Why does chemisorption increase with rise in temperature, even though it is an exothermic process?

To answer that first think about a reaction like combustion reaction. Even though combustion reactions are exothermic you need to provide heat to the system by increasing its temperature. See, every reaction or chemical change happening requires a certain minimum amount energy called the "activation energy" for the reaction to start. This energy is required to break the bonds of the reactants so that they can react to form product. That's why we say that at low temperatures, kinetics controls the reaction rather than thermodynamics. Now, once you are at a temperature which is able to provide the sufficient activation energy to the reactants, the thermodynamics comes into play which essentially says in this case that if a reaction is exothermic, increasing the temperature will shift the reaction in backward direction.

Now coming to chemisorption, in this case too you need to increase the temperature to provide sufficient activation energy so that the reaction can start. So increasing the temperature increases the rate of reaction. After a certain temperature when thermodynamics become more important, the reaction shifts backward.

Hope it helps!


The enthalpy change of the reaction can be calculated using the enthalpies of formation of the compounds in the reaction.  To find the temperature change of the reaction, divide the quantity of heat released by the heat capacity of the solution in which the reaction occurred.

This won't give the exact answer.  Some of the heat will be absorbed by the surroundings including the flask.  The reaction produces gas and some of the energy released by the reaction is used up by the work needed for the gas to be released into the atmosphere.

Rough estimate:

Let's say the teacher used a 500 ml bottle of peroxide.  The solution contains about 0.5 mol of peroxide.  The enthalpy of formation of peroxide is about −190 kJ/mol.  The enthalpy of formation of water is about −240 kJ/mol and that of oxygen is 0.  So, 0.5 mol of reaction gives about 25 kJ of energy released by the reaction.  We'll assume all of it went into the 500 ml of water.  The 500 ml of water has a heat capacity of around 2100 J/K.  So, the solution gets about 12 K hotter.


Yup, it's $[\ce{Kr}]4d^{10}$.  Post-transition ($p$-block) elements rarely if ever ionize or even covalently bond their $d$ subshell electrons.  Although we use the bracketed core designation only for noble gas cores, by the time we get to Group 12 or 13 the $d$ electrons are also effectively part of the core.


With molecule (2), you have two possible cations from addition of Br+; one is tertiary, one is secondary. Tertiary is much more stable so you will get very little product resulting from water reacting the secondary cation = high selectivity observed.

With molecule (1), Br+ addition gives 2 possible secondary cations; nothing much to choose between them = poor selectivity, ~1:1 mixture observed.


Is it possible to disable the splash screen of ChemDraw?

I assume the question seeks out for a way to use ChemDraw more quickly, for example while being confronted with a larger number of chemical structures already generated, and to be processed further.  Then

If it is to browse through a number of structures aiming "to find the right file", to gain an overview, then Thumbfish (Windows only) may be helpful.  Chemdoodle (Win/Mac/Linux) and MarvinSketch offer in their file-open menu a preview as well.
On the other hand, provided the extensive documentation especially of ChemDraw seconds native, the *.cdxml format, the (batchwise) conversion into other formats is easily done with OpenBabel (supported file formats) and Apple Script ChemBot (application here).


Why is it hard to reduce titanium dioxide?

Reasons why Titanium is hard to extract and purify:

The +4 oxidation state of $Ti$ is the most stable (due to its noble gas configuration) and so reducing $TiO_2$ to $Ti$ is quite difficult.
$Ti$ has chemical properties which are similar to its neighbouring elements (Vanadium, Zirconium, Chromium) and the presence of compounds containing these elements in its ores makes purification even more difficult. For example: Vanadium and Zirconium also form volatile halides ($ZrCl_4$)and oxyhalides ($VOCl_3$) which have to be separated from $TiCl_4$ by fractional distillation in Kroll's Process.

I have tried to explain why the chlorine route is taken using only thermodynamics (I'm not absolutely sure about this but these reasons are what I could come up with).

Consider these reactions which could take place in the fluidized bed reactor:

$\ce{TiO_2(s) + 2Cl_2(g) + 2C(s) -> TiCl_4(g) + 2CO(g)}$
$\ce{Ti(s) + C(s) -> TiC(s)}$

Comparing the Gibbs' Energies of these reactions can tell us which one is spontaneous since if the change in Gibbs' Energy is negative, the reaction will be spontaneous.
And the equation for change in Gibbs' Energy is:

$\Delta{G} = \Delta{H} - T\Delta{S}$

At normal temperatures, reaction 2 would occur more readily than reaction 1 since $TiC$ is extremely stable (reflected by its very high melting and boiling points) while $TiCl_4$ is unstable (it hydrolyses immediately on contact with air or oxygen to form $TiO_2$ and $HCl$). And so it is almost impossible to reduce Ti using C at low temperatures.

At higher temperatures (Kroll's Process is carried out at around $1000^oC$), $T\Delta{S}>>\Delta{H}$ so we can neglect $\Delta{H}$.

For the first reaction, since all the products are gaseous (while some of the reactants are solid), $\Delta{S}$ is positive. Therefore $\Delta{G}$ will be negative and the reaction is spontaneous and will occur easily.

For the second reaction, since the number of molecules is reduced, $\Delta{S}$ is negative. Therefore $\Delta{G}$ will be positive and the reaction becomes unfavourable.

Notes:

For the states of the reactants and products, I compared the melting and boiling points of the reactants and products with $1000^oC$ which is the temperature at which Kroll's process is carried out.
Actually, from what I read, there are many possible products for the first reaction. depending on the reaction conditions, a mixture containing $CO$, $CO_2$ and $COCl_3$ can be formed. But all these products are still gaseous so it doesn't affect the explanation.


What is the mechanism for the hydrolysis of the boron-alkoxide complex in NaBH4 reduction?

I don’t know whether the mechanism of hydrolysis of the boronic acid esters has been studied. However, the dissociative mechanism immediately suggests itself if only by comparison to similar structures.

Since the workup is typically written $\ce{H+/H2O}$, you may consider some buffered source of external protons. These can protonate any of the alcoholates which can then dissociate. (The $\ce{B=O}$ double bond I drew in should be considered no more than additional stabilisation from oxygen lone pairs; boron is electron-deficient by definition.) Another water molecule can associate: the first bond is cleaved. Repeat three times and end up with four molecules of alcohol and tetrahydroxidoborate $\ce{[B(OH)4]-}$.


Because an oxygen ion is $\ce{O^2-}$ and a chlorine ion $\ce{Cl^-}$, you know that the salts have to be of the form $\ce{M_xO_y}$ and $\ce{M_xCl_{2y}}$.

Let $o$ be the atomic mass of oxygen, $c$ of chlorine and $m$ of the unknown metal M. The problem states that

$$0.6(xm+yo)=xm$$

which implies

$$0.6yo = 0.4xm \implies m = \frac{3y}{2x}o$$

If you recall some formulas for salts, you know that $x$ can only be 1 or 2 (e.g. $\ce{M_5O_3}$ makes no sense). This means that $m$ is a multiple of $\frac34o$ which is (approximately) 12. So we only need to check the elements with an atomic weight which is a multiple of 12:

12: Carbon (C) isn't a metal
24: Magnesium (Mg) is, and you can check the oxide/chloride satisfy the equations
36: n/a
48: Titanium (Ti) checks out
60: n/a
72: n/a
84: Krypton (Kr) isn't a metal

and so on. Even if the atomic weight corresponds to a metal, you can verify the value of $y$ doesn't match with the oxide that's formed.


Why H2S2 is hydrogen disulfide and not hydrogen persulfide?

TL;DR: "Hydrogen persulfide" is technically a very crude mixture of various hydrogen polysulfides ($\ce{H2S_n ; n\geq2}$). It has long sulfur chains and is formulated as $\ce{H-(S)_{n}-H}$. To avoid confusion, each members of the mixture is named according to number of sulfur atoms. So, $\ce{H2S2}$ is hydrogen disulfide.
Long answer:
It is prepared by heating sodium sulfide ($\ce{Na2S.9H2O}$) for three hours at 100° C. The crystalline sulfide melts and the sulfur is gradually dissolved to a deep-red solution of which the solute has various kinds of sodium polysulfides. This solution is mixed with rather less than its own bulk of water and is then introduced as a thin stream into a mixture of dilute hydrochloric acid and ice which dissolves the alkali and gives a crude oily mixture which is referred to as "hydrogen persulfide".
Individual constituent of the crude mixture can also be separated. Treatment of hydrochloric acid to the distillate amounting to approximately one-sixth of the original bulk yields a solution which is called hydrogen trisulfide, $\ce{H2S3}$. A more volatile liquid, hydrogen disulfide, $\ce{H2S2}$, can be condensed in a second receiver cooled by a mixture of ether and solid carbon dioxide.
The crude persulfide can also be formed if a large proportion of sulfur or sulphur dioxide either as gas or in solution is reduced by hypophosphorous acid.
The so-called hydrogen persulfide, are yellow liquids at r.t. They are unreactive towards acids but are very sensitive towards alkalis causing vigorous decomposition. It is stable only for few hours if stored in vessel. Even at r.t., they readily decompose into hydrogen sulfide and sulfur. The compounds burn with a blue flame, giving water and sulfur dioxide. The addition of alcohols induces rapid decomposition producing sulfur. Ketones, nitrobenzene, aniline and pyridine also catalyze the decomposition. It is miscible in benzene, toluene, chloroform, bromoform, carbon disulfide, ether and heptane, giving relatively stable solutions. As such, it finds its use in vulcanization.
Source: http://sulphur.atomistry.com/hydrogen_polysulphides.html


Does sodium cyanide react with water, and can it cause an explosion?

Usually, I dislike conspiracy theories.1 However, this one (if it even is one) is true. Looking at the list presented on Wikipedia of the chemicals that were stored in that warehouse, why would you even suspect sodium cyanide as being a cause of the explosion? It is a stable ionic compound that can be put on the shelf, wet with water, dissolved in water, and won’t react with air with any measurable half-life. (Except that you possibly shouldn’t put it on a shelf because it is slightly volatile and highly poisonous.)

What was also in the warehouse and is an explosive was ammonium nitrate. In the 1921 explosion in Oppau (now part of Ludwigshafen), BASF workers tried to use dynamite to loosen a plaster of ammonium nitrate (it is hygroscopic and baked together). Mixtures containing at least $40~\%$ ammonium sulphate were considered stable and the mixture in question was a $50~\%$ mixture of the two according to the production data. After approximately 20,000 uses of dynamite where nothing happened, the entire 4,500 tonne lot exploded killing some 500 or 600 people and leaving thousands injured. The theory $40~\% = \text{safe}$ was later disproven in careful examinations. Apparantly, the explosion could be heard in Munich, some $300~\mathrm{km}$ away.

Another substance stored there and dangerous is calcium carbide $\ce{CaC2}$. When calcium carbide gets wet, it produces acetylene gas (and calcium hydroxide). The acetylene is flammable and can also explode in certain acetylene–oxygen mixtures.2 Luckily, no acetylene or carbide explosions are known enough to make it onto Wikipedia.

So far more likely than sodium cyanide having any primary effect would be one the following theory:

Fire (something must have ignited because the compounds I read there aren’t self-flammable).
Attempts to extinguish fire with water by the fire brigade.
Water reacts with calcium carbide liberating acetylene.
Acetylene burns.
The critical heat is reached, ammonium nitrate explodes.
Toxic sodium cyanide is liberated into the environment causing further severe health risks for man and animal alike.

1: Okay, that was a lie. I love listening to them and making fun of them.

2: A teacher at school talked about his time at uni and mentioned a professor that let a balloon of acetylene explode — he made a fuse long enough to ignite it from outside of the lecture hall (nobody was allowed inside), had the doors closed and my teacher said ‘it was the loudest bang I ever heard.’


Is a metal hydroxide fully or partly oxidised in insufficient base?

What you are worried seems to be in your comment, which I'm explaining[maybe not]

Oxidations:

  $$\ce{NiOOH + H2O + e- -> Ni(OH)2 + OH-} \quad E^{\circ}=+0.49\\
\ce{Ni2O3(s) + 6H+ + 2e- -> 2Ni^2+ + 3H2O} \quad E^{\circ}=+1.753\\$$

I suppose if you know about electrode potential you might be able to understand now,If you wish I might explain it, if you can't on your own.


Bridging Hydride Covalent Model Electron Count - Hartwig

A bridging hydride is both an X- and an L-type ligand. Since hydrogen has a valency of 1, the hydride itself can only be an X-type ligand for one of the metal centers. Then we treat the M-H σ bond as an L-type ligand to the other metal center. In the ionic method, this is 4 electrons. In the covalent method, this is 3 electrons.
EDIT:
It was noticed that in the table you provided, the example is actually for a bridging halide, not a hydride. This case is similar, but not exactly like the case of hydride. The main difference is that the halide has lone pairs, so those act as the L-type ligand for the second metal center instead of the M-X σ bond.


Calculate the root-mean-squared speed of free electrons in the Sun's corona given pressure, temperature and mass of an electron

I'll go out on a limb here and say your answer is right and theirs is wrong.

Note that the answers differ by a factor of the square root of 3, so I'm guessing they forgot to include the factor of 3.

There would be a small correction for relativistic effects, but not enough to explain the differences in the two answers.


Defintion of Solubility in Different cases

There are two ways of dealing with this.

Solubility product

The more general way is to state the solubility product. Taking your first example, this would have the same value no matter how much acetate is present in solution before adding the silver ions. The disadvantage is that for highly concentrated solution (in fact, in the presence of ions, even for moderate concentrations) this gets inaccurate unless you switch from concentrations to activities.

Solubility (of a solid)

The other way is to specify the solubility of $\ce{AgCl}$ in pure water, in 0.1 M sodium acetate, in 0.5 M sodium acetate and so on. In this case, because silver chloride is much less soluble than silver acetate, no silver acetate would precipitate (the silver ion concentration will be sufficiently low in solution so that silver acetate will not be exceeding the solubility product).

Similarly for AgCN, you could specify its solubility in pure water, in aqueous solution buffered at pH 7, or buffered at any other pH. For the latter example, you should probably also specify ionic strength or the entire makeup of the solution prior to adding AgCN. In this case, there will be a strong dependence of the solubility on the pH because HCN is soluble and its formation will decrease the concentration of cyanide ions while the concentration of silver ions will remain the same up to the point where it precipitates.

In both of these cases, there is no ambiguity in terms of concentration because there are no sources of the ions forming the precipitate other than what was added as a (stoichiometric) solid.

On the other hand, if you add NaCl to a solution already containing LiCl, the apparent solubility of NaCl will be lower than in pure water because of the common ion effect. You could still talk about the solubility of NaCl in 2 M LiCl, and the quantity would refer to the concentration of sodium and chloride ions due to the addition of NaCl, not due to ions already present before addition.

Concentration of ions vs. concentration of added solid

With any salt that does not have 1:1 stoichiometry, we give the concentration of the salt, not that of the ions. For example, we talk of a 1 M $\ce{MgCl2}$ solution even though it is a strong electrolyte containing 1 M magnesium ions and 2 M chloride ions. The concentration given is based on the formula of the solid added, not on the concentration of ions after dissolution. In a similar way, we would say we added acetic acid to a concentration of 1 M even though we know full well that some of it will be ionized.


A block of Osmium is quite immune to atmospheric attack, due to the low surface area of the osmium, but when heated it can form Osmium tetroxide. I wouldn't like to risk wearing an osmium ring just incase Osmium tetroxide is formed. The oxide has a pungent smell so if you do wear a ring of Osmium, at least you'll have some form of warning to take it off. So really, I wouldn't recommend it, but do it at your own risk.


How different concentration of HCl on magnesium affect the enthalpy change

In fact all three reactions produced the same amount of heat energy and hence had the same enthalpy change. By changing the concentration of the HCl only effects the reaction rate and how much energy is produced in a certain time. However eventually, each reaction will produce the same amount of energy provided that all the Mg reacts which does happen in your experiment as HCl is in excess. The reason for this can be explained quick easily using the formula: $$\Delta H=\Delta U+pV$$ Then using the ideal gas equation, we can substitute $pV$ with $\Delta n(g)RT$. Note that the only variable that actually changes is n(g) which is the change in the number of moles of gas molecules. So the equation becomes: $$\Delta H= \Delta U + \Delta  n(g)RT$$ So the only thing that effects the enthalpy change of the reaction in the change in internal energy (U) and the number of moles of gas molecules formed which are both constant for all three reactions.


Do molecules with bridges through rings exist?

I'm not sure about the existence of molecules with bridges through rings. However, there are several publications of synthesis of molecules mimicking wheels and axles ([2]rotaxanes; The “[2]” refers to the number of interlocked components) as one shown below (Ref. 1):

(The diagram is from Reference 1)

This specific molecule (8; an “impossible” [2]rotaxane) represents a macro-cycle with a straight-chain molecule with bulky end groups going through its center. The inclusion of two bulky end groups prevents the straight-chain molecule leaving the macro-cycle (mechanically interlocked) as depicted in the diagram (See Ref. 2 for the total synthesis of the molecule).

Note that Ref. 1 also cited articles for the synthesis of [2]catenanes, which contain two interlocked rings (instead of one axle and one macrocycle). Keep in mind that there are some advanced catenanes and rotaxanes that exist (e.g., [3]catenanes and [3]rotaxanes).

(The structures are from Reference 1)

References:

Edward A. Neal, Stephen M. Goldup, "Chemical consequences of mechanical bonding in catenanes and rotaxanes: isomerism, modification, catalysis and molecular machines for synthesis,"  Chem. Commun. 2014, 50(40), 5128-5142 (https://doi.org/10.1039/C3CC47842D).
Jeffrey S. Hannam, Stephen M. Lacy, David A. Leigh, Carlos G. Saiz, Alexandra M. Z. Slawin, Sheila G. Stitchell, "Controlled Submolecular Translational Motion in Synthesis: A Mechanically Interlocking Auxiliary,"  Angew. Chem., Intl. Fd. 2004, 43(25), 3260-3264 (https://doi.org/10.1002/anie.200353606).


Is these a pair of diastereomer or identical compound?

Maybe the answer of the book is wrong? For the fisher projection, it is Diastereomer definitely.


Gattermann-Koch reaction - Role of cuprous chloride

According to this site:

  The formylation of aromatic compound can be accomplished by treatment with $\ce{CO}$ and $\ce{HCl}$ in the presence of Lewis acid. The reaction is performed either under pressure or using copper(I) chloride or anhydrous aluminum chloride whose role may be to aid the reaction between $\ce{CO}$ and $\ce{HCl}$ via the complex which it forms with $\ce{CO}$.

As you can see, it does not involve in the mechanism (provide $\ce{Cl-}$ via dissociation of $\ce{CuCl}$) but act as a promoter to speed up the reaction.


Accurate method to calculate the pH of a salt from a weak acid and weak base

Ignoring activities, activity coefficients, ionic strength effects, and so on, I get a pH of around 9.3. Given the 1 M concentration, I would not bet a lot of money on this, though. My solution, assuming a more typical solution concentration, is given in the two figures below. Sorry these are figures rather than proper formatting: I have been away for 3 months and am beyond rusty at this point. If the mods wish to delete this answer, they should feel totally free to do so.

There are obviously other ways to do this problem and what I have posted is simply an approximation: any pH problem solution, of this type, that does not explicitly involve the formal concentration of the salt, i.e., "F", always gives me pause. But maybe this helps a bit.


Selectivity of diazo-coupling with p-amino phenol

I think the right way to approach this question isn't "we don't know the product, let's try to reason it out from first principles"; rather it should be more like "we found that this is the product when we ran the reaction in the lab, how can we rationalise this observation using a model of nucleophilicity that we know?"

So, in that previous comment which you quote, I did not mean to say "phenol will be deprotonated, hence this product will definitely be formed"; rather I meant something more like "if this is indeed the correct product, then a possible reason is because the phenol is deprotonated, making it (or the positions ortho to it) more nucleophilic than the aniline".

That said, I did just look it up, and there are indeed some references from the primary literature which support the claim that the product above is formed.

So, since we have experimental evidence (and insofar as this evidence is sound), you should not doubt what the product is; the only question left is why it is formed. And so I offer you the same explanation: that phenol is deprotonated under the basic conditions, and that phenoxide is more nucleophilic than aniline (at the ortho positions).

As for why phenoxide is more nucleophilic than aniline: I could offer you some wishy-washy explanations, like how a negative charge makes the oxygen atom more electron-donating towards the ring (and that outweighs the natural nucleophilicity ranking of N > O), but I don't think there is a simple one-line answer which is sufficiently rigorous for me to accept. I tried to look for some other evidence, but it is very difficult to compare C-reactivity of phenoxide vs aniline, since most EAS reactions take place in acidic media. There are some studies which conclusively show that phenoxide-O is more nucleophilic than aniline-N, but that isn't what we're looking for either.

So... all in all, it is better to use this experimental observation (that the azo coupling occurs ortho to oxygen) as proof that phenoxide-C is more nucleophilic than aniline-C, rather than to try to do it the other way round, i.e. "phenoxide-C is more nucleophilic than aniline-C, hence it must be true that this reaction will form this product". After all, chemistry is first and foremost an experimental science.

It's a subtle difference, but adopting the correct scientific approach will be better for you as a chemist in the future.


What is the name of "cis/trans" isomerism when all four groups are different?

When all four substituents are different cis-trans type of isomerism cannot be applied.  Moreover the Cis trans system is often ambiguous because configurational descriptions have not been defined according to any general and clear set of rules. For these types,  there is system called E-Z system of nomenclature and is based on the sequence rules of Cahn, Ingold  and Prelog (CIP) rule, originally developed for naming optical isomers on R-S system


What is the actual voltage to the standard hydrogen electrode and why is it used?

The voltage given for any reduction is totally meaningless in the absence of either a comparison to another reduction potential or an oxidation potential. Absolute energies are totally meaningless. Only differences matter.

As usual then, we need references. For example, heat of formation of standard Gibbs energy if formation are set to 0 for elements in their standard forms at standard state.

Likewise, for reduction potentials, we need a reference, and that reference if the standard hydrogen electrode, whose standard reduction potential is 0 by fiat. All other potentials can be determined by comparison.


chemical equilibrium and exothermic reaction

Since, H2 is added at the equilibrium state, it's initial pressure will be more and starts decreasing as it goes into the reaction with N2. This is clearly depicted by graph A. While the pressure of N2 starts decreasing from the level it was in equilibrium state, represented by graph C. Also, the reaction shifts forwards in order to gain equilibrium, pressure of NH3 will increase. So, the correct answer would be,

  A = H2, B = NH3, C = N2


Dangers of Colbalt Chloride Paper Burning

Not a good story to share ...

It is not surprising that a rapid heating of a Petri dish provided by a Bunsen burner cracks the former.  By application of the heat, the glass wants to expand, to dilate; hence you swirl test tubes over a Bunsen, provided your lab still has access to natural gas.
A more gentle heating of the glass, to dry the paper impregnated by $\ce{CoCl2}$ were gradual heating on a heating plate.  By this you equally would obtain the colourless, waterfree form, in a more safe manner (no open flames, no ignition).
Yes, $\ce{CoCl2}$ is considered toxic, and dangerous for the environment.  It is suspected carcinogen. You set yourself unnecessarily in danger, not only because of the flames, but that the dry salt, too: finely dispersed in air, it is then easily inhaled.  You won't die on the spot, however, do not repeat this. And ChemSE is not a subsitute for a physician.
Depending where you did this, disposal of heavy metal, and aqueous solutions of heavy metals (like $\ce{CoCl2}$) is regulated by law, and hence the rinse with water was probably fine to extinguish the fire, but the waste water generated is not normal household waste.  Especially if it were on a larger scale.
For future reference: look up safety informations of chemicals you intend to work with, chemical suppliers include them in MSDS sheets. In chemistry lab classes, students are requested to get familiar with these information prior to performing the experiments.


The self-accepted answer given here is wrong.

Each line in a bond block of a MOL file represents one bond between two atoms.

A line in a bond block has the following compact format:

111222tttsssxxxrrrccc

With other words, each entry (=different number or letter) in such a line can take up to three spaces. If it needs less. e.g. when the atom number is smaller than 100, the data is right-aligned.

111 is the number of the first atom

222 is the number of the second atom

These atom numbers refer to the position of the atom in the atom block in the same file, where the cartesian coordinates and the atom type (element) are given.

ttt is the number decoding the bond type: 1=single, 2=double, 3=triple, etc.

sss describes the stereochemistry, 1 is a single bond pointing up, 6 is a single bond pointing down, etc.

The latter are used in the flat 2D MOL file of morphine cited in the question, e.g. for the only hydrogen atom in the file: a bridgehead H-atom.


for a polydisperse polymer, why is its weight average molecular weight always bigger than its number average molecular weight?

Mathematics has many examples of identities and concepts which are extremely powerful due to their simplicity, such as the pigeonhole principle and the triangle inequality. One such identity will guide us through the problem.

We start with the very innocent identity, that for a real number $z$, the following is true:

$$z^2 \geq 0$$

Now let $z = x-y$. We have:
$$(x-y)^2 \geq 0$$
$$x^2 -2xy + y^2 \geq 0$$
$$x^2 + y^2 \geq 2xy $$
$$\frac{x^2 + y^2}{2} \geq xy $$

Now this last inequality has a very powerful interpretation. To show it more clearly, we can replace $x^2$ with $a$ and $y^2$ with $b$, yielding:

$$\frac{a + b}{2} \geq \sqrt{ab} $$

In words, the arithmetic mean of two positive real numbers will always be greater than or equal to their geometric mean. This is called the AM-GM inequality, and it often shows up in the most surprising of places.

Let's get started. We define the number average molecular weight ($M_{n}$) and the weight average molecular weight ($M_{w}$) of a polydisperse polymer mixture as the following, respectively:

$$M_{n}=\frac  {\sum\limits_{i} N_{i}M_{i}}{\sum\limits_{i} N_{i}} \quad , \quad M_{w}=\frac  {\sum\limits_{i} N_{i} M_{i}^2 }{\sum\limits_{i} N_{i}M_{i}}$$

We then define the dispersity index $Đ$ (or $PDI$) as:

$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2}$$

To make things easier, let us start by analyzing the first non-trivial case ($i=2$):

$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2)(N_{1} + N_{2}) }{(N_{1} M_{1} + N_{2} M_{2})^2}$$

We multiply the bracketed terms and expand the binomial to obtain:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{1}N_{2}M_{1}^2 + N_{1}N_{2}M_{2}^2}{N_{1}^2 M_{1}^2 + 2N_{1}N_{2}M_{1}M_{2} + N_{2}^2 M_{2}^2}$$

Rearranging the terms slightly:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}$$

Notice that the terms in the numerator and denominator match except for the coloured parts.

Here is where the magic happens. From the AM-GM inequality, we must have:

$$x^2 + y^2 \geq 2xy \quad \Rightarrow \quad M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2}$$

We then replace the term $(M_{1}^2 + M_{2}^2)$ in the numerator with the term $2M_{1}M_{2}$. By doing so, the new numerator is smaller than or equal to the old numerator. Thus:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}} \geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}=1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$

What about the next non-trivial case ($i=3$)? We have:

$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2 + N_{3} M_{3}^2)(N_{1} + N_{2} + N_{3}) }{(N_{1} M_{1} + N_{2} M_{2} + N_{3} M_{3})^2}$$

Multiplying everything out and performing the appropriate arrangements yields:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}$$

This time, we can use the AM-GM inequality thrice:

$$M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2} \quad , \quad M_{1}^2 + M_{3}^2 \geq 2M_{1}M_{3} \quad , \quad M_{2}^2 + M_{3}^2 \geq 2M_{2}M_{3}$$

The end result is the same:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}} $$
$$\geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}} = 1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$

Now stating the general case:

$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2} = \frac{\sum\limits_{i} N_{i}^2 M_{i}^2  + \color{#ff0000}{\sum\limits_{i,j\ (j>i)} N_{i}N_{j}(M_{i}^2 + M_{j}^2)}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}} \geq \frac{\sum\limits_{i} N_{i}^2 M_{i}^2  + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}=1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$


When we talk about Sn1 or Sn2 reactions (and any other reaction to be honest), not only the reactants are crucial to describe what's going on but also the solvent you used. It is perfectly possible that the second reaction will be favored in the direction shown, as long as you use the right solvent for this.

There are two types of solvents one can use in an Sn2 reaction: polar protic (which has a reasonably acidic hydrogen in its structure) and polar aprotic (has a net dipole moment but no acidic hydrogens).

Examples of polar protic solvents: Water, methanol, ethanol, ammonia etc

Examples of polar aprotic solvents: DMSO, acetonitrile, acetone etc

So what does the solvent has to do with the reaction?

Nucleophiles are negatively charged species or dipoles, they will attack a sufficiently positive center (usually a carbon atom connected to an electron withdrawing group)
The solvent is commonly present in significant excess compared to the
reactants.

When you use a polar protic solvent, such as methanol or water, the nucleophile species (ions such as Cl- or molecules such as ethanol) will be surrounded by an overwhelming number of solvent molecules that will interact strongly with them by dipole-dipole interactions or hydrogen bonding. This creates a barrier around the nucleophile, making it considerably more difficult to approach the positive center and consequently slows the reaction.

Smaller nucleophiles (such as F-) will be surrounded by a bigger number of solvent molecules than bigger nucleophiles (such as I-), so this surrounding effect is stronger in smaller species.

If you consider only the nucleophile, Br- is stronger than I- since it has a smaller ionic radius and, therefore, has a more 'concentrated' electron cloud (less polarizable). If no solvent effect was considered, the reaction as shown in your textbook would be incorrect since the equilibrium would be towards the formation of methyl bromide.

HOWEVER, if you use a polar protic solvent, like methanol, the order of nucleophilicity is inverted (because of that surrounding barrier of solvent molecules) and I- becomes a better nucleophile than Br-.

So, to be consistent and correct, your textbook should also provide the solvent in which this reaction is done. If it is a polar protic one then it is correct.

(I took these pictures from master organic chemistry, they have a good text talking about it here: https://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/ )


Spectrometry in Chemistry & Flame Test Lab

... two colors that are two similar (dark yellow and orange) in a flame test ...

Do you know what the two compounds might be?

   The yellow colour typically indicates sodium, the orange flame colour calcium.

In order to distinguish the two compounds, you have several options:

Use a different method

   You can try to precipitate the calcium ions in the form of an insoluble salt, such as its oxalate.

Refine your flame test

   View the flames though a piece of cobalt glass! This filters of the yellow flame colour typical for sodium. Note that the flame colours for the other metals will look different now!


What is the effect on humans of using dry ice and liquid nitrogen as food additives?

Nitrogen has a boiling point of 77K (-196 °C), CO2 melts at 195K (-78 °C). You don't cool down the ice cream or whatever else you want to eat to those temperatures, that would be very unsafe. Both will evaporate or sublimate quickly when they are in contact with warmer material, so there won't be much left once you eat e.g. the ice cream.

Air is composed of 78% N2 and 0.04% CO2, nitrogen gas can only be dangerous when it replaces all the oxygen in the air. Carbon dioxide is toxic at around 7-10% concentration. Both conditions are not realistically achievable by the methods you describe.

The biggest danger I see would be ingesting a piece of dry ice shortly after it was put in a drink, which would be cold enough to hurt.


What happens to the carbon in carbon steel when it rusts?

Cast iron has considerable free carbon as graphite, but carbon steel has iron carbide in cementite, $\ce{Fe3C}$ in a ferrite ($\ce{α-F}$e) matrix, along with other phases and additives. In a corrosive environment, the ferrite and cementite form a galvanic couple, causing increased corrosion of the $\ce{α-F}$e while effectively protecting the $\ce{Fe3C}$. According to Acta Metallurgica Sina,, "So, the ferrite phase diminishes gradually, and the accumulated cementite multiplies on the surface of the specimens along with the development of corrosion."

That said, what now happens to the remaining $\ce{Fe3C}$? Wikipedia states, "While cementite is thermodynamically unstable, eventually being converted to austenite [$\ce{γ-Fe}$] and graphite at higher temperatures, it does not decompose on heating at temperatures below... 723 °C..."

So, at room temperature, the remaining carbon would be specks of cementite amid the rust.


Shifting of groups in carbocation rearrangement

In general, migratory aptitude of different groups follow following order:
$\ce{-Ph} > \ce{-C(CH3)3} > \ce{-CH(CH3)2} > \ce{-CH2CH3} > \ce{-CH3} > \ce{-H}$

For example, 2,3-diphenylbutane-2,3-diol (1) was allowed to react under homogeneous (25 mol% of p-toluenesulfonic acid) or heterogenious (20 mol% of graphene oxide) catalytic conditions using toluene as a solvent at $\pu{100 ^\circ C}$. Under both conditions, only pinacol rearrangement with phenyl migration was observed affording 3,3-diphenylbutan-2-one (2) in a 90 and 95% conversion. Methyl migration has not been observed [Reference].

Reference:

Pinacol Rearrangement and Direct Nucleophilic Substitution of Allylic Alcohols Promoted by Graphene Oxide and Graphene Oxide $\ce{CO2H}$: M. Gomez-Martinez, A. Baeza, D. A. Alonso, ChemCatChem, 2017, 9, 1032–1039 (https://doi.org/10.1002/cctc.201601362)


What happens when prop-2-en-1-ol reacts with concentrated hydrogen bromide

Rather than trying to rationalise what the question wants to know, I ran a quick SciFinder search. One of the first results was the following reaction, reported for the perdeuterated allyl alcohol by Thiem, Mohn and Heesing in 1985:[1]

Scheme 1: Reaction of perdeuterated allyl alcohol with $\ce{HBr}$. All implicit hydrogens in this scheme are deuterium atoms.[1]

Thiem, Mohn and Heesing do not go into details regarding the mechanism, but as I commented it really doesn’t matter whether the reaction follows a $\mathrm{S_N1}$, $\mathrm{S_N2}$ or $\mathrm{S_N2'}$ mechanism, since the final product will always be the same. In any case, the first step will be protonation of the hydroxy group to turn it into a better leaving group.

Another one of the first results was the following, reported by Karki and Magolan in 2015:[2]

Scheme 2: Reaction of normal allylic alcohol with $48~\%\ \ce{HBr}$ in DMSO/chloroform.[2]

Their proposed mechanism includes the in situ creation of $\ce{Me2S^+-Br}$ which will then react with the double bond giving a bromonium ion; this will then be captured by a second bromide to give 2,3-dibromopropan-1-ol.

I was unable to find single-step reactions of allyl alcohol with hydrogen bromide leading to either 2-bromopropan-1-ol (which would imply a single Markovnikov addition of $\ce{HBr}$ onto the double bond), 1,2-dibromopropane (which would imply Markovnikov hydrobromination of the double bond of allyl bromide) or 1,3-dibromoprop-1-ene (where I have no clue how exactly that would be generated from allyl alcohol in the first place). Funnily enough, you state that this last product, C according to your scheme be the correct answer. I challenge the exam.

References:

[1]: J. Thiem, H. Mohn, A. Heesing, Synthesis 1985, 775. DOI: 10.1055/s-1985-31344.

[2]: M. Karki, J. Magolan, J. Org. Chem. 2015, 80, 3701. DOI: 10.1021/acs.joc.5b00211.


Concerning the first part of your problem: In the general case, which you requested, we start with the mass action law:

$$
K_\text{b} = \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]} = \frac{[\ce{OH-}]^2}{[\ce{B}]}
$$

That’s essentially what you just reformulated to obtain your expression for $\mathrm{pH}$ above. At this point, you made an error in the sign of the second term on the right-hand side – it should be added, not subtracted, i.e. $\mathrm{pH} = 14 + \frac{1}{2} \lg \left( [\ce{B}]K_\text{b} \right)$.

Now comes the important thing: The mass action law uses the concentrations of our components in the equilibrium! What is given as 0.1 M is, however, the starting concentration $[\ce{B}]_0$ before equilibrium, which will of course decrease; $[\ce{HB+}]$ and $[\ce{OH-}]$ will increase by the same amount, which we will denote $x$. Thus, we rewrite:

$$
K_\text{b} = \frac{[\ce{OH-}]^2}{[\ce{B}]} = \frac{x^2}{[\ce{B}]_0 - x}
$$

Now, you just solve for $x$, which is the concentration $[\ce{OH-}]$.

One more thing. You will see that $x$ is actually pretty small in this case, so $[\ce{B}]_0 - x \approx [\ce{B}]_0$. By using $K_\text{b} = \frac{x^2}{[\ce{B}]_0}$, we essentially obtain the same result. You can’t always rely on this, however, and if you have stronger acids or bases, this will get important.


Peanuts contain peanut oil. When you crush the peanut you also compress some of the peanut oil out out of the nut particles.


I fear that without additional information it will be difficult to provide you with a definitive answer. The $x$ usually represents the mole fraction and in the context of a 2-component alloy this seems plausible. The $C$s might indeed be specific heat constants, i.e. $C_{3}$ would be the specific heat of the alloy and $C_{1}$ and $C_{2}$ would be the specific heats of the components making up the alloy, whereby component 1 would have a mole fraction of $x$  and component 2 would have a mole fraction of $(1-x)$. For a mixture of ideal gases such a formula would be correct but I doubt that it will work well with alloys in the real world. But if this came up in the context of a school lesson then the approximation might be ok.


Is there a free, super simple app or website for IUPAC names from structures?

A bit more intuitive to interact with may be Chemdoodle Webcomponents which is a JavaScript library set-up to integrate well within webpages and mobile devices like phone or tablet, too.  It comprises 2D/3D sketcher, a NMR predictor, and is freely available (open source GPL license).  Derived from Chemdoodle, there is a structure -> name utility speking multiple dialects (e.g., IUPAC / CA, Hantzsch nomenclature), too.

Example:

The interaction with Chemdoodle is nicely explained by Pritt Balagopal here on ChemSE.


How many litres of water will be sufficient to generate 1600 psi when heated?

First lets' state the Universal Gas Constant: R = (0.08206 x L x atm) / (mol x K)

And the Ideal Gas Law: PV = nRT

Now, lets find a handy calculator to do everything for us: http://www.ajdesigner.com/idealgas/ideal_gas_law_mole_equation.php

Now we can rearrange the gas law to solve for moles: n = PV/RT... and do some conversions... 320 °C = 593.15 K, and 1600 psig = 108.87 atmospheres (or did you mean psia?)... or, just plug the numbers into the calculator.

So, if the pressure is 1600 psi at 320 °C in a 20 L vessel you need 44.735912618737 moles of gas. Now, we multiply by the molar mass of water (18.01528 g/mol) to obtain more familiar values... it appears that you will need 805.93 grams of water (more or less). And let's use the density of water at room temperature to find the volume...  805.93g / 998.23 g/L = 0.80736 liters.

☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺

My advice, for what it's worth, is- if you are actually going to do this, you should probably use more water, and let the steam boil for a few minutes to evacuate the air from the vessel before locking it shut. Oh, and be sure to wear eye protection while you run for cover.

☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺

Edit based on comment:

Question: ...for how long would the pressure be mainted if [water vapor] is exiting also? I mean, I will start release pressure as soon as it hits 1600PSI, but I'll keep on heating the water too. So, for how long will I be able to maintain that 1600PSI exit force?

2nd Answer: The loss of pressure would be instantaneous... concurrent with the loss of water. To maintain the pressure you would need to increase the temperature as the water exits. The time it would last, as it empties and as the temperature increases, depends on the rate/amount of water being released and/or the rate of temperature increase. If you want to maintain 1600 psi, then you either need more water or more heat; if you use more water, then you could maintain the 320 °C because you would no longer have 20 L of "head-space" in the vessel. The amount of vapor in the head-space of the vessel would be less than 805.93g. Not that it matters, but to calulate the amount of vapor, you would subtract the liquid volume from the total vessel volume because water is practically  incompressible (maybe around 1% at 320 °C). The way to maintain 1600 psi would be to fill the tank (maybe 2/3 full), and use a pressure relief valve (like all pressure cookers do). A pressure relief valve would automatically control the pressure regardless of the change in temperature and volume (as long as it's sufficient or in excess- to supply the minimum amount of pressure to the valve).

Supposing that you had an infinite amount of heat on tap, the pressure would (theoretically) be maintained until the last molecule of water escaped the vessel. Now, let's forget that steel melts at about 1370 °C and water actually breaks down at around 2500 °C (says google- not sure if that chages because pressure would be 1600 psi) and steel burns at about 4000 °C; releasing one molecule of water per hour, the pressure would be maintained until you depleated 44.736 moles (so, multiplying by Avagadro's number: 6.022×10^23), you will have 2.694×10^25 hours.

To find the necessary temperature for 1 molecule of H2O to supply 1600 psi in 20L (the final temperature at hour 2.694×10^25), convert 1 molecule to moles and rearrange the gas law to solve for T.

n = 6.022×10^-23 : T = PV/Rn

T(final) = (108.87atm x20L)/(0.08206 x6.022×10^-23mol) = (about) 4.406×10^26 K. To convert K to °C, subtract 273.15 from 440,600,000,000,000,000,000,000,000 K.


In Parent chain choosing should one consider longest chain or chain with more substituents?

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

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

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

  (…)

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

  (…)

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

  The senior ring, ring system, or principal chain:

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

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

  (…)

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

  (…)

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

  (…)

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

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

  (…)

You have already found the principal characteristic group, i.e. the carboxylic acid group $\ce{{}-(C)OOH}$ that is expressed as suffix ‘oic acid’.

According to Rule P-44.1.1, the principal characteristic group (suffix) has to be considered first in order to decide the parent chain for this compound. Only the second rule (P-44.3.2) is looking at the longest chain.

Therefore, the name of the first example that is given in the question is 2-ethylbutanoic acid rather than ‘3-carboxypentane’ since the principal characteristic group has to be considered before the longest chain.

Note that, according to Rule P-44.1.1, the parent chain has the maximum number of substituents corresponding to the principal characteristic group (suffix). Also note that the second example that is given in the question contains two carboxylic acid groups.

Therefore, the name of the second example is 2-ethylbutanedioic acid rather than ‘3-carboxypentanoic acid’ since the maximum possible number of substituents corresponding to the principal characteristic group has to be considered before the longest chain.


Lennard-Jones potential and vibrational energy level diagram explanation

This diagram is conceptually horrible, because it is a mix of quantum and pseudo-classical physics.

Forget for a while the vibrations. If you calculate single point energies (within Born-Oppenheimer approximation) for different internuclear values, you'll get something like blue curve.

Now, for the $r_e$ distance between nucleus (again, within Born-Oppenheimer approximation) you can calculate the "vibrational (normally harmonic) frequency" in a pseudo-classic fashion. For this frequency, you can assign energies (of a quantum harmonic oscillator having this frequency). This energy values correspond to the difference between energy values for the horizontal blue lines, and the well (blue).

The widths of the horizontal lines have not physical meaning.

For the sake of completeness, the $D_e$ would be the energy of equilibrium state if no vibrations are considered. $D_0$ when the fundamental vibrational state energy is add. The former energy is not measurable because this is not a real energy for the system.


Every can of soda, coke etc. is supersaturated with carbon dioxide after opening, i.e. relieving the pressure in the can.

The surface tension of a gas bubble adds to the outside pressure, and more pressure always increases solubility of gases, as you know from Le Chatelier.

Same with gas supersaturated with a condensable gas or solvent with too much solute, you have to overcome the surface tension of small particles or droplets before precipitation can take place.


How to understand the (non-)degeneracy of states of singlet Σ and Δ terms for a ππ configuration of two nonequivalent electrons?

The setup here is a diatomic molecule with two $\pi$ electrons in different subshells. A typical scenario would be one in a $\pi$ orbital and the other in a higher energy $\pi^*$ orbital.

In a diatomic molecule aligned with its molecular axis on the z-axis, the $\pi$ subshell consists of two orbitals, one with a node in the x-z plane and one with a node in the y-z plane. The $\pi^*$ subshell likewise consists of two orbitals, which are the antibonding counterparts to the two $\pi$ orbitals.

In the cited text, "+1" and "-1" indicate the two orientations, so it might be simpler to replace them with x and y. Likewise, the "$\pi'$" can be changed to $\pi^*$ for clarity.

Now consider the first $^1\Sigma$ term:
$$^1\Sigma^+: \pi_{+1}(1)\pi'_{-1}(2) + \pi_{+1}(2)\pi'_{-1}(1)+\pi_{-1}(1)\pi'_{+1}(2)+\pi_{-1}(2)\pi'_{+1}(1)$$

We can rewrite this in the friendlier notation and use only every other term, since the others are just exchange within the same orbitals. (What follows here is conceptual, not mathematically rigorous.):
$$^1\Sigma^+: \pi_x\pi^*_y + \pi_y\pi^*_x$$

The second $^1\Sigma$ term likewise simplifies to:
$$^1\Sigma^-: \pi_x\pi^*_y - \pi_y\pi^*_x$$

Now we further separate the contributions from each axis:
$$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y $$
$$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y $$

We see that what we have done is to "undo" the formation of the $\pi$ and $\pi^*$ orbitals and essentially restored the original atomic $p$ orbitals:

$$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y=p_{Ax} + p_{Ay} $$
$$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y=p_{Bx} + p_{By}$$

where A and B are the two atoms in the molecule. If A and B are different elements, these terms will have different energy.

If you do the same exercise with the $^1\Delta$ terms, you'll get something like:
$$^1\Delta^+: p_{Ax} + p_{By}$$
$$^1\Delta^-: p_{Bx} + p_{Ay}$$

Since both terms have equal contributions from A and B, they will have the same energies whether or not A and B are the same element. And since the orbitals on the two axes are orthogonal to each other, the change in sign does not affect the energy.

Conceptually, the sigma terms are diradicals where either both electrons are on A or both electrons are on B in orthogonal p orbitals. The delta terms are diradicals in which one electron is on each atom.


How many valence electrons do elements in the d block have?

The simplest answer is: d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell". This works if you are using the definition of valence shell to be the outermost shell.

But it is wrong to apply the concept to determine the valency of the element, because that varies with the compound it is present in. For example, iron can show two valencies, $2$ and $3$ in the compunds iron(II) sulphate ($\ce{FeSO4}$) and iron(III) sulphate ($\ce{Fe2(SO4)3}$) respectively. But from the above definition, iron has $8$ valence electrons.

As a matter of fact, this applies even to many elements outside d-block too, for example chlorine typically shows a valency of ($-1$) in compunds like $\ce{NaCl}$, but it can also show a different valency (more appropriately, oxidation state) in ions like perchlorate anion ($\ce{ClO4^-}$), where its valency is $\mathbf 7$ (while the oxidation state is $\mathbf{+7}$, notice the difference that there is no plus or minus sign in valency).


How do biochemists seperate nucleotides from DNA to add to the PCR

You just buy the nucleoside triphosphates from your favorite supplier, they're commercially available.

As far as I know the NTPs you can buy are still produced biologically and not synthesized (see this example from Sigma Aldrich, which states a microbial source). So you grow a large amount of bacteria, lyse the cells to get at the contents and then digest the DNA or RNA with an appropriate enzyme that cuts it into single nucleotides. To get the triphosphate from that you have to phosphorylate the monophosphates again with the appropriate kinase.


Which chemical reactions don't involve bond breaking?

The reaction of a Lewis acid with a Lewis base results only in bond formation. Your question gets at the essence of chemistry, though. A chemical reaction is the rearrangement of electrons between atoms. So generally, most reactions will involve breaking bonds. Even for monomers (e.g. alkenes) to polymerize, $\pi$-bonds within the monomers have to be broken to form $\sigma$-bonds in the polymer. You seldom have one without the other.


Chemists need terminologies. NO terminology is perfect and you will always find exceptions. In general, in an n-component mixture, the component which has the highest concentration is the solvent.*

Therefore your point no. 3 is invalid because pouring does not determine the solute-solvent label. Only point 2 is relevant. Point 1 is irrelevant.

*Corollaries & Exceptions:

If we mix 50 mL of hexane and 50 mL of heptane, which one is the solvent and which one is your solute? It is up to you to label one as solvent and solute. Both components were liquids.

If I take 100 gram of water, I can dissolve about 140 grams of potassium iodide! The mixture remains a liquid. Now which one is the solute and which one is the solvent? Traditionally, you may call the liquid component as the solvent and the solid phase as the solute, because here water molecules are helping to solvate potassium and iodide ions.

The same goes with sugar in water.

So how to overcome these paradoxical labeling issues?

Physical chemists and analytical chemists prefer to use the term mole fraction x when they want to extra precise to what they are saying about mixtures. It is defined as "Amount of a constituent divided by the total amount of all constituents in the mixture. It is also called mole fraction. Amount fraction is equal to the number fraction: the number of entities of one constituent divided by the total number of entities in the mixture."

Just as nobody cares about "physical" vs. "chemical" changes taught in school, no one will chastise you for identifying one component as solvent and the other component as solute in complicated cases.


Synthesis of 2-Phenylcyclopentanone from cyclopentanol

Step 1 dehydrate cyclopentanol to cyclopentene with conc Sulfuric acid

Step 2 epoxidise cyclopentene with m- Chloroperbenzoic acid

Step 3 open the epoxide with phenyl grignard to give 2-phenyl cyclopentanol

Step 4 oxidise to required product


Why does de Broglie wavelength lambda = c / frequency?

$v$ is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also, there should be rather $p$ instead of $mv$, unless $v \ll c$.
As $$ E = \sqrt{(mc^2)^2 + (pc)^2}$$
then
$$p = \sqrt{\left({\frac {E}{c}}\right)^2 - {(mc)}^2}$$
For a massless object like a photon:
$$p = \frac Ec$$.
For an object with mass, there is also $$p  = \frac {mv}{\sqrt{1-(v/c)^2}}= \gamma mv$$
See Matter wave, paragraphs De Broglie relations and Special relativity, where is clear c is the phase speed of light in vacuum:

$$\begin{align}&\lambda =\,\, \frac {h}{\gamma m_0v}\, =\, \frac {h}{m_0v}\,\,\,\, \sqrt{1 - \frac{v^2}{c^2}}\\
& f = \frac{c}{\lambda} = \frac{\gamma\,m_0vc}{h} = \frac {m_0c^2}{h} \bigg/ \sqrt{\frac{c^2}{v^2}-1}
\end{align}$$


What is responsible for the odor of gasoline?

Many of the components of gasoline (petrol) have notable odours

A simple starting point for understanding the odour of gasoline is the typical composition of the product. That composition varies a lot and always has done as different refineries use different processing techniques and different crude oil sources have different components to start with. And regulators have changed the rules over time (for example to reduce the amount of benzene, a carcinogen). All that matters, in practice, is the octane rating which is what engines care about and this is achievable in a variety of ways by varying the composition and the additives that modify the combustion characteristics.

This (old) paper gives a crude overview of a typical composition.

  The typical composition of gasoline hydrocarbons (% volume) is as follows: 4-8% alkanes; 2-5% alkenes; 25-40% isoalkanes; 3-7% cycloalkanes; l-4% cycloalkenes; and 20-50% total aromatics (0.5-2.5% benzene).

Modern blends will now have far less benzene (but will still have a lot of other aromatics). Without getting too specific on the individual chemicals we can already tell some of the sources of the distinctive odour of the mixture. It is pretty meaningless to try to pin it to a single component.

Aromatics as a class are actually named because of their distinctive smell. even if we substitute other aromatics like toluene or xylene for the benzene, there will be a distinctive "aromatic" component in the product. But plenty of the other hydrocarbons also have notable odours. Cycloakenes (maybe 7% of the mixture) often have sharp smells (cycopentene is even said to have "a petrol-like odour"). Even linear and branched alkanes have some smell (though it is far less intense than the unsaturated components of gasoline).

In short, trying to pin down the typical smell of gasoline to a single compound is a little silly when there are often scores of components with well-known and strong odours and the specific composition varies a lot. aromatics are a big part with other unsaturated compounds providing some other odiferous components with a smaller but notable contribution even from unsaturated hydrocarbons.

Also, gasoline is not "odourised" as it doesn't need to be. But it sometimes has specific ingredients added to make tracing the source easier. But this is done with complex instrumentation not the human nose.


You may not necessarily need to start your procedure beginning with phenol. You may start it with readily available (for cheap price compared to the trouble you are going to have by handling methyl iodide, allyl bromide, and $\ce{AlBr3}$ or $\ce{AlCl3}$) 4-anisaldehyde (4-methoxybenzaldehyde). That is the exact product you are getting after your step 3. Although, steps 4-6 are some very thoughtful organic synthesis steps, I doubt it will give you the expected product. Therefore, I suggest following steps to follow:

Buy 4-anisaldehyde (4-methoxybenzaldehyde) from Sigma-Aldrich (available in 98% purity.
Prepare Grignard reagent using chlorocyclohexane/$\ce{Mg}$ in diethyl ether.
React Grignard reagent (cyclohexyl magnesium chloride) with 4-anisaldehyde.
Perform dehydration of resultant $2^\circ$-alcohol to give double bond.
Perform epoxydation of double bond with 3-chloroperbenzoic acid.
React epoxide with nitromethane.
Reduce nitro group to $1^\circ$-amino group.
Finally, convert $1^\circ$-amino group to dimethyl amino group by reductive amination with formaldehyde.


What substance has the lowest solubility product?

The $K_\mathrm{sp}(\ce{Co2S3})$ value of magnitude of $10^{-124}$ appears in paper by Goates et al. [1].
The refined "thermodynamic" value of $\pu{2.6E-124}$ that you've listed and is used by numerous textbook up to these days has been proposed by Waggoner [2].
However, thirty years later (late 1980s) there's been another study by Licht [3], which showed significant deviation from the previous studies due to two factors:

A new value of the free energy of sulfide ion formation for aqueous solutions  has been used:
$$Δ_\mathrm{f}G^\circ(\ce{S^2-(aq)}) = \pu{(111 ± 2) kJ mol-1}$$
Previous value [1] was
$$Δ_\mathrm{f}G^\circ(\ce{S^2-(aq)}) = \pu{20.64 kcal mol-1} \approx \pu{83.68 kJ mol-1}$$

Free sulfide activity $a(\ce{S^2-})$ used for $K_\mathrm{sp}$ determination

$$
\begin{align}
\ce{M_xS_y &<=> x M^{$2y/x$+} + y S^2-} &\qquad K_\mathrm{sp} &= a(\ce{M^{$2y/x$+}})^x \cdot a(\ce{S^2-})^y \\
\ce{HS- &<=> H+ + S^2-} &\qquad K_2 &= \frac{a(\ce{H+})\cdot a(\ce{S^2-})}{a(\ce{HS-})}
\end{align}
$$

has also been misinterpreted in early studies by acidification of hydroxyl $\ce{OH-}$ instead, erroneously substituting $K_2$ by $K_\mathrm{w}.$

Table I from [3] lists more recent $\mathrm{p}K_\mathrm{sp}$ values, among which the one for $\ce{Co2S3}$ that has increased significantly $(\mathrm{p}K_\mathrm{sp}(\ce{Co2S3}) = 49.9,$ $K_\mathrm{sp}(\ce{Co2S3}) \approx \pu{1.26E-50}).$
Five least soluble sulfides from that table are:
$$
\begin{array}{lrc}
\hline
\ce{M_xS_y} & \mathrm{p}K_\mathrm{sp} & K_\mathrm{sp} \\
\hline
\ce{Ir2S3} & 196.3 & \pu{5.0E-197} \\
\ce{Bi2S3} & 115.1 & \pu{7.9E-116} \\
\ce{Mo2S3} & 107.8 & \pu{1.6E-108} \\
\ce{Ni3S4} & 104.5 & \pu{3.2E-105} \\
\ce{In2S3} & 96.3 & \pu{5.0E-97} \\
\hline
\end{array}
$$
So that it seems like the new champion (at least among sulfides) is iridium(III) sulfide $\ce{Ir2S3}$ with $K_\mathrm{sp} = \pu{5.0E-197}$.
References

Goates, J. R.; Gordon, M. B.; Faux, N. D. Calculated Values for the Solubility Product Constants of the Metallic Sulfides. Journal of the American Chemical Society 1952, 74 (3), 835–836. DOI: 10.1021/ja01123a510.
Waggoner, W. H. Textbook Errors: Guest Column. The Solubility Product Constants of the Metallic Sulfides. Journal of Chemical Education 1958, 35 (7), 339. DOI: 10.1021/ed035p339.
Licht, S. Aqueous Solubilities, Solubility Products and Standard Oxidation-Reduction Potentials of the Metal Sulfides. Journal of The Electrochemical Society 1988, 135 (12), 2971. DOI: 10.1149/1.2095471.


Difference between exothermic and exergonic

The classifications endothermic and exothermic refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$.

If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you're going to use reaction enthalpy $\Delta{}_{\mathrm{R}}H$.

Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}H < 0$, an exothermic reaction that releases heat to the surroundings (temperature increases)
$\Delta{}_{\mathrm{R}}H = 0$, no net exchange of heat
$\Delta{}_{\mathrm{R}}H > 0$, an endothermic reaction that absorbs heat from the surroundings (temperature decreases)

In 1876, Thomson and Berthelot described this driving force in a principle regarding affinities of reactions. According to them, only exothermic reactions were possible.

Yet how would you explain, for example, wet cloths being suspended on a cloth-line -- dry, even during cold winter? Thanks to works by von Helmholtz,  van't Hoff, Boltzmann (and others) we may do. Entropy $S$, depending on the number of accessible realisations of the reactants ("describing the degree of order") necessarily is to be taken into account, too.

These two contribute to the maximum work a reaction may produce, described by the Gibbs free energy $G$. This is of particular importance considering reactions with gases, because the number of accessible realisations of the reactants ("degree or order") may change ($\Delta_\mathrm{R} S$ may be large).  For a given reaction, the change in reaction Gibbs free energy is $\Delta{}_{\mathrm{R}}G = \Delta{}_{\mathrm{R}}H - T\Delta{}_{\mathrm{}R}S$.

Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}G < 0$, an exergonic reaction, "running voluntarily" from the left to the right side of the reaction equation (react is spontaneous as written)
$\Delta{}_{\mathrm{R}}G = 0$, the state of thermodynamic equilibrium, i.e. on a macroscopic level, there is no net reaction or
$\Delta{}_{\mathrm{R}}G > 0$, an endergonic reaction, which either needs energy input from outside to run from the left to the right side of the reaction equation or otherwise runs backwards, from the right to the left side (reaction is spontaneous in the reverse direction)

Reactions may be classified according to reaction enthalpy, reaction entropy, free reaction enthalpy -- even simultaneously -- always favouring an exergonic reaction:

Example, combustion of propane with oxygen, $\ce{5 O2 + C3H8 -> 4H2O + 3CO2}$. Since both heat dissipation ($\Delta_{\mathrm{R}}H < 0$, exothermic) and increase of the number of particles ($\Delta_{\mathrm{R}}S > 0$) favour the reaction, it is an exergonic reaction ($\Delta_{\mathrm{R}}G < 0$).
Example, reaction of dioxygen to ozone, $\ce{3 O2 -> 2 O3}$. This is an endergonic reaction ($\Delta_{\mathrm{R}}G > 0$), because the number of molecules decreases ($\Delta_{\mathrm{R}}S < 0$) and simultaneously it is endothermic ($\Delta_{\mathrm{R}}H > 0$), too.
Water gas reaction, where water vapour is guided over solid carbon $\ce{H2O + C <=> CO + H2}$. Only at temperatures $T$ yielding an entropic contribution $T \cdot \Delta_{\mathrm{R}}S > \Delta_{\mathrm{R}}H$, an endothermic reaction may become exergonic.
Reaction of hydrogen and oxygen to yield water vapour, $\ce{2 H2 + O2 -> 2 H2O}$.  This is an exothermic reaction ($\Delta_{\mathrm{R}}H < 0$) with decreasing number of particles ($\Delta_{\mathrm{R}}S < 0$). Only at temperatures at or below $T$ with $|T \cdot \Delta_{\mathrm{R}}S| < |\Delta_{\mathrm{R}}H|$ there is a macroscopic reaction.  In other words, while the reaction works fine at room temperature, at high temperatures (e.g. 6000 K), this reaction does not run.

After all, please keep in mind this is about thermodynamics, and not kinetics. There are also indications of spontaneity of a reaction.


Does the current of an electrolytic cell increase if the surface area of the Cathode is increased

Just remember two major rules in electrolytic electrochemistry:

If you fix voltage, current cannot be controlled. This is called constant potential electrolysis
If you fix the current, voltage cannot be controlled. This is called constant current electrolysis

Now your electrolysis cell has resistance:

(i) It will depend on electrolytes

(ii) Surface area of the electrodes

(iii) Distance between the electrodes

So let us know your results and findings. Plot a graph systematically by choosing condition 1 or 2.


Only one unique adiabatic path between two states

For an infinitesimal adiabatic process $dS_\mathrm{surroundings} = 0$ since no heat is exchanged with the surroundings. But $dS_\mathrm{system} = 0$  is only true for a reversible process, since only then  $dS_\mathrm{system} = dq/T$. For an irreversible process $dS_\mathrm{system} \neq dq/T =0$. Therefore
$$dS_{\mathrm{irrev}}\neq dS_{\mathrm{rev}}=0$$
This means that the state accessed with a reversible process through an integral change $\int dS_{\mathrm{rev}}=0$ cannot be accessed directly through an irreversible adiabatic process, for which $\Delta S = \int dS_{\mathrm{irrev}} \neq 0$.
Now since S is a state function, the terminal states of the two processes must therefore differ (QED).
As regards the second part (how do we know which is the reversible), you can use the approach just outlined to conclude that for the reversible process $\Delta S =0$.


Bond dissociation energy order for C−C, N−N, O−O and F−F single bonds

In chemistry there is nothing like always. You have to see any element characteristics than decide its property. The case you are asking about:

Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., $\ce{N, O, F}$), which are unusually weak compared with single bonds between their larger congeners. It is likely that the $\ce{N–N, O–O,}$ and $\ce{F–F}$ single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms.
$\ce{N-N}gt \ce{F-F}$, in $\ce{F}$ there are greater number of lone pairs than in $\ce{N}$ so it experiences strong repulsion between its electrons that make it vulnerable to be broken.
$\ce{F-F} \gt \ce{O-O}$ it's so because $\ce{F}$ atom is very small that enables strong electrostatic attraction between nucleus and electrons
For further information on 3) you can follow this link.


Formation of water vapour during combustion

Water vapor produced during the burning of hydrogen-containing fuels should not steal any heat from the fuel, unless they condense on solid surfaces. In such cases, it may require a small amount of heat to evaporate them again. However, in my experiences, there is a stage during burning that there is a lot of water vapor on the solid object, and as the object is warmed up, there is no more vapor condensed.


In August 2018 IBM launched a free cloud service called IBM RXN. This neural network is quite young, but it is constantly being trained, and the prediction should get better over time. I haven't used it all that extensively, but it seems it works quite adequately for simple reactions, including the OP's case. For example:

They also published a great paper [1] which describes on which base their AI operates.

References

Schwaller, P.; Gaudin, T.; Lányi, D.; Bekas, C.; Laino, T. “Found in Translation”: Predicting Outcomes of Complex Organic Chemistry Reactions Using Neural Sequence-to-Sequence Models. Chemical Science 2018, 9 (28), 6091–6098. https://doi.org/10.1039/C8SC02339E.
(Open Access)


How many stereoisomers are possible for 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane?

In this structure the left hand methyl groups can be on the same side of the ring. This is one stereoisomer. This molecule is not chiral and there is only one isomer with the methyl groups on the same side of the ring.

Initially I assumed that there was only one other isomer: the two methyl groups on the opposite side of the ring. But building models sometimes wrecks intuition. The picture below shows the two possible version with trans (relative to the plane of the ring) methyl groups:

So my initial view was wrong: the molecule is chiral and that gives three total stereoisomers.


How can I find the heat that will be released in kJ per gram for the reaction of Al with Fe2O3?

When you converted the heats to kJ/g, you inadvertently "broke" the Hess's law equation that you plugged them into.

Each of the enthalpies of formation in that equation is multiplied by the number of moles of the matching species, and so the units of the enthalpy of formation are kJ/mol.

If you assume one mole of product and convert that to grams, you then would have to convert all of the other species to grams (two moles of Al, two moles of Fe, etc).

It's easier to just do the whole calculation in moles, then divide the final answer by grams. However, you need to specify which species is the gram basis. In other words, you are asking for the heat per gram, but per gram of what?

Whatever the answer is, then find the total heat output in terms of kJ/mol of that substance, then divide it by the molar mass of the substance (in g/mol).

One way to avoid these kinds of problems is to always explicitly include the units in any calculation that you do. In this case, you would have quickly seen that the Hess's Law equation would not give you the correct units.


How do I turn room temperature water into boiling water for drinking using an chemical/additive?

Adding something to water that heats it to 100°C and leaves only water is of course nonsense.

A self-heating cup is easy: You take a double-walled beaker, fill the space between with two compounds that react with great heat but no gas formation, separated by an extra wall with a valve.

Google the self-cooling beer keg. It works very similar, only it gets hot outside, not inside.

Or just google "self-heating coffe cup".


Is it possible to graph the equivalence of p[Cu²⁺] with pH?

TTbulaski, I gather that you are developing a sensor for copper ions whose output is proportional to -log [Cu(II)]? I believe this is what you are trying to say.

What you can test is whether the output of your sensor gives any linear relationship between actual Cu concentration and the antilog of your sensor signal or not. Plot the Cu ion concentration on the x-axis and the anti-log of your sensor signal on the y-axis? Do you see a linear relationship?

pH and -log [Cu(II)] connection is pretty tricky and it depends what level of depth you want to go to. There is no point in looking for a similarity between the two measurements.
People have done calculations to show what species of Cu is present as a function of pH. It is called speciation diagram of copper.

P.S. The most important thing to test any metal ion sensor is something called as "selectivity". No analytical chemistry reviewer will accept the paper until and unless you show how selective is your sensor. For example if you have Zinc or Cd or other transition metals, do you still get a signal for copper in a correct way?

Not long time ago, NASA did a mistake in ion-selective electrodes. They designed or used a nitrate selective electrode and it turned out that is more selective towards perchlorate ion. It was an interesting news.


Reaction of unsaturated lactone and alcohol

You are correct that the alcohol atacks the carbonyl then the ring oxygen leaves in a manner analagous to ester transesterification. The intermediate generated is an enol which picks up a proton to for the ketone.

image from here: 1


Using RDKit to determine "nonring, nonterminal atoms"

I think you want to use SMARTS patterns to perform searches. These are something like "regular expression" patterns for molecular graphs.
At a rough guess, your query converts to [*R0!D1]:

* matches any atom (i.e., you don't care if it's a carbon or an oxygen)
R0 matches an atom in a ring with size 0 (i.e., not a ring atom)
D1 matches an atom with explicitly one bond (so !D1 matches atoms with more than one bond).

It's often helpful to use SMARTSview to test patterns.
More on RDKit's SMARTS support can be found in the documentation


How much does water expand when heated, and does salt affect that?

The volumetric expansion of a liquid is given by the following equation
$$\Delta V=V_0\beta\Delta T$$
where $\beta$ is the coefficient of thermal expansion and $\Delta T$ is the change in temperature.

Here's a link to a nice table containing coefficients of expansion for water, both pure and salinated.  Wikipedia shows the maximum solubility of salt in water to be around 36% (w/w).

From the first link we find that

for pure water at $25\ \mathrm{^\circ C}$: $\beta = 257\times10^{-6}/\mathrm K$

for water with 35% $\ce{NaCl}$ dissolved at $25\ \mathrm{^\circ C}$: $\beta= 297\times10^{-6}/\mathrm K$

Now you should be able to calculate the values you are interested in.


Organic molecules with fast non-radiative decay rates

Anything that has potential energy surfaces that cross or get close in energy. Offhand things like s-indacene and the indenofluorene family of compound, there are many others as well (azulene mentioned below). In general, if the $S_0$ and $S_1$ state for a $\pi$-conjugated organic come close in energy there is the possibility for all of that energy to be used on vibrations instead of luminescence. As far as predicive rules, that is a difficult thing to generalize. Perhaps you could argue that a low $S_0 \ce{->} S_1$ transition energy opens the possibility for non-radiative decay since these potential energy surfaces are not seperated by much energy, biradicaloid $\pi$ systems fall in this category.This is the same reason there is typically not emission from higher energy state, $S_n \ce{->} S_1$ transitions are usually quite close in energy and have crossings to get to $S_1$ efficiently through molecular vibrations. For further examples and explanations see, for example:
http://www.sciencedirect.com/science/article/pii/S0009261414008847?via%3Dihub
Or Josef Michl's excellent book "Excited States and Photochemistry of Organic Molecules" for more details on the topic.


Conjugation effects are not very important in this case.
The main reason is steric hindrance. If the more substituted alkene is formed, then both the methyl group and the nitrogen ring will be on the same plane, and repulsions will destabilize the molecule.

In the case of A, the methyl group is more free to move in a pseudo-chair conformation of the cyclohexene ring.

But there are exceptions, for example:

In the second case the more substituted alkene is formed, because is more stable and there is essentially no difference in regard to steric hindrance.


Assuming your concern is on $\mathrm{S_N1}$ type nucleophilic substitution reactions, you are in the correct path with your first explanation. However, everything has gone south from that point on. As you found out, the stability of carbocations in $\mathrm{S_N1}$ reactions are: $3^\circ \gt 2^\circ \gt 1^\circ$. However, reaction rates of $\mathrm{S_N1}$ reactions do not depend on electron density (or lack of it) on carbocation, because rate determine step is rate of carbocation formation step:

As depicted in the diagram, $3^\circ$-carbocation from 2-chloro-2-methylbutane forms faster than $2^\circ$-carbocation from 2-chloro-3-methylbutane because the stability of $3^\circ$-carbocation is much larger than that of $2^\circ$-carbocation as you predicted in your first explanation. Also, you need to understand that in these reactions, 1,2-hydride (or methide) transfer to give more stable $3^\circ$-carbocation from $2^\circ$-carbocation is possible if there are available $\alpha$-hydrogen or $\alpha$-methyl group in the sought $2^\circ$-carbocation, alike in the case of 2-chloro-3-methylbutane here. Regardless of this rearrangement, the rate of carbocation formation of 2-chloro-2-methylbutane is still faster than that of 2-chloro-3-methylbutane, because it is strictly the stability of $3^\circ$- versus $2^\circ$-carbocations.

The second reaction (attack of the nucleophile to available carbocation) is very fast compared to chloride dissociation, thus it is not affecting the overall rate of the reaction. Thus, the product formation from 2-chloro-2-methylbutane is faster than the product formation from 2-chloro-3-methylbutane, and hence, 2-chloro-2-methylbutane is more reactive than 2-chloro-3-methylbutane.


A cell has an internal ohmic-equivalent resistance.  The salt bridge is a resistive path for the flow of ions.  Ions, plus their hydration shells and possible coordination shells, have low drift velocities through a medium.  The salt bridge can constrain the ion current of the cell and its material equilibrium rate.  A poor salt bridge by length, diameter, concentration, and permeability will increasingly adversely affect a cell's current and voltage as discharge rate increases.


Gaussian Error: Zero Point Energy Calculation on Fluoroform fails

As you can see in the error message

#T B3LYP/6â€311+G(3df,2p)

there is some Unicode garbage in your input file; the dash between the 6 and 3 isn't an ASCII dash. Deleting and replacing it should fix the problem.


Is the VSEPR theory correct in determining the bond angle of sulfur dioxide?

First off, your structure violates the octet rule on sulphur and is thus not a good representation of the molecule. A better one is this one:

Figure 1: electronic structure of $\ce{SO2}$.

All methods of approximation of bond angles you have learnt or will learn about are no more than educated guesses. VSEPR gives very good approximations for e.g. $\ce{NH3}$ and $\ce{H2O}$ but fails horribly at $\ce{PH3}$ and $\ce{H2S}$. For the latter two, an estimation based on the most likely orbitals used for bonds (all p) does a much better job. To accurately determine a bond angle, one needs to perform a structure elucidation experiment such as X-ray diffraction.

This has indeed been performed with $\ce{SO2}$ (likely multiple times using different methods; I am too lazy to check the references) and the results give a bond angle $\angle(\ce{O-S-O})\approx119^\circ$. This is indeed very close to $120^\circ$ so you can say that the VSEPR method gives a good approximation of the $\ce{SO2}$ bond angle.

Figure 2: Structural data of $\ce{SO2}$. Taken from Wikimedia, where a full list of authors is available.


The glassware isn't particularly valuable. Finding someone who wants it will be the problem. If you were going to buy it new, I'm thinking maybe $150.

It is indeed a distillation setup. The large flask is the pot in which you'd put the stuff that you wanted to boil. The small flask would collect the stuff that you boil off. The nipples on the condenser are for rubber hoses to connect to a water supply so that the condenser jacket can be water cooled. Water goes in on the bottom and out on the top. In order to use this you're missing some parts.

You need a couple of ringstands and clamps to hold the whole thing together.
Two thermometers would go into the top of the condenser. One to monitor the pot temp, and the other to monitor the condensate temp.
You need some metal clamps to hold the flasks to the condenser.
You need something to heat the pot, a Bunsen burner would do.


Why do Alpha particles not collide with electrons during alpha decay?

Actually they do collide.  But, the emitted alpha particle carries much more energy than the binding energy of the electron(s) in a helium atom or $\ce{He^+}$ ion.  Therefore such a collision scatters the electrons (which carry away some of the original energy of the alpha particle) instead of forming an atom.


One reason cations and anions exist is due to the stability of a full or half-full valence shell. The stability from those electronic configurations means that the atom or molecule does not require protons to "hold" the extra electron.

Recall also that nitrogen has three (or five) valence electrons, rather than seven. The 1 s shell is full and is not considered part of its valency. The three 2 p electrons are the valence electrons although they hybridize with the 2 s electrons to produce the trigonal pyramidal structure of ammonia with its lone pair. The single anion $\ce{N^{-1}}$ could exist, but would not be stable because it puts four electrons in the p shell. The p shell would prefer to have three electrons as it does in the nitrogen atom or no electrons as it does in the $\ce{N^{3+}}$ cation.


How is a side-group that contains a double bond named?

When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain.

You are right in so far as functional suffixes that express the characteristic group of a compound (e.g. the suffix ‘ol’ for alcohols) are always attached to the name of the parent structure (i.e. the principal chain). However, since the given structure does not include any characteristic group that is expressed as suffix, the order of seniority of suffixes is not relevant in this case.

The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain (see Subsections P-44.3 and P-44.3.2 in the Blue Book).
Only if this rule does not effect a choice of a senior parent structure, the principal chain has the greater number of multiple bonds (see Subsections P-44.4.1 and P-44.4.1.1).
(Note that the order of seniority between length of chain and unsaturation given in earlier IUPAC recommendations was reversed. Thus, the first criterion to be considered in choosing the preferred chain was unsaturation; the length of the chain was the second criterion.)

In this case, the longest chain has eight carbon atoms; i.e. the unmodified parent hydride is octane. (Note that two equivalent chains with eight carbon atoms exist in the given structure.)

Then

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Therefore, the correct name of the parent hydride is oct-3-ene.

The preferred names of the substituent groups are derived from the respective parent hydrides. The presence of free valences formally derived from the loss of one, two, or three hydrogen atoms from a parent hydride is denoted by replacing the ending ‘ane’ of the parent hydride name by ‘yl’, ‘ylidene’, and ‘ylidyne’, respectively (see Subsection P-29.2).

Therefore, the preferred names for the substituent groups of the given structure are
methylidene $(\ce{=CH2})$,
ethylidene $(\ce{=CH-CH3})$, and
ethyl $(\ce{-CH2-CH3})$.

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes.

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Note that Rule e takes precedence over Rule f.

Therefore, the structure given in the question is named as 3-ethyl-5-ethylidene-6-methylideneoct-3-ene rather than 6-ethyl-4-ethylidene-3-methylideneoct-5-ene since low locants are given first to multiple bonds (i.e. ‘ene’) and the locant ‘3’ is lower than ‘5’.

Furthermore, note that the double bond in the ethylidene group gives rise to cis-trans isomerism. As drawn, the configuration is ‘E’.


Why do resorcinol and iodine react to give 2-iodobenzen-1,3-diol?

Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer plethora of subjective questions, which should be answered and cleared to show the way to right direction.

I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which the Resorcinol (Benzin-1,3-diol) acts as the electrophile and Iodine as nucleophile.

First, resorcinol is not called benzin-1,3-diol (may be in an old book). It's correct name is 1,3-dihydroxybenzene. It is true that the reaction is an electrophilic aromatic substitution. However, the electrophile here is polarizable $\ce{I2}$ molecule because it exsists in $\ce{I^{\delta +}-I^{\delta -}}$ form under reaction conditions (where $\ce{I^{\delta +}}$, which lacks electrons is the reactive center). Aromatic nucleus with two hydroxy group is electron rich (hydroxy groups activate the ring), hence acts as the nucleophile.

But isn't the +M Effect stronger?

Yes, mesomeric effect (old resonance effect) is stronger than inductive effect. Here you have two hydroxy groups to give strong $(+M)$-effect to activate the aromatic nucleus. Hence the reaction is o,p-directive. As a result, you can expect either 2-iodo-1,3-dihyroxybenzene or 4-iodo-1,3-dihyroxybenzene or 2,4-diiodo-1,3-dihyroxybenzene or etc. as your product or mixture of them based on the reaction condition used (Ref.1 & 2).

Why is the Product 2-Iodo-1,3-diol and not 4-Iodobenzen-1,3-diol?

As I explained before, the reaction conditions govern the product ratio (see the PDF of Ref.1 I attached below for better understanding). The iodination of resorcinol is not a straight forward reaction. Slight changes of reaction conditions gives you different product. For example, the major product given in the reaction using $\ce{I2/NaHCO3}/\pu{0 ^\circ C}$ condition is 2-iodo-1,3-dihydroxybenzene as indicated in the OP's question. In a large scale reaction, the yield obtained was only 66% (Ref.3). Using the same conditions, Ref.1 reported 77% isolated yield. Keep in mind that, some of 2-iodo-1,3-dihydroxybenzene isomerized to 4-iodo-1,3-dihydroxybenzene (major) and 4,6-diiodo-1,3-dihydroxybenzene in an acidic medium at high temperature (Ref.1). Thus, temperature plays an important role in percent yield in the reaction.
On the other hand, when the electrophile ($\ce{I+}$ source is changed to $\ce{ICl}$, the main product is 4-iodo-1,3-dihydroxybenzene (4-iodoresorcinol) with 4,6-diiodo-1,3-dihydroxybenzene (4,6-diiodoresorcinol) as minor product (Ref. 1 & 2). Ref.2 also stated that use of nascent iodine $(\ce{KIO3/KI/HCl})$ has resulted in the regiospecific 2,4-diiodonation of resorcinol at room temperature (which is a rapid reaction). The $\ce{^1H}$-NMR spectrum of this reaction mixture has revealed only two products, 2,4-diiodoresorcinol as major product (56% isolated yield) and 2,4,6-triiodoresorcinol as minor product in $89:11$ ratio.
References:

Ib Thomsen, and Kurt B. G. Torssell, "Iodination of Resorcinol, 5-Methoxyresorcinol, Phloroglucinol and Resorcyclic Acid," Acta Chemica Scandinavica 1991, 45, 539-542 (DOI: 10.3891/acta.chem.scand.45-0539)(PDF).
Frederick L. Weitl, "Regiospecific 2,4-diiodination of resorcinol with nascent iodine," J. Org. Chem. 1976, 41(11), 2044–2045 (DOI: https://doi.org/10.1021/jo00873a040).
Shin-ichiro Tsujiyama and Keisuke Suzuki (submitted); David B. Guthrie, Heather M. Gibney, and Dennis P. Curran (checked), "Preparation of Benzocyclobutenone Derivatives Based on an Efficient Ggeneration of Benzynes [Bicyclo[4.2.0]octa-1,3,5-trien-7-one, 5-(phenylmethoxy)-]," Org. Synth. 2007, 84, 272-284; Coll. Vol. 11 2009, p. 488-497 (DOI: 10.15227/orgsyn.084.0272).


Geometry optimization: dependence on initial geometry

Starting with two moieties, the final "optimized" geometry is not
  unique and depends highly on how they are placed initially. Is there a
  way to avoid this?

Yes and no. No, because, as OP anticipated, starting from an input geometry the optimisation algorithm will indeed find the closest (from the algorihtm point of view) local minimum and not necessarily the global one. And yes, because by performing an extensive conformational analysis, one in principle can find structures that correspond to all possible local minima and then identify the global one.


Does Friedel-Crafts reaction fail with benzoic acid?

It could be that $\ce{-COOH}$ group is electron withdrawing hence there is an increased electron density in $\ce{-COOH}$ group due to -R resonating effect of electron from benzene ring.

Since friedel craft catalyst is a Lewis acid (electron deficient) the catalyst (anhydrous $\ce{AlCl3}$) reacts with electron rich COOH group of benzoic acid and gets bonded.


How do we experimentally know the strength of bonds?

You measure the energy of any bond by carrying out reactions and seeing how much energy (usually in the form of heat) has to be invested.  For example, if you react one mole of $\ce{C}$ with one mole of $\ce{O2}$, you get $\ce{CO2}$ and heat.  Measure the energy with a calorimeter and divide by how many atoms and you get the energy you invested to produce one bond.  Not coincidentally, bonds are generally measured in units of energy/mol (e.g. kJ/mol).

Whether a bond is ionic or covalent has to do with whether the bonding electron resides on one atom or both.  A perfectly covalent bond finds the electron half on one atom and half on the other.  A perfectly ionic bond has the electron 100% on one atom and not at all on the other.  Real systems are a mix of both.  When we say a bond is ionic (such as $\ce{NaCl}$), we are really referring to the fact that we have measured a circumstance where the electron truly favors one atom.  There are various ways to measure this, one of the most common is to melt the solid and see if it conducts electricity.  In the case of $\ce{NaCl}$, the bond is so ionic that a small voltage will cause the ions to wander from their neighbor.

With a covalent bond such as you find in $\ce{SiO2}$, atoms can't wander without breaking the bond, so a small voltage doesn't cause a current.  A large voltage may break the covalent bonds and cause a current, but it also changes the substance (in this case releasing oxygen gas and producing silicon metal).


Francium has isotopes so how can its atomic mass be a whole number?

For elements with no stable isotope (i.e. Francium, Radium, and Actinium), the atomic mass is chosen to be that of the longest lived isotope.

More generally, the masses for stable elements are reflective of the natural abundance of each isotope in a sample of the element. Sodium has more than one isotope, so the statement is not really true, though only one of them is stable and included in the averaging. In this case, it's important to remember that the mass of protons and neutrons are only approximately 1 amu, so even if there is only one isotope there's no requirement that it have an integer atomic mass.