How to justify this contradiction in acid-base equilibrium of methanol and water?

We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. [...]

[...] It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

You are pointing out competing factors that, in abstract, should lead to either methanol or water being more acidic than the other (that is. lead to either hydroxide or methoxide to be more basic than the other, respectively). Once we move from considering properties in abstract to an actual system in which an acid-base equilibrium is present, the competing factors are superimposed, with a non-contradictory result: either methanol will be more acidic (and hydroxide more basic) or water will be more acidic (and methoxide more basic), depending on which factors dominate. (In particular, as pointed out in Why is methanol more acidic than water? and Is methanol really more acidic than water?, the relative acidities depend on the solvent.)


Note that I have never heard about Varsol until now, living in the Central Europe. By searching Wikipedia ( what you could do as well before asking ), I realized the below:

Varsol alias White spirit alias many other synonyms, used as a painting solvent, is based on C7-C12 oil hydrocarbons, with density 0.765(type T1), 0.780(type T2), 0.795(type T3)
( divided by volatility, T1 the most volatile )

Methanol has density 0.792. So it would rather sink under T1, T2, or being indecisive with T3 ( this may be temperature dependent )

Note that methanol has very limited solubility in non-aromatic hydrocarbons.

Note: with reported partial aromaticity, methanol will be somewhat soluble.


What are the products of the reaction of 1-methoxycyclohexene with dilute hydrochloric acid?

Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.


In pharmaceutical industries, $56\%$ of the drugs currently in use are chiral molecules and $88\%$ of the last ones are marketed as racemates (or racemic mixtures), consisting of an equimolar mixture of two enantiomers.

More recently, drugs originally marketed as racemic mixtures are reintroduced using the active isomer. Examples include racemic citalopram (Brand Name: Celexa among others) and its S-enantiomer, escitalopram (Brand Name: Cipralex and Lexapro among others); racemic omeprazole and its S-enantiomer esomeprazole; and racemic modafinil and its R-enantiomer armodafinil.
Sometimes, the difference in pharmacological activity between stereoisomers is dramatic. The dextrorotatory isomers in the morphine series are cough suppressants with less risk of substance abuse, whereas the levorotatory isomers contain the analgesic activity and significant risk of substance abuse. Although the direction of optical rotation is opposite to that of the morphine series, dextropropoxyphene contains the analgesic activity, and the levo-isomer contains antitussive activity.

Some were originally approved as racemic mixtures, and later a specific isomer was marketed with claims of having fewer adverse reactions in patients.
An example of the latter is the local anesthetic levobupivacaine, which is the S-isomer of bupivacaine. Both the R- and S-isomers have good local anesthetic activity, but the R-isomer may cause depression of the myocardium leading to decreased cardiac output, heart block hypotension, bradycardia, and ventricular arrhythmias. In contrast, the S-isomer shows less cardiotoxic responses but still good local anesthetic activity. Escitalopram is the S-isomer of the antidepressant citalopram. There is some evidence that the R-isomer, which contains little of the desired selective serotonin reuptake inhibition, contributes more to the adverse reactions than does the S-isomer.

The reintroduced S-enantionmer of citalopram with less pronounced side effects
The second part is a bit tricky, but there are different possibilities.
i. Absolute configuration
The two enantiomers of a chiral drug are best identified on the basis of their absolute configuration or their optical rotation. The absolute configuration at a chiral center is designated as R- or S- to unambiguously describe the 3-dimensional structure of the molecule. A chiral drug may have more than one chiral center, and in such cases it is necessary to assign an absolute configuration to each chiral center within the molecule.
Although the enantiomers of chiral drugs have the same chemical connectivity of atoms, they exhibit marked differences in their pharmacology, toxicology, pharmacokinetics, metabolism etc. Therefore, when chiral drugs are synthesized, as much effort goes towards the rigorous separation of the two enantiomers. This ensures that only the biologically active enantiomer is present in the final drug preparation.

The difference between two enantiomers of a drug is illustrated below using a hypothetical interaction between a chiral drug and its chiral binding site. In this case, one enantiomer is biologically active while the other enantiomer is not. The portions of the drug labeled A, B, and C must interact with the corresponding regions of the binding site labeled a, b, and c for the drug to have its pharmacologic effect. The active enantiomer of the drug has a 3-dimensional structure that can be aligned with the binding site to allow A to interact with a, B to interact with b, and C to interact with c. In contrast, the inactive enantiomer cannot bind in the same way no matter how it is rotated in space. Although the inactive enantiomer possesses all of the same groups A, B, C, and D as the active enantiomer, they cannot all be simultaneously aligned with the corresponding regions of the binding site:

Here is another example of epinephrine:

A postulated fit to epinephrine’s receptor can explain why (-)-epinephrine exhibits 12 to 15 times more vasoconstrictor activity than (+)-epinephrine.
This is the classical three-point attachment model. For epinephrine, the benzene ring, benzylic hydroxyl, and protonated amine must have the stereochemistry seen with the (-) isomer to match up with the hydrophobic or aromatic region, anionic site, and a hydrogen-bonding center on the receptor. The (+) isomer (the mirror image) will not align properly on the receptor.
ii. Another possibility
It is difficult to conclude that one isomer is superior to the other. For instance, S-verapamil is a more active pharmacological stereoisomer than R-verapamil, but the former is more rapidly metabolized by the first-pass effect. First-pass refers to orally administered drugs that are extensively metabolized as they pass through the liver.

This illustrates that sometimes absolute configuration has little impact on pharmacological activity and sometimes makes little difference whether a racemic mixture or one isomer is administered.
The popular nonsteroidal anti-inflammatory drug (NSAID) ibuprofen is sold as the racemic mixture. The S-enantiomer contains the anti-inflammatory activity by inhibiting cyclooxygenase. The R-isomer does have centrally acting analgesic activity, but it is converted to the S-form in vivo.

In some cases, the portion of a molecule containing the chiral center(s) may be in a region that does not play a role in the molecule's ability to interact with its target. In these instances, the individual enantiomers may display very similar or even equivalent pharmacology at their target site. Even in these cases, the enantiomers may differ in their metabolic profiles as well as their affinities for other receptors, transporters, or enzymes.

The body with its numerous homochiral compounds being chiral selective, will interact with each racemic drug differently and metabolize each enantiomer by a separate pathway to generate different pharmacological activity. Thus, one isomer may produce the desired therapeutic activities, while the other may be inactive or, in worst cases, produce undesired or toxic effects.
Although slightly off topic, it is important to note that in pharmacology area, activity of racemic drugs can be divided into three main groups:

Group 1. Racemic drugs with one major bioactive enantiomer

Group 2. Racemic drugs with equally bioactive enantiomers

Group 3. Racemic drugs with chiral inversion

Overall, it is critical to distinguish the single enantiomer from the racemic form because they may differ in their dosages, efficacies, side effect profiles, or even indicated use.
The decision to use a single enantiomer versus a mixture of enantiomers of a particular drug should be made on the basis of the data from clinical trials and clinical experience.
The two enantiomers of a chiral drug may differ significantly in their bioavailability, rate of metabolism, metabolites, excretion, potency and selectivity for receptors, transporters and/or enzymes, and toxicity. The use of single-enantiomer drugs can potentially lead to simpler and more selective pharmacologic profiles, improved therapeutic indices, simpler pharmacokinetics due to different rates of metabolism of the different enantiomers, and decreased drug interactions.

References

Chiral drugs

Pharmaceutical Analysis: David C. Lee, Michael L. Webb, Eds,, Pharmaceutical Analysis; 1st Edn., CRC Press LLC (Behalf of Blackwell Publishing Ltd.): Boca Raton, FL, 2003.

Chemistry for the Biosciences: Jonathan Crowe and Tony Bradshaw, In Chemistry for the Biosciences: The Essential Concepts; 3rd Edn., Oxford University Press: Oxford, United Kingdom, 2014.

Chiral Drugs: Lien Ai Nguyen1, Hua He, Chuong Pham-Huy, "Chiral Drugs: An Overview," International Journal of Biomedical Science 2006, 2(2), 85-100 (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3614593/); Also availble at: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3614593/).

Stereochemistry in Drug Action: Jonathan McConathy, Michael J. Owens, "Stereochemistry in Drug Action," Primary Care Companion to The Journal of Clinical Psychiatry 2003, 5(2), 70–73 (doi:10.4088/pcc.v05n0202); Also available at: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC353039/.

Organic Medicinal Chemistry and Pharmaceutical Chemistry: John M. Beale, Jr., John H. Block (Eds.), Wilson and Gisvold's textbook of Organic Medicinal Chemistry and Pharmaceutical Chemistry; 12th Edn., Lippincott Williams & Wilkins (a Wolters Kluwer business): Baltimore, MD, 2011.


When you opened the beer, carbon dioxide was lost to the atmosphere, increasing the mole fraction of water in the beer and raising its freezing point to a point above the temperature of the beer.  Not all of the water froze, leaving it a slush.

Edit:  There's also the possibility that the loss of CO2 from the solution is not the issue.  It could just be that when you opened the beer you disturbed a supercooled solution enough to start it freezing.

2nd edit:  Instead of continuing the comments I'll add here.  The freezing on the outside of the can shows that the temperature of the beer is below the freezing point of water.  Is the temperature of the beer lower than the freezing point of the beer?  If it were, the beer should freeze instantly upon being disturbed.  If the beer is above its freezing point, but the loss of CO2 raises the freezing point it should freeze gradually as it sits out.


18 Electron Rule For Determining the Stability of Transition Metal Complexes

A brief description of the 18 electron rule

A valence shell of a transition metal contains the following: 1 $s$ orbital, 3 $p$ orbitals and 5 $d$ orbitals; 9 orbitals that can collectively accommodate 18 electrons (as either bonding or nonbonding electron pairs). This means that, the combination of these nine atomic orbitals with ligand orbitals creates nine molecular orbitals that are either metal-ligand bonding or non-bonding,  and when metal complex has 18 valence electrons, it is said to have achieved the same electron configuration as the noble gas in the period.

In some respect, it is similar to the octet rule for main group elements, something you might be more familiar with, and thus it may be useful to bear that in mind. So in a sense, there's not much more to it than "electron bookkeeping"
As already mentioned in the comments, 18 electron rule is more useful in the context of organometallics.
Two methods are commonly employed for electron counting:

Neutral atom method: Metal is taken as in zero oxidation state for counting purpose

Oxidation state method: We first arrive at the oxidation state of the metal by considering the number of anionic ligands present and overall charge of the complex

I think this website does a good job of explaining this: http://www.ilpi.com/organomet/electroncount.html (plus, they have some practice exercises towards the end)
Let's just focus on Neutral Atom Method (quote from the link above)

The major premise of this method is that we remove all of the ligands from the metal, but rather than take them to a closed shell state, we do whatever is necessary to make them neutral. Let's consider ammonia once again. When we remove it from the metal, it is a neutral molecule with one lone pair of electrons. Therefore, as with the ionic model, ammonia is a neutral two electron donor.
But we diverge from the ionic model when we consider a ligand such as methyl. When we remove it from the metal and make the methyl fragment neutral, we have a neutral methyl radical. Both the metal and the methyl radical must donate one electron each to form our metal-ligand bond. Therefore, the methyl group is a one electron donor, not a two electron donor as it is under the ionic formalism. Where did the other electron "go"? It remains on the metal and is counted there. In the covalent method, metals retain their full complement of d electrons because we never change the oxidation state from zero; i.e. Fe will always count for 8 electrons regardless of the oxidation state and Ti will always count for four.

Ligand Electron Contribution (for neutral atom method)
a. Neutral Terminal (eg. $\ce{CO}, \ce{PR_3}, \ce{NR_3}$) :  2 electrons
b. Anionic Terminal (eg. $\ce{X^-}, \ce{R_2P^-}, \ce{Ro^-}$) :  1 electron
c. Hapto Ligands (eg. $ \eta^2-\ce{C_2R_4},  \eta^1-\text{allyl}$): Same as hapticity
d. Bridging neutral (eg. $\mu_2-\ce{CO}$) : 2 electrons
e. Bridging anionic (eg. $\mu_2-\ce{CH_3}$) ( no lone pairs): 1 electron
f.Bridging anionic (eg. $\mu_2-\ce{Cl}, \mu_2-\ce{OR}$) (with 1 lone pair)): 3 electrons
(with 1 lone pair)
or, $\mu_2-\ce{Cl}$(with 2 lone pairs): 5 electrons
g. Bridging alkyne 4 electrons
h. NO linear 3 electrons
i. NO bent ( lone pair on nitrogen): 1 electron
j. Carbene (M=C): 2 electron
k.Carbyne (M≡C): 3 electron
Determining # Metal-Metal bonds
Step 1: Determine the total valence electrons (TVE) in the entire molecule (that is, the number of valence electrons of the metal plus the number of electrons from each ligand and the charge)-- I'll call this T (T for total, I'm making this up )
Subtract this number from $n × 18$ where $n$ is the number of metals in the complex, i.e $(n × 18) – T$ -- call this R (R for result, nothing fancy)
(a) R divided by 2 gives the total number of M–M bonds in the complex.
(b) T divided by n gives the number of electrons per metal.
If the number of electrons is 18, it indicates that there is no M–M bond; if it is 17 electrons, it indicates that there is 1 M–M bond; if it is 16
electrons, it indicates that there are 2 M–M bonds and so on.
At this point, let's apply this method to a few examples
a)  Tungsten Hexacarbonyl (picture)
Let's use the neutral atom method, W has 6 electrons, the carbonyls donate
12 electrons and we get a total of 18. Of course there  can be no metal metal bonds here.
(b) Tetracobalt dodecacarbonyl (picture), here let's figure out the no. of metal-metal bonds. T is 16, R is 12, Total # M-M bonds is 6, # electrons per metal is 15, so 3 M-M bonds.
A few examples where the "18 electron Rule" works
I. Octahedral Complexes with strong $\pi$ - acceptor ligands
eg.  $\ce{[Cr(CO)_6]}$

Here, $t_{2g}$ is  strongly bonding and is filled and,  $e_g$ is strongly antibonding, and empty. Complexes of this kind  tend to obey the 18-electron rule irrespective of their coordination number. Exceptions exist for $d^8$, $d^{10}$ systems (see below)
II. Tetrahedral Complexes
e.g. $\ce{[Ni(PPh_3)_4]}$ ($\ce{Ni^0}$, $d^{10}$ 18-electron complex

Tetrahedral complexes cannot exceed 18 electrons because there are no low
lying MOs that can be filled to obtain tetrahedral complexes with more than 18 electrons. In addition, a transition metal complex with the maximum of 10 d-electrons, will receive 8 electrons from the ligands and end up with a total of 18 electrons.
Violations of 18 electron rule
I. Bulky ligands : (eg. $\ce{Ti(\text{neopentyl})_4} $ has 8 electrons) Bulky ligands prevent a full complement of ligands to assemble around the metal to satisfy the 18 electron rule.
Additionally, for early transition metals, (e.g in $d^0$ systems),  it is often not possible to fit the number of ligands necessary to reach 18 electrons around the metal. (eg. tungsten hexamethyl, see below)
II. Square Planar $d^8$ complexes (16 electrons) and Linear $d^{10}$ complexes (14 electrons)
For square planar complexes, $d^8$ metals with 4 ligands, gives 16-electron complexes. This is commonly seen with metals and ligands high in the spectrochemical series
For instance, $\ce{Rh^+}, \ce{Ir^+}, \ce{Pd^2+}, \ce{Pt^2+}$) are square planar. Similarly,  $\ce{Ni^2+}$ can be square planar, with strong $\pi$-acceptor ligands.
Similarly, $d^{10}$ metals with 2 ligands, give 14-electron complexes. Commonly seen for for $\ce{Ag^+}, \ce{Au^+}, \ce{Hg^{2+}}$
III. Octahedral Complexes which disobey the 18 electron rule, but still have fewer than 18 electrons (12 to 18)
This is seen with second and third row transition metal complexes, high in the spectrochemical series of metal ions with $\sigma$-donor or $\pi$-donor ligands (low to medium in the spectrochemical series).
$t_{2g}$  is non-bonding or weakly anti-bonding (because the ligands are either $\sigma$-donor or $\pi$-donor), and $t_{2g}$ usually contains 0 to 6 electrons. On the other hand, $e_g$ are strongly antibonding, and thus are empty.

IV. Octahedral Complexes which exceed 18 electrons (12 to 22)
This is observed in first row transition metal complexes that are low in the spectrochemical series of metal ions, with $\sigma$-donor or $\pi$-donor ligands
Here, the $t_{2g}$ is non-bonding or weakly anti-bonding, but the $e_g$ are only weakly antibonding, and thus can contain electrons. Thus, 18 electrons maybe exceeded.
References:
The following weblinks proved useful to me while I was writing this post, (especially handy for things like MO diagrams)
http://www.chem.tamu.edu/rgroup/marcetta/chem462/lectures/Lecture%203%20%20excerpts%20from%20Coord.%20Chem.%20lecture.pdf
http://classes.uleth.ca/201103/chem4000b/18%20electron%20rule%20overheads.pdf
http://web.iitd.ac.in/~sdeep/Elias_Inorg_lec_5.pdf
http://www.ilpi.com/organomet/electroncount.html
http://www.yorku.ca/stynes/Tolman.pdf
and obviously, the wikipedia page is a helpful guide https://en.wikipedia.org/wiki/18-electron_rule


Glycolysis is a process of which converts Glucose(Described as C$_{6}$ in your diagram), to two Pyruvate (2C$_{3}$). Then in downstream of glycolysis, Pyruvates are then converted to AcetylCoA to be used for many things in body. Glycolysis is part of a metabolic pathway, and in total it uses 1 Glucose, 2 ATP, 2 NAD+ to produce 2 Pyruvate, 4 ATP and 2 NADH. So those curve lines represents the Total input and output of energy. Each step is facilitated by different enzymes, so you could search for enzymes in each step for more information about each reaction. Wikipedia has table of each reactions.


What results from the reaction of 5-bromopentan-1-ol and NaOH?

The three reactions are:

An $\mathrm{S_N2}$:

Deprotonation followed by an intramolecular Williamson ether synthesis ($\mathrm{S_N2}$):

An $\mathrm{E2}$:


Which method can be applied to separate ferrofluid liquid from oil?

Typical ferrofluids are suspension of nano-sized magnetic particles in oil. If you add more of the same oil and mix thoroughly, separation is impossible.

If you add a chemically different oil, you might be able to remove this compound from the mixture again by distillation in vacuum.


Why proton concentration is divided by 10⁻⁷?

The textbook is precisely correct.
The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
One must be very fastidious with units when finding the equilibrium constant.
For example, the reaction

$$\ce{aA + bB <=> cC + dD}$$

equilibrium constant $K_c$ is exactly

$$K_c = \frac{([\ce{C}]/c^⦵)^c\cdot ([\ce{D}]/c^⦵)^d}{([\ce{A}]/c^⦵)^a\cdot ([\ce{B}]/c^⦵)^b}$$

For pure water in its standard state $c^⦵ = [\ce{H+}] = \pu{1e-7 M}$.
It also correlates with so-called biological standard state of $\mathrm{pH} = 7$.
You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
In this case those cannot be cancelled out, and must be written explicitly.

In fact, your own textbook contains extensive explanation [1, p. 91]:

  For chemical reactions the standard state for solutes is defined as $\pu{1 M}$; however, in living cells the concentration of $[\ce{H+}]$ is roughly $10^{-7}~\mathrm M$, much lower than the standard value of $\pu{1 M}$.
  It is therefore appropriate to define the reference concentration of $\ce{H+}$ in biochemical reactions relative to the $\ce{H+}$ concentration found in the living state (i.e., $10^{-7}~\mathrm M$), rather than the value $\pu{1 M}$ defined by the chemical standard state.
  Recall that when a solute in a dilute solution has a concentration of $\pu{1 M}$, the activity of that solute is unity.
  For the biochemical standard state we define the activity of $\ce{H+}$ to be unity when $[\ce{H+}] = 10^{-7}~\mathrm M$.

  [...]

  The mass action expression $Q$ is unitless.
  We strip the units from each concentration term in $Q$ by dividing each by its
  proper standard concentration (e.g., $\pu{1 M}$ for all solutes
  except $\ce{H+}$; $10^{-7}~\mathrm M$ for $\ce{H+}$; $\pu{1 bar}$ for gases, etc.).

Refrences

Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.


In which nucleophilic aromatic substitution rate is faster m-fluoronitrobenzene or p-fluoronitrobenzene?

The rate of reaction depends on stability of intermediate. The given reaction follows $\ce{S_N2Ar}$ mechanism through "Meisenheimer intermediate". So, the stability of "Meisenheimer" primarily drives the reaction.

ortho and para isomer are almost same except their position, which is the main reason why is the former reacts much slowly than the latter.

In the ortho isomer, the negative charge of the ring isn't delocalised with $\ce{NO2}$, whereas in the para isomer, it is delocalised, as seen in the following image,

(Source: Master Organic Chemistry)


Why is no MS-gas phase NMR coupling possible?

(The magnets are not really getting stronger. The > 20 T magnets are still too big and expensive to be widely used. First 900 MHz was from 2001, the 1GHz Bruker magnet came out five years ago. Since then seemingly nothing... except this 1020 MHz hybrid prototype by JEOL et al. Looks a bit unwieldy. There is a reason it is a hybrid (inner coils made of Bi-2223, i think): The classic superconducting materials (Nb3Sn, NbTi) have an absolute upper limit of ~25 T. HTSC can go higher, but they are very difficult in manufacturing. )
Fundamental problem is T1 relaxation in the gas phase. It is much too slow. You need hyperpolarisation techniques, that you have to interface with the GC. Difficult. (Impossible with mass spectrometry, because there the sample is in vacuum.)
The amount of analyte would be tinytiny. Likely impossibly small for NMR, even with hyperpolarisation. Certainly too small with mass spectrometry.
Other point: Why? GC is not useful for complex molecules, because the have very little vapour pressure. For less complex molecules, GC-MS is fine. There  are a number of working NMR techniques to separate different analytes. HPCL-NMR, GPC-NMR, DOSY, etc.
I guess nobody has seen a problem that needs GC-NMR to solve it. Or at least he has not been able to convince anyone it would be worth financing it.
(I initially misread the original question to be about a GC-NMR combination. Hence all the GC in my answer.)


Why exactly does high pressure (72 bar) nitrous oxide's pseudocritical¹ temperature drop so far when oxygen is added to make 50%v/v mixture? (Entonox)

Dissolution(solvation) is solvation of solute molecules by molecules of solvents. This decreases chemical potential $\mu={\left(\frac{\partial G}{\partial n}\right)}_{T,p}$ of the solute, comparing it (often hypothetically ) to the chemical potential of the solute at the same concentration in gaseous phase.
Different solvents cause different chemical potential drop, leading to different solubility. As dissolution equilibrium means the equal solute chemical potentials in the solid phase and the solution.
Chemical  potential and fugacity

It [fugacity] is equal to the pressure of an ideal gas which has the same temperature and molar Gibbs free energy[=chemical potential] as the real gas.

are linked by relation $\mu=\mu_0 + RT \ln{\frac{f}{f_0}} \tag{1}$
Molecules in gaseous phase move independently on each other, unless there is a strong chemical affinity, what is not this case. Pointing effect does not depend(*) on the used gas, as there is no molecular wrapping in gas.
Pointing effect does not directly affect chemical potential of vapor. But  pressure affects the chemical potential of liquid instead, leading to higher saturated vapor pressure at the same temperature. The vapor pressure therefore increases to balance chemical potentials of the  substance in both phases.
For liquids near their critical conditions, like for the case of nitrous oxide with or without oxygen, there are big changes of state variables like molar volume or density with small change of conditions like pressure and temperature. This leads to relatively big change of saturated vapor pressure with the change of pressure of the system.
Note that for the liquid at its critical temperature $T_\mathrm{c}$ and near its critical pressure $p_\mathrm{c}$, $$\lim_{p \to p_c}{\ \left(\frac {\partial V}{\partial p}\right)}_{T }= -\infty \tag{2} $$
to get a picture what these changes are like.
As we know from the thermodynamics, for stable composition and zero non-volume work:
$$\mathrm{d}G=-S \cdot \mathrm{d}T + V. \mathrm{d}p  \tag{3}$$
Therefore $${\ \left(\frac {\partial G}{\partial p}\right)}_{T }=V \tag{4}$$
$${\ \left(\frac {\partial \mu_\mathrm{liq}}{\partial p}\right)}_{T }=V_\mathrm{m, liq} \tag{5}$$
If we assume ideal vapor behavior, we can consider $p_\mathrm{vap} = f_\mathrm{vap}$ and
$$RT \cdot {\left(\frac{ d \ln{p_\mathrm{vap}}}{\mathrm{d}p}\right)}_T =  V_\mathrm{m, liq} \tag{6}$$
$$RT \cdot \ln {\frac{p_\mathrm{vap}}{p^{\circ}_\mathrm{vap}}} =  V_\mathrm{m, liq} \cdot (p - p^{\circ}_\mathrm{vap}) \tag{7}$$
$$ p_\mathrm{vap}  =  p^{\circ}_\mathrm{vap} \cdot \exp {\left(\frac {V_\mathrm{m, liq}}{RT} \cdot (p - p^{\circ}_\mathrm{vap})\right)} \tag{7}$$
(7) is a simplified saturated vapor pressure dependency rather for "normal" liquids with $T_\mathrm{c} \gg T$, when the liquid compressibility can be neglected. Otherwise, $V_\mathrm{m,liq}=f(p)$ would have to be included to the integration, as the liquid molar volume and compressibility start steeply grow toward the critical point. E.g. water vapour and liquid densities converge to about $\rho = \pu{0.21 g/cm}$ at the water critical point.
For illustration, we can take water at the boiling point with
$V_\mathrm{m} = M/\rho = (\pu{18.02 g/mol})/(\pu{0.9584 g/cm3})=\pu{18.8 cm3/mol}$, $T=\pu{373 K}$ a $p^{\circ}_\mathrm{vap}=\pu{e5 Pa}$. It can be seen the pressure effect can be neglected for small and medium pressure.
$\begin{array}{cc}\small \\
\text{pressure [MPa]} & \text{H2O vapor pressure [kPa]} \\
0.1 & 100.0 \\
0.2 & 100.1 \\
0.5 & 100.2 \\
1.0 & 100.5 \\
2.0 & 101.2 \\
5.0 & 103 \\
10 & 106 \\
20 & 113 \\
50 & 135 \\
100 & 183 \\
\end{array}$

(*) Special cases are systems with significant chemical interactions, including gas solubility in liquid. E.g. well soluble gases cause additional effect of the opposite sign than the Pointing effect. They decrease the saturated vapour pressure by decreasing solvent molar fraction and eventually its chemical potential in both phases by chemical interaction. E.g. $\ce{NH3(g + aq),H2O(l + g)}$.


There are many tricks. Since I'm not a developer, I'm not privy to all of them, but there are some simple things to start.

Some integrals are predictably close to zero: Even in fully first-principals calculations, there are parts of the code that prune out integrals that will clearly be negligible. This cuts out a lot of work. Consider, for example, overlap or Coulomb integrals between atoms that are very far apart.
Gaussian Orbitals: Using Gaussian-type orbitals are incredibly fast, because most quantum integrals can be derived as simple recurrence relationships. That is, you almost never need to do the integration. See, for example, Obara and Saika and the recent PRISM algorithm from Gill and Pople.

I believe there are some actual quantum developers on here and perhaps they'd be willing to share other tricks. But these are big - eliminate any integral you don't need, and don't actually take the integrals in the first place. ;-)

Yes, there are numeric tricks (using BLAS, LAPACK, etc.) and use of multi-core processing. But these are actually minor optimizations.

UPDATE

I realized that I forgot to talk about convergence of the self-consistent field which is also optimized. There are various convergence strategies, most notably DIIS, which helps to extrapolate the final orbital coefficients and electron density.


The mixing of two compounds is a process which requires consideration of three types of interactions: solute-solvent, solvent-solvent and solute-solute. You make a good argument for the solute-solvent interaction being stronger in mixtures of butanamide and water, compared to butylamine and water. However, as you say, this argument alone would predict a higher solubility for butanamide, which goes against the observed data. Also, in both cases the solvent-solvent interaction is the same (water with water), so this can't really be the source of an explanation.
By exclusion, we're guided towards considering the solute-solute interactions. At a first glance, both butylamine and butanamide display a combination of dipole-dipole and hydrogen bonding interactions, making it tricky to compare how their relative strengths change between compounds.
To make things easier, we can refer to additional experimental data. A nice way to gauge the strength of intermolecular interactions in a substance is to look at its melting and boiling points. According to Wikipedia, the melting and boiling points of butylamine are approximately -49 °C and 78 °C respectively. Meanwhile, for butanamide, the values are a whopping 115 °C and 216 °C! Clearly the mutual interactions between butanamide molecules are much stronger than those between butylamine molecules.
And there is our answer. You are correct that, compared to a butylamine molecule, a butanamide molecule should interact more strongly with water molecules. However, it turns out that butanamide molecules also interact very strongly with themselves, far more so than butylamine molecules. Taking everything into account, the effect of the solute-solute interactions is more prevalent; the water molecules have more difficulty keeping butanamide molecules apart, to the point that after a critical value of 163 grams of butanamide in a liter of aqueous solution (at 15 °C), the water molecules just can't stop the butanamide molecules from meeting and packing into a solid. Even if you add more butanamide, it doesn't visibly dissolve. Some butanamide molecules in the solid continuously manage to break loose from each other and get pulled into the aqueous solution, but they are replaced just as fast by molecules of butanamide already in solution which get too close and pack into the solid state.
You will find that this can be made into a very rough rule of thumb: when comparing the solubility of two roughly similar substances in a same solvent, if one of the substances has significantly higher melting and boiling points, they are likely to be less soluble.
Note that I have discussed only the influence of the enthalpic contributions to mixing. An entropic contribution also exists (favouring dissolution for both compounds), but in this case it can be safely neglected due to its comparatively small effect.


$$\ce{X + 2Y -> 2Z}$$

\begin{align}
\ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\
\ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2}
\end{align}

$(-1) \times 2$ + $(2)$ gives:

$\ce{4Y +2W + 3X->2W + 2X + 2Z + 2Y} \equiv \ce{2Y +X -> 3Z}$

Using Hess's Law,
the enthalpy change for the required reaction is given by  $-2\times \Delta_rH_1+\Delta_rH_2= 400-150 = \pu{250 kcal}$


Can hyperconjugation take place between lone pairs of an atom attached to an sp2 carbon?

Hyperconjuation is the interaction of electrons in sigma orbitals (in $C-H$ or $C-C$) with an adjacent empty p orbital or $\pi$ orbital. What you are describing is not hyperconjugation, but resonance; more specifically, the chlorine shows the $+R$ effect.

Now if you draw the resonance structures of this molecule you will realize that the $C_3$ has a negative charge on it in one canonical form. This means, in the actual molecule, that $C_3$ carbon has a partial negative charge, which is why the hydrogen from $HBr$ preferentially bonds to that carbon.

A more acceptable explanation would be that, when the hydrogen bonds to $C_3$, the resultant intermediate carbocation (positive charge on $C_2$) is stabilized by the $+R$ effect of chlorine, resulting in a racemic mixture of $2-bromo-2-chlorobutane$ to form as the major product.


Two possible chains - isopropyl or ethyl and methyl?

The first important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

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

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

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

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

Since the compound that is given in the question does not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-44.1.1 is not relevant in this case.

The second criterion given in P-44.3.2 is the longest chain. However, both possible names, (3Z)-3-ethyl-2-methylhex-3-ene and (3Z)-3-(propan-2-yl)hex-3-ene, indicate a principal chain with six skeletal atoms.

Also, both possible principal chains include the double bond. Therefore, the criteria given in P-44.4.1 (a) (greater number of multiple bonds) and (b) (greater number of double bonds) cannot be used to reach a decision.

The criterion given in P-44.4.1 (h) (lower locant for the suffix) is not relevant since the compound that is given in the question does not have any suffix.

Finally, the criterion given in P-44.4.1 (j) stipulates that the principal chain shall have the lower locant for the double bond, i.e., for the ‘ene’ ending. However, both names, (3Z)-3-ethyl-2-methylhex-3-ene and (3Z)-3-(propan-2-yl)hex-3-ene, have the locant 3 for the ‘ene’ ending.

Further criteria can be found in Section P-45:

  P-45 SELECTION OF PREFERRED IUPAC NAMES

  P-45.0 INTRODUCTION

  Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. (…) Application of
  the criteria in this section will generate the preferred IUPAC name.

  (…)

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

  (…)

In accordance with P-45.2.1, the preferred principal chain corresponds to (3Z)-3-ethyl-2-methylhex-3-ene, which has a principal chain with two substituents (3-ethyl, and 2-methyl), whereas the principal chain of (3Z)-3-(propan-2-yl)hex-3-ene has only one substituent (3-(propan-2-yl)).


How to compare the energies of sigma and pi bonds in molecular orbitals of diatomic heteronuclear species?

Actually, the gist idea is the difference of energy between the  orbitals $E_{2s(\mathrm{A})}-E_{2p(\mathrm{B})}$ and $E_{2s(\mathrm{B})}-E_{2p(\mathrm{A})}$.
If the difference is less than 10 eV in absolute value, there are interactions between the orbitals $2s$ and $2p$ and the diagram for diatomic homo-nuclear species with $Z<8$ is valid. (with a slight difference is that the levels of energy of the atomic orbitals of the  more electronegative atom are more negative than the those of the other one). This is the case for the molecules you propose, i.e. $\ce{CO}$ and $\ce{NO}$.


Coulomb force and potential energy in water vs vacuum

Yes, much larger in vacuum. The interaction is proportional to the reciprocal of the dielectric constant (relative permittivity) $1/\epsilon$ so is less in water. Think of it as an attenuation of the electric field around an ion, the larger $\epsilon$ is the more the field is attenuated and is true whatever the charges on the ions are making them attractive or repulsive.  The force at distance r between charges $q_1, q_2$ is

$$F=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in SI units; $\epsilon_0 $ is the permittivity of free space. The charges are $q=ze$ where e is the charge on the electron and z the ionic valency, $\pm 1, 2 \cdots $ etc.

The interaction energy is

$$U=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r} \qquad \mathrm{Joule}$$

The electric field around charge $q_1$ is

$$E=\frac{q_1}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in V/m. The force acting on  a second charge $q_2$ is $F=Eq_2$

Its worth plugging in some numbers for the energy for different $\epsilon$ and comparing this to thermal energy $k_BT$.


In zero order why we don't need the coefficient of reactant in integration of rate law?

You are right, for consistency the stoichiometric coefficient should be retained.
Keeping track of the stoichiometric coefficients is essential to convert between rates for different species.

If you want to compare the rate of formation of reagent A and product, you write

$$\text{Rate}=-\frac{\mathrm d[A]}{a\,\mathrm dt} = \frac{\mathrm d[product]}{\,\mathrm dt}= k$$

and so:

$$\ce{[A]-[A]_0=-a*k*t} $$

or

$$\ce{\frac{[A]-[A]_0}{a}=-k*t}$$

and

$$\ce{[product]-[product]_0=k*t}$$

However you would say that the reagent A is consumed with an effective rate constant $\ce{k' = a*k}$.


Is scandium considered a transition metal?

It may depend on who you ask, and maybe even when you ask.  The tl, dr of what is below is that we are moving towards counting scandium as a transition metal.
Wikipedia follows the IUPAC definition and accepts the label for scandium because it forms atoms with partially filled $d$ orbitals, and also copper because it forms ions with partially filled $d$ orbitals in common settings.  Either possibility is admitted by IUPAC and thus by Wikipedia.
On the other hand, Britannica does not use the label "transition metal" for Group 3 elements, because they do not retain $d$ electrons in their ions.  Instead they render a separate classification, the "rare earth metals", in this group.
Britannica is based in the UK, of course, and we may be seeing a difference based on different countries.  Let's look at a couple other sources:
https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/trans.php, based in the United States, says yes, again following the IUPAC definition.  https://www.chemguide.co.uk/inorganic/transition/features.html, based in the UK, reports that most UK syllabuses use a definition that excludes scandium from the transition metals.  The latter source does state, however, that "the most recent IUPAC definition" would include scandium because of its neutral atom configuration.
The development of organometallic chemistry may have something to do with this evolution.  Many elements, including magnesium and more recently calcium, show lower oxidation states in some organometallic environments.  In the case of scandium this shows up as an atom retaining its $d$ electron while using the $s$ electrons and/or vacant $d$ orbitals for bonding to organic ligands.  Going back to the Wikipedia article on scandium, this gives references to several organometallic examples in which scandium has reduced oxidation states, from +2 all the way down to 0.  For example, Cloke et al. [1], show a single scandium atom in a low oxidation state in a setting similar to how a later transition metal might appear:

Cocondensation of scandium vapour with 1,3,5-tri-tert-butylbenzene affords the first example of a scandium(0) complex, the sandwich compound bis(η-1,3,5-tri-tert-butylbenzene)scandium; ESR and UV studies also the reveal the presence of a second compound from this reaction, formulated as a similarly unique example of scandium(II) arising from activation of one of the methyl groups of a tert-butyl ring substituent.

Such developments favor, and may have motivated, the broader, scandium (and yttrium) inclusive definition of a transition metal now used by IUPAC.
Reference
1.
F. Geoffrey N. Cloke,   Karl Khan  and  Robin N. Perutz, "η-Arene complexes of scandium(0) and scandium(II)", J. Chem. Soc., Chem. Commun. 1991, 1372-1373.


How do I round a calculation involving both addition/subtraction and multiplication/division to the correct number of significant figures?

Approach 2 is better than approach 1.

This approach corresponds with methodology employed in both web links you provided.

The basic 'rule' is to do your calculation keeping all digits as you go right until to the very end (without rounding parts as you suggest in approach 1). This ensures your calculation retains maximum accuracy. You leave the 'rounding' til the very end, that is when you report your answer, only 'so many' digits are 'significant'.

To use your example, when calculating the subtraction part: 2.378-1.2 = 1.178, in approach 1, by 'rounding' this down to 2 significant digits (1.2), you are reducing the accuracy of your calculation before you finish.

The correct (most accurate) way to calculate the result is to keep all significant digits in your calculation until the very end, that is, 1.178 / 1.03 = 1.14368932 and then report only the 'significant' digits. In this case, since the number 1.2 only has 2 significant digits, your final answer can only have 2 significant digits. That is, the final answer is only as precise as the least precise measurement (in your case, 1.2), so in reporting the final answer for your calculation, you can only rely on 2 significant digits, so you report 1.1 as your calculated answer.

If you re-examine the answer provided in your link (http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm) in which you state problem 3 follows approach 1, you will find that the solution provided in that link actually follows approach 2, similar to what I described in the above paragraph. That is, all decimal places are retained in the calculation until the very, when the final answer is reported to the number of significant figures corresponding to the least significant measurement.


Could micrograms of polonium really change the viscosity of a whole teapot?

I think it is highly unlikely polonium makes water notably more viscous. The concentration is just too low, and radiation doesn't make water more viscous, controlrods are cooled using water and they produce a lot more radiation.

what can happen in the case of tea is that the radiation induces the formation of free radicals that in turn initiate polymerization or other reactions that form larger molecules which would make the solution more viscous, this still sounds highly improbable though.

there are multiple studies on radiation initiated polymerisation:
http://large.stanford.edu/courses/2012/ph241/weil1/


Most drinking water has residue like what you are showing.  Boiling water shows there are ions in the water, but it does not show what kind of ions are present.  Boiling water is practically useless when determining scaling potential and it is not safe to do using cooking pots and pans.  Part of the reason laboratory equipment is more expensive than cooking pots and pans is because lab equipment has much better fire prevention measures.  Boiling down water in this manner has the potential to start a fire.
pH is a better measure for scaling potential.  If you want to get technical, water pH is used to determine the Langelier Saturation Index or the Ryzner Index which are indicators of scaling potential.
Your water is probably too alkaline. Alkalinity and acidity need to be balanced because if water is too alkaline then too much scale forms; however, if the water is too acid pipes corrode.
For a household, filtering water with an activated carbon filter first, then an ion exchange filter might work.  Ion exchange filters contain a resin which exchange scale forming ions like calcium with ions like sodium. These filters can be easily fouled by organic material, so the organic material will need to be removed by something like an activated carbon filter.


How to interpret the Maxwell-Boltzmann distribution to find the activation energy?

The marked location corresponds to a level of kinetic energy in the reactants sufficient to result in a successful collision (energy wise, it says nothing about orientation).  The energy required for a successful collision is the gap in potential energy between the reactants and transition state.
A heterogeneous system is a system where the reactants are in different phases.  A common example is a gaseous reactant that collides with a solid catalyst.  In that case the Maxwell Boltzmann distribution can be applied to the gaseous reactant.
Can you clarify this?  What reactions have negative activation energies?  The reactions you'll commonly encounter have positive activation energies.


How to calculate a molar mass based on density?

What we are talking here are 3 measures of concentration:

mass fraction is about $m_{solute}/m_{solution}$
mole fraction is about $n_{solute}/n_{solution}$
molar mass is about the mass of one mole solution $m_{solution}/n_{solution}$

What we know that the mass fraction is $0.35w/w$ and the density is $1.3g/mL$. You can count the molar mass of water and HClO4 I assume, let's call them $M_{solvent} (M_{H_2O})$ and $M_{solute} (M_{HClO_4})$.

I will use the following equations by the calculations:

$$m_{solvent} = m_{solution} - m_{solute}$$
$$n_{solute} = \frac{m_{solute}}{M_{solute}}$$
$$n_{solvent} = \frac{m_{solvent}}{M_{solvent}}$$
$$n_{solution} = n_{solute} + n_{solvent} = \frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}$$

From the mass fraction you can get the mole fraction this way:

$$\frac{n_{solute}}{n_{solution}} = \frac{\frac{m_{solute}}{M_{solute}}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{1}{1+\frac{\frac{m_{solvent}}{M_{solvent}}}{\frac{m_{solute}}{M_{solute}}}} = \frac{1}{1+\frac{m_{solvent} \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(m_{solution} - m_{solute}) \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(1 - \frac{m_{solute}}{m_{solution}}) \cdot M_{solute}}{M_{solvent} \cdot \frac{m_{solute}}{m_{solution}}}}$$

You can count the molar mass the following way:

$$\frac{m_{solution}}{n_{solution}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solution} - m_{solute}}{M_{solvent}}} = \frac{1}{\frac{\frac{m_{solute}}{m_{solution}}}{M_{solute}} + \frac{1 - \frac{m_{solute}}{m_{solution}}}{M_{solvent}}}$$

You don't need the density for these calculations.

$$M_{HClO_4} = 100.46g/mol$$
$$M_{H_2O} = 18.02g/mol$$
$$m_{HClO_4}/m_{solution} = 0.35$$
$$\frac{n_{HClO_4}}{n_{solution}} = \frac{1}{1+\frac{(1 - \frac{m_{HClO_4}}{m_{solution}}) \cdot M_{HClO_4}}{M_{H_2O} \cdot \frac{m_{HClO_4}}{m_{solution}}}} = \frac{1}{1+\frac{(1 - 0.35) \cdot 100.46g/mol}{18.02g/mol \cdot 0.35}} = 0.088$$
$$M_{solution} = \frac{m_{solution}}{n_{solution}} = \frac{1}{\frac{\frac{m_{HClO_4}}{m_{solution}}}{M_{HClO_4}} + \frac{1 - \frac{m_{HClO_4}}{m_{solution}}}{M_{H_2O}}} = \frac{1}{\frac{0.35}{100.46g/mol} + \frac{1 - 0.35}{18.02g/mol}} = 25.28g/mol$$


Simple and cheap anti-graffiti layer based non-paint solvent

The "professional" solutions here in the US are often fluorinated paints. My understanding is that the graffiti literally slides off and people give up after a few tries.

Given the costs, my suggestion would probably be a thick layer of polyurethane. These are often applied to seal wood floors and can be transparent or stained. If you apply multiple layers (say 2-3 on a floor) the surface is very smooth and slick. It's hard for paint or other stains to stick.


Viscosity in gasses is generally different than liquids.  Like liquids exhibiting fluidic properties, the viscosity is related to internal friction; however, the friction mechanism is different than inter-molecular interactions experienced by closely packed molecules. The molecules in a gas are far apart resulting in the influence of an entirely different friction mechanism.

The friction mechanism dominating gas viscosity is related to the number of collisions between gas particles. More collisions means more internal friction. In gasses, two dominant factors affect the number of collisions: (1) the mean distance a gas particle has to travel before it collides with another particle and (2) the average velocity of the particle within the gas.

Some noble gasses (He, Ne, Ar) are monatomic particles with a lighter mass than the diatomic elemental gasses. The light mass results in a much higher velocity particles at a given temperature (consider  $E=\frac{3}{2}kT=\frac{1}{2}m \bar{v^2}$) resulting in more collisions. Xenon has a high molecular diameter, so the shear increase in size of the molecule increases the probability of a collision, thus reducing the mean distance it travels before a collision.

More information is available from the following link.


Since this sounds like a homework question, I will not provide a full solution!
Is $\ce{SiO2}$ an oxidizing or reducing agent in the reaction?
Figure out the oxidation state for the $\ce{Ca}$ and $\ce{Si}$ in the starting materials and the product. As a rule of thumb, consider the oxidation state of (-2) for $\ce{O}$. Unless peroxides are concerned, this works well.

 $$\ce{Ca^{(+2)}O + Si^{(+4)}O2 -> Ca^{(+2)}Si^{(+4)}O3}$$
 Apparently, there is no change - this is not a redox reaction!

Is $\ce{SiO2}$ a basic or an acidic oxide?

How are basic and acidic oxides defined?
Are there any rules (of thumb) for the oxides of metals and non-metals?

 Metal oxides are usually basic oxides. These can be considered anhydrides of the corresponding metal hydroxides, $\ce{CaO}$ is a typical example: $\ce{CaO + H2O -> Ca(OH)2}$
 What about $\ce{SiO2}$?


1 mole of substance is equal to atomic/molecular mass in grams?

The statement is more or less correct.

But there are two caveats here:

There is no proving of these concepts since there are definitions.
In general, it's a bit strange to use the ratio of amu to atom. An atom has a well defined mass based on it's isotope, but the number of grams per mole is defined based on natural abundance and weighted average mass. Therefore, there is some discrepancy here. Of course, here, you're referring specifically to carbon-12, so there is no confusion, but just watch out in the general case.


The plasmacluster supposedly works by ionizing water molecules in the air into $\ce{H+}$ and either $\ce{O^2-}$ or $\ce{HO-}$ ions. These ions then, supposedly, actively attach themselves to and break down pollutants, neutralizing their charges by pulling them apart. After they achieve this, the ions come back together to reform water.

At least, this is what their website says. They are correct that water can be split into positive and negative ions as well as the fact that these ions can come back together to reform water. I am unsure that the ions will stay separated from each other long enough to break up pollutants, and it is not very clear how positive and negative ions break up "pollutants." I find it unlikely that these ions will kill bacteria, which is one of the claimed benefits of this type of ionizer (I am not a biologist, so I could be wrong about this). Even if these ions are fatal for bacteria, odds are the plasmacluster cannot produce concentrations sufficient to do this. Also, if it could produce such concentrations, I might be somewhat concerned about what those concentrations of ions could do to humans.


What is a "period" of the periodic table?

I agree that the "periods" in the periodic table are not mathematically regular.  The simplest definition is that a period begins when a new s-subshell starts to fill.  Recall that in the s, p, d, f subshells there are 2, 8, 10, and 14 electrons, respectively, so the periods have to get bigger over time and thus cannot be regular.  As to the lack of predictability of stability, that may be true, but the periodic table is pretty good at predicting the chemistry of elements by collecting together similar elements into groups (the columns).  This similarity is mostly expressed by the s and p subshells, which is where a lot of the reactivity occurs.


Determining the exact charge of ions that will be formed by an atom using the periodic table

Each atom tends to imitate the noble gases which have a full valence shell. Looking at your table you can see that boron is 3 spaces above helium, but 5 spaces behind neon. Boron would "prefer" to have a configuration like the noble gases, and because losing 3 electrons seems easier than gaining 5, boron, as boring as he is, decides to do the obvious thing and three electrons walk the plank... Now he has 5 protons and only 2 electrons therefore imitating helium's electron configuration. This means boron has a charge of $+3$.

Oxidation states are common on most periodic tables, which show you the result of less than obvious behavior of atoms when this trend does not always hold. In the case of hydrogen, it is common than it will bond with another hydrogen to imitate helium, otherwise it will tend to form a $+1$ ion.


What do we learn from an isotope peak (or M+2 peak) in a mass spectrum?

Each element has different natural abundances for its isotopes. Two peaks with approximately 1:1 ratio indicates the presence of bromine. $\ce{^{79}Br}$ is the most common isotope, corresponding to $0.5069$ of naturally occurring bromine.  The isotopic composition for $\ce{^{81}Br}$ is $0.4931$.

What this means is that, for a molecule contaning a single bromine atom, approximately, half of the molecular ions will have $\ce{^{79}Br}$ and the other half will have $\ce{^{81}Br}$, resulting in two peaks with roughly the same intensity with $m/Z$ differing by two units.

References:

[1] https://www.nist.gov/pml/atomic-weights-and-isotopic-compositions-relative-atomic-masses

[2]https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumental_Analysis/Mass_Spectrometry/Organic_Compounds_Containing_Halogen_Atoms


What is the purpose of UV-VIS Spectroscopy?

The broad absorption bands of the electronic transitions observed in UV/Vis spectroscopy/spectrophotometry, as well as the myriad types of electronic transitions that might lead to absorption at a given wavelength, indeed make the technique much less powerful for identification of unknown compounds as compared to, e.g., IR, NMR, microwave, and mass$^*$ spectroscopies.

Jon Custer is correct in his comment to the question that UV/Vis spectra can provide key information in identifying unknown species.  However, I know of only two approaches to obtaining predictions of a given UV/Vis spectrum:

Experimental spectra of compounds suspected to be structurally related to the unknown species
Quantum chemical computation of the electronic transition spectra of candidate species

The former approach requires the relevant experimental data to exist; the latter is not trivial to perform accurately. Both suffer from significant uncertainties: How similar is a reference experimental spectrum to that of the unknown species? Can I trust that my quantum computation will give realistic band locations and strengths?

In my experience, the strength of UV/Vis lies in its facility for inexpensive, convenient, routine, non-destructive quantitation of known solutes, with known absorption spectra and/or extinction coefficient(s) at given wavelength(s).  The following list of strengths is almost certainly incomplete:

The instrumentation and consumables are relatively inexpensive.
No auxiliary cooling, gas supply, etc. are required.
Preparation and analysis of fixed-volume samples is trivial: pour liquid into cuvette; place cuvette into instrument; measure.
Continuous analysis of dynamic systems (e.g., chromatographic eluates and ongoing reactions) is possible using a flow cell.

A simple example is described here, unfortunately paywalled.
A wide variety of configurations is presented at a commercial vendor here.
(No endorsement intended. There are numerous other suppliers out there; I chose this one because it had a good variety shown in schematic.)

Interference from most solvents is minimal, particularly in the visible range.
Beer's Law in its linear form has a relatively high dynamic range for most UV/Vis-active solutes$-$that is, the concentration where absorbance deviates from linearity is much greater than the minimum concentration detectable:

$$
\frac{C_{\max}}{C_{\min}} \gg 1
$$

Another implication of Beer's Law is that interference(s) from a small number of absorbing co-solutes are relatively straightforward to handle, either by measuring at isobestic points or by measuring multiple wavelengths and solving the linear absorbance equations for the unknown concentrations. For example, if two species $\ce{X}$ and $\ce{Y}$ are present and absorbances are measured at wavelengths $\lambda_1$ and $\lambda_2$, the concentrations can be found by solving the following two expressions:

$$
A_{\lambda_1} = \epsilon_{\ce{X},\lambda_1}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_1}\cdot C_\ce{Y}
$$
$$
A_{\lambda_2} = \epsilon_{\ce{X},\lambda_2}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_2}\cdot C_\ce{Y}
$$

The steady-state nature of the measurements allows for 'deep' examination of a particular sample.

The full UV/Vis spectrum for a given volume of sample can be measured, giving a rich "fingerprint" of the particular solution and the solutes in it
The absorbance at a particular wavelength can be tracked over time, allowing for detailed kinetic analyses

At least in the visible range, the quality of the substance being measured is directly perceptible to the human senses. This is actually quite viscerally satisfying, to see that a sample anticipated to contain significantly more of a solute does indeed produce a noticeably stronger coloration of the solution.

While IR and microwave spectroscopies share some of the above advantages, they are far less suited for analysis of solutes, as the solvent absorbance is often much more likely to confound measurements.  Further, the instrumentation and sample preparation are generally much more involved.

Mass spec is terrifically expensive and complicated, is an intrinsically destructive method, and is less well suited for direct quantitation since there is no way to know a priori what fraction of the injected sample is actually making it to the detector and thus careful calibration is required.

NMR, though non-destructive, also requires quite expensive equipment that is complicated to operate properly, and interpretation of the spectra requires quite a bit of practice.

I know too little about EPR/ESR, EXAFS, XANES, and others to comment on them here. I believe the above strengths hold in comparison to those methods as well, however.

$^*$ Technically, MS is mass spectrometry, not spectroscopy, since the measurement is of discrete ion impacts on a detector, as opposed to absorbance of portions of the electromagnetic spectrum by a detector.


Why do (did) they make trans fat instead of using saturated fat?

Catalytic hydrogenation tends to make trans fats... unfortunately catalytic hydrogenation is the easiest (read lowest cost) way of hydrogenating fats so it became the industry preferred way of hardening fats before anyone realized there was a health risk to trans fats. Cheap plant oils can be hardened by hydrogenation... and hydrogenated plant oils are generally cheaper than naturally saturated fats.

Because they are unnatural, trans fats also don't go rancid so quickly... so goods made with trans fats can have longer shelf lives. Which cuts costs. Given so much of mass food production is severely cost conscious, it was inevitable that trans fats became widely used and are now having to be removed as ingredients.


Am I understanding buffering capacity against strong acid/base correctly?

Further to the answer by Poutnik the buffer capacity $\beta$ of a weak acid - conjugate base buffer is defined as the number of moles of strong acid or base $C_B$ needed to change the $p$H by $\pm1$ unit, where

$$\displaystyle \beta=\frac{d[B]}{d\text{pH}}$$

and the equilibrium concentration base present is  $\displaystyle \mathrm{[B]}=\frac{k_w}{\mathrm{[H^+]}} - \mathrm{[H^+]}+\frac{C_BK_a}{\mathrm{[H^+]}+K_a} $.

(See answer How to set up equation for buffer reaction?)  where $K_w$ is the water ionization equilibrium constant $K_w =\mathrm{ [H^+][OH^-]} = 10^{-14}$, $K_a$ is the acid dissociation constant, and $C_B$ the total concentration of buffer.

After performing the calculation $\displaystyle \beta=2.303\left[\frac{K_w}{[\text{H}^+]}+[\text{H}^+]+\frac{C_BK_a[\text{H}^+]}{ ([\text{H}^+]+K_a)^2 }  \right]$

This can be analysed to find its maximum, but can be simplified first by ignoring the first two terms, because $K_w$ is tiny as is $[H^+]$ compared to the last term close to $K_a$. Plotting this last term with $C_B=0.02$ produces the following curves

where the max buffering is seen to be very close to the $pK_A$. You can see that buffering will be ok only when the $p$H changes by $\approx \pm 1$ and when the $pK_A$ is in the range from about $pK_A $ from $\approx 4 \to 9$. The red curve is the last term in the equation, the blue curve the full equation.

To answer your first question the $p$H should be close to the $pK_A$ and as the total concentration of base ($C_B$) increases the maximum just increases but does not shift.

To find the maximum differentiating the last term only with $[H^+]\equiv x$ produces, after a little algebra, $\displaystyle \frac{d\beta}{dx}=C_BK_a\frac{K_a-x}{(K_a+x)^3}=0$ where the maximum is found when $[H^+]=K_a$


Why do we call a benzene ring attached to an hydroxyl functionality an alcohol and not a benzene?

In your first example, heptylbenzene, the benzene ring is the parent structure for the preferred IUPAC name because the ring is always the parent structure if the ring and the chain have the same number of substituents that are cited as the principal characteristic group (by means of a suffix or a class name, or implied by a trivial name). The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

In heptylbenzene, neither the benzene ring nor the heptane chain has any substituents that could be cited as the principal characteristic group. Therefore, the ring is selected as the parent structure.
In 1-phenylpropan-1-ol, however, the propane chain has an $\ce{-OH}$ group, which is the principal characteristic group of this compound and expressed as suffix (‘-ol’). Thus, Rule P-52.2.8 cannot be used. If the ring and the chain have different numbers of substituents corresponding to the principal characteristic group, another rule comes first, which stipulates that the senior parent structure is the one that has the maximum number of substituents cited as the principal characteristic group.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Therefore, the propane chain is the parent structure in 1-phenylpropan-1-ol.


I don't know if anyone is still looking for the answer but here I go anyway because I spent 40 minutes researching this for an assignment
TL;DR - The better solubility of potassium salts is the key factor, but not in the way one would initially suspect. Industrial processes and the efficiency of large-scale soap making explains the choice.
The reason why potassium-based soaps are 'soft' or liquid is to do with the manufacturing process and the difficulty in separating them (R-COO-K) from the glycerin/glycerol after reaction with the soap lye (The alkaline salt solution used to make the soap e.g. KOH). The better solubility of the potassium salts contributes to this significantly.
Glycerin/glycerol has a lot of good properties for making a liquid soap too. It's colourless, viscous, non-toxic to humans and even has some antiviral and antimicrobial properties. It makes sense to utilise these and make the soap liquid instead of trying to get it out. Also it is often cost ineffective to dissolve out the glycerin with large amounts of expensive KCl or other K salt. Comparing this to sodium-based soaps, it takes smaller amounts of cheaper Na salts to separate the by-products.
It all comes down to money I guess.
Here are my sources, read them if you want to know more:

https://pubs.acs.org/doi/pdf/10.1021/ed025p379
http://www.madehow.com/Volume-2/Soap.html
https://pubchem.ncbi.nlm.nih.gov/compound/Glycerol


What are the reaction products of sulfur and potassium nitrate?

There are at least two videos (first and second) showing the reaction between sulfur and potassium nitrate. For the chemicals:

\begin{array}{cc}
\rm Species&\rm Melting~point\\\hline
\rm Sulfur &  \rm 115~^\circ C     \\
\rm Potassium~nitrate & \rm 334~^\circ C
\end{array}

It isn't clear if nitrogen, some nitrogen oxides, or both would be produced. My guess is that it would depend on particle size, mixing and so forth, thus that a mixture of nitrogen oxides would be very likely. Also the test tube starts with a tiny amount of oxygen gas from the air. One possible balanced reaction would be:

$$\ce{4KNO3 + 5S -> 2K2O + 5SO2_{(g)} + 2N2_{(g)}}$$

Looking at the products of black powder burning $\ce{K2SO4}$ and $\ce{K2S}$ can also be formed.


Why does copper have a high enthalpy of atomisation when it has a half filled d sub shell?

When we look up the Wikipedia articles for various elements, we find that the heat of fusion and heat of vaporization of copper are modestly below those quantities from iron, cobalt and nickel, but hugely exceeding zinc.

In the first-row transition elements when they're condensed as metals, we don't really have separate $3d$ and $4s$ subshells because the metallic bonding spreads these subshells into bands and the bands overlap in energy.  This article in Chem Libretexts shows this overlap between $s$ and $p$ orbitals in sodium; a similar concept involving $s$, $d$ and even$p$ orbitals in transition metals.  So we can't separately consider $3d$ and $4s$ subshells in a first-row transition element such as copper.  With contributing orbitals from different subshells mixed together we have a significant number of filled and empty states mixed together in the valence band, leading to relatively strong metallic bonding as long as we keep that energy overlap.  Among elements in this period, only when we get to zinc do we see the $3d-4s$ band overlap drop off and the nonbonding character of the filled $3d$ subshells emerge as a major factor.  Therefore, metallic bonding and with it, enthalpy of atomization drop off sharply at zinc, not so much at copper.


Why are oxides always of the form AyOx with A representing any atom except oxygen and fluorine

Generally the least electronegative element is written first. And while naming also, +ide is attached as prefix to the more electronegative element.
For Ex- $ICl$ is written as Iodine Chloride not Chlorine Iodide. Similarly $OF_2$ is written as Oxygen Difluoride.
Coming back to your question, Oxygen is the second most electronegative element in the periodic table after Fluorine. Therefore it is generally written after all the other elements (except fluorine).


Is there any explosion hazard with an enclosed graphite crucible melting silver?

Silver will absorb oxygen readily from air while molten and spatter as it cools. This is hardly an explosion, but some care should be taken to avoid being spattered with hot particles ejected from the crucible or plugging the holes.


Effect of Catalysis on Reaction Rate in elementary Reaction

If it is in equilibrium, you are already looking at it from a macroscopic scale. It can therefore not be an elementary reaction. For example, $$\ce{A + B -> C + D}$$ is an elementary reaction, the reverse $$\ce{C + D -> A + B}$$ is a different elementary reaction. The rates of both reactions
\begin{align}
r&=k_\mathrm{fwd}[\ce{A}][\ce{B}]\\
r&=k_\mathrm{rev}[\ce{C}][\ce{D}]\\
\end{align}
would determine the location of the equilibrium
\begin{align}
\ce{A + B &<=> C + D}& K&=\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}.
\end{align}

For a true termolecular reaction, i.e. three species react in one concerted transition state,
$$\ce{A + B + C -> D + E + F},$$
the rates are always given as
\begin{align}
r&=k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]\\
r&=k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{F}].\\
\end{align}
You cannot simplify a rate law of an elementary reaction.$%edit$


EMF measured while verification of Nernst equation keeps on increasing

You should arrive to a plateau value, the EMF can't increase perpetually. The ideal potential difference is around 1.1 V given the experimental arrange, this is also known as the Daniell Cell. I think the low initial value you get is because of the time response of the system, until it reaches a steady state.
There are basically two types of connections:

i) Electrode/Solution: Some specie B transfers an electron from the solution to electrode surface (by receiving or by giving). The electron transfer rate is very high, specially on batteries, so this can not limit the whole process.

ii) Ion migration through solution: This step is the limiting process. The ions are the charge carriers in bulk solution, which move much slower compared with electrons in a conductive metal. A way to overcome this is using strong acid solutions, as sulfuric acid in the Lead-Acid Battery used on cars.

The response you have will depend on the time of measure (immediately after closing the connection, 5 seconds after, etc.) and the time interval of such measure. It will depend also on the initial concentrations of the solutions (high or low). If the concentration is low, the ionic strength of solutions is weak and few ions are to carry the charge; adding table salt (NaCl) is a possible solution. The ionic bridge may be highly conductive but, if the water permeation is also limited, you have an extra limitation there, which will be higher as the phase thickness increases (ie using too much agar, or long bridge). A solution may be, using an U-Tube with NaCl 4 M and covered in the ends with commercial water-purifier filters well sealed to the glass.

A delay on response could be due to the hydration time of the agar gel, which can take a while. Once water has permeated the agar, the ions begin to move faster, until steady state is reached.


Are 1st Electron Affinities positive or negative?

This is sadly, one of those matters of "convention" that chemistry is so plagued by.

For most neutral atoms, the magnitude of $\Delta H$ for the following reaction is negative (i.e. energy is released upon the addition of an electron).

$$\ce{A (g) + e- (g) -> A- (g)} \tag{1}$$

The value of $\Delta U$ for the reaction is, in some places, called the "electron affinity". In other places, $\Delta H$ for the reaction is called the "electron-gain enthalpy".[1]

The convention that is more common (as far as I know) is to define the electron affinity as the negative of $\Delta U$ for the above reaction. This means that electron affinities, as defined in this way, are usually positive. This convention is adopted by most major introductory inorganic chemistry textbooks and is also what I learnt in undergrad. However, if you are supposed to use a different convention for your examinations etc., then do so.

The moral of the story is: in every book you read, always double check what is the sign convention. It can and will differ from book to book.

[1] To make things even more confusing, $\Delta H$ (the change in enthalpy) and $\Delta U$ (the change in internal energy) is used in different contexts. The difference between the two is given by

$$\Delta H(T) = \Delta U(T) + \frac{5}{2}RT$$

which may be understood using Kirchhoff's law since $\Delta H(0) = \Delta U$. (Alternatively there is a derivation in Housecroft & Sharpe's Inorganic Chemistry 4th ed., p 25.) Since $5RT/2$ is only roughly $6~\mathrm{kJ~mol^{-1}}$ at $298~\mathrm{K}$, this contribution is usually negligible compared to the usual magnitudes of electron affinities.


Could we reach a pH of 4.9 by blowing through straws?

See here for partial pressure of $\ce{CO2}$ from exhaled breath that I used (28 mmHg = 0.037 atm). I am assuming Henry's law is applicable to calculate:

${7.4*10^{-4} M} =  [\ce{CO2}_{(aq)}]$

Used Henry's Law Constant of $2*10^{-3} \frac{M}{atm} $

See here for equilibrium definition and equilibrium constant value ($K=4.45*10^{-7}{M}$) I used:
$$K = \frac{[H^+]^2}{[\ce{CO2}_{(aq)}]}$$
I then solved for pH from the above equation after rearranging to solve for $[H^+]$, and got a pH of 5.2; so I don't think we can achieve pH of 4.9 from only breathing air into the beaker.


The vinyl coated lead rings suggested in the comment are a great solution. They are, however, expensive and take time to order. There is also the cost of shipping.

A quick and less expensive route is to use a PVC flange and attach what ever heavy metal you have (steel, cast iron, lead fishing weights). Epoxy to flange or drill holes and tie. You may want to paint the metal to avoid rust or other reactions with the metal.

Home Depot has inexpensive 3" flange or 4" flange.


Why is the melting point of hydrogen fluoride lower than that of ammonia?

The additional energy involved here is the stabilization energy, a form of potential energy that is stored in the more stable arrangement of a solid compared to the liquid form. This energy is determined by many factors:

Size and shape of the individual particles.
How they can 'pack' into a solid arrangement.
etc...

SO, there is clearly something different about how Ammonia is able to arrange in solid form that makes the solid stable for a greater range of energy added.
When we look at a phase change diagram the slope and horizontal component of the line determine total energy needed for that phase change. This will be different for each substance considered, and this is where your discrepancy comes from.


Does the free energy of the universe decrease for every process?

Didn't you ask the last question about this, too?

The issue here is that when $\Delta G > 0$, the reaction is indeed non-spontaneous as written. If you wish to add energy to the system, for example, via electrolysis, you should include the additional changes in the computation of the free energy change. You will find that after you sum up everything $\Delta G < 0$.

Also, note that $\Delta S_{\mathrm{universe}} < 0$ does not mean that the process will not happen; it only means that it's (fairly) unlikely to happen.


Which of the following species has an enthalpy change of formation = 0 kJ/mol?

The correct answer is indeed c. $\ce{Na(s)}$

Since for the heat of formation to be zero, the molecule should be the must abundant chemical formula of the element.

$\ce{Na}$ is a metal and is predominantly found in solid state (I dont think it exists as gas anywhere in the atmosphere of earth).

$\Delta_\mathrm{f}H^\circ$ is a quantity which is taken relative to the heat of formation of the most abundant naturally occuring form of an element.


How is NaOH able to clear a solution of Paraformaldehyde?

para-formaldehyde consists of long chains of the following type:

$$\ce{HO-CH2-O-[CH2-O]_{$n$}-CH2-OH}$$

When being dissoluted in water, this needs to be broken down into formaldehyde monomers (and then further into formalin):

$$\ce{(CH2O)_{$n$} + H2O <=> $n$ CH2O + H2O <=>> CH2(OH)2}$$

Acids and bases are both able to speed up the first step. Bases by deprotonating one end and causing a domino effect of bonds being broken and formed with a hydroxide released on the other end. Acids by protonating one end and then the domino occuring in the other direction.

$$\ce{HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-OH ->[\ce{OH-}] \\HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-O- ->\\ HO-[CH2-O]_{$n$}-CH2-O-CH2-O- + H2C=O -> \\HO-[CH2-O]_{$n$}-CH2-O- + 2 H2C=O ->}\\ \dots\\\ce{HO- + ($n$+3) H2C=O}$$


Would Chlorine or Chloramine gas show up on a VOC test?

Chlorine, no. Chloramine, maybe.

Monochloramine, $\ce{NH2Cl}$, is an inorganic compound, and not very volatile (when dissolved in water it forms more or less of an azeotrope, evaporating slightly faster than the water itself). So, not a VOC.

However, the way it does its job as a disinfectant is to basically shred organic compounds, like cell walls, key enzymes, etc, forming "organic chloramines" where one or more of the remaining hydrogens is replaced with an organic functional group. These now-organic compounds have a higher vapor pressure, so by design, chloramine is used to create VOCs. The idea, when used in tap water or pool/spa chlorination, is to allow the organic chloramines to boil off and disperse in the air, leaving clean water behind. That "chlorine smell" of a chlorinated pool is actually mostly organic chloramines evaporating out. In a more enclosed space like an interior room, these compounds can't disperse and so the smell builds.


Transition metal order of filling electron

This is the order of electrons filling for all elements but typically this diagram is only used for transition elements because when writing electron configurations for any other element its simple and intuitive but transition metals are somewhat irregular so a helpful diagram like that is useful.

Its transition metal compounds or complexes that are coloured not the metal on its own. Its coloured because when it forms compounds or complexes the orbitals in the 3d subshell aren't the same energy as they would be if it wasn't part of a compound or complex. The electrons make transitions between these different energy 3d orbitals when they absorb photons of visible light. The colour of the compound will be the complementary colour of the colour absorbed.

The 3d orbital is typically seen as the higher energy orbital but when a transition metal is ionized the 4s electrons are taken first.


Charge consistency in fragment qm/mm methods

If you're doing any sort of computational run (either quantum chemistry, solid state, or molecular dynamics), you have to set the charge on the overall system.

You said in your comments that you're doing a distributed multipole analysis across multiple fragments of a protein.

In this case, one of the constraints on the calculation is exactly that the overall charge must match that specified. The program won't arbitrarily change the charge to be an anion or a cation if you specified a neutral system.

Now, that's not to say that the individual fragments won't have fractional or integer charges. Depending on the method, it's entirely reasonable that there's some amount of partial charge transfer between fragments.

It's just that there's absolutely no way that the sum of the charges will be different than the constraint.


Preparation of alkyl halide from alcohols

It can happen.

It practice, you avoid it happening by keeping the temperature low. $\ce{Br-}$ is a better nucleophile than $\ce{H2O}$ is a base, so at lower temperatures, the substitution is faster.

If the elimination happens, alkenes react with $\ce{HBr}$ to form alkyl bromides. As long as acid is present, the carbocation can reform. For example, with isopropyl alcohol

Direct SN1 Substitution
$$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$

Substitution by elimination-addition
$$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + H2O <=> CH3-CH=CH2 + H3O+}$$
$$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$

In some cases, two processes may appear to generate different alkyl halides, but this would be no different than if a carbocation rearrangement occurred.


Relative fraction of population of different enzyme conformations, knowing the rate constants

If we use the equilibrium constants:

$$\ce{A <=>[$K_1$] B <=>[$K_2$] C <=>[$K_3$] D}$$

And we call $[\text{E}]$ the total enzyme concentration, we have:

$$ [\text{A}] + [\text{B}] + [\text{C}] + [\text{D}] = [\text{E}]\label{a}\tag{1}$$

It's not hard to see that $\displaystyle K_1 K_2 K_3 = \frac{[\text{D}]}{[\text{A}]}$. From here, we have that $\displaystyle [\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3}$. We also get:

$$[\text{B}] = K_1[\text{A}] = \frac{[\text{D}]}{K_2 K_3}$$
$$[\text{C}] = K_2[\text{B}] = \frac{[\text{D}]}{K_3}$$

Hence, substituting these last three equalities in $\ref{a}$, we get:

$$ \frac{[\text{D}]}{K_1 K_2 K_3} + \frac{[\text{D}]}{K_2 K_3} + \frac{[\text{D}]}{K_3} + [\text{D}] = [\text{E}]$$

So:

$$[\text{D}] = \frac{K_1 K_2 K_3 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

From here we can also express the concentrations of the other species in terms of $[\text{E}]$:

$$[\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3} = \frac{[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{B}] = K_1 [\text{A}] = \frac{K_1 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{C}] = K_2 [\text{B}] = \frac{K_1 K_2[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

With these equations we may compare the concentrations of each species at equilibrium, thus we are able to determine which one is the highest based on the values for the equilibrium constants; and we can also obtain their mole fractions (divide each by $[\text{E}]$, which will eliminate this term from all the numerators).


How to draw a reaction coordinate diagram for SN1 mechanism?

Yes, that diagram would be correct for the generic reaction. However in the specific case of an SN1 reaction the intermediate is almost always higher in energy than the reactants or the products, as they usually involve a carbocation. It is also typical of SN1 reactions for the first step to be rate-limiting, as carbocation formation from a neutral reactant is thermodynamically unfavorable unless specific reaction conditions are present. The three general rules for reaction coordinate diagrams are as follows:

The number of “peaks” should be equal to the number of steps in the reaction.

The height of each peak relative to the valley to its left is representative of the activation energy of that step. Larger vertical distance between peaks = greater activation energy = slower step. The rate-limiting step of any reaction will have the greatest distance between its peak and the valley preceding it.

The height difference between your reactant(s) and product(s) is equal to your total change in enthalpy (delta H) for the reaction. If the products are below the reactants then enthalpy change will be negative, and the reaction is exothermic. The inverse is true of endothermic reactions.

It may be worth clarifying that this diagram is supposed to represent an SN1 mechanism and the second step is rate-limiting. The diagram of these reactions should look like this:

Hopefully this is helpful!


Density calculation with cubic equation of state

First off, in the two-phase region you have three roots not two.

Smallest volume = liquid

Largest volume = gas

Middle volume = an unstable state you can't really reach

To determine whether a given temperature and pressure  gives the liquid or the gas you must compare the Gibbs free energy, $G$, between the liquid and gas roots as determined from solving the cubic equation.

The change in Gibbs free energy from liquid to gas is given by

$\Delta G = \int_{liquid}^{gas}{V dP}$

where $P$ and $V$ are pressure and volume along a path of constant temperature.  Note that the need for constant temperature; you can't just hold pressure constant and change the volume because that is not constant temperature.

The proper way to do it is to render

$dP=(dP/dV) dV$

where the derivative $dP/dV$ is obtained as a function of temperature and volume by differentiating the equation of state at constant temperature.  Then you have

$\Delta G = \int_{liquid}^{gas}{V(dP/dV) dV}$

with $dP/dV$ known as a function of volume at constant temperature.  You carry out the integral, which requires knowing how to integrate rational functions, and check whether $\Delta G$ is positive, negative, or zero:

$\Delta G>0$ means liquid has lower free energy, so liquid is more stable.  You plot the point for the liquid phase root.

$\Delta G<0$ means gas has lower free energy, so gas is more stable.  You plot the point for the gas phase root.

$\Delta G=0$ means both phases have the same free energy, you hit the equilibrium between liquid and gas. You plot both points and connect them with a line segment corresponding to both phases in equilibrium.  Jackpot!


Pheomelanin has a very broad absorption spectrum in the visible, with monotonically increasing absorption as the wavelength decreases until a maximum is reached at 300 nm.

I think that this would correspond to a dark red/brown colour.

Source: Phys. Chem. Chem. Phys. 2011, 13 (20), 9119–9127


Simple answer to your question should be - None. Whoever designed this question for students does not understand the basic principles of analytical chromatography.
Under low concentration of analyte, which is the usual case, peak tailing has kinetic origins, or it could be from secondary interactions or due to the structure of the packed bed. All peaks in real chromatography tail to some extent i.e., none of them are 100% perfect Gaussians.
In rare cases, you can inject a lot of sample on an analytical column. There peak tailing  originates due to non-linear isotherms. This is what the test question is about. You have to read this section on isotherms Distribution Isotherms in Chromatography to choose the right answer.
There is another form of chromatography, called preparative mode, where you really inject a lot and this is where isotherms come in.


Maybe it would be best to reorganise that to

$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right] \frac{\ce{[Br_2]_0}\ce{[F]}}{\ce{[F]_0}}$

and as there is 40 times more $\ce{F}$ than $\ce{Br_2}$, it is safe to assume that $\ce{[F]}\approx\ce{[F]_0}$. I.e. even if all the $\ce{Br_2}$ is gone there is still $\ce{[F]}=3.9\times10^{-9}$. This simplifies the above equation to

$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right]\ce{[Br_2]_0}$

which is the familiar form of the expression of concentration. It is also easy to see what the pseudo first-order rate constant is.


Images are from Wikipedia.
The first obvious thing that you can see is that N2O4 is planar but N2H4 is not. And the primary reason starts here.
N2O4 can be viewed as two nitrogen dioxide molecules joined together. If you draw a Lewis structure of NO2, you will recognize two things. Firstly, it has one unpaired electron. So imagine when two NO2 molecules come near together, they will share their two unpaired electrons (like how radicals react that you might be familiar with in organic chemistry) to form a single sigma bond. This is the very bond you are asking about.
And secondly, it is sp2 hybridized at the nitrogen atom. Thus, when you join the two NO2 molecules together, taking the relative orientation of the sp2 orbitals and how they should overlap to form a sigma bond in mind, you can see why N2O4 is planar. Because the N=O bonds (I write the "=" symbol for double bond for simplicity, it should be about 1.5 order due to resonance) heavily repulse each other when they are in the planar position, it weakens the N-N bond. Indeed this bond is so weak that it is homolytically dissociated at near below room temperature or above, creating basis for the popular equilibria between NO2 and N2O4.
The situation is even worse for N2O3. N2O3 can be viewed as formed from a NO (itself has an unpaired electron and a sp2 nitrogen) and a NO2 molecule, with the exact mechanism as above. And guess what? N2O3 has a lone electron pair on the nitrogen (in place of the missing oxygen compared to N2O4). Lone pairs repulse much, much more heavily than the bonding pairs do, and this further weakens the N2O3 N-N bond. This is illustrated by the easily disproportionation of N2O3 into NO and NO2.
Then why N2H4's bond is stronger? Because the nitrogen atoms are sp3 hybridized in this case (note that oxygen is divalent but hydrogen is only monovalent), thus they form a tetrahedral electronic structure. Thus they can comfortably adopt a skew conformation to minimize the repulsion. Either N2O3 or N2O4 could achieve this because the sp2 hybridization confines them to planar geometry.
Using Pauling's formula with the 1.14 A triple bond in N2 as a reference:
D(n) = D(m) - 0.6log(n/m)
With D(n, m) is the length of the bond with order n, m respectively.
The N-N bond order in N2O4, N2O3 and N2H4 can be estimated to be 0.44, 0.19 and 0.85 respectively.


To solve a problem like this, we need to know the solubility product constant $K_\mathrm{sp}$ for $\ce{AgCl}$, where at equilibrium,

$$K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]=1.8\times 10^{-10}\ \mathrm{M^2}$$

This relationship stipulates that under equilibrium conditions at all times, the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ must satisfy this relationship.

If the solution contains just $\ce{AgCl}$, we can determine the solubility. The concentrations of both ions must be equal based on the chemical formula.

$$\begin{array}{lcl}
x&=&[\ce{Ag+}]=[\ce{Cl-}]\\
x^2 &=& 1.8\times 10^{-10} \\
x &=& \sqrt{1.8\times 10^{-10}}=1.342...\times 10^{-5}
 \end{array}$$

We have $1.342\times 10^{-5}\ \mathrm{mol/L}=1.92\times 10^{-3}\ \mathrm{g/L}$ of $\ce{AgCl}$ in soltion.

However, if we already have some $\ce{Cl-}$ in solution, then things are a little different. Let's say we have $3.8\ \mathrm{g/L} = 6.5\times10^{-2}\ \mathrm{mol/L}\ \ce{NaCl}$ in solution, now we have a different concentration of $\ce{Cl-}$. When $\ce{AgCl}$ is added, some will dissolve, and the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ will both increase by some amount $x$.

We solve these problems using the ICE (Initial, Change, Equilibrium} approach:

$$\begin{array}{c|r|l}
\ & [\ce{Ag+}] &[\ce{Cl-}] \\ \hline
\mathrm{I} & 0 & 6.5\times 10^{-2} \\
\mathrm{C} & +\ x & +\ x \\
\mathrm{E} & x & 6.5\times 10^{-2} + x\\ \hline \end{array} \\
\ \\
K_{\mathrm{sp}}=1.8\times 10^{-10}=x(6.5\times 10^{-2} + x)\\
1.8\times 10^{-10}=x^2 +(6.5\times 10^{-2})x \\
0 = x^2 +(6.5\times 10^{-2})x - 1.8\times 10^{-10}\\
x = 2.77\times 10^{-9}, -0.065$$

One of the values for $x$ is nonsensical. The concentration of $\ce{Ag+}$ cannot be negative. Thus the amount of $\ce{AgCl}$ that dissolved in one liter was $9.77\times 10^{-9}\ \mathrm{mol}=1.38\times 10^{-6}\ \mathrm{g}$.

Note that since $x\ll 6.5\times 10^{-2}$, we could have simplified to $6.5\times 10^{-2}+x \approx 6.5\times 10^{-2}$ and then our maths would not have needed to invoke the quadratic theorem while getting the same answer within significant figures. This approximation is often valid. However, unless you are taking a test, you should always be able to Wolfram Alpha.


Timescale of Protonation and Deprotonation across micelle layers

Permeation of protons, potassium ions, and small polar molecules through phospholipid bilayers as a function of membrane thickness

  Reported values for proton permeability coefficients vary over
  a
  wide
  range,
  from
  $10^{-2} \, \mathrm{cm/s}$ to $10^{-7} \, \mathrm{cm/s}$
  (Nichols
  and
  Deamer,
  1980;
  Nichols
  et
  al.,
  1980;
  Biegel
  and
  Gould,
  1981;
  Elamrani
  and
  Blume,
  1983;
  Grzesiek
  and
  Dencher,
  1986;
  Perkins
  and
  Cafiso,
  1986;
  Norris
  and
  Powell,
  1990).
  This
  variation
  relates
  to
  the
  use
  of
  different
  techniques,
  lipids,
  and
  bilayer
  systems
  (liposomes
  and
  planar
  bilayers)
  or
  different
  pH
  ranges
  (Perkins
  and
  Cafiso,
  1986).
  The
  proton
  data
  of
  this
  study
  agree
  with
  the
  higher
  reported
  values.

  As revealed by Fig. 4, proton permeability coefficients strongly
  depend on membrane thickness, starting at a maximum value of $1.3 \cdot 10^{-2} \, \mathrm{cm/s}$ for the shortest lipid and steadily decreasing to a value of $4.9 \cdot 10^{-5} \mathrm{cm/s}$ for the longest lipid. It follows that proton permeability decreases by a factor of approximately 250 as the thickness of the
  hydrophobic region is increased from $20 \, \mathring{\mathrm{A}}$ to $37 \, \mathring{\mathrm{A}}$.


What happens to a molecule while it is reacting?

Well, a lot of things happen to the reactants. Some bonds stretch (and maybe eventually break), the others shrink, and your molecules morph into different molecules, which are the products.

(source)

As for staying at the very peak, that would be kinda unnatural, but luckily, not every peak looks like this; sometimes there is a tiny dent near the top, and with some effort you might be able to stop your reaction right there, that is, to isolate the activated complex. The discovery of its structure may shed some light on the reaction mechanism and earn you a lot of likes from your colleagues.


Does turning propane or butane gas tank upside down help to use the remaining gas?

In principle it won't make any difference, but in practice it sometimes can.

Since what you want to emerge from the tank is gas, in principle nothing you do will make more gas come out. The liquid inside will vaporise until there is no gas left and the pressure won't change while there is still liquid in the vessel (both gases liquefy easily under pressure and the pressure will be relatively constant while there is still liquid.) All this is true at equilibrium and constant temperature.

However, in practice the assumption that the inside of the vessel is in equilibrium and constant temperature is not always true. If you use a lot of gas, the vessel will cool as the liquid evaporates (and you sometimes see a ring of frost on the outside of the vessel corresponding to the level of the liquid). If this happens then shaking the vessel may have the effect of warming the liquid and increasing the pressure of the gas. So shaking it might help a little.