How to determine amount of CO₂ in soda water

First of all, after degasing by heating you have to be sure that no $\ce{CaCO_3}$ is formed, so you can minimise this problem by transfering the water in another flask.

More accurate would be to start directly with demineralised water which you can buy cheap at filling stations.

Secondly, you shouldn't wait to much while titrating, otherwise the water is taking up $\ce{CO_2}$ from the air.

You should use a bulb which doesn't contain much air (which contains $\ce{CO_2}$ and can interact as well. You can use a round flask and directly connect it with a stopper.

Third, Phenolphtalein is a good indicator because you titrate a weak acid with a strong base.

Make sure, you don't use too much Phenolphtalein.

Last but not least be sure that the $\ce{CO_2}$ from flask doesn't go away in the air or into the NaoH of the burette.

To cool will help, because $\ce{CO_2}$ is more soluble in cold water.

If you want to be a super-purist, do a potentiometric measurement and plot the point of change using a simple voltmeter.


Using non-food grade NaOH for pH control

Iron hydroxide is insoluble, so it will precipitate out. If there are other metals that aren't  alkali metals or alkaline earth metals heavier than calcium, they will also precipitate out. Most of these reactions will release some heat, but if the concentration of the metal ions is low, that shouldn't matter too much.

If you plan to drink the water, using non-food grade $\ce{NaOH}$ might cause a problem, because you won't know what else is in there.

If you add more details, we can probably tell if there are any other things to be concerned about.


Why is 5,6-dimethylidenecyclohexa-1,3-diene non-aromatic?

In this question, conjugation of p-orbitals in the system is misrepresented. It should be noted that, a π-system to be an aromatic or anti-aromatic, there should be a closed loop(s) of p-orbitals (conjugated), which is planner. What makes it an aromatic or anti-aromatic whether it follows Hückel's
rule of (4n+2) π-electrons or not.

Aromatic (most stable): If a planer, closed loop(s) of p-orbital system has (4n+2) π-electrons (in the closed system), it is aromatic. Example:, benzene (a closed p-orbital system with six π-electrons) and naphthalene (a closed p-orbital system with ten π-electrons).

anti-Aromatic (least stable): If a planer, closed loop(s) of p-orbital system has (4n) π-electrons (in the closed system), it is anti-aromatic. Example:, cyclobutadiene (a closed p-orbital system with four π-electrons) and cyclopentadiene carbocation (a closed p-orbital system with four π-electrons).

non-Aromatic (stability depends on the system): Every other systems are non-aromatic. Examples: 5,6-dimethylidenecyclohexa-1,3-diene and heptafulvene (In both cases, each p-orbital system is not closed).

See following diagram:


How to determine the concentration of a base through indirect titration?

It says that the end-point occurred at $21.30\ \mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3\ \mathrm{mL}$ or $0.0113\ \mathrm{L}$ of $\ce{HCl}$ must have been used.

I think, this is where you went wrong. If a question says something along the lines of

  The end-point occurs at $\dots~\mathrm{ml}$

then you should generally assume that that volume of whatever you are titrating with was used. In this case, that would mean a full $21.3~\mathrm{ml}\ \ce{HCl}$. The same thing of course happened at your point 7, where you write:

  It says that for the next reaction, $25\ \mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10\ \mathrm{mL}$. From this information, I worked out that $0.0021\ \mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.

Working through this bit by bit, and starting at your point 5:

$$c(\ce{HCl}) = \frac{n(\ce{HCl})}{V(\ce{HCl})} = \frac{0.003976~\mathrm{mol}}{0.0213~\mathrm{l}} = 0.1867~\mathrm{M}$$

For the titration of $\ce{Ba(OH)2}$:

$$n(\ce{HCl}) = c(\ce{HCl}) \times V(\ce{HCl}) = 0.1867~\mathrm{M} \times 27.10~\mathrm{ml} = 5.059~\mathrm{mmol} = 0.005059~\mathrm{mol}\\
n(\ce{Ba(OH)2}) = \frac{n(\ce{HCl})}{2} = \frac{5.059~\mathrm{mmol}}{2} = 2.529~\mathrm{mmol} = 0.002529~\mathrm{mol}\\
c(\ce{Ba(OH)2}) = \frac{n(\ce{Ba(OH)2})}{V(\ce{Ba(OH)2})} = \frac{2.529~\mathrm{mmol}}{25~\mathrm{ml}} = \frac{0.002529~\mathrm{mol}}{0.025~\mathrm{l}} = 0.1012~\mathrm{M}$$

The difference still there can be attributed to a rounding error somewhere along the way.


Why is N2 enthalpy zero, entropy 191.5, yet G is listed as zero and not [-T*0.1915]?

Enthalpies of formation and Free energies of formation are expressed relative to the standard state of 25 C and 1 bar.  So, in the standard state, they are taken to be zero.  The entropy value you cited is the absolute entropy, relative to absolute zero.  If you wanted the entropy of formation of nitrogen relative to the standard state of 25C and 1 bar, that would be 0.


Differentiating between Prins reaction / ene reaction experimentally

The traditional Prins reaction was run in aqueous acid (H2SO4, HCl) with formaldehyde. The carbocation formed by addition of the alkene was trapped by a nucleophile, or if E1 elimination occurred, the most stable alkene formed. The Lewis acid-catalyzed ene reaction is intramolecular without nucleophile capture and the less stable alkene is formed, at least when an isopropenyl group is present in the reactant. Two examples follow:

Malonaldehyde 1, which has two enantiotopic aldehyde groups, reacts via transition state 2 to produce alcohol 3 as the major product. The cyclopentadiene ring is in effect a t-butyl group that prefers being equatorial in the transition state. Note the exocyclic double bond as opposed to a more stable endocyclic position. The hydroxyl group is axial. Aldehyde 4 follows the same type of pattern forming axial alcohol 5. If an open transition state were operable, the equatorial alcohol would be anticipated.
[PS: You can Google much of this information.]

1) J. Am. Chem. Soc., 1990, 112, 2149-2158

2) J. Am. Chem. Soc. 1984, 106, 718-721


How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia?

You are on the right track. Dimensional Analysis is your friend:

$$\frac{10\,\text{L  $\ce{N2}$}}{1} \cdot \frac{1\,\text{mol  $\ce{N2}$}}{22.4 \,\text{L  $\ce{N2}$}}\cdot \frac{3\,\text{mol  $\ce{H2}$}}{1\,\text {mol  $\ce{N2}$}}\cdot \frac{22.4\,\text{L  $\ce{H2}$}}{1\,\text {mol  $\ce{H2}$}}=\,?\,\, \text{L $\ce{H2}$}$$

If you do this, you will realize that the quantities cancel exactly and give you a nice clean number.


Why don’t all objects melt and liquefy when heated sufficiently?

The problem with melting most substances is that you would be doing it in a non-innocent atmosphere and that sometimes even the functional groups inside a compound are non-innocent.

In a perfectly innocent world, everything could be molten given sufficient temperature. Given real world paper, remember that reaction $(1)$ is exothermic and the only reason why it does not spontaneously occur is the lack of activiation energy.

$$\ce{C_nH_mO_l + \frac{$2n + m - l$}{2} O2 -> n CO2 + \frac{$m$}{2} H2O}\tag{1}$$

Once you supply sufficient heat, the activation energy will be overcome and the reaction suddenly becomes spontaneous; for paper, I am told that this happens at around $451~\mathrm{^\circ F}$, which is $233~\mathrm{^\circ C}$ in proper units.

Now you could suggest melting things in an inert argon atmosphere — or maybe even take helium for additional inertness. But then still you have potential reactions that can occur with organic materials. What then will happen might be called smouldering. Technically, this is no more than burning without oxygen.


Both approaches are correct.

Avogadro's number is $6.02214129\times 10^{23}$ and represents the number of carbon-12 atoms in 12 grams of unbound carbon-12 in the ground electronic state.

$12$grams$/6.02214129\times 10^{23} = 1.9926467\times 10^{-23}$grams

The unified atomic mass unit (u) is $1.660538921 \times 10^{-24}$ grams

$12 \times 1.660538921 \times 10^{-24}$ grams $ = 1.9926467\times 10^{-23}$grams


Why is the electronegativity of potassium and rubidium same?

There are a few ways that electronegativity can be measured/calculated, the original being that of Pauling, followed by others scales such as the Allred-Rochow and Mulliken electronegativity. When looking at the values obtained by Pauling, one must keep in mind that these values aren't grounded in quantum mechanics (i.e. electronegativity does not have an observable that can be measured). However, these values seen on the Pauling electronegativity charts are relative values estimated from bond energies.
Anyways, defining Pauling's electronegativity as the "ability of an atom to attract electrons" (Matthew Salem, Chem Libre Texts), one can look at two factors to estimate these values.
1. Nuclear Charge/Atomic radius - As we know from Coulombs law,
$\large F=k\frac{q_{1}q_{2}}{r^2}$
the attraction between to oppositely charged points is a function of the magnitude of each charge (q) and the distance between them ($r^2)$. So the number of protons in the nucleus along with the distance of the electron from the nucleus determine the attractive force it feels from the nucleus.
2. Electron Shielding –  The electrons that are closer to the nucleus can shield bonding electrons from the  attractive forces of the nucleus. The ability to shield outer electrons is greatest in the s orbitals, a little weaker in the p orbitals, and poor in the d/f orbitals.
With these factors in mind, one can approximate the effective nuclear charge $(Z_{eff})$ of an atom, which is an estimation of the attraction certain electrons will feel. Effective nuclear charge is estimated with the equation below,
$(Z_{eff}) = Z - S$
where Z is the atomic number and S is the shielding constant (number of non-valence electrons).
If you calculate the $(Z_{eff})$ for rubidium and potassium, one would find that they are equal, suggesting that the electronegativity of each atom is are approximately the same.
This link discusses $(Z_{eff})$ in more depth if you are interested.


Why is heating increasing the solubility of sodium hydroxide?

The question to the Newton - Ask a Scientist program by Danna C Griffiths (to be found via the Internet Archive) has an answer for it.

What I understood from the explanation on that site:

Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of $\ce{NaOH}$ in a saturated solution of $\ce{NaOH}$.

When you dissolve $\ce{NaOH}$ in pure water, the process is highly exothermic, but as the concentration of $\ce{NaOH}$ increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more $\ce{NaOH}$ to dissolve).

So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.


Is the axial position more suited for a lone pair or an electronegative element?

So in $\ce{XeF2}$ there will be three lone pairs on the xenon, and the xenon will have two covalent bonds to fluorines. The xenon is of course the central atom, on to that we apply VSEPR-theory. We count five electron dense areas (two bonds and three lone pairs) and so VSEPR-theory predicts a trigonal bipyramidal shape for $\ce{XeF2}$

The lone pairs want to stay as far away from each other as possible (in VSEPR). Indeed would there have been just two lone pairs putting them axially would have introduced the maximal possible angle of 180°. Instead we have three lone pairs, so the equatorial positions introduce the best possible angle of 120°. Compare this to the 90° angle introduced if we had put two lone pairs axially and just one equatorial.

What is left now are the axial positions to which we allocate the fluorines, obtaining a linear molecule. In this case the VSEPR prediction has been confirmed experimentally.


Does solute concentration vary significantly with depth?

(1)If I dissolve glucose in an ordinary beaker filled with water, are the concentrations of glucose at the top and bottom significantly different?
(2) What if we use NaCl as the solute?

1 & 2 are basically the same question. The difference in gravity between the top of the beaker and the bottom of the beaker is real but insignificant. So due to gravity alone there won't be a concentration gradient.
However there are solutions which can be centrifuged which will create a concentration gradient. Essentially the centrifuge creates a large gravity gradient. Close to center gives lower gravity, far from center gives higher gravity. Once you stop the centrifuge however the concentration gradient is not at equilibrium. So over time the concentration gradient would disappear.

(3) Is there a molecule large enough to ``sink'' to the bottom, and if so would it still be classified as a solute?

The above discussion about a centrifuge confirms the basic notation, but there is an unstable gradient, not a sinking per se.
Obviously if you drop a marble in a glass of water it sinks. But the marble isn't dissolved in the water.

(4) Likewise, are there solutes which are lighter than H2O, so they float upwards?

This is a bit more interesting. Open a plastic bottle of soda and put a balloon over the neck of the bottle.  Carbon dioxide was dissolved in the soda under pressure. When the soda is opened, the pressure is released and the excess carbon dioxide will "float up" out of the soda and blowup the balloon. Since the gas phase is less dense than the liquid phase (soda) it is on top. The gas doesn't really "float" on the soda however. It is also important to realize that once equilibrium is established the carbon dioxide remaining in the soda is homogeneously distributed throughout the soda.
A different sort of twist are soap molecules which will "float" to the surface, not because they are lighter, but because of the surface tension of water. The famous soap/pepper experiment illustrates this.
Another sort of twist here is a bartender who makes fancy cocktails by floating one layer of a liquor over another due to the higher layers being less dense. The segregation of the layers is unstable however. One good swirl with a straw and the layers mix.

For 3 and 4 there is what is called a phase diagram. For example it shows if two liquids will mix, or if there will be two phases where one liquid would float on the other.
Think of gasoline and water. There is a gasoline layer floating on the water so there are two liquid phases because there is an interface. However a very tiny amount of gasoline will be dissolved in the water and a very tiny amount of water will be dissolved in the gasoline. In this case the gasoline that is dissolved in the water is distributed homogeneously throughout the water. Likewise the water dissolved in the gasoline is distributed throughout the gasoline.

I used gasoline as a conceptual example. Gasoline however isn't a pure compound but rather a mixture. So for those that want to be picky substitute n-hexane for gasoline.


Define, (actually define) the "stability" and "energy" of a compound

As a chemist, I would agree that textbooks are not clear about the term stability and energy. It is not your fault and undergraduate organic chemistry books make the situation worse. I cannot recall the text, it was an old book, however it clearly said that stability of a compound does not mean anything. We should always ask, stability with respect to what? Are we talking about thermodynamic stability or kinetic stability? For example H2+O2 mixture is not stable. This statement is meaningless because we are not saying stability with respect to what? This mixture should form water right away but people actually waited for 30 years and did not find much water in a mixture of H2 and O2. Thermodynamically, this mixture is not stable and it will react violently once you provide the right activation energy. When you are reading about stability, ask this question yourself.

In the same vein, the term energy should also be clarified by authors. As a crude analogy is that you can think of energy as the potential energy of the whole molecule when the authors talk about conformers. Have a overall picture of the molecule. Imagine two springs, A) one which is less stretched and the other one (B) is more stretched. You can say spring B has more (potential) energy than (A). An eclipsed conformer has higher potential energy (overall, no need to think locally) than a staggered one.


Ultrasound Vegetable Glycerin homogenization

Depending on intensity, ultrasound can produce intense heat at some locations. In fact, cavitation can induce temperatures far exceeding 5,000 K. Even a small ultrasonic room humidifier can burn through a plastic comb held in the upwelling stream. The sound intensity is decomposing the oil and/or propylene glycol.

Try lower frequencies and lower sound intensity. You might experiment with different shape containers, as well, to avoid focusing the sound into a high-intensity spot.


Calculation of empirical formula from elemental analysis

A sad Ph.D. student had been working on the synthesis of a natural product, with empirical formula $\ce{CH2N}$, for several years. He finally succeeded in synthesising it - or so he thought - because his NMR data seemed to fit the target structure. Excited about his breakthrough (in fact, he was so excited that he didn't even bother taking a mass spectrum!), he sent a sample for elemental analysis, expecting to get some results along the lines of 42.8% carbon, 7.2% hydrogen, and 50.0% nitrogen.

Not long after this, the results came back, and he discovered that his compound inexplicably had 40.0% carbon, 6.7% hydrogen, and 53.3% nitrogen. "That's not what I expected," he thought to himself - "I wonder if I can publish it and try to gloss over the data."

Upon checking the Author Guidelines for the Journal of Organic Chemistry (section 2.2.2), he found that they were actually pretty strict.

  Found values for carbon, hydrogen, and nitrogen should be within 0.4% of the [calculated] values for the proposed formula.

Thus was the story of the sad Ph.D. student who had to go back to the drawing board. (Thankfully, he had synthesised enough of the immediate precursor and by varying the conditions, he eventually succeeded in making his target.)

TL;DR Rounding down 1.14 to 1 is a gross error and won't ever get published in a reputable journal. Considering that this is not even a real-life scenario, there's absolutely no reason why standards should be lowered.


Normalization of the wavefunction φ(x) = A

Your derivation is mostly correct, but you made a small mistake when evaluating the integral

\begin{equation}
\int\limits_{-b}^{3b} | \psi (x) |^2 \,\mathrm dx = \int\limits_{-b}^{3b} | A |^2 \,\mathrm dx = 1
\end{equation}

The antiderivative of a constant $C$ is the product of this constant with the integration variable, e.g. $x C$.
So

\begin{align}
\int\limits_{-b}^{3b} \lvert A \rvert^2 \,\mathrm dx = \Bigl[ x \lvert A \rvert^2 \Bigr]_{-b}^{3b} &= \lvert A \rvert^2 \bigl((3b - (-b)\bigr) = 4 b \lvert A \rvert^2 = 1 \\
&\Rightarrow  A = \pm \frac{1}{2 \sqrt{b}}
\end{align}

As for the $\pm$ sign ambiguity: In quantum mechanics, we know that a wave function can always be multiplied by a constant phase factor:

\begin{equation}
\psi'=\mathrm e^{-\mathrm i\alpha}\psi,
\end{equation}

where $\alpha \neq \alpha(x,t)$, because the probability distribution of a system is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation.
So - since $-1 = \mathrm e^{\mathrm{i} \pi}$ and $+1 = \mathrm e^{\mathrm{i} 2 \pi}$ are examples of such constant phase factors - it doesn't make a difference, from a physics point of view, whether you use $+\frac{1}{2 \sqrt{b}}$ or $-\frac{1}{2 \sqrt{b}}$ since both lead to the exact same physical results and are thus physically equivalent.


Both cases are equivalent in the case of    molar ratio 1:1. It would lead to a double buffer of $\ce{HAc/Ac-}$ and $\ce{NH4+/NH3}$. But the buffer capacity would be inferior, as the resulting $\mathrm{pH}$ is too far from the both respective $\mathrm{p}K_\mathrm{a}$.


Constitutional isomers for four carbon thioesters

One way to break things down in by the number of carbon atoms on either "side" of the thioester.

There are 4 carbon atoms to work with, of which one will be taken up by the thioester moiety.  Thus there are three left to work with.

On the carbonyl "side" of the thioester, between zero and three carbons are possible.  However, on the sulfur "side" of the thioester, at least one carbon atom is required; otherwise the compound is not a thioester but a thioacid.  Thus the possibilities for carbon atom distribution are:

one carbon on S side, two on CO side.
two carbons on S side, one on CO side.
three carbons on S side, zero on CO side.

You have drawn the only structures for the first two possibilities.

For the last possibility, there are two possible ways to attach a three-carbon alkyl chain to the thioester sulfur:

as a $n$-propyl group
as an isopropyl group

Thus, the four possible thioesters are:

methyl thiopropanoate
ethyl thioacetate
n-propyl thioformate
isopropyl thioformate


Reasons which explain why the efficiency of a reaction increases as the average rate decreases?

Well it depends on the reaction. A notable example is the Haber process. The Haber process is the production of ammonia gas from nitrogen and hydrogen gas: $$\ce{2N2 + 3H2 -> 2NH3 + heat}$$ This reaction is extremely important in the industry sector as ammonia gas is used for a wide range of products such as fertiliser, bleach, cleaning agents and in WW2, it was used to make explosives.

Lets go off on a tangent for a moment:

  In WW1 the Germans had very little resources due to British ships blocking Germany's allies from providing them materials. This meant that they had to produce great quantities of ammonia themselves as they desperately needed it to make explosives. Thankfully German chemistry, Fritz Haber, created the Haber process which allow the Germans to mass produce ammonia. It is estimated that WW1 was prolonged by 2 years due to the creation of the Haber process. Haber later received the Nobel Prize in 1918 for his production process of ammonia and 'contribution to agriculture'.

So the question you might ask is why was creating ammonia so hard? After all it is a pretty simple reaction. Well this has to do with your question.

You see, producing ammonia gas is actually quite easy, however the main issue is that it is very difficult to get a high yield at a fast rate. The reason is due to the extremely high activation energy of this reaction. This is because for the reaction to occur, the extremely strong triple bonds of nitrogen must first be broken, which requires quite a lot of energy. This high activation energy means that the reaction is going to be really slow. This shown by the Arrhenius equation:
$$k = A\mathrm e^{-\frac{E_\mathrm A}{RT}}$$ However we are able to increase the rate of the reaction. If we were to increase the temperature, according to Arrhenius  equation, the rate of the reaction will increase. However now there is another problem, this time about the yield of the reaction.

Note above in the chemical equation, the Haber process is an exothermic reaction, meaning that heat is released. Now if we were to increase the temperature, according to Le Chatelier's principle, the equilibrium will shift to the left. This means that less ammonia gas will be produced. This is shown by the graph below:

So the above conundrum describes your question. By increasing the temperature of the sytem, you can increase the rate of the reaction, however by doing so you are going to decrease the yield.

Note: this is only true if the reaction is exothermic. If it is endothermic, then the yield will increase with temperature.


How does the use of inert gas in fluorination of saturated aliphatic hydrocarbons make it less violent?

Let's say the gas mixture has something like 10:1 molecules ratio of inert gas : fluorine. The alkanes in the gas would have 10 times more chance of colliding with an inert gas molecule as compared to a fluorine molecule. This reduces the rate at which alkanes and fluorine react together, making the reaction a lot safer (fluorination is highly exothermic to the point where it can be explosive).

The inert gases are basically a necessity, without which it's impossible to do this reaction safely and responsibly.


Finding the pressure equilibrium constant of the reverse reaction

They are actually just looking for the equilibrium constant for the reverse reaction.

$$K_p(-\xi) = K_p^{-1}(\xi) = \frac{1}{1.85} = 0.541$$


Is the rate of a zero order reaction always independent of the concentration of its reactants?

When you look at the overall reaction, it is if you have an equal concentration of each reactant, then by the reaction progressing or adding equal amounts of both, the reaction rate won't change. For example, if you have 2M of both A and B, $[A]^{-1}[B]^1 = 1/2*2=1$, and increasing the concentration to 3M for both won't change the rate since $1/3*3=1$. The differential rate law applies to only one single species, so unless you have equimolar amounts of each reactant or make an approximation using an excess of one, you cannot apply the differential rate law.


Can activated charcoal be used to filter bacteria out of air?

Activated charcoal has been shown to be able adsorb certain species of bacteria, but the effectiveness varies between species [1] (which is potentially useful for medical applications), and has not been tested for many, so it is unlikely that activated charcoal could be used to adsorb all the bacteria from the air or a gas.

Activated charcoal general usage is in the adsorption of toxins.

[1] Naka, Katsumi, et al. "Adsorption effect of activated charcoal on enterohemorrhagic Escherichia coli." Journal of Veterinary Medical Science 63.3 (2001): 281-285.


Liquid that looks like smoke when bottled?

So you want to make something that looks like smoke. What makes smoke look the way it does, anyway?

Smoke is composed of extremely fine particles, often below $\mathrm{1\ µm = 0.001\ mm}$ in diameter (check here for comparisons with other particulate materials). These particles are large enough to scatter light, which allows them to be seen, but small enough to be extremely light and stay aloft with even very weak currents in air or liquids.

One interesting and very beautiful consequence of these combined properties is that they can clearly reveal fluid flows, which is what gives smoke its wispy, delicate and complex appearance. That's what fluid flows are like all the time, we just almost never have a direct view of it.

So fundamentally your request is quite simple; all you need is a common, non-toxic source of very fine particulate. Fortunately there are good ways to achieve this in practice.

One of the most popular seems to be using fine mica dust. These non-toxic silicate minerals can be ground to form tiny, almost perfectly smooth crystalline sheets. When the particles align properly, light is reflected into a viewer's eyes giving a shiny, glittery appearance. If the mica dust is suspending in a liquid, then the many millions of particles twinkle momentarily as they tumble chaotically under the action of currents in the fluid.

Liquid mixtures containing similar materials are called rheoscopic fluids. There are many many sources online for ready-made fluids, or you can make your own from mica powder or other fine solids. Suspensions in water eventually settle, but last for a long time without degradation, needing only a shake to awaken. With a myriad of water-soluble dyes at your disposal, you can adjust the fluid to the desired colour.

Though rheoscopic fluids based on water and mica are non-toxic and quite safe, that doesn't mean they pose no hazard whatsoever. Very fine particulate can be an irritant if inhaled or exposed to eyes and mucous membranes (especially a problem if handing the dry powder). Just take a bit of care and you can have fun too! Try putting the fluid on a warm surface and see the convection currents!


Precipitating silver(I) salts from the solution of barium(II) cyanate and iodide

Consider a $\pu{10.0 mL}$ solution containing $\pu{1.0e-10 M}$ each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$. If $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0e-17}$; $K_\mathrm{sp}(\ce{AgI}) = \pu{8.5e-17}$.

Assuming that $\ce{Ba(CN)2}$ and $\ce{BaI2}$ dissociate completely.
$\ce{[CN-]_i = [I-]_i =} 2\cdot10^{-10}$ molar
Neglecting any volume change of solution the initial concentration of $\ce{Ag+}$ will be
$\ce{[Ag+]_i} = \dfrac{3.5\cdot10^{-9}\pu{mol}}{0.010\pu{L}} = 3.5\cdot10^{-7}\pu{M}$
Now if both the $\ce{CN-}$ and $\ce{I-}$ are quantitatively removed then the same amount of $\ce{Ag+}$ must be removed.
$\ce{[CN-]_i + [I-]_i =} 4\cdot10^{-10}$ molar
$\ce{[Ag+]_f} = 3.5\cdot10^{-7}\pu{M} - 4\cdot10^{-10}\pu{M} \approx 3.5\cdot10^{-7}\pu{M}$
So the final concentration of $\ce{Ag+}$ is essentially the same as the initial concentration. The concentration of $\ce{Ag+}$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $\ce{CN-}$ is
$\ce{[CN-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{6.0\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{1.7e-10}$
The the final concentration of $\ce{I-}$ is
$\ce{[I-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{8.5\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{2.4e-10}$
Conclusion:
Since $\ce{[CN-]_i > [CN-]_f}$ some $\ce{AgCN}$ will ppt.
Since $\ce{[I-]_i < [I-]_f}$ no $\ce{AgI}$ will ppt.


Dissociation of silver chloride in a solution

The answer is yes and no. Silver chloride doesn't "dissociate", it decomposes into silver metal and dissolved chlorine.  The chloride reacts with the water to form hydrochloric acid and hypochlorous acid as shown below:
$$\ce{2AgCl_{(s)} +H_2O_{(l)} ->[h\nu] 2Ag_{(s)} + Cl_2_{(aq)} +H_2O_{(l)}->2Ag_{(s)} + HCl_{(aq)} + HClO_{(aq)}}$$


Which conformations of 2,3-dimethylbutane are more stable?

The correct arrangement (from least stable to most stable) is probably: $4 > 2 > 3 \gtrapprox 1$ based on my calculations for the four gas phase rotamers you can see in the figure below.

Your reasoning for 4 over 2 sounds good, that would also be my point. For the two minima 1 and 3 (with Boltzmann weights of 46%:54% at the highest calc. niveau (red), 33%:66% at the lowest (blue)) you might also be right. But from the really small energy difference I don’t want to argue too much about it.

My calculation methods seem a bit random, but at first I only wanted to make that fast wB97X-D3/def2-TZVP//HF-3c scan and then decided to calculate the bigger one.


What do (2S,3S) and similar notations mean?

When naming a organic compound, S/R-notation comes if there is a chiral atom present. The S/R-notation always accompanies a numerical such as (2S,3R)- where the numerical refers to the chiral carbon position in the molecule according to the IUPAC numbering while R and S representate the orientation of the chiral atom according to the Cahn–Ingold–Prelog priority rules. For example, let's look at glyceraldehyde molecules where all this stereochemistry started from:

The IUPAC name of glyceraldehyde is 2,3-dihydroxypropanal. However, it has a one chiral carbon at $\ce{C}$2 and hence has two stereoisomers. The priority order of groups attached to chiral carbon is $\ce{-OH \gt -CHO \gt -CH2OH \gt -H}$. Hence you can look at the $1 \rightarrow 2 \rightarrow 3$ priority group rotation through the bond of chiral carbon and least priority group, which is $\ce{H}$ here (though $\ce{C-H}$ bond). Accordingly, the rotation for the first compound is clockwise as shown in the image, thus it carries (R)-notation. Since chiral carbon is 2 according to the IUPAC rules, the complete notation is (2R)-. Similarly, the mirror image is caring (2S)-notation since it has counterclockwise rotation as depicted in the image. Therefore complete names of these two isomewrs are (2R)-2,3-dihydroxypropanal and (2S)-2,3-dihydroxypropanal (Note that these isomers are called enantiomers).
In addition, if a $\ce{H}$ in $\ce{C}$3 is replaced by another different group, say it is $\ce{CH3}$ group, it will also become chiral and can have either (S) or (R) rotation. Suppose you replace $\ce{H}$ in $\ce{C}$3 of (2R)-2,3-dihydroxypropanal by a $\ce{CH3}$ group making it has (S) rotation. By addition of $\ce{CH3}$ group make it 2,3-dihydroxybutanal. With correct stereochemistry, its correct name is (2R,3S)-2,3-dihydroxybutanal.


How do we calculate a character table from its basis functions

I think now I answer my question.

Although in principle, I can calculate the full character table totally from the orthogonality theorems of characters. Namely, I need to use the two relations:

$$\sum_{k} N_k\chi^{(\Gamma_i)}(C_{k})\chi^{*(\Gamma_j)}(C_k) =\delta_{ij}$$

$$\sum_{\Gamma_j}N_k\chi^{(\Gamma_j)}(C_{k})\chi^{*(\Gamma_j)}(C_{k'})= \delta_{kk'}$$

But in practice, it is not always easy to use these two relations to calculate the characters. Even for $D_3$ group which has only two three irreducible representations, we need proper initial guess if we don't want to solve the set of equations.

On the other hand, if we know the basis functions in the first place, then we can effectively get the characters.

For example, if we want to calculate the characters of $A_1$ of $C_{2v}$ group. First, I need to choose a basis function. Let's say I choose $z$ as my basis function (I have listed the basis functions of $A_1$ in my question), since the irreducible representation is of one dimension, I simple need to see how the operations will change the sign of $z$.

There are four classes for $C_{2v}$ group, and the representatives are $E$, $C_2$, $\sigma_v(xz)$ and $\sigma_{v}^{'}(yz)$

$E$
It is evident that $E$ won't change the sign of $z$, so the character is 1.
$C_2$
The principle axis is $z$, so the character is also 1.
$\sigma_v(xz)$
It is a reflection across $xz$ plane, but it doesn't change the sign of $z$, so the character is still 1.
$\sigma_v^{'}(yz)$
It is a reflection across $yz$ plane, the character is also 1.

So, based on the above discussion of using basis function to calculate characters, we see that basis functions can be readily used for calculating characters. Although the example is really trivial (we can write down the characters of $A_1$ even without thinking), but the example indicates that if we need to calculate the characters of $A_2$, we just don't need to solve the set of equations.


How to predict ease of oxidation or reduction for a given oxidising or reducing agent?

This goes to the heart of organic chemistry and most of it is just about learning the properties of the most common reagents, then understanding where to find the detailed information. A good Organic chemistry text book is the starting point, then there are lists to help, for examples list of oxidising agents, list of reducing agents or entire series of books on Organic chemistry reagents e.g. Fieser & Fieser which a good library should have. For individual transformations databases such as Reaxys or Scifinder are searchable by structure and can find the conditions under which a transformation has previously been done. If the transformation has not previously been reported then you have to use your judgement as to which conditions to try, that is where the real art of synthesis comes in. It is important to note that solvent and temperature play a huge role in such transformations. Reactions that proceed in one solvent may give a different outcome in another.

Dibal-H does reduce carboxylic acids, but require multiple equivalents and elevated temperature or additional reagents such as TMSiCl paper here


Where am I going wrong here?

Your chemical reaction is not balanced. You wrote

$$\ce{1/2 H2(g) -> H(g) -> H+(g) -> H+(aq)}$$

but it should be

$$\ce{1/2 H2(g) -> H(g) -> H+(g) + e- (?)-> H+(aq) + e- (?)}$$

I put a question mark for the physical state of the electron because it's not clear where that electron should go or what enthalpy of formation we would assign it. You could instead involve a second element in your reaction, e.g.:

$$\ce{H2 + Cl2 -> 2H+ + 2Cl-}$$

You could assign an enthalpy of formation for the mixture of hydrogen and chloride ions, but you would not know how to assign it to the individual components of the mixture.

  In my book it is given that by convention heat of formation of aqueous H+ is taken to be zero.

Yes, that's what we do. It would equally be possible to assign it a value of 100 kJ/mol. All the enthalpies of formation of ions would change (for example, that of chloride ions would by 100 kJ/mol lower to compensate). However, all the calculations of enthalpies of reactions would have the same result because it all cancels out in a balanced equation. Setting the enthalpies of formation of elements to zero is also a bit arbitrary (not once we specify that to measure them, we are making the compounds from the elements).

Updating the definition

Instead of saying the enthalpy of formation of a compound is the enthalpy of making that compound from the elements, we could widen the definition to say:

"The enthalpy of formation of a species is the enthalpy of the reaction making the species where all other reactants and products are either elements or the hydrogen ion""

This definition would include ions. For chloride ions, the following reaction would work:

$$\ce{1/2H2 + 1/2Cl2 -H+ -> Cl-}$$

For sodium ions, the following reaction would work:

$$\ce{Na + H+ - 1/2H2 -> Na+}$$

Why the hydrogen ion?

Again, it is arbitrary and any other ion would have been fine as well. The hydrogen ion is special for two reasons, though. It is the smallest ion, and it is fairly common in reactions, so it saves some arithmetic effort when doing calculations.


Reaction of Hydrogen and Dinitrogen Tetroxide?

A likely reaction between dinitrogen tetroxide and hydrogen would be as follows with nitrogen and water being the products

$\ce{N2O4 + 4H2 → N2 + 4H2O~(I)}$

$\ce{\Delta H_{rxn}=[0+(4 x -241$.$ 8)-9$.$2 - (4 x 0)] ~=~ -976$.$4~ kJ}$

For the record, dinitrogen tetroxide and hydrazine (usually in combination with methylated hydrazines) is currently used as a rocket fuel.  The equation describing this reaction is

$\ce{N2O4 + 2N2H4 → 3N2 + 4H2O~(II)}$

$\ce{\Delta H_{rxn}=[(3 x 0) +(4~ x ~-241$.$ 8) - 9$.$2 - (2 x 50$.$6)] ~=~ -1077$.$6~ kJ}$


Predict melting point of water CaCl2 solution

I think there are three issues here. The first is that your purple line of calculating the freezing point doesn't appear to be correct. Let's arbitrarily select $b=1 ~ \mathrm{mol/kg}$ to compare. $i=3$ and $K_\mathrm{f}=1.86 ~ \mathrm{^{o}C/mol}$. The molecular weight of $\ce{CaCl2}$ is $111 ~ \mathrm{g/mol}$. Then, our mass concentration is $b·\mathrm{MW}/(b·\mathrm{MW}+1 ~ \mathrm{kg})=0.1 ~ \mathrm{kg}_\mathrm{salt}/\mathrm{kg}_\mathrm{total}$. The freezing point is $b·i·K_\mathrm{f}=1·3·1.86=5.58 ~ \mathrm{^{o}C}$. This is much closer to the red/black line than the purple one.

The second problem is that that relation is only valid at low concentrations. As the concentration of salt goes up, it has an increasing effect on the activity of the water. As we can see from the phase diagram, once you get to a 6:1 mol ratio of $\ce{Ca^{2+}}$ to $\ce{H2O}$, the interaction is extremely strong. I'm just speculating here, but I would guess six water molecules are forming a weak octahedral coordination complex with the $\ce{Ca^{2+}}$, with the electronegative oxygen ends oriented toward the ion. This effect becomes so strong, that the hydrate becomes the relevant element for consideration above that concentration.

Third, you're ignoring the existence of hydrates. If you find the melting points and the cryoscopic constants of the hydrates, you can actually calculate the right side of the plot the same way.


Why is carbon dioxide considered a Lewis acid?

If you've learned how to draw resonance structures, they will help you greatly here (and in general). In the attached figure, 1 is the relevant resonance, implying the carbon in $\ce{CO2}$ is partially positive and therefore may be prone to accept electrons (see the resonance hybrid).

In theory, any Lewis base (electron-donor) can react with $\ce{CO2}$ (Lewis base abbreviated B in 2 & 3). We can draw the reaction mechanism using the charged resonance structure (2) or the neutral, most "stable" structure (3). In reality, these are not two discretely different forms but just "snapshots in time" of a true $\ce{CO2}$ molecule, so it doesn't really matter which structure we as humans choose to represent on paper.

I hope you can see how $\ce{CO2}$ is a Lewis acid because it is able to accept electrons. Although in the principal resonance structure you see all atoms have full octets of electrons, the double bonds can easily break, rearranging the electrons. If full octets prevented atoms from receiving electrons, a great percentage of all known reactions would not occur. The key is that the electrons in the octet must have an alternative place to go if new electrons are donated to the molecule (in this case, the electrons migrate to the $\ce{O}$ atom). In more advanced chemistry, we would say $\ce{O}$ acted as an electron sink in this process.

It is interesting to note that any $\ce{C=O}$ structure can react as a Lewis acid in exactly the way your $\ce{CO2}$ molecule did, and this reaction of the $\ce{C=O}$ group (known as a carbonyl functional group) is among the most important patterns in the biochemical reactions occuring in our bodies.


Why does freezing point decrease on adding impurities?

No, but good question. Freezing is a thermodynamic process. Nucleation is a kinetic process.

Kinetic Explanation: At the freezing point of water, water wants to become ice (thermodynamically driven). In order for freezing to initiate and ice to form a temperature dependent minimum number of water molecules must arrange them selves in a crystal or else the particle will dissolve and no freezing will occur.  Impurities or containers with rough surfaces can help arrange and hold molecules which allows freezing to occur with a fewer number of molecules.

Thermodynamic Explanation:  Freezing is a thermodynamic process dictated by $\Delta G_f = 0 = \Delta H_f -T\Delta S_f$.  When water freezes it must dispel the impurities to form an ice crystal (reverse mixing). Since mixing increases entropy, this means that for water to freeze the negative change in entropy ($\Delta S_f$) is larger than that of pure water.  This must be compensated by a lower temperature ($T$) neglecting and differences in ethalphy. The enthalpy ($\Delta H_f$) of mixing will contribute as well to the freezing temperature of water and must be accounted for.  In the case of Sodium chloride though the enthalpy mixing is positive which would favor a higher melting point, but the entropy contribution is large and magnified by the temperature such that entropy dominates and reduces the temperature of freezing.

If you look at the phase diagram of water and salt you will see that the mixing of the salt and water depresses the melting temperature due to the increased change in entropy from mixing water and salt to make brine from ice and salt.


What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state?

Since comments caused certain level of confusion, I guess I'll try to provide a further illustration.
You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$.
Electron doesn't get stuck forever on any of the levels with $n > 1$.

Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms.
In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$.
The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this.

I put together a rough drawing in Inkscape to illustrate all possible transitions*:

I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots).
To determine $N$, you need to sum the states, as Soumik Das rightfully commented:

$$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$

For $n = 6$:

$$N = \frac{6(6-1)}{2} = 15$$

Obviously the same result is obtained by taking the sum directly.

* Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$.


Approximating strength and hardness of alloy

Look at the free resource provided at MatWeb.

Under the Physical Properties tab, you can choose up to 3 of them from a rather extensive list, including hardness. For example, I chose "Carbon" and "Hardness, Rockwell R," and 25 results were returned.

Under the Alloy Composition tab, you can choose up to 3 elements and percentages for your alloy, and see if you get a hit. I chose "Nonferrous Metal" and carbon (minimum 5%, maximum 75%) and silicon (minimum 5%, maximum 75%), and got one hit.

I believe for your purposes, you can assemble a list of materials by hardness and strength (I'm not a metallurgist either, so you might have to do some research on which of the hardness and strength metrics you'll need to choose) and/or by selecting by composition (again, you might need to research which 2- and 3-element combinations make sense and which ones don't).


Why oxygen is enclosed in parenthesis in some chemical reaction?

It represents oxygen from an oxidising agent, as opposed to from diatomic oxygen. It's normally done in square brackets like this:
$$\ce{CH3CH2OH + 2[O] → CH3COOH + H2O}$$
Is the reaction between ethanol and potassium manganate (VII) to form ethanoic acid. The oxygen in the reaction is from the permanganate, not from diatomic oxygen. It kind of simplifies the reaction, so you only see the reaction of what you want -the ethanol and oxygen - instead of the whole reaction with the manganate there. (Manganate is an oxidising agent).

In your reaction the alcohol:
$\ce{RCH2OH}$ is reduced by the oxygen to form an aldehyde and water, using oxygen from an oxidising agent - the [O].


Reason for double acylation of ferrocene only occurring with certain reagent combinations

One factor:  since ferrocene is covalently bonded, the rings "communicate" with each other via the molecular orbitals passing through the iron atom.  Thus the acetyl group placed on one ring not only deactivates that ring, but to some extent deactivates the second ring as well.  The second ring would still be attacked preferentially in the second stage but may require a stronger electrophile, which acetyl chloride is compared with acetic anhydride.


Why is the product of the reaction between iron and steam is iron (II, III) oxide and not iron (II) oxide or iron (III) oxide?

Recall that $\ce{Fe3O4}$ is basically just an equimolar mixture of $\ce{FeO}$ and $\ce{Fe2O3}$. So, in that way, both of the reactions that you mentioned do infact happen, and the result is what was mentioned in your textbook.


Whether or not the ice cube will adhere to something is dependent on the temperature of the ice, the state of the surface of the ice, and the moisture content of the material.

The temperature of an ice cube can range from the freezing point of water ($0\ \mathrm{^\circ C}$) to well below this temperature. Typically most materials have a small degree of moisture regardless of whether they are “wet” or not. If the ice cube is cold enough, it will freeze the water present in the material. The giant covalent network structure of the ice extends into the material, accounting for why it sticks to the ice—the moisture in the material has become part of the ice cube.

If the surface of the ice cube is has melted, it will not be able to freeze the water in the material any more than it will be able to freeze the ice already melted on its surface. No giant covalent network solid will form, and the ice will not stick to the material.


What happens to gasoline when it lands on your hand?

It evaporates, but it will first take some hydrocarbon compounds in your skin

Gasoline is a mixture of a variety of volatile hydrocarbons, some branched chain alkanes and some aromatics. The warmth of your hand is enough to cause small quantities to evaporate. But it will also dissolve some of the hydrocarbon-soluble oils from your skin. Then when it evaporates, they will be left but they will form a microcrystalline powder rather than returning to the parts of the skin they were extracted from: this is the white residue. This is a common when volatile non-polar solvents are spilled on the skin (don't do it too much or your skin will be deprived of the oils that keep your skin flexible and impervious to outside contamination). Small quantities are nothing to worry about, just don't get into the habit or it will hurt your skin.


Effect of nuclear spin statistics on the IR spectrum of CO2

Because the degenerate bending mode has vibrational angular momentum, it is a vibrational $\Pi_\text{u}$ state and just as in electronic $\Pi$ states there exist levels with $+$ and $-$ parity. As a consequence, the degenerate bending mode has $\Pi_\text{u}^\pm$ symmetry and the vibrational wave functions can be both symmetric and antisymmetric.

See also Herzberg Vol 2 (p372 and Fig 99 on p373).


Mysterious precipitate in reaction of copper sulfate and calcium hypochlorite

The creation reaction for copper hypochlorite per Mellor:
$$
CuSO4+Ca(OCl)2 → CaSO4(s)+Cu(OCl)2
$$
However, copper hypochlorite is not stable and on standing breaks down. I believe Mellor is likely accurate on describing the reaction and products:

Reference: "A comprehensive treatise on inorganic and theoretical chemistry", Volume 2, by Joseph William Mellor, page 271. To quote:

"R. Chenevix notes the ready solubility of cupric oxide in chlorine water, and P. Grouvelle found that the soln. obtained by passing chlorine into water with cupric oxide in suspension possessed bleaching properties, and these were retained even after the soln. had been boiled for a quarter of an hour. A. J. Balard found that the distillation of P. Grouvelle's liquor furnished some hypochlorous acid and a green oxychloride, 3CuO.CuCl2.4H20, was formed in the retort. A. J. Balard prepared a soln. of cupric hypochlorite by dissolving cupric hydroxide in hypochlorous acid. It is also made by the action of cupric sulphate on calcium hypochlorite. A. J. Balard found that copper filings are partially dissolved by hypochlorous acid, the soln. after standing some time contains cupric chloride, and deposits a green pulverulent cupric oxychloride."

Link: https://books.google.com/books?pg=PA271&lpg=PA271&dq=zinc+hypochlorite&id=7XoGAQAAIAAJ&ots=R0byHnamKL#v=onepage&q=zinc%20hypochlorite&f=false

So, the cupric oxychloride product maybe mixed with CaSO4.

Note, Copper oxychloride, per Wikipedia, is now referred to as dicopper chloride trihydroxide with the formula Cu2(OH)3Cl (or, more informatively, I would write as Cu(OH)2Cu(OH)Cl).

[I have personally prepared dibasic magnesium hypochlorite on another forum by mixing MgSO4 with NaOCl (aq) and adding NaOH. A white precipitate of dibasic magnesium hypochlorite results (which I have collected and experimented with). It is also NOT Mg(OH)2. Link: https://www.sciencemadness.org/whisper/viewthread.php?tid=26741#pid303723 There is also the reputed compound 'copper carbonate' which actually is also always basic copper carbonate per Wikipedia https://en.wikipedia.org/wiki/Basic_copper_carbonate ].


What type of isomers are cis-but-2-ene and 2-methylpropene?

(2Z)-But-2-ene and 2-methylprop-1-ene are constitutional isomers as they have different connectivities, i.e. the sequence in which the atoms are bonded to each other is not the same. For example, the carbons highlighted here are different: in (2Z)-but-2-ene it is only bonded to two other carbons (through a single bond and a double bond), but in 2-methylprop-1-ene it is bonded to three other carbons:

Cis/trans isomers are a subset of stereoisomers, which need to have the same constitution, or molecular connectivity. The two isomers of but-2-ene satisfy this:

Note how both highlighted carbons are now bonded to the same partners, but with a different spatial arrangement.


what is the physical significance of ∆H=∆U for a chemical reaction?

This equation holds for a reaction involving ideal gases at constant T. Under these conditions, $\Delta (pV) = \Delta n RT$. Assume now that a gas phase reaction occurs at constant pressure and temperature. Then  $\Delta (pV) = p\Delta V$. This is equal to the negative of the pressure-volume work done by the system during the reaction since $w = -p\Delta V$. Therefore, at constant p and T, $\Delta H$ is equal to the change in energy of the system minus the pV work it did. This is consistent with the fact that we equate the enthalpy change for such a process at constant p and T with the heat exchanged, that is, $\Delta H = \Delta U -w = q_p$. Now if $\Delta n =0$ no expansion occurs at constant p and T, and so $w=0$ and $\Delta H = \Delta U = q$.


Rate of catalytic hydrogenation of alkenes

Catalytic hydrogenation is more or less a surface phenomenon so sterics play the deciding role in the rate of hydrogenation.

From Wikipedia:

Here's a newman projection of the three Alkenes.

 A = cis-dimethylbutene, B = trans-dimethylbutene, C = 2-methylpropene

From here it is easy to see that the cis-dimethylbutene will have the least hindered approach towards the surface of the catalyst as both the methyl groups can be made to face away from the surface.

The rates of hydrogenation in the case of trans-dimethylbutene and 2-methylpropene will be decided by which one of them is more strained.

2-methylpropene is more strained than trans-dimethylbutene.

So, we can now establish the order.

$$\text{A > C > B}$$

References

Catalytic Hydrogenation:A Core Technology In Synthesis; Ryoji Noyori
Wikipedia article on Hydrogenation


Internal Energy Definition, Why is dU zero when dT is zero?

The second term in each of your two equations can be expressed solely in terms of the equation of state for the solid, liquid, or gas:$$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
In the case of an ideal gas, the right hand sides of both these equations are equal to zero.


amphoteric

  A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: $\ce{H2SO4}$ is an acid when studied in water, but becomes amphoteric in superacids.

from the IUPAC goldbook.

For ampholytes like $\ce{NaH2PO4}$ the $\ce{pH}$ is independent on the concentration (in first approximation). Consider the reactions of that are possible:
\begin{align}\ce{
H2A-    + H3+O &~<=> H3A     + H2O  \tag1\\
H2A-    + H2O  &~<=> HA^{2-} + H3+O \tag2\\
HA^{2-} + H2O  &~<=> A^{3-}  + H3+O \tag3\\
}\end{align}

As the first approximation, we will assume that everything happening in equation $(3)$ is negligible, as is will only very little influence the concentration of $\ce{HA^{2-}}$. Therefore we can focus on the acidity constants $K_{a_1}$ of $(1)$ and $K_{a_2}$ of $(2)$:
\begin{align}
K_{a_1}  &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})}  \tag4\\
K_{a_2}  &= \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \tag5\\
\end{align}
We will just go ahead and multiply these equations, since they are coupled, simultaneously happening processes. Then we will cancel whatever cancels.
\begin{align}
K_{a_1} \cdot K_{a_2} &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})} \cdot
  \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \\
K_{a_1} \cdot K_{a_2} &= c^2(\ce{H3+O})\cdot  \frac{c(\ce{HA^{2-}})}{c(\ce{H3A})} \tag7\\
\end{align}
Now another major assumption is that the reactions $(1)$ and $(2)$ are happening to the same extent and therefore we can approximate
$$c(\ce{HA^{2-}})=c(\ce{H3A})\tag8$$
and rewrite $(7)$ as
\begin{align}
c^2(\ce{H3+O}) &= K_{a_1} \cdot K_{a_2}\\
c(\ce{H3+O})   &= \sqrt{K_{a_1} \cdot K_{a_2}}\\
\ce{pH}        &= \frac12\left(\mathrm{p}K_{a_1}+\mathrm{p}K_{a_2}\right).
\end{align}

However, this equation only works as a first approximation. Usually the coupled equilibria are very complex and in most instances a computer needs to be involved to calculate it accurately. More information on polyprotic acids.


Is there a convention for breaking chemical formulae across lines?

I remember a couple of books from 1970-80s, where long formulas were split with what would be a $\rm \LaTeX$ \cdot "$\cdot$" notation, I believe. This is an example from Pope, M. T.; Jeannin, Y.; Fournier, M. "Heteropoly and isopoly oxometalates" in Inorganic Chemistry Concepts; Springer: Berlin, 1983; Vol. 8.:

The reason for that, it seems, is that the book was printed in A5 format. Personally, I find an option for mhchem to be able to split formulas across pages very useful.


What do the different grades of chemicals mean?

Sigma-Aldrich gives a very useful table outlining what the different purity levels are and suggested applications. I was less successful at finding equal documentation from some of the other suppliers, but the analysis for the purity of the chemicals they sell is in the catalog and on the bottle.

In general, technical grade or laboratory grade are the lowest purity. ACS Reagent grade means that the chemical conforms to specifications defined by the Committee on Analytical Reagents of the American Chemical Society (but Aldrich "ReagentPlus" means >95% pure). So, "ACS Reagent grade" chemicals should be comparable from different suppliers.  Analytical grade is generally the most pure.

For some of the other grades, such as HPLC grade solvents, the issue is less about overall purity than about being free of substances that would interfere with a particular application. I've reproduced a piece of the table below to give you an idea of the information available.


Calculating partial pressures from mixture of gases

First calculate the total pressure from the ideal gas law, explicitly keeping track of units to avoid confusion and mistakes:

$$\text{Ideal Gas Law:}~~~~{P=\frac{n}{v}*R*T}$$

where:
$\frac{n}{v}=\pu{\frac{1}{35}\frac{mol}{m^3}}$
$R=\pu{8.31\frac{m^3~Pa}{mol~K}}$
$T=\pu{322K}$

Then plug the values in to the ideal gas equation to get the total pressure:

$$P_{total}=\pu{\frac{1}{35}\frac{mol}{m^3}*8.31\frac{m^3~Pa}{mol~K}*322~K}$$

$${P_{total} =\pu{76.5Pa}}$$

Then just multiply the total pressure by the fraction of each gas as stated in the problem:

${P_{\ce{N2}} = \pu{76.5Pa * 0.02 = 1.5 Pa}}$
${P_{\ce{CH4}} = \pu{76.5Pa * 0.79 = 60 Pa}}$
${P_{\ce{C2H6}} = \pu{76.5Pa * 0.19 = 15 Pa}}$


The ion product of water is usually expressed as
$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$
At a temperature of $25\ ^\circ\mathrm C$, its value is approximately $K_\mathrm{w}=10^{-14}$, or $\mathrm pK_\mathrm{w}=14$.

However, the ion product of water is not to be confused with the acid dissociation constant of water.

Generally, the dissociation constant for the simplified reaction
$$\ce{HA <=> A- + H+}$$
is defined as
$$K_\mathrm{a}=\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$$
Thus in case of water
$$K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}$$
The concentration of pure water at a temperature of $25\ ^\circ\mathrm C$ is $c=55.345\ \mathrm{mol\ l^{-1}}$.
Therefore,
$$\begin{align}K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}=\frac{10^{-14}}{55.345}=1.807\times10^{-16}=10^{-15.74}\end{align}$$
or
$$\mathrm pK_\mathrm{a}=15.74$$


Low-cost nontoxic corrosion inhibitor for steel exposed to hydrochloric acid

An extremely wide variety of plant extracts have been shown to serve as corrosion inhibitors, including black tea, chamomile, orange peels, black pepper, mustard seed, neem, the list goes on an on.

I have skimmed almost 100 papers so far and handfuls of review papers, and it is not clear to me that any of these "results" can be trusted. Most of these papers (all?) are from "PhD Factories" in developing countries, and zero of them give actionable insights. However, apparently almost every plant extract any author has ever tested offers protection against corrosion by a factor of roughly 5x in grams-per-liter concentrations.

A plant extract is generally made by taking about 10 grams of dried plant, and heating it in 0.5-1 liter of water (or some other solvent), and removing all but ~20ml/20g of liquid, leaving a very concentrated solution of plant compounds ("plant tea"). Or you can mix the dried material directly with the hydrochloric acid if you can then filter it.

Thus a food-grade hydrochloric-acid / "plant-tea" mixture might be an effective and cheap alternative to extremely expensive commercial products. 1 gallon of muriatic acid is \$10, producing 5 gallons of plant-extract/HCl descaling solution at a cost of \$2/gal. It may have comparable performance to commercial descaling solutions at more than \$50/gal.

However, these extracts are poorly understood and are extremely complicated bio-chemical solutions. My search continues for a single well-understood compound with low-toxicity and good corrosion inhibiting properties for steel in acids like HCl.


Propagating uncertainty with the slope of a tangent line

You have not propagated your errors properly. Note that the sum in your nominator evaluates to a small number with an error of roughly the same size (99.73±0.01 - 99.72±0.01 = 0.01±0.01). Also, error propagation cannot be done by simply adding up the errors.
Instead of calculating the error propagation manually (cumbersome, and prone to errors, hehe ;-) I think it's better to use suitable software.  Both R and Python (both are freely available, commonly used tools) can handle error propagation.
Here's how to do it with R (the code below should be reproducible, assuming you have installed R).
install.packages("errors")
library(errors)
options(errors.notation="plus-minus", errors.digits=1)
y1 <- set_errors(99.726, 0.01)
y2 <- set_errors(99.71753, 0.01)
t1 <- set_errors(0.076, 0)
t2 <- set_errors(0.061, 0)
(y1 - y2) / (t1 - t2)
# 0.6 ± 0.9

The value of the slope is 0.56 kPa/s, but since the error is 0.9 kPa/s, we should report just a single significant digit, so that makes it 0.6 ± 0.9 kPa/s.
And not rounding till the end is not wrong, in fact, it's good practice. As long as you handle your uncertainties correctly, too many significant digits (due to intermediate calculation steps or whatnot) become practically irrelevant, since your value cannot be more precise than the size of your uncertainty anyway.
Good luck!


Can I break starch down into glucose units?

Starch is what plants primarily use as a glucose storage. As such, it is essential that they can break it back down into glucose otherwise it would defeat its own purpose and be removed by evolution very quickly. Think about it: Requiring a lot of energy to build up something that serves no further purpose — not exactly giving a plant an advantage in survival.

There are different methods how to break it down. Plants, of course, use enzymes. Humans can also break down starch with the help of enzymes (which is why potatoes give us energy), much like most other animals, fungi and many single-cell organisms.

The first amylose-breaking enzyme is present in saliva, so it’s that early that the breakdown begins. However, distinct glucose usually isn’t liberated until the gut.

Chemically, the easiest method would be acidic hydrolysis, i.e. use hydrochloric acid and heat. The downside is, that this method might destroy the individual glucose units, too; so using enzymes is preferred if you want glucose.


Formal Charge in Coordinate covalent bonds

The coordinate covalent bonds are still counted as 1 electron, since the pair is shared between both atoms. That means the Carbon will have a formal charge of -1 and Oxygen a formal charge of +1 on Carbon Monoxide.

Carbon FC: $\ce{(4 valence electrons on neutral atom) - (1*2e- from lone pair + 3*1e- from covalent bonds) = -1}$

Oxygen FC: $\ce{(6 valence electrons on neutral atom) - (1*2e- from lone pair + 3*1e- from covalent bonds) = +1}$


In cleaving ethers with both SN1 and SN2 mechanisms possible, which occurs first?

One "side" of a molecule doesn't react by one mechanism and the other "side" by another mechanism. Generally speaking, a molecule reacts by a single mechanism.  Now, the mechanism can change depending upon the other reactants that are present as well as the reaction conditions.  In the case of t-butyl ethyl ether, if the reaction were carried out with $\ce{KI}$ under non-acidic conditions an $\mathrm{S_N2}$ reaction would occur and your answer (iodoethane and tert-butanol) would be correct.  On the other hand, if the reaction were carried out under acidic conditions then an $\mathrm{S_N1}$ reaction would occur.  As you pointed out a tertiary carbocation is more stable than a primary carbocation so the expected products in this case would be ethanol and 2-iodo-2-methylpropane.  Apparently, the reaction conditions in your reaction must have been such that the reaction occurred by the $\mathrm{S_N1}$ mechanism.


Is there a pressure dependent function for freezing points of water and carbon dioxide?

For ice I melting pressure as a function of temperature:

$$\pi = 1- 0.626000 \times 10^6 (1- \theta^{-3}) + 0.197135 \times 10^6 (1- \theta^{21.2})$$

where

$\pi =$ (pressure in MPa)/0.000611657 MPa

$\theta =$ (temperature in kelvins)/273.16K

Source is International Equation of the Pressure along the Melting and along the Sublimation Curve of Ordinary Water substance (no paywall).

For carbon dioxide:

$$\frac{p_m}{p_n} = 1 + a_1 (\frac{T}{T_t}) + a_2 (\frac{T}{T_t})^2$$

Where:

$T$ is temperature in kelvins

$T_t =$ 216.592K

$p_m$ is the melting pressure in MPa

$p_n =$ 0.51795 MPa

$a_1 =$ 1955.5390

$a_2 =$ 2055.4593

Source is A New Equation of State for Carbon Dioxide... (no paywall)

Check original sources before using equations in case I made typo(s).


Can the inverse of a micelle be formed with non-polar solvents?

Yes, they are aptly enough called inverse micelles, eg such as formed by the detergent AOT. While you retain the hydrophobic/philic balance typical of a water-soluble detergent, wherein you have a long hydrophobic tail and hydrophilic headgroup, you generally have to alter the headgroup area (making it smaller relative to the tails) to favor inverted phases of this type. In addition if charged the headgroups will strongly repel, requiring a cosolvent or some counterions to reduce such interactions.


What is the point of using molality when we have molarity?

One of the key attractions of molality is that changing the temperature of a solution does not change the molality, while it may change the molarity. This is because the volume of the solution changes as a result of expansion or contraction of the solvent upon changing the temperature, and thus the molarity changes, since $$M=\frac{n}{V}.$$

The downside, of course, is that the amount of solute required to reach a given molality depends on the nature of the solvent; ambiguity may also arise when a mixed solvent system (e.g. a water-ethanol mixture) is used, as the choice of which compound is the solute changes the molality of the resultant solution.1 Molarity avoids this problem by requiring only the volume of solvent.

As for why it's introduced in school, that's a pedagogical question beyond the scope of this answer, but both molarity and molality have their uses in the laboratory.

1 Consider, for example, a solvent made from 95% water and 5% ethanol. Molality could be determined with respect to only the water, or the combined water-ethanol solution, leading to different calculated concentrations.


If you look at what happens in the equilibrium dissociation of water, both a hydronium ion (Arrhenius acid) and a hydroxide ion (Arrhenius base) are formed. Since these two ions are  formed in equal quantities ($\ce{10^{-7} mol/L}$ each in pure water), the water will be neither acidic nor basic.

$$\ce{H2O_{(l)} + H2O_{(l)} <=> H3O+_{(aq)} + OH^{-}_{(aq)}}$$  (from wikipedia)

Wikipedia can explain more about acids and their definitions if you are interested: http://en.wikipedia.org/wiki/Acid.


Why doesn't HCl form when you dissolve NaCl in water?

If you dissolve NaCl in water you will get some HCl molecules but there's definitely not going to be a significant concentration of HCl formed.

The reaction that you propose -

$\ce{Cl- + H2O -> HCl + HO-}$

is highly thermodynamically unfavorable.

We can ascertain this fact through consultation of any pKa/pKb table. In the equation above, the product acid (HCl) is a much (as in almost a trillion trillion times) stronger acid than water.

Given that HCl is several trillion times stronger than water as a acid, then naturally, HCl will want to protonate hydroxide ion, a byproduct of HCl formation from chloride ion. This is ignoring the fact that hydroxide ion is also a strong base in water, so it has a high proton affinity in water.

So even if the products were formed - again, very unfavorable from a thermodynamic standpoint because the reactant base and reactant acid are both so weak - then the products would certainly react with each other and form the reactants again, resulting in no net change in solution contents and pH.

As a result, the reaction that you propose is more like this (except that the bottom/reverse arrow should be a lot bigger).

$\ce{Cl- + H2O  <<=> HCl + HO-}$


How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?

It is like others have already noted that phenol is very strongly ortho-para directing, while bromine is only slightly deactivating (or sometimes slightly activating). It is so activating that you do not even need a Lewis acidic catalyst or elevated temperature to let the reaction go fast and in good yield. Moreover, if you want the mono substituted product you will actually need to cool the reaction down to < 5 °C and use the rather dangerously flammable solvent $\ce{CS_2}$, yielding 4-bromophenol in 85% yield (source: Vogel). I assume the para product is formed due to the fact that there is no significant difference in electron density between ortho and para carbon atom while ortho is slightly sterically hindered.


Using Arrhenius equation to calculate the ratio of two reaction rates

To a first approximation, the activation energy of a single given reaction is a constant at all temperatures (see reasons here, although more precisely, there is some temperature variation). However, the activation energies for both reactions need not be the same.

If you consider both reactions at $\pu{321 K}$,

$$\begin{align}
\ln k_1 &= \ln k_2 \tag{1}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{(\pu{321 K})R} &= \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{(\pu{321 K})R} \tag{2}\\[4pt]
\ln A_1 - \ln A_2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{(\pu{321 K})R} \tag{3} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{321 K})(\pu{8.31446 J K-1 mol-1})} \tag{3} \\[4pt]
&= -26.56 \tag{4}
\end{align}$$

Now, denote the desired temperature at which $k_1  = 2k_2$ as $T^*$. At this new temperature, we have

$$\begin{align}
\ln k_1 &= \ln 2 + \ln k_2 \tag{5}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{T^*R} &= \ln 2 + \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{T^*R} \tag{6}\\[4pt]
\ln A_1 - \ln A_2 - \ln 2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{T^*R} \tag{7}
\end{align}$$

Here we can reuse the previously calculated value for $\ln A_1 - \ln A_2$, and we can also plug in the same activation energies:

$$\begin{align}
T^* &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{R(\ln A_1 - \ln A_2 - \ln 2)} \tag{8} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{8.31446 J K-1 mol-1})(-26.56 - 0.693)} \tag{9} \\[4pt]
&= \pu{313 K}. \tag{10}
\end{align}$$


How can I build the wyoming like montmorillonite unit cell in avogadro for a molecular dynamics simulation?

I would suggest you try and find the unit cell (.cif file) of the compound in question from the CCDC database, and then use Mercury (a free version is available at the CCDC website) to create the system.
You can of course use Avogadro if that is what you like, but in my experience the coordinate files it produces for super cells are quite messy.


TL;DR Planar structure is definitely possible, and more so in solid state as a statistically averaged geometry in space and time (in gaseous state ring is bent by $35^\circ$). Cyclobutane can be approximately attributed with $\mathrm{sp^4}$/$\mathrm{sp^5}$ hybridization, and the strain caused by non-bonded $\ce{H...H}$ repulsions ("eclipsing") can be reduced to minimum by introducing ring substituents:

substituing $\ce{H}$ with $\ce{OH}$ (octahydroxycyclobutane, planar);
substituing $\ce{C}$ with $\ce{O}$ (oxetane, dihedral angle $10^\circ$).

Single-crystal diffraction studies of octahydroxycyclobutane, an $\ce{OH}$-substituted derivative, in late 1960s showed that unit cell symmetry implies a planar ring; further it's been supported by possible H-bonds formation [1]. A hypothesis that a statistically disordered nonplanar ring may exist has been rejected as it possessed worse agreement with experimental data.

20 years later crystal structure of cyclobutane at low temperature ($\pu{117 K}$) has also been elucidated [2]. From [2] and references therein:

  The gas-phase equilibrium conformation, as derived from electron diffraction and spectroscopic experiments, is nonplanar with a dihedral angle of about $35^\circ$ and an inversion barrier of only $\pu{6 kJ mol^-1}$. The puckering of the molecule is observed in Raman-spectra, and the corresponding absorption band
  is still present in the solid at $\pu{113 K}$. Theoretical calculations using ab initio methods also result in a bent conformation as the energetic minimum.

  As one would expect from the shallow potential of the ring inversion, there are various planar and bent derivatives, and it does not seem possible to predict the ring conformation from the substitution pattern.

In contrast with octahydroxycyclobutane, a statistically disordered model was considered more reliable. Although no particular close $\ce{H-H}$ contacts were observed, an averaged ring is considered planar:

Most importantly, electron density map suggests rather increased $\mathrm{p}$-character in hybridisation (like $\mathrm{sp^5}$ in case of cyclopropane) with higher $\pi$-character of $\ce{C-C}$ bonds in comparison with the one expected for expected for alkanes:

  The residual electron density shows significant maxima in positions between the carbon atoms. These maxima are displaced tangentially by $\pu{0.10(3) A}$ to the outside of the $\ce{C-C}$ bond vectors of the time- and space-averaged molecule, indicating a "banana bond" type electronic arrangement which is frequently found in highly strained small ring systems. In cyclopropane the covalent bonding maxima are displaced by $\pu{0.2 A}$ from the $\ce{C-C}$
  vectors of the carbon triangle. Since in cyclobutane there is reduced ring strain our observation is in good agreement with the cyclopropane findings.

This can be considered as an experimental confirmation of predicted earlier  hybridization of cyclobutane ($\psi(\ce{C-C}) = \mathrm{sp^{4.28}}$, $\psi(\ce{C-H}) = \mathrm{sp^{2.19}}$) [3]:

  The difference between the bond overlaps of the planar and non-planar models is very small. [...] This clearly indicates that other interactions (even if small), must be dominant in causing non-planarity of the molecule. Non-bonded repulsions make a dominant contribution in determining conformations in acyclic and larger cyclic systems (e. g. ethane, cyclohexane), and they favour staggered arrangements of $\ce{CH}$-bonds. In a planar model of cyclobutane all hydrogen atoms are in eclipsed positions, this is a cause of additional strain and is energetically less favourable. A relieve is obtained by bending the skeleton of the molecule, since the $\ce{CH}$ bonds are then approaching a staggered conformation.

Cyclobutane, just as cyclopropane, is a strained cycle with the bent "banana" bonds with intermediate character between $\sigma$ and $\pi$, which allows for mimicking the behavior of double-bond compounds, such as decolorization of bromine water [4, p. 195], even though protons in cyclobutane are less acidic  than in cyclopropane:

  Four-membered rings also exhibit angle strain, but much less than three-membered rings, and for that reason are less easily opened. Cyclobutane is more resistant than cyclopropane to bromination, and although it can be hydrogenated to butane, more strenuous conditions are required.

Bibliography

Bock, C. M. J. Am. Chem. Soc. 1968, 90 (11), 2748–2751. DOI 10.1021/ja01013a002.
Stein, A.; Lehmann, C. W.; Luger, P. J. Am. Chem. Soc. 1992, 114 (20), 7684–7687. DOI 10.1021/ja00046a012.
Maksić, Z.; Klasinc, L.; Randić, M. Theoretical Chemistry Accounts: Theory, Computation, and Modeling (Theoretica Chimica Acta) 1966, 4 (3), 273–286.
Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013.


What is the reason for the acidic character in phenols?

You are confusing the two terms electronegativity/electropositivity and higher/lower electron density.

The source you read states that the $\ce{sp^2}$ hybridised carbon in phenol is more electronegative than the $\ce{sp^3}$ hybridised one in (say) ethanol. This means, that this carbon has a greater desire to draw electrons towards it.

Therefore, even though any carbon is still more electropositive than oxygen, the phenolic oxygen will be able to draw the electrons towards it less effectively and therefore has a lower electron density than in ethanol.

But the oxygen atom in itself hasn’t changed; specifically, it is still hybridised the same way (assuming that one can speak of hybridisation for oxygens). Thus, oxygen’s electronegativity is still the same.

The combination of still having the same electronegativity but having less electron density means that any other electrons will be drawn towards it more strongly — specifically, this concerns the electrons in the $\ce{O-H}$ bond.

In a nutshell: The oxygen’s electronegativity didn’t change (it sounds weird to talk of electropositivity for oxygen if there is no fluorine around) it only has less electron density on one side due to more pulling. It combats that by pulling more from the other side.

This can be called an inductive effect.


Where does the thermal energy come from in a chemical change when two substances react with one another?

The energy was in the chemicals, in the form of bonds. Energy is liberated when there is bond formation. As the reaction between $\ce{Cu}$ and $\ce{HNO3}$ is feasible at room temperature so it happened.

And another thing, room temperature is a nice temperature for that reaction because first we need to break bonds of $\ce{HNO3}$ to make new bond, which is easily done at this temperature.

Now coming to your question, the answer is 'no'. The first law is still safe. You know it wrong. The truth is (from wikipedia) : "the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed".

Give special attention to the word 'total energy', you said the 'thermal energy' in your question. And the total energy (atleast in this case) is conserved in the form of summation of thermal energy and bond energy.


You seem to be using a very simplified version of the index of hydrogen deficiency. A more complete expression is:

$$IHD = \frac{2C + 2 - H - X + N}{2}$$

Where $C$, $H$, $X$, and $N$ represent the number of carbon, hydrogen, halogen, and nitrogen atoms, respectively. The logic of the IHD is that any simple acyclic alkane has a formula of $\ce{C_{n}H_{2n + 2}}$, meaning that given $n$ carbons, a simple acyclic alkane will have $2n + 2$ hydrogens. Every increment in the IHD represents either a degree of unsaturation due to $\pi$-bonds, or the presence of a ring. In your formula, $H_{reference}$ should be twice the number of carbons in your molecule, plus two. $H_{molecule}$, on the other hand, is the actual number of hydrogens in the molecule.

For example, if your molecule has the formula $\ce{C6H8}$, $H_{reference} = 2(6) + 2 = 14$ and $H_{molecule} = 8$. The IHD is then $\frac{14 - 8}{2} = 3$, suggesting that the total number of $\pi$-bonds plus rings in the molecule is three. Note that this isn't sufficient to determine the actual structure of the molecule, but it's helpful when combined with spectroscopy data (that is, it could have one ring and two $\pi$-bonds, or just three $\pi$-bonds, or some other combination).


Internal Cannizzaro reaction of phenylglyoxal

I agree with the other answers that in general intramolecular reactions are faster than intermolecular reactions. Let's look at why this intramolecular reaction might be fast.

We can describe the Cannizzaro reaction as the delivery of a hydride from an anionic hydrate to a carbonyl group. The reaction shown certainly meets those criteria. I see this reaction as a carbocation rearrangement or a (semi-)pinacol rearrangement. In the latter class of reactions, a group undergoes a 1,2-migration to quench a carbocation and generate a carbonyl.

In the phenylglyoxyal rearrangement, if the ketone is considered as its ionic resonance structure (2), then the ketone carbon is a carbocation. We know that hydrogens can and will undergo 1,2-migration if the resulting carbocation is more stable. In this case (3->4), the resulting carbocation is very stable, because it's really a carboxylic acid (5). Furthermore, the carboxylic acid will be deprotonated rapidly under basic conditions (5->6).


Why is a TTT diagram called isothermal transformation diagram?

They are two names for the same procedure.  A steel is heated to a temperature in the austenite range then quenched/quickly cooled to some specific temperature. Held at that constant ( isothermal) temperature ( T) for a specific time (T) then quenched to room temperature for examination of the transformation (T) products.  You may be thinking of CCT ; continuous cooling diagrams ; where a steel sample is austenitized  then continuously cooled at some specific rate.  Then examined for transformation products. The results can have similarities depending on alloy content , etc.


Why is nitric acid a stronger oxidising agent than sulfuric acid?

The nitrate ion is a strong oxidant whereas the sulfate ion is a milder oxidant that only shows oxidizing properties in high concentrations. The oxidizing atom in question is the oxygen atom in the acid, not the hydrogen ion. This is because hydrogen ion is not very electrophilic, which is why non-oxidizing acids cannot dissolve copper: even copper has a higher affinity to electrons than hydrogen.

In a nitrate ion, the molecular structure is inherently not very stable, as the nitrogen atom has to bear a positive charge to maintain the molecular structure. Besides, nitrogen is in itself a very electronegative element, so that makes the nitrate ion even more unstable, hence its oxygen atom readily reacts with other atoms or molecules that oxygen can oxidize, including chloride ions:
$\ce{Cl- + O -> OCl-}$
$\ce{H+ + OCl- <=> HOCl}$
$\ce{H+ + Cl- <=> HCl}$
$\ce{HCl + HOCl <=> H2O + Cl2 }$


Your questions:

Q1) How the rearrangement from (G) to non-1-yne can happen (i.e., its mechanism)?

For example, see the following mechanism suggested in Ref.1 as the Acetylene Zipper reaction:

Q2) Why Sodium Amide in particular yields terminal alkyne, unlike the fused Potassium hydroxides why almost always yield internal alkyne?

To answer your question, I have to dig down literature to the 19th Century. To my knowledge, sodium amide ($\ce{NaNH2}$) is not the best base to use in this purposes. The best base, in particular, is the monopotassium salt of 1,3-diamminopropane or simply, KAPA (see mechanism), a difunctional “superbase.” According to the Ref.1:

  KAPA produces exceptionally rapid migrations of triple bonds from the interior to the terminus of the carbon chain in seconds at $0^{\circ}$.

On the other hands, from a synthetic standpoint, $\ce{KOH}$ as a dehydrohalogenating agent, tends to promote the shift of the triple bond away from the end of a chain, which is sometimes considered as a disadvantage (Ref.2).

Q3) Is there any limitation by how much this alkyne formation site can shift?

There are no limitations. Yet, if one substitution of disubstituted acetylene is branched, the triple bond migrate through the straight chain to the end (Ref.2). Mean time, the reaction has some interesting facts:

An equilibrium between alkynes and allenes can be established using basic catalysts (see the mechanism). For example, when very strong bases such as sodium metal ($\ce{Na}$) or sodium amide ($\ce{NaNH2}$) are used, the rearrangement of a triple bond of a disubstituted acetylene tends to migrate to a terminal position. This isomerization equilibrium shifts toward forward direction in favor of 1-alkyne formation due to the fact that formation of the terminal metal acetylide salt, which is stable, precipitated out from the solution, thus automatically removing itself from the reaction mixture.
Favorskiî was the first to discover that 2-alkyne could be irreversibly removed from solution by isomerization to the 1-alkyne and resultant precipitation of the sodium salt in 1887 (Ref.3). This procedure used metal $\ce{Na}$ as the strong base, and thus, has the disadvantage that some olefin would produce  as a result from hydrogenation of the alkyne by the hydrogen liberated upon acetylide formation. On the other hand, $\ce{NaNH2}$ when used in inert solvents, was shown to accomplished the same rearrangement but avoids the reduction problem (Ref.3,4). Temperatures greater than $\pu{100 ^{\circ}C}$ are employed with $\ce{Na}$ metal while the best results with a $\ce{NaNH2}$ catalyst are obtained at $\pu{150-160 ^{\circ}C}$. The data are summarized in the Table (vide infra). The rates of the rearrangement differ with the structures of the compounds isomerized. In general, 2-alkynes rearrange faster than 3-alkynes, and 1,2-alkadienes (allenes) rearrange faster than 2-alkynes.  The amount of 1-alkyne formed is proportional to the reaction time, and equilibrium is obtained earlier at higher reaction temperatures (Ref.3).
Later studies have found that $\ce{NaNH2}$ brings about the propargylic rearrangement in liquid ammonia at room temperature with essentially quantitative hydrocarbon recovery. (Ref.3). The ratio of 1-alkyne, 1,2-alkadiene, and 2-alkyne varied with the amount of $\ce{NaNH2}$ used. High $\ce{NaNH2}$ concentrations favored the formation of 1-alkynes. Equivalent or excess amount of $\ce{KNH2}$ in liquid ammonia, sometimes in admixture with HMPA, have also been shown to convert 2-alkynes to 1-alkynes (Ref.3,5).
The advent of newer strong bases has made the isomerization to terminal acetylenes a very facile reaction. The addition of dialkylacetylenes to a slightly excess of KAPA (the monopotassium salt of 1,3-diamminopropane) results in extremely rapid, quantitative isomerization of the triple bond to the terminal position where it is trapped as the acetylide ion (Ref.1).

Table:
$$
\begin{array}{ccc}
\hline
\text{Disubstituted alkyne} & \text{Base} & \text{Temperature} & \text{1-Alkyne} & \text{Disubstituted alkyne} & \text{Ref} \\ \hline
\ce{C2H4C#CCH3} & \ce{Na} & \pu{100 ^{\circ}C} &  \pu{4\! -5 h} &  n\text{-}\ce{C3H7C#CH} & 3\\
n\text{-}\ce{C3H7C#CCH3} & \ce{Na} & \pu{100 ^{\circ}C} &- &  n\text{-}\ce{C4H9C#CH} & 3\\
n\text{-}\ce{C13H27C#CCH3} & \ce{Na} & \pu{200\! -220 ^{\circ}C} &  \pu{15\! -20 h} &  n\text{-}\ce{C14H29C#CH} & 3\\
n\text{-}\ce{C4H9C#CCH3} & \ce{NaNH2} & \pu{160 ^{\circ}C} &  \pu{12 h} &  n\text{-}\ce{C5H11C#CH} & 3,4\\
n\text{-}\ce{C5H11C#CC5H11} & \ce{NaNH2} & \pu{210 ^{\circ}C} &  \pu{7 h} &  n\text{-}\ce{C10H21C#CH} & 3,5\\
n\text{-}\ce{C4H9C#CC2H5} & \ce{NaNH2} & \pu{170 ^{\circ}C} &  \pu{9 h} &  n\text{-}\ce{C6H13C#CH} & 2\\
n\text{-}\ce{C6H13C#CC6H13} & \ce{K(NH2C3H6NH)} & \pu{20 ^{\circ}C} &  \pu{1 min} &  n\text{-}\ce{C12H25C#CH} & 1\\ \hline
\end{array}
$$
Entry 4: The catalyst used was finely divided $\ce{NaNH2}$ suspended in mineral oil.

References:

C. A. Brown, A. Yamashita, “Saline hydrides and superbases in organic reactions. IX. Acetylene zipper. Exceptionally facile contrathermodynamic multipositional isomeriazation of alkynes with potassium 3-aminopropylamide,” J. Am. Chem. Soc. 1975, 97(4), 891-892 (DOI: 10.1021/ja00837a034).
T. L. Jacobs, “The Synthesis of Acetylenes,” Organic Reactions 1949, 5, Chapter 1 (78 pages) (https://doi.org/10.1002/0471264180.or005.01).
Hermon Pines, Wayne M. Stalick, In Base-Catalyzed Reactions of Hydrocarbons and Related Compounds; Academic Press, Inc.: New York, NY, 1977, “Chapter 3: Isomerization of acetylenes and allenes,” pp. 124–204.
H. H. Guest, “Rearrangements of the Triple Bond,” J. Am. Chem. Soc. 1928, 50(6), 1744–1746 (DOI: 10.1021/ja01393a036).
L. Brandsma, In Preparative Acetylenic Chemistry (Studies in Organic Chemistry Series: Vol. 34); 2nd Ed.; Elsevier Science: New York, NY, 1988, “Chapter XI. Base-Promoted Interconversions of Acetylenes,” pp. 231–246.
T. H. Vaughn, “A New Reaction of 1-Iodoacetylenes and Some New Mercury Acetylides,” J. Am. Chem. Soc. 1933, 55(8), 3453–3458 (DOI: 10.1021/ja01335a073).


What legal guidelines exist for completeness of a Materials Safety Data Sheet?

OSHA's FAQ says that chemicals that make up less than 1% (0.1% if it is a carcinogen) of a product do not need to be listed on the MSDS. So it looks like 3M could be in compliance if the oderants are present in trace amounts (in that case, I have no idea why they feel the need to include the thyme oil).

The only relevant thing OSHA seems to say about consumer products specifically is that employers are not required to provide an MSDS for consumer products used in the workplace, as long as the consumer product is used in the same manner as a consumer would.

The Consumer Product Safety Commission specifically disclaims jurisdiction over chemical products, and the Chemical Safety Board doesn't appear to have anything related to consumer chemical products.


The splitting described doesn't actually happen in bulk water. There is just the autoionization reaction:

$$\ce{H2O + H2O <=> H3O+ + OH-}$$

The net transfer of a proton from one water to another creates two ions. This is somewhat disfavored. Hence, the autoionization constant is quite small:

$$\mathrm{K}_{w} = \ce{[H3O+][OH-]} = 10^{-14}$$

Neutralization is a somewhat related process, whereby one species is able to create an excess of hydronium, and another is able to create an excess of hydroxide. This creates an imbalance in the autoionization equilibrium, and it rapidly shifts left to consume excesses of both ions.


Kinetics of converting a water + heavy water mixture to HDO

[Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast.

If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.

$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$

With one deuterated water, it leads to formation of the desired product:

$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$

If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.

This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.

So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.

The analogous mechanism would work for $\ce{H3O+}$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on temperature and pH. However, even at neutral pH the rate is fast.

  Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Here is a paper where they used computational chemistry to look at this question: http://omh.umeche.maine.edu/pdfs/JChemPhys_135_124505.01pdf.pdf. I would not consider this "easy", though.


What are "O-hydrocarbons"? A pharma translator's question (Russian to English, "О-углеводороды")

My fellow translator sent a letter to the authors, and they replied that they used this term to stand for "oxygen-substituted hydrocarbons", just like andselisk suggested.


What does the subscript n on methylene groups in a cyclic alkene mean?

Yes, n denotes the number of $\ce{CH2}$ groups between the parentheses.

Example:

n = 1, 2 or 4 means cyclopentane, cyclohexane, or cyclooctane.


Is this the correct relative Brønsted acidity in these four acids?

Fluorine being the most electronegative element in the periodic table, it doesn’t like bearing a positive charge. Thus $\ce{H2F+}$, which has the positive charge on the fluorine nucleus, is not very stable and willing to expel the extra proton. So, it is indeed a strong acid.

$\ce{H2F+}$ is formed, e.g., by autoionization of liquid $\ce{HF}$, but only because that reaction also involves creating a $\ce{F-}$ ion (associated with another molecule of $\ce{HF}$ to form $\ce{HF2-}$), which is very favorable:

$$\ce{3 HF <=> H2F+ + HF2-}$$


When is Raoult's Law valid for solid-liquid solution?

It is valid for both volatile and nonvolatile solutes, as it refers to the partial vapour pressure, not total vapour pressure.
Saying that, note that  it applies only on mixtures with  ideal behaviour, as there are many more or less significant  positive and negative deviations from the law. These deviations relate to preference of intermolecular bonding to the same, or to the other molecules.
Typical examples are ethanol water mixture forming an azeotrop of the minimal boiling point,  or hydrochloric acid forming an azeotrop with the maximal boiling point.


No indicator is suitable when titrating weak acid with weak alkali

An indicator works when the solution's pH is in its range. It's not necessarily the equivalence point.  The point is called the end point where it actually changes color.

If the difference in volume is small between the equivalence and end points, then the indicator can be used.  Since a gradual change implies more volume must be added to obtain the same pH change in other acid base titration, the difference between equivalence point and end point(if you use indicator) will be more and therefore indicator will not be suitable for it.


Identification of aqueous solutions of sodium hydroxide, sodium acetate and potassium thiocyanate using nickel(II)

I know that you would expect a precipitate to form when $\ce{Ni^2+}$ ions are
exposed to $\ce{OH−}$ ions

Yes, you are right, they will indeed form a green precipitate, nickel(II) hydroxide:
$$\ce{NiCl2 + 2NaOH -> Ni(OH)2 + 2NaCl}$$
Nickel(II) hydroxide does not dissolve in excess $\ce{NaOH}$.

I know that you would not expect a precipitate to form when $\ce{Ni^2+}$ ions
are exposed to $\ce{CH3COO−}$ ions

I haven't found any reaction like that to form nickel(II) acetate as it is generally formed from the reaction of nickel(II) carbonate and acetic acid but a little bit of searching gave me a site which shows the formation of nickel(II) acetate from various nickel(II) salts and acetic acid and various salts of acetic acid. The closest reactions I could find from that site is the reaction between acetic acid/lead acetate and nickel(II) chloride and reaction between sodium acetate and nickel(II) nitrate. Technically, the reaction can be written and balanced on pen-paper as it is a simple double displacement reaction but the reaction hasn't been performed practically. Also, I don't know the credibility of the site, so take that site as a reference in your discretion.

I have done some research and concluded that $\ce{NiCl2 + KSCN}$ will not form
a precipitate

Theoretically, it can be formed from a double displacement reaction, but practically it is found to be unfeasible as the end products get contaminated with unreacted reactants and byproducts to the point where yield is somewhat negligible. From a paper1:

Compounds containing nickel(II) thiocyanate are prepared mostly by metathesis between nickel(II) chloride or nitrate with alkali metal
thiocyanates. However, in such syntheses if precautions are not
taken, the final product may be contaminated with starting
materials. Methods known for preparing this compound include:
treatment of a dilute thiocyanic acid solution with nickel hydroxide
or carbonate, followed by evaporating the mixture at $\pu{150 ^\circ C}$; mixing and boiling aqueous solutions of ammonium thiocyanate and
barium hydroxide, followed by addition of an aqueous nickel sulphate.

Conclusion
A: $\ce{NaOH}$
B: $\ce{CH3COONa}$, as it is somewhat a weak base, pH = 8.9
Reference

Sarma, K.P., Poddar, R.K. A convenient method of preparing nickel(II) thiocyanate and its use in synthesis. Transition Met Chem 9, 135–138 (1984). DOI: 10.1007/BF00935928 (PDF)