Why use a salt bridge at all in an electrochemical cell?

The use of a salt bridge is to maintain the electrical neutrality throughout the cell.

A salt bridge will allow the movement of ions.
Electrons flow from anode to cathode.Due to this a negative charge will be developed in the cathode and a positive charge in the anode.This charge developed will oppose the electron flow (current from anode to cathode) and at one level the electron flow will be ceased.So the cell's action will be stopped.
So in order to prevent this, salt bridges are used.

There is a similar question in this site.Please refer that.

Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?


Why are combustion reactions spontaneous?

Clarification on the definition of spontaneity must be considered first, as it's meaning is often misinterpreted (and through colloquialisms is further confused). In order to do so, we must separate reaction concepts into thermodynamics (where free energy and spontaneity are discussed) and kinetics (where activation energy and rate are discussed).

Thermodynamics: looking only at the reactants' and products' Gibbs free energies, and the difference between them ( &Delta G ), a reaction is defined as (a) spontaneous if the difference is negative, (b) at equilibrium if zero, and (c) non-spontaneous (a sign that it is spontaneous in reverse) if positive. Thus, thermodynamics is interested only in which form is more stable energetically: the reactants or products. No details about the path (see kinetics below) is included. Check out Khan Academy's "Gibbs free energy and spontaneity" page for some great figures and further detail.

Kinetics: interested in the pathway between reactants and products, and how much energy change is required to obtain an arrangement that leads to one or the other. In other words, this is where the activation energy becomes important, as it is the energy barrier that must be overcome before a reactant could even think about becoming more stable. This barrier impacts the rate at which reactants and products can be formed. A "spontaneous" reaction in kinetics would be a "fast" reaction, one with a nearly zero or negative activation barrier (no representative "hump" in those internal reaction coordinate diagrams).

In summary, thermodynamics is interested in stability of reactants or products, and is not directly discussing the path by which it would arrive at one or the other. Hence, a reaction can be "spontaneous" and still not occur without further prompting because it is kinetically too difficult (read "high activation energy") to get there. However, a "spontaneous" kinetic reaction implies the path is "downhill" (so to speak) and thus is also a "spontaneous" thermodynamic reaction.

Following that understanding, you mention temperature, which is where I believe the confusion builds as it is mixed with the "spark" in combustion. Temperature does effect the Gibbs free energy by the following relationship:

ΔG = ΔH−TΔS

Thus, reactants can be "pushed" with temperature to become products by adding the necessary energy to a non-spontaneous or equilibrium system to breach the activation energy "hump," and be "spontaneous" at that temperature. In other words, they can indeed be triggered just by increasing the temperature. Examples of this are found in spontaneous combustion events as temperatures are slowly raised to a point at which the energy barrier is breached, and reactants are becoming products while releasing heat (because these spontaneous reactions are exothermic, recall negative &Delta G). Heat release then causes surrounding reactants to overcome the same barrier energetically, and result in what we know as fire/burning events.

What about a spark, then? A spark contains the energy needed to begin the combustion process (e.g. from gasoline and oxygen to carbon dioxide and water) but in a sudden shock that triggers this cascade of heat release much more rapidly. A CDC study illustrated the impact of this latent heat given off by the initial sparked reaction by observing a stoppering of combustion by removing the heat with conduction (i.e. blowing out a candle).

So, all in all, "spontaneity" as a definition perhaps isn't wrong, but is used in various circumstances to describe different processes; here, either stability between states or rate at which the states will change.

Be sure to check out the other related stack exchange page mentioned by Mithoron in the comments, with a great answer regarding spontaneity by Brian and Melanie Shebel. Covers more detail about the thermo/kinetic descriptions, whereas here we cover a more broad scope to address your questions about "spontaneity" in general.


Intramolecular lactonisation in an unsaturated carboxylic acid

The reaction as you are given turns up no results in SciFinder, meaning it has not been reported in any scientific journal. It is thus the imagination of whoever set the test.

However, closely related reactions are reported and these may come as no surprise to the advanced chemist. They mainly differ in that iodine $\ce{I2}$ is added to the reaction mixture. The mechanism is called iodolactonisation.

Basically, iodine is attacked by the double bond to form an iodonium ion (whether or not this is actually a three-membered ring shall be deemed irrelevant in the context of this answer). This is then rear-side attacked in an intramolecular $\mathrm{S_N2}$ reaction by the deprotonated carboxyl group to form the five-membered lactone kinetically. In a subsequent step, the iodo substituent is removed in an $\mathrm{E2}$ type reaction with a base to regenerate the double bond; due to steric constraints relating to the required anti-periplanar transition state, this newly-formed double bond is shifted along the six-membered ring.

This is described for your compound multiple times in the literature using different bases, for example by Khanjin et al.[1] They use DBU as a milder alternative.

Reference

[1]: N. A. Khanjin, J.P. Snyder, F. M. Menger, J. Am. Chem. Soc. 1999, 121, 11831–11846. DOI: 10.1021/ja992453d.


In laboratories, ethyl acetate is usually stored in special ventilated cupboards that are furthermore typically designed to withstand fire. This is because you will typically store not just one but many flammable solvents/reagents in that type of cupboard.

At home, you typically do not have that possibility. However, the basic advice remains the same: store in the original container (if possible) and in a well-ventilated area away from spark sources. The original bottle should, in theory, be airtight enough — remember that the ethyl acetate must have somehow found its way to you in that bottle without having lost too much content.

I would strongly advise against storing in a closed drawer. As you noticed, a few months were enough for enough solvent to evaporate but stay in the vicinity for you to smell it. The amount of solvent you lose per unit of time is typically very small and if it can diffuse away it does not do much harm. But the closed drawer prevented it from efficiently diffusing away and allowed it to concentrate.


Is the term "carboreduction" interchangeable with "carbothermic reduction"?

The term "carboreduction" has an IUPAC definition:

  Process in which a metal oxide is reduced in the presence of carbon or a carbon-containing compound.

So technically the term does not require heating and isn't limited to elemental carbon.

Carbothermic definitely implies heating.


How can dish-soap make glass less misty?

The fogging is caused by tiny droplets of water. These condense onto the glass when hot humid air comes in contact with a cold surface . The dish-soap acts as surfactant. The dish soap prevents droplets from forming. It forces any water molecules that condense onto the surface to be evenly spread across the surface. Thus, the water molecules form a thin film instead. This film is usually thin enough to not impede vision.


What you're specifically describing is typically known as UV degradation.  The color change is due to the oxidation of the polymers that make up the plastic.  These chemical reactions are demonstrated in satisfying detail here.  Under the section titled "Photodegradation" is where you can find the intricacies of the polymer-oxidation process.  And yes, such damages are reversible.  As you will read, the oxidation process results in the formation of free radicals, which are "atoms or groups of atoms with an odd (unpaired) number of electrons," to quote Wikipedia's definition.  The formation of these radicals is the reason for the color change (since certain light gets reflected because of these new molecules, in many cases yellow light, etc...)

To get rid of the radicals, you can add hydrogen peroxide, which allows hydrogen to bond with the radical (hydrogen has an additional electron), thus de-yellowing the plastic. This is demonstrated here.

As a further note, UV degradation is not the only process that makes plastics turn yellow, as you will probably read in that paper.  There are also processes like biodegradation and mechanical degradation that are similar to photodegradation but still have their own cool processes.

Your initial question was "What makes ABS plastic turn yellow."  I should point out that the first few words of the second link read "Manufacturers add bromine to ABS plastic to act as flame retardant. Over time, those chemicals react to the plastic's basic polymers and turn yellow."  Ergo, there is probably more to the yellowing of ABS plastic in particular than just photodegradation.

Lastly, I should point out that the yellowing of plastic needn't happen in the presence of oxygen.  But again, that is addressed beautifully in the paper.


There isn't a good overall formula that I know.
You can look at C/H ratio and easily tell a bit here.
There must be at least one of the following, but not two:

carbon ring
carbon double Bond

(1) 1-Butene

(2) 2-Butene which has two stereoisomers
cis-2-Butene
trans-2-Butene

(3) Isobutylene

(4) cyclobutane

(5) methylcyclopropane


Pyridine synthesis by tin(II) chloride reduction of 5-nitronorbornene

This reaction was conducted by Ho and Liao1. Nitro compound 1 is reduced to the nitroso compound 2 which undergoes a [3+3] electrocyclization to form 3 (4). Tautomerization of 4 to 5 is followed by a retro-"Diels-Alder" reaction to open to 6. Conjugate addition of the nitrogen to the unsaturated aldehyde moiety of 6 to afford dihydro pyridine 7. Tautomerization of 7 to 8 is followed by the loss of acetaldehyde. [Structures are from the paper; my numbering.]

ADDENDUM: The mechanism provided in the paper does not address a number of stereochemical issues. Nitro compound 1 is obviously formed as an endo-Diels-Alder product between (E)-(2-nitrovinyl)benzene and 1,3-cyclopentadiene. The bicyclic compound 3 (4) must have a cis-fusion given the mechanism of the [3+3]-electrocyclization (2 → 3) (vide infra). However, the stereochemistry arising from the retro-Diels-Alder reaction would not be expected to kinetically form the (Z,Z)-diene 6 but rather the (E,Z)-isomer 6a. Clearly, 6a is incapable of cyclization as prescribed. But 6a is a masked, vinylogous malonaldehyde that is susceptible to enolization and isomerization to 6b or 6c. While 6c can cyclize by the reported procedure of conjugate addition, there is also the probability that ring closure occurs by electrocyclization of 6b.

Tse-Lok Ho and Po-Yau Liao, Tetrahedron Lett., 1994, 35, 2211. DOI:10.1016/S0040-4039(00)76799-2


Unit analysis for neutralization reaction

There is no special unit analysis.
It's a recommended practice to solve the problem algebraically first using proper notations for physical quantities, and plug the numeral values at the end minding the units — this way you reduce the chance of making the erroneous calculations and keep track of all units; as a bonus, you simplify handling significant figures.

The majority of problem solving in chemistry begins with writing down the balanced chemical reaction:

$$\ce{HCl + NaOH → NaCl + H2O}$$

So, to find the volume of hydrochloric acid $V(\ce{HCl})$, you may use its molar concentration $c(\ce{HCl})$ and the amount $n(\ce{HCl})$:

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{c(\ce{HCl})}$$

Unknown amount $n(\ce{HCl})$ can be found from the reaction's stoichiometry:

$$n(\ce{HCl}) = n(\ce{NaOH}) = \frac{m(\ce{NaOH})}{M(\ce{NaOH})}$$

where $m(\ce{NaOH})$ and $M(\ce{NaOH})$ are mass and molecular mass of sodium hydroxide, respectively.
Putting everything together and plugging in the appropriate numerical values, you arrive at the volume of approx. $\pu{100 mL}$:

$$
\begin{align}
V(\ce{HCl}) &= \frac{m(\ce{NaOH})}{M(\ce{NaOH})\cdot c(\ce{HCl})} \\
  &= \frac{\pu{2.02 g}}{\pu{40.00 g mol-1}\cdot \pu{0.500 mol L-1}} \\
  &= \pu{0.101 L}
\end{align}
$$


Is there such thing as too much fermentation

Continued fermentation of wine produces vinegar, which is not necessarily a bad thing. However, further fermentation may encourage growth of "vinegar eels" (harmless worms, actually, but not very appetizing).

Vinegar itself can be metabolized by microorganisms. According to the Compendium of the Microbiological Spoilage of Foods and Beverages, "several molds, yeast and bacteria are capable of spoiling vinegar... [such as the appropriately named] Lactobacillus acetotolerans."


Are the orders of reactants with respect to a reaction different for different starting concentrations?

A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate of consumption of  reactants $$\dfrac{-d[A]}{dt}= k[A][B]$$ and $$\dfrac{-d[B]}{dt}= k[A][B]$$
It seems that this reaction is a second order reaction since the rate order appears to be:$$\text{Rate} = k[A][B] $$
As @Andrew astutely pointed out in the comments, a reaction can be of a lower order if one of the reactant's concentration is much higher than the other's.
Let's suppose, that one reactant is in great excess (say $10 \text{M of B vs } 0.01\text{M A}$)
So what we're observing now is that the reactant B is barely consumed, yet the reaction terminates since reactant A, the limiting reagent is exhausted.
In other words, the rate equation looks something like this: $$\text{Rate}=k'[A]$$, where $k^{'}=k[B]_o$ .Now, the reaction appears to be a first order reaction. This is known as a pseudo first-order reaction, since the rate of the reaction ultimately depends only on one reactant.
An example that I found here:
$$ \rm{CH_3Br + OH^- \rightarrow CH_3OH + Br^-}$$

[...]"Imagine we had an initial concentration of CH3Br of 100 μM and and an initial concentration of OH- of 10 mM.   Now even if all of the CH3Br has reacted the concentration of OH- will be essentially unchanged.  Therefore during the course of the reaction, the concentration of OH- will be essentially constant.  This makes the reaction "like a first order reaction", thus the name pseudo-first order."

Effectively, the rate now is: $$\ce{rate= k'[CH_3Br]}$$
This can also be mathematically represented in the integrated rate law:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} = kt \quad \ce{where [OH-]\neq[CH3Br]}$$
Now, since $\ce{[OH-]_0 \gg [CH3Br]_0}$, we can write the equation as:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} \approx \dfrac{1}{\ce{[OH-]}}\ln \dfrac{\ce{[CH3Br]_0}}{\ce{[CH3Br]}}=kt $$
Remember, however that the rate and order of a reaction usually depends upon its mechanism.
For further reading, check here and here as well.


What kind of system is required for dynamic equilibrium to be set up?

If energy can be transferred, it does not mean it is transferred.

If there is e.g. zero net thermal energy transfer, a system is in thermal equilibrium with the neighborhood, even if it is not isolated.

As Ivan says, an equilibrium is possible even in an open system, if net transfer of particular matter kinds is zero.

The thermodynamic equilibrium consists of thermal, radiative, chemical and mechanical equilibrium.

Note that thermodynamic equilibrium happens at following conditions:

  For a completely isolated system, the entropy $S$ is maximum at thermodynamic equilibrium.
  For a system with controlled constant temperature and volume, the Helmholtz free energy $A$ is minimum at thermodynamic equilibrium.
  For a system with controlled constant temperature and pressure, the Gibbs free energy $G$ is minimum at thermodynamic equilibrium.

$$\begin{align}
G&=H-TS \\
\Delta G&=\Delta H-T\Delta S \\
\end{align}$$

Total entropy of the not isolated system and it's surrounding increases in time.

Entropy of the system alone may decrease with time, if the Gibbs energy of system decreases, as this latter decrease also means
$$\Delta S_{\rm surrounding} +  \Delta S_{\rm system}  \gt 0$$


If the analyte happens to be in a transparent container, shine light on it, see if the light is scattered (Tyndall effect) -- if so, you have either suspension, or colloidal matter.  Instead of a laser pointer, a good pocket light (like the ones with halogen bulb halogen or LEDs) may be used as light source, too.

(source)


Spins spin coupling on proton NMR for an aldehyde?

I cannot decide whether the H in the aldehyde group would be split into a doublet or remain as a singlet […]

What would be the argument against $^3J$ coupling between these protons? The aldehyde proton will appear as a doublet.

Don't forget that coupling works both ways! $\ce{(CH3)2C\color\red{H}-}$ will (at least) split to a doublet too. Coupling with this proton will also  split the signal for the methyl protons to a doublet.

What does this mean for the proton at C-2?

Formally, a doublet (from coupling with the aldehyde proton) of a septet (from coupling with six magnetically equivalent methyl protons) is to be expected.

In really, you will observe a smaller number of lines! I suggest to try this with a pen on squared paper to figure out what happens when the two coupling constants are identical or one is twice as big as the other.


effect of temperature on position of equilibrium

Well, For endothermic reaction the Number of moles at the product side increase with increase in temperature.
And for exothermic, it decreases with Increase in temperature..
And exactly vice versa with respect to the reactant side..
It is in Accordance with the Le chaitliers Principle..


Gold's atomic structure and bonding in relation to the separation techniques used

There are differences in gold extraction for different mines - depending on the chemical nature of the raw ore, the impurities present, the availability of resources etc.

A good example of gold extraction comes from the Super Pit mine in Kalgoorlie, Western Australia; their gold extraction process is outlined in their website Mineral Processing, which is outlined below:

Crushing and grinding: this is a physical process where the mined raw ore is crushed, ground and water added to make a slurry.

Flotation: chemicals are added to the slurry to remove everything other than gold and pyrite.  Water is also removed at this stage.  The slurry is agitated forming air bubbles, which then rise through the slurry, carrying mineral grains.  From Introduction to Mineral Processing:

  The ability of a mineral to float depends upon its surface properties. Chemical modification of these properties enables the mineral particles to attach to an air bubble in the flotation cell. The air bubble and mineral particle rise through the pulp to the surface of the froth or foam that is present on the flotation cell.

Roasting: water is re-added and the concentrated slurry is roasted to remove $\ce{SO2}$

CIL: or carbon in leach, chemicals are added to dissolve the gold and have close to pure gold precipitate on the carbon.

Elution: A mixture of caustic soda and cyanide is used to liberate the gold from the carbon.

For your specific request for resources about flotation, the example process below is for a sulphide copper ore containing iron bearing pyrite:

(Image Source)

Suggested additional resources about flotation:

Froth Flotation – Fundamental Principles

A great resource is the book Handbook of Flotation Reagents: Chemistry, Theory and Practice (chapters are behind a paywall).


If you think about the quantities involved in this reaction, you'll realize why does the product contain the methoxy group and not the halide. Methanol is present in large excess since it is the solvent of this reaction. Cl- ions will also be present and a very small amount of the chlorohydrin may be formed as well. The reason why methanol is a reagent and not only act as a solvent in this reaction is because its oxygen atom acts as a nucleophile, just like halides or any other molecule containing electronegative atoms (such as O, N, S etc).

I found the following mechanism on Google and it is draw as if the reaction was being performed in water, however you can replace the blue water molecule for methanol or any other alcohol since the mechanism is the same for both.

Whenever you look at a reaction schemes like the one you've posted, you should always address the reactivity of solvents and other reaction components, sometimes they end up reacting with your substrate rather than your desired reagent, this is a very good example of it. So if you intend to synthesize a chlorohydrin, you cannot use a nucleophilic solvent, otherwise, it will react with the epoxide and form the correspondent adduct.

I hope this answers your question.


No. The $n-$factor is still $1$ because one mole of $\ce{S2O3^2-}$ furnishes 1 mole of electrons.

In your balanced reaction, 2 moles of $\ce{S2O3^2-}$ gives 2 moles of electrons. To find n-factor, you need to calculate the amount of electrones given by one mole of $\ce{S2O3^2-}$.


Why does SO2 have a larger bond angle than H2O, H2S, and NH3

Here are the bond angles for each molecule (data from wikipedia):

\begin{array}{|c|c|}\hline
\mathrm{Molecule} & \mathrm{Bond \space Angle \space (^\circ)} \\ \hline
\ce{H2S} & 92.1 \\ \hline
\ce{H2O} & 104.5 \\ \hline
\ce{NH3} & 107.8 \\ \hline
\ce{SO2} & 119 \\ \hline
\end{array}

  So $L \propto \frac{1}{BA}$ where $L$ is the number of lone pairs and $BA$ is bond angle.

This is true but only in very specific situations; when dealing with molecules that have a central atom in the same period and outer atoms of the same element (e.g $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$). It breaks down as soon as you start comparing molecules with central atoms from different periods (e.g $\ce{PH3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle) or when you compare molecules with different outer atoms (e.g $\ce{NF3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle). For the reasons behind this I direct you to an excellent previous answer from @ron.

Additionally, I think you need to reconsider the number of lone pairs in $\ce{SO2}$, which can be described by these two resonance structures.

As you can see there is only one lone pair, but unfortunately this doesn't help us very much as we have no other molecule to compare it to.

  Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.

This is incorrect. In fact the reverse is true - that $BA \propto \frac{1}{ENC}$ - but only in the same situations as mentioned before. In this case the trend is only really pronounced in the second period; it is very slight in the third period and virtually non existent in the fourth. In fact, this 'rule' doesn't really have anything to do with electronegativity (that's just a coincidence) and it's essentially just a result of the first rule.

  Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom.

This is sort of true - it's known as Bent's rule and it can be very useful but it's not really applicable here. It's been discussed many times on this site but here and here are some good introductions.

So how should you answer this question?

The first thing to note is that $\ce{SO2}$ only has three 'groups' on the central atom (sometimes called 'effective electron pairs' in VSEPR theory) - two bonds intermediate between a double and a single bond and a lone pair - whereas all the other molecules have three. Therefore we expect $\ce{SO2}$ to have the largest bond angle of the four molecules, and this is indeed the case. $\ce{H2O}$ and $\ce{NH3}$ are hydrides of the same period so we can use the first rule to determine that $\ce{H2O}$ has a smaller bond angle. Now we just have to decide whether $\ce{H2O}$ or $\ce{H2S}$ has a smaller bond angle. We can apply the hybridisation arguments given by @ron in the answer I linked earlier to determine that $\ce{H2S}$ has the smallest bond angle, and indeed we find that it is almost unhybridised with a bond angle very close to $\mathrm{90~^\circ}$.


Why do batteries specifically vent *hydrogen* in the event of abuse?

Lead storage batteries use an electrolyte of sulfuric acid and water. On overcharging you are electrolysing the water with emission of $\ce{H2}$ in one electrode and $\ce{O2}$ in the other.

Other batteries use molecules with H and Li. The idea is to use light elements so you can accumulate more charge for the weight.


Is it correct to say H2SO4 behaves like an acid in this reaction?

In general, this depends on the definition of acid that you are using.

For this specific case, the Broensted definition would suffice.
A Broensted acid is a reagent that donates a proton ($\ce{H^+}$) to its reaction partner.

The protons of the sulfuric acid are transfered (donated) to the carbonate anion, which gives carbonic acid. Carbonic acid decomposes into water and carbon dioxide, as you will experience when drinking sparkling water.

The key thing now is the affinity to donate said proton, which leads to the concept of acid strength and can be expressed in terms of $K_\mathrm{a}$ values.

If the anions were reversed and $\ce{ZnSO_4}$ would react with $\ce{H_2CO_3}$, one would find no reaction occuring.

This is a very easy explanation of the things going on, and I'd suggest you read a bit into

Acid definitions: Broensted and Lewis acids
Acid strength: Equilibrium of dissociation reactions
Conjugated Acid / Base pairs


First the question doesn't really make sense. $\ce{CaF_2}$ is virtually insoluble, so a "solution of $\ce{CaF_2}$" is a really odd phrase.

Skipping this nit-picking, the reactants are 1 L of a $2.0 \times 10^{-5}$ M $\ce{Ca^{2+}}$ and a $0.1$ M $\ce{F^-}$ solution  and $\ce{CaF_2}$ is to be precipitated.

This is a sloppy use of significant figures since volume of 1L and the molarity of the $\ce{F^-}$ solution are only given to one significant figure.

Ignoring the significant figure problem too, let's work backwards from the given answer of 946 ml.

For the initial solution we can solve for the moles of $\ce{Ca^{2+}}$ present.

$\text{moles }\ce{Ca^{2+}} = 1 \rm{L} * 2.0 \times 10^{-5}\frac{moles}{L} = 2.0 \times 10^{-5}\text{ moles}$

Now if we add 946 ml of 0.1 M NaF, then the moles of $\ce{F^-}$ in solution is 0.0946 moles which vastly more than the calcium. So we can assume the precipitated $\ce{CaF2}$ has a negligible effect on the $[\ce{F^-}]$, and calculate the final concentration of $[\ce{F^-}]$ as:

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0946}{1.946} = 0.0486 \text{M}$

Now from the $\rm{K}_{\text{sp}}$ we have

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0486^2} = 7.20 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (7.20 \times 10^{-8})(1.946) = 1.40 \times 10^{-7}$

The initial moles of $\ce{Ca^{2+}}$ is $2.0 \times 10^{-5}$ which is +/- $0.05 \times 10^{-5}$. Thus so far as I am concerned the solution is a "magic number" and there is no way to get from the problem statement to that value. In other words why not $2.00 \times 10^{-7}$ or $3.00 \times 10^{-8}$ for the final $\ce{Ca^{2+}}$ concentration?!?

Thus you have to assume that you're going to precipitate some fraction of the $\ce{Ca^{2+}}$. So are you going to precipitate 99%, 99.9%, 99.95% or what? There just isn't anything in the problem statement on which you can pin a value.

Just out of curiosity the % $\ce{Ca^{2+}}$ recovered is:

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.40 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.3 \%$

Note that we can also assume that the original solution is saturated in respect to $\ce{CaF2}$. As shown in OP's post, this leads to a concentration of $2.91 \times 10^{-3} \rm{M}$ for $\ce{F^-}$ and given the 1L volume, $2.91 \times 10^{-3} \rm{moles}$ of $\ce{F^-}$. Adding that to the number of moles from the 946 ml of $\ce{F^-}$ solution gives a slightly different value for the total moles of $\ce{F^-}$:

total moles of $\ce{F^-}$ = $0.0946 + 0.00291 = 0.0975$

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0975}{1.946} = 0.0501 \text{M}$

So the magic value for the textbook solution is to assume that: $\ce{[F^-]}_{\text{final}}= 0.0500$

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0501^2} = 6.77 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (6.77 \times 10^{-8})(1.946) = 1.32 \times 10^{-7}$

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.32 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.34 \%$

So we get a slightly different value, but it still seems like a magic number of some sort.

Now if I were to write the problem I would have stated:

  You are given $1.00$ Litre of solution which is saturated with  $\ce{CaF_2}$ and has a $\ce{Ca^{2+}}$ concentration of $2.00 \times 10^{-5}$ M. How many mililitres of $0.100$ M $\ce{NaF}$ would you need to recover 99% of the $\ce{Ca^{2+}}$ as a precipitate of $\ce{CaF_2}$? The solubility product of $\ce{CaF_2}$ is $1.7\times10^{-10}$ .

Curious I played with this some more using the assumption of the $\ce{F^-}$ saturated starting solution. Obviously if you add vast amounts of $\ce{F^-}$ solution then any $\ce{CaF_2}$ which formed would dissolve agin. So there is some amount of $\ce{F^-}$ solution which precipitates the maximum amount of $\ce{CaF_2}$.

The max % $\ce{Ca^{2+}}$ recovery = 99.44% with 1913 ml of $\ce{F^-}$ solution. To +/- 0.01% any volume between about 1400 ml and 2500 ml would yield effectively the same result.


How to make dichloromethane human-friendly and safe for storage?

According to A Comprehensive Guide to the Hazardous Properties of Chemical Substances at pages 445-446:

Disposal/Destruction
...
Small amounts may be destroyed in the laboratory by hydrolyzing with ethanolic solution of potassium hydroxide


An organiker would look at it this way.

An inorganiker might look at it as literal N(+5) with two double-bonded oxygens.  It does redox as N(+5). Coordination structures are varied,


Hydrogen peroxide and ThickenUp (potassium chroride)

Hydrogen peroxide is an oxidizer and unstable.  It can react with a variety of common materials to exothermically release oxygen gas, exothermically oxidize the material, or both.  Additionally it can damage skin.  Fortunately these safety concerns are only really a major consideration at higher concentrations, where the amount of oxygen released can be problematic, and the heat is sufficient to pose a burning or fire hazard, made worse by the oxygen.  At lower concentrations (such as the 12% you're talking about), it poses mild hazard to skin, and significant hazard to clothes and eyes, but shouldn't present any major safety concerns.
The potassium chloride shouldn't do anything, but the hydrogen peroxide may react with the maltodextrin, or other unlisted ingredients.  To test for this, you can dilute down your hydrogen peroxide (or buy some medical hydrogen peroxide that is already lower concentration), and mix it with the TickenUp.  If bubbles or foam forms, that's an indicator that the hydrogen peroxide is reacting with the maltodextrin or some other ingredient in the TickenUp, and you shouldn't use it, because it will destroy your peroxide before it can bleach your plastic.
Of note, additional care should be taken to wash the plastic after using this form of Retrobright, because the maltodextrin and potassium chloride may leave an undesirable residue otherwise.  Additionally, the xantham gum is only required if you wish to apply the Retrobright as a gel, if you are planning on submerging the plastic, the xantham gum is unnecessary.


Equilibrium constant for heterogeneous equilibria having aqueous as well as gaseous reactants

Your conception of the equilibrium constant is flawed. $K_\mathrm{p}$ is preferably used when the reaction only contains gaseous and solid components whereas $K_\mathrm{c}$ is used primarily for  a reaction taking place in water.
In reality, however, both of these equilibrium constants are special cases of equilibrium. This is because in the actual scenario, what we use to calculate the equilibrium constant is a concept known as activity.
$K_\mathrm{p}$ and $K_\mathrm{c}$ are said to be special cases of $K$ since, in $K_\mathrm{p}$, the activity is expressed in terms of the partial pressure of the components and in $K_\mathrm{c}$, the activity is expressed in terms of concentrations of the products and reactants.
In case of reaction as stated above:
$$\ce{A(aq) + B(aq) <=> C(g) + D(aq)}$$
The equilibrium constant would be expressed as follows:
$$K = \frac{a_\mathrm{C}\cdot a_\mathrm{D}}{a_\mathrm{A}\cdot a_\mathrm{B}}$$
Now, for the aqueous components, the activity is expressed in terms of their concentration in the solvent. For the gaseous component, we express it in terms of its partial pressure. Substituting these conditions, we get:
$$K = \frac{P_\mathrm{C}[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}$$
An example of where this is used is in electrochemistry.
A galvanic cell may be made which has a heterogenous reaction in which solid, aqueous and gaseous components may be given.
The Nernst equation used to calculate the EMF of such a  cell is given as:
$$E_\mathrm{cell} = E^\circ_\mathrm{cell}-\frac{2.303RT}{nF}\log_{10}Q$$
Here, $Q$ is the reaction quotient. Using this, we calculate the  EMF taking all the activities into consideration.
Note: The activity of a solid or the solvent is taken to be $1$ in such cases.


Why sodium is the worst reducing agent in group 1?

The general trend in the reactivity of the group 1 elements is that it increases down the group. This can be attributed to the decrease in 1st ionisation energy as the atomic radius increases.

Reducing power can be measured by the standard electrode potential E0. This refers to the half - cell:

$$M^+(aq)+e⇌M(s)$$

If you look at the values for group 1 we see a different pattern:

Electrode Potential (V) Li. -3.02 Na. -2.71 K. -2.92 Rb. -2.99 Cs. -3.02

Lithium has an anomalously large and negative electrode potential. After Li the values become more large and negative, indicating, as you rightly say, that sodium has the lowest reducing power.

The reason for the unusual value for lithium is that E0 values are measured in solution whereas ionisation energies are measured in the gaseous phase.

To turn a metal atom in the solid state into an aqueous ion we can think of a 3 step process:

The metal is atomized. $\ce{Li(s)→Li(g)}$

The metal is ionised. $\ce{Li(g)→Li+(g)+e}$

The ion is hydrated. $\ce{Li+(g) +(aq)→Li+(aq)}$

Steps 1 and 2 are both endothermic, i.e. they require energy. Step 3 is exothermic as it is a bond forming process. It is referred to as the enthalpy of hydration.

The small size of the lithium ion gives it a large charge density and so water molecules are strongly attracted giving a large, negative enthalpy of hydration.

This is the main factor responsible for lithium's large, negative electrode potential.

In aqueous conditions it is as reducing as cesium though much less reactive in anhydrous conditions.

This anomaly accounts for sodium having the lowest reducing power, but only in aqueous conditions.


How much pressure is needed to make dry ice?

Isn't that a bit different?

At 293 K, you need about 56 bar to liquefy carbon dioxide. If you then release the pressure, a part of the liquid evaporates again while cooling down the rest until it solidifies at around -80 °C.


Why are entries missing on a solubility data chart for ionic compounds?

In the case of magnesium chromate, good solubility in water has been known since 1931.  From Hill et al. [1], citing [2]:

The only figure on the solubility of the salt [as of 1940] appears to be that of Kohlraush[2], who reported that at 18°[C] 100 ml of saturated solution contained 60 g of magnesium chromate.

Most of the other compounds listed, as described in comments to the question, do not exist or are highly unstable because the ions are incompatible in the proposed solid. Iodide and sulfide ions are oxidized by iron(III), iodide is also oxidized by copper(II), chromate ions are reduced by iron(II) and tin(II), and tin(II) also reduces nitrate.  Carbonate ion is not redox-active, but fails to form a stable salt with some weakly-basic or amphotetic oxide forming metal ions like tin(II) or iron(III); a similar failure occurs with proposed silver hydroxide.
References
1.
Arthur E. Hill, Glenn C. Soth, and John E. Ricci, "The Systems Magnesium Chromate—Water and Ammonium Chromate—Water from 0 to 75°", J. Am. Chem. Soc. 1940, 62, 8, 2131–2134. https://doi.org/10.1021/ja01865a059
2.
Joseph Mellor, Treatise on Inorganic and Theoretical Chemistry, Vol. II (London: Longmanns, 1931), p. 275.

Besides missing the reference on magnesium chromate, the chart has some errors associated with oxides and sulfides.
Magnesium oxide is listed as insoluble in water, which it is if it is calcined at a high temperature, but otherwise it becomes (kinetically) reactive forming the hydroxide; see this answer. Barium oxide is not "soluble", but reacts in the same way as calcium, strontium, and low-temperature calcined magnesium oxides.
Among sulfides, the $\ce{S^{2-}}$ ion is a relatively strong base, so the alkali and alkaline earth metal sulfides that "dissolve" in water are in fact all prone to hydrolysis.  Calcium and magnesium sulfides do undergo more complete hydrolysis, as the corresponding hydroxides have limited solubility and their precipitation limits the basicity of the solution. Sulfide ion is also too strongly basic to coexist with ammonium ion, so ammonium sulfide as $\ce{(NH4)_2S}$ properly does not exist.


Mechanism for conversion of imide to ester with NaH

NaH is a strong base but a bad nucleophile. This is one possible mechanism.


Reaction between Iron (III) Oxide and CO?

The reaction products depends on stoichiometric ratio of the reactants as well as reaction conditions (i.e temperature, pressure etc.)

$$\ce{Fe2O3 + CO ->[500-600 C] 2FeO + CO2}$$

  Iron(III) oxide and carbon monoxide to produce iron(II) oxide and
  carbon dioxide. This reaction takes place at a temperature of
  500-600°C. (source)

$$\ce{3Fe2O3 + CO ->[400 C] 2Fe3O4 + CO2}$$

  Iron(III) oxide react with carbon monoxide to produce iron(II,III)
  oxide and carbon dioxide. This reaction takes place at a temperature
  near 400°C. (source)

$$\ce{Fe2O3 + 3CO ->[700 C] 2Fe + 3CO2}$$

  Iron(III) oxide react with carbon monoxide to produce iron and carbon
  dioxide. This reaction takes place at a temperature near 700°C. (source)


Spraying aluminium hydroxide nanoparticles onto PLA as fire retardant?

I suspect this is too late. The whole point of embedding the nanoparticles inside PLA is that it remains inside until needed (when temperatures rise above ~300 °).

By applying it as a coat, it can be easily scratched or rubbed off because PLA is so soft or flexible. You are also running into the problem of shedding of Al(OH)3 NP everywhere which may be a problem.

A coat of sprayed Al(OH)3 NP is only going to be a few micrometres thick, or you will need to use some glue to make them stick, which complicates thing and changes the appearance of your product. On the other hand, embedded Al(OH)3 NP are present throughout the entire thickness of the product. During fire, your sprayed NP might dehydrate just from the heat of radiation without even coming into contact with fire.

This is a bad solution.


Confusion about direction of dipole arrow in alpha-helices and other molecules

I (accidentally) stumbled upon the following statement in Atkins' "Elements of Physical Chemistry" (p378):

  We represent dipole moments by an arrow with a length proportional to $\pmb{\mu}$ and pointing from the negative charge to the positive charge (1). (Be careful with this convention: for historical reasons the opposite convention is still widely used.)

Unfortunately he does not go into more detail. And I know this does not really answer your question.

The definition from the IUPAC is the same as the one used by Atkins:

  electric dipole moment, $\mathbf{p}$
  Vector quantity, the vector product of which with the electric field strength, $\mathbf{E}$, of a homogeneous field is equal to the torque.  $\mathbf{T} = \mathbf{p} \times \mathbf{E}$. The direction of the dipole moment is from the negative to the positive charge.

The source quoted there is from 1993, so you can probably understand my surprise, when I did a little more searching and found in C. Párkányi's "Theoretical Organic Chemistry" (1997, p239):

  [...] in organic chemistry the positive direction of the dipole moment is normally defined as the direction from the center of the positive charge towards the center of the negative charge. This convention prevails in physical organic chemistry and in inorganic chemistry. However, while the dipole still points from the positive charge to the negative charge, in physical chemistry and in chemical physics the positive direction of the dipole moment is defined in the opposite way, i.e., from the negative charge to the positive charge.

There are also a few sources given, but I currently have not enough time to look them up. I find this statement (definition) highly confusing and it also does not give another reason why this is the direction it uses. Please forgive me for not going into more detail here, I really don't want to add any more to the confusion.

Conclusion
Don't add to the confusion. Use
$$\huge\ominus \overset{\mathbf{p}}{\longrightarrow}\oplus$$
as your definition from now on. Popular use does not make it right. Help using it consistently in the correct way and flush out the historic remnants that are still being taught. However, keep Atkins' warning in mind when you read books, paper, etc.. In the literature you will find both versions.

If anyone wants to argue with you about this, give the following derivation. The dipole moment operator $\mathbf{P}$ is a vector operator  that is the sum of the position vectors $\mathbf{r}$ of all $N$ charged particles weighted with their charge $q$.[goldbook]
$$\mathbf{P} = \sum_i^N q_i \mathbf{r}_i$$
For a molecule (neutral by definition)
we find
$$\begin{align}
  0& =\sum_i^N q_i &\Leftrightarrow&&
  0&=\sum_{i_+}^{N_+}q_{i_+} + \sum_{i_-}^{N_-}q_{i_-} &\Leftrightarrow&&
  \sum_{i_+}^{N_+}q_{i_+} &= - \sum_{i_-}^{N_-}q_{i_-} &.
\end{align}$$
This can be transformed into
$$\sum_{i_+}^{N_+}|q_{i_+}|\mathrm{e} =
- \sum_{i_-}^{N_-}|q_{i_-}|(\mathrm{-e}).$$
Therefore we can write
$$\begin{align}
  \mathbf{P} &= \sum_{i_+}^{N_+} q_{i_+} \mathbf{r}_{i_+}
             + \sum_{i_-}^{N_-} q_{i_-} \mathbf{r}_{i_-}
             &\Leftrightarrow&&
  \mathbf{P} &= \mathrm{e}\left(
                \sum_{i_+}^{N_+} |q_{i_+}| \mathbf{r}_{i_+}
              - \sum_{i_-}^{N_-} |q_{i_-}| \mathbf{r}_{i_-}
                \right)&.
\end{align}$$
In the parenthesis the first term is a linear combination of vectors for all positive charges with the resulting vector $\mathbf{r}_+$ and the second term is a linear combination for all negative charges with the resulting vector $\mathbf{r}_-$. The dipole operator is therefore equivalent to
$$\begin{align}
  \mathbf{P} &= \mathrm{e}\left(\mathbf{r}_{i_+} - \mathbf{r}_{i_-}\right),
\end{align}$$
which is a linear combination of vectors that points from negative to positive.


Why is the gram no longer a function of the atomic mass number?

Unless I misunderstand the question, the gram was never defined that way.

For historical reasons the SI unit for mass is an oddity:

The base unit is not the gram, but the kilogram. This is the only case where the base unit has a prefix. The gram is defined as $\frac{1}{1000}$ of a kilogram.
The kilogram is the only remaining SI unit that is based on a man-made object, the International Prototype of the Kilogram.

At its 25th meeting in 2014, the General Conference on Weights and Measures (CGPM) has (again) declared its intention to abandon this atificial standard and link

  the definitions of the kilogram, ampere, kelvin, and mole to exact numerical values of the
  Planck constant $h$, elementary charge $e$, Boltzmann constant k, and Avogadro constant $N_A$ […]

(Source)

One of the possible approaches for the redefinition of the kilogram would be a side-effect of the redefinition of the Avogadro constant with in the Avogadro project.


When disolving a iron pill for titration, why do I use sulfuric acid instead of nitric or hydrochloric acid?

In this case, cerium(IV) sulfate is a strong oxidising agent, hence, using $\ce{HCl}$ would likely be oxidised to form $\ce{Cl}$ of higher oxidation state. Therefore, the titre would be vastly higher than expected.

$\ce{HNO3},$ on the other hand, is an oxidising agent and would oxidise $\ce{Fe}$ to a higher oxidation state than expected (of $\ce{Fe^3+}$). Hence, not as much cerium(IV) sulfate would be needed to reach the endpoint, resulting in a titre lower than expected.

Therefore, $\ce{SO4^2-}$, being a relatively unreactive species, is the best reactant in this case.


Does the amine group participate in resonance in aniline?

The lone pair of nitrogen is also part of the extended π-system; the molecule itself is still aromatic. As I already explained in the question about fulvene, aromaticity is nothing you can determine based on simple rules, especially not counting electrons.
Structural indicators like bond lengths and angles give a good idea how well established resonance is. Additionally you can draw a qualitative molecular orbitals diagram and look if it is similar to benzene. Eventually the determination of aromaticity is an experimental one, for example with NMR spectroscopy.

From the crystal structure (M. Fukuyo, K. Hirotsu, and T. Higuchi, Acta Cryst. 1982, B38, 640-643. DOI: 10.1107/S056774088200363X) we know that the carbon-carbon bond lengths are all similar and between 136 and 141 pm. The NMR spectrum clearly shows the protons have aromatic shift, see chemicalbook.com.

The molecular orbitals of aniline (calculated at DF-BP86-D3BJ/def2-SVP; Gaussian 09 D.01) show that the nitrogen is part of the π-system. They are also very similar to benzene. Here are the orbitals of the π-system in decreasing energy. Virtual (unoccupied) orbitals are red/yellow, and occupied orbitals are blue/orange.


What is the right quantum numbers of Copper?

The correct quantum numbers for last electron of Copper ($\ce{Cu}$) would be $B. n=3,l=2,m=+2,s=-1/2$.

  if I use B quantum numbers, it is the same with the last electron of zinc.

Yes, last electrons of Both can have same quantum numbers because Pauli Exclusion law says that no two electrons in an atom can have all four quantum numbers same but it doesn't say that two electrons of different atoms can't have all four quantum number same.

Refer this for more.


Half-life equation for 2nd order kinetics

Roughly speaking, half-life is not a thing at all for the second order (or any order other than first, for that matter).

What would they do after finding the number of half-lives? Raise 2 to that power? Too bad, because that only works for the exponential decay, which is the solution of the first-order kinetics and no other order. Really, what if the number of half-lives equals 2? After spending the first half-life, we now have a different half-life ahead of us, because it is concentration dependent, and the concentration has changed.

Following your link, we see the same statement, except they don't put enough emphasis on it, as to my taste:

  For this reason, the concept of half-life for a second-order reaction is far less useful.


Would cooling of a battery using cooler affect its life time?

Let's see what I know. Not really sure whether this is a terrific answer to your question, but it's what I have.

Through knowledge of using laptops, I know that lithium-ion batteries will decrease in performance when overheated. This probably applies to most redox reaction batteries for the reason that the reaction proceeds more quickly when temperature is increased, and as a consequence, kinetic energy of the molecules are increased.

Increasing the temperature of the battery also increases internal resistivity, which is somewhat more of a physics question. The electrons collide with the material that they are flowing through much more frequently in a higher temperature situation as opposed to a lower temperature situation. This doesn't mean, however, that you can dip your phone in LN2 and have it work forever. There's obvious physical limitations to everything.

What you're doing to your phone (a.k.a. cooling it) shouldn't have a problem at all on your phone's performance, mainly because you're doing to it what others do to their laptops - placing it on a cooling pad.


Why are oxyacids strong oxidizing agents?

Your premise is incorrect. Some oxyacids are oxidising agents. Some are not. Take e.g. carbonic acid. You cannot oxidise anything with carbonic acid, really.

Also, the choice of words is bad. Sulphurous acid (or hydrogen sulphonic acid; it doesn’t really make a difference; both are structural isomers) is also an oxyacid ($\ce{H2SO3}$) however it is a pretty good reducing agent oxidising itself to sulphuric acid. Heck, technically $\ce{H4SiO4}$ is an oxyacid, too, but it won’t oxidise anything at all.

So reducing the question to where it is correct we arrive at:

  Why are some oxyacids where the central atom is in its highest possible oxidation state strong oxidising agents?

And it basically answers itself. Depending on the electronegativity of the central atom it will be more or less ‘happy’ to cede all its valence electrons to oxygens and be the centre of an oxyacid. For those that don’t like that lot in life, most notably nitric acid, they will often find species whose electrons they can take back (the atoms are electronegative after all) and thus oxidise them.


Is it possible to synthesise ethers from carboxylic acids?

This is an interesting question. I can tell you that Wolff-Kishner won't work - both the hydrazine and the hydroxide required will attack the ester in a substitution fashion (with a loss of -OR'). Other reductions might work.

For example a Google search for "reduce ester to ether" brought up the following:

The reaction appears to be complete within 3 hours with moderate to good (but not stellar) yields. The abstract and key figures are available for free at www.organic-chemistry.org. The paper is published in the Journal of Organic Chemistry.

It appears that $\ce{Et3SiH}$ is the terminal reductant, but most of the work is being done by indium. The combination of $\ce{Et3SiH}$ and $\ce{InBr3}$ produces the active reductant $\ce{HInBr3}$, which provides the equivalent of $\ce{H.}$. The second $\ce{H.}$ comes from $\ce{Et3SiH}$. The proposed mechanism from the paper is as follows:

Other mild sequential one-electron reductions with a source of $\ce{H.}$ would also likely work. The two-electron reductions, like Wolff-Kishner ($\ce{NH2NH2 +NaOH}$) and hydride-transfer (e.g. $\ce{LiAlH4}$), are all heavy-handed and nucleophilic with the tendency to take apart or over-reduce the ester.


How to make solution of specific concentration with non-stoichiometric hydrate?

Anhydrous Amiloride HCl is commercially available and is stable in air.  This suggests that you should be able to remove the waters of hydration, however many that may be, by repeatedly heating in a drying oven until a constant weight is reached.  If you know the weight before drying you could of course calculate the original degree of hydration after the final drying, if that is of interest.

If you need to further verify dryness, you could have the dried product analyzed by Karl Fischer water anaysis (~$50). Just be sure to seal the sample under a dry atmosphere and label "HYGROSCOPIC" all over the sample and submittal paperwork!

Once you have the constant-weight product, stored under a good desiccator, just use the formula weight of the hydrochloride salt, 266.1, to calculate the concentration of your Amiloride HCl solutions.


Usual way to work with air sensitive compounds is to work on Schlenk-line (search Wiki). When you transfer your substance you can leave it in Schlenk tube, or if it's really sensitive and you need to keep out of the line for longer period of time you have in general three possibilities: (1) use glassware with J. Young valve (available even for NMR tubes); (2) store your standard Schlenk glass in glove box filled with inert gas and (3) the oldest is to melt your glass.

Now back to your question. Usual way is to prepare glass with narrow place on it prior to doing experiment (that's were later you will melt it). By narrow I mean the hole inside should be ~2-3 mm (thinner glass, smaller hole). I used to use Schlenk lines, so on opened end I have standard 14 female cone that was connected to a vacuum line. I used to set vacuum ~0.1 Torr and with propane/butane/oxygen torch one can easily close it. The huge advantage of this old way is that you can prepare your glass even with 3mm thick walls that allows you to work with substances with very low boiling points (in this case to close the glass one need to freeze the stuff with liquid N2 or if not available at least cool it with dry ice bath). Of course later opening is quite tricky and usually also had to be done after freezing of chemicals.

Last part if it's advisable. That one is as usually hard to answer. I have no idea how experienced you are. Judging by question not to much (sorry). So I wouldn't start with anything to reactive in the beginning. Generally try to play first with glass and torch and get a bit used to it. Do it on proper chemical glass (not just bottle glass).
If you are "advanced guy" and you want to make yourself f.e. PfBF2 borane you have to use this technique - any other gives just a miserable yield.


Why is the major product of the reaction of 1-naphtol with benzene diazonium chloride not the ortho product?

When benzene diazonium chloride attacks the ortho position with respect to -OH it(the benzene ring) experiences a lot of steric hindrances,along with that,the triple bond also repels the lone pairs of oxygen,therefore the nitrogen atom will attack para with respect to the -OH group.If the temperature is increased i think that an ortho product will be formed as it is the thermodynamicly stable product,while the para product is the kinetically stable product.


What is the correct way to transfer large volumes of chemicals?

Given that the chemicals in question aren't air sensitive (i.e. theres no need for an inert atmosphere), a measuring cylinder would suit your needs.

They're available in glass and various plastics, in sizes upto around 3000 mL. The glass ones are generally better as theres less risk of the chemicals reacting with the glass compared to with plastic, but it really does depend on what you're using.

Once you're done with the measuring cylinder, just rinse it with water (into a suitable toxic waste container if the chemicals cannot go down the drain, which I suspect they can't), and leave it to dry somewhere (else put it in a warm oven for an hour or so).


Why is oxygen more stable than oxygen dication?

You have to consider the system as a whole - you can't directly compare $\ce{O_2}$ and $\ce{O_2^2+}$ because they have different numbers of particles. To put it another way when you consider the relative stability of two interconvertible specifies you really have to write down a chemical reaction that connects them, and then consider which direction is thermodynamically favoured, and then consider whether it is kinetically feasible. Stability is a really tricky and more often than not misused word!
So here we have to look at $$
\ce{O_2 -> O_2^2+ +2e^-}
$$
As you note $\ce{O_2^2+}$ has a stronger bond than $\ce{O_2}$. But that is not all that is going on. We are also binding two electrons to the Oxygens - how does that affect the energetics? Well the reaction above is simply the sum of the following processes
$$
\ce{O_2 -> O_2^+ +e^-}
$$
$$
\ce{O_2^+ -> O_2^2+ +e^-}
$$
In other words the energy required is the sum of the first two ionisation energies of molecular oxygen. Do we know these? Well after a good 10 seconds looking I can't find the second energy, but the first is very well known, it is after all related to the discovery of compounds of Xenon and the start of noble gas chemistry. It is 12.0697 eV (from https://webbook.nist.gov/cgi/cbook.cgi?ID=C7782447&Mask=20) which in more familiar units is 1165 kJ/mol - a huge energy, bigger than any bond energy I am aware of, and note there is at least this much to come as the second ionisation energy will be even larger, as it is pulling a negative electron away from an already positively charge centre. Thus the energy gained by attaching the two electrons will massively out weigh any increase in bonding in the molecule.
Can we relate this to MO theory? Well as Koopman's theorem tells us the ionisation energy is an approximation of the orbital energy, this is telling us that relative to having a free electron it is more energetically favourable to put the electron in the lowest energy unoccupied orbital in the molecule. That this orbital is antibonding is irrelevant. Antibonding simply means it weakens any bonding in the molecule; it doesn't say anything about the energy of that orbital relative to the free electron. And you know this must be true - many molecules have electrons in antibonding orbitals and those electrons don't just rush away to infinity, hence occupying the orbital must be a lower energy proposition for the electron compared to being unbound. Thus $\ce{O_2}$ is thermodynamically more stable than $\ce{O_2^2+}$ and two free electrons because the decrease in energy caused by the electrons occupying the highest unoccupied molecular orbitals is (much) greater than any possible decrease in the bond energy.
That's the thermodynamics. Kinetics is difficult in general, but very unlikely to be an issue here. At this point I chicken out.


Trend in ionisation potential in period 2

Emperically speaking, Be is on the "metallic" side of the table. Lower IE than non-metals is obviously expected.

A quick glance at the $IE_1$ trends for period II tells us that like the OP noted, Be certainly has an abnormally high $IE_1$, deviating significantly from the expected plot, due to the fully filled configuration.

However, Carbon (Z=6) has a higher nuclear charge(=more protons) than Be (Z=4), a smaller radius and hence a stronger hold on its valence electrons.

Even Nitrogen displays an exception, due to half-filled $2p$ subshell, as evident in the plot.

Just because of a tiny bump in the plot, expecting N and Be to have higher IE than all the elements of the periodic table is clearly an exaggeration.


You have two errors in your code, one minor and one major.
The minor error is in your multiplication function. The value you return should have $\pi^{3/2}$, but the way it is arranged now the power is only being applied to the denominator of that expression.
The major error is how you are normalizing your overlap integrals. You can't simply divide $\langle1|2\rangle$ by $\langle1|1\rangle$ in order to normalize it. You have to explicitly normalize each of the Gaussian basis functions. To do this you can include a factor of ((4*self.alpha*other.alpha)/np.pi**2)**(3/4) in the value you are returning from your multiplication (in terms of readability of your code, it may be wiser to define variables within the function like proportionality_constant, gauss_integral, and normalization and then return the product of these). Or you could simplify and just multiply the exponential by ((4*self.alpha*other.alpha)/(self.alpha+other.alpha)**2)**(3/4) This also means you only need the second part of your overlap function, since the multiplication will take care of the normalization.
The reason your code works for STO-1G is that dividing by $\langle1|1\rangle$ accidentally gives the correct normalization since there is only one normalization constant associated with the one Gaussian function of STO-1G.


Why are s orbitals more “penetrating” than p orbitals?

I found an old slide from when I was teaching.
These are plots of the 2s and 2p orbitals for hydrogen (assuming 1 electron). The radial wave function is on the left and the probability density as a function of distance from the nucleus is on the right ($4\pi r^{2}R(r)dr$ if you understand the calculus).
You can see that there's a blob of density that comes closer to the nucleus for the s orbital. There's your penetration.


What are the practical differences between the various DMSO oxidations (Swern, etc.)?

tl;dr:
Swern and Parikh–Doering are by far the most practically useful and widely used oxidations. Whilst both have slight differences, the reality is that most research groups have a preference. This may be because they just have the reagents on hand (distilled oxalyl chloride vs good quality sulfur trioxide–pyridine), or because someone had a bad experience with one over the other.
The Anderson group (in the post you linked) commonly uses the Parikh–Doering conditions, whilst other groups appear to favour the Swern conditions using oxalyl chloride in place of sulfur trioxide–pyridine.

A brief overview of various DMSO oxidations and their drawbacks:
One thing to mention about all of these DMSO based reactions is the formation of stoichiometric dimethylsulfide (stench) as a by-product, which increasingly makes them avoided, especially by process chemists. That said, the reagents are cheap (compared to a Dess–Martin, for example), and the reaction is generally very reliable and amenable to use on moderate scales.  The DMSO and triethylamine must also be distilled for optimal results (which can be a drawback, especially if large scales are needed).
1. Moffatt oxidation
Reagents: DMSO, DCC, PPA.
Reference: Pfitzner, K.; E.; Moffatt, J. G. A New and Selective Oxidation of Alcohols. J. Am. Chem. Soc. 1963, 85 (19), 3027–3028. DOI: 10.1021/ja00902a036.
The Moffatt oxidation was one of the earliest examples of a DMSO-mediated oxidation. Two issues generally exist. Firstly, a strong acid (polyphosphoric acid, TFA, etc.) is required to get the reaction going, and, although the reaction becomes basic as it proceeds, the initial conditions using polyphosphoric acid/similar quite often have unwanted side effects (stripping of protecting groups, from experience). Secondly, and more importantly from a scale point of view is the stoichiometric N,N'-dicyclohexylurea produced as a by-product (similar to the Steglich esterification).
Due to the difficult-to-remove stoichiometric by-products, and the harsher conditions, Moffatt oxidations have largely been forgotten/ignored. Reagents such as EDCI have been introduced, however the high cost relative to oxalyl chloride/sulfur trioxide–pyridine have limited the appeal of Moffatt oxidations.
2. Albright–Goldman oxidation
Reagents: DMSO, acid anhydride
Reference: Albright, J. D.; Goldman, L. Dimethyl Sulfoxide-Acid Anhydride Mixtures. New Reagents for Oxidation of Alcohols. J. Am. Chem. Soc. 1965, 87 (18), 4214–4216. DOI: 10.1021/ja01096a055.
The Albright–Goldman oxidation commonly uses acetic anhydride, and goes at room temperature - although with significant reaction times which can give side reactions time to occur.
The biggest drawback, and reason why the Albright–Goldman is rarely used, is due to a competing Pummerer reaction (which is often significant), along with protection of the alcohol as its acetate ester if unhindered.
Some improvements have been made using TFAA in place of acetic anhydride. However, mixing TFAA and DMSO causes a violent exotherm unless the temperature is carefully and cautiously controlled.
3. Parikh–Doering
Reagents: DMSO, sulfur trioxide–pyridine, triethylamine
Reference: Parikh, J. R.; Doering, W. v. E. Sulfur trioxide in the oxidation of alcohols by dimethyl sulfoxide. J. Am. Chem. Soc. 1967, 89 (21), 5505–5507. DOI: 10.1021/ja00997a067.
Along with the Swern (see below), the Parikh–Doering is one of the most widely used DMSO based oxidations, with many examples in the synthetic literature.
The Parikh–Doering is able to be ran at mild temperatures (0 °C to room temperature), and the by-products are easily removed (crashing out and filtration). One advantage to the Parikh–Doering over the Swern is the fact that sulfur trioxide–pyridine is a bench stable solid, compared to oxalyl chloride which is a liquid that must be distilled and stored in the freezer to prevent degradation.
4. Swern oxidation
Reagents: DMSO, oxalyl chloride, triethylamine
Reference: Omura, K.; Swern, D. Oxidation of alcohols by “activated” dimethyl sulfoxide. a preparative, steric and mechanistic study. Tetrahedron 1978, 34 (11), 1651–1660. DOI: 10.1016/0040-4020(78)80197-5.
The Swern is the newest of the widely used DMSO oxidations, and often represents the best combination of reliability, cost of reagents, and ease of reaction.
Unlike the Parikh–Doering, the initial stages of the reaction must be ran at −78 °C to avoid unwanted decomposition. However, the conditions are equally as mild. One advantage over the Parikh–Doering is that oxalyl chloride decomposes into gaseous by-products, meaning that the only thing needed to be removed is the hydrochloride salt of triethylamine. Swern reactions can often be filtered, concentrated, and telescoped into the subsequent reaction without attrition in yield.

References:
The majority of the chemical information here comes from Comprehensive Organic Synthesis, Vol. 7. A whole chapter is dedicated to DMSO based oxidations and their examples in total synthesis: see Lee, T. V. Oxidation Adjacent to Oxygen of Alcohols by Activated DMSO Methods. DOI: 10.1016/B978-0-08-052349-1.00191-8.


If we use Magnesium instead of the sodium with alkyl halide in dry ether then will a reaction similar to Wurtz reaction will occur

In 1930, Palmer and Kipping1 reported the formation of cyclohexyl magnesium bromide in ether in 50-60% yield. They did not make an effort to detect the coupling product, "bicyclohexyl" [choice (b)] in the Grignard reagent itself. Subsequent reaction of the Grignard reagent with SiCl4 led to the isolation of bicyclohexyl and some cyclohexene. Although the authors did not speculate on the source of these products, it is likely that they are part of the 40-50% of the "missing" Grignard reagent formed by Wurtz-like coupling and elimination, respectively.

The following year, Gilman and Zoeller2 reported their results on the relative merits of forming cyclohexyl magnesium bromide and chloride. Cyclohexyl chloride gave generally higher yields than the corresponding bromide, and under their conditions, higher yields than what Palmer and Kipping had obtained.

To answer your question, it appears that both bicyclohexyl and cyclohexene may be by-products during the formation of cyclohexyl Grignard reagents.
1) K. W. Palmer and F. C. Kipping, J. Chem. Soc., 1930, 1020.
2) H. Gilman and E. A. Zoellner, J. Am. Chem. Soc., 53, 1931, 1945.


@bon is right in that you are talking about molecular orbital theory. Speaking broadly, the redistribution of electrons the way you mention happens if the orbitals actually become energetically non-equivalent. This can happen if an atom is surrounded by ligands in certain way. Because orbitals have certain orientations, certain complexes (f.e, tetrahedral) will energetically be more favorable than others (f.e, octahedral).

A simple example is the Jahn-Teller effect: have a look here and here. Essentially, for low-spin d$^7$ configurations you will have something resembling your example, although those are d orbitals; I actually do not know if this happens with p orbitals that you use in your example.

This Wikipedia article on crystal field theory also provides a nice overview of the different ways the orbital energy levels can distort in. However, for better understanding, I suggest a good textbook (I can give a few suggestions if you're interested).


Which software is suitable to calculate the root mean square deviation between two protein structures?

All of these are freely available packages:

Visual Molecular Dynamics - good visualization tool that can also do quick-and-dirty calculations from a graphical interface. The scripting interface has a learning curve, but yes, it can be quite powerful.

Avogadro - not as powerful as VMD, but can be useful for certain things.

Rasmol - Similar to VMD, although I've never used it.

LAMMPS, Gromacs, or NAMD - If one of these is installed on a computer you have access to, each of them can do this sort of computation and will allow you to do more advanced things if needed (they are actually molecular dynamics simulators). Each of these packages requires some familiarity with running code from the command-line and will have a learning curve.

Alternatively, you could always write your own code. RMSD is a pretty straightforward computation, and it's always nice to have more control over your analysis!


"Raw" latex consists of long hydrocarbon chains. The chains themselves have covalent bonds and are therefore very strong; imagine them like hemp threads. A volume of latex is basically a huge mess ("ball of wool") of those chains, which are weakly held toghether by van der waals forces.
This raw latex is therefore a thermoplast: It is soft and sticky when hot and rather brittle when cold. This must have been rather annoying when only this form was known. It is also not very elastic but malleable,  (is that the word?)

But now it was discovered that sulfur likes to form cross-connections between said chains. This basically makes the latex a single, huge molecule (hence a duroplast). But there are not too many of those cross connections, and the material looks rather like a loose net. If this vulcanized latex is streched, it can regain its form rather well; it is therefore called an elastomer. The more sulfur-bridges there are, the harder is the material; i.e. a hair band has less bridges than an ice hockey puck.

I think this view should explain the behaviour of latex quite well and on an intuitive level.

Testing for crosslinks seems rather difficult in general; but in the case of latex you can test for sulfur however. If you only have an emulsion this is very easy, if it smells repugnant there is sulfur in it. If you have solid material you might as well just test for elasticity.

In general it seems a save bet that elastic materials have crosslinks.


What are protecting groups in organic chemistry?

Functional groups become protecting groups out of practicality.

I would define an ideal protecting group (PG) as a functional group that is

easy to introduce/attach to a given, other functional group (the protection target)
stable/non-reactive, i.e. protecting, under a wide range of reaction conditions
easy to remove by well-defined and mild reaction conditions
with high yields throughout

Clayden/Greeves/Warren/Wothers have referred to PGs as functional groups with an Achilles' Heel.[1] Obviously, the usefulness of a given PG is always dependent on the target molecule and the reactions it will undergo. The "easy" part of the definition also encompasses factors such as price and toxicity.

As a corollary, one can introduce the concept of orthogonal PGs. Since the removal conditions can be very different for different PGs, one can successively attach PGs to, say, the hydroxyl groups of a sugar. By later removing them one by one, a chemist is then able to target them one at a time.[2]

Why do protecting groups tend to be large?

I am not convinced this is entirely the case, because I have seen applications of the methyl group as a PG. However, a larger group can introduce more steric hindrance or a point of attack for the removal that is different from the point of attachment to the protection target.

[1] Clayden, Greeves, Warren, Wothers: Organic Chemistry. Oxford University Press, 2004 (page 635).

[2] Werz, Seeberger, Angew. Chem. Int. Ed., 44, 6135, (2005). DOI:10.1002/anie.200502615.


Why isn't pressure filtration popular compared to vacuum filtration?

It's much easier to assemble a system that's safe under external pressure than one that's safe under internal pressure. Convex flasks and tubes resist external pressure better than internal pressure. If something leaks or breaks, you have air rushing into the system, rather than reactants and products blasting out. For that matter, if you've put things together with ground glass joints or rubber stoppers or corks, internal vacuum (external pressure) works to keep them together, while pressurizing the system's interior tends to force them apart.

Also note that a vacuum system will never be under more than one atmosphere of external pressure. With a pressurized system, you need a way to limit the pressure to safe levels, and you need to make sure you know what those levels are.


How can I get big bubbles when I electrolyse water with detergent?

This might be tricky. The demos I've seen mostly rely on collecting the hydrogen and oxygen, separately or together, and then bubbling them through a soap solution.
If you've got the soap in the electrolyte, you might have a hard time getting anything more than two piles of foam, one of which is slightly flammable. I doubt the hydrogen foam would even rise in air; the volume percentage of gas is too low.


Neglecting any resistance in the connecting wires, this situation is no different than having one large copper plate. If everything is connected together, you can't measure any voltage because the fermi levels of all the metals will be equal (at equilibrium). This is essentially shorting the terminals of a battery and the zinc will start to dissolve in the electrolyte and hydrogen will be produced at the copper electrodes. You can measure a potential difference between the copper plates and the zinc plate, if they're not shorted, but it won't be any different with one copper plate or two for an open circuit.

If you're actually using the cell to do work, you may get more work out of the cell with more copper area, as the larger surface area will result in less solution resistance and better diffusion.


Why, in titration, is universal indicator not commonly used?

The thing with universal indicators is that you have 6 different colors or so differentiating certain pH-ranges.

Now if you want to titrate to a certain pH-Value, it is easier to have an indicator like phenolphtalein that changes from colorless to a color at a certain point (phenolphtalein being colorless from 0 to 8 and red for $< 0$).

So you can titrate more accuratly to a certain pH-Value with specialized indicators. On the other hand you can't see where on the pH scale you are.

With an universal indicator you can get a feeling for where you are approximatly, but titrating to a certain value is very difficult.


Can Ferrofluids vaporise to gaseous state and Condense back to liquid state and do it many times over?

Part of the ferrofluid could theoretically be vaporized. As ferrofluid is made from tiny magnetic particles(eg. iron powder coated with a surfactant to inhibit clumping) suspended in a fluid. The fluid can be organic solvents, water and most commonly ink. Some of these fluids can evaporate and recondense. However, as a whole ferrofluid, cannot be vaporized. You can imagine ferrofluid as a cup of mixed sand and water that wouldn't settle(suspended).


Is coupled cluster variational for two electrons?

No. The reason for this is not to be found in the excitations, but in the evaluation of the method, i.e. the working equations.
$$%Introducing some shortcuts
\require{cancel}
\newcommand{\op}[1]{\mathrm{#1}}  %\op{H}
\newcommand{\bracket}[2]{\left\langle#1\middle|#2\right\rangle}
\newcommand{\bra}[1]{\left\langle#1\right|}
\newcommand{\ket}[1]{\left|#1\right\rangle}
\newcommand{\order}[1]{^{(#1)}}         %E_n\order{1}
\newcommand{\overlap}[3]{\mathcal{#1}_{#2}\order{#3}} %\overlap{S}{m}{1}
\newcommand{\integral}[3]{\mathcal{#1}_{#2,#3}} %\integral{V}{i}{j}
$$

The variational principle

In short: a trial wave function always has a higher energy than the exact wave function.

Assume we know the exact solutions to the Schrödinger equation, which form a complete basis and there is one wave function $\Psi_0$ which leads to the lowest energy $E_0$.
\begin{align}
\op{H}\Psi_i &= E_i\Psi_i &
\{i \in \mathbb{N}, i\geq 0\}\tag{1}\label{schrodinger}
\end{align}

We choose the solutions to be orthonormal (boundary conditions).
$$\bracket{\Psi_i}{\Psi_j} = \delta_{ij} =
\begin{cases}1, & i=j \\ 0, & i\neq j \end{cases}
\tag2\label{orthonorm}$$

The trial wave function $\Phi$ can be expressed as a linear combination of the complete set of wave functions.
$$\Phi = \sum_i^\infty a_i\Psi_i\tag3\label{expansion}$$

The expectation energy of the trial wave function is given through
$$W = \frac{\bra{\Phi}\op{H}\ket{\Phi}}{\bracket{\Phi}{\Phi}}.
\tag4\label{expen}$$

With $\eqref{expansion}$, $\eqref{orthonorm}$, and $\eqref{schrodinger}$ we can expand and simplify.
\begin{align}
W &=
  \frac{
    \sum_i^\infty\sum_j^\infty a_i a_j \bra{\Psi_i}\op{H}\ket{\Psi_j}
  }{
    \sum_i^\infty\sum_j^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty\sum_j^\infty a_i a_j E_j\bracket{\Psi_i}{\Psi_j}
  }{
    \sum_i^\infty\sum_j^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty a_i^2 E_i\bracket{\Psi_i}{\Psi_j} +
    \sum_{j\neq i}^\infty a_i a_j E_j\bracket{\Psi_i}{\Psi_j}
  }{
    \sum_i^\infty a_i^2 \bracket{\Psi_i}{\Psi_j} +
    \sum_{j\neq i}^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty a_i^2 E_i
  }{
    \sum_i^\infty a_i^2
  } & | -E_0\\
W - E_0 &=
  \frac{
    \sum_i^\infty a_i^2 (E_i - E_0)
  }{
    \sum_i^\infty a_i^2
  } \\
W - E_0 &=
  \frac{
    \sum_{i>0}^\infty a_i^2 (E_i - E_0)
  }{
    \sum_{i>0}^\infty a_i^2
  } + \frac{a_0^2 (E_0 - E_0)}{a_0^2}\tag{4'}\label{notnegative}\\
W - E_0 &\geq 0\tag5\label{varprin}
\end{align}

In $\eqref{notnegative}$ the last term is zero. Since $E_i > E_0$ (by definition) and $a_i^2$ is always greater or equal zero, the expectation value of the trial wave function must always have a higher energy than the exact solution of the Schrödinger equation, i.e. $\eqref{varprin}$.

Configuration Interaction

The CI wave function is set up as a linear combination of determinants from a reference calculation. This reference calculation can in principle be any other method the only requirement is that it also forms a complete (orthonormal) basis.
When Hartree-Fock is chosen, we know that we obtain a complete set and that this method itself is variational. The CI trial wave function can then be expressed as a linear combination of solutions of the HF wave function.
$$\Psi^\mathrm{CI} = \sum_i^\infty a_i\Phi_i =
a_0\Phi^\mathrm{HF} + \sum_{i>0}^\infty a_i\Phi_i$$
An alternative expression of the above is using an "excitation operator" $\op{T}$ to generate the higher energy determinants from the reference.
$$\Psi^\mathrm{CI} = (1 + \op{T})\Phi^\mathrm{HF}$$

The energy of the CI trial wave function can be evaluated with the Lagrange multiplier method, which requires, that the wave function is normalised (adding zero).
$$L = \bra{\Psi^\mathrm{CI}}\op{H}\ket{\Psi^\mathrm{CI}}
     - \lambda\left[\bracket{\Psi^\mathrm{CI}}{\Psi^\mathrm{CI}}-1\right]$$

Since the boundary conditions did not change, the variational principle still holds.

Coupled Cluster

Because of the inherent complexity of the CC ansatz, one key element to evaluate to CC energy is missing: the requirement of the normalised CC trial wave function.
We generate the trial wave function with an exponential approach (rather than a linear) expanded into a Taylor series.
\begin{align}
\Psi^\mathrm{CC}
&= \mathrm{e}^{\op{T}}\Phi^\mathrm{HF}\\
&= \left[\op{1} + \op{T}_1 + \left(\op{T}_2 + \frac12\op{T}_1^2\right)
   + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
   + \cdots \right] \Phi^\mathrm{HF}\\
&= \Phi^\mathrm{HF} + \left[\op{T}_1 + \left(\op{T}_2 + \frac12\op{T}_1^2\right)
   + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
   + \cdots \right] \Phi^\mathrm{HF}\\
\end{align}

Or to shorten the journey
$$\Psi^\mathrm{CC} = \Phi^\mathrm{HF} + \sum_i^\infty t_i \Phi_i.$$

One can easily see what kind of mess it would become requiring the wave function to be normalised $\bracket{\Psi^\mathrm{CC}}{\Psi^\mathrm{CC}}=1$, so the Lagrange ansatz won't work. Instead we choose the wave function to be orthonormal to the reference, which in the Hartree-Fock (complete orthonormal basis) takes away a lot of work.
\begin{align}
\bracket{\Phi^\mathrm{HF}}{\Psi^\mathrm{CC}}&= 1&
\implies\quad \bracket{\Phi^\mathrm{HF}}{\Phi_i} &= 0
\end{align}

We can write the CC-Schrödinger equation as
$$\op{H}\mathrm{e}^{\op{T}}\Phi^\mathrm{HF} =
  E^\mathrm{CC}\mathrm{e}^{\op{T}}\Phi^\mathrm{HF}$$
and evaluate the energy as
\begin{align}
\bra{\Phi^\mathrm{HF}}\op{H}\mathrm{e}^{\op{T}}\ket{\Phi^\mathrm{HF}}
  &= E^\mathrm{CC}
     \bracket{\Phi^\mathrm{HF}}{\mathrm{e}^{\op{T}}\Phi^\mathrm{HF}}\\
E^\mathrm{CC}
  &= \bra{\Phi^\mathrm{HF}}\op{H}\mathrm{e}^{\op{T}}\ket{\Phi^\mathrm{HF}}\\
E^\mathrm{CC}
  &= E^\mathrm{HF} +
  \bra{\Phi^\mathrm{HF}}\op{H}
    \ket{\left[\op{T}_1 + \left(\op{T}_2 +
     \frac12\op{T}_1^2\right)
     + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
     + \cdots \right]\Phi^\mathrm{HF}}.\\
\end{align}

From this we can see that the evaluation of the expectation value of the energy is significantly different from what is required for the variational principle $\eqref{expen}$.

Without going further into details, the dropping of the requirement of a normalised trial wave function leads to not obeying the variational principle.

Additional considerations

It is often said that Full-CC is equivalent to Full-CI, which is a convenient half truth. It is extremely important that this gets more true with a better description of the reference wave function, i.e. with larger basis sets. Full-CI is exact at the complete basis set limit (for the electronic Schrödinger equation). This is not necessarily true for Full-CC.

Since we are in computational chemistry hardly ever concerned about absolute energies, the non-variational nature of the method is by far outperformed by the "infinite" approach generating the "excited" determinants. Relative energies don't suffer significantly (compared with the size consistency and extensivity) under the non-variational approach.


Deriving the equation for oxygen estimation by Winkler’s method

[Concepts]

Normality, N: The normality is the relation between the sustance equivalence and the liters of a solution. The sustance equivalents are referred to the charges per mole of a sustance. The number of equivalents (NOE) of a sustance is obtained as follows:

Acid equal to number of hydrogens.
Hidroxide equal to number of hydroxils.
Salt equal to number of cations.

To determine the normality of a solution, these are the steps you must follow:

Weight Equivalent Gram, WEG:
\begin{equation}
    WEG=\frac{Molecular\ weight\ of\ solute}{NOE}
   \end{equation}
Weight Equivalent Number, WEN:
\begin{equation}
    WEN=\frac{Mass\ of\ solute}{WEG}
   \end{equation}
Normality, N:
\begin{equation}
    N = \frac{WEN}{L}
   \end{equation}
Doing some basic algebra, you find that:
\begin{equation}
   N=M\cdot NOE
  \end{equation}
Where M stands for Molarity.

Dilution factors, DF: It is employed when the solution concetration of your sample is too concentrate. In order to determine the concentration of the substance you want to find, you have to dilute the initial solution. This is done through the dilution factors. I.e. you have 25 ml of a solution 10 M and you are asked to dilute it 100 times. To do so, you have to take an aliquot (A portion of a large whole) of 0.250 ml, pour into a volumetric flask of 25 ml and then fill it with destilled water to capacity. Then, you have a diluted solution of 0.1 M. In general, a dilution factor is expressed as:
\begin{align}
   Dilution\ factor=\frac{Total\ volumen}{Sample\ volume}
  \end{align}
Dilution: Dilution is the process of decreasing the concentration of a solute in solution, usually simply by mixing with more solvent. To dilute a solution means to add more solvent without the addition of more solute. This can be represented as:
\begin{align}
   V_1\cdot M_1 &= V_2\cdot M_2\\
   M_2&=\frac{V_1}{V_2}\cdot M_1
  \end{align}

[Answer]

Now, if you put all together:

If you look at the first parenthesis at the right of the equa, having in mind that E is WEG and doing some basic algebra:
\begin{align*}
   V_1\cdot N\cdot E\cdot 1000&=V_1(L)\cdot\frac{WEN}{L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{\frac{Mass\ of\ solute}{WEG}}{L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{Mass\ of\ solute}{WEG\cdot L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{Mass\ of\ solute}{L}\cdot 1000\\
   &=V_1(mL)\cdot\frac{Mass\ of\ solute}{mL}\\
   &=Mass\ of\ solute\cdot\frac{V_1(mL)}{mL}\\
   &=Mass\ of\ solute
  \end{align*}
The square bracket is just a dilution factor.
If you put together the last two steps with the dilution section, you find a similarity.

[History]

If you are interested in the origins of this method, I recommend you the following paper:

Winkler, L. "Die Bestimmung des in Wasser Gelösten Sauerstoffes". Berichte der Deutschen Chemischen Gesellschaft. (1888) 21, 2, 2843–2855.

I cannot give you a brief handout of it, my german is basic. But, I want to point out that in page 2845, you will find an equation similar than the indicate in dilution section.

[How to understand the Winkler method]

  The Winkler method for the determination of dissolved $\ce{O2}$ involves adding an excess of ‘white’ $\ce{Mn(OH)2}$ which contains $\ce{Mn^{2+}}$ ions.

  For every molecule of $\ce{O2}$ two molecules of $\ce{I2}$ are formed. The quantity of iodine present can be found though titration of the sample with sodium thiosulphate.

[Bibliography]

Shashi,  B.P. A complete review on: effluent testing and treatment in pharmaceutical industry.


Are there reactions where the stirring direction would make a difference?

No, for three reasons:
Firstly, you need a relatively large molecule and/or a high viscosity and/or a strong shear/extensional flow, before the molecule even notices that it´s in a nonuniform environment. The molecule, if it has an elongated form, will then align perpendicular to the normal vector of that flow field. That only occurs for sufficiently concentrated solutions of polymers, or liquid crystals.
Secondly, you need three vectors to be able to differentiate a left and right hand. You only have the normal vector of the flow field, and the preferred direction of your molecule. Or in other words, your reaction will be indifferent to you watching it from the top of the beaker or through the bottom. ;)
Thirdly, to get a third vector into your system, you need the normal of the flow field to change significantly, in a specific direction (using, say, a propellor instead of a stirring bar), on the lenght scale of your molecules. That is impossible with a macroscopic stirrer.


Energy absorbed by the gas from a PV-diagram

I suspect you forgot to take the work done by the gas into account. From the first law of thermodynamics:

$$Q = \Delta U - W \label{eq:1}\tag{1}$$

For the monoatomic gas $U=\frac{3}{2}PV$, so

$$\Delta U = U_2 - U_1 = \frac{3}{2}(P_2V_2 - P_1V_1) = \frac{3}{2}(3P_0 \cdot 2V_0 - 2P_0 \cdot V_0) = 6P_0V_0 \tag{2}$$

On the other hand, since the gas expanded ($V_2 > V_1$), the work is negative and can be found geometrically as the area under the line:

$$W = -(\color{blue}{2P_0(2V_0-V_0)} + \color{orange}{0.5 \cdot (3P_0 - 2P_0)(2V_0 - V_0)}) = -2.5P_0V_0 \tag{3}$$

Now back to \eqref{eq:1}:

$$Q = 6P_0V_0 -(-2.5P_0V_0) = 8.5P_0V_0$$


Why are the hydrogen-carbon bonds bent in a graphical depiction of an alkene, but are straight horizontally and vertically in an alkane?

Fundamentally, it depends on how you put a three-dimensional structure onto a 2D surface.
The Valence Shell Electron Repulsion (VSEPR) model can predict the structure of most molecules. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsion. So with the first example (propane), each bond is equally 90 degrees from each other, and in the second (propene), the bonds are 120 degrees from each other. If the bonds were placed 90 degrees away from one another, at least one gap between electron groups would be 180 degrees, thus a less stable arrangement of electrons, regardless of if it is a double bond (c=c) a or a single bond (c-c)


Production of hexavalent chromium during electrolysis using a stainless steel as the cathode

In anode, things get oxidized. That is why the chromium in anode would become hexavalent chromium (hexavalent chromium is the oxidized form of chromium).

In cathode, things get reduced (opposite of oxidized). That is why you could remove rust, because rust is the oxidized form of iron, so if you reduce rust, you would get iron.

You will not obtain hexavalent chromium, because that is not possible through reduction of chromium; that is only possible through oxidation of chromium, which is impossible at a cathode.

You can reduce hexavalent chromium to trivalent chromium with sulphite in acidic environment:

$$\ce{Cr2O7^2- + 8H+ + 3SO3^2- -> 2Cr^3+ + 3SO4^2- + 4H2O}$$

Then you can precipitate out the trivalent chromium with sodium hydroxide:

$$\ce{Cr^3+ + 3NaOH -> Cr(OH)3(s) + 3Na+}$$


Is there an alternative to the half-reaction method to solving red-ox reactions?

There is one other way that I have used, using a linear system to solve this reaction. Although it can be used for other more casual reactions too, such as the combustion of ethanol for example.

As you can see, in this special case (because the reaction is relatively easy to solve already) we see that there must be $H_2O$ molecules on the product side of the equation, because there is no oxygen on the products already, but there is oxygen in the reactants. So start by assuming the products include $H_2O$ with a coefficient of at least 4 (because there is 4 oxygen in permanganate ion).

Also, there is no hydrogen anywhere in the equation, so we can assume that to balance the water (which has at least 4 moles) there must be some hydrogen ions added as reactants to balance the reactants side (it is acidic conditions in this case). As it turns out, since there must be at least 4 moles of water, then there must be at least 8 moles of hydrogen ions to satisfy the convention of having all mole coefficients in the lowest terms.

Now we can start building the linear system:

Since we obviously do not know the coefficients yet, we will assign variables to them. Let a b c d e and f be the molar coefficients of each respective molecule in the following reaction we have established so far:

a$H^+$ + b$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + e$Mn^{2+}$ + f$H_2O$

Now how do we build the linear system for this reaction? We might do it as follows:

Starting with the hydrogen atom, simply because it comes first when reading it from left to right. We write this equation to represent the hydrogen atoms in our reaction:

a = 2f

It is as simple as it looks, there is one mole of hydrogen atoms per mole of hydrogen ions, and there is two moles of hydrogen atoms per mole of water molecules.

We can keep doing this for the rest of the elements in our reaction, until we have three more equations along with the one for hydrogen, because there are three more different elements to account for.

b = e This one is pretty straightforward... This accounts for Manganese atoms.

4b = f Oxygen atoms are now accounted for.

c = d Also straightforward, now iron atoms will be accounted for as well. Notice that c and d are lone variables, they do not show up with the other equations, but at least they must be equivalent

Remember that we established some mathematical axioms when we said that there must be at least four moles of water molecules to satisfy the one mole of permanganate ion. An axiom is just something that we assume to be true, such as f = 4 in this case.

Now look at the equation we established for hydrogen atoms, the first equation a = 2f. Since we are assuming that f = 4 we can therefore safely assume that a = 8 without going against our original axiom. Also if we assume that f = 4 we can also go ahead and solve the equation to account for oxygen atoms 4b = f and solving it gives us b = 1 and the rest is pretty straight forward. If b = 1 and b = e then e = 1.

Given what we have found out so far, lets take a look at our equations again:

a = 8

b = 1

c = d

e = 1

f = 4

So,

8$H^+$ + 1$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + 1$Mn^{2+}$ + 4$H_2O$

So now we can look at the charges to figure out the correct coefficients for iron(II) and iron(III) ions, which should be the same of course (c = d). As it stands currently (ignoring the iron ions), there is a total charge of 7+ on the reactants side, and 2+ on the products side. These should be the same for the equation to be balanced, so we can start by creating another axiom and saying that c > 1. If there are 2 moles iron then our charges would be 11 on the reactants and 8 on the products. Incrementing it to 3 would increment the charge on the reactants side by 2+ while the product's total charge would increment 3 due to the charge of the ions themselves. Eventually things are balanced when the moles for iron reach 5 making both sides have a total charge of 17 and everything is nicely balanced.


Potassium Vanadate can be prepared by any of the three process:

  Potassium Metavanadate, $\ce{KVO3}$, is obtained as white or colourless crystals by dissolving vanadium pentoxide in hot, strong
  caustic potash solution. By varying the concentrations several
  hydrates have also been prepared: $\ce{2KVO3.3H2O}$;
  $\ce{KVO3.2H2O}$; $\ce{2KVO3.5H2O}$; $\ce{KVO3.3H2O}$. On being
  heated, all these hydrates lose their water and leave a white,
  nacreous mass of the anhydrous salt, which melts at 495° C. Two
  other hydrates were prepared by Rammelsberg: $\ce{KVO3.H2O}$ and
  $\ce{KVO3.7H2O}$. (source)

  [...] the reaction of vanadium(V)
  oxide from waste catalyst with potassium chloride in the presence of
  steam.
  $$\ce{2KCl + V2O5 + H2O(g) → 2KVO3 + 2HCl↑}$$
  (source)

  Ammonium metavanadate react with potassium carbonate to produce
   potassium metavanadate, ammonia, carbon dioxide and water. Potassium
   carbonate - concentrated solution. The reaction takes place in a
   boiling solution.

  $$\ce{2NH4VO3 + K2CO3 → 2KVO3 + 2NH3 + CO2 + H2O}$$ (source)


As orthocresol pointed out, $\ce{Ba_2S_2}$ does not exist as it's own substance. Perhaps you've misread the question and it instead asks for something else? Otherwise it is wrong, or at least not simplified. Your answer is in effect the correct one, although I'm not sure if sulfur and barium would react like that exactly.

It might in addition generate multiple polysulfides, such as $\ce{BaS_2}$, $\ce{BaS_3}$, following the rule of $\ce{BaS_{$x$}}$. However, I believe your answer is correct in that the initial reaction between barium and sulfur would produce $\ce{BaS}$, while further reactions might add more sulfurs onto it.


There was this guy who recently protected a ketone in the presence of an ester. You might have heard about him ;)