface area.

A branched 6-chain carbon, tumbling in solution cannot align as much of its surface area with another molecule as the linear hexane.

At some point the larger molecules are becoming able to be 'entangled' and form structures somewhat similar to knots.  When the molecules get this large, viscosity increasing branching becomes more likely.

Finally, when you get to true polymers, the chains are so long that branching again begins exclude other molecules from structure and again linear molecules are more likely to have higher viscosities (although for polymers, these are solution viscosities).


Any molecule composed of exactly two atoms is diatomic, but I will answer this question assuming that you mean elemental diatomic molecules. Only H, N, O, F and Cl are diatomic elemental gases under STP. Br and I are diatomic liquid and solid respectively, under STP. Oxygen remains diatomic in liquid state, while nitrogen and all halogens(F, Cl, Br, I) are diatomic in all three classical physical states. Gaseous sulfur can be diatomic at or below standard pressure. Structurally similar to diatomic nitrogen, diphosphorus is an unstable allotrope of phosphorus. Diatomic carbon is a gas that exists under very high temperature. There may also be other diatomic allotropes.
Some elements form stable diatomic molecules because of the octet rule (or duplet for hydrogen). Generally a bond can share up to three pairs of electrons, so elements which are three or less electrons short of a stable electron configuration can pair up and share up to three electrons to become stable. Beyond that, elements will find other ways to become stable.
However, chemical stability isn't governed only by the octet rule, there are other factors such as geometry, bond length and nucleophilicity/electrophilicity that affect molecular stability. Therefore, some diatomic allotropes are not the most stable allotropes of the elements. Examples from above are disulfur, diphosphorus and dicarbon.


Why is Nitrogen 15 found almost exclusively in marine environments?

Let's start that the sentence found in the source you provided:

  the nitrogen isotope 15, which is found almost exclusively in marine environments

is misleading and simply incorrect.

Now to some basics. When we discuss isotopes in environmental sciences, we never use the absolute values of an isotope content. Instead we use isotope ratios:

$$\text{Ratio (R)} = \frac{\text{heavy isotope abundance}}{\text{light isotope abundance}}$$

However, the ratios themself are not very useful, so the delta ($\delta$) notation is used:

$$\delta[^0\!\!/\!_{00}] = \left(\frac{R_{\mathrm{sample}}}{R_{\mathrm{standard}}}-1\right)\times1000$$

All standards have the delta value equal $0$ by convention. There is an international standard material for each isotope, for example for $\ce{^13C/^12C}$ ratio it is VPDB (Vienna PeeDee Belemnite), and for $\ce{^15N/^14N}$ it is atmospheric nitrogen. The latter standard was chosen because the content of the $\ce{^15N}$ isotope in atmospheric nitrogen is uniform throughout the world and it does not vary significantly over time. Many chemical, biochemical and physical processes cause the so-called isotopic fractionation to occur - it results in the increased (enrichment) or decreased (depletion) isotope ratios.

OK, we know the terminology, so now to the point. In general, the metabolic processes of living organisms cause enrichment of nitrogen. The extent of the enrichment depends on the trophic level (position in the food chain) and dietary sources of nitrogen. The highest enrichment is found in terrestial herbivores, while marine organisms usually show lower $\delta\ce{^14N}$ values. Given those facts the cited sentence makes no sense, and is probably a result of a misinterpretation of some scientific paper.

There is a lot of literature on the subject if you are interested. I can give you one good reference as a starting point:

M.A. Vanderklift, S. Ponsard. Sources of variation in consumer-diet
$\delta\ce{^15N}$ enrichment: a meta-analysis. Oecologia (2003) 136:169–182 DOI 10.1007/s00442-003-1270-z.


How to calculate the mass of iron with a temperature of 325 °C needed to melt ice and raise the temperature of water from 0 to 22 °C?

I can not see anything wrong with your calculations at all.  For posterity, the following is a rough method to confirm:

Melting ice

Enthalpy of fusion $\Delta H_\text{ice} = 6.01\ \mathrm{kJ\ mol^{-1}}$
Mass of ice $m = 75\ \mathrm g$
Molar mass of $\ce{H2O}$ $M = 18\ \mathrm{g\ mol^{-1}}$

Hint: usually, when you are given mass and molar mass, then the amount of the substance is required.

Note that the Enthalpy of fusion ($\Delta H_\text{ice}$) is measurement in $\mathrm{kJ\ mol^{-1}}$, and as the ice is measured in grams, and you are provided with the molar mass, you'll need to use the formula:

$$n = \frac{m}{M}$$

hence:

$$n_\text{ice} = \frac{75\ \mathrm g}{18\ \mathrm{kJ\ mol^{-1}}}$$

$$n_\text{ice} = 4.1666\ \mathrm{mol}$$

Therefore, the enthalpy of fusion:

$$\Delta H_\text{fus} = 6.01\ \mathrm{kJ\ mol^{-1}} \times 4.1666\ \mathrm{mol} = 25.04167\ \mathrm{kJ} = 25041.67\ \mathrm J$$

Heating water

Specific heat capacity of water $c_p = 4.184\ \mathrm{J\ g^{-1}\ ^\circ C}$
Mass of water $m = 175\ \mathrm g$
Temperature change $\Delta T = 22\ \mathrm{^\circ C}$

So:

$$Q = mc_p \Delta T$$

$$Q = 175\ \mathrm g \times 4.184\ \mathrm{J\ g^{-1}\ ^\circ C} \times 22\ \mathrm{^\circ C} = 16108.4\ \mathrm J$$

Total energy used by the water/ice mixture

$$Q_\text{total} = 25041.67\ \mathrm J + 16108.40\ \mathrm J = 41150.07\ \mathrm J$$

Iron

Specific heat capacity of iron $c_p = 0.45\ \mathrm{J\ g^{-1}\ ^\circ C}$
Mass of water $m = X$ (unknown)
Temperature change $\Delta T = 325\ \mathrm{^\circ C} - 22\ \mathrm{^\circ C} = 303\ \mathrm{^\circ C}$ decrease

As, according to conservation laws:

$$Q_{\ce{H2O}} = Q_{\ce{Fe}}$$

Then:

$$41150.07\ \mathrm J = X \times 0.45\ \mathrm{J\ g^{-1}\ ^\circ C} \times 303\ \mathrm{^\circ C}$$

Or:

$$41150.07\ \mathrm J = 136.35\ \mathrm{J\ g^{-1}}\times X$$

Then:

$$X = \frac{41150.07\ \mathrm J}{136.35\ \mathrm{J\ g^{-1}}}$$

$$X = 301.8\ \mathrm g$$


Choice of lower bound of curve for PV work

The error comes from reading the graph: the crossing of the axes corresponds to the point (2,2) and not (0,0)


Is Jet Fuel/Kerosene still toxic after drying out?

Kerosene is not a pure compound. To quote Wikipedia:

Regardless of crude oil source or processing history, kerosene's major components are branched and straight chain alkanes and naphthenes (cycloalkanes), which normally account for at least $70~\%$ by volume. Aromatic hydrocarbons in this boiling range, such as alkylbenzenes (single ring) and alkylnaphthalenes (double ring), do not normally exceed $25~\%$ by volume of kerosene streams. Olefins are usually not present at more than $5~\%$ by volume.[10]

[10]: American Institute of Petroleum (September 2010). "Kerosene/Jet Fuel Assessment Document" (PDF). EPA. p. 8. Retrieved October 28, 2016.

Because kerosene is a complex mixture, it is hard to say something specifically. However, the compounds noted are basically long chain alkanes and substituted cycloalkanes, a smaller fraction being alkylated aromatic rings. Benzene and naphthalene do not seem to be significant contributors.
None of these are classified as toxic. Most importantly, while alkylated benzenes and alkylated naphthalenes may still intercalate into DNA making them potential carcinogens, they are metabolised by a completely different pathway from benzene proper, making their metabolites much less dangerous. (Rather than the very reactive 1,2-epoxybenzene, various benzoic acids are the primary oxidation products.) That doesn’t mean they’re harmless, but toxic is the wrong designation and should be reserved for more dangerous compounds.

A general rule for all compounds is:

As long as you smell them, they can affect you.

The desired effect of painting a door with kerosene is, however, that the kerosene penetrates the door and stays there. The least-volatile constituents will remain on the door itself, protecting it from damage. If all the kerosene diffused away, the method would not be used because it would be worthless. Therefore, after the door has vented properly, letting the volatile components disappear, there is practically nothing left that will be harmful to any noteworthy extent.
In fact, I am include to proclaim that your average coloured paint may well contain more harmful chemicals that the kerosene does.
Yes, you can touch it, no you do not need space suits to be safe.


Determine the rate equation (intermediate reaction)

You can't write $[\ce{B}] = [\ce{C}]$ because  $\ce{C}$ is consumed in the second step but $\ce{B}$ isn't. However, you were pretty much on the right track.

Since $\ce{E}$ is produced from $\ce{D}$, you can write $[\ce{E}] = [\ce{D}]_i - [\ce{D}]$ where $[\ce{X}]$ denotes the concentration of $\ce{X}$ at a certain point in time and $[\ce{X}]_i$ denotes the initial concentration of $\ce{X}$. This is assuming that the initial concentration of $\ce{E}$ is $0$, of course.

On top of this, $\ce{C}$ is either produced from $\ce{A}$, or consumed in the production of $\ce{E}$. Therefore you can write $[\ce{C}] = [\ce{A}]_i - [\ce{A}] - [\ce{E}] = [\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]$. (There is the reverse step in which $\ce{C}$ produces $\ce{A}$, but the equation still holds, since you can just consider this to be "negative" production of $\ce{C}$ from $\ce{A}$.)

Your rate law is therefore: $r = k_1[\ce{D}]\big([\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]\big)$. The initial concentrations are merely constants, so you could view this as being of the form $r = k_a[\ce{D}] + k_b[\ce{A}][\ce{D}] + k_c[\ce{D}]^2$. I can't think of any way to simplify this (the steady-state approximation is not very valid here since $\ce{C}$ is produced quickly and consumed slowly, and in any case it does not yield a nicer expression). I imagine that one could use some form of an integrated rate law to find expressions for $[\ce{A}]$ and/or $[\ce{D}]$, but it would just make things even uglier.


How are electrons in conjugated molecules excited?

This first picture shows what happens to the molecular orbitals of ethylene if we bring two ethylene molecules (the right and left sides of the molecular orbital vs. energy part of the diagram) close together and form 1,3 butadiene (center of the figure)

The next figure shows the result of adding a third ethylenic group to the molecule and form 1,3,5-hexatriene. Note how in each case the molecular orbitals combine (split) and generate a new set of molecular orbitals; and in each case the separation between the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO) decreases.  When light is absorbed an electron is promoted from the HOMO to the LUMO.  If we make the HOMO-LUMO separation small enough by extending conjugation, we can shift absorbtion into the visible region (longer wavelength, smaller HOMO-LUMO energy separation). We can add other groups too, like cyano, nitro, phenyl, carbonyl, etc. to extend the conjugation in our system.


How can an excited helium dimer be formed?

The electronic configurations are $\mathrm{(1s)^2}$ for $\ce{He}$ and $\mathrm{(1s)^1(2s)^1}$ for $\ce{He^*}$.

Put this information in an energy diagram and see what the molecular orbitals look like, especially how they are filled. (Further hint: Filled-unfilled orbital interactions are generally favorable.)

I have attached a graphic:

This leads to a bond order of $\frac{1}{2}(2-1+1) = 1$, and therefore $\ce{He2^*}$ is stable with respect to $\ce{He + He^*}$. The excited dimer is known as an excimer.

As an aside: The relative energy levels of the molecular orbitals are very hard to get right out of the blue, so take them with a grain of salt. As was kindly pointed out by Martin, the excitation of helium will initially form a singlet excited configuration, depicted above. This should be able to relax to a triplet excited state.


Why does soap reduce surface tension of water

Apoorv did not give the right explanation, because this would imply, that surface tension goes especially down, when micelles are formed, but the opposite is the case. Compare e.g.: first chapter of The Colloidal Domain by D.Fennell Evans and Håkan Wennerström.

First thing to note is that pure water has a higher surface tension than hydrocarbon chains which is due to the strong dipole dipole interaction.

If there is a very dilute solution of surfactant in water no micelles are formed but you have monomers "floating around". In contrast to ions in water they are not homogenously distributed, but they concentrate especially on the surface where their polar head is directed to the water and their hydrophobic chain is directed to the air. In the end you substituted water molecules with hydrocarbon chains which leads to a lower surface tension.

If you increase the concentration of surfactant, they start to form micelles in the water (and not on the surface). So after reaching this critical micelle concentration (called CMC) the surface tension stays quite constant. This constant behaviour is used experimentally to estimate the CMC.

See the figure below Equation 11:
http://infohost.nmt.edu/~jaltig/SurfaceTension.pdf

Please note that Apoorv described some kind of reversed micelle. The normal way is that they stick their polar head to the water and have their hydrocarbon chain inside the micelle.

In the picture the white group is the polar head:
https://en.wikipedia.org/wiki/Micelle#/media/File:Phospholipids_aqueous_solution_structures.svg


Synthesis of Diazepam with (2-amino-5-chlorophenyl)(phenyl)methanone

While amines may undergo multiuple methylations to form quaternary ammonium salts, the nitrogen atom being akylated in the diazepam synthesis is an amide. Amides are different.

The lone pair on the amide nitrogen atom in diazepam is thoroughly delocalized into the carbonyl, into the aromatic ring, and all the way to the imine (see image below - some resonance contributors not shown). Due to this delocalization, that amide nitrogen atom is not nucleophilic and will not be methylated a second time. I would be much more worried about alkylating the imine nitrogen atom, which is made more nucleophilic by delocalization.

This delocalization is the reason why you need to add a base in order to do the methylation step in the first place. The amide in N-demethyldiazepam is likewise a poor nucleophile due to delocalization.

However, once a base deprotonates the amide, the amide becomes the better nucleophile due to the concentration of negative charge. Even though one of the lone pairs is delocalized, the other is not.


Basically canola oil is obtained from a cultivated form of Rapeseed. Canola oil is just what rapeseed oil is called in North America.  From a marketing perspective it was thought that having the term "rape" in the product name would not help sales.

If you look for articles on rapeseed oil pyrolysis you'll find a lot of information since rapeseed (canola) oil is used in biodiesel production.

This article states that the major products produced from pyrolysis in a reactor include

gases: $\ce{CO2, CO, CH4}$ and $\ce{H2S}$
oils: oleic acid, 1H-indole, 2,3,5-trimethoxy toluene, toluene, (Z)-9-octadecanamide, psoralene, phenol and phenol derivatives

I didn't know what "psoralene" was, so here is a picture and a link


Why is distillation not a viable way to separate ammonia from water?

Having embarrassed myself with my first answer, let me try this again. Ammonia and water hydrogen bond exceptionally well. In fact, the $\ce{O-H...N}$ value is 29 kJ/mol while the $\ce{O-H...O}$ value is 21 kJ/mol. (Ref) So, if there is any water around, ammonia would be happy to become solvated as they recondense.

Also, ammonia is a weak base and reacts with water (which is not usually the case when you're doing a distillation). $$\ce{NH3 + H2O <=> NH4+ + OH-}$$ So, if there's any water around as the ammonia is distilling and they recombine, they would also react which would make it difficult to remove the last of the water. (Some reaction would occur because the lowest energy point of the equilibrium is not at reactants or products.)

So, as noted in the comments, distillations to prepare dry ammonia use a scrubber to remove the water.


Does the reaction of 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one with NaOH involve ring contraction?

Drawing on the suggestion of @user55119 I propose the mechanism shown below.

Whether the first intermediate bicyclic [4.1.0] heptanone arises from an intramolecular electrocyclic process or an intramolecular aldol is unclear but either seems possible. Once the highly strained bicyclic [4.1.0] heptanone intermediate is formed, the reaction proceeds with OH- attack on the carbonyl in a process analogous to a Favorskii mechanism here, with the second intermediate anion stabilised by the nitro group which also directs the direction of ring opening. Finally loss of OH- in a manner similar to base-catalysed aldol and driven by re-aromatisation gives the observed product.


What is the 'ancestral term' in the LS coupling scheme?

"Ancestral Terms" are terms such as parent and grandparent terms that couple to form the final term.

For example, if you see $3d^7(^4F)4s4p(^3P°) ^6F°_{9/2}$

this means 7 3d electons couple to form a $^4F$ grandparent term

the 4s and 4p electrons couple to form a $^3P°$ parent term and

the grandparent and parent terms couple to form the final $^6F°$ term.

The above is one of several example given at the NIST page Notations for Different Coupling Schemes


Resonance structures of some aromatic compounds

Resonance structures can be drawn for all aromatic compounds.  Tropolone, furan and thiophene are considered to be aromatic compounds because

we can draw resonance structures
and they are cyclic compounds with 4n+2 (n=1) pi electrons

Here are the resonance structures we can draw for tropolone, the positive charge can be delocalized to every carbon in the ring.

Just like in benzene, where instead of drawing 2 resonance structures (top of following figure) we usually just draw the one resonance structure shown at the bottom of the figure;

instead of drawing all 8 resonance structures for tropolone, we usually represent them by just drawing the one resonance structure shown at the bottom of the tropolone figure.

and here are the resonance structures for furan

and for thiophene, they are the same as furan, just replace the oxygen with a sulfur atom in the resonance structures.


Why are the concentrations in the equilibrium constant multiplied?

As discussed in comments, you can find a good derivation in a standard university level physical chemistry or thermodynamics textbook.

Let's instead consider why it should be true.

From a microscopic perspective, what we're trying to balance, in essence, is the density of states on both sides of the equation, in essence, the count of ways to rearrange the various pieces. These are heavily dependent on the relative spacing of energy levels and degrees of freedom on both sides. Fortunately, this complex information is captured in the value of $K$.

But the point is that if you're trying to find the total number of permutations of two different collections, you multiple the values. For example, if there are two dice, the total number of outcomes is 36.

And concentrations essentially measure the number of things to rearrange, so the total number of possibilities should scale with the product, not the sum.

This is all a bit hand-wavy, but again, I'm not trying to derive the answer. I'm just trying to suggest that even without a formal derivation, it seems very reasonable that the correct operation to use is multiplication.

EDIT:

After reading this again, I realized I can extend this with another hand-wavy justification.

Remember that we have:

$$\Delta G = -RT \ln K$$

If you have two reactions that you combine together, you want the energies to sum. But that translates into products inside the log! Once again, the proper operation for equilibrium constants is multiplication.


Why does NaBH4 reduce double bonds conjugated to carbonyl groups, while LiAlH4 does not?

Neither $\ce{LiAlH_4}$ nor $\ce{NaBH4}$ are able to reduce an isolated $\ce{C=C}$ bond. But if you have an enal (a conjugated aldehyde) it can react (as an electrophile) either at the β-carbon or at the carbonyle group's carbon. According to the HSAB Principle the β-carbon is a “soft” center and would react preferably with “soft” nucleophiles while the carbonyle carbon is a “hard” center and prefers to react with “hard” nucleophiles.
Now, $\ce{NaBH4}$ is a rather soft nucleophile and thus it reacts with an enal at the β-carbon. After this reaction the $\ce{C=C}$ bond is gone. But what is left behind is still a very reactive simple aldehyde that gets reduced to the alcohol by $\ce{NaBH4}$ in a second step.
$\ce{LiAlH4}$ on the other hand is a rather hard nucleophile and thus reacts with an enal at the carbonyl carbon. After this reaction there is just an isolated double bond left that can't be reduced by $\ce{LiAlH4}$ in a second step.


Slightly askew of what was asked, but still relevant.  TLC is not dependent on conjugation.  That reasoning suggests to me You are using UV light to visualize the TLC.  UV light is used because it is clean, quick and non-destructive.

There are destructive visualization techniques, such as vanillin stain.  In this case, after elution one dips the tlc slide into the stain, and then bakes the coated slide, and various compounds appear in vivid colours, that are normally somewhat related to the functional groups present in the molecules.  Myriad stains exist, vanillin was just my favorite when I was practicing.

(picture taken from http://www.ochemonline.com/images/thumb/5/5d/Tlc_stains.JPG/1000px-Tlc_stains.JPG)

These stains are not quite as clean or quick as UV visualization, but are still quite convenient.  There are other monitoring techniques, but nothing beats TLC for generality, cost and speed.


In principle, there are no zeroth-order reaction. Experimentally, however, zeroth-order rate law occurs because of the reaction condition. For example, in catalytic reactions such as hydrogenation of an alkene using platinum as a catalyst, the hydrogenation occurs on the surface of platinum. There is a finite surface area of platinum, so at a some point, it doesn't matter at all how much more hydrogen is put in. This is because no reaction can occur if the hydrogen cannot bind to the platinum.

An example of inverse first order reaction can be found here.

Basically, you have a two-step reaction:
$$\ce{A + B <=> C}$$
$$\ce{C <=> D\ (slow)}$$
Increasing concentration of $\ce{C}$ would decrease the rate of reaction.


Help understanding the pressure's effect on equilibrium and Le Chatelier's principle

What the textbook means when it is saying one occupies more volume than the other is that the reactants (in this case) will occupy less volume than the product under standard conditions (~ atmospheric pressure). Therefore you have to use more force (pressure) to confine the products to the given volume than you need to do so for the reactants.
Therefore there is less enforcement on the reactants than there is on the products. Therefore it is energetically more favorable to be in the reactants-form.

In order to understand this a little better you can have a look at the ideal gas law:
$$
p = \frac{nRT}{V}
$$
As you can clearly see the pressure is linearly dependent on the amount of particles in the system. Given that during the reaction (in the way you stated it in your question) this amount increases (doubles if the reaction did run through completely). Thus the pressure increases as well. However a higher pressure is connected to a higher energy state of the particles in the "box" as they repulse one another.
Why don't the reactants stay as they are then as this is the state that produce the least pressure? Because by forming the products, energy is released ($\Delta_rG < 0$). Therefore the reaction can take place until the increase in pressure results in a "raise of energy" that is equal to the energy released during the reaction. Any further reaction would be energetically unfavorable.
And this point is the one that can be influenced by the volume. Let's say that the reaction can take place until the pressure reaches $1.5p_0$ and $p_0 = \frac{n_0RT}{V_0}$ (and for simplicity let's assume a isothermal reaction). This means that 50% of the the reactants can react until that limit is reached. If we increase the volume, this point is reached only after having a higher particle count than with a lower volume which is why the equilibrium shifts to the side with more gaseous particles if the volume is increased and vice versa.


Does the shared electron in Covalent bonds revolve around nucleus?

I am guessing that you have an understanding of the atomic orbitals of an atom such as 1s and 3p atomic orbitals. Atomic orbitals are the probability distribution of where an electron is going to be $90$% of the time. Technically, electrons don't revolve around the nucleus but are rather quantum particles that are superimposed in several positions at the same time. However the description of electrons revolving randomly around the nucleus is suffice at this level.

When a covalent bond is formed, the electrons do still continue to revolve around the nucleus in orbitals, however a new kind of orbital is formed which are called molecular orbitals (rather than atomic orbitals). Here is a simple explanation of what molecular orbitals are.

Molecular orbitals are formed by constructively and destructively overlapping the atomic orbitals together. The simplest example of this is when $\ce{H2}$ is formed. Here, each H atom has a 1s atomic orbital. When they form, these atomic orbitals form 2 molecular orbitals. Where one molecular orbital is due to constructively overlapping the 2 atomic orbitals and the other molecular orbital is due to destructively overlapping the 2 atomic orbitals. This shown in the diagram below:

As you see that these 2 molecular orbitals that are formed are quite different to each other. The MO that is formed due to constructive overlap has greater electron density between the 2 positive nucleus. Meanwhile the MO that is formed due to destructive overlapping has decreases electron density between the 2 positive nucleus and in fact have a node between them. Therefore the first MO is lower in energy than the second MO.

Like AOs, the same rules apply to MOs. Each MO is able to hold 2 electrons and the electrons fill the lowest energy orbital first. So in $\ce{H2}$ the 2 electrons will fill the first MO and will revolve around the nucleus in that orbital (well technically only for $90$% of the time).

This theory can be applied to any compound, however it gets more complicated as the number of AOs increases and the number of atoms increases. However essentially, when atoms form covalent bonds, the electrons continue to revolve around the nucleus in MOs rather than AOs.


Why does the $V_{\mathrm{max}}$ of an enzyme not change with competitive inhibition? Shouldn't it decrease since there are fewer active sites?

You can think of $V_{max}$ as describing the nature of the active site - it's deftness at converting substrates to the product, the rest is just probability. Competitive inhibitor does not change properties of the active site - they just hang there for some amount of time until the E-I complex dissociates hence they don't affect the maximum theoretical conversion rate of that enzyme.
Competive inhibitors only decrease the chance of inhibitor binding to the enzyme. Thus you can always raise the concetration of your substrate to the state that probability (now the other way around) of inhibitor binding the enzyme will become negligible with regard to the substrate allowing it to work at his maximum rate.


Why does the weaker C–H bond have a higher wavenumber than the C=O bond?

A property of the harmonic oscillator is that the oscillation frequency, $\omega$, is dependent not only on $k$ (the spring constant) but also on the mass $m$ of the object:

$$\omega = \sqrt{\frac{k}{m}}$$

We can crudely model a chemical bond as a two-body harmonic oscillator, which largely obeys the same rule, except that the mass must be replaced with the reduced mass $\mu$, defined by:

$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$$

where $m_1$ and $m_2$ are the masses of the two bodies. So, for a C–H bond, we have

$$\frac{1}{\mu_\ce{CH}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{1 u}} \implies \mu_\ce{CH} = \pu{0.923 u}$$

and for a C=O bond, we have

$$\frac{1}{\mu_\ce{CO}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{16 u}} \implies \mu_\ce{CO} = \pu{6.86 u}$$

From J. Chem. Phys. 1946, 14 (5), 305–320 the force constant $k$ for the C=O bond in formaldehyde is approximately $2.46$ times that for the C–H bond in methane (the units are in the old cgs system, so I am lazy to quote the actual values).

So, we can come up with a very rough theoretical estimate. The wavenumber used in IR spectroscopy, denoted by $\bar{\nu}$, is directly proportional to the (angular) frequency $\omega$.

$$\begin{align}
\frac{\bar{\nu}_\ce{CH}}{\bar{\nu}_\ce{CO}} &= \frac{\omega_\ce{CH} / 2\pi c}{\omega_\ce{CO} / 2\pi c } \\
&= \frac{\omega_\ce{CH}}{\omega_\ce{CO}} \\
&= \sqrt{\frac{k_\ce{CH}}{k_\ce{CO}} \cdot \frac{\mu_\ce{CO}}{\mu_\ce{CH}}} \\
&= \sqrt{\frac{1}{2.46} \cdot \frac{\pu{6.86 u}}{\pu{0.923 u}}} \\
&= 1.74
\end{align}$$

This is not too far from the experimental value of $\pu{2900 cm-1}/\pu{1730 cm-1} = 1.68$ (using values in the middle of the typical ranges for alkanes and aldehydes).


As with any experimental measurement of physical properties, the techniques involved improve with time, and sometimes new, better ones appear. That also means some values for some physical properties sometimes are slightly adjusted to reflect these better measurements and less error.

In some very rare occasions, the basic definition of some measurement is changed, like the recent change in the definition of the kilogram (see https://en.wikipedia.org/wiki/Kilogram), and it sometimes reflect in some minute adjustments made to another measurements based on the old definitions.

In other, more mundane measurements, the standard (for the measurement) is based on a set of conditions (pressure, atmosphere composition, purity, and a lot of other parameters) and a specific procedure for the measurement that sometimes change in newer versions of the procedure, sometimes resulting in a slight change of the expected measured values.

That doesn't mean the material changed, just that the observed value for the physical property using the described set of parameters and procedures were adjusted.


If $pK_a = 4.76$, then $K_a = 1.74 \times 10^{-5}$

Then approximating that $K_a$ can be written in terms of concentration rather than the true definition which is in terms of activities:

$K_a = 1.74 \times 10^{-5}M = \frac{[A-][H+]}{[HA]} = \frac{[H+]^2}{c-[H+]} $

$1.74 \times 10^{-5}M = \frac{[H+]^2}{c-[H+]} $

where $c$ is the molar concentration of the vinegar

If you want pH explicitly in the equation, then approximating pH = -log[H+], then:

$1.74 \times 10^{-5}M = \frac{(10^{-pH}M)^2}{c-(10^{-pH}M)} $

So, for example if pH is 3, $10^{-3}$ is 0.001, so

$1.74 \times 10^{-5} = \frac{10^{-6}M}{c-(10^{-3}M)}$

$1.74 \times 10^{-5}c -1.74 \times 10^{-8}M = 10^{-6}M$

$1.74 \times 10^{-5}c = 1.02 \times 10^{-6}M$

$c = 0.059M$


Is the new equilibrium total pressure of this equilibrium reaction always more than its initial total pressure?

The higher pressure of $\ce{NO2}$ even after reaching equilibrium after adding some amount of $\ce{NO2}$ can be shown mathematically.

Suppose initially before adding $\ce{NO2}$,the pressure of $\ce{NO2}$ was $p^0_{\ce{NO2}}$, and that of $\ce{N2O4}$ is $p^0_{\ce{N2O4}}$.So, the equilibrium constant is $$K_\mathrm{p} = \frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}$$ Now assume that the addition of $\ce{NO2}$ just increases the pressure instantaneously to $p_x$ but immediately after that due to dissociation to achieve equilibrium, the pressure decreases by $\ce{p_y}$ atm.

$$\ce{2NO2 <=> N2O4}$$
$$p^0_{\ce{NO2}} + p_x - p_y \ce{<=>} p^0_{\ce{N2O4}} + \frac{p_y}{2}$$

So, the increase of pressure $\ce{N2O4}$ will be $p_y/2$ atm. So, now a new equilibrium is achieved where the equilibrium constant $K_\mathrm{p}$ will be still the same assuming that during addition of the gases, temperature is not changed( or the change is very much negligible in real cases). Thus, equating $K_\mathrm{p}$ on both the sides we have,

$$\frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}= \frac{p^0_{\ce{N2O4}} + \frac{p_y}{2}}{(p^0_{\ce{\ce{NO2}}} + p_x - p_y)^2}$$

Taking the squares on numerator and denominator and taking square root we will find:

$$1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}}= \sqrt{1+\frac{p_y/2}{p^0_{\ce{N2O4}}}} > 1$$

So, we see that $1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}} > 1$. Hence it is proved that $\ce{p_x - p_y > 0}$. So, the net pressure of $\ce{NO2}$ after achieving equilibrium is $p^0_\ce{NO2} + p_x - p_y > p^0_{\ce{NO2}}$. So, its pressure has increased than the previous conditions even after reaching equilibrium.


No, you cannot. There is no way to reset the transient pulses in the neurons that carry the essence of thought processes. The information is already gone.

You will find the neural synapses all in the discharged state. All thought process, all memory, all of what makes a person is gone.


What decides the nature of emulsifying agents?

An emulsifier is a surfactant. A surfactant is characterized by the fact that its molecule has one or more hydrophilic parts and one or more hydrophobic parts. The ratio between the molar mass of the hydrophilic parts and the molar mass of the hydrophobic parts in the molecule is expressed by its HLB value (Hydrophilic-lipophilic balance).

Higher HLB means higher water solubility, and lower HLB means lower water solubility.

The emulsifier should have a certain solubility in the outer phase of the emulsion to be distributed in the emulsion.

An emulsifier for forming o/w-emulsions should have higher water solubility, that means higher HLB, and an emulsifier for forming w/o-emulsions should have lower water solubility (= higher oil solubility), that means lower HLB. The corresponding range of HLB is given in the Wikipedia artivle above.


Comparing wt% to at% which is normalized to 100%

This question is difficult to answer without knowing more about the data.

First is the issue of "normalization". Do you mean that in the at% study, all measurements added together were only 96% or 90% or 80% or something, and so to get all measurements to add to 100%, the simply divided each measurement by 0.96 (or 0.9 or 0.8)?  This is what I infer from your comments.

Second is the issue of measuring at%.  Are they measuring the raw number of total atoms in their OES experiment? Or are they just comparing their results to some calibration standard for each element measured?  Your comments make me think they are using calibration standards.  From a description of those standards, you should be able to convert the standards to wt%, and thus map from wt% to at%.

Whether you want to map to the "normalized" at% or the unnormalized at% is up to you.


In the bicarbonate ion, why can the hydrogen not bond to the carbon?

First of all I appreciate your effort to think out of the box.

Now, as you are thinking of a new structure for the bicarbonate ion, you just have to compare the stability factors between the two different structures and you'll get your answer!

Here let's compare the structures:

Stability factors for the conventional structure

Octet saturation in case of each of the element.
Resonance stability (moreover, two equivalent canonical structures give more stability) in the molecule.
The negative charge resides on the more electronegative $\ce{O}$ atom.

Unstability factors for your proposed structure

Octet unsaturated in case of the positively charged $\ce{O}$ atom.
No resonance stabilisation in the molecule.
(your point) Energetically disfavoured due to numerous charge accumulation.
The positive charge resides on the more electronegative $\ce{O}$ atom.

That's why your proposed structure isn't the actual structure of bicarbonate ion.


What is the dissociation percentage of a polybasic organic acid at a certain pH?

Reused my answer to how-to-numerically-model-a-phosphoric-acid-titration-curve, edited for diprotic cases:
First, calculate the common denominator $CD$:
$$a_1 = [\ce{H+}]^2$$
$$a_2 = [\ce{H+}] \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$
$$a_3 = K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$
$$CD = a_1 + a_2 + a_3 $$
.. and then reuse it and its additive terms:
$$[\ce{H2A}]=c_0 \cdot \frac {a_1}{CD}= c_0 \cdot\frac{[\ce{H+}]^2 }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$[\ce{HA-}]=c_0 \cdot \frac {a_2}{CD}= c_0 \cdot\frac{ [\ce{H+}] \cdot K_\mathrm{a1} }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$[\ce{A^2-}]=c_0 \cdot \frac {a_3}{CD}=c_0 \cdot \frac{K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$c_0$ is the total concentration of respective particular acid forms.
For the general degree of "N-proticity", the expression terms span from $\ce{[H+]}^N$ to $K_1 \cdot ... \cdot K_\mathrm{aN}$


Iodine monochloride is usually highly specific for anti addition to alkenes. While bromonium ions form more readily than do iodonium ions, the latter do form. Additionally, the large size of the iodine atom further reduces the chance of syn addition. To answer your question in particular, it is unlikely that a planar carbocation is formed, and the anti-addtion would be expected to be the main product.

The more typical issue with iodine monochloride is that while the chloride is typically added to the Markovnikov carbon, the regioselectivity is sometimes relatively mediocre. For example, the addition of ICl to propene gives 2-chloro-1-iodopropane and 1-chloro-2-iodopropane in a 69:31 ratio; the yields of the expected regioisomer are better when styrene is used, giving approximately 95% of the expected 1-chloro-2-iodo-1-phenylethane. (J. Chem. Soc., 1931, 2742) This is likely due to the greater stability of the benzylic stabilization of the cation.


I have worked in natural product isolation a little. Although I have never witnessed tests with HeLa cells, I can provide some insight into the general screening procedure.

In the lab I worked in which dealt with isolation of myxobacterial natural products, a novel strain was first tested as a crude extract in a preliminary screening via serial dilution assay. Tests were performed against a set of standard screening organisms including S. aureus, Candida albicans, E. coli and a few others I forgot.

If an isolated showed enough inhibitory activity at this stage, a semipreparative HPLC was performed to locate the peak(s) which caused said activity. The HPLC fractions were collected in a 96-well plate, the solvent removed by nitrogen flow and then the well plate was infected with the organism of interest. Hopefully, one or two wells would remain with out bacterial or fungal growth and hopefully these would not be in the fatty acid realm at the end of the 96 wells.

After that, a slightly larger separation was performed with the goal of gaining a few milligrams of the active compound (after confirming via HR-MS and dereplication that it was not already known) to perform another serial dilution assay to identify the minimal inhibitory concentration (MIC) — the isolationists’ equivalent to an $\mathrm{LD_{50}}$. This was typically performed around three times.

After that, the paper was drafted and submitted to peer review.

(Okay, there are a few steps I left out, but they contain other, unrelated stuff like making sure the myxobacterium was properly classified, elucidating the structure, hoping for crystals and maybe even performing some low-scale fermentation of the relevant myxobacteria to gain enough sample for proper structure elucidation. Ths last step could also preceed the MIC value determination.)


Structures of Thiosulfate, Dithionate & Tetrathionate Ions

Here you go...

Thiosulfate $\ce{[S2O3]^{-2}}$

There ia a negative charge on the apical sulfur and another negative charge distributed across the 3 oxygens by resonance

Dithionate $\ce{[S2O6]^{-2}}$

There is one negative charge distributed by resonance over each group of 3 oxygens.

Dithionite $\ce{[S2O4]^{-2}}$

There is one negative charge distributed by resonance over each group of 2 oxygens.

Tetrathionate $\ce{[S4O6]^{-2}}$

There is one negative charge distributed by resonance over each group of 3 oxygens.


How can I dispose of chlorine and sodium hydroxide as table salt and water?

It looks as if the simplest way would be to mix the products, and let the mixture warm up, so we get sodium chlorate in water. We can then have that crystalize and heat it until it disintegrates into $\ce{NaCl}$ and $\ce{O2}$. That would need no additional substances and have a clean end result. The only remaining question would be the energy budget, but the forming of the chlorate is exothermic and additional heat might be available.

Alternatively, we might find something useful to oxidize with the $\ce{NaClO3}$.


Memorizing polyatomic ions? Using Periodic Table

Number of oxygens:

If your anion is in:

the 2nd period;
the VIIth main group/the 17th group/a halogen; or
is silicon:

then the -ate anion will have three oxygens.

All other -ate elemental anions will have four oxygens.

(Take note of ‘aluminate’ which I haven’t physically seen in that form yet but your teacher seems to insist be $\ce{AlO2-}$; disregarding this general trend.)

From the -ate anion, remove one oxygen to arrive at the -ite, remove two for hypo-ite. Add one for per-ate.

Charge:

Start at an anion with the most oxygens. (Note: Usually this is the -ate but because the halogens and permanganate are special they have the per-ate anion which is more oxygen-rich and important here.) Assume the maximum possible oxidation state for the non-oxygen atom. Assume $-\mathrm{II}$ oxidation state for every oxygen. Add up and take the negative value to arrive at the charge.

A hydrogen- will add $\ce{H}$ and lower the charge by one.

A thio- means one oxygen is replaced by sulphur.

A di- means take two of the (hydrogenated) anion and subtract water. Pyro means the same thing.

Memorise all the remaining! That is most importantly:

Chromite. Similar to aluminate I’  not sure if I ever saw it out in the real world.
Cyanide/cyanate. Thiocyanate is just like thiosulphate.
Oxalate, acetate and tartrate. Organic ions that will never fit into this scheme nicely.
Peroxide and superoxide. But those are -ides anyway.
Permanganate; which is almost like a perhalogenate. Luckily, molybdate and chromate fall into a category described above.

Periodic Table


Why are gas laws only applicable to gases?

Gases, liquids and solids have different interactions between their components

Ultimately they are applicable to gases because they only apply to gases. The reasons why they work are worth understanding, though, because they are critical to the development of chemistry which might not exist if not fro some of the convenient properties of gases.

The most important law is that a given number of atoms or molecules as a gas always occupies the same volume if temperature and pressure are fixed. This was what allowed the generation of data that formed the basis of modern chemistry: things combine in fixed ratios. This led to the experimental proofs of the existence of atoms which would be a lot harder to do if we had to rely on measuring the volumes of solids or liquids where the volume tells us nothing useful about how much stuff makes up the substance. It is easy to measure the volume of a gas.

But why is this true? The essence of the reason why gases make up a fixed volume whatever they are made of, is that, to a good approximation, in gases the component parts are small, far apart and have only elastic interactions. If you do the statistics of such a system a whole bunch of apparently relevant factors cancel out (like the mass and velocity of the particles) to give a system which obeys the key gas laws independently of the molecule or atom making up the gas.

But this only works in gases. Liquids have strong interactions between their molecules; not strong enough to prevent individual molecules wandering around randomly but strong enough to hold them to a fixed volume without needing a container (gases will expand to infinite volume without a container to prevent them diffusing into the atmosphere). Solids have even stronger interactions that hold their individual molecules or atoms in a well defined location. The specific properties of solids and liquids depend on the details of those forces and, since there are a wide variety of forces and force strengths, the collective properties are a lot more complex.

In gases, the ideal gas-law approximations are good because the molecules of the substance have enough energy to break the attractive forces that would otherwise make them solids or liquids. Once free of those attractive forces they tend to occupy a much larger volume and their behaviour becomes better explained by the (simpler) gas laws as we can start ignoring the complexity of the attractive forces.

But the gas laws are an approximation. If they were not, nothing would ever liquefy or solidify. But the hotter a gas gets the more it behaves ideally as the less the attractive forces are relevant to the behaviour of the gas.


Pre: My views may be clashing with some people. I express them as far as I know.

To start with, hybridization is a hypothetical concept$^{*1}$ just to explain all facts and make our work easier, otherwise we would have to take into account every single interaction of electron-proton and also other phenomenons; many theories developed which explained all the major/important facts at a macro-level. And what I have been saying to take into account was all micro-level analysis. Attributing hybridization is purely personal choice because neither does it account all facts (for some molecules) nor it can be dropped altogether (for its $\cdots$), if then we would look for another theory$^{*2}$ to explain observed phenomenon. Parallels similarity between this and the situation of Ideal-Gas equation, in general no perfect theory has been developed, but the most easy, practical and useful are the hybridization$^{*3}$ and van-der-Waals equation.

As we move downwards in the periodic table the hybridization concept fails at many places due to overpowering of other factors. I wonder you didn't mentioned the bond angle$^{*4}$ ($93.5^\circ$) for a $\ce{sp^3}$ hybridized (ideally $109.5^\circ$).

In general atom/molecule doesn't seek to follow a theory, it does what it is best comfortable with, it is us, who mould the theory to the observations.

As a general rule, for sake of convenience, assume all molecules show hybridisation and the tendency to hybridise decreases down the group.

In conclusion, I would say it depends on the need of the situation and such dubious statements such as "Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?" should be replaced with "Does hybridization explain all (actually not all) properties of $\ce{PH3}$"

Also there are weights in $\ce{sp^\alpha}$ where $\alpha$ ranges from 1 to 3(even upto 4$^{*5}$ and maybe beyond) and is in between these for $\ce{PH3}$ and some may continue to treat it has a hybridised molecule with suitable weights.

$^{*1}$(as far as I know)
$^{*2}$(probably the MOT)
$^{*3}$(many may argue for other theories which is genuine, but in context this follows)
$^{*4}$(One may argue as we move down the group electronegativity decreases and atomic size increases. In case of $\ce{NH3}$ due to higher bond pair bond pair repulsion (since electronegativity of $\ce{N}$ atom is very high hence it attracts bonded electrons of $\ce{N-H}$ bond towards itself) bond pair moves away from each other and hence shows greater bond angle. This may be considered one of the major factors for less hybridization.)
$^{*5}$[$\ce{O}$ in $\ce{H2O}$ which means that they have 20% s character and 80% p character, but does not imply that they are formed from one s and four p orbitals]


Why do we say that a transition state that resembles a stable species in free energy also resembles it in structure?

EDIT for a different approach:

The energy of a quantum mechanical system of nuclei and electrons is governed by the positions of the pieces. So, we're looking at the energy associated with the wave functions of all the pieces.

For two species that are similar in structure, the idea is that those two structures lead to relatively similar solutions of the wave functions. If the wave functions are similar, the energies of the particles described by those wave functions must be similar as well.


Reaction of vicinal diol in presence of benzaldehyde

This reaction is similar to the one where you use ethylene glycol to form a temporary protecting group for an aldehyde that you do not want to react in further steps. The answer therefore should be (b).


Setup for Sensing change in concentration of NaCl

This will work with the correct set up. You're constructing a galvanic cell, which will result in a cell voltage. The cell voltage will change when you alter the concentration of NaCl on either side. Important though that your two half cells (each side) are connected both electronically and ionically. Do you have a salt bridge or a membrane between them? https://en.wikipedia.org/wiki/Salt_bridge