A Van-der-Waals surface Area (VdW) is a surface constructed by spheres around atoms with the radius of each sphere being the van-der-Waals radius of the respective atom.

The Solvent accessible surface area (SAS) is an surface normally being constructed by "rolling" a sphere the size of the vdw-radius of the solvent molecule around the solute and taking the middle of the solvent sphere as the surface of the solute.

Picture is taken from the original GEPOL publication of Pascual-Ahuir and Silla J. Comp. Chem., 1990, Vol. 11, No. 9, 1047-1060

So the difference is just the size. Both have the same questionable physical meaning. They are used for example in continuum solvation models like COSMO/PCM.

I used the SAS for Frozen-Density-Embedding-models to close unphysical holes (in PCM the surface constitutes an infinite continuum of solvent, so there would be an infinite volume of solvent between two subsystem molecules) between molecules of different subsystems in a QM/QM/Cont. multilayer method. Nonetheless, they are both only viable for qualitative results.

For the PSA, just read the wikipedia article.

The PSA is only the surface of the polar atoms of a molecule and it is used in medicine to calculate permeability of cell membranes of molecules in the body.


Calculating the cell potential of a half cell from two half cell equations

Josh, you and others have missed the key point. It is not a simple question! They are asking you to draw a potential diagram not Ecell, basically a Latimer diagram.
So take the approach of using a Latimer diagram. The book answer is correct. Nitrous acid when goes to nitrous oxide, the potential must be +1.29 with respect to SHE. Check the web for solving Latimer diagrams.
You would work on the portion starting from nitrous acid to nitrous oxide.


How to apply the Henderson–Hasselbalch equation when adding KOH to an acidic acid buffer?

If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html

and Does the number of H+ ions in solution go up on dilution of a weak acid?

Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted.

pH = 14 - pOH

pH = 14 + log(0.025) = 12.4

In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example.


Why can mercury(I) exist, but not zinc(I)?

On the contrary, zinc(I) compounds do exist, though they are rare, and relatively unstable. Most zinc(I) compounds contain a $\ce{[Zn2]^{2+}}$ core, which is analogous to the $\ce{[Hg2]^{2+}}$ cation. The $\ce{[Zn2]^{2+}}$ ion does, however, rapidly disproportionate into zinc metal and zinc(II), and has only ever been obtained by cooling a solution of metallic zinc in molten $\ce{ZnCl2}$ and in a few compounds such as decamethyldizincocene.
This should show you that the chemistry of zinc and mercury really aren't all that different. The $\ce{[Hg2]^{2+}}$ ion's much greater stability relative to $\ce{[Zn2]^{2+}}$  is due to the atypically large ionization enthalpy of the $\ce{Hg}$ atom ($100.7\:\mathrm{kJ\:mol^{-1}}$ greater than $\ce{Zn}$), which in turn due to relativistic stabilization of the $\mathrm{6s}$ orbital. Relativistic effects account for a myriad of the properties that are characteristic of heavy metals, including why mercury is the only metal that is a liquid at room temperature, and even gold's color.


Gaseous state: Boyles, Charles and Lusaac's law

Actually, it's a big NO.

Why?

Because you need to consider the statement of the laws carefully. Boyle's law states that pressure is inversely proportional to volume at isothermal conditions. Boyle's law is not valid if temperature isn't constant. The same goes for Charles' law, where you must have isobaric conditions. The conditions required for both laws aren't the same, and so you can't really make a logical conclusion that way.


α-Halo ketone and strong base: Favorskii vs SN2

I would comment however I am new and have yet to unlock that feature.

Do you have an exact substrate of interest? I have studied the Favorskii rearrangement a tiny bit in a graduate mechanisms course and if your seeking a very general answer to a general question; The favorskii seems to be favorable when your substrate is containing rings that are conformationally rigid where our alpha-halo-ketone moiety is found.

Check out Eaton's 1964 synthesis of Cubane to find two really elegant Favorskii reactions in the synthesis of Cubane. Notice for almost all of our examples of favorskii reactivity we notice a common pattern in the substrate scope?

Rings! My theory is conformational rigidity of a ring is partially responsible and we can confidently point our fingers to Favorskii where cyclopropanone intermediates are possible AND energetically favorable (conformation argument).

If our substrate scope is including non-ring alpha-halo-ketones.. We may want to consider alternatives like SN2 to predict reactivity.


What is (approximately) the minimum exchange current for a potentiometric measurement?

Pfeiffer at al. defines sensor effective (SERC) and sensor ineffective redox couples (SIRC) in relation to the biggest net exchange current found in the system $\text{i}_{net}$.

$\text{SERC} \geq 0.01~\text{i}_{net}$

$\text{SIRC} < 0.01~\text{i}_{net}$

DOI: 10.1016/0169-7722(92)90041-C


How to identify if a material is perovskite or not?

As MaxW indicated in the comment, perovskites are a huge family of natural and synthetic compounds. Following is a breif description of the group of compounds named perovskites:

The original Perovskite is a mineral of formula $\ce{CaTiO3}$. It was first discovered in 1839 by the Prussian minaralogist Gustav Rose in the Ural Mountains, and is named after Russian mineralogist Count Lev Aleksevich von Perovski (1792–1856). The natural crystals have a hardness of 5.5-6 and a density of $\pu{4000-4300 kgm-3}$. The crystal structure of this mineral was initially thought to be cubic, but was later  shown to be orthorhombic (Ref.1).

As with many minerals, Perovskite has given its name to a family of compounds with the same type of crystal structure, which are called perovskites that have a general formula close to or derived from the composition $\ce{ABX3}$. In these mineral type with formula $\ce{ABX3}$, $\ce{A}$ and $\ce{B}$ are usually two different sized cations, while $\ce{X}$ is an anion that bonds to both (e.g., $\ce{KNbO3}$). For instance, $\ce{PbTiO3}$) is a perovskite, crystals of which are Tetragonal (space group: $P_4mm$) with $a=\pu{0.3902 nm}$ and $c=\pu{0.4143 nm}$ of its unit cell. In contrast, the original Perovskite, $\ce{CaTiO3}$ crystals are Orthorhombic (space group: $Pbmn$) with $a=\pu{0.54035 nm}$, $b=\pu{0.54878 nm}$ and $c=\pu{0.76626 nm}$ of its unit cell (Ref.1).

To some extent, the multiplicities of phases that belong to the perovskites family can be rationalized by assuming that perovskites  are simple ionic compounds where $\ce{A}$ is usually a large cation and $\ce{B}$ is usually a medium-sized cations, and $\ce{X}$ is an anion (if the charges of the ions are written as $q_\ce{A}$, $q_\ce{B}$, $q_\ce{X}$, then: $q_\ce{A} + q_\ce{B} = 3q_\ce{X}$). At present, many hundreds of compounds are known that adapt the perovskite structure. Few examples are $\ce{NaMgF3}$ (Orthorhombic; space group: $Pbnm$), $\ce{CsPbI3}$ (Cubic; space group: $Pm \overline{3} m$),  $\ce{KCuF3}$ (Tetragonal; space group: $I4/mcm$), $\ce{LaAlO3}$ (Trigonal; space group: $R3c$), $\ce{KNbO3}$, $\ce{KNbO3}$ (Orthorhombic; space group: $Amm2$), and $\ce{(Fe,Mg)SiO3}$ (Orthorhombic; space group: $Pnma$). In fact, a perovskite structure mineral, Bridgmanite $\left(\ce{(Fe,Mg)SiO3}\right)$ is the most abundant solid phase in the Earth’s interior, making up $38\%$ of the total (Ref.1).

Figure A in following diagram illustrates an ideal cubic perovskite, $\ce{SrTiO3}$ (space group: $Pm \overline{3} m$):

The traditional view of the perovskite lattice is that it consists of smaller $\ce{B}$ cations within oxygen octahedra, and larger $\ce{A}$ cations which are nine fold coordinated by oxygen. According to this Princeton website where I abstracted the structure A (Note: I have modified to some extent):

  The easiest way to visualize the structure ($\ce{ABO3}$) is in terms of the $\ce{BO6}$ octahedra (e.g., $\ce{TiO6}$), which share corners infinitely in all 3 dimensions, making for a very nice and symmetric structure. The $\ce{A}$ cations ($\ce{Sr}$) occupy every hole which is created by 8 $\ce{BO6}$ octahedra, giving the $\ce{A}$ cation a 12-fold oxygen coordination, and the $\ce{B}$-cation a 6-fold oxygen coordination. In the example shown below, ($\ce{SrTiO3}$) the $\ce{Sr}$ atoms sit in the 12 coordinate $\ce{A}$ site, while the $\ce{Ti}$ atoms occupy the 6 coordinate $\ce{B}$ site. Note that there are many $\ce{ABO3}$ compounds for which the ideal cubic structure is distorted to a lower symmetry (e.g., tetragonal, orthorhombic, etc.).

Based on these facts, your last two entries, $\ce{Pb(Zr_{0.45}Ti_{0.55})O3}$ and $\ce{Pb(Zr_{0.871}Ti_{0.129})O3}$, are of course belong to perovskite family. The first one have adapted to the structure of $\ce{PbTiO3}$ (Tetragonal; space group: $P_4mm$), which is a perovskite (Ref.1). The second one have adapted to the structure of $\ce{LiNbO3}$ (Trigonal; space group: $R3c$), which is a synthetic perovskite (Wikipedia).

There is another family of compounds with formula $\ce{A2BB'O6}$, which is called double perovskites. The crystal structures of double perovskites feature an ordered rock-salt like arrangement of corner-sharing $\ce{BO6}$ and $\ce{B'O6}$ units in the crystal structure (Figure B). This is what researchers in Max Plank Institute where I have extracted the Figure B have to say about double perovskites:

  Proper choice of the metal ions $\ce{B}$ and $\ce{B'}$ and of the alkali, alkaline earth, or rare earth ions $\ce{A}$ allows to realize a large variety of physical properties in double perovskites. Prototype compounds which have found much interest are $\ce{Sr2FeMoO6}$ and $\ce{Sr2FeReO6}$ as they are half-metallic ferromagnets with Curie temperatures above room temperature and large magnetoresistance effects at ambient conditions.

The double perovskite structure is so named because the unit cell of double perovskite is twice that of usual perovskite. An example of crystal structure of a synthetic double perovskite, $\ce{Sr2FeMoO6}$, is shown in Figure C, which has a Orthorhombic (Phase Prototype: $\ce{Sr2WNiO6}$; space group: $Cmme$) unit cell (Ref.2). It has the same architecture of 12 coordinate $\ce{A}$ sites and 6 coordinate $\ce{B}$ sites, but two cations are ordered on the $\ce{B}$ site. The structure of $\ce{Sr2FeMoO6}$ shown here,the $\ce{Fe}$ and $\ce{Mo}$ atoms have ordered in a 3D chessboard type fashion.

Hence, it is safe to suggest that the compound used for first three entries, $\ce{Pb(Zr_{0.52}Ti_{0.48})O3}$, is an example of double perovskites, solely because it has a prototype structure of $\ce{Pb2FeNbO6}$ (Monoclinic; space group: $Cm$), which resembles a double perovskite ($\ce{A2BB'O6}$ type). The crystal structure of $\ce{PbFe_{0.5}Nb_{0.5}O3}$ is slightly distorted cubic ($a = b = c = \pu{4.0056 \mathring A}$; and $\alpha = \beta = \gamma = \pu{89.664 ^\circ C}$) and declaired to be a perovskite (Ref.3). On the hind site, $\ce{PbFe_{0.5}Nb_{0.5}O3}$ resembles more of $\ce{Pb(Zr_{0.52}Ti_{0.48})O3}$ than that of $\ce{Pb2FeNbO6}$.

Note: The best advice to OP is start with the book (Ref.1), which is written in basic language, understandable for any novice student learning about Perovskite materials. In addition, since OP is conveniently in England, I believe he can consult the author of the book, Prof. Richard J. D. Tilley, Ph.D., the Emeritus Professor in the School of Engineering at the University of Cardiff, Wales, United Kingdom, if OP has additional questions on Perovskite materials.

References:

Richard J. D. Tilley, In Perovskites: Structure-Property Relationships; John Wiley & Sons Ltd.: Chichester, West Sussex, United Kingdom, 2016 (ISBN: 978-1-118-93566-8).
Y. C. Hu, J. J. Ge, Q. Ji, B. Lv, X. S. Wu, G. F. Cheng, "Synthesis and crystal structure of double-perovskite compound $\ce{Sr2FeMoO6}$," Powder Diffraction 2010, 25(S1), S17-S21 (https://doi.org/10.1154/1.3478711).
Renbing Sun,  Wei Tan,  Bijun Fang, "Perovskite phase formation and electrical properties of $\ce{Pb(Fe_{1/2}Nb_{1/2})O3}$ ferroelectric ceramicse," Physica Status Solidi A: Applications and Meterial Science 2009, 206(2), 326-331 (https://doi.org/10.1002/pssa.200824432).


How do you melt magnesium without causing its combustion?

I have seen molten magnesium; it was covered with charcoal and other fluxes and in a furnace that had a gas shroud. (I think methane that burned any oxygen that got into the shroud atmosphere.) When an immersion thermocouple was put into the magnesium, the open spot immediately burst into flames. I recommend that you not try to melt magnesium; you certainly must get more information than you can get on the internet.


Why is the atomic radius of krypton smaller than that of potassium?

Your concept is close to the real answer. You should think it as no. of protons per electron over all. This is the way things have been done in literature and if you are not satisfied then consider the fact that not all electrons are shielding the nucleus to same extent. There are certain rules [slaters rule] made and which works in most cases by which you can actually predict how much of nuclear charge is acting on the electron. Even with that you will see valence electron for K is more shielded than for Kr. But the numbers you came up with don't make that much sense. It does a bit but better if you go by the way it has been discussed.


Why is solid phase peptide synthesis limited to 70 amino acids?

The issue is one of yield. Each step has a certain yield percentage of yield, that the synthesizers will maximize as much as possible. However, given that it is fairly impossible to generate 100% yield all the time, the amount of desired protein you get over time will deteriorate exponentially with length.

For example, say that each step gives you 95% yield. If you have a protein that is ten amino acids long, your final step will give you .95^10 yield, which is about 60%.

However, with the same yield per step, but a 70 amino acid chain, the amount of protein you want drops to 2%.

Even if you are at 99% yield per step (a difficult feat), your final yield will be under 50% for a 70 amino acid chain.


S orbitals are lower in energy than P orbitals.  Electrons prefer to be in as low an energy orbital as possible.  Therefor, when we mix S and P orbitals to make hybrid orbitals, the more S character the orbital has the lower the energy of electrons occupying that orbital. Let me use chloroform, $\ce{CHCl3}$, as an example to make my point (I have data for it and the $\ce{C-Cl}$ bonds are analogous to those in methyl chloride).  The chlorine atom is electronegative, so electrons in the $\ce{C-Cl}$ bond will spend more of their time closer to chlorine - away from the carbon, then they would in say a $\ce{C-H}$ bond.  So if the electron density in the hybrid orbital contributed by carbon to the $\ce{C-Cl}$ bond is going to be reduced, why put as much S character in it?  Instead save that S character for other hybrid orbitals emanating from the carbon that have higher electron density in them.  It turns out in the case of chloroform, knowing the various bond angles and the $\ce{C_{3V}}$ symmetry of the molecule, that the carbon portion of the $\ce{C-Cl}$ has a hybridization of about $\ce{sp^4}$ (more P character and less S character just as we predicted since it has a lower electron density) and the $\ce{C-H}$ bonds are approximately $\ce{sp^{2.75}}$ (using that S character we saved from the $\ce{C-Cl}$ bond to stabilize the electrons in these 3 orbitals, just as we predicted).  The same effects play out in methyl chloride, just not as dramatically (e.g. the hybridization index of the $\ce{C-Cl}$ bond will be greater than 3 but less than 4).


Why is dichloromethane immiscible in water?

The physical causes of solubility of substances in each other can semiquantitatively be expressed by the Hansen solubility parameters.

$\begin{array}{}
\text{Water}:&&\delta_d=15.6&\delta_p=16.0&\delta_h=42.3&R_0=47.8\\
\text{Dichloromethane}:&&\delta_d=18.2&\delta_p=6.3 & \delta_h=6.1&R_0=20.2
\end{array}$

$$R_0=20.2,\ \ \ R_a=37.8$$

$$RED=\frac{R_a}{R_0}=1.87$$

RED < 1 would show solubility of both substances in each other. But RED > 1, so both substances are little soluble in each other.

The solubility parameters above show, the energies from polar and from hydrogen bridging bond intermolecular attractions are higher in water.

Generally, chlorinated hydrocarbons have lower energies from polar and hydrogen bridging bond intermolecular attractions.

The strongest hydrogen bridging bonds are formed between the atoms of the strongly electronegative elements $\ce{F}$, $\ce{O}$ and $\ce{N}$. But the atoms of other electronegative elements, e.g. $\ce{Cl}$-, $\ce{S}$- and $\ce{C}$-atoms, are capable of weak hydrogen bridging bonds at suitable molecular structure.

The atomic radii of the third period elements ($\ce{P}$, $\ce{S}$, $\ce{Cl}$) are much larger than those of the second period elements ($\ce{N}$, $\ce{O}$, $\ce{F}$). $\ce{Cl}$-atoms have lower charge density therefore. They are weaker proton acceptors therefore.

Hydrogen bridging bonds can be proved e.g. by molecular spectroscopy.

Li Bian: Proton Donor Is More Important Than Proton Acceptor in Hydrogen Bond Formation: A Universal Equation for Calculation of Hydrogen Bond Strength. J. Phys. Chem. A 107 (2003) (51) 11517–11524

Weinhold, F.; Klein, R. A.: What is a hydrogen bond? Mutually consistent theoretical and experimental criteria for characterizing H-bonding interactions. Mol. Phys. 110 (2012) (9-10) 565-579

Gilli, G.; Gilli, P.: The Nature of the Hydrogen Bond: Outline of a Comprehensive Hydrogen Bond Theory. Oxford University Press, Oxford, 2013


What is the name of the compound Sc(HCO2)3?

The formula you have is definitely scandium formate. There is one of four reasons why someone (like your instructor or your online homework service) might tell you that scandium formate is wrong.

Someone wants you to call this compound scandium(III) formate because scandium can have multiple oxidation states.
Someone wants you to call this compound scandium methanoate (or scandium(III) methanoate) since methanoate is another name for the formate ion.
You are correct and your instructor is wrong (or your assignment software has the wrong answer keyed in).
There is a typo in the formula you were given.


Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?

As I understand this, there are basically two effects at work here.

When you populate an $\mathrm{s}$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the $\mathrm{d}$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in $\mathrm{d}$-orbitals and putting one in an $\mathrm{s}$-orbital increases as you fill the $\mathrm{d}$-orbitals.

Additionally, pairing electrons in one orbital (so, adding the second $\mathrm{s}$ electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron.

I'm assuming that the effect isn't strong enough to avert fluorine having a $\mathrm{2s^2}$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the $\mathrm{s}$ from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the $\mathrm{5d}$ instead of the $\mathrm{4f}$ orbitals.

Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $\mathrm{6s^2}$ occupation, but is for gold - more $\mathrm{d}$ electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an $\mathrm{s}$-orbital.


Preparation of p-chlorobenzoic acid from p-chloronitrobenzene

This is simply the summery of all comments involved:

$$\ce{Cl-C6H4-NO2 ->[Sn/HCl] Cl-C6H4-NH3+Cl-} \tag1$$

$$\ce{Cl-C6H4-NH3+Cl- ->[\text{a. } NaNO2/HCl \ (\pu{0-5 ^\circ C})][\text{b. } CuCN] Cl-C6H4-CN} \\ \tag2$$

$$\ce{Cl-C6H4-CN ->[H+/H2O][\Delta] Cl-C6H4-COOH} \\ \tag3$$

Note: The reaction $(1)$ is a typical reduction reaction, which reduce nitro group to amino group. The reaction $(2)$ is called Sandmeyer Reaction. Finally, The reaction $(3)$ is a hydrolysis condition to convert cyano group to carboxylic acid group. That hydrolysis can be done under basic conditions as well (then, final product is carboxalate).

Also note that even though it is not necessary, I put this answer up just to avoid people asking same kind of questions again and again. When they have a visual answer they might not do so.


Do I need to include lattice energy when calculating the energy required to vaporize NaCl?

Lattice energy

  Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol.

First, we have to understand the term lattice energy.
Here is the textbook explanation (Fleming: Physical Chemistry): The lattice energy is the energy required to separate the ions in an ionic lattice so that they are at infinite distance (but still ions). This would be difficult to do experimentally, but the value may be determined by using a Born Haber cycle. As the diagram below shows, we know the enthalpy of reaction for:

$$\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}$$

We can get the same energy in a thought experiment (path 2 in the diagram), first turning all species into gas, then into atoms, then transferring the electron and - finally - forming the lattice from the individual anions and cations. All these processes are experimentally accessible except for the last, so you can determine the lattice energy this way.

Source: diagram adapted from the cited textbook.

Enthalpy of vaporization

  But do we need to add this energy again when we are trying to vaporize an ionic compound?

You problem is not well-defined. What are the initial and final temperatures, what is the pressure? What does it mean to vaporize solid NaCl or any ionic compound, i.e. what is the product? If we boil sodium chloride at 1 atm pressure, it produces mostly NaCl monomers in the gas phase, see https://chemistry.stackexchange.com/a/14560. If you are interested in this process, you could write it as:

$$\ce{NaCl(s) -> NaCl(g)}$$

So to get the required energy, you could first separate all ions (lattice energy) and then calculate the energy of forming individual pairs of sodium and chloride ions (assuming the interaction is purely ionic).

$$\ce{NaCl(s) -> Na+(g) + Cl-(g)}\tag{1}$$

$$\ce{Na+(g) + Cl-(g) -> NaCl(g)}\tag{2}$$

It should be obvious that process (1) cost much more energy than what you get out of process (2) because you are going from 6 nearest neighbors to just one. If you look up the atomic distance of $NaCl(g)$, you could estimate the enthalpy of (2) pretty accurately and get a numeric answer.

  [Comment from https://chemistry.stackexchange.com/a/14177]: Roughly, somewhere around half of the lattice enthalpy of a salt comes solely from the binding of the smallest electrically neutral agglomerate; that is, it takes about as much energy to break a macroscopic solid NaCl crystal into a gas of ion pairs as it does to break all the ion pairs and create a true plasma. Thus, it is rather unlikely that, at reasonable temperatures, an ionic gas will break down any further than the smallest possible electrically neutral aggregates. – Nicolau Saker Neto Jul 8 '14 at 16:39

According to the comment above, the heat of vaporization should be about half of the lattice energy (with opposite sign).

  I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.

Because enthalpy is a state function, it does not matter if you choose the actual (realistic) path or take a different path. You are right that if you step through what actually happens, you need to know heats of fusion and vaporization as well as heat capacities of all phases for the relevant temperature ranges.


The Celcius scale was originally based on the freezing and boiling points of water, so 0 °C was chosen as the freezing point until 1954. But now, the size of one degree on both Celcius and Kelvin scales is defined as 1/273.16 of the difference between absolute zero and the triple point of VSMOW (for various reasons relating to the actual measurement), but the 0 point of the Celcius scale was left as the freezing point of water which means that even though both scales are defined at 0 K and 273.16 K (the triple point), the Celcius scale still starts at the freezing point and 0.01 °C was defined as the triple point of VSMOW, which means 0 °C = 273.15 K.

Absolute zero is the coldest possible thermodynamic temperature, 0 K. The triple point is the temperature and pressure at which solid, liquid, and gas phases can exist in a stable equilibrium. If you look at the phase diagram below, it's the point at which the solid, liquid, and gas regions all meet. For water, the temperature of the triple point (273.16 K) is remarkably close to its freezing point at atmospheric pressure (273.15 K). Because 0 °C was originally defined as the freezing point of water, instead of changing the zero point when they redefined the scale based on absolute zero and the triple point, they decided to define the triple point as 0.01 °C.

We add 273.15 to °C to get kelvins because that's the difference between 0 K and 0 °C. We don't add 273.16 because the triple point is not 0 °C.

Source


Atomic mass ratios - Is the problem providing enough info?

Are you asking whether we can figure out the ratio of the mass of an oxygen atom to a hydrogen atom from this information? If so, then the answer is yes, assuming that you know that $\ce{H2O}$ and $\ce{H2O2}$ are the formulas for each compound, and that atoms exist and are indivisible in chemical reactions.

$$2n\space \rm{atoms \space H} = 1 \space g \space H$$
$$1n \space \rm{atoms \space O} = 8 \space g \space O$$

Divide the second equation by the first:

$$\frac{1n \space \rm{atoms \space O}}{2n\space \rm{atoms \space H}} = \frac{8 \space g \space O}{1 \space g \space H}$$

Cross multiply to get a 16:1 mass ratio. You can confirm this by looking at $\ce{H2O2}$, where the ratio of atoms is 1:1 and mass is again 16:1.

If you didn't know the formulae, then you could assume that the ratio was some integer multiple of 8:1, but without more data you wouldn't know which. This is because you would not know whether water had a 1:1 atom (or mole) ratio of $\ce{H}$ to $\ce{O}$, or if it was something else.

Incidentally, what you are describing is the law of multiple proportions, and it is one of the key findings that led to the development of atomic theory.


Chemistry is a field with far too many exceptions and edge cases to be making sweeping categorical statements. There are numerous examples of ionic compounds that are isolable and stable, the components of which will spontaneously decompose under certain conditions (this is where reaction kinetics is relevant). Two that readily come to mind are ammonium nitrate ($\ce{NH4NO3}$) and ammonium bicarbonate ($\ce{NH4HCO3}$). Both of those can be isolated as pure salts, and both also can and will chemically decompose when circumstances are favorable (the former by multiple possible pathways depending on conditions).

  Isn't it theoretically possible to have two substances that would otherwise undergo a large extent reaction with each other but don't due to an activation energy barrier?

This isn't just theoretically possible, it's ubiquitous, it's the norm. If it weren't, human life couldn't exist. Combustion reactions are probably the most obvious and mundane example; just about every organic compound imaginable is usually more thermodynamically stable as a mixture of $\ce{CO2}$, $\ce{H2O}$, $\ce{N2}$, etc., but the kinetic barrier to the reaction (along with physical hindrances and the availability of an oxidizer, e.g., $\ce{O2}$) prevents this from occurring spontaneously without some trigger and the appropriate environmental conditions.


Gibbs free energy=potential free energy? (Transition state theory)

Yes both mean exactly the same thing.

By definition gibbs free energy is the total amount of non pv work that can be extracted from the system. Non Pressure-Volume work refers to the work which does not allow change in both volume and pressure of the system. An example of non pv work is electrical work which is done by cells and batteries in creating a potential difference about its ends.

Now pv work is related to kinetic energy of the molecules of the system. We can see this by two ways. First is that pressure and volume both are dependent on movement of molecules. Hence increasing pressure at constant volume is the same thing as increasing kinetic energy of the gas due to which there are more collisions of higher magnitude. Similarly increase in volume at constant pressure also means increase in kinetic energy of system. Second is that pressure and volume are both related and dependent on Temperature by gas equation where temperature of a system is directly related to kinetic energy of the gaseous system.

So whenever pv work is done by or on system changes take place only in kinetic energy of the molecules of system and not in the potential energy of system.

Hence whenever non pv work is done by or on system, changes in only potential energy of the system takes place. So Gibbs free energy is same as potential energy of system.


The influence of lemon juice & sugar on the ph-value

You're right about the lemon juice and the milk but wrong about the sugar. Lemon juice is weakly acidic (hence its sour taste) and milk is slightly basic (or alkaline, the terms are used interchangeably). A sugar solution is completely neutral since sugar can't take hydrogen ions out of the water or donate them in.

If you don't have background in chemistry, then let's not talk about pH - I thought pH was confusing when I learned it because "lowering the pH" makes a solution more acidic, and "raising the pH" makes it more basic! (Ironically, the pH scale was originally devised to make the concepts of acidity and alkalinity more accessible to the general public)

Your observations about tea probably have nothing to do with acidity or alkalinity since the foods you've described are not significantly acidic or alkaline (except lemon juice). Tea is slightly bitter because of tannins in the leaves. You've evolved to think of things that taste bitter as unappetizing because they often do not have nutritional value and may even be poisonous.

Sugar (and sweet things in general), on the other hand, is very tasty because you've evolved to know that consuming sugars (carbohydrates) will give you energy. Milk also contains sugars, as well as lipids (fats) which give it its creamy taste, which you've also evolved to find appealing because they give you energy - more than simple sugars.

The creamy taste of milk or sweet taste of sugar are usually enough to cover up the bitter taste of tannins (if the tea is brewed right, that is, but that's an issue for another discussion) and make your tea tasty.


Electrophilic attack of X+ on double bond

These kind of reactions are called Electrophilic addition reactions. These are usually given by alkenes and alkynes. In your example, the electrophile is $\ce{Cl+}$. The alkene will act as a source of π electrons and will donate it to the electrophile which is in need of electrons. So, alkene is called as nucleophile in this reaction.

  Is it the double bond attacking the X+ ion or the other way around?

It depends on how you look at it. We can say that alkene polarizes the $\ce{Cl-Cl}$ bond by pushing electrons into the anti-bonding molecular orbital. This weakens the bond and hence electrophile is getting generated. In the next step,the electrophile attacks the double bond [or double bond attacks the electrophile] and hence electrophile gets added across across the double bond

Why does the 3-membered cyclic transition state form?

The cyclic transition state is just a temporary relief for the electrophile that's just been added. The π molecular orbital donates electrons to the electrophile.

While the p-orbitals in the π-bond overlap with the electrophile, each p-orbital forms a σ-bond.

  But would the same thing happen if I put any other electrophile?

Generally,electrophilic addition reactions take place through carbocation intermediate.
But this reaction can take place through cyclic 3-membered intermediate,if the electrophile has a lone pair.

Apart from $\ce{Cl+,Br+}$ even $\ce{Hg^2+}$ (when the reagent is $\ce{Hg(OAc)2/H3O+}$) and $\ce{NO+}$ (when the reagent is $\ce{NOCl}$) can form the 3-membered cyclic intermediate.


One expects [poly(ethylene oxide)-co-(polypropylene oxide)-co-poly(ethylene oxide)] triblock surfactants will micellize it (e.g., Pluronics); possibly Tergitols.  It looks too bulky to fit inside even gamma-cyclodextrin.

Hang a suitable side chain off the phthalic anyhydride and the product will dissolve in pretty much anything.  What does pH mean outside water?


Do any substances in a liquid state work as well as water to put out fires?

Carbon tetrachloride was once used for this purpose as it has practically no flammability at lower temperatures .


Alcohols can act as both.  Many reactions of alcohols are believed to start with their protonations or deprotonation. However, isolating salts where they act as bases is much harder, while alcoholates are quite easy both to make and isolate. Protonated alcohols are very active alkylating agents and if fitting $\beta$-hydrogen is present, are prone to beta-elimination.


Is viscosity proportional to the number of hydroxyl groups?

It is a reasonable rule of thumb, but certainly not always true.

Compare for example the viscosities of Dodecane ($\ce{C12H26}$, $\mu=1.374 \ce{\; mPa\cdot s}$, no $\ce{OH}$ groups) and Ethanol ($\ce{C2H5OH}$, $\mu=1.095 \ce{\; mPa\cdot s}$, 1 $\ce{OH}$ group) (source). There you can see that ethanol has a higher number of $\ce{OH}$-groups, but a lower viscosity.

That said, I do think you are on to something regarding the strenght of intermolecular interactions, it is just not purely the number of $\ce{OH}$-groups that matters for that. The length of the carbon chain for example (like in the example I mention above) also plays a role.


Most non-ferrous alloys are heated in order to form a solution.
Techniques:
Annealing:
Most non-ferrous alloys that are heat-treatable are also annealed to relieve the hardness of cold working. These may be slowly cooled to allow full precipitation of the constituents and produce a refined microstructure.
Different annealing techniques are subjected with non-ferrous alloys like  recrystallization annealing, partial annealing, full annealing, and final annealing.

Quenching
Quenching is a process of cooling a metal at a rapid rate. This is most often done to produce a martensite transformation. In ferrous alloys, this will often produce a harder metal, while non-ferrous alloys will usually become softer than normal.

Reference: Wikipedia


Why does Vapor Pressure decrease on addition of solute?

For solvent at temperature $T$, surface area $A$ and  having the molar fraction $x$, the rate of molecules leaving liquid is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=-C_1 \cdot A \cdot x \cdot f(T)$$
Lower molar fraction means lower surface density and lower evaporation rate per surface area.
Factor $C_1$ includes pure solvent molecular surface density and $f(T)$ includes probability of a molecule leaving liquid.
For vapour of the partial pressure $p$, the rate of molecules coming to liquid from vapour is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=C_2 \cdot A  \cdot p \cdot  g(T) $$
The $g(T)$ involves temperature dependent relation of pressure and collision rate with liquid surface. Higher temperature means higher kinetic energy and less collisions needed for the same pressure.
For equilibrium, the net rate must be zero, therefore:
$$p= \frac{C_1 \cdot x \cdot f(T)}{C_2 \cdot  g(T)}$$
Therefore the equilibrium pressure is proportional to the molar fraction.
It is quite analogical to the steady state of water level(=pressure), when it is simultaneously being filled up(=evaporation) and drained(=condensation). When you dilute a non-volatile solute, you decrease the filling up (= surface concentration of molecules is lower), so draining (=condensation), brings the water level(=pressure) to lower value to match draining rate with the filling.


How to calculate molarity of a solution by solving the percentage?

Both of them are pretty straightforward. Don't go by the formulae, but instead, internalize the steps required here.

$2\%$ w/v solution has $\pu{2g}$ of solute in $\pu{100 ml}$ solution. That implies $\pu{20g}$ of solute in $\pu{1L}$ solution. $\pu{20g}$ solute is 0.5 moles of $\ce{NaOH}$ (why?). So, molarity is simply moles divided by volume of solution.

The $2\%$ w/w solution case is trickier, and you'll also be given the density of solution in this case.  $2\%$ w/w solution has $\pu{2g}$ solute in $\pu{100g}$ solution, or $\pu{20g}$ solute in $\pu{1000g}$ solution. If the density is $\pu{2gm L^-1}$ (assumed), then the volume of solution would be $\ce{500 mL}$. Now, you can proceed by the same logic as above.

In fact, the conversion of $\pu{100ml}$ volume to $\pu{1L}$ in case 1 wasn't even necessary, but I only did that to make the step-by-step calculations more approachable for you.


Why is only one proton transferred in the reaction between sulfuric acid and sodium chloride

Sodium bisulphate (pKa ~ 2 - https://en.wikipedia.org/wiki/Sulfuric_acid) is a much weaker acid that hydrochloric acid (pKa = ~ -7), which means that chloride is not a strong enough base to deprotonate sodium bisulphate.

However all acid pbased reactions are at equilibrium.  Roughly one part in 10^9 of the species involve sulphate anion, which for all practical purposes is negligible.

Table of pKa charts: http://www.lauher.com/che142/acidbase.pdf


Which of the following isomers of 2,3‐dihydroxy‐4‐methoxy‐4‐oxobutanoic acid are identical?

On rotating the compound B by 180°, we get the two -OH groups on the left side and the two H atoms on the right. In other words, B would then be an enantiomer of A.

Note: on rotating a Fischer projection by 180°, the compound remains the same; it's just that all the atoms have to be rotated, i.e. the atoms on C3 in this case would go to C2 and vice versa.


How to liquefy or extract water from superabsorbent polymer

This may be true but I suggest another type of explanation.  If a super absorbent polymer absorbs its maximum in distilled water it is called a fully swollen hydrogel.  There are two ways of decreasing the amount of water stored without using heat.  The first one is to increase the salt load of additional water and place the hydrogel within this salted water.  The salt in the water will decrease the total water absorbed by the hydrogel and the hyrogel will shrink, giving up some of its water.  A more complete and full way to make the hydrogel to give up its water is the addition of an acid.  Lowering the pH of the hydrogel to below 3.5 will convert all the sodium carboxylates to the free acid form which does not absorb water.  These hydro gels are pH sensitive and if you plot pH vs absorbed fluid you would find a bell curve that starts around 4.5 or 5.0, tops out around 7.5 and decreases as you increase the pH until around 10.5 to 11.0.


Does imidazole and hydrochloric acid yield imidazole hydrochloride salt?

The correct structure of the salt is as shown above. This structure is more stable than the one that has two hydrogens on the same nitrogen - due to electron delocalization.

If you have a water solution of about 50/50 imidazole base and the hydrochloride salt, you get the maximum buffer capacity, due to the fact that both acid and base forms are present, and a small amount of a strong acid or base subsequently added to the mixture will react with these imidazole species, and therefore have little effect on the pH.


Well, sodium does have non-zero electron affinity, so it surely can acquire another electron, provided it can get one for free. That's what happens when a lone sodium atom meets a lone electron somewhere in deep space. But here on Earth, with all that condensed matter around, we don't have many free electrons (*). They all belong to some other atoms, and if you want to get an electron (that is, to get reduced), you have to oxidize something else.

(*) Then again, some might recall metals in which the electrons are kinda free, but wait: they are not really free. They are bound by the collective field of metal cations, and you need to apply certain energy to take one out.

It turns out that $\ce{Na}$ and other alkali metals can actually be reduced to form alkalides, but those are exotic compounds. They are not particularly stable, and get oxidized with pretty much anything. It is much easier for $\ce{Na}$ to form a cation, thus achieving the stable octet, and call it a day.

There is a whole different dimension to the problem: why wouldn't two sodium atoms share their electrons, establish a bond, and form a diatomic molecule? Again, they would, but only in a vacuum! When you have many such molecules, the sodium atoms would share their other empty orbitals, find out they don't have enough electrons to fill them all, and just resort to being a metal.


Finding 1/V from absorbance in Lineweaver-burk plot

First, let me tell you about the pitfalls of Lineweaver-Burke plots.  The tranformations of your data (i.e. taking the reciprocal) distort the error structure of your data and can lead to erroneous fits.  It is better to directly use nonlinear fitting techniques to find kinetic parameters.  These techniques do not require transformation of the data and thus do not distort error structure.

Second, let me answer your question.  The classical Lineweaver-Burke plot puts inverse reaction rate on the $y$ axis, i.e. $1/V$.  Your question is how to find $V$ from absorbance data.  The answer is, you don't need to.  Let $A_i(t)$ be the absorbance data for tube $i$ as a function of time.  What is $V_i(t)$ for that tube?  We won't know unless you know the molar absorption coefficient of the substrate (or product -- I'm a little unclear from your question what you are hoping to detect by measuring $A_{\textrm{530 nm}}$).  Assume that it is the product that is absorbing, and that it has a molar absorption coefficient of $\epsilon$ and that you are using a photometer with path length $L$.  Then the reaction rate for the $i$th tube will be

$V_i(t)= \frac{1}{\epsilon L}\frac{dA_i(t)}{dt}$

This means take the slope of the absorbance measurement and divide it by $\epsilon$ and $L$ -- numbers you don't know -- to get $V$.  But the key assumption is simply that $V \propto \frac{dA}{dt}$.  Try making up any numbers you like for $\epsilon$ and $L$.  Find $K_m$.  Then make up new numbers and find $K_m$ again.  Did the result change?


This would become much clearer if you show the lone pair on the oxygen atom.

One $p$ orbital of the Oxygen atom forms the $\pi$ bond with Carbon, while the other three (remaining one $s$ and two $p$) orbitals are $sp^2$ hybridized.

Two $sp^2$ orbitals are used to form the $\sigma$ bonds while the last $sp^2$ orbital is occupied by the lone pair.


What are the chemical process responsible for the warping of wood?

Warping (be it a bow, crook, cup, or twist); as well as loosening of knots, checking, splitting, and collapse; of wood is caused by uneven shrinkage.

To explain uneven shrinkage chemically, consider wood as 50% cellulose.  Cellulose, which is a polymer consisting of linked glucose monomers, is arranged into fibrils.  Fibrils are part of larger structures which make up the cell walls of wood fiber.  Binding these cell walls together is lignin, which is a phenylpropanol polymer.  Cellulose and lignin (along with hemicellulose) are the major structural compounds in wood.

Cellulose in wood is a tangle of rigid crystalline cellulose and flexible amorphous cellulose.  While the structure of crystalline cellulose makes it water insoluble, amorphous cellulose has sites like its base monomer glucose which are capable of absorbing water. When water is absorbed both the shape and volume of the amorphous cellulose change.  While amorphous cellulose changes its shape, the crystalline cellulose (as well as lignin) retains its shape and volume.

Cellulose may also change shape and volume due to other factors such as temperature fluctuation and biological degradation.  Because (1) wood is not uniform; (2) moisture, temperature, and biological changes are rarely uniform throughout the bulk of a piece of wood; and (3) most wood is exposed to some degree of these changes over time; uneven shrinkage occurs.  Because of uneven shrinkage in wood, it warps.


Using Transition State Theory (TST) one has to be aware of at least some of the many flaws of it. And may also often lead to wrong assumptions. However, it is a wonderful tool to describe basic kinetics in a theoretical manner.

The most important/ fundamental assumptions are

separation of electron and nuclear movement (comparable to Born-Oppenheimer approximation)
(all possible) transition states (TS) are in equilibrium with the reactants  $$\ce{A + B <=> [AB]^{\ddagger}}\tag1$$
combined Boltzmann statistic for reactants and TS
All molecules that reach the transition state from the reaction coordinate of the reactants will leave the TS to the reaction coordinate of the product(s) - and vice versa
\begin{align}
\ce{A + B <=> &[AB]^{\ddagger}->P}\tag{2a}\\
\ce{P <=> &[AB]^{\ddagger}-> A + B}\tag{2b}
\end{align}
at the TS the transition vector can be separated from all other motions and hence may be treated classical as translation

In other words TST reduces a very complicated multidimensional problem to a one-dimensional translation. That implies that all treated molecules must be connected on a potential energy hypersurface with one vector. (Borrowing from the amazing wikipedia)

Hence reactions that are described with TST are of first order by default. One would have to make various adjustments to treat higher orders (and I would assume that not matching units would be of minor concern).


About the way chemicals and molecules are represented

Below is 2 types of sugars; glucose and mannose.

If you know how to interpret the lines, then you can determine which one is which.


CaSO4 precipitate in mixture of Na2SO4 and CaCl2 solutions

As you've already noted, there are 3 substances of interest here: $\ce{Ca^{2+}}, \ce{SO4^{2-}}, \ce{H2O}$.

Let $V$ be the additional volume of sodium sulfate solution. Then:

The total volume is $V + 0.2\ \mathrm{L}$.

The total amount of sulfate is $(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})$.

The total amount of calcium is $(0.1\ \mathrm{L}) (0.02\ \mathrm{M})$.

We have then the following relationship:

$$\ce{[Ca^{2+}][SO4^{2-}]}=\frac{(0.1\ \mathrm{L}) (0.02\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\frac{(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\leq K_{\mathrm{sp}} = 2.5\times 10^{-5}$$

Now, it's just algebra to solve for the threshold value of $V$, keeping in mind that there are two solutions for $V$ because this is a quadratic equation. In this case, the correct solution is the positive root:

$$V \approx 0.48\ \mathrm{L}$$

The mistake you made is that the calcium and sulfate ion concentrations are definitely not equal, since the additional volume of solution contains only sulfate and not calcium.


Why do nonmetal oxides react with water to form acidic compounds?

Well, in terms of this reaction, sulfur in $\ce{SO3}$ acts as a Lewis acid for an incoming $\ce{OH-}$ nucleophile. The reaction mechanism seems to be any old nucleophilic addition reaction:

$\ce{H2O <=> H+ + OH-}$ : (Water can disassociate into its ions)
$\ce{SO3 + OH- -> SO2(OH)O-}$ : (I couldn't decide on whether this is the right
way to write this intermediate, but I'll be happy to add a diagram for clarity)
$\ce{SO2(OH)O- + H+ -> SO2(OH)2}$ : (This is the final step of the mechanism to yield $\ce{H2SO4}$)


How does entanglement explain myoglobin's preference for O₂?

So, there are two effects at play here which they account for. One they denote by a $\textbf{U}$ and the other by a $\textbf{J}$. $\textbf{U}$ represents many-body coulomb repulsions between the electrons. That is, they are accounting not just for two electrons repelling each other, but up to three or more. $\textbf{J}$ represents valence fluctuations in spin. Specifically, iron in hemes are in a high-spin triplet state. What they are saying then is that it is necessary to include spin fluctuations which result in changes in the magnetic moment of the system in order to properly capture the binding of ligands. If you look at figure 4 of this paper, you'll find that they have included a comparison of spin-fluctuations in $\ce{MbO2}$ and $\ce{MbCO}$ which are used as model systems and known to be analogous in ligand binding to hemes. Basically what this means is that the increased spin-fluctuations cause $\ce{O2}$ to bind more strongly, but this has basically no effect on $\ce{CO}$. They describe these effects in the following quote:

  In this case, the effect of J on the
  binding energy of O2 may be regarded as a balance between two
  competing effects. The charge analysis (Fig. 3) reveals that
  metal-to-ligand charge transfer is higher for J =0 eV, which is
  expected to enhance ligand–protein interactions for small values
  of J. However, NBO analysis reveals a larger ligand-to-metal
  back charge transfer for J = 0.7 eV, which is consistent with the
  increased occupancy of the Fe $\ce{d3z^2−r^2}$ orbital (Table 1), and is
  expected to cause variational energetic lowering at higher values
  of J due to electronic delocalization.

I won't pretend to fully understand their argument (frankly it seems to me like they are waving their hands at a solution and are basically saying this is some complicated effect having to do with spntaneously changing magnetic moments).

So, in that paper they only ascribe the effect to strong electron correlation, many-body coulomb effects, and spin-fluctuations.

So, they definitely use the word entanglement but don't define exactly what they mean or even use it as an explanatory device. I found in another one of their papers though, Weber, C., O’Regan, D. D., Hine, N. D., Littlewood, P. B., Kotliar, G., & Payne, M. C. (2013). Importance of many-body effects in the kernel of hemoglobin for ligand binding. Physical review letters, 110(10), 106402., where they do actually define what they mean by entanglement and it depends on these parameters $\textbf{U}$ and $\textbf{J}$. You can find their discussion of this on page 11 in an appendix.

Their use of the word entanglement is derived from the Von Neumann Entropy which can be used as a measure of the amount of entanglement in a given system. This is because if two particles are entangled in some way, their entropy will be lower than if the two particles were not entangled. Note that this use of the word of entanglement is related to the usual descriptions of entanglement, but it is more of a measure of the average of spontaneous entanglement and disentanglement. That is, particles can become entangled due to some way in which they interact, but almost immediately interact with something else that breaks their entanglement so that it's very rare to find systems where this has any meaningful effect. This spontaneous entanglement in these heme systems is a result of the many-body effects which I've been describing. This manifests itself in the entropy they calculate. This highlights the importance of using dynamical mean-field theory (which they use) as follows:

  With this definition of the entropy, suitable for DFT
  calculations, we found that L = 0.757 in FeP-p and
  L = 0.765 in Fep-d. The entropy at the DFT level
  is hence much smaller than the DFT+DMFT entropy,
  which is in the range Λ = 2.5 − 4.2. This is explained by
  the absence of the many-body excitations induced by the
  correlations (U and J) which contribute significantly to
  the entropy.

There's no doubt they aren't being very forthcoming in their terminology or definitions, but they are also doing some very complicated and interesting stuff.


According to the Franck-Condon principle which you are implicitly referring to, absorption is indeed very fast compared to all other processes.

If you are looking for time scales of photophysical processes, I can suggest you for example B. Valeur's book "Molecular Fluorescence: Principles and Applications", chapter 3 (Characteristics of fluorescence emission). You'll find this list of characteristic times:

absorption: $10^{-15}$ s
vibrational relaxation: $10^{-12} - 10^{-10}$ s
$\ce{S_1}$ lifetime (fluorescence): $10^{-10} - 10^{-7}$ s
intersystem crossing: $10^{-10} - 10^{-8}$ s
internal conversion: $10^{-11} - 10^{-9}$ s
$\ce{T_1}$ lifetime (phosphorescence): $10^{-6} - 1$ s

as well as examples of molecules with their photophysical data.

For chemical processes, it depends on many things, such as concentration, diffusion rates... and you would need to precise your question (type of photochemical reaction, type of quenching, etc.).


How to remove nitrose gasses from ozone gas mixture?

There is no need to use caustic since $\ce{NO2}$ reacts with water itself. As per [1] it is even better absorbed in water than in $\ce{NaOH}$.

Ozone is soluble in water depending on the concentration of $\ce{O3}$ in the gas phase. It is much more soluble than oxygen so you will loose some, but it will have much lower concentration of $\ce{NO2}$. Of course, the washed ozone will be moist. If you need it dry, you can use a column of molecular sieves or some other desiccant.

[1] Absorption of Nitrogen Dioxide by Aqueous Solutions,
F. S. Chambers Jr., T. K. Sherwood,
Ind. Eng. Chem., 1937, 29 (12), pp 1415–1422


Producing hydrogen sulfide at home - how dangerous is it?

On the question as to how dangerous is H2S let me detail some facts that I find problematic. Here is a good paper.

I summarize some of my findings:

First, H2S after initial exposure can deaden your sense of smell (10-minute exposure at 50-100 ppm impairs the sense of smell). So, one may not really have an accurate conception of the magnitude of one's time exposure to the gas.

Second, skin exposure may be a path to ingestion, especially if moist I suspect. Note, some sources actually disagreed on skin exposure poisoning of Hydrogen sulfide with one study alluring to skin toxicosis from high concentrations over a long period.

Third, following a significant exposure event, the very fact that you are still alive should not be comforting. People having incurred a lethal dose may be a de facto walking dead with a slow hourly progression to death.

Fourth, the ppm considered fatal for H2S is not that far removed from some known very toxic gases (like HCN). However, my opening reference doubts if one can even assign accurate figures on ingestion via the lungs. Here is an extract addressing dosing level ranges:

  Findings in the canary, cat, dog, goat, guinea pig, rabbit and rat (2) are consistent as to the effects of hydrogen sulfide: at 150-225 mg/m3 signs of local irritation of eyes and throat after many hours of exposure; at 300-400 mg/m3 eye and mucous membrane irritation in 1 hour and slight general effects with longer exposure; at 750-1000 mg/m3 slight systemic symptoms in less than 1 hour and possible death after several hours; at 1350 mg/m3 grave systemic effects within 30 minutes and death in less than 1 hour; at 2250 mg/m3 collapse and death within 15-30 minutes, and at 2700 mg/m3 immediate collapse, respiratory paralysis and death.

Viewing all these together, if one has a choice to avoid working or experimenting with the gas, I would do so.


As $\ce{NaOH}$ and  $\ce{HCl}$ are strong base and acid respectively, they are completely dissociated in water. The salt $\ce{NaCl}$ is a  strong electrolyte and completely dissociated in water. So, the ionic equation is:

$\ce{(Na^+ +OH^- )}+\ce{(H^+ +Cl^- )}\rightarrow \ce{(Na^+ +Cl^- )}+\ce{H2O}  $

As the ions $\ce{Na^+}$ and$\ce{Cl^-}$ do not combine and remain in the solution as spectator ions. Their role is to keep the solution electrically neutral, the net ionic equation is:
$\ce{OH^- }+\ce{H^+ }\rightarrow \ce{H2O}  $.

On the other hand, we have to notice that the ion $\ce{H^+}$ is not stable in water, it combines with $\ce{H2O}  $ to form the ion hydronium $\ce{H_3O^+}$ . We usually write the complete dissociation of $\ce{HCl }$ as follows:

$\ce{HCl} +\ce{H_2O }\rightarrow$ $\ce{Cl^-} +\ce{H_3O^+}$

and the ionic equation is

$\ce{(Na^+ +OH^- )}+( \ce{H_3O^+}$$ +\ce{Cl^- })\rightarrow \ce{(Na^+ +Cl^- )}+2\ce{H2O}  $

And the net ionic equation is

$\ce{OH^- }+\ce{H_3O^+ }\rightarrow 2\ce{H2O}  $.

As for your exam questions:
$2\ce{HNO_3} + \ce{Ca(OH)_2 }\rightarrow \ce{Ca(NO_3)_2} + 2\ce{H_2O}$
 Now you can try to write the ionic and the net ionic equation to see if you understand what I have explained.


How does sulfuric acid react to heating?

Sulfuric acid, $\ce{H2SO4}$, having an enthalpy of formation of $-814\ \mathrm{kJ/mol}$, is quite stable and won't decompose easily.
According to A Kinetic Study of the Decomposition of Spent Sulfuric Acids at High Temperature, Dominique Schwartz, Roger Gadiou, Jean-François Brilhac, Gilles Prado, and Ginès Martinez:

The decomposition of $\ce{H2SO4}$ to $\ce{H2O}$ and $\ce{SO3}$ is predominant between $400$ and $700\ \mathrm K$. The formation of a small amount of gaseous sulfuric acid can be observed. Above $673\ \mathrm K$, the equilibrium constant of the reaction R1 becomes higher than 1 and increases rapidly.
$$\ce{H2SO4 <=> H2O + SO3}\tag{R1}$$
The second process is the reduction of sulfur trioxide to $\ce{SO2}$. This endothermic reaction needs a high temperature to take place, the equilibrium constant of the reaction R2 being higher than 1 above $1050\ \mathrm K$.
$$\ce{SO3 <=> SO2 + \dfrac12O2}\tag{R2}$$

So, you only have to worry when the temperature reaches $400\ \mathrm K$, or $127\ \mathrm{^\circ C}$ (false precision).


D-Threose: reduction and optical activity

When you reduce the aldehyde (from the open chain threose, since the cycle cannot be reduced as it has only acetal-like functionality) you introduce a symmetry element (both ends of the molecule become $\ce{CH2OH}$).

The presence of a symmetry element means the reduced molecule is not chiral, but meso.


Will unused spare razor blades rust if stored for a long time?

Aside from storing a in dry location, you might use a vapor corrosion inhibitor (VCI) in a container with the blades. Amazon offers a similar product for ~US$11.

My experience has been that VCI is effective and does not leave a noticeable film that needs to be removed. Whether this is cost-effective for your intended use is another question (perhaps you can use the remaining VCI on other objects or share with ten other people).

And then, beards are even more cost-effective ;-)


Intermolecular forces between carbon dioxide and water

According to the this PubChem data table for the physical and chemical properties of carbon dioxide, it contains two hydrogen bond sites (each of the oxygens). This largely accounts for the high solubility of $\ce{CO2}$ compared to that of non-hydrogen bonding compounds like methane, which has about $\mathrm{1/25^{th}}$ of the Henry's law solubility in water as does $\ce{CO2}$.


Why can't electrolysis utilize specific ions for reduction/oxidation

See, this is pretty much the same reaction. The potential is not always exactly what they say in the reference book. It depends on concentrations. The potential of $\ce{2H+ + 2e- -> H2}$ is $0 V$ only in standard conditions, that is, when the concentration of $\ce{H+}$ is 1 mol/L. When it is lower, the potential is different (see Nernst equation). Indeed, $\ce{H+}$ is present in pure water, but the concentration is quite low ($10^{-7}$ mol/L). If you plug that concentration into the Nernst equation, you'll get the same value that is listed for $\ce{2H2O + 2e- -> H2 + 2OH-}$.


I have just started working with sand columns and was also surprised that I could not find any valid and trustworthy guidelines or process breakdown for acid-washing sand.
However, I found this paper, which may answer some of your questions. In short, acid washing (or leaching) is supposed to remove impurities (mostly iron impurities) from the sand.

From the introduction of the paper:

  Iron is one of the most troublesome impurities in silica
  sand because it impairs the transparency of high-quality
  glass, e.g. for tableware, and optical glass; it impairs
  transmission in optical fibers; it adversely affects the production
  of pure silicon products, e.g. silicon metal for
  photovoltaic application, silicon carbide. One of the main tasks in the purification of silica sand
  is to further reduce iron compounds to a required level
  by a highly efficient method.

In table IV in the results section you can see a comparison between several acids with regard to their efficiency in removing iron and aluminum impurities. According to the data, sulfuric and phosphoric acids should be preferred. From what I understand, however, diluted HCl solution may be preferred by some, probably due to better price and availability, but that's just an assumption.

As for the procedure itself, I heard about adding a solution of 1% HCl to the sand, stirring and letting it stand for a few hours, and then washing with distilled water and decanting ~10 times, and then drying in an oven overnight.  I don't have any valid reference for it, though.


Seeking safe, water-soluble chemicals for an IR spectroscopy experiment

Disclaimer: I assume you refer to compounds which may be «soluble in water», not necesarily to constrained to «recording the spectra in aqueous solution» for limitations of IR transmission spectroscopy which are a bit lesser present with IR ATR spectroscopy, and basically not present for Raman (M. Farooq's answer). I assume you mention water soluble because you want to clean your setup after recording the data simply with water.  Note: often, the recordings mentioned below are literally about thin films of theses substances, i.e., an optical pathlength across the sample much less than $\pu{1 mm}$.
Complementary to NIST webbook chemistry offering a search by the vibrational energy here is the Japanese Spectral Database for Organic Compounds, SDBS. (Querries may be either in Japanese, or English.)  It equally offers a «reverse search» starting with spectral information you already have in hand:

(credit)
I'm not aware about the spectral width of the sensors you mention, if these record e.g., at $\pu{(1200 \pm 10) cm^{-1}}$, or say $\pu{(1200 \pm 50) cm^{-1}}$.  Obviously, the larger the allowance, the list of compounds indexed in the database matching this criterion will increase. In a query with the former threshold, browsing across the list of hits however were entries like formic acid (attention, potential danger because it is corrosive), or acetone  (attention, potential danger because it is flammable).
By back-and-forth between the listing (to rule out compounds not water soluble, requiring some intuition / knowledge) and the spectra, you may identify if there are other bands of (dis)interest to you, too, e.g.

(example acetone, loc. cit.)


Significance of 4π in the uncertainty principle

The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of "uncertainty" used. Without rigid definition of this quantity one ofthen just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; since they are proportional to each other it does not matter)
$$ \Delta x \Delta p \sim \hbar \, .$$
To make the statement more precise, one have to define what is actually meant by "uncertainty" and usually uncertainties are defined as the standard deviations.
$$
\Delta x = \sigma_x = \sqrt{\langle (x - \langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \, , \\
\Delta p = \sigma_p = \sqrt{\langle (p - \langle p \rangle)^2 \rangle} = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \, , \\
$$
where angle brackets $\langle \phantom{x} \rangle$ stand for the average value. And with these definitions one can indeed prove that
$$
\Delta x \Delta p \geq \hbar / 2 \, .
$$
Actually, we do even favour the usual notation for the standard deviation to avoid ambiguity
$$
\sigma_x \sigma_p \geq \hbar / 2 \, .
$$

So, from this perspective the 2 is nothing but the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the reduced Planck constant. $4 \pi$ is the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the Planck constant itself.

In fact, you can prove a more general Robertson uncertainty relation that for any two observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$
$$
    \sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|.
$$
Now, for $x$ and $p$ it is postulated by so-called canonical commutation relation that $[\hat{x},\hat{p}]=i\hbar$, thus, leading to
$$
    \sigma_x \sigma_p \geq \hbar / 2 \, .
$$

From that perspective, the Heisenberg uncertainty relation is equivalent to canonical commutation relation, and $\pi$ actually comes to the first one from the last one (as part of $\hbar$). So now the question really is why do we have $\pi$ in canonical commutation relation? And as user3371583 noted, it indeed has something to do with Fourier transform, but the connection of position and momentum through Fourier transform is not trivial.

Intuitively, however, you can get a sense of it by looking at the simpliest case of a one-dimensional de Broglie plane wave, in which $\pi$ factor in a connection between $x$ and $p$ naturally comes from de Broglie relation
$$
    p = \hbar k \, ,
$$
where $p$ is the momentum and $k = 2 \pi / \lambda$ is the wavenumber. So, wave function for plane wave can be written as follows
$$
    \psi(x) = \mathrm e^{\mathrm i(kx-\omega t)} \, ,
$$
and its derivative with respect to $x$ is
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i k \mathrm e^{\mathrm i(kx-\omega t)} = \mathrm i k \psi(x) \, .
$$
If we substitute then $p / \hbar$ for $k$ from de Broglie relation we get
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i \frac{p}{\hbar} \psi(x) \, ,
$$
or
$$
    p \psi(x) = - \mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, ,
$$
which suggest that
$$
    \hat{p} \psi(x) = -\mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, .
$$
$\hat{x}$ is simple multiplicative operator in this picture
$$
    \hat{x} \psi(x) = x \psi(x) \, ,
$$
so, the commutator between $\hat{x}$ and $\hat{p}$ is indeed $[\hat{x},\hat{p}]=i\hbar$ and through this commutator $\pi$, in a sense, propogates to Heisenberg uncetainty relation.


Why isn't sodium hydrogen phthalate used instead of KHP?

For classical analytical problems you have to check at least 100 year old literature. It is easy in Google Scholar. Set the date limits to 1920s or 1950s whatever you wish to try.  In older times, the conventional wisdom, which still persists, was that one should use a higher molecular weight standard. The rationale was that the relative error will be small on a balance. KHP has a higher molecular weight than NaHP. This is one advantage. Second, advantage is that it available in extremely high purity. It can be dried as well.
Both KHP and NaHP work perfectly. There is a caveat with NaHP, that it crystallizes as a hemihydrate i.e., for two NaHP units, there is a water molecule. KHP crystallizes as a normal anhydrous salt. Analytical chemists run away from hydrated salts because you would never know what is the hydration level, what was the humidity level at which they were stored. We would never know the "true" molecular weight of the substance which is being weighed out. There are salts such as ferrous ammonium sulfate with water crystallization, but it is very stable with respect to storage. NaHP, who knows how it loses its water of hydration.
Search "Acid potassium and acid sodium phthalates as standards in acidimetry and alkalimetry" on Google Scholar. The reference is Journal of the American Chemical Society 37, 10 (1915): 2352-2359.


How to list elements according to their radius? For example Ag, Cd, Au Can you check the accuracy of these sentences I've created?

This is the radius general trend on the periodic table:

The main reason that explains how it changes is the nuclear effective charge experimented by the electrons when atracted by the nucleos. Briefly speaking, the more electrons you add as you go down a group (in different energy levels), the lower the nuclear effective charge that the electrons will experiment, so the electron denisity will be less "compresed", resulting in a larger radius.
Going from the right to the left, you add one by one, a new electron that experiments almost the same nuclear effective charge, so it will be atracted with a similar force, and also, you are adding another proton to the nucleus, so the overall atraction will be higher, the electron density will decrease and the atomic radius will decrease.


What are the precise conditions under which Benzene undergoes ozonolysis?

In Über das Ozobenzol, an early German publication on the reaction of benzene with ozone, Ber. Dtsch. Chem. Ges., 1904, 337, 3431-3433 (DOI)
, the authors write

Leitet man in solches Benzol 1–2 Stunden lang bei einer Temperatur von 5–10° einen 5-procentigen Ozonstrom, so scheidet sich eine gelatinöse Masse aus, welche stark opalisirt. […]
Lässt man das unangegriffene Benzol verdunsten, so bildet es eine weisse, amorphe Masse von furchtbar explosiven Eigenschaften. Selbst beim Uebergiessen mit warmem Wasser detonirte diese Substanz heftig.
Uebergiesst man aber das gelatinöse Product mit eiskaltem Wasser,
so entsteht eine krystallinische Modification, die wohl möglich noch
heftiger als die amorphe bei der geringsten Berührung, ähnlich wie
Jodstickstoff, explodirt.

To sum this up:

The authors prepared the primary triozonide of benzene by passing a stream of ozone though it at a temperature of 5–10 °C.

Once the remaining benzene evaporated at room temperature, a white, amorphous mass with frightfully explosive properties was obtained.

When warm water was poured over it, this mass detonated.

When cold water was poured over the amorphous, jelly-like mass, a crystalline modification was obtained, which detonated on the slightest touch, similar to to nitrogen triiodide.

In the 1950s, a group of Dutch researchers published a series of articles about their kinetic experiments on the ozonization reactions.
The second article in the series, Recl. Trav. Chim. Pays-Bas, 1951, 70, 1005–1019, (DOI) notes that the ozonisation of benzene was performed in chloroform at a temperature of 242 K (–31 °C).

In the late 1960s, ozonization of benzene again gained some interest as a possible source for glyoxal in industrial synthesis.
In Glyoxal from Ozonolysis of Benzene, published in Ind. Eng. Chem. Prod. Res. Dev., 1969, 8, 89–92
DOI), the authors describe the ozonolysis of benzene in a mixture of acetic acid and nitromethane at temperatures between –20 °C and 15 °C.

Summary

All anyone can tell me is that it undergoes ozonolysis under "vigorous" conditions.

Apparently, your sources never bothered to check the literature! While the ozonide itself may detonate on touch, claims on the "vigorous" conditions necessary for its synthesis seem to be fully unsubstantiated by the literature!


That many types of clouds have flat bottoms is a function of decreasing air temperature with increasing height, and the water vapor pressure, commonly in this context called the dew point, of the air. When some force causes a non-cloudy parcel of air to rise, moisture will condense out as cloud particles just as soon as the air temperature reaches the dew point of the air mass.  The colder air (higher up) then contains a cloud, where the warmer air (even just slightly above the dew point) below the cloud remains clear.


Freely available software for generating 3D coordinates of molecules

Summary: RDKit ETKDG is currently the best open source tool

This is an interesting question, since a reviewer just asked us to do such a benchmark.

The trick is that there's a fine distinction between "generating 3D coordinates" (i.e., one conformation) and "generating 3D coordinates and doing some conformer sampling" presumably with the latter helping to find a low-energy conformer, thermally-accessible ensemble, or a pose that might match an experimental structure or drug docking pose.

I'm not sure there are many papers on the first (i.e., generate one 3D structure - how good is it)? For one, it would depend a lot on your definition of 'good.' I personally prefer 'low energy conformer' because it's easy to define. Most people usually mean 'close to the experimental crystal structure' which is not usually the same as 'lowest energy'.

I digress. There are a few benchmark papers - largely grabbing a few conformers, usually including a mix of free or open source tools and commercial tools.

Ebejer, J.-P.; Morris, G. M.; Deane, C. M. "Freely Available Conformer Generation Methods: How Good Are They?" J Chem Inf Model 2012, 52 (5), 1146–1158.
Friedrich, N.-O.; Meyder, A.; de Bruyn Kops, C.; Sommer, K.; Flachsenberg, F.; Rarey, M.; Kirchmair, J. "High-Quality Dataset of Protein-Bound Ligand Conformations and Its Application to Benchmarking Conformer Ensemble Generators." J Chem Inf Model 2017 57 (3), 529-539.
Nils-Ole Friedrich, Christina de Bruyn Kops, Florian Flachsenberg, Kai Sommer, Matthias Rarey, and Johannes Kirchmair. "Benchmarking Commercial Conformer Ensemble Generators." J Chem Inf Model 2017, 57 (11) , 2719-2728.
Naruki Yoshikawa, Geoffrey R. Hutchison. "Fast, efficient fragment-based coordinate generation for Open Babel" J. Cheminf. 2019 11, art. 49.

While the newest method in Open Babel (development for v3) is faster than RDKit, particularly without using force field minimization, RDKit's ETKDG is very, very good - competitive with most commercial packages.


Hyperconjugation in bridged carbon systems

Hyperconjugation is indeed very difficult to practically in cases involving bridgehead protons. The underlying reason is essentially the same that also explains why these protons cannot be enolised if there is a carbonyl group adjacent to them. This is best explained by drawing out the compound in a three-dimensional manner, which I have done for you in the figure below.

Figure 1: approximate three-dimensional representation of the cation in question; cation represented by the p orbital.

If you look at figure 1 closely, you can see that the empty p orbital and the bridgehead hydrogen σ-bond are almost at a $90^\circ$ angle towards each other. For hyperconjugation, you want approximately parallel arrangements, so this obviously fails.

However, you should always be aware that this type of compounds, essentially a norbonyl cation, is essentially the example species for non-classical cations. In this case, instead of a hyperconjugation involving the $\ce{C-H}$ bond one essentially expects the $\ce{C-C}$ bond of the ‘back half’ of the six-membered ring to provide that hyperconjugation. For a more accurate and detailed explanation, please consult the Wikipedia page or the textbook of your choice.


Experimentally finding order of a reaction

There are a couple of odd things here.

the time for a reaction to be complete is usually infinite in theoretical kinetics, so I don't see how it could be useful and/or relevant. One usually looks at the initial rate of a reaction.
if bleach is in large excess, you may have a degenerate order. That means that the concentration of bleach barely varies during the reaction and can be considered a constant.
you use an expression with an infinitesimal amount of time ($\mathrm{d}t$), when $\mathrm{d}t$ is everything but infinitely small...

Let me note bleach $\mathrm{ClO^-}$, and bromocresol green "BCG".

I assume that the reaction is monitored through the changes in absorbance, that is, the concentration of BCG. Its rate $r$ is the rate at which BCG is disappearing.

$$r = - \frac{\mathrm{d}[\mathrm{BCG}]}{\mathrm{d} t} = k \times [\mathrm{ClO^-}]^\alpha \times  [\mathrm{BCG}]^\beta $$ where $\alpha$ and $\beta$ are the partial orders.
If you use two different concentrations of bleach ($C_1$ and $C_2$), the ratio of rates will be :
$$\frac{r_1}{r_2} = \frac{k \times (C_1)^\alpha \times  [\mathrm{BCG}]^\beta}{k \times (C_2)^\alpha \times  [\mathrm{BCG}]^\beta} = \left(\frac{C_1}{C_2}\right)^\alpha$$

A simple way to get the initial rate is to measure the slope of the tangent to the graph of  ${[\mathrm{BCG}]}$ along time (dotted line).

This line crosses the time axis at $t=\Delta t$.

In this case,
$$ r_i(t=0) = \frac{\mathrm{[BCG] (t=0)}}{\Delta t_i}$$
Therefore, one can write  :
$$\frac{r_1}{r_2} = \left(\frac{C_1}{C_2}\right)^\alpha =  \frac{\Delta t_2}{\Delta t_1}$$
which kind of looks like the formula you were given.


Acetoxy silicone (I assume this is silicone caulk) is a bit of a different thing to the RTV silicone used for low-melt metal casting.

Really you want to find some high temperature mould silicone, this is fairly widely available from sculpture and model making suppliers and is used for things like casting white metal miniatures and jewellery.


Per the NIH:

  ANTICOAGULANT CITRATE DEXTROSE A...
  ...
  Citric Acid, anhydrous, USP      0.073 g
  Sodium Citrate, dihydrate, USP   0.220 g
  Dextrose, monohydrate, USP       0.245 g
  Water for Injection, USP         q.s.
  pH: 4.5 – 5.5

They used the citrate ion in the form of citric acid and sodium citrate. The citric acid is for chleating calcium while the sodium citrate does this and acts as a buffer to prevent the citric acid from raising your blood pH too much when injected as this would be painful and hazardous. Sodium is naturally in your blood and will not interfere with action potentials and is a abundant and cheap ion to use in manufacturing so it makes a logical counter ion for the citrate buffer.


The Principle of Internal Standardization (in the context of ion chromatography)

This ratio has nothing mystical, and will not work in every case.

So let's first comment on the reasons to use an internal standard. As you mention, when you inject multiple times solutions containing the same concentration of IS:

  but the IS peaks remain more or less stationary

More or less is a major point here: this means that repeatability of the measurement process is an issue, when multiple samples with the same concentration lead to a variation in the response. If the accuracy required for the measurement is such that these small variations might not be an issue, there is no need for an IS. In chrmatography measurement, sources of variability include repeatability in the volume injected, detector sensitivity. If there is some sample preparation steps involved, loss of analyte (in transfer from one tube to the other, in the filtration steps etc...) can also lead to variations from one analysis to the other.

So how will an IS solve this variability issue? The basic point is the assulmption that these external variability issues will affect both the analyte and the IS in a similar way. So one has to look at the sources of errors. For instance, let's consider an error in the measurement of a sample volume (let's call it $\epsilon$ and assume it is close to 1) . We have:

$$m_{analyte} = V_{real} c_{analyte} = \epsilon V_{set} c_{analyte}$$
$$m_{IS} = V_{real} c_{IS} = \epsilon V_{set} c_{IS}$$

If the measurement system accurately measures $m_{analyte}$ and $m_{IS}$, it is quite obvious that by dividing both equations we can remove this variability parameter and we get:

$$R_{analyte/IS} = {m_{analyte} \over{m_{IS}}} = {c_{analyte} \over c_{IS}}$$

This is the point were the ratio comes out: taking a ratio allows to remove the variability parameter ($\epsilon$) in this case.

Now this is a bit too simplistic, and a measurement procedure is not as simple. The general concept will still be to try to factor out the variability parameter as a proportion of the measured value, and then to take a ratio which will remove this variability from the measured values.

If the IS is correctly chosen so that it behaves as the analyte through the entire analytical procedure (both in terms of physico-chemical properties, as well as in the response of the detector), and if the chromatographic properties as well as the detector are such that peak area is proportional to concentration for the analyte as for the IS, then one can write:

$$A_{analyte} = \epsilon(analyte) R_{method}(analyte) c_{analyte}$$
$$A_{IS} =  \epsilon(IS) R_{method}(IS) c_{IS}$$

where $A_{analyte}$ and $A_{IS}$ are the measured peak areas for both analyte ans IS, $R_{method}(x)$ is the method response factor, which is dependent on each compound. In general the variability factor $\epsilon$ should depend on each compound, but under the above assumption of similar properties, one can assume that $\epsilon(analyte) \approx \epsilon(IS) \approx \epsilon$. Without IS, one would determine $R_{method}(analyte)$ through a calibration procedure. With an IS, one takes the ratio and the above equation can be written as:

$$R_{area} = {A_{analyte} \over A_{IS}} = {R_{method}(analyte) \over R_{method}(IS)} {c_{analyte} \over c_{IS}}$$

The ratio:

$$R_{analyte/IS} = {R_{method}(analyte) \over R_{method}}$$

is the ratio of the response factors of the method for analyte and IS. This is probably the ratio that is referred to. And it is not mystical at all, it is obtained through a standard calibration procedure with the IS.

One caution word to conclude this answer: I have tried to be quite precise in the answer. I have often times heard some people state that the use of an, internal standard is a general solution for analytical issues. Thus I want to insist on a warning: the solution is good as long as the assumptions stated above hold. If any one of these fails, then the whole analytical procedure will also fail. And of course, the use of an IS will only protect from errors after addition of the IS, and not from variability in the addition of the IS (you mention pipetting it in the analyte solution, which will limit the repeatability to that achieved by the pipettes used).


The way that a meso compound is often described in introductory organic chemistry courses is a compound that has stereocenters but is not chiral (due to a symmetry element). This description works for most cases, such as those in the previous questions (see this or this). By this description, the stereoisomers of 1,2,3-trimethylcyclopropane are meso.

However, the IUPAC Gold Book definition of a meso compound is:

  A term for the achiral member(s) of a set of diastereoisomers which
  also includes one or more chiral members.

As is pointed out in the question, none of the diastereoisomers of 1,2,3-trimethylcyclopropane are chiral, so by this definition, none of these compounds are meso.

The Gold Book also has a description of the meso prefix:

  A prefix to indicate the presence of a symmetry element of the second
  kind (see chirality) in a chemical species which is usually one of a
  set of diastereoisomers that also includes a chiral member.

The usage of meso in this way, to make a particular call out for a member in a stereoisomeric set of compounds, fails for 1,2,3-trimethylcyclopropane. Because both of the stereoisomers are achiral, neither can unambiguously be called "meso-1,2,3-trimethylcyclopropane." Since that's the case, neither are meso.


Rapidly cooling Aluminum vapour to solid state?

There are a number of actual production processes that involve coating surfaces with metals via what are effectively vapour-phase metals.

Sputtering seems to be preferred to simple evaporation because it gives better results and allows for better control over the properties of the deposited metal coating (or non-metal, a very wide variety of compounds can be deposited this way including some created by vapour phase reactions). Sputtering creates aluminium (the same applies to many other elements or compounds) vapour by bombarding solid aluminium with inert atoms and allowing the atoms to hit the coated surface.

There don't appear to be any unusual crystal structures in the resulting coatings, though the grain size and other factors can be controlled by altering the conditions.


Feasibility of using ionic liquid fuels in an MHD type reactor

If fuel is oxidized at high temperature, the conductivity of the fuel itself is irrelevant -- only the conductivity of the combustion products, which are ionized because of the high temperature and are therefor conductive. Brittanica states that temperatures greater than ~2,500 K are needed for sufficient ionization. Alkali metals can be used to increase ionization, or higher temperatures have been used with inert gases.

Note that the temperature of ions and electrons in the gas are not necessarily the same, due to the difference in mass and time to reach equilibrium. This impacts on conductivity and the rate of erosion of the MHD components.


Propose a synthesis of indane from benzene

To prepare indane (6) from benzene, make the procedure as simple as possible and amenable to scale-up. Propiophenone (2) is readily available by Friedel-Crafts acylation of benzene with propionyl chloride (1). Removal of the carbonyl group of propiophenone and the creation of a carboxyl moiety can be achieved via the Willgerodt reaction to produce amide 3. Vigorous  acid or base cleavage of the amide is possible but nitrous acid affords a mild route to carboxylic acid 4. One can use the AlCl3 F-C route again but hot polyphosphoric acid is a clean method for the intramolecular acylation without having to form the acid chloride. Finally, Clemmensen reduction of indanone (5) gives indane (6). Wolff-Kishner reduction is also a good option for the final step.


Am I visualizing/understanding atomic form factor correctly?

In your illustration, there seems to be a little «almost step».  But more importantly, if displaying a heavy scatterer (iron, Fe) next to a weaker one (C), you should put a bit more emphasis on the descent for iron.  This is one of the reasons why the atomic position of the heavier atoms are easier to attribute in the Fourier map, than the lighter ones.
Take a look at IUCr's illustration from their freely accessible on-line dictionary:

(credit)
Maybe your school has the printed or / and electronic version of the IT (the International Tables of Crystallography) which goes into further detail (e.g., listing $f$ and $f'$ for the elements, obviously dependent on the wavelength of radiation used during the diffraction experiment).
Because currently many schools' libraries may restrict access and borrow the books in person, note that IUCr moderates a dedicated section of teaching pamphlets, too.  Topics include e.g., crystal packing and symmetry.


Ions are hygroscopic because they form very stable, low energy hydrates, which are thermodynamically favorable over water vapor. I.E. salt attracts water vapor because the hydrated salt is lower energy than the water vapor. Other very porous materials are hygrosopic due to capillary action. Water vapor is attracted to the surface initially due to a variety of reasons, then the narrow channels in the porous material cause the water to spread out to lower their energy.


What are the assumption of Gibbs free energy equation referring to the system and the surrounding?

The only restriction on your equation is that the system is at constant temperature.  The system can be open or closed, the pressure does not have to be constant, and any type of work is allowed.

Let's start with the basic definition of G in terms of H, which is completely unrestricted:

$$G \equiv H-TS \Rightarrow dG = dH - d(TS) = dH -TdS -SdT$$

Then, to get your equation, the only restriction we need to add is constant $T$:

$$dG= dH -TdS$$

Why, then, might you be associating this equation with the more extensive restrictions of constant $T$ and $p$, and no non-$pV$ work?  I suspect it's because, to use $dG = dH -TdS = 0$ as a condition for equilibrium, these other restrictions are required. For more on this, see the end of my answer at What is wrong in this argument that dG must always be zero?


Which electron is knocked off a Helium atom?

An interesting question! Does the polarization of ionizing radiation interact with electron spin? Apparently, that has been investigated to some extent, and some spin information can be obtained from an entangled electron and photon.

So the answer to your question is not a clear-cut "yes" or "no", but depends on the parameters of the incident photon.


Will charcoal briquettes (not activated) absorb water out of the air in the house in a subtropical climate?

The body of your question shows that your primary concern isn't charcoal's hygroscopic character per se but it's suitability as a practical means to lower the relative humidity inside your house.

Let's start with it's hygroscopic character: It seems that charcoal indeed displays hygroscopic properties (the propensity to absorb water to itself) to some degree (at least to the extent that it affects it's burning properties).

The subject was treated by this paper, wherefrom the following citation was selected as part of this answer to your question:

  The charcoal samples placed in the environment with high relative humidity showed an
  increase in humidity by the heightened final carbonisation temperature, with the highest average detected for charcoal produced
  at 750 °C. This was the reason why the charcoal produced in
  this temperature showed higher moisture related to higher porosity
  presenting a larger area available for water condensation inside the
  pores.

  The difference between the moisture adsorbed by the charcoal packed in
  two environments can be attributed to more water molecules present in
  the humid atmosphere (RH = 98%) and a greater ease of inducing the
  porous region of the charcoal. This means that the more humid the
  environment, the greater the adsorption of moisture by charcoal.

  These two different moisture conditions represent different possible
  situations of charcoal storage where high relative humidity is not
  ideal.

Now if one of the reasons you would like to use charcoal over other means of lowering the humidity inside your house is that you were planning to re-use the charcoal later for it's primary purpose (as a fuel), this means that it wouldn't suit that purpose very well, at least if you don't plan to dry your charcoal as an intermediary step in the process (dehumidify > dry > burn). Water requires a lot of energy to evaporate which would waste fuel if you wanted to use it directly as such after using it for your house.

Now for absorbing humidity, there are other (much better) substances you could use out there: You could for instance use the very safe and well known SiO2 form frequently referred to as "silica-gel", which works very well at higher RH levels, and can be conveniently "recycled" by ~ 1h pass in the oven at ~100-120C. The main weakness of silica gel in your case is that it quickly loses efficiency at higher temperatures that are not uncommon in Florida. If you'd like to be a slightly less conventional, one could recommend Calcium Chloride (CaCl2), a fairly cheap substance which will hold up to ~4 molecules of water for each of it's own up to a temperature of 45°C, and at the very minimum 2 molecules of water for each of it's own at any temperature naturally encountered in Florida.

I hope this helps!


Carbon dioxide
"is" neither organic nor inorganic, it's just carbon dioxide.
This distinction is nothing but a convention to separate things
between the textbooks and professors.
Carbon dioxide forms carbonic acid in water and the usual salts.
This was/is reason to think of it as inorganic.
On the other hand, urea, the diamide of carbonic acid was thought
to be organic when Wöhler synthesized urea for the first time
in lab, today it is categorized as inorganic.


Why is formic acid considered an acid if it is a reducing agent?

A molecule might be an oxidising agent under the electron transfer definition, and simultaneously be a reducing agent under the alternative definition: i.e. that reduction is a loss of oxygen or a gain in hydrogen.

For your specific question, formic acid can be deemed to be an oxidisation agent since it loses a hydrogen during protonation of another molecule. The really silly way to think about this is following deprotonation, formic acid has a higher percentage of oxygen than it did before, i.e. it possesses more oxygen than before.

The case of formic acid being a reducing agent is due to the more modern definition of electron transfer. To see the mechanism of this for clarity, see here. It's clear that the formic acid is donating an electron pair (i.e. itself is being oxidised) and is making the other molecule accept those electrons (i.e. that molecule is reduced). The formic acid is therefore a reducing agent.

So your original issue is due to two different definitions of oxidisation and reduction which still exist to this day, and which are by no means mutually exclusive!

EDIT: Of course, the case of formic acid is more simple because it isn't the acid doing the reducing, it's its conjugate base i.e. formate. So formic acid is an oxidising agent under the proton definition, yet formate is an reducing agent under the electron transfer definition.


What is the difference between primitive Hexagonal unit cell and Hexagonal closed packing HCP

With the help of these Bravais lattice can HCP be formed?

Yes. One choice of unit cell for HCP is a primitive hexagonal cell with two atoms per unit cell. If you place the first atom on the origin, the second will be at (1/3 2/3 1/2). Here is a 3D view of the situation and some more information: http://lampx.tugraz.at/~hadley/ss1/crystalstructure/structures/hcp/hcp.php

  Should HCP also be placed among 14 Bravais lattice?

No, because it is already in there, just hiding a bit. If you look further into it, the space group is #194, P$6_3$/mmc, see http://img.chem.ucl.ac.uk/sgp/large/194az1.htm


Net direction of dipole moment in oxygen difluoride

In H2O, as O is more electronegative than hydrogen, the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and the dipole moment of H2O is high. (1.84 D)
On the other hand,

In the case of F2O, the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are acting in opposition resulting in a low dipole (0.3 D)
So, basically, the dipole moment is towards fluorine atoms (which are more electronegative than oxygen atom). The reason why H2O has a higher dipole moment is that the difference between the electronegativity of O and H is higher than that of O and F, so water has a higher dipole moment even tho both water and oxygen difluoride have a similar bent molecular geometry.


How to derive van 't Hoff equation for Henry's law constant?

Assuming Henry's coefficient $\mathcal{H}_i$ is defined by the following relation:
$$
\mathcal{H}_i = \gamma_i^\inf \cdot p_i^\mathrm{sat}
$$
Where $\gamma_i^\inf$ is the activity coefficient for an infinite diluted solution and $p_i^\mathrm{sat}$ is the vapor pressure of pure component $i$.
One can try to extrapolate Henry's coefficient at another temperature by assessing the following ratio (this is a common approach in Chemistry, not always valid):
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\gamma_i^\inf(T_1) \cdot p_i^\mathrm{sat}(T_1)}{\gamma_i^\inf(T_0) \cdot p_i^\mathrm{sat}(T_0)}
$$
The RHS maybe rearranged in order to have the well know form of ratio of  mixed equilibrium constants $K_i$ considering that bulk concentration are close and ratio of activities also simplifies (over a small range of temperature).
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\frac{\gamma_i(T_1)\cdot p_i(T_1)}{\eta_i(T_1)\cdot x_i(T_1)}}{\frac{\gamma_i(T_0)\cdot p_i(T_0)}{\eta_i(T_0)\cdot x_i(T_0)}} =\frac{K_i(T_0)}{K_i(T_0)}
$$
That is, we are stating that ratio of Henry's constants approximates to the ratio of equilibrium constant because we assume that only ratio of pressure is significant and other ratios tend to unity. This is, off course, not always acceptable, but in some restrained conditions it might hold.
Where the concerned reaction is the converse of  gas solubilization:
$$
\ce{X_{i,(aq)} <=> X_{i,(\mathrm{g})}}{\quad \Delta_\mathrm{R}H = -\Delta_\mathrm{sol}H}
$$
And its mixed equilibrium constant:
$$
K_i = \frac{\gamma_i\cdot p_i}{\eta_i \cdot x_i}
$$
Then one may apply van 't Hoff relation using enthalpy $\Delta_\mathrm{sol}H$ of the concerned reaction, from its integrated form, it comes:
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \exp\left[\frac{\Delta_\mathrm{sol}H}{R}\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\right]
$$
This formulae is generally a good approximation around $\pu{20°C}$ over a small range of temperature. Do not forget that $\Delta_\mathrm{sol}H$ is not a constant but rather a function of temperature.
If you wish to get your definition, you just have to differentiate the last relation with respect to temperature:
$$
\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}T} \approx \frac{\Delta_\mathrm{R}H}{RT^2} = -\frac{\Delta_\mathrm{sol}H}{RT^2}
$$
Or equivalently:
$$
\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}\left(\frac{1}{T}\right)} \approx -\frac{\Delta_\mathrm{R}H}{R} = \frac{\Delta_\mathrm{sol}H}{R}
$$