What do positive, negative and zero overlap of atomic orbitals mean?

The terms in the lecture slides you refer to describe the situation in a certain, very specific subset of molecular orbital theory: the idea of linear combining single atomic orbitals to form bonding and antibonding orbitals. While one can arrive at the same bonding/antibonding orbitals with elaborate molecular orbital theory through the localisation of orbitals approach, the simplified view these lecture slides (and many others) present is not consistent with molecular orbital theory. (However, a subset of the result can be obtained by a proper symmetry analysis.)

Let’s consider a rather simple case such as $\ce{HCN}$. It is obvious that the carbon atom’s atomic orbitals will in some way interact with the atomic orbitals of both hydrogen and nitrogen to form bonds. In the method presented in the lecture slides, you would look at certain pairs of orbitals and then determine what type of interaction would be between. I will only present a few representative examples. For consistency, I will define the molecule axis to be my coordinate system’s $z$ axis.

The $2\mathrm p_z$ orbital of carbon and the $2\mathrm p_z$ orbital of nitrogen.

These two p-type orbitals are aligned in the same way. Without taking their phases into account, we can see that the $-z$ lobe of the nitrogen orbital and the $+z$ lobe of the carbon orbital can interact. Depending on whether the phases match or do not match, we would arrive at a bonding and an antibonding orbital, respectively. The resulting bonding orbitals would be a $\sigma$ and $\sigma^*$ orbital.
The $2\mathrm p_x$ orbital of nitrogen and the $2\mathrm p_x$ orbital of carbon.

These orbitals are parallel and still well aligned. However, at any one time both lobes of one will interact with both lobes of the other. There is a nodal plane that completely contains the bond axis. If the $+x$ lobe of both orbitals feature the same phase, the interaction is bonding. Otherwise it is again antibonding. The bonding orbitals are $\pi$ and $\pi^*$, respectively.
The $2\mathrm p_x$ orbital of carbon and the $1\mathrm s$ orbital of hydrogen.

Now, we are trying to mix an orbital that only has one phase (the s orbital) and one that has two phases (the p orbital). This would be okay if only one of the two phases of the p orbital were pointing towards the s orbital; however, because the p orbital is perpendicular to the bond axis, both its phases would have to interact at the same time. Thus, we have one interaction on the $+x$ half of the coordinate system which is bonding and at the same time an interaction on the $-x$ half which is antibonding. Mathematically, the bonding and antibonding contributions cancel out perfectly. Thus, there is no interaction between the two orbitals. This is zero overlap.
The $2\mathrm p_x$ orbital of carbon and the $2\mathrm p_y$ orbital of nitrogen.

Now, we are attempting to mix two different p orbitals. However, they are not oriented in the same direction. The $+x$ direction of one orbital and the $+y$ direction of the other have the same phase while the $-y$ and $-x$ directions have the other. If we start at $+x$ and go around in a circle, we will start in an area of no interaction (the $\mathrm p_y$ orbital has a nodal plane in the $xz$ plane). Then, turning towards $+y$ we reach an area of bonding overlap. After crossing $45°$, the bonding overlap decreases until we reach zero again at $+y$. From there to $-x$, we experience an increasing and decreasing antibonding interaction. We then rinse and repeat what we just saw for the remaining $180°$.

It would probably not be impossible to consider this interaction a curly bond that rotates around the central axis by $90°$ between the two atoms. However, traditional analysis would note that at the same time there is something like an antibonding second part displaced by $90°$ to this. Therefore, the analysis would consider these two orbitals nonbonding with zero overlap.

In a proper MO or LCAO approach, molecular orbitals will stretch across the entire molecule. Thus, the analysis of whether an orbital is bonding or antibonding with respect to a certain bond reduces itself to the question of ‘does this orbital feature a nodal plane along the bond axis or does it not?’. From the $+,-,0$ picture we had above, we arrive at a black and white picture. This is because nonbonding interactions are typically separated out before the molecular orbitals are even calculated by a symmetry analysis. If you do that, you would realise that the $2\mathrm px$ orbital and the $1\mathrm s$ mentioned in case three are representative of two different irreducible representations. (Strictly speaking, $2\mathrm p_x$ and $2\mathrm p_y$ both belong to the same degenerate irreducible representation.) Only orbitals that belong to the same irreducible representation can be combined in bonding or antibonding orbitals.


Boiling of water does not involve temperature change, so why is there a change in internal energy during the process?

The change in internal energy $U$ is dependent on the heat transferred $q$ and the work done $w$.

$$\Delta U = q + w$$

$\Delta U = 0$ in two cases: when $q=-w$ and when $q=w=0$.

The heat transferred is not necessarily zero if there is no temperature change. The most common expression of heat transfer is $q=C\Delta T$, where $C$ is the heat capacity of the system and $\Delta T$ is a temperature change. We can also write $q=ms\Delta T$, where $m$ is the mass of the system and $s$ is its specific heat capacity (per unit mass). However, heat is being transferred during boiling (else how does the liquid boil?). We need a new equation for heat that does not require temperature.

In a constant pressure system (most boiling happens in a constant pressure,  open to the atmosphere, system), change in enthalpy is equal to $q$ since all of the work and other pressure-volume terms cancel. The enthalpy change of vaporization is usually given in $\mathrm{kJ}$ per mole or unit mass. Now we have a new equation for heat:

$$q_{vap}=m\Delta_{vap}H$$

Let's say we are boiling 100.0 g of water. Water's enthalpy of vaoprization is $2257\ \mathrm{kJ/kg}$. Thus the heat transfer is:

$$q=(0.1000\ \mathrm{kg})(2257\ \mathrm{kJ/kg})=225.7\ \mathrm{kJ}$$
Note that in this case, the work done is also not zero. There is a volume change of the system. Let's take our 100 grams of water.

$$V=0.1000\ \mathrm{kg}\times \dfrac{\mathrm{m^3}}{958.4\ \mathrm{kg}}=1.040\times 10^{-4}\ \mathrm{m^3}$$

At the boiling point, the density is $958.4\ \mathrm{kg/m^3}=0.9584\ \mathrm{kg/L}$. Using the ideal gas law (water vapor is reasonably ideal at atmospheric pressure and its boiling point), we can determine the volume of 100.0 g of water vapor, given that

$$100.0\ \mathrm{g}\ \ce{H2O}\times \dfrac{1\ \mathrm{mol}\ \ce{H2O}}{18.02\ \mathrm{g}\ \ce{H2O}}=5.549\ \mathrm{mol}\ \ce{H2O}$$

$$V=\dfrac{nRT}{P}=\dfrac{(5.549\ \mathrm{mol})(0.080257\ \mathrm{\frac{L\cdot atm}{K\cdot mol}})(373.15\ \mathrm{K})}{1\ \mathrm{atm}}=169.9\ \mathrm{L}$$

We can now calculate work. So that we get joules out, I am going to convert pressure to pascals and volume to $\mathrm{m^3}$.
$$P=1\ \mathrm{atm}=101,325\ \mathrm{Pa}\\
V_{gas} = 169.9\mathrm{\ L} =0.1699\ \mathrm{m^3}\\
\Delta V= 0.1699\ \mathrm{m^3} - 1.040\times 10^{-4}\ \mathrm{m^3} =0.1698\ \mathrm{m^3}\\
w=-P\Delta V=-101325\ \mathrm{Pa}\times 0.1698\ \mathrm{m^3}=-17.20\ \mathrm{kJ}$$

Now we calculated $\Delta U$:

$$\Delta U = q +w = 225.7\ \mathrm{kJ} - 17.20\ \mathrm{kJ} = 208.5\ \mathrm{kJ}$$


Is there a systematic approach to determing structural formula from NMR spectra?

Of course there is a systematic approach to solving structure. The key is that it is an iterative process, and it is not uncommon to have to repeat the cycle several times. For complicated molecular structures (for example, alkaloids with C35+ frameworks) you might get 20 or more possible candidate structures that you have to work through very closely.

Of course, the method you use will depend on the class of compound, but for unknown organic molecules, I would start with a molecular mass and NMR data from 1H, COSY, HSQC, HMBC. Subsequent iterations or refinements might require NOESY, TOCSY and if there is enough material maybe even an ADEQUATE.

The general approach that I teach is as follows. The first step is to identify the molecular formula from mass spec data and calculate the double bond equivalence. Subsequently:

1. Environments

From 1H and 13C data, determine number of carbon environments, and account for all carbon atoms, as these will represent the individual jigsaw pieces
Determine proton environments attached to 13C - tally up CH3, CH2, CH groups.
Tally up integrals, identify symmetry and equivalence.
This should account for almost all of your C and H atoms

2. Chemical shifts

Establish functional group types based on chemical shifts (aromatic, alkene etc).
Look for possibility of heteroatom inclusion that ties in with the molecular formula

3. Couplings and spin systems

From obvious 1D coupling and COSY data, start assembling spin system fragments
From HMBC start assembling C–C connectivities
Consider drawing out a correlation map to show connectivities, especially for complicated systems
You should be reasonably confident about the spin system fragments leading into the next stage

4. Structure hypothesis

Assemble spin system fragments into allowable possible structures, and test the structure against the data - that is rationalise the chemical shifts and couplings for all signals to see that they fit the model. If not, refine and try again.

A number of structure elucidation software packages exist, and they essentially base their method on building up a molecular correlation map, and then doing a chemical shift prediction for the proposed structure to give an overall confidence score. The sort of molecular map you might get looks something like the picture below. On the left are all the explained correlations, and on the right the incorrect correlations. What is important is look closely and the incorrect correlations, and rationalise them. In this case, all of the incorrect correlations would be allowable, and therefore we would have good confidence in this structure (which is understandable, as we know exactly what it is!). If we couldn't rationalise some of the correlations, we need to either reconsider our molecular structure, or re-analyse our data.


Why do post-Hartree-Fock methods fail to predict the direction of the dipole moment of carbon monoxide?

The question itself is void, as all methods with the exception of Hartree-Fock predict the direction of the dipole moment correctly. Deathbreath found a Full CI calculation in Jeremy P. Coe, Daniel J. Taylor, and Martin J. Paterson. J. Comp. Chem. 2013, 34 (13), 1083-1093 of which luckily an open version exists (arXiv:1301.4904 [physics.chem-ph]). The following passage is taken from the latter:

  The groundstate dipole moment of carbon monoxide, although fairly small, when calculated using HF strikingly has the incorrect sign compared with experimental results. Previous work has suggested that the accuracy of the dipole calculation
  depends on the amount of correlation accounted for.14  The bond length (2.1316 Bohr) and the experimental dipole value (0.122 Debye) are taken from Ref. 15. The positive value for the dipole here signifies a polarity of $\ce{C^−O^+}$.
  With a cc-pVDZ basis, two frozen core orbitals and a cut-off value of $c_\mathrm{min} = 5 \times 10^{−3}$, we see in Fig. 1 [omitted] that the MCCI method, starting from close to the incorrectly signed result of the HF single SD, quickly reaches a correctly signed value which converges at around half of the FCI value.

  14 G. E. Scuseria, M. D. Miller, F. Jensen, and J. Geertsen. J.
  Chem. Phys. 1991, 94, 6660.
  15 J. S. Muenter. J. Mol. Spectrosc. 1975, 55, 490.

The problem indeed is just a technical one, resulting from analysing the wrong density. Gaussian 09 by default only analyses the SCF density (see population, density). In order to obtain values for the post-HF methods, the density=current keyword needs to be specified. The updated table therefore becomes:
\begin{array}{lr}
\text{Method} &
\mathbf{q}(\overrightarrow{\ce{CO}})/\mathrm{D}&
\mathbf{d}(\ce{CO})/\mathrm{10^{-10}m}\\\hline
\text{Experimental}^{15} & 0.122 & 1.128\\\hline
\text{HF} & -0.14 & 1.102\\\hline
\text{MP2}  &   0.26 & 1.135\\
\text{MP3}  &   0.08 & 1.118\\
\text{MP4(SDQ)} & 0.11 & 1.118\\
\text{MP4(SDTQ)//MP4(SDQ)} & \text{n.a.} \\
\text{CCSD} &   0.08 & 1.124\\
\text{CCSD(T)//CCSD} & \text{n.a.}\\
\text{CISD} & 0.05 & 1.118\\
\text{QCISD} & 0.08 & 1.127\\
\hline
\text{BP86} &    0.19 & 1.135\\
\text{PBE0} &    0.12 & 1.122\\
\text{M11}  &    0.06 & 1.119\\
\text{B3LYP} &   0.10 & 1.124\\
\hline
\text{B2PLYP} & 0.15 & 1.129\\
\hline
\end{array}

Sample Input for Gaussian 09 for analysis
(requires the addition of %chk=mp2qzvpp.chk to the sample input of the question)

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
Geom=check
Guess=(Read,Only)
Density=(Check, Current)

dens-curr

0 1

I apologise for getting everyone worked up over a stupid beginner mistake.


Determine the oxidation number of sulfur in SF6

In the case of $\ce{SF6}$, sulfur would have the oxidation number of +6 because the charge being applied to the fluorine is +6. Similarly fluorine would consequently have an oxidation number of -1 since $$6x + 6 = 0 \Rightarrow x = -1$$

(the right-hand side is equal to zero since that happens to be the net charge on the overall chemical formula).


Change of ionic packing upon crystals are pressurized

Your termination of BCC is incorrect about the $\ce{NaCl}$ crystal structure upon pressurization. It is actually simple cubic (SC as $\ce{CsCl}$) where eight $\ce{Cl-}$ ions ate the corners of unit cell and one $\ce{Na+}$ in the center or vise versa. Although either $\ce{Na+}$ or $\ce{Cl-}$ in the center, it is not BCC because ion possessing the center of unit cell is different from the ions at the corners of unit cell:

Thus, lattice constant $a$ can be expressed as:
$$a_\mathrm{SC} = \frac{2(r_\ce{Na+} + r_\ce{Cl-})}{\sqrt{3}} \tag1$$
For FCC $\ce{NaCl}$ crystal structure:
$$a_\mathrm{FCC} = 2(r_\ce{Na+} + r_\ce{Cl-}) \tag2$$
Since $a_\mathrm{SC} \lt a_\mathrm{FCC}$, we can conclude that upon pressurization, $\ce{NaCl}$ crystal structure would be  contracted.


How does impurity of distilled water vary with time in air?

Gases like oxygen and nitrogen from air are absorbed quite easily. As you pour the water to the beaker it is most likely already saturated with those gases. Other air components and contaminants (carbon dioxide, sulfur dioxide, nitrogen oxides, ammonia) will also get quickly absorbed. How fast they will contaminate your sample depends on their concentration in the air, temperature of the water, size of the container and so on. Other possible contaminants are dust particles, impurities present inside the beaker, and even glass itself (depending of its type) is slightly soluble in water. The concentration of solid impurities will usually not change very quickly. Of course, as the water evaporates, solid impurities will concentrate.


Could a magnet pull oxygen out of the air?

I'm a physicist, so apologies if the answer below is in a foreign language;  but this was too interesting of a problem to pass up.  I'm going to focus on a particular question:  If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region?  The TL;DR is that thermal effects are going to make this idea basically impossible.

The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$, where $\vec{B}$ is the magnetic field.  Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point:  $\vec{m} = \alpha \vec{B}$, where $\alpha$ is what we might call the "molecular magnetic susceptibility."  Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$.  But potential energy is given by $\vec{F} = - \vec{\nabla} U$;  which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$.

Now, if we're talking about a sample of gas at a temperature $T$, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor:
$$
\rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT}
$$
In the limit where $kT \gg \alpha B^2$, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample.  To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$;  in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.

So how high does this have to be?  The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$; and so we have1
$$
\chi_\text{mol} B^2 \gtrsim \mu_0 RT
$$
and so we must have
$$
B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}.
$$
If you believe Wikipedia, the molar susceptibility of oxygen gas is $4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$;  and plugging in the numbers, we get a requirement for a magnetic field of
$$
B \gtrsim \pu{258 T}.
$$
This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines.  Even at $\pu{91 Kelvin}$ (just above the boiling point of oxygen), you would need a magnetic field of almost $\pu{150 T}$; still well out of range.

1 I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules.  A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation;  and if the gas gets sufficiently dense, then all bets are off.


Nitration of tri-substituted benzene with Acetic acid and HNO3

Your reaction goes way back to 1930s when Ingold and coworkers have done pioneering work on  nitration of aromatic compounds using nitric acid in organic solvents in place of sulfuric acid (vide infra). Let's see how this has been developed:

Historical perspectives: The earliest recorded experiments on the kinetics of aromatic nitration has been performed in 1884, when the nitration of benzene and some of its simpler derivatives with various amounts of nitric acid have been examined at various temperatures (Ref.1). No solvent has been used (just nitric acid), and thus, the reaction mixtures were heterogeneous. The first nitration of aromatic substances with nitric acid in sulfuric acid was obtained in 1904, when the velocity constants in the nitration was studied (Ref.1). In their punlication in 1938, Ingold and coworkers have reported the kinetics of nitration using nitric acid in organic solvents (other than sulfuric acid), specifically nitromethane as the solvent (Ref.1). In this studies with nitromethane as the solvent and nitric acid in constant excess, they demonstrated zeroth order kinetics for the nitration of benzene and activated aromatic nucleous such as toluene. They have observed that, when less reactive compounds such as chlorobenzene were used, there were some dependence of the rate on the compound. Hence, they have continued these experiments with considerably less reactive compounds such as 1,4-dichlorobenzene and ethyl benzoate to show that less reactive compounds have established first-order kinetics on nitration.

Originally, Benford and Ingold thought that their results were peculiar to nitromethane, yet they have observed ail the same phenomena with acetic acid as solvent (your case!). Based on these data, Ingold and coworkers proposed the formation of nitronium ion ($\ce{NO2+}$), which may be the rate-determining if it is sufficiently rapidly removed by the aromatic compound used as the substrate. They proposed the actual mechanism of the formation of nitronium ion as follows (Ref.2):

$$\text{In sulfuric acid:} \: \: \ce{HNO3 + H2SO4 <=> H2NO3+ + HSO4-}\\
\text{and} \: \: \ce{H2NO3+ + H2SO4 <=> NO2+ + H3O+ + HSO4-}$$

$$\text{In organic solvents:} \: \: \ce{2 HNO3 <=> H2NO3+ + NO3-}\\
\text{and} \: \: \ce{H2NO3+ <=> NO2+ + H2O}$$

This mechanism is well established now. Based on these fact, I hope I answered your question. In their further work, Ingold and coworkers have proposed the mechanism for nitration of aromatics based on the nitronium ion formation (Ref.3). In that publication, they proposed existence of nitronium ion through rapid equilibrium with concentrated nitric acid, and that mechanism is believed to be the case to date.

References:

Gilbert A. Benford, Christopher K. Ingold, “177. Influence of directing groups on nuclear reactivity in oriented aromatic substitutions. Part V. A dilatometric investigation of the kinetics of aromatic nitration especially in nitromethane solution,” Journal of the Chemical Society 1938, 929-955 (DOI: 10.1039/JR9380000929).
E. D. Hughes, C. K. Ingold, R. I. Reed, “Kinetics of Aromatic Nitration: the Nitronium Ion,” Nature 1946, 158(4013), 448-449 (https://doi.org/10.1038/158448c0).
R. J. Gillespie, E. D. Hughes, C. K. Ingold, D. J. Millen, R. I. Reed, “Kinetics and mechanism of aromatic nitration,” Nature 1949, 163(4146), 599-600 (https://doi.org/10.1038/163599b0).


How can we measure the rate of reaction of EDTA chelation with zinc carbonate?

I think that the reaction will be too fast for you to measure with the tools in a high school lab. I think that you are not measuring the zinc binding to EDTA but instead making a measurement of the dissolution rate of zinc carbonate in an aqueous solution of EDTA.

If you were to want to measure the rate of the binding rate of zinc by EDTA the best way might be to use a radioactive tracer. I would start with radioactive zinc in an anionic EDTA solution. I would bring this into contact with a non-radioactive solution of zinc. I would then keep on pulling out small samples from the mixture and filtering them through an ion exchange resin which is able to remove either the anionic EDTA complex or the cationic form of zinc.

I would use zinc nitrate or some other form of zinc which does not tend to form anionic complexes when in solution.

I know that this experiment has been done as a teaching lab experiment in a nuclear chemistry unit with radioactive europium, but it has a reputation as a difficult experiment to do.

I suggest you read Williams, K.R.; Choppin, G.R., 1974,  J. Inorg. Nucl. Chem., v. 36, no. 8, pp. 1849-1853. This has details of an experiment on the kinetics of metal exchange between EDTA complexes and "free unsolvated metal" for trivalent metals.

An alternative non radioactive method would be to use stopped flow experiments with a metal senstive indicactor present in the mixture. If we assume that the exchange of zinc between the solution and the indicator is very fast then this method could be of use.

Another method could be to take a solution of the zinc complex of EDTA and then treat it with C13 KCN, you could measure the C13 NMR spectrum again and again if it is a relatively slow process. But I think for you that you will be unable to deal with this problem with the chemcials and tools you have in a school lab. Sorry for having to pour some cold water around.


I see where I made my mistake. I set $[\ce{L}] = 5$, but this "$L$" is referring to the equilibrium concentration of free ligand -- not the total ligand concentration. In order to do that I should have constrained the system to $[\ce{L}] + [\ce{LR}] = 5$. Working through what this would do currently.

So in the first case, my assumption actually dictated that there be $\pu{10 mol}$ ligand total instead of the $\pu{5 mol}$ I erroneously stated. I incorrectly equated total ligand and free ligand the first time I did this.

So I see where I messed up, but now I'm curious if the values of $[\ce{L}]$, $[\ce{R}]$, and $[\ce{LR}]$ can be solved as functions of $[\ce{L_{total}}]$, $[\ce{R_{total}}]$ and $K_\mathrm{D}$.


Why does acetone have a lower boiling point than hexane?

Hexane is a larger, longer-chain molecule so there are more London dispersion forces between n-hexane molecule which outweighs the dipole forces in acetone.


Calculate the total number of valance shell electrons ($n_1$) in all constituent atoms or ions. For example, in $\ce{NH_4^+}$, $n_1$ will be 8 (5 of N, 4 from all H's and -1 because the molecule has +1 charge).
Now, calculate the total number of electrons that are required to fill valance shell of every constituent atom or ion ($n_2$). In other words, calculate the total number of electrons that will be required to fill valance shell of each atom or ion individually. We can use following formula to find $n_2$$n_2 = [\text{# of Hydrogen atoms} \times 2] + [\text{# of all other atoms except Hydrogen} \times 8]$
In case of $\ce{NH_4^+}$, $n_2$ will be 16 ($[4 \times 2] + [1 \times 8]$).
Now, calculate the total number of electrons that atoms or ions will need to share among each other in order to complete their octets ($n_3$).
$n_3 = n_2 - n_1$
For $\ce{NH_4^+}$, $n_3$ will be 8.
Now calculate the number of bonds that will be present in our molecule
$n_4 = \frac{n_3}{2}$
For $\ce{NH_4^+}$, $n_4$ will be 4.
Now let's find the number of unshared electrons, $n_5$
$n_5 = n_1 - n_3$For $\ce{NH_4^+}$, $n_5$ will be 0.
Now find the number of lone pairs, $n_6$, which is nothing but $n_5$ divided by 2._For $\ce{NH_4^+}$, $n_6$ will be 0.
Now find the central atom. Central atom is the one which is either least in count or more electropositive.
Always remember that Hydrogen and Fluorine can never be placed at center.
So in our case, Nitrogen gets to be the central atom.
After allocating the central atom, place the surrounding atoms such that the molecule is symmetric.
In our case, geometry of molecule will be tetragonal.
Distribute the bonds between atoms. Remember that the total number of bonds, that you distribute, should be equal to $n_4$.
In our case, we simply place a bond between Nitrogen and each Hydrogen atom.
Now use the lone pairs to complete the octet of atoms. Remember that, by now, every atom will already have 2 electrons in their valance shell because of the bond.In our case, which is $\ce{NH_4^+}$, we don't have any lone pairs. OOPS!
Now comes the Formal Charge.

  Formal Charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in a chemical bond are shared equally between atoms, regardless of relative electronegativity. It is not an actual electrical charge but just a mathematical construct.

  $\text{Formal Charge of an atom} = \text{# of valence shell electrons (in Ground state)} - \text{# of unshared/non-bonded electron pairs or lone pairs} - \text{# of bonds}$

Please note that you must place formal charge on each and every atom when it is a non-zero value for that atom.
This step is little confusing so let's get back to our example. Nitrogen has 5 electrons in its valance shell in ground state. It made 4 bonds in our molecule and has none unshared electron pairs or lone pairs (please note that a single unpaired electron doesn't count here). So formal charge on Nitrogen will be 1 (5 - 0 - 4), whereas formal charge on each Hydrogen atom will be 0 as every Hydrogen atom has 1 electron in its valance shell and made one bond in our molecule.

In a sense, you actually obtain the structure in $10^{th}$ step but Formal charge, if a non-zero value, should be represented for each atom in molecule in order to complete its Lewis Dot Structure.


Tap water pH adjustment using citric acid

Maybe citric acid have had a negligible effect on the $\mathrm{pH}$. Everybody knows that in contact with air, pure water is carbonated by $\ce{CO2}$ from the atmosphere. And the $\mathrm{pH}$ may go down to $5.5$. Later on, the water may loose its $\ce{CO2}$, and the $\mathrm{pH}$ goes back to $7$.


How much water vapour can a gram of Uppsalite absorb at NTP?

One fallacy in your question is assuming that the particles 1) are spherical and 2) only accessible on the outside. The press release linked says that this material is nanoporous - the particles are perforated by pores "all having a diameter smaller than 10 nano meters". The extra surface area is inside these pores. Spheres have the lowest surface-area-to-volume ratio. Other shapes have higher ratios (in other words, more surface area for less volume). These nanoporous materials maximize the surface area to volume ration - minimizing the volume. If these particles are $\ce{MgCO3}$, then their density is $\pu{2.958 g/cm^3}$ (Wikipedia). One gram of this material has a volume of:

$$\pu{1 g}\times \pu{1 cm^3 g^-1}\times \left(\frac{\pu{1 m}}{\pu{10^2 cm}}\right)^2=\pu{3.38E-7 m^3}=\pu{3.38E-3 L}$$

The surface-area-to-volume ratio is then:
$$\frac{A}{V}=\dfrac{\pu{800 m^2}}{\pu{3.38E-7 m^3}}=\pu{2.37E9 m^-1}$$

A second fallacy is that the volume of water that can be absorbed is the same as the volume of the particle. The particle is not absorbing water per se. The water is not taken into the interior volume where the $\ce{MgCO3}$ is. That would require displacement of the magnesium carbonate, and where would it go? The water goes into the pore of the material - it adsorbs to the surface inside those pores.

You can read more about the science at the original research article:  A Template-Free, Ultra-Adsorbing, High Surface Area Carbonate Nanostructure. PLoS ONE 2013, 8 (7). DOI: 10.1371/journal.pone.0068486.


Why can we use standard entropy when determining temperature at which a reaction becomes spontaneous?

The question implicitly assumes that $\Delta H^\circ$ and $\Delta S^\circ$ are independent of temperature.

The temperature dependences depends on the change in heat capcity ($\Delta C_p$) are given by $\Delta H^\circ(T_2) = \Delta H^\circ(T_1) + \Delta C_p(T_2-T_1)$ and $\Delta S^\circ(T_2) = \Delta S^\circ(T_1) + \Delta C_p \ln(T_2/T_1)$.  (These equations assume $\Delta C$ is constant.)

So if $T_2-T_1$ or $\Delta C_p$ is small then $\Delta H^\circ$ and $\Delta S^\circ$ will be fairly constant.


Connection of term symbols with specific microstates for atomic carbon

To add on to what Feodoran has mentioned, you might be interested in looking at the Clebsch–Gordan coefficients, which are essentially the coefficients of the microstates in the term symbols. You can look these up in e.g. the appendices of Atkins Molecular Quantum Mechanics, 5th ed. Note that each term symbol is a collection of states, so a $D$ term (for example) which has $L  = 2$ would comprise five states: one each with $M_L = -2, -1, 0, +1, +2$. These five states can in general be denoted as $|L, M_L\rangle$.

$L$, which is a total angular momentum, is obtained by a coupling of two individual angular momenta. For two electrons in p-orbitals, these individual angular momenta are $l_1 = 1$ (with $m_{l1} = -1, 0, +1$) and $l_2 = 1$ (with $m_{l2} = -1, 0, +1$). The microstates in the table specify both values of $m_{l1}$ and $m_{l2}$, so they are simultaneous eigenstates of $m_{l1}$ and $m_{l2}$, i.e. they can be denoted by $|m_{l1},m_{l2}\rangle$.

Now, in general each coupled representation $|L, M_L\rangle$ will be a linear combination of the microstates, or uncoupled representations $|m_{l1},m_{l2}\rangle$. It is in general not possible to construct a state which is simultaneously an eigenstate of $L, M_L, m_{l1}$, and $m_{l2}$, because the relevant operators do not commute (refer to Atkins or any other QM text on coupling of angular momentum). There are examples of one-to-one correspondence between the coupled and uncoupled representations – for example, microstate 13 perfectly corresponds to $|L = 2, M_L = +2\rangle$, but it is not general, so:

$$ |L, M_L\rangle = \sum_{m_{l1},m_{l2}} c^{L, M_L}_{m_{l1},m_{l2}} |m_{l1},m_{l2}\rangle$$

The Clebsch–Gordan coefficients are simply the $c^{L, M_L}_{m_{l1},m_{l2}}$ in the expansion above. Usually the notation is slightly different – so instead of $(L, M_L, m_{l1}, m_{l2})$, it is usually $(j, m_j, m_{j1}, m_{j2})$ – but the concept is entirely the same.

Using the coefficients for $j_1 = j_2 = 1$ in Atkins (which is relevant to the $\mathrm{p^2}$ configuration of atomic carbon), we find that the $^1S$ term is given by

$$|L = 0, M_L = 0\rangle = \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 1, m_{l2} = -1\rangle + \frac{1}{\sqrt 3} |m_{l1}= -1, m_{l2} = 1\rangle}_{\text{a combination of microstates 8 and 11}} - \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 0, m_{l2} = 0\rangle}_{\text{microstate 14}}$$

It's slightly more confusing because there is a spin component as well, which the coefficients do not cover, and this is why I've given a handwavy "combination of microstates 8 and 11". This "combination" is needed to generate the singlet spin wavefunction $2^{-1/2}(|\alpha\beta\rangle - |\beta\alpha\rangle)$.

As an aside, the familiar singlet and triplet spin wavefunctions

$$\underbrace{|\alpha\alpha\rangle, |\beta\beta\rangle, \frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{triplet spin wavefunctions}}, \underbrace{\frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{singlet spin wavefunction}}$$

are also coupled representations. These four wavefunctions are, respectively,

$$|S=1, M_S=+1\rangle, |S=1, M_S=-1\rangle, |S=1, M_S=0\rangle, |S=0, M_S=0\rangle$$

(Different notation again, but it is the same concept.) The uncoupled representations are the states in which each electron has been assigned a spin, i.e. $|\alpha\alpha\rangle, |\alpha\beta\rangle, |\beta\alpha\rangle, |\beta\beta\rangle$. For example:

$$|m_{s1} = +1/2, m_{s2} = +1/2\rangle = |\alpha\alpha\rangle$$

and so we can see a simpler example of how the coefficients link the two representations:

$$\begin{align}
|S=1, M_S=+1\rangle &= |m_{s1} = +1/2, m_{s2} = -1/2 \rangle \\
|S=1, M_S=-1\rangle &= |m_{s1} = -1/2, m_{s2} = -1/2 \rangle \\
|S=1, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle + \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle  \\
|S=0, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle - \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle \\
\end{align}$$


Peculiar behaviour of tea boiling as milk+Water

Original Hypothesis

As mentioned in this post, milk will overflow thanks to the effect of surfactants, which allow for the formation of stable bubbles. Since tea contains both milk and water, it contains surfactants from the milk, but at a lower concentration thanks to the water. This would result in some bubble formation, but not as much as you would find in pure milk, explaining why the tea filled up the container, but did not overflow. Either it was a coincidence that the bubbling stopped right at the top of the container, or it relates to the fact that more surface tension is required to create bubbles without the support of the container walls. In addition, other components in the tea could be contributing to this effect, but I think it is a result of the fact that milk and water are mixed. If you want to experiment with this, I would try a different volume of your water/milk mixture and see if that also makes it to the top of the container without overflowing.

Results of Additional Experiments

The results of your additional experiments seem to indicate that something else is going on. It is interesting that you found a different ratio that filled the container when there was no tea involved. This seems to indicate that the tea is at least somewhat involved in this interaction. Also, the fact that past the ratio of 1:5 (milk:water) the container did not fill up with bubbles seems to indicate that my hypothesis about surface tension was wrong, and that there is nothing special about filling the container. In this case, you probably just got lucky that your original tea did not overflow.

These results are dependent on the fact that you used the same vessel for your new experiments and allowed adequate time for bubbles to form. More information on your experimental procedures will help me narrow down on a possible explanation.


Is the mixture of alkyl dimethyl benzyl ammonium chloride and vinegar dangerous?

Please note that I worked in medical research, but I am not a medical doctor!

I think that you're safe, and I cannot imagine a reaction between acetic acid, the main component in vinegar, and a quarternary ammonium salt that would release harmful products.

Benzalkonium chloride is a long known disinfectant and has been used as a biocide and preservant in medical products. However, it may cause problems. The European Medicines Acency has published a Q&A on Benzalkonium chloride.

In general, it is not the worst idea to have a look at MSDS (Material Safety Data Sheets) for compounds and products, in order to be aware of what to do if something spills on your skin, gets into your eyes, is swallowed, or inhaled. Here is an example of a MSDS for benzalkonium chloride.

To sum it up:

Mixing household chemicals can release harmful products, as known for the reaction of bleach and acid, from which chlorine gas is released.
I seriously doubt that you have created a harmful mixture in your case.
Manufacturers are obligated to provide MSDS, wich have to include first aid measures! Take a look at this vital information.


What is the correct definition of the Gibbs free energy of activation?

There are a lot of possible definitions and it possibly depends very much on the case and the context you are looking at. However, there are the definitions provided by the IUPAC goldbook, which can be considered the official ones.

The first definition you suggested is correct so far and known as (standard) Gibbs energy of activation (standard free energy of activation) $\Delta^\ddagger{}G^\circ{}$:

  The standard Gibbs energy difference between the transition state of a reaction (either an elementary reaction or a stepwise reaction) and the ground state of the reactants. It is calculated from the experimental rate constant  $k$ via the conventional form of the absolute rate equation:
  $$\Delta^\ddagger{}G^\circ{} = \mathcal{R} T \left[\ln\left(\frac{\mathcal{k}_\mathrm{B}}{h}\right) − \ln \left(\frac{k}{T} \right)\right]$$
  where  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant and  $h$ the Planck constant ($\frac{\mathcal{k}_\mathrm{B}}{h} = 2.08358\cdot 10^{10}~\mathrm{K^{−1}s^{−1}}$). The values of the rate constants, and hence Gibbs energies of activation, depend upon the choice of concentration units (or of the thermodynamic standard state).

Your second definition refers to the definition of enthalpy of activation, $\Delta^\ddagger{}H^\circ{}$:

  The standard enthalpy of activation $\Delta^\ddagger{}H^\circ{}$ is the enthalpy change that appears in the thermodynamic form of the rate equation obtained from conventional transition state theory. This equation is only correct for a first order reaction, for which the rate constant has the dimension reciprocal time. For a second order reaction, for which the rate constant has the dimension (reciprocal time) × (reciprocal concentration), the left hand side should be read as  $k c^\circ$, where $c^\circ$ denotes the standard concentration (usually  $\mathrm{1 mol\cdot dm^{−3}}$).
  $$k = \frac{\mathcal{k}_\mathrm{B} T}{h} \exp\left\{\frac{\Delta^\ddagger{}S^\circ}{\mathcal{R}}\right\} \cdot\exp\left\{\frac{-\Delta^\ddagger{}H^\circ}{\mathcal{R}T}\right\}$$
  The quantity  $\Delta^\ddagger{}S^\circ$ is the standard entropy of activation, and care must be taken with standard states. In this equation  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant,  $T$ the absolute temperature,  $h$ the Planck constant, and $\mathcal{R}$ the gas constant. The enthalpy of activation is approximately equal to the activation energy; the conversion of one into the other depends on the molecularity. The enthalpy of activation is always the standard quantity, although the word standard and the superscript $^\circ$ on the symbol are often omitted. The symbol is frequently (but incorrectly) written  $\Delta{}H^\ddagger{}$, where the standard symbol is omitted and the  $\ddagger$ is placed after the  $H$.

The activation energy (Arrhenius activation energy) $E_a$ itself is defined as:

  An empirical parameter characterizing the exponential temperature dependence of the rate coefficient, $k$,  $$E_a = \mathcal{R} T^2 \frac{\mathrm{d}(\ln k)}{\mathrm{d} T}$$, where $\mathcal{R}$ is the gas constant and $T$ the thermodynamic temperature. The term is also used for threshold energies in electronic potential surfaces, in which case the term requires careful definition.

And while we are at it, entropy of activation $\Delta^\ddagger{}S^\circ$:

  The entropy change that appears in the thermodynamic form of the rate equation obtained from conventional transition-state theory.

Additional note and general warning
Although there are the official definitions, the term activation energy is often used as a synonym for many kinds of different barriers. One has to be very careful when reading articles, which definition is used.
In computational chemistry articles you may also find the terms electronic energy of activation, zero-point corrected energy of activation, solvent corrected energy of activation, ..., etc., pp., and many more.


Since Avogadro's Law is speaking about ideal gases then it's true for all gases that behave ideal.
Since ideal gases don't interact with each other and don't have any volume for single molecules (they only have a volume as a whole) Avogadro's Law holds for different gases.

Since for real gases the approximations of the ideal gas don't hold then Avogadro's Law doesn't hold. Yet a ratio similar can be achieved by dividing V over n of different gases to get the molar Volume and by using different equtions to describe real gases a ratio factor can be achieved for a given pressure and temperature.


How do I build a thermochromic thermometer?

We calculate the $K_c$ values. Ratios that give $K_c$ greater than 1 are product favored and will turn from blue to pink at low temps. The opposite is true for the ratios that give $K_c$ values less than 1. These are reactant favored, and will turn from pink to blue at the low temp. First, look to see what is chaging color most drastically, meaning the reaction is proceding to a greater extent ($K_c$ value a lot greater than or less than 1), then look at the cost. If it's too expensive, a thermometer that is less effective but still OK, may be better.

Overall, think of a thermometer that you stick on the side of a fishtank (sort of).


Changes in UV-Vis spectra in host-guest complexes

Can anybody clarify which serves as evidence for formation of a host-guest complex?

There's no single, general answer to this.

As you aptly described in the body of your post, there is a wide variety of changes to the UV-Vis absorption of a host-guest system on binding of the host and guest.  The changes to the recorded spectrum could fall into any of these categories, depending on the particular structural features of host and guest, and the exact manner that they bind to one another.

AFAIK, the only way to know how the spectrum changes when host and guest bind, is to perform careful experimentation and observe the changes that occur.


Should the anode potential be increased or decreased to generate current?

Use Nernst equation. If you increase anode potential, the cell potential becomes more negative, thus decreasing Gibb's energy. This makes the reaction less feasible.


Let me try to answer the part 1. of your question.
The process of compacting powdered Aluminium is also called Sintering. And as you can guess, the thin oxide layer of Alumina inhibits the binding of Aluminium grains between the powder particles.

So, while sintering, we need to either:
1) Break part of the Alumina film or,
2) Reduce part of it to Aluminium,
 in order to facilitate formation of a metallic bond between the powder particles.

For 1), we use a nitrogen atmosphere as it enhances the destruction of Alumina.
For 2), we use because magnesium as a alloying element because magnesium, even in small amounts, concentrates on the surface of powder particles
and supports sintering by local reduction of Alumina.

In conclusion, I would say that compacted Aluminium does have Alumina inside the molded object. But metallic bonds between powders are formed after partly removing the oxide film by the above mentioned methods.


Which compound will have the most stable enol form?

The most questions about keto-enol tautomerism can be answered by their NMR spectral data (e.g., evidence of existing enol-form of acetylacetone at room temperature). According to its $\ce{^1H}$-NMR spectral data, it was declared that phloroglucinol, in its neutral form, exists nearly exclusively in its aromatic form (Ref.1). However, their dianion form (e.g., disodium phloroglucinolate) favors keto structures. These results were further analysed and confirmed by kinetic studies (Ref.2).

Wikipedia states that:

  4-Pyridone is an organic compound with the formula $\ce{C5H4NH(O)}$. It is a colorless solid. The compound exists in equilibrium with a minor tautomer, pyridin-4-ol.

Martin - マーチン has given here a nice answer using computational evidence for this preference  of 4-Pyridone (data are from Ref.3).

Based on these, it is clear that in their neutral forms, phloroglucinol has the most stable enol form, existing in predominantly in phenol form (1,3,5-trihydroxybenzene).

References:

R. J. Highet, T. J. Batterham, “The Structure of the Phloroglucinol Dianion,” J. Org. Chem. 1964, 29(2), 475-476 (https://doi.org/10.1021/jo01025a501).
Martin Lohrie, Wilhelm Knoche, “Dissociation and keto-enol tautomerism of phloroglucinol and its anions in aqueous solution," J. Am. Chem. Soc. 1993, 115(3), 919-924 (https://doi.org/10.1021/ja00056a016).
Peter I. Nagy, Giuliano Alagona, Caterina Ghio, “Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution,”  J. Chem. Theory Comput. 2007, 3(4), 1249–1266 (https://doi.org/10.1021/ct6002252).


Can silver metal be used to electrochemically reduce copper?

In principle, electrically depositing copper(II) ions onto a silver electrode should be possible. As it should also be possible to deposit those ions on just about every other cathode in existence as long as there are no more noble cations in solution.

However, depending on what you want to do with the copper, you may run into problems. Do you want copper-plated silver? That should be fine. Do you want to redissolve the copper to get a pure solution by acidifying? In that case, you cannot use the typical $\ce{HNO3}$ method because that would oxidise the silver, too, and you would end up with a mixture of copper(II) and silver(I) nitrates in solution.


What is the velocity of an ejected electron in lithium when subjected to UV radiation?

The energy is still the sum of the kinetic and the binding energy. The only difference is how you calculate the binding energy, but you could just look it up


What is the reaction mechanism between boric acid and 2-ethyl-1-hexanol

The research paper sighted in the question (Ref.1) is about the extraction of boron (as boric acid: $\ce{H3BO3}$) from salt lake brine using 2-ethylhexanol/kerosene as organic phase. It is true that the given diagram in this paper (as shown in the question) is showing an esterification, but the authors refer it to two references, one of them is Ref.2. However, the association used to calculate the distribution of boric acid between aqueous and organic phases in Ref.1 is not the esterification. Accordingly to the paper, the association reaction is as follows:
$$\ce{H3BO3 + nROH <=> H3BO3.nROH}$$
Since there are no water molecule elimination during the reaction, this is not an esterification. It is a solvolysis reaction as OP correctly recognized.
Extraction of boron from existing sources is a challenging problem (Ref.3). Boron can be extracted as neutral ester $(\ce{B(OR)3})$, as boric acid complexes $(\ce{(H3BO3)_x.(ROH)_y})$ or as tetrahydroxy borate $(\ce{R+B(OH)4-})$. However, the extraction of boric acid by monohydric alcohols is caused by purely physical actions and the amount of extracted boric acid is roughly proportional to the concentration of an extractant (Ref.3). Accordingly, it is safe to say the extraction process is based on complexation of boric acid with intended alcohol.

References:

Ran Zhang, Yingming Xie, Jianfeng Song, Lixin Xing, Dingfeng Kong, Xue-Mei Li, Tao He, "Extraction of boron from salt lake brine using 2-ethylhexanol," Hydrometallurgy 2016, 160,  129–136 (DOI: https://doi.org/10.1016/j.hydromet.2016.01.001)(PDF).
Jianhua LÜ, Jidong LIU, Yujie SUN, Chunli LI, "Kinetics of Forward Extraction of Boric Acid from Salt Lake Brine by 2-Ethyl-1,3-hexanediol in Toluene Using Single Drop Technique," Chinese Journal of Chemical Engineering 2014, 22(5), 496-502 (DOI: https://doi.org/10.1016/S1004-9541(14)60059-8).
Adam Balinski, Volker Recksiek, Norman Kelly, "Solvent Extraction of Boric Acid: Comparison of Five Different Monohydric Alcohols and Equilibrium Modeling with Numerical Methods," Processes 2021, 9(2),  398 (15 pages) (DOI: https://doi.org/10.3390/pr9020398).


Order of Basicity of Nitrogen Trihalides

In case of Boron, boron has vacant orbitals which leads to pi back bonding from halides to boron and that concept is used to explain the basicity of Boron. However Nitrogen doesn't have vacant orbitals, hence that pi bonding is not possible from halide to Nitrogen. Consequently electronegativity is used for explanation for basicity of Nitrogen Halides.


What happens to the lost mass in radioactive decay?

On the nuclear physics side, it goes in to kinetic energy of the various particles as well as any gammas. If it is a simple reaction (D+D -> 3He+n or 3H+p) you can determine the energy of each particle in the center of mass by simple kinematics. In something resulting in more particles (most fission events result in 2 nuclei and multiple neutrons) you get a spread of kinetic energies of the various bits. And, if one or more of the nuclei are left in an excited state than you can get emission of gammas, energetic photons.

As these daughter particles interact with the rest of the world, they will be thermalized through interactions with electrons and nuclei surrounding them, transferring kinetic energy, resulting in heat to the surroundings. Any gammas will eventually be absorbed/scattered, also resulting in energy transfer to the surroundings. If this did not happen, fission reactors would have a hard time making steam to turn turbines.


How is the "quasi-steady-state" assumption of Michaelis-Menten kinetics different from the steady state assumption of ordinary kinetics?

Is Wikipedia's use of "quasi" in the title for the enzyme kinetics reflective of [...] the further assumption that $[S]$ is constant[...]?

No I don't think the "quasi-steady state" hypothesis is related to whether $[S]$ varies with time.

  [W]hy add a further descriptor that denotes uncertainty or inaccuracy when we already know that steady state is an assumption?

Different fields use different terms for the same concept.  For example, some folks call it the "pseudo-steady state hypothesis" and abbreviate it "PSSH".  Some people just say "steady state assumption".  It's all different vocabulary for the same thing.

I don't think there's a principled reason to prefer one term over another.  If you solve the full differential equations for kinetic models that would normally invoke the PSSH, you'll see one problem.  Before the reaction starts, the concentration of intermediate $I$ is zero.  But when the reaction is going on, we assume the intermediate is at steady state, i.e. that $\frac{dI}{dt}=0$.  But if $I$ starts at zero and has an initial concentration of zero, then it is everywhere zero, which violates the prinicple that intermediates actually occur in real-life chemical reactions.  So there are two phases to such models: the first is the "burst" phase, also known as single-turnover conditions, or unsteady state kinetics.  In that phase $\frac{dI}{dt}$ is not 0 but increases rapidly.  That phase ends quickly and is replaced by the pseudo-steady phase for $I$.  If you want to get precise about the strict definitions of the two phases, you'll need to invoke singular perturbation and asymptotic matching, as described in this thorough book chapter from Mathematical Biology by James D. Murray.

So you could argue that the "pseudo" or "quasi" is good to have in there to remind you of these complexities.  Or you could argue that all those extra letters aren't doing a whole lot of good.  Either way, it's just a semantic issue and is unrelated to differences in enzyme kinetics and more "traditional" chemical reactions.


Additive for water to increase the viscosity to a value between 5 and 20 cP

You wasn't clear about your purpose to increase viscosity of water (water has viscosity of $\pu{1 mPa\cdot s}$ or $\pu{1 cP}$). However, whatever your reason, xanthan gum would be the materiel you are looking for. Kyoto Electronics has published a note on measurement of viscosity of xanthan gum solution. According to their results (average of 5 tests), 0.1% xanthan gum solution has shown viscosity of $\pu{19.1 cP}$. Five sample averages of 0.2% and 0.3% solutions have given viscosity readings of $\pu{57.54 cP}$ and $\pu{134.6 cP}$.
Accordingly, you can use range of percentages (0.01-0.1%)of xanthan gum to get water viscosity of $\pu{5-20 cP}$. The amount of material to add to water is minimal as you wanted. In addition, Casson plot (shear rate versus shear stress) of 0.1% xanthan gum solution is a straight line. Hence this solution is believed to be a Newtonian solution.


How to distinguish mass concentration and density?

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

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

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

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

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

References

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


What does 'preconcentration techniques' mean?

Preconcentration means to increase the concentration of a sample prior to analysis or detection. For example you can do preconcentration for an non-volatile organic sample with evaporation its solvent.

An operation (process) as the result of which microcomponents are transferred from the sample of larger mass into the sample of smaller mass, so that the concentration of the microcomponents is increased. Examples include the decrease in solvent volume during distillation or evaporation, and the transfer of microcomponents from an aqueous solution into a smaller volume of organic solvent by extraction.(IUPAC)

There are lots of methods for preconcentration: solvent evaporation, using a different phase (ex: a solid or liquid that could absorbs sample from the solvent) and ...


How to selectively convert benzene to p-nitrotoluene?

I'd doubt that exclusive nitration of toluene in para position is possible in homogenous solution.

But imagine to stuff a toluene molecule into a very narrow "tube". This tube will block any attack at C-2 (and C-3), while a nitrating agent may approach C-4.

Exactly this has been done in the vapour phase nitration of toluene with nitric acid over a zeolite (ZSM-5). See: Kalbasi, R.; Ghiaci, M.; Massah, A. Highly selective vapor phase nitration of toluene to 4-nitro toluene using modified and unmodified H3PO4/ZSM-5. Applied Catalysis A: General 2009, 353 (1), 1–8. DOI: 10.1016/j.apcata.2008.10.013.


UV-Visible Spectroscopy in the analysis of sodium chloride in potato chips

NaCl solution as you know is almost transparent in the VIS region, UV-VIS spectroscopy can be used to determinate concentration of soluble salt see J. Phys. Chem. A 2008, 112, 2242-2247 however UV part of the spectra is used instead of VIS part, so this does not fit the restrictions of your question.

Surely AAS has a greater sensibility and so is the best technique to use in this case.


I can't point to a definitive authority, but in most of the papers I read, #5 other is used most often.  Specifically for ethylamine:

$\ce{^{13}C\,^{12}CH_7N}%edit$

I've always assumed the reason for this choice is to accentuate the presence of the isotopically unusual atoms.  I suppose a good test for my theory would be formulas with multiple, isotopically distinct atoms of an element such as iron, for which the most abundant isotope is not the lightest.

Hopefully other answers will be along shortly to point to something more authoritative, such as a IUPAC report, guideline, or document.


Here is the proton nmr spectrum for pyridine.

Here is the proton nmr spectrum for furan

For comparison, benzene has a single proton nmr signal at 7.27 ppm.  In both pyridine and furan there are different types of protons, so multiple signals are seen in both cases.

Here are the resonance structures for pyridine

and furan.

Pyridine removes electron density from the alpha and gamma positions by resonance while furan adds electron density to all four ring positions by resonance.

In addition to resonance effects there are also inductive effects and ring currents that will modify the proton chemical shifts.  Inductively, the alpha positions in both molecules will be deshielded.  Ring currents will also deshield the protons as you've noted.

All of the signals are in the general vicinity of benzene.  The alpha-protons in both molecules are the most deshielded due to inductive effects, with furan's alpha signal upfield from pyridine's due to the electron donating resonance effect of the furan oxygen.  The beta proton in furan is the most shielded since it is not inductively deshielded and, again, the oxygen in furan is pushing electron density into the ring via resonance which will cause shielding.


Are Organic Solvents used in Addition Reactions catalysts?

In the example you give, the dichloromethane is acting as a solvent. Solvents are usually chosen to be inert under the reaction conditions so that they do not react and produce unwanted products.

There are examples of reactions in which no solvent is used, or in which the solvent is a reagent. An example could be forming an ester from a carboxylic acid, using an excess of the alcohol as both solvent and reagent, in the presence of an acid catalyst.  That said, solvents are usually used as they severely effect the outcome of a reaction. Polar solvents (THF, diethyl ether) promote very different reactions to non polar ones (hexane) due to the stabilisation they provide for intermediates during the reaction, dilution can also play a role, deciding whether inter or intra-molecular reaction occurs.


How could I create a machine that as rapidly as possible creates greenhouse gases from air?

Current concentrations of $\ce{CO2}$ and $\ce{CH4}$ in the atmosphere are around 390 ppmv and 1.75 ppmv, resp. These concentrations and the wish to obtain large quantities of both gases rapidly seem mutually exclusive, unless you use (at least one) additional carbon source.

Taking the options for interconversion into account, $\ce{CH4}$ in pressurized bottles would be an obvious choice ;).


From Wikipedia:

  A chiral compound can contain no improper axis of rotation ($S_{n}$), which includes planes of symmetry and inversion center. Chiral molecules are always dissymmetric (lacking $S_{n}$) but not always asymmetric (lacking all symmetry elements except the trivial identity). Asymmetric molecules are always chiral.

That means that a dyssymetric molecule lacks of $S_{n}$ axis(planes of symmetry and inversion centers) but can have $C_{n}$ axis of rotation. For example the following molecule is dissymetric: it does not have any plane of symmetry or inversion center but it has a $C_{2}$ axis of rotation(the "axis of rotation" is a synonyms of the "axis of symmetry" you mentioned in your question).


Does liquids that emit light when vibrated exist?

Find compounds that emit light under photolysis.

As that may involve radical production pathways, one can also try to apply sonolysis to the same compounds as:

H2O (with dissolved O2, N2,..) + Sonolysis -> *H + *OH

Reference source: 'Free radical formation from sonolysis of water in the presence of different gases'), to quote:

"In this case, O2 reacts with •H to form •OH, indicating that the presence of dissolved O2 in water sonolysis gives additional amount of •OH. "

Also:

H2O2 + Sonolysis -> *OH + *OH

Apparently, with applications, see, 'Bactericidal effect of hydroxyl radicals generated by the sonolysis and photolysis of hydrogen peroxide for endodontic applications'.

where it is likely that one may be able to replicate to varying degrees the action of light treatment.

Here is a source example citing the action of reactive oxygen species (ROS), which can be created by the UV photolysis of, for example, H2O2 (or sonolysis per reference above):

"In biological systems much attention is focused on the two free radicals, superoxide anion and hydroxyl radical....The generation processes of ROS can be monitored using luminescence. Because of the very weak native luminescence of ROS, both luminol and lucigenin have been used in the past to give measurable signals. However, the luminescence of these two substances is very small compared to that of Pholasin®, which allows precise analysis on very small samples."

So, perhaps try performing a sonolysis of, say, luminol or lucigenin, in H2O2 might be interesting.


Benzylic Oxidation of Cyclic Molecules Using KMnO₄

For the homogenous permanganate oxidation of aromatic side chains, as long as there is at least one benzylic hydrogen, then the side chain will be oxidized to the corresponding benzoic acid.  Here are some examples.

(image source)

Look at the third example, the permanganate oxidation of tetralin to phthalic acid.  It is very analogous to your molecule.  Your molecule would also be oxidized to phthalic acid.


What method is used to discover oxidation state of drug binding cysteine residue?

NMR could likely be used to characterize a bulk sample of the drug and identify the presence of cysteine residue oxidation products, but would not be of much use for establishing that cancer cells oxidize (inactivate) the drug, nor precisely how it was inactivated, given the tiny amounts of material involved. Maybe LC-MS could be used. Probably worth pulling up the original paper to see how they did the work.


The Wikipedia page on ether cleavage says:

  Depending on the specific ether, cleavage can follow either $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms. Distinguishing between both mechanisms requires consideration of inductive and mesomeric effects that could stabilize or destabilize a potential carbocation in the $\mathrm{S_N1}$ pathway. Usage of hydrohalic acids takes advantage of the fact that these agents are able to protonate the ether oxygen atom and also provide a halide anion as a suitable nucleophile.

So in essence, the mechanisms for such reactions are usually a trade-off between $\mathrm{S_N1}$ and $\mathrm{S_N2}$ pathways, with adequate carbocation stability playing a major role in the determination of the pathway (as intermediate cases like secondary cations often lead to a "confusing" case where it is difficult to say qualitatively whether the path is majorly $\mathrm{S_N1}$ or $\mathrm{S_N2}$).

In particular, the Wikipedia page cited above has pretty much answered your question under the $\mathrm{S_N2}$ ether cleavage tab:

  If the potential carbocation can not be stabilized, ether cleavage follows a bimolecular, concerted $\mathrm{S_N2}$ mechanism. In the example, the ether oxygen is reversibly protonated. The halide ion (here bromide) then nucleophilically attacks the less sterically hindered carbon atom, thereby forming methyl bromide and 1-propanol.

So for your case, the upper oxygen after protonation can cleave the $\ce{C-O}$ via a $\mathrm{S_N1}$ pathway to generate a reasonably stable potential carbocation which is delocalized over the benzene ring.
Hence,it can simply remove a molecule of methanol(which is a good leaving group by itself) on it's own without needing an iodide to assist in the removal.

The strongly acidic solution of $\ce{HI}$ will anyway ensure it that the methanol undergoes $\mathrm{S_N2}$ to form methyl iodide later. Also, from the point of view of a nucleophile,it would be much easier to perform an $\mathrm{S_N2}$ reaction on a separated methanol molecule than on the entire substrate molecule initially given in the question due to steric considerations,so that also becomes a contributing factor.

But as for the lower oxygen,the poor fellow does not have such a choice. The potential carbocation that can be generated by cleaving the lower $\ce{C-O}$ bond is a phenyl cation,which is pretty unstable as explained here. So, like the Wikipedia page said and illustrated in the example above, the pathway has to be a concerted $\mathrm{S_N2}$ mechanism now so as to take the reaction forward.

Note: Even if the said $\mathrm{S_N2}$ starts out slow, Le Chatlier's principle will ensure that ultimately, most of  the reactant goes into the product as equilibrium is attained.


How to determine the relative contribution of resonance structures when different rules give contradictory outcomes?

The Rules

Actually the rules for determining the relative stabilities (more accurately, contributions) of resonance structures do have their own priorities. Here is one version of them, with descending order of importance listed below (as is indicated by the bold words). The rules can be found in common organic chemistry textbooks (like [1], also supported by [3]).

  In any case, the more the octet rule is fulfilled (for appropriate atoms), the better.
  Resonance structures with charge separation are usually higher in energy than those in which charges can be neutralized, unless rule 1 is not obeyed.
  If charge is separated, then the best structure is consistent with the electronegativity of the atoms (e.g., a negative charge is best on the most electronegative atom).
  The more and stronger the bonds, the better. Adjacent atoms should not have the same sign of charges.

Examples

The first rule is undoubtfully the most important. The example of $\ce{CH2^+-O-CH3 <-> CH2=O^+-CH3}$ is already mentioned. Here I would like to provide an example to justify the importance of rule 1.

  Consider the Lewis acids $\ce{BX3}$ where $\ce{X}=\ce{F},\ce{Cl},\ce{Br}$. We know that $\ce{BF3}$ is the least Lewis acidic despite the fact that $\ce{F}$ is the most electronegative. This fact can be explained by considering the resonance structures $\ce{X2B-X <-> X2B^-=X+}$. According to rule 1, the second resonance structure is more important, so the electron deficiency of $\ce{B}$ is determined by the orbital overlapping between $\ce{B}$ and $\ce{X}$. Therefore, since $\ce{F}$'s $2\mathrm p$ orbitals interact with $\ce{B}$'s best, $\ce{BF3}$ is the least Lewis acidic.

Another interesting example illustrating the relative importance of rule 3 and 4 comes from Grützmacher's work[2].

  The phosphaethynolate anion $\ce{OCP-}$ may attack $\ce{TMSOTf}$ using $\ce{O}$ and $\ce{P}$. At 298K the product is $\ce{TMS-O-C\bond{3}P}$, and at 318K the product is $\ce{TMS-P=C=O}$. To account for this, let's analyze $\ce{OCP-}$'s resonance structures:
  $$\ce{P\bond{3}C-O- <-> O=C=P-}.$$
  The latter has stronger bonds and the former has its negative charge on a more electronegative atom. According to rule 3, the former dominates (actually a ratio of former vs latter 40:52), hence at ambient temperature $\ce{O-}$ attacks more quickly to obtain the kinetic product, while $\ce{TMSPCO}$ is the thermodynamic product.

Theoretical Justifications

You can find theoretical justifications using Hückel Theory as always. I will not show this here because it may be to tedious to do the calculations. However, you can read this article for more information J. Chem. Educ. 2007, 84, 6, 1056.

Ref.

[1]  Grossman, The Art of Writing Reasonable Organic Reaction Mechanisms, Springer, 2002.

[2] Grützmacher et al., Dalton Trans., 2014,43, 5920-5928.

[3] Humbel, J. Chem. Educ., 2007, 84, 6, 1056-1061.


The transition energy, or wavelength if you prefer, are governed by the nature of the molecules involved. Whole body free rotational motion (in the gas phase) gives rise to microwave spectra ( fractions of wavenumbers cm$^{-1}$  or THz frequencies). The frequency is determined by the molecules moment of inertia. At higher energy comes vibrational motion, stretching and bending of chemical bonds, a few hundred to a few thousand cm$^{-1}$. These bond frequencies are approximately given by $\sqrt{k/\mu}$ where $k$ is force constant which relates force to bond extension as in Hook's law, and $\mu$ is the reduced mass. There are 3N-6 types of vibration for N atoms, (3N-5 for linear molecules). At higher frequencies are electronic transitions where and electron gets promoted from a bonding to anti-bonding orbital. Typically these transitions are in the visible and uv regions.

The exact number of transitions is difficult to estimate because each vibrational level has rotational levels superimposed on them and there are also vibrations that are combinations of levels. There are, however, fewer electronic transitions because there are generally few molecular orbitals than types of vibrational modes, and each excited state has its own set of vibrational and rotational levels.

One of a chemist's essential tools, arguably more important that any other spectroscopy, is NMR spectroscopy which is caused by transitions between nuclear spin states when in a magnetic field and occurs at very low frequency, 100-800 MHz range.


What is the proper term for lines of constant volume and constant enthalpy on a thermodynamic diagram?

But what do you call a line of constant volume? An Isochore?

Correct.

  And what do you call a line of constant enthalpy?

The process is isenthalpic. More than one engineering/thermodynamic text I found by Googling indicates they are simply referred to as "isenthalpic lines." See this for example.


Reaction of anisole and succinic anhydride in presence of aluminium chloride

Aluminium chloride ($\ce{AlCl_3}$) is mainly used for this Friedel-Crafts alkylation or acylation, where this compound acts as catalyst, and also controls the production of the desired compound.

For example, in this case, the reactants are anisole ($\ce{Ph -OMe}$) and succinic anhydride. So, the main reaction which drives these reactants towards product is the nucleophilic attack by the para carbon (w.r.t the $\ce{OMe}$ group) of the phenyl ring at the carbonyl carbon of the succinic anhydride, and an addition-elimination of the attached oxygen of the ester part in succinic anhydride.

For this, we need the state of the reactants in a proper way. We have to make sure that, the oxygen of $\ce{OMe}$ does not attack prior to the phenyl ring, and also, the electrophilicity of the carbonyl carbon needs to be increased to much extent, which is generally not so much due to present resonance in the compound.

$\ce{AlCl_3}$ plays a special role over controlling these two phenomena. Being a Lewis acid, it forms a Lewis acid-base interaction with the oxygen of $\ce{OMe}$ and also with the oxygen atom of $\ce{C=O}$ group in succinic anyhdride. Due to this interaction, a δ+ charge is introduced on both oxygen atoms described before. This increased formal charge helps to increase the electrophilicity of the carbon due to increased -I effect of the attached oxygen and also reduces the chances of attacking by $\ce{OMe}$ group. This catalyses to form our desired product with the proper reaction between the reactants.


Why isn't my galvanic cell making enough voltage?

Just keep in mind that Eo is determined under highly controlled conditions and concentrations. I sense two key problems here. The major problem is already stated above that the resistance of a wet filter paper must be high, and this is giving you lower voltage. Evaporation of water from the filter paper would make the situation worse. Since you are not repeating the experiment anyway, make sure to identify the causes of lower voltage. Secondly, both lead and aluminum nitrate tend to form hydroxides in water if you don't add little bit of acid in them. Did you notice that lead solutions were cloudy or they were clear and transparent?

As a side note, negative voltage reading of the battery just means that your connections of the lead are the other way round on the voltmeter. Test your voltmeter with a battery keeping the signs on the battery in your mind. Once you get a positive reading of a simple AAA cell you have correctly made the connections.


Is the enthalpy of a molecule determined by the energy of the electrons?

I find that enthalpy is one of the concepts that students struggle with the most and I feel your pain in trying to understand it!

  I understand that the strength of the bond of a molecule is determined by the potential and kinetic energy of the electrons

The strength of a bond is best understood in terms of potential energy. A strong bond involves a deeper potential energy "well" that electrons and bonded atoms can exist in. The atoms involved in the bond (and the electrons too) will have kinetic energy and that can explain how the bond behaves over time, but the overall intrinsic strength of the bond is a reflection of how low the potential energy of the system is at the equilibrium bond length. We call this the bond dissociation energy and it directly tells us how strong the bond is.

https://en.wikipedia.org/wiki/Morse_potential

  My textbook states that in a reaction where NaOH dissolves in water, the energy of the solid NaOH is greater than the energy of the dissolved ions in water. Is the "energy" of the reactants and products determined by the bond strength?

Yes, this refers to the potential energy of the system that is decreasing when the stronger attractions are formed in the solution vs the weaker ones broken in the undissolved ionic lattice and bulk water. Essentially, you have to break weaker bonds/IMF's in the reactants then you form in the products. Please note that this is a general statement and includes all "bonds" (covalent, ionic or metallic) plus other types of attraction (IMF's, ion-dipole forces etc.) that exist before or after a process or reaction occurs.

  Also, my textbook states that enthalpy is "the amount of energy, in the form of either kinetic or potential energy, a substance has." I'm a bit confused about enthalpy. My textbook enthalpy changes:

The key part here is the potential energy. When this decreases (as in your example of NaOH dissolving), the potential energy turns into thermal energy. By itself, this doesn't change the enthalpy, as that is the sum of all internal energy at constant P. However, it will raise the temperature of the system and this makes it out of thermal equilibrium with the surroundings. Heat now flows from the system and this is when the enthalpy decreases as the system is losing energy. It is best to think of the enthalpy decreasing when the potential energy of the system decreases as the heat flow is what then tends to result. The reason this happens, as you are suggesting, is because stronger bonds are forming than existed before.

  Are these enthalpy changes determined by the bond strength also? And is the enthalpy of a certain molecule also from the bond strength?

Hopefully this last bit makes sense now, and the relationship between bond strength and the enthalpy of the system is a but less mysterious!!


I'm fairly sure this question is merely testing one on their ability to count atoms. If the structure has a chemical formula of $\ce{(C6H10)_m}$ where $m \in \mathbb{Z}^+$, then it is – at least in principle – a possible polymerisation product of cyclohexene $\ce{C6H10}$. Whether it is formed in real life is a different question, but my belief is that that is not part of the question, so you're probably overthinking this if you're wondering about how exactly 1 is formed.

1 ($\ce{C18H30} \equiv \ce{(C6H10)3}$) and 2 ($\ce{(C24H40)_n} \equiv \ce{(C6H10)_{4n}}$) fulfill the above criterion, but 3 ($\ce{(C21H36)_n}$) does not.


The carbon atom in question is called pseudo-asymmetric. Check it out on the IUPAC page.


how do you do the others?

You do it like a detective. Even the slightest details matter, and you've to link a lot of concepts together.

The main theme you're missing in your question is this equilibrium:

$$\ce{ CrO4^2- <=>[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$

Dichromates are supposed to be stable in an acidic medium. So using this you can figure out that ($7$) is $\ce{H+}$ and ($8$) is $\ce{OH-}$. Again for ($5$) you're adding a base which is converting it to a chromate. So ($5$) should be a dichromate $\ce{Cr2O7^2-}$.

So we're only left with ($9$) and $\ce{Cr^3+}$ and they are bound to go with each other. Since this is not an exam here's a reasoning too,

$\ce{K2Cr2O7}$ is a strong oxidizer in acidic mediums. If it's going to oxidize others, it will itself get reduced. To be precise here's the reaction that's taking place:

$$\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$$


Will (3R,4S,5R)-1,3,4,5,6-pentahydroxyhexan-2-one tautomerise in nitric acid?

Well I am just a high school student, so I may be wrong, and I beg pardon in advance, but I would like to share my notions about this topic.

@ChakravarthyKalyan I would like to ask you to review your last step in the conversion of 3d to 1. 3d on oxidation should yield this product (in the image)--which is by my knowledge not same as 1.

So even your theory does not yield (d) as the correct option, I assume. Also I doubt the double tautomerism of the compound in option (d) in acidic medium, though your mechanisms were convincing. This closely resembles Lobry-Debryn-van-Eckenstein rearrangement that occurs only in presence of alkaline medium.

As per my knowledge, ketoses in presence of strong oxidizing agents like conc. HNO3, undergo oxidative cleavage forming 2 carboxylic acids(acc. to Popoff's rule), or as @user55119 had pointed out to me, that alpha ketoacids can undergo oxidative decarboxylation and form CO2 and an acid(though its mechanism is quite unclear to me as well).


Standard Half-Cell Reduction Potential for Periodate to Iodine

Here is the standard half-cell reduction potential $\ce{IO4^{−}/I2}$ in acid medium depicted using Latimer diagram.

(source)

Note: Here $\ce{IO4-}$ (periodate ion) gets hydrated to form orthoperiodic acid which is $\ce{HIO4.2H2O}$  or $\ce{H5IO6}$.


Unfortunately, your reasoning is wrong, because you forgot to take into account the acidity of water. While in most cases we can ignore the acidity of water as the hydrogen ion concentration contribution by $\ce{HCl}$ dominates, at an extremely low concentration of $\ce{HCl},$ water becomes the main $\ce{H+}$ contributor.

It is well known that water has a $\pu{10^{-7} M}$ concentration of hydrogen ions at $\ce{25^\circ C}.$ Therefore, water's effect on the acidity of the solution is significant when looking at solutions of $\pu{10^{-6} M}$ or lower concentrations of strong acids. For example, if there is $\pu{10^{-9} M } \ce{HCl,}$ then the total $\ce{H+}$ concentration would still nearly be $\pu{10^{-7} M}.$ Therefore, no matter how much you dilute the acid, it can never turn into a basic solution.


In biochemistry, when do rate constant units matter in calculating free energy changes?

$K$ should be a unitless value. Otherwise, you'd get a nonsensical unit when you try to take the log of the unit. Each component of the product $K$ is usually an activity, which is frequently similar in value to the molar concentration, or a fugacity, which is frequently similar in value to the partial pressure of a gas.


Equations for internal energy and their relevance to the derivation of the equation for entropy for perfect gases

Your equation $dU = TdS - pdV$ is valid for all pure substances, not just ideal gases.  It is telling you that if you knew an equation that explicitly defined $U$ as a function of $S$ and $T$, then you could calculate all other thermodynamic properties of that substance, i.e. if you knew that $U=f(S, V)$, then you could easily calculate $T$ and $p$ because $T=\frac{\partial f(S, V)}{\partial S}$ and $p=\frac{\partial f(S, V)}{\partial V}$.  This is because $S$ and $V$ are canonical variables for $U$.

Now let's apply the ideal gas equation $dU=C_V dT$.  You can integrate this to yield $U=C_V T + K$.  Unfortunately, this does not express $U$ as a function of $S$, but rather of $T$.  So how do we proceed?  This physics.se question has a good general answer.

But in this specific case things are bit easier.  Let's find $dS$ from the equation you were given.

$dS =\frac{1}{T}(dU + p dV)=\frac{1}{T}(C_V dT + \frac{RT}{V}dV)=C_V \frac{dT}{T}  + R \frac{dV}{V}$

Since you now know $dS$ and from the ideal gas law you also know $p dV$, you have an expression for $dU$ from which you can derive all thermodynamic quantities.


Why does dry boiling discolor steel cookware?

Stainless steel have many different compositions. If you heat it up the surface undergoes oxidation. Diffusion at higher temperatures plays also its roll - shifts surface composition of the kettle. The various shades of yellow and purple are metal oxides. The vinegar removes the oxides, thus removes the coloration. Vinegar should remove the oxides even without boiling. It will just take more time.


It's simply a protonolysis:

Except in the picture D is used instead of H. Hydroboration based reactions always proceed via this boron-ate complex (negatively charged) then migration and loss of leaving group. Search google for carbonylation, amination, cyanidation using boron.


Acetic acid or ethanoic acid - trivial names vs systematic names and correct formula representation

While I agree that the systematic name ethanoic acid makes a lot more sense, the name acetic acid has been used for so long that it's the accepted IUPAC name and is used almost universally. Both refer to the same compound.

Molecular formulas like $\ce{C2H4O2}$ are often seen when doing elemental analysis and the like, but give no structural information. $\ce{CH3COOH}$ makes it clear that it's an acid and shows the carbonyl group.


Products of radical bromination and mechanism

My prof says that I am missing my R/S. I can clearly see my chiral centres, but how do I determine the orientation of the H? Red is the major product, blue is the minor product, and magenta is the very minor product.

You don't determine the orientation of the H you just acknowledge the formation of both the R and S stereo isomers in your products in 50/50 proportions.

  My prof says that I am missing a propagation step with another bromine-bromine molecule. I would greatly appreciate your input!

All you need to show is the organic radical reacting with diatomic bromine to form you brominated molecule and another bromine radical.


Why does ionization energy increase as we go from left to right in a period?

Crash Course on Ionization Energy:

As we all know, atoms prefer a full valence shell. So as we go right in a period, we are increasing $e-$. And also ADDING PROTONS. Because we are adding protons, the size of the atom gets smaller because the nuclear charge will be more powerful. Adding protons in a period trumps the addition of electrons. At the end of the day, we have a small atom with many electrons in it's valence shell that does not want to let go of them. Especially the Noble Gasses and Halogens.
Ionization energy decreases as we move down a group because:

As we move down, a new full energy level is being added. More electrons means more repulsion.
This creates the shielding effect where the addition of the shells, shields the outer electron from receiving the nucleic charge. NOTE: Here, however the addition of another energy level trumps the addition of protons.

This is just a piece of the whole picture.


How to distinguish between Na2CO3 and NaHCO3 by a chemical test?

Heating salt is an important test in determining the anion in the salt by checking the gas evolved. This test is called dry heating test and is a important part of qualitative analysis of salt. But this test goes under physical examination of salt (checking color, appearence etc.).

One important chemical test to differentiate carbonate and bicarbonate is $\ce{MgSO4}$  test.

$$\ce{Na2CO3 + Mg^2+ -> MgCO3↓(white) + Na+}$$

$$\ce{NaHCO3 + Mg^2+ -> MgCO3↓(white) + H2O + CO2 ^}$$

When carbonate salt is reacted with $\ce{MgSO4}$, a white salt precipited while bicarbonate salt on reacting with $\ce{MgSO4}$, not only a white salt precipited but also a brisk effervescence occur. The gas  is detected  by passing it through lime water.


What is the proper Lewis structure for HCOOH?

With familiarity you will recognize that $\ce{COOH}$ in a formula generally refers to a carboxylic acid group. Sometimes you will also see $\ce{CO_2H}$. Either is acceptable. Vinegar, or acetic acid, is a carboxylic acid. Its formula is $\ce{CH3COOH}$ or $\ce{H3CCOOH}$.

It is important to realize that a carboxylic acid group is NOT a peroxide. Peroxides have  involve O-O bonds. You may also generalize that the oxygens in peroxides have a negative one oxidation state. A common peroxide is hydrogen peroxide, $\ce{H2O2}$ or $\ce{HOOH}$.

Also note that carbon is generally tetravalent - i.e. it is commonly found in stable molecules as having an octet of electrons. In addition, carbon generally exhibits no overall formal charge. So this suggests that most of the time you will see carbon forming four bonds - and this is the case the majority of the time. This configuration gives carbon no net formal charge and fills its octet.

Of course, you may find carbocations or carboanions, but carbocations are highly unstable and exist only ephemerally. Carboanions also exist; see the cyanide ion for an example.

The above, in addition to keeping in mind that a carboxylic acid is a Bronsted/Lowry acid - i.e. proton donor - should enable the construction of an acceptable Lewis structure.


Do I have to consider the molecular mass of the oxygen atom or the diatomic oxygen molecule when determining the empirical formula of an iron oxide?

When determining the empirical formula of a compound, you only consider the molar mass of the element. In the compound itself, it is only the atom (or ion) that exists.

If you would assume the molar mass of dioxygen, then you would also assume, that the iron oxide is made out of iron and dioxygen molecules.

I do not know, and also do not understand, why the question states "dioxygen" as that is - in my opinion - false.