Mass of sugar that was added to the water (Raoult's law)

Your question

  Here it says that after the sugar was added to water, vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.

  Does it mean that before sugar was added, there was only one element $\ce{H2O}$?

One component, not element. Water isn't an element. But yes. The question does actually say it: "sugar is added to water".

  I used Raoult's law: [...] So can I use this to find the partial pressure before the sugar was added?

Why do you want to find the partial pressure before it was added? It's already given to you. When there's only one component in the system, the partial pressure is the same as the total pressure, which is the vapour pressure of pure water: $15.5~\mathrm{mmHg}$.

In fact, the partial pressure after it was added is also given to you, it's just $(15.5 - 0.13)~\mathrm{mmHg}$. You already have all the pressure data you need. This question is giving you partial pressure data and asking you to find the mole fraction.

  I know that $1.31~\mathrm{kg}$ of water is $72.78~\mathrm{mol}$.

Yes. Although if you use a more accurate molar mass for water, you will get a more accurate answer.

  And if my assumption is right and before sugar was added there was only water, will the mole fraction before sugar was added just be $72.78$ since there was only $\ce{H2O}$?

No, the mole fraction would be $1$. The mole fraction of a component is the number of moles of that component, divided by the total number of moles.

The answer

The vapour pressure of pure water, which is $P_{\ce{H2O}}^*$ in Raoult's law, is $15.5~\mathrm{mmHg}$.

After you add a certain amount of sugar, the vapour pressure is decreased to $P_{\ce{H2O}} = 15.37~\mathrm{mmHg}$.

So, if you just plug these numbers into Raoult's law, you find:

$$\begin{align}
x_{\ce{H2O}} &= \frac{P_{\ce{H2O}}}{P_{\ce{H2O}}^*} \\
&= \frac{15.37~\mathrm{mmHg}}{15.5~\mathrm{mmHg}} \\
&= 0.9916
\end{align}$$

$x_{\ce{H2O}}$ is the mole fraction of water after sugar is added. So:

$$\begin{align}
x_{\ce{H2O}} &= \frac{n_{\ce{H2O}}}{n_\mathrm{total}} \\
0.9916 &= \frac{n_{\ce{H2O}}}{n_{\ce{H2O}} + n_\mathrm{glucose}}
\end{align}$$

The number of moles of water is:

$$\frac{1310~\mathrm{g}}{18.02~\mathrm{g~mol^{-1}}} = 72.70~\mathrm{mol}$$

This is what I said earlier about using a more accurate value of the molar mass. So:

$$0.9916 = \frac{72.70~\mathrm{mol}}{72.70~\mathrm{mol} + n_\mathrm{glucose}}$$

Rearranging:

$$n_\mathrm{glucose} = 0.6149~\mathrm{mol}$$

The molar mass of glucose is $180.16~\mathrm{g~mol^{-1}}$. So:

$$\begin{align}
m_\mathrm{glucose} &= (0.6149~\mathrm{mol})(180.16~\mathrm{g~mol^{-1}}) \\
&= 110.8~\mathrm{g}
\end{align}$$


Purification of Solid Phase Synthesized Peptide

If you are a using reversed-phase (RP) column, it is okay to directly inject your crude sample in the running HPLC solvent system.  But, make sure to filter the injecting sample through a membrane filter to remove minute particles (from glass beats, etc.) from the original sample after dissolving it in suitable chosen solvent. Please also note that your chosen solvent for peptide separation usually contain small percentage of TFA. Therefore, you may need at least a rotary evaporator  for final concentration.


Is (cycloprop-2-en-1-ylidene)propanedinitrile aromatic or not?

At least formally, the target compound is the product of a Knoevenagel reaction of cyclopropenone with malononitrile.

For cyclopropenone, a zwitterionic mesomeric structure is conceivable. This species, a cyclopropenyl cation with a negative charge on the oxygen atom attached, is aromatic. Think that this is the little brother of tropone.

For the compound in question, charge separation is possible as well. The resulting species would again be an aromatic cyclopropenyl cation with a dicyanomethyl substituent over which the negative charge can be delocalized.


Reaction between bromobenzene and methyllithium to yeld toluene: possible?

MeLi certainly does lithium-halogen exchange and in the presence of MeI takes bromobenzenes to toluenes   Synth. Comm. 2001 2323

If you're asking whether the MeBr produced by the exchange can act as the methylating agent? I have never seen this reported but maybe someone else can pull up a reference.


Redox reaction: Iron(III) and Iodide ion

Iodine exists as a diatomic molecule. This reaction indicates how the iodide ion is converted into diiodine. Your reaction just indicates the electron transfer: you still need to balance the reaction.
You need an $\ce{I_2}$ molecule on the right hand side, which would need two iodide ions on the left hand side:
$$\ce{2 I- -> I_2 + 2e-}$$
To accept two electrons, we would need two $\ce{Fe^3+}$ ions:
$$\ce{2 Fe^3+ + 2 e- -> 2 Fe^2+}$$
Now we have a balanced reaction. All the electrons donated by iodine are accepted by iron. Adding up the two reactions and cancelling the electrons gives us:
$$\ce{2 Fe^3+ + 2I- -> 2 Fe^2+ + I_2}$$


Entropy change upon mixing cold and hot water

The change in entropy doesn't "equals to" q/T.  For a specified closed system, the change in entropy between an initial state and a final state is given by $$\Delta S=\int{\frac{dq_{rev}}{T}}$$ where the subscript "rev" refers to a reversible path between the initial and final states.  The process you described is not reversible.  You need to devise and employ a reversible path to get the change in entropy.

Here is a hint:  For the system comprised of the 454 gm of water initially at 94.4 C, the change in entropy is $$75.291\frac{454}{18.015}\ln{[(273.2+41.1)/(273.2+94.4)]}$$
If you would like a cookbook recipe for how to determine the change in entropy for a system experiencing an irreversible process, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/


What is the difference between a glassy, amorphous and a vitreous phase?

After some reading I have decided to answer this question myself for anyone else with the same confusion. The info is from the downvoted answer of alphonse and the wiki-link, but with a better overview.

Both vitreous and glassy are amorphous. Amorhous is the broadest term to rule them all.

Some amorphous materials have a Glass Transition Temperature (GTT). Above this temperature the material becomes a viscous liquid. note that both below and above the GTT the material is amorphous.

An amorphous solid that has a GTT is called a glass.
The transition (heating) over the GTT results in a glassy phase.
The reverse transition (cooling) corresponding to the GTT is called vitrifcation. The result is a vitreous phase.

If I made mistakes please comment and I will update the answer.


Does a non-terminal alkyne react with sodamide?

By the given question conditions you should easily guess that $(A)$ is $\ce {Hex-2-yne}$. When it is reacted with $\ce{NaNH_2}$ in presence of heat it can isomerise to terminal alkyne by following this mechanism and then $\ce{CH_3I}$ addition will simply follow an $\ce{S_N2}$ mechanism.

Secondly,Ozonolysis will yield an aldehyde which can get oxidised to acid in oxidative medium.


Flame Test: Why does the flame colour of calcium disappear quickly?

A flame is a constantly upward "moving" chemical reaction. The velocity is dependent on the flow rate of the gases entering the burner. There is an extensive upward motion (on Earth). In a gravity-less flame, the flame is round, still there is an upward movement. Now another important thing to note is that the Group II flame emission is not due their atoms but their hydroxides, such as CaOH. In an ordinary Bunsen burner flame you can expect "weird" molecules, which cannot exist in an ordinary bottle. The typical flame temperature is not hot enough to cause atomic excitation and completely break calcium compounds into calcium atoms. If this were true you would see a violet flame (422 nm) after introducing calcium. We see reddish color which is always contaminated with sodium's yellow color. The flame color is then dependent on the thermal stability and volatility of the emitting compound being formed in the flame.


What is the empirical formula of Haemoglobin?

Unfortunately this is only half an answer. I'd be thankful if someone could chip in with the rest.

When you want structural information about proteins, the first place to look is the Protein Data Bank, a freely accessible repository of protein structures determined by techniques such as X-ray crystallography, protein NMR and neutron diffraction, among others.

The problem now is being sure about what you want to find. A basic search for human hemoglobin currently provides 190 hits, so you need to browse a bit to find the most adequate entry. To keep things simple, I looked for a structure for adult hemoglobin, with no mutations, ligands or associations with other proteins. There are some database entries which seem to be right on target, namely 2DN2 for deoxyhemoglobin and 2DN1 for oxyhemoglobin.

There is a lot of information in each entry page. The whole structure can be downloaded as a .pdb and opened in appropriate structure visualization programs, of which there are probably several free options. Depending on the program used, retrieving the empirical formula can be trivial.

Forgoing the use of molecular visualization software, there is an "Experimental Data & Validation" section in the protein page, containing a "Full Validation Report" link to a PDF file. There, you can find a section called "Entry Composition" with useful information. For example, in the deoxyhemoglobin structure there are 4777 non-hydrogen atoms. The protein is composed of two copies each of two unique subunits (the alpha and beta subunits), totalling four subunits. The alpha subunit contains 1069 non-hydrogen atoms (685 $\ce{C}$, 187 $\ce{N}$, 194 $\ce{O}$ and 3 $\ce{S}$), while the beta subunit contains 1123 non-hydrogen atoms (724 $\ce{C}$, 195 $\ce{N}$, 201 $\ce{O}$ and 3 $\ce{S}$). Additionally, each subunit also contains one molecule of protoporphyrin IX coordinated to an iron atom. One important thing to note is that the structure also contains solvent molecules, in this case a total of 221 molecules of water, which must be removed from the atom count.

The only flaw is that hydrogen atoms are not shown. This is because X-ray crystallography measurements generally cannot detect hydrogen atoms, and so they are usually added separately to the backbone of the protein, in such a way to complete all atom valencies. I don't know how this can be done other than using a program.

A last thing to keep in mind is that no matter how good the measurement or the data processing, you can very well end up a few atoms off, especially regarding hydrogen atoms. Hopefully perfect precision is not required.


What is the product of the reaction of 2-methyl-5-phenyl-2-pentene with sulfuric acid?

With the help of user SendersReagent in comments, the answer to the question was a simple intramolecular Friedel–Crafts reaction:


Alcoholic potassium hydroxide and aqueous potassium hydroxide

Why is alcoholic KOH used in dehydrohalogenation of alkyl
  halides?

One possible explanation is lack of solubility of the alkyl halides in water. Thus, one switches to a less polar, presumably alkyl side chains containing alcohol. (H2O is formed in minute amounts from OH and the proton from the hydroxyl group of the alcohol and can thus be neglected.) Boiling points vs. reaction conditions can also be relevant.

KOH forms homogenous mixtures with H2 at any ratio, this may not be true for some alcohols (For the amounts needed for the reaction this should not be relevant). In terms of ionisation, one can suspect that K+ is not as well solvated in alcohols as in water because the alcohol molecules will have a harder time arranging optimally around the ion. Also, the solvent shell structure may be a lot less pronounced. For OH-, one basically expects the same. In alcohol, one might also see some ion pair formation (or rather retention?) for KOH.


Why does ZnO turn yellow when it is hot?

The thermochromism of $\ce{ZnO}$ results from a minor loss of oxygen upon heating to temperatures around 800 °C, i.e. a non-stoichiometric $\ce{Zn$_{1+x}$O}$ with $x = 7 \times 10^{-5}$ is formed.

Under air, this effect is reversible. Heating (and cooling) of the material while hooked up to a vacuum pump might result in a more persistent colour change.


Is acetate a weak base due to its resonance structure?

Consider this generic acid-base reaction in water:
$$\ce{HA <=>[K_\text{a}] A- + H+ }$$

The (thermodynamic) dissociation constant $K_\text{a}$ depends mainly on two things:

The stability of the product $\ce{A-}$
The instability of the starting material $\ce{HA}$

The higher one of those factors is, the higher the dissociation constant. So in effect, what a chemist states is the following:

The carboxylate anion is the conjugated base of the carboxylic acid, whose strength is determined by the stability of said carboxylate ion and the instability of the acid. It may be reasoned that carboxylic acids would be much less strong if it was not for the additional stability via conjugation.


The boiling points of Alkali metal group

You can explain the trend in terms of increased charge density, and ensuing charge repulsion between nuclei and between electrons as you go down the group.
An alternative explanation is provided in a libretext, which draws information from a popular educational chemistry site:

When any of the Group 1 metals is melted, the metallic bond is weakened enough for the atoms to move more freely, and is broken completely when the boiling point is reached. The decrease in melting and boiling points reflects the decrease in the strength of each metallic bond.
The atoms in a metal are held together by the attraction of the nuclei to electrons which are delocalized over the whole metal mass. As the atoms increase in size, the distance between the nuclei and these delocalized electrons increases; therefore, attractions fall. The atoms are more easily pulled apart to form a liquid, and then a gas. As previously discussed, each atom exhibits a net pull from the nuclei of +1. The increased charge on the nucleus down the group is offset by additional levels of screening electrons. As before, the trend is determined by the distance between the nucleus and the bonding electrons.

Although the two perspectives (my opening explanation and that in the libretext) would seem to be equivalent, they aren't identical, since a description in terms of Coulomb's law is more fundamental and does not require invoking concepts such as polarizability or moments. I would argue that appealing to nuclear screening to explain bonding is not the best way to understand the trend.


Balancing chemical systems - searching and sorting preliminary information

From your diagram, $R_G$ is the recycled gas - the hydrogen that has not been consumed plus the methane that has been produced. $R_L$ is the recycled liquid - the unreacted toluene. Both go back into the mixer with fresh toluene added at the rate given and fresh hydrogen added to bring the mole ratio of hydrogen to toluene (and hydrogen to methane) to what is given.

$R_L$ is easier. The problem tells you that 80% of the toluene is converted into benzene and 8% of the toluene is converted into biphenyl. What percentage of toluene is left over? The original feed flow of toluene is 3450 lb/hr, but the amount of toluene entering the gross feed to the reactor is higher (includes $R_L$).

$$G_{\text{tol}}=F+R_L$$
$$R_L = 0.12\times G_{\text{tol}}$$

For $R_G$, you need to know how much hydrogen you have going into the mixer. You are told that the mole ratio of hydrogen to toluene in the gross feed is 4:1. Now that you have $R_L$, you can figure out the total toluene in the gross feed, and then do the stoichiometry calculation to get to total hydrogen in the gross feed. Since you have the balanced equations, you can determine how much hydrogen is consumed. If 80% of the toluene is consumed in the generation of benzene, how much hydrogen is consumed? How much methane is produced? Do the same with the conversion of toluene to biphenyl.

Once you are able to obtain the molar flow rates of all molecules out from the reactor, you should be able to calculate the flow rate of the gas exiting from the separator. This gas stream (Let's call it $K$) contains only $\ce{CH4}$ and $\ce{H2}$ of mol fractions $y$ and $1-y$ respectively. The mol fractions can be calculated considering that all $\ce{CH4}$ and $\ce{H2}$ that exists the reactor leaves the separator in $K$.

The molar composition of the $R_G$ and $P$ streams that leave $K$ is the same. This means that you are now able to use a molar flow rate balance on the mixing point between $M$, $R_G$ and the stream to the mixer (Lets's call this $Y$).

$$Y = M + R_G$$

By doing a balance on methane, you can find $R_G$.

$$(0.5)*(Y) = (1-y)(R_G) $$

Since you know $Y$ and $y$, you will then be able to find $R_G$.


Change in water vapor with change in pressure and volume

If water is introduced in an empty container, the water will slowly evaporate. The pressure $p$ will slowly increase from $0\ \mathrm{Pa}$ to a maximum which is $3167\ \mathrm{Pa}$ at $25\ \mathrm{^\circ C}$. This corresponds to a molar concentration of water vapor equal to: $${c = p/RT = \frac{3167\ \mathrm{J\ m^{-3}}}{8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298\ \mathrm K} = 1.278\ \mathrm{mol\ m^{-3}}}$$
Expressed in gram per cubic meter, the concentration of water is:  $ c = 23.0\ \mathrm{g\ m^{-3}}$.
This means that between $0$ to $23\ \mathrm{mg}$ water can be evaporated in a $1$ liter container at $25\ \mathrm{^\circ C}$, and that a maximum amount of $2.3\ \mathrm{mg}$ water can be evaporated in a $0.1$ liter container. As a consequence, if the gas volume of your water vapor is compressed from $1$ liter to $0.1$ liter without changing the temperature, $23\ \mathrm{mg} - 2.3\ \mathrm{mg} = 20.7\ \mathrm{mg}$ water will be condensed as a liquid on the inner wall of the container.


Weird Wikipedia Section on Oxidizing Behavior of Nitric and Sulfuric Acids

It is chemistry that is being weird not Wikipedia

The trouble with simple explanations of oxidising acids is that there is no simple explanation that covers all the cases. The specific reactions leading to oxidation are complicated and have a variety of different mechanisms.

So copper in sulphuric acid can be oxidised by a reaction that involves SO2 production. That has a completely different mechanism to the reaction with nitric acid and a different concentration dependence.

And, given the variety of reactions, the need to have protonated molecules of the acid is not a good generalisation: it depends on the specific reaction that is happening.

The oxidising reactions depend on the substrate being oxidised as well as the acid. Which mechanism will dominate isn't easy to determine without trying the reaction. So don't trust simple generalisations.


Are the breaking of bonds simultaneously endothermic and exothermic?

Bond breaking does NOT release energy!!

In the course of a chemical reaction, you typically break some bonds and form some new bonds. The bond breaking is always endothermic. The formation of new bonds is exothermic, so depending on whether the old bonds or the new bonds were stronger, the reaction overall can be either endothermic or exothermic.

When somebody says chemical bonds “contain potential energy” it is a HIGHLY misleading statement. You justifiably interpreted this as: when you break bonds, they release the potential energy that was stored within. This is, unfortunately, wrong. That phrase should honestly be banned, but somehow it has always persisted.

What that phrase really means is more nuanced. It means that some chemical bonds can be broken in a reaction which subsequently produces more energy via bond formation.

Bond breaking is always endothermic!

For further reading see: Is Bond Formation "Strictly" Exothermic?


If something contains large amounts of cyclopentane and cyclohexane, what is it called?

The most likely name would be naphthenes. It could also be called cycloalkanes.

Added:
Since the reference to the question is about the coal and oil production, the answer would be naphthenes. When the coal and oil were distilled, a layer (called naphthene) was removed which was found to contain mainly cyclopentane and cyclohexane.

I would complete the statement as:
If a fraction of distillate contains large amounts of cyclopentane and cyclohexane, it is called naphthene.


Is the carbon atom in the carbon dioxide molecule partially positive?

You're correct in assuming that the carbon atom in $\ce{CO2}$ has a partial positive charge. This is because the oxygen atoms are much more electronegative, so they pull the electrons away from the carbon atom. However, this molecule is still nonpolar. This is because, when you draw a dipole moment, you have to take all bonds into account. Take water for example:

In this molecule, there are two bonds, each with their own dipole. But these cancel out like any other vectors, leaving you with a vertical net dipole. The dipoles in carbon dioxide cancel out in a similar way; however, they cancel each other completely, because the bond is linear, not bent like in water:

This produces a zero net dipole, rendering the molecule nonpolar.


Which is the correct way to name pentan-3-one?

Whether you write pentan-3-one or 3-pentanone both are correct.

But according to IUPAC rule pentan-3-one this is correct format of writing the name of organic compound. So I prefer you to write pentan-3-one rather than 3-pentanone.

You should also read this.


Why is hot concentrated sulfuric acid preferable to cold?

A really short answer is kinetics. Heating up chemicals adds more energy to the system, allowing molecules to colide with more force increasing the probably of a reaction taking place.


Why is no dynamical correlation recovered in the CASSCF method?

The difference between dynamical and static correlation has always been ill-defined, and there is no clear-cut delineation between them.

Correlation simply means that the probability of finding an electron somewhere at position $a$ and another electron somewhere at position $b$ is not simply the product of the two probabilities.

(In that respect, even Hartree-Fock has some correlation since it forbids two same spin electrons occupying the same state (Fermi correlation).)

Correlation was originally thoguht of as the instantaneous repulsion of electrons. This was dynamic correlation. The origin of the difference between static and dynamical correlation came from studies of bond dissociation in molecules. As you pull a bond apart, the electrons should be further apart, and correlation energy should decrease. However, the opposite was found, and in many cases the correlation energy increases. Therefore, there must be some sort of "other" correlation occurring. This was static correlation.

This is because there are additional (near-)degenerate configurations that contribute strongly to the nature of the wave function during dissociation. You can think of this as a manifestation of entanglement, of which Hartree-Fock, being single determinant, fails miserably.

In short, static correlation came to describe situations where the underlying Hartree-Fock reference was a qualitative and quantitative failure.

Dynamic correlation, on the other hand, was used to describe the cases where the underlying Hartree-Fock reference has the correct qualitative behavior, but was not quantitatively accurate.

CASSCF does a better job at calculating static correlation because most of the important near degenerate configurations are multi-electron excitations within a small valence space. Thus it generally captures the configurations necessary to describe bond breaking, etc. with a well-chosen active space.

Single reference coupled cluster, does better at capturing dynamic correlation because it is usually truncated to few-pair excitations, like the double excitations of CCSD, but cover the whole virtual space.

For high levels of coupled cluster, e.g. CCSDTQ, the distinction between static and dynamic correlation really breaks down, and some high level coupled cluster calculations actually do quite a good job at otherwise "multireference problems." Multireference coupled cluster --- combining the best of CASSCF and CC --- is an active area of research today.


In your first test, the $\ce{KMnO4}$ probably reacted with the $\ce{H2O2}$ excess remaining after your preparation. The ionic equation for the reaction is

$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O +3O2 + 2OH-}$$

The $\ce{MnO2}$ formed is an insoluble black precipitate. Although there is an $\ce{OH-}$, the reaction apparently occurs in neutral and faintly acidic medium as well.

The result of your second test doesn't seem to be very clear. Sodium perborate apparently reacts very quickly with any water so if you added any water during the second test, it must have reacted to form oxygen gas bubbles (which appear white).


Why does Coca-Cola Zero Sugar Contain Sodium Despite no Sodium Ingredients?

Two sources: water and natural flavors

According to the Wisconsin department of healt natural sodium levels in water is $\pu{40-60 mg/L}$ meaning a $\pu{20 fl oz (\! 591 mL)}$ Coke has about $\pu{25 mg}$ of sodium from water. Thus the other $\pu{50 mg}$ of sodium must come from natural flavors. This makes sense since Coca-Cola uses plant products like coca leaves which would naturally add sodium to the carbonated sugar water.


Does resonance overrule hyper conjugation in terms of stability in alkenes?

We can hydrogenate unsaturated compounds and measure the heat given off, this is called the heat of hydrogenation. As the following figure illustrates, if we hydrogenate the various butene isomers they all wind up producing the same molecule, butane.

image source

By comparing their heats of hydrogenation we can see which butene was lowest in energy to begin with, that is, which butene is the most stable. The heats of hydrogenation reported in the figure tell us that trans-2-butene is more stable than 1-butene by about 2.3 kcal/mol. This extra stability can be attributed to the additional hyperconjugative resonance structures involving hydrogen present in trans-2-butene (6) compared to 1-butene (2).

Next let’s consider 1,3-butadiene. Here when we hydrogenate it, butane is again produced and we might estimate that it will have a heat of hydrogenation equal to twice that reported for 1-butene (1-butene is the best model because its double bond has the same substitution pattern as found in 1,3-butadiene) or 60.6 kcal/mol of heat given off. When we actually run the experiment we find that only 57.1 kcal/mol of heat is given off, 3.5 kcal/mol less than we estimated. In other words 1,3-butadiene is about 3.5 kcal/mol more stable than isolated double bonds and this is due to the resonance interaction between the two double bonds in 1,3-butadiene.

The resonance stabilization in 1,3-butadiene (3.5 kcal/mol) is larger than the hyperconjugative stabilization in trans-2-butene (2.3 kcal/mol).

Your question actually asked about 2-methyl-1,3-butadiene. In this case we have the resonance stabilization found in 1,3-butadiene plus some hyperconjugative stabilization from the additional methyl group. It will be even more stable than 1,3-butadiene. Further, 2-methyl-1,3-butadiene is more stable than any butene, so answer #1 is correct. Resonance stabilization is larger than hyperconjugative stabilization in this series of compounds.


Is every substance in equilibrium with its vapours?

If they were maintaining an equilibrium with their vapour you wouldn't be smelling it.

In theory any equilibrium in the case of vapor pressure occurs in a closed container, in reality a vapor pressure equilibrium is rarely if ever reached.


Let me add to the confusion:

https://books.google.de/books?id=imBzjH6SHi8C&pg=PA303&lpg=PA303&dq=solubility+product+constant+aluminum+hydroxide&source=bl&ots=jXoPiVihnF&sig=eSK8saszXshTXVEKGOlThUiKVr0&hl=de&sa=X&ei=Xl5TVcepDuj_ywOXmYDQBA&ved=0CG8Q6AEwCTgK#v=onepage&q=solubility%20product%20constant%20aluminum%20hydroxide&f=false
$2\times10^{-32}$
http://faculty.ncc.edu/LinkClick.aspx?fileticket=smTKdcQlPY4%3D&tabid=1903
$1.8\times10^{-33}$
http://www.austincc.edu/dlyon/gen_chem_1_summer04/Chapter%2020%20solubiliity%20product%20f08.pdf
$1.9\times 10^{-33}$
http://www.csudh.edu/oliver/chemdata/data-ksp.htm
$4.6\times10^{-33}$
http://www.periodni.com/solubility_product_constants.html
$3\times 10^{-34}$
https://users.stlcc.edu/gkrishnan/ksptable.html
$3\times 10^{-34}$

Usually I'd first have a look at the CRC Handbook of Chemistry and Physics. Unfortunately, I couldn't find the data for aluminium hydroxide.


The acetal formation in question is actually an example of a 5-endo-trig cyclisation, which, according to Baldwin's rules should be unfavourable.

The first thing to realise about Baldwin's rules is that they are largely empirical, that is, based on observation of data rather than theoretical proof (though they can be rationalised based upon orbital overlap arguments).

The second thing to realise is that, as with any rule, there are many exceptions.

  From Wikipedia: Baldwin's rules discuss the relative rates of ring closures of these various types. These terms are not meant to describe the absolute probability that a reaction will or will not take place, rather they are used in a relative sense. A reaction that is disfavoured (slow) does not have a rate that is able to compete effectively with an alternative reaction that is favoured (fast). However, the disfavoured product may be observed, if no alternate reactions are more favoured.

The 5-endo-trig acetal formation is actually one of the most well known exceptions to Baldwin's rules (Baldwin himself acknowledged this).

In an 1981 JOC paper (J. Org. Chem. 1981, 46, 2196), Mike Perkins and Jack Barrett attempted to understand why it was that these acetalizations were so 'favourable'.

Using very basic computational methods (MNDO methods), they studied the energy needed in order to get the molecule into the correct conformation for the hydroxyl to attack the oxonium (this is the ring closing step, and the one which you might expect to he the disfavoured step).

The conclusions of the paper are that the barrier is actually much smaller than expected, making formation of the acetal acceptable, in spite of baldwin suggesting otherwise.


What is the difference between electric current and electric charge?

Your question is very broad. The first thing. No they are not the same. Simple explanation for these terms can be:
Charge: Its the "amount" of unbalanced electricity in a body.  Its usually measured in the form of electrons (either excess or deficiency). Its units is coulomb.
Current: It is the "flow" of electrons from higher electric potential towards the lower. Its unit is ampere.

To visualize how they are not the same is by considering two situations (without going into any calculations). Two bodies (A & B) are connected via some copper wire. A has 1 million Coulombs charge & the second one has 1million & 1 coulombs charge. Now individually these charges are enormous, but as the difference between the two is only 1 coulomb, its a very low difference in charge, so a little current will flow between them.

Now if you change the bodies' charge. A has 100 Coulombs & B has 10 .. much more current will flow between the bodies, because there is a way bigger potential difference.

ps. and yes its a physics questions.


Why Does Electrical Work Cause Changes in Internal Energy of the System?

If water is electrolyzed producing $\ce{H2}$ and $\ce{O2}$ gases, a great amount of energy has to be added to this water. The obtained substances are not hotter than the original water. But they possess a much higher potential energy. Their internal energy (and enthalpy, and free enthalpy) is much higher than the same atoms in water. This internal energy may be liberated into heat if these substances are sent in a flame, or into electric energy in a combustion cell.
It is no use thinking about the work done by an electric field on an electron in the atoms. The electric field inside an atom is difficult to calculate, because of all other electrons interfering with the field of the nucleus inside the atom.


The structure as drawn indeed contains 5 chiral carbon atoms.  However, the one carbon that is bound to two oxygen atoms is called the anomeric carbon.  It is part of a hemiacetal moiety in the structure you have drawn.  That means that at room temperature and in aqueous solution, it will equilibrate reasonably quickly between:

One enantiomer of hemiacetal
An open-chain non-cyclic molecule where the anomeric carbon is an aldehyde moiety and is thus not chiral.
The other enantiomer of hemiacetal.

Thus, for many applications it is probably better to think of this molecule as having only four "fixed" stereocenters.


How would you go about synthesising amino sugars from sugars?

First, the anwer is “Would glucose and ammonia work?” is: no, definitely not. The $\ce{OH}$ group is a poor leaving group, and alcohol + ammonia definitely won't yield an amine. Moreover, in your case, you have many hydroxyl groups and you only want one transformed into an amine, which is tricky.

I don't know how the synthesis of the above compound can be achieved, but I know one way to help you start a literature search: your expected compound is called an amino sugar. In this particular case, it's the 6-amino-D-glucose.


Which causes of quenching increase specifically in electrochemical experiments?

I wonder if, when you say UV/Vis spectroscopy, you are also including fluorescence in that. If you are, then the soft iodide / triiodide ions are pretty good reductive quenching agents and they may be responsible for the reduction in fluorescence emission you observe.

Typically, the $\text{I}^-$ will react with the excited state of your dye $\text{D*}$ and quench it by photo-induced electron transfer:

$$\ce{I^{-} + D}^*\ce{\quad ->[\text{photo-induced electron transfer}]\quad I^{\ .} + D^{\ .\ -}\quad ->[\text{back electron transfer}]}\quad \ce{I- + D}$$

As you can see, this process provides a non-radiative pathway for the excited state of the dye to decay back to its ground state. The process is essentially catalyzed by the iodide anion, which is not consumed in the process. The more iodide you have, the more prevalent this process, the more likely it is for the excited state of your dye to go through this process rather than fluorescing, so ultimately the more quenched your dye.

Such a process can be modeled by the Stern - Volmer equation:

$$ \frac{I_0}{I}=1+k_q\tau_0[\text{Q}]$$

Where $I_0$ is the fluorescence intensity in the absence of quencher Q; $I$ the intensity in the presence of a certain concentration [Q] of quencher; $k_q$ is the kinetic constant of the quenching process; $\tau_0$ is the intrinsic lifetime of the fluorophore, i.e. the lifetime you would measure in the same conditions, but in the absence of quencher Q.


First, why would one use ethanol for a Grignard reaction? If a Grignard is destroyed by water (as suggested by the poster) then wouldn't a Grignard also be destroyed by ethanol?

As well as being excellent nucleophiles, Grignard reagents are also very strong bases because they essentially function as alkyl anions. Therefore they will react with ethanol to form magnesium ethoxide, the alkane, and magnesium bromide.

Grignard reactions are usually carried out in ether solvents such as diethylether and tetrahydrofuran because they are unreactive towards the reagent and they actually stabilize it by coordinating to the magnesium ion to form a tetrahedral complex.

(image source)


Excited helium gas from the colision of alpha particles with beta particles?

This is very much a non-issue. Alpha particles run at great speed, so do beta particles, and as soon as they cool down, they cease to be called alpha and beta particles. When that happens, they may react all right, but there is more to it.

Everything on Earth (look around you! yes, that too) is made of atoms. Helium has greater ionization energy than any other atom. So an alpha particle, as soon as it cools down enough to react, will snatch electrons from whatever else is nearby and become a neutral helium atom. Beta particles (provided they are cooled down, too) are as good as anything else.

But wait, they both cool down by bumping into matter, and matter contains electrons. You will have your helium atoms, but you'll never be able to tell whether the electrons come from beta particles or from something else. Nor does it really matter, after all.


Which is the strongest base among the given anilines?

In my opinion, option 4 should be the answer. This is because too much crowding near the nitrogen atom will throw it out of the plane, as a result of the which the lone pair of electrons of nitrogen atom get localized and are better available for accepting the protons. Moreover, presence of $\ce{-CH3}$ groups also increase basicity due to +I effect.


POCP: How many ozone molecules can be created by one ethene molecule?

I think this question cannot be answered so easily - your proposed reaction mechanism, however, looks good. A detailed explanation and "main reaction mechanism" can be found here, page 192 (just search for "ethene" in the book). As you already indicate ethene reacts with a hydroxyl radical, followed by the addition of oxygen. The reaction with NO generates the HOCH$_2$CH$_2$O radical which can either dissociate or react with another oxygen molecule.

Further reactions of the oxidized products occuring during this pathway can also be found in the above reference, see page 372 (unfortunately only some parts of the book are available for free).


Which of the following reaction will have greater enthalpy and why?

Maurice has stated why equation 1 is much more exothermic than equation 2. A visualization of the processes may help you understand the details. Equation 1 involves the reaction of a carbon atom with four atoms of hydrogen to form methane in the gas phase. The bond dissociation energy (BDE) of the hydrogen molecule is +104 kcal/mol. Accordingly, the heat of formation (ΔHfo) of the hydrogen atom is half this value or +52 kcal/mol. The ΔHfo of the carbon atom from graphite is +171 kcal/mol. Both hydrogen and graphite are taken as 0 kcal/mol at the Standard State. The right hand side of the chart shows that the sum total of the necessary atoms is +379 kcal/mol above the Standard State. This is your starting point for equation 1. To form methane in the gas phase the reaction will be exothermic by -379 kcal/mol plus the ΔHfo of methane (-18 kcal/mol) for a total of -397 kcal/mol for equation 1. On the other hand, equation 2 is nothing more than the ΔHfo of methane as -18 kcal/mol below the Standard State as shown in the left hand diagram. This diagram also shows the stepwise BDE's of C-H bonds starting from methane. The same value for equation 1 is obtained, -397 kcal/mol.


How can I change the bond type (single, double) of an existing bond in Avogadro?

As discussed in the Avogadro website about the draw tool you can change the bond order:

By clicking on the bond - the bond order will go from single, to double, to triple, back to single.
By setting the bond order and "drawing over" the bond. The new setting (e.g., double) will replace the existing bond.


Stability comparison of 2-methylpropene and trans-but-2-ene

Since these two compounds are constitutional isomers, their heats of formation may be compared directly. 2-Methylpropene (isobutene) has a heat of formation = -4.0 kcal/mol. Trans 2-butene's heat of formation is -2.7 kcal/mol. Isobutene liberates 1.3 kcal/mol fewer kcal/mol than trans-butene (E-2-butene) upon combustion. The more negative the heat of formation, the more stable. The book gets an A.

Resource: http://webbook.nist.gov/chemistry/


Why do Magnesium and Lithium form *covalent* organometallic compounds?

The character of the bond is determined by the electronegativity of the atoms.

Speaking of bonds as purely ionic or covalent is not always correct - usually it is more correct to say that a bond has ionic or covalent characteristics.

So comparing the difference in electronegativities gives us the following:

$$\begin{array}{cc}\hline
\text{Difference in electronegativity} & \text{Type of bond} \\ \hline
< 0.4 & \text{Non-polar covalent} \\
0.4 \mathrm{-} 1.7 & \text{Polar covalent} \\
>1.7 & \text{Ionic} \\ \hline
\end{array}$$

At the upper end of the polar covalent spectrum, the bonds frequently have both covalent and ionic characteristics.

For organometallic compounds, the difference in electronegativity between Li and C is $1.57$.  So while this is still in the polar covalent range, it is also close to ionic.  Similarly, the difference between Mg and C is $1.24$ - again, a very polar covalent bond.

Compare this to the difference between H and C ($0.35$) - a non-polar covalent bond.

So to answer your question, the thing that is "special" about carbon is that it has a fairly low electronegativity compared to the chalcogens and halogens.  Granted, bonds with carbon are also going to be weaker than in say LiCl, but that's what makes organometallic compounds actually work to form carbon-carbon bonds.

(The electronegativity values came from wikipedia's great chart)


Why is the order of reflection simplified to n = 1?

The common form of the Bragg equation is $\lambda=d_{hkl}\sin(\theta)$ but in crystallography it is used as $\displaystyle \lambda =\frac{2d_{hkl}}{n}\sin(\theta)=2d_{nh,nk,nl}\sin(\theta)$ where the indices are divided through by $n$ to give $\lambda=d_{hkl}\sin(\theta)$. The indices $hkl$ have to be integers (from derivation via Laue equations) but can also be multiples of integers, i.e. they can have a common factor. When the $hkl$ do have a common factor they are not strictly lattice planes but planes parallel to $(h/n, k/n,l/n)$ with a spacing $1/n$ of the lattice. Suppose a plane is (210) then each lattice plane passes through a lattice point (atom). If the indices were (420) then not all planes are equivalent as only half of the points (atoms) pass through a lattice plane so they are not equivalent. Thus sticking with $n=1$ ensures we always use for example, (210), and not (420) or (630) etc., and so always calculate the correct value for $d_{hkl}$


Sulfuric acid is an excellent dessicant and is also very non-volatile itself (it boils at 337 °C).  In other words it would likely still have been there in yet another month.

Your instructor was correct to suggest that you rinse the sulfuric acid with some cold water.  Benzoic acid does have some non-negligible solubility in water (1.7 g/L at 0 °C).  Unfortunately, it is even more soluble in most organic solvents so cold water may well be your best bet.  The solubility in 25 °C water is 3.4 g/L, twice the solubility that it has at 0 °C, so definitely keep it cold.

Additionally, use as little water as necessary to get the sulfuric acid out.  If you are working with a gram or so of benzoic acid, I would start with a dropwise addition of about 10 mL. If it’s still gooey, or still appears wet the next day, you need to repeat the rinse. It’s best if you can use a vaccum filter system of course, and spread the material thin and even before rinsing.


How does common salt speed up the setting of the plaster of Paris?

My guess would be that it absorbs the water in the plaster of paris. Plaster of paris normally has what's called 'water of crystallization' in it. However, it sets too fast with just this, so extra water is added. The salt absorbs that extra water, thereby making it set faster.
Here's a website with details about plaster of paris chemically: http://collections.infocollections.org/ukedu/en/d/Jg16foe/7.4.html
and a book called 'Science in the Real World: A simplified story of how technology using chemistry and physics is used in the real world of industry' has good information on it as well.


Percentage of carbon in stainless steel to avoid brittleness

All  standardised grades of alloys have definite limits of composition.
If its common, then a quick search will find it. 0.15% is the specified limit for SS202.

http://www.azom.com/article.aspx?ArticleID=8209

Whether its brittle or not depends on what you mean by brittle and its thermal processing. Its a martensitic stainless steel so note the details in azom and that it will be more brittle when hardened as described.


The molar heat of fusion for water is 6.02 kJ/mol (http://www.chemteam.info/Thermochem/Molar-Heat-Fusion.html), while the molar heat of vaporization for water is 40.7 kJ/mol (http://www.chemteam.info/Thermochem/Molar-Heat-Vaporization.html). That is equal to 6 times difference, even a little more. I suggest this energy is used to break existing hydrogen bonds and barely for anything else. The energy of hydrogen bonds is "6-30 kJ/mol" according to this source: http://book.bionumbers.org/what-is-the-energy-of-a-hydrogen-bond/.
Let's consider for simplicity, that breaking 1 mole of hydrogen bonds will require 6.02kJ of energy.
So, when water melts, there is maximum one mole of hydrogen bonds is broken per one mole of water. That is totally different for vaporization process: during vaporization of 1 mole of water it looks like 6 moles of hydrogen bonds are broken.
This invariably leads one to answer E:
More hydrogen bonds are broken during vaporization.


Acid-base reaction vs. salt dissolving reaction

tl;dr

  What wasn't clear to me that the second reaction is going to be that of a salt dissolving reaction that is dissolved completely as is claimed in the video.

Remember the rule: "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution"

  I guess there is a way to determine that $\ce{NH4NO3}$ is a salt that is dissolving completely in water and not an acid or base

Realize that all such types of species of the form $\ce{A+B-}$ are ionic salts. If you try draw out the Lewis structure of $\ce{NH4NO3}$, you'll quickly realize it to be very different from that of, say, water or acetic acid.

In ammonium nitrate, you have two separate ions - ammonium cation and nitrate ion - held together by strong electrostatic forces of attraction (often called an "ionic bond"). Water, having a high dielectric constant, is able to break these electrostatic forces easily and dissolve the salt in 100% concentration.

Hence, we say that "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution" i.e. when dissolved in water, they'll dissociate completely.  It seems you already knew the latter part, while the part for ionic salts is new for you.

As I said before, note that ionic salts are different from weak acids or weak bases like $\ce{CH3COOH}$. The $\ce{O-H}$ bond here is covalent and breaking it is an altogether different job as compared to breaking ionic bonds. Here, the reaction (or "equilibrium") has to favor the dissociation of acetic acid. This is ensured by the high stability of the conjugate base. If you've taken GOC classes, you would quickly realize that the conjugate base $\ce{CH3COO-}$ is much weaker than say $\ce{HSO3-}$ or $\ce{NO3-}$ (guess which acids the latter two are? ;) )

Again, realize that the rule is actually not completely correct. Later down the equilibrium road, you'll come across the equilibrium constant $K_\mathrm{sp}$ that defines exactly how much soluble certain salts are in aqueous solutions. Certainly, many ionic salts would be in that list, and that too with very low solubilities instead! Again, the way these people simplify calculations at an elementary level is to implicitly assume that "All ionic salts whose $K_\mathrm{sp}$ is provided are only partially soluble, while those whose $K_\mathrm{sp}$ is not provided are completely soluble (i.e. 100% dissociated)"


Which of the following compounds possess more than one percent enol content?

All the compounds that have a relation $\alpha-\beta$ between the carbonyl groups. The hydrogen between two $\ce{CO}$ is very acid because two electrowithdrawing remove electron density from the $\ce{C-O}$ bond. Therefore for this kind of compound the most stable for is the enol. In Organic Chemistry this is very useful in creating new carbon-carbon bonds in mild conditions like in aldoholic reactions, Michael reactions and so on. See the very good book of Marc Loudon, Organic Chemistry.


Why is coulomb integral is called local while the corresponding exchange integral is called non-local

The definitions of local and non-local operators are given at p. 12 of the book. Basically, an operator $\hat{A}$ is called local if the value of $\hat{A} f(x)$ at some point $x=x'$ depends only on the value of $f(x)$ at the very same point $x=x'$ and not on the values of $f(x)$ at any other points. Otherwise an operator is called non-local.

For instance, any $j$-th Coulomb operator in Hartree-Fock theory is local since the value of $\hat{J}_{j} \psi_{i}(q)$ at $q=q'$ depends only on the value of $\psi_{i}(q)$ at $q=q'$, while any $j$-th exchange operator is non-local since the value of $\hat{K}_{j} \psi_{i}(q)$ at $q=q'$ depends on the values of $ψ_{i}(q)$ at all points $q$.


Flammable lighter than air gases (safeish)

I would propose to use hydrogen, as it is pretty easy to obtain right there where you need it and it's the lightest gas ever.
You can even watch a  video where the whole setup for the bubble experiment is described  (at the end).
Besides hydrogen, methane looks good to me, and it is also in our everyday life, being the natural gas. But the setup for using household methane is imagined more complicated to me than the one with hydrogen.
Just be very careful with any flammable gas, don't make it another Hindenburg :)


Are there any actual uses of isodiaphers?

In practical radiochemistry, the term is rarely needed but it is not useless. In particular, isodiaphers are used in radiochemistry to describe chains of alpha decays.
I would expect, the relative frequencies of the practical use of the related terms is about
"isotopes" > "isobars" > "isodiaphers" > "isotones".
Isobars are used in radiochemistry to describe chains of beta decay, therefore their importance might be similar to isodiaphers for alpha decay. However, isobars are also very important to describe fission yields so that in total they might be used more often than isodiaphers.


Verifying data acquired from a molecular dynamics simulation

With the simple molecules in your system, using the Gibbs Ensemble Monte Carlo would be the way to go. Each phase gets its own simulation box, and particles are swapped back and forth until equilibrium is reached.

Cassandra, DL_MONTE and other programs can do this easily for you. Once equilibrium is reached you can sample each box in the NVT or NPT ensemble at that composition.


Deriving an expression for the constant pressure heat capacity

You could start by dividing your whole equation by $T$:

\begin{align}
 \frac{G}{T} = \frac{H}{T} - S
\end{align}

and then take the derivative with respect to $T$ at constant $p$:

\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} = \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \ .
\end{align}

That way you've already introduced the desired derivative of the entropy with respect to $T$ at constant $p$.
Then you can use that

\begin{align}
 \frac{G}{T} = \frac{H}{T} - S  \qquad &\Rightarrow& \qquad \mathrm{d} \!\left(\frac{G}{T}\right) &=  - \frac{H}{T^2} \mathrm{d} T + \frac{1}{T} \underbrace{\mathrm{d} H}_{= T \mathrm{d}S + V\mathrm{d}p} - \mathrm{d} S \\
&&&= - \frac{H}{T^2} \mathrm{d} T + \frac{V}{T} \mathrm{d}p \\
&\Rightarrow& \qquad \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= - \frac{H}{T^2}
\end{align}

and simply using the product rule

\begin{align}
 \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} = \frac{1}{T} \underbrace{\left(\frac{H}{\partial T}\right)_{p}}_{= C_{p}} + H \underbrace{\left(\frac{\partial \frac{1}{T}}{\partial T} \right)_{p}}_{=-\frac{1}{T^2}} = \frac{C_{p}}{T} - \frac{H}{T^2} \ .
\end{align}

Putting it all together gives you the desired result:

\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
- \frac{H}{T^2} &= \frac{C_{p}}{T} - \frac{H}{T^2} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
\Rightarrow \left(\frac{\partial S}{\partial T} \right)_{p} &= \frac{C_{p}}{T} \ .
\end{align}


Can you combine elemental gold and boron?

Yes, gold and boron can bond with each other to form alloy clusters. They are very important in semiconductor industry and used in fabrication process. Here(1) is a short note on gold-boron alloy:

  Photoelectron spectroscopy and density-functional theory are combined to investigate the electronic and structural properties of a
  series of $\ce{B-Au}$ alloy clusters: $\ce{B6Au_{n}^{−}}$ and
  $\ce{B6Au_{n}}$ ($\ce{n = 1−3}$). Rich spectral features are
  observed for each species, and vibrational structures are resolved for
  numerous detachment transitions of $\ce{B6Au^−}$ and
  $\ce{B6Au^{2−}}$. Electron affinities of $\ce{B6Au_{n} (n = 1−3)}$ are
  evaluated to be 2.70 ± 0.03, 2.91 ± 0.02, and 3.21 ± 0.05 eV,
  respectively. Global structural searches are performed for both the
  anions and their neutrals. The calculated electronic binding energies
  are compared with experimental measurements to establish the anion
  global-minimum structures and their corresponding neutral states.
  The ground-state structures of these clusters are shown to be planar
  or quasi-planar with an elongated B6core, to which the first and
  second Au atoms are bonded terminally and the third Au in a bridging
  position. All three anion clusters are π antiaromatic.
  Natural bond orbital analyses show that the $\ce{B-Au}$ bonding is
  highly covalent, providing new examples for the $\ce{Au/H}$ analogy
  in Au alloy clusters.

You can find more information about gold-boron alloy clusters in the reference below.

Reference

The Journal of Chemical Physics 138, 084306 (2016); doi: http://dx.doi.org/10.1063/1.4792501
https://www.google.com/patents/US3211550
https://www.ncbi.nlm.nih.gov/pubmed/23901981
http://aip.scitation.org/doi/10.1063/1.4816010
https://www.researchgate.net/publication/235886590_On_the_structures_and_bonding_in_boron-gold_alloy_clusters_B6Aun-_and_B6Aun_n_1-3
http://www.google.co.in/patents/US3137595


Does an orbital have its own energy, separate from the electron? Can I picture them as a entity independent from the hosted particle?

An orbital of current quantum models of atoms and molecules has 3 major interpretations:

orbital(1) as  a wave function, being a solution of the quantum wave equation describing behaviour of electrons around an atomic kernels. Functions do not have energy, even if they describe it.

orbital(2) is the 3D region describing probability of occurance of an electron, related to orbital(1) . Space regions do not have energy either.
(Not exactly true, considering radiation energy and quantum vacuum energy, but nothing comparable to the electron energy at usual conditions.)

orbital(3) is a quantum state of an electron, described by orbital (1) and orbital (2), consisting of set of discrete values of physical quantities as functions of quantum numbers, including energy. In the case this quantum state is not occupied, the orbital energy means "an electron would have this energy".

orbital (4) from the obsolete Bohr model means a path of an electron, which is a crude approximation of the quantum model of an atom.

Last, but not least: Orbitals are not physical things, they are part of the quantum theoretical model of atoms and molecules. Within the given quantum model, an electron cannot have different energy than energy of it's quantum state.

As models of multielectron cases are imperfect, real electron energy usually more or less differ from its energy predicted for the given quantum state.


How to find the lifetime of sodium atoms in the 3p using information on the spectral line width?

Half the fwhm has to be added/subtracted  to the central wavelength and then this change in wavelength converted to an energy. Thus the limits on the wavelength are $\lambda \pm \Delta \lambda /2) $ and the energy change $\Delta E=hc/(\lambda - \Delta \lambda /2)-hc/(\lambda + \Delta \lambda /2)$ Joules with $c$ in m/s and $h$ in J s. Plugging in the numbers gives a lifetime of $\approx 10^{-9}$, i.e. $1$ ns.

As a rule of thumb, lifetimes of excited electronic states are rarely less than $10^{-12} $s and if the transition is allowed, as in this case often of the order of nanoseconds to microseconds. If transitions are forbidden then the lifetimes can extend from microseconds to seconds.


Is a carbon-fluorine bond stronger than a carbon-chlorine bond?

For free radical reactions, the most important parameter in assessing bond strength is bond dissociation enthalpy (BDE). Typical values for $\ce {C-F}$ bonds are around $\mathrm {100\ kcal/mol}$, while for $\ce {C-Cl}$ bonds are around $\mathrm {80\ kcal/mol}$; $\ce {C-Cl}$ bonds are therefore weaker.

This can be rationalized by considering the poorer orbital overlap between the $\ce {Cl}$ bonding orbital (3 s/p character, more diffuse than 2s/p) and the $\ce {C}$ $\mathrm{sp^3}$ hybrid (predominantly $\text{2p}$ character) than that between the $\ce {F}$ bonding orbital (2 s/p character) and the $\ce {C}\ \mathrm{sp^3}$ hybrid.

Some data on typical BDE values in $\ce {F/Cl}$ compounds can be found here.


Is there any way to chemically check sweetness level in any drink or water or tea etc?

As asked by the OP, I think the answer to the question has to be no. If it was known that a drink contained only glucose as a sweetener, then it would presumably be possible to measure accurately the glucose concentration in the drink. But maybe other substances in the drink would be interferences in the measurement. Following on, as @Alchimista commented, perhaps a Scoville-type sweetness test could be used. But the possible interferences problem complicates this as well. In addition, perceptions of sweetness may also be affected by seemingly innocuous substances in the drink, e.g., I find that a dash of table salt makes watermelon taste sweeter and I have no clue how that might arise.

Then there is the matter of inorganic substances. The old fashioned name for lead acetate was ‘sugar of lead’. Poisonous though it is, it earned its deadly sweet name. As well, some salts of beryllium ('glucine'; all berylium salts are highly poisonous) and yttrium are sweet tasting: see wiki. So it is not just various organic substances that can taste sweet and who knows what else is sweet tasting: no sane chemist (nowadays) tastes chemical substances.

My guess is that there are quite a few chemicals that would taste sweet, but we will probably never know what they are. Then how would sweetness levels of two lead acetate aqueous solutions be compared? I don’t see how it can be assumed, a priori, that sweetness is directly proportional to lead ion concentration and no one should do such an human subjects experiment anyway. And what about a drink containing glucose and lead acetate? Even if the lead acetate did not interfere with the quantification of the glucose concentration, the drink would be sweeter than expected from just the glucose measurement result.


Accuracy benchmarks for atomic multipole moments

To the best of my knowledge, there is no definition of "chemical accuracy" for higher-order (electric) multipole moments, in part because they are difficult to measure experimentally, with relatively large uncertainties. I found one recent paper that has a small comparison, prefaced with:

  There are several techniques to determine experimentally the dipole moments, [4,5] but it is still very difficult to obtain precise experimental values of higher multipole moments such as quadrupole or octupole moments, [1,6,7] independently of the experimental conditions. Theoretical calculations are therefore essential but challenging for quantum chemistry methods. The accurate calculation of these properties is highly dependent on the method employed, [8] either regarding approximate density functionals [9] or methods based on wavefunctions. [10,11] Consequently, calculating the multipole moments is a way to assess any electronic structure method.

The authors use this work as a method development showcase for the Piris natural orbital functional 6 (PNOF6), the details of which I will generally leave out, but they provide valuable data for Hartree-Fock (HF), coupled cluster with single and double excitations (CCSD), and multireference configuration interaction with single and double excitations (MR-CISD, they abbreviate as MRSD-CI after the old abbreviation SDCI). I will leave in the PNOF data, since it (to my understanding) is a natural orbital-based almost black-box correlated method when the number of occ-virt correlated pairs can be controlled. For the molecules studied (small, organic, closed-shell), CCSD in a triple-$\zeta$ basis set is most likely a reasonable choice.

  Table III. $\Theta_{zz}$ component of the quadrupole moments, in atomic units, computed with the Sadlej-pVTZ basis set at the experimental equilibrium [42] geometries.

  \begin{array}{lllll}
\hline
                     & \text{HF} & \text{CCSD} & \text{Expt.}             & \text{PNOF6} \\
\hline
\ce{H2}              & 0.4381    & 0.3935      & 0.39 \pm 0.01            & 0.3935 \\
\ce{HF}              & 1.7422    & 1.7156      & 1.75 \pm 0.02            & 1.6939 \\
\ce{BH}^{a}          & 2.6772    & 2.3388      & 2.3293~\text{(FCI/aVTZ)}^{b} & 2.3706 \\
\ce{HCl}             & 2.8572    & 2.7233      & 2.78 \pm 0.09            & 2.7753 \\
\ce{HCCF}            & 3.3530    & 2.9335      & 2.94 \pm 0.10            & 3.2482 \\
\ce{CO}              & 1.5366    & 1.4889      & 1.44 \pm 0.30            & 1.4562 \\
\ce{N2}              & 0.9397    & 1.1712      & 1.09 \pm 0.07            & 1.0530 \\
\ce{NH3}             & 2.1258    & 2.1661      & 2.45 \pm 0.30            & 2.1080 \\
\ce{PH3}             & 1.7217    & 1.5695      & 1.56 \pm 0.70            & 1.6507 \\
\ce{ClF}             & 0.9413    & 1.0514      & 1.14 \pm 0.05            & 1.1122 \\
\ce{CH3F}            & 0.3482    & 0.3002      & 0.30 \pm 0.02            & 0.3269 \\
\ce{C2H2}            & 5.3655    & 4.6850      & 4.71 \pm 0.14            & 5.1531 \\
\ce{C2H6}            & 0.6329    & 0.6234      & 0.59 \pm 0.07            & 0.6275 \\
\ce{C6H6}            & 5.5418    & 5.6653      & 6.30 \pm 0.27            & 6.3571 \\
\ce{CH3CCH}          & 4.2913    & 3.6939      & 3.58 \pm 0.01            & 4.1146 \\
\ce{CO2}             & 3.8087    & 3.1966      & 3.19 \pm 0.13            & 3.6012 \\
\hline
\text{MAE}           & 0.2646    & 0.0902     &                           & 0.1517 \\
\text{MAE}~(\mu_{z}) & 0.0843    & 0.0177     &                           & 0.0309 \\
\hline
\end{array}

  a. Calculations performed with the aug-cc-pVTZ basis set.
  b. Full CI calculation reported by Halkier [29].

I included the mean absolute error (MAE) for the dipole results from Table II to give an idea of the relative difficulty compared to calculating quadrupole moments. The quadrupole MAEs are about 3, 5, and 5 times larger than the dipole MAEs for each method.

Plotted, it looks like

which is admittedly not very useful other than to show pictorially that CCSD is generally in at least qualitative agreement with the experiment. I personally would consider HF in qualitative agreement to the level that HF usually gets geometries qualitatively correct, though bond lengths are usually too short due to lack of electron correlation. Mean-field electrostatics are the dominant contributor.

For percent errors,

\begin{array}{lrrr}
\hline
&  \delta_{\text{HF to CCSD}} &  \delta_{\text{CCSD to Expt.}} &  100 * \text{err}/\text{Expt.} \\
\hline
\ce{H2}        &                       11.33 &                           0.90 &                           2.56 \\
\ce{HF}        &                        1.55 &                           1.97 &                           1.14 \\
\ce{BH}        &                       14.47 &                           0.41 &                           0.00 \\
\ce{HCl}       &                        4.92 &                           2.04 &                           3.24 \\
\ce{HCCF}      &                       14.30 &                           0.22 &                           3.40 \\
\ce{CO}        &                        3.20 &                           3.40 &                          20.83 \\
\ce{N2}        &                       19.77 &                           7.45 &                           6.42 \\
\ce{NH3}       &                        1.86 &                          11.59 &                          12.24 \\
\ce{PH3}       &                        9.70 &                           0.61 &                          44.87 \\
\ce{ClF}       &                       10.47 &                           7.77 &                           4.39 \\
\ce{CH3F}      &                       15.99 &                           0.07 &                           6.67 \\
\ce{C2H2}      &                       14.53 &                           0.53 &                           2.97 \\
\ce{C2H6}      &                        1.52 &                           5.66 &                          11.86 \\
\ce{C6H6}      &                        2.18 &                          10.07 &                           4.29 \\
\ce{CH3CCH}    &                       16.17 &                           3.18 &                           0.28 \\
\ce{CO2}       &                       19.15 &                           0.21 &                           4.08 \\
\hline
\end{array}

this is further confirmation that CCSD is probably a good qualitative benchmark.

  Table IV. $\Theta_{zz}$ and $\Theta_{xx}$ components of molecular quadrupole moments, in atomic units, computed using the Sadlej-pVTZ basis set at the experimental equilibrium [42] geometries.

  \begin{array}{lrrr}
\hline
                & \text{HF}  & \text{MR-CISD} & \text{Expt.}  & \text{PNOF6} \\
\hline
\ce{H2O}~(xx)   &  1.7966    & 1.8050         & 1.86 \pm 0.02 & 1.7808 \\
\ce{H2O}~(zz)   &  0.0981    & 0.0950         & 0.10 \pm 0.02 & 0.0869 \\
\ce{H2CO}~(xx)  &  0.1019    & 0.1100         & 0.04 \pm 0.12 & 0.0516 \\
\ce{H2CO}~(zz)  &  0.0921    & 0.2230         & 0.20 \pm 0.15 & 0.1255 \\
\ce{C2H4}~(xx)  &  2.7819    & 2.3700         & 2.45 \pm 0.12 & 2.5892 \\
\ce{C2H4}~(zz)  &  1.4942    & 1.1700         & 1.49 \pm 0.11 & 1.3266 \\
\ce{O3}~(xx)    &  1.1175    & 1.2830         & 1.03 \pm 0.12 & 1.2426 \\
\ce{O3}~(zz)    & -0.2387    & 0.1680         & 0.52 \pm 0.08 & 0.3606 \\
\hline
\text{MAE}      &  0.1772    & 0.1448         &               & 0.1066 \\
\hline
\end{array}

The ozone result for Hartree-Fock most likely has the wrong sign due to its multiconfigurational ground state, with the MRCI result still in relative error of ~67%.

For octupole moments, they only present results for methane ($\ce{CH4}$), for which the first non-vanishing multipole moment is the octupole moment, due to its $T_{d}$ symmetry.

  \begin{array}{ll}
\hline
                 & \Omega_{xyz}~(\text{a.u.}) \\
\hline
\text{CCSD}      & 2.0595                     \\
\text{PNOF6(14)} & 2.1142                     \\
\text{Expt.}     & 2.95 \pm 0.17              \\
\hline
\end{array}

Considering the lower bound of the experiment, the PNOF6 result is in 23% error, and considering the upper bound of the experiment, the CCSD result is in 36% error.

I won't draw any more conclusions about the data other than to say that in general, there may be large errors in the experimental data that are hard to predict (non-systematic). In that case, if you were to perform a study looking at convergence of a result with respect to some other parameter (say, cluster size), I would first look at convergence of said result with respect to method (HF, MP2, CCSD) and basis (aVDZ, aVTZ, ...) at the largest cluster size, just to check that CCSD/aVTZ is a reasonable benchmark if you're after absolute accuracy. Since experimental error is large, it may be more reasonable to just compare against higher-level, full cluster calculations.

References

0: Mitxelena, Ion; Piris, Mario. Molecular electric moments calculated by using natural orbital functional theory. J. Chem. Phys. 2016 144, 204108, DOI: 10.1063/1.4951685
1: A. D. Buckingham. Direct Method of Measuring Molecular Quadrupole Moments. J. Chem. Phys. 1959 30, 1580,  DOI: 10.1063/1.1730242
6: W. H. Flygare and R. C. Benson. The molecular Zeeman effect in diamagnetic molecules and the determination of molecular magnetic moments (g values), magnetic susceptibilities, and molecular quadrupole moments. Mol. Phys. 1971 20, 225, DOI: 10.1080/00268977100100221
7: E. R. Cohen and G. Birnbaum. Far infrared collision‐induced absorption in gaseous methane. II. Determination of the octupole and hexadecapole moments. J. Chem. Phys. 1975 62, 3807, DOI: 10.1063/1.430932
8: R. Glaser, Z. Wu, and M. Lewis. A higher level ab initio quantum-mechanical study of the quadrupole moment tensor components of carbon dioxide. J. Mol. Struct. 2000 556, 131, DOI: 10.1016/S0022-2860(00)00658-X
9: A. J. Cohen and Y. Tantirungrotechai. Molecular electric properties: an assessment of recently developed functionals. Chem. Phys. Lett. 1999 299, 465, DOI: 10.1016/S0009-2614(98)01317-7
10: P. Bundgen, F. Grein, and A. J. Thakkar. Dipole and quadrupole moments of small molecules. An ab initio study using perturbatively corrected, multi-reference, configuration interaction wave functions. J. Mol. Struct. 1994 334, 7, DOI: 10.1016/0166-1280(94)03974-P
11: J. M. Junquera-Hernández, J. Sánchez-Marín, and D. Maynau. Molecular electric quadrupole moments calculated with matrix dressed SDCI. Chem. Phys. Lett. 2002 359, 343, DOI: 10.1016/S0009-2614(02)00722-4
29: A. Halkier, H. Larsen, J. Olsen, P. Jørgensen, and J. Gauss. Full configuration interaction benchmark calculations of first-order one-electron properties of BH and HF. J. Chem. Phys. 1999 110, 734, DOI: 10.1063/1.478042


Why does one of Dalton's compound elements with oxygen and phosphor consists of four atoms instead of two?

The plate is from Dalton's book A New System of Chemical Philosophy.

The explanation of the plate is on pages 546-547 of the book.

The explanation says that "51" is "phosphorous acid" with weight 32.

(keep in mind that Dalton is basing this on O=7 and P=9)

"52" is "phosphoric acid" with weight 23.

The discussion of phosphorous acid is mostly on pages 408-410.

Initially, he contradicts the plate saying "I am inclined to believe" that phosphorous acid is one O and one P with a weight of 7 + 9 = 16.

But then he says:

  If this be the case, it may appear singular that none of the other elements exhibit acid properties when combined with 1 atom of oxygen ; but it should be observed, that the phosphoric oxide is in a liquid form, and disposed to separate into phosphorus and phosphoric acid, circumstances that do not combine in regard to the other oxides.  In fact, phosphorous acid may be considered as phosphoric acid holding phosphorous in solution, rather than a distinct acid.

So, if phosphoric acid (one phosphorous atom and two oxygen atoms accord to Dalton) "holds" another phosphorous atom, the result is what is show on the plate.


Is there any difference between getting dissolved and getting dissociated?

This is a simple logical and semantic problem, which is not a problem at all. Your professor is right and wrong- both at the same time. He is creating a classification which does not exist and which is meaningless.

Look at the word origin of electrolyte:

  Etymology from OED: < electro- comb. form + ancient Greek λυτός that may be
  dissolved, soluble (see -lyte comb. form). Compare electrolysis n.,
  electrolyse v.

  and the portion -lyte "< ancient Greek -λυτος, combining form (in e.g.
  εὐδιάλυτος : see eudialyte n.) of λυτός that may be dissolved, soluble
  < λύειν to loose (see lysis n.) + -τός, suffix forming verbal
  adjectives. Compare French -lyte, German -lyt.

  A substance which gives rise to ions when dissolved (typically in
  water) or fused; a liquid or gel which contains ions and can be
  decomposed by electrolysis; spec. the conducting fluid used in a
  battery.

Does this solve the semantics problem?

By definition, an electrolyte is soluble in a given solvent. It is also understood among electrochemists that an electrolyte is added to a solvent to impart conductivity to that solvent. I will never call sugar an electrolyte because nobody would use sugar to increase conductivity of water. You should ask: Does a given substance produce ions in a solution? If yes, it can be called as an electrolyte.

Look at this logic: grapes is a fruit but not all fruits are grapes.

Electrolytes are soluble in the solvent, but not all soluble substances in that solvent behave as electrolytes. Anything which produces ions in that solvent, can be labelled as an electrolyte.

Q-1 If solubility and dissociation are different then what would be the criteria for getting dissolved and what will be criteria for getting dissociated (for me the criteria for getting dissolved is dissociation).

The measure of dissociation is conductivity. How much electrical resistance is offered by that dissolved (and dissociated) substance in that solvent

Q-2 how would you arrange the following in increasing order of 1.solubility 2.dissociation

NaCl, MgCl2, FeCl2, ZnCl2

There is no correlation between solubility and dissociation.  If I want to study their dissociation I will measure the conductivity of their solutions systematically.


What's the most effective way to reliably trigger and observe hard water scale buildup at a micro level?

You have to find out what kind of hardness is present in your water. Is it permanent hardness or temporary hardness? The latter goes away with boiling.

The easiest and visual demo for children is to show the behavior of hard water with a soap bar. Soap when added to hard water does not lather very well. Watch the video here:

https://www.youtube.com/watch?v=jDgSohKVLio

https://www.youtube.com/watch?v=bBYsfEHKYCk

As a control, you would need to buy distilled water.

Does your electric kettle develop crusts with time?


The English names are not very systematic, they just give hints.  In principle you have to either memorize the name-structure correspondence, or know "all" acids of sulfur, so based on the name you can guess what the structure it corresponds to.

In addition, the hypo having one oxygen less is just a shortcut. What the names refer to is the oxidation state of central atom from lowest to highest.
Broadest range of acids are for chlorine: hypchlorous - chlorous - chloric - perchloric.

Answer 1: Please refer to the Wikipedia page on sulphur oxo-acids
.
The $\ce{H2S2O5}$ acid drawn is hypothetical one, but it's fine to have a name for it. Note, that sulfur is in low oxidation state.
In the pyrophosphoric acid, the phosphorus is in high oxidation state and has extra oxygen in between them.

For the summary formula $\ce{H2S2O3}$ you try to mix these properties, putting "extra" oxygen between sulfurs in low oxidation state. This won't work. The acid you should get when you attempt to obtain this one will be the thiosulfuric acid, where one sulfur mimics oxygen as $\ce{S^2-}$.

Answer 2: Chemistry of phosphorus is distinctly different from sulfur.
Phosphoric acid and Phosphorus acid for the acids with one phosphorus atom.
Your formula $\ce{(H3PO3)_{$n$}}$ with $n=2$ is not really justified, the name corresponds just to the formula on hypophosphoric acid.


Does standard electrode potential of a reaction depend on stoichiometric coefficients?

If you think of the Nernst equation, you'll see that: $$\Delta G° = -nFE°$$

Since $n$ corresponds to the number of electrons transferred, the ratio $\frac{\Delta G°}{n}$ remains constant, because any change in the stoichiometric coefficients will multiply both terms by the same value.

Therefore, we see that the potential is an intensive property, not depending on the dimensions of the system under study.


Active transport of Ca2+ ions against chemical and electrostatic gradients, assisted by ATP hydrolisis

At equilibrium, the entire process (transport of $2~\ce{Ca^2+}$ ions and hydrolysis of $1~\text{ATP}$) involves no change in G.

\begin{align}
0 &=\Delta{G_\mathrm{total}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+},\text{ concentration gradient}}} + 2\Delta{G_{\ce{Ca^2+},\text{ electrostatic}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi\\
\end{align}

So basically I'm saying you made two mistakes, one a sign error and one a factor of 2 error.


No, you do not evaporate the metals in question.  The use of aqua regia in the analysis aims to dissolve them, to form salts, which in turn are significantly more volatile than the metals or will decompose in the gaseous state.  (Just have a look alone at the melting points of the pure metals, which still are considerably lower than the boiling points, too.)

Without further detail about the colour of the "smoke", it remains a speculation that the sample placed on the L'vov plate still might have been a bit wet (from the acidic digestion).  If so, it were not good for the instrument, as it may lead to corrosion to metal parts of the housing around the graphite tube.


Does everything exist as a gas at p = 0?

Regardless of whether a substance is in a vacuum, it will vaporize if some fraction of its particles have enough energy (due to their temperature) to exceed their binding energy.  Binding energy is the energy contained in the intermolecular bonds holding a substance together.  It is definitely possible for the particles of a substance to have a binding energy that is substantially higher than thermal energy will readily provide.

No.  While most liquids have fairly low binding energies (and thus, relatively high vapor pressures) there are liquids that will not evaporate in a vacuum at an appreciable rate, one example being ionic liquids.  As I wrote above, the attraction between the ions of these liquids is strong enough that they do not measurably evaporate over time.
If you know the Periodic Videos YouTube channel, they've had a couple of videos about these fluids over the past couple of years.  (Here's a link to one, if you're interested.)  One of them featured the behavior of liquids in a vacuum, but I can't seem to find the link; it may have been on a sister channel.
No, consider all of the asteroids, space ships, satellites, and such floating around in space.  For all intents and purposes they will float around until the end of time without evaporating.  Again, the reasoning is as I listed above.  We also don't see the vast majority of solid objects here on Earth slowly evaporating away, which we would expect if sublimation were a concern in a vacuum.  The reasoning is the same as above.
If the solid is very, very cold, it is less likely to experience sublimation since its particles' average energies would be very low.  Even dry ice would not vaporize if its temperature were very close to 0 K.

Here is a phase diagram of both water and carbon dioxide that might help illustrate the situation.  You can see that as the temperature of the substance decreases the vapor pressure of the substance also decreases.  At a sufficiently low temperature the vapor pressure is effectively zero (below the x-axis of this chart.)  There is always the possibility of a single atom or molecule gaining enough energy to vaporize or doing something weird like tunneling off of the surface of the substance, but the process is so slow as to be unmeasurable.  Again, asteroids in the solar system have been hanging around for several billion years without subliming.

Picture from Wikipedia.

Note: Be careful if you look up phase diagrams of common substances.  They often involve only a few points that are actually referenced and the rest is drawn with quite a bit of artistic license.  Compare the phase diagrams of water here and here to see what I mean.  They are often drawn without scales (or with minimal scales) and are sloppy at best.


Does hydrogen bonding strength correspond to Brønsted basicity in a given medium?

@Dissenter, you really need to stop asking questions like this.  Once again you point out that the emperor has no clothes!  :)

It turns out that the mechanism of simple proton transfer is an active area of investigation.  There is not a single yes or no answer that would adequately address your questions.  But if this truth were told in the introductory texts who would continue on when the answer to seemingly simple questions leads to such a morass?  Instead, particularly in introductory circumstances, there is a gentlemen's agreement of sorts that we will simply say that the emperor's clothes are the finest we've seen.

Let's look at your two questions in reverse order.

  why do we depict acid base reactions as concerted?

In many introductory texts, and even some that are for advanced courses, proton transfer is presented as a concerted process based largely on Occam's razor (the simplest explanation is often the correct explanation).  However, enough experiments in this area have been run that I think it would be fair to say that a stepwise mechanism is now favored, but by no means mandated for all proton transfers (we'll see why some proton transfers proceed in a concerted fashion below).  The stepwise mechanism proceeds as follows:

step 1, formation of the encounter complex $\ce{A-H}$ and $\ce{B-}$ diffuse through solution until they are close enough to form a hydrogen bond

$$\ce{A-H + B- -> A-H..B-}$$

step 2, the $\ce{A-H}$ bond elongates while the $\ce{B-H}$ bond contracts and a transition state is reached.

$$\ce{A-H..B -> [A..H..B]}$$

The transition state is not necessarily linear or symmetric.  There could also possibly be an intermediate accompanying the transition state.  This depends upon your definition of an intermediate.  How long does a species need to survive in order to be called an intermediate; is it 10 vibrations, 1 vibration or what?  Obviously it is difficult to detect such short-lived molecules, but ultra-fast spectroscopy has made great strides and we may soon know whether an intermediate exists along this multi-step reaction coordinate.

step 3, the transition state elongates in the opposite direction

$$\ce{[A..H..B] -> A..H-B}$$

step 4, the hydrogen bond is broken and the complex diffuses apart

$$\ce{A..H-B -> A- + B-H}$$

  Does hydrogen bonding strength correlate with Brønsted basicity in a
  given medium?

If we throw out cases where geometry interferes with hydrogen bonding (e.g. one example would be where steric factors preclude close approach or proper alignment of the species involved), then it would not be unreasonable to expect some correlation between hydrogen bond strength and the pKa difference between the two species involved.  There is certainly experimental evidence to support this view.  However, it has been found that if we step back and examine a larger range of pKa differences, the correlation breaks down.

Here is a diagram showing how the rate of proton transfers might change as a function of the pKa difference between the two species involved.

If we look at a small portion of the plot, the part at low pKa, we could fairly easily convince ourselves that there is a linear correlation.  But eventually as the pKa becomes large, the rate of proton transfer becomes diffusion controlled and insensitive to any further increase in the pKa.  The same concept applies to hydrogen bonding.  There could be reactions (acid-base proton transfers for example) where there is no activated process, the reaction is diffusion controlled; as a consequence we cannot expect to accurately measure or compare hydrogen bond strengths as a function of Brønsted basicity differences in such a regime.

Summary Answers

  why do we depict acid base reactions as concerted?

To keep it simple in introductory classes.  It turns out that many acid-base proton transfers are not concerted, but rather involve a multi-step process.  Some other proton transfers, for example those involved in diffusion controlled reactions, are probably concerted.  This is still an active area of research.

  Does hydrogen bonding strength correlate with Brønsted basicity in a
  given medium?

The answer is "yes" if we 1) examine only a small range of basicities and 2) look in the range of basicities where the hydrogen bonding is described as an activated process.


Why is density of water maximum at 4 degree celsius?

The answer appears to be fairly complicated, but fundamentally it results from a competition between two different options for water molecules interacting.

One place to start would be The structure of water; from ambient to deeply supercooled, Lars G.M. Pettersson and Anders Nilsson, J. Non-Crystalline Solids 407 399-417 (2015) where they discuss the 'inhomogeneous structure hypothesis'. At its heart they define high density liquid (HDL) and low density liquid (LDL) (not connected to cholesterol!) structures. LDL is tetrahedral, HDL is more distorted. LDL is directional H-bonded, while HDL is more van der Waals influenced. LDL minimized enthaly, HDL maximizes entropy. These are all consistent with a variety of experimental and simulation work given in the paper.

A recent paper applying DFT to the problem, Interstitial Voids and Resultant Density of Liquid Water: A First-Principles Molecular Dynamics Study, Sohag Biswas et al., ACS Omega 3 2010-2017 (2018), suggests a similar general picture. Again, the tradeoff of hydrogen vs van der Waals bonding appears to be important. Different exchange/correlation terms provide somewhat different answers. One suggested reason for the HDL structure is that, by incorporating an 'interstitial' water molecule into the LDL, one disrupts the directional hydrogen bonds and enhances the breaking and making of them, blurring out the structure (while being higher density).

So, it isn't a simple one-phase picture, but still seems to be a tradeoff of entropy and enthalpy of different configurations in the water.


Evaporation isn't a good option since it costs too much energy to heat the big amounts of water.

Is it? I've seen prices around USD 130-150 per ton (minimum order of 25 tons) for ammonium sulfate with a purity around 99%. You might want to calculate the costs to operate a production plant using natural gas (or other fuels) for evaporation yourself.

There is however a different energy source that you might want to take into consideration: the sun.

Evaporation of water from solutions of sodium chloride in open ponds by solar heating alone has a long tradition in the production of sodium chloride in the form of sea salt.

Update

Ammonium sulfate will not evaporate from the open ponds. At temperatures between 235-280 °C, it will start to decompose.

However, if surface temperatures reach that level under irradiation with unconcentrated sunlight, the fate of the ammonium sulfate is the least of your worries ;)


External and internal pressure

To study the effect of pressure on properties of a solid, is equivalently to learn how changes in volume transform physical parameters. For external pressure at constant temperature, this relationship manifests through compressibility $\kappa$.

$$\kappa =-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$$

An alteration in volume will modify the energy of a system. The relationship is internal pressure, something chemists first encounter when discussing real gases.

$$\pi_T=\left(\frac{\partial U}{\partial V}\right)_T$$

In solids, one might discover different notations. For example[1],

$$\alpha =\left(\frac{\partial W}{\partial V}\right)_T \sim \frac{\mathrm{d}W}{\mathrm{d}V} $$

where $W$ is potential energy, and $\alpha $ is called energy–volume correlation. Disregarding possible sorption, varying external pressure is the purest option of probing internal pressure.[2] Unfortunately, this might not always be feasible.[2][3] Most actinides, for instance, are highly radioactive, thus limiting their study to very specialised laboratories.[3] Many actinides are too scarce to perform necessary large scale measurements.[3] Here is where chemical pressure comes in.

Chemical pressure

Instead of external pressure or thermodynamic internal pressure, we now discuss chemical pressure. Much akin to external pressure, scientists take interest in how energy is dependent on volume. Chemical pressure comprises of substituting in or adding chemical entities to a solid network.[2] [4] [5] [6]

External pressure presses down on the solid. Volume is modified via a compressibility relationship.[1] A more direct or pure method.[2]
Chemical pressure is induced by changing cell volumes via the introducing of new chemical entities, such as atoms or compounds, in the lattice. An indirect, less pure approach.[2]

Obviously, chemical doping modifies a system in other ways besides volume alone.[2] [5] [7] By choosing an element with similar electronic configuration, these deviations from ideality are minimised.[2] [7] Another advantage of chemical pressure over extrernal pressure is that the former may be positive and negative.[2] [8] Positive pressure implies that pressure from chemical substitution is directed inwards, and adds to external pressure. Negative pressure pushes the solid apart, hence acting against external pressure.[2]

Immunisation against terminological confusion

Condensed matter science, like other fields of study, is entitled to its own nomenclature. For chemists specialising elsewhere, this might seem a superfluous obfuscation. This is especially true for cutting-edge chemistry where terminology is fresh and perhaps not entirely settled. For this reason, synonyms are widespread, and overlap with other terms may occur.

External pressure is also called physical pressure[4] to establish a physical pressure – chemical pressure dichotomy.
Chemical pressure is often enclosed between quotation marks. More confusingly, internal pressure is used as an alternative to chemical pressure.[5] Perhaps this is to emphasise an external pressure – internal pressure dichotomy. Or, it could be, that since chemical pressure is used as a tool to measure internal pressure, the concepts have blurred lines because the distinction seemed unnecessary.

Units and experiments

Internal chemical pressure will probably have concurring units with external physical pressure. Pascals ($\ce{Pa}$) when operating in the SI framework. Very few studies explicitly mention this fact[5] because more often the changes in some concrete parameter are discussed[1] [9]. Because of what chemical pressure constitutes, we do not apply a chemical pressure of $10\ \mathrm{kPa}$. We might, however, bring about (or apply) a chemical difference in pressure ... equivalent to several $\mathrm{kbar}$ of external pressure[5]. For solids where external pressure experiments are currently unthinkable, this equivalence cannot be discussed. But the goal is not to compare chemical pressure to external pressure anyway; as before, we wish to learn more about how volume affects (potential) energy.

As far as I am aware, chemical pressure cannot be measured directly. Chemical pressure is induced inside a solid material by substituting or introducing new chemical entities. Where would one put the barometer? If it is possible to carry out external pressure measurements alongside chemical doping, we may compare one to the other. To emphasise, the effects need not match. In these and other cases, volume changes in cells are estimated or measured, and their effect on the system as well.

/(short overview/list of experiments in progress, next week)/

In my opinion, this subsection deserves a question on its own.

TL; DR

/(in progress, next week)/

Additional reading

For an overview of highly-correlated electron systems in metallic solids, I recommend C. Marini. Pressure-induced metallization process in Strongly Correlated Electron Systems [Online]; G. Stefani, G. Altarelli, P. Postorino, Eds. Università degli Studi Roma TRE: Rome, Italy, 2009–2010. Archived link: bit.ly/2yPJcCo

References and bibliography

[1] G. Borelius. 'Internal Pressure in Solids and Liquids'. Physica Scripta, 1970, 1, 2-3, pp 141–147. DOI: 10.1088/0031-8949/1/2-3/011
[2] J-M. Fournier. 'Chemical pressure in actinide systems'. Physica B: Condensed Matter, 1993, 190, 1, pp 50–54. DOI: 10.1016/0921-4526(93)90441-8
[3] U. Benedict. 'Properties of Actinide Metals Under High Pressure'. Journal de Physique Colloques, 1984, 45, C8, pp C8-145–C8-148. DOI: 10.1051/jphyscol:1984826
[4] D. M. Friedman. New Research on YBCO Superconductors; Y. Itoh, Ed. Nova Science Publishers: 2008. Chapter 1: Recent Advances of NMR and NQR Studies of $\ce{YBa2Cu3O_{7-δ}}$ and $\ce{YBa2Cu4O8}$, pp 25–69 Subsection 6: 'Chemical Pressure, Physical Pressure, and Site Disorder', pp 59–59. ISBN: 978-1-60456-084-8
[5] A. Hauser, A. Nahid, S. Delahaye, A. Sadki, S. Schenker, R. Sieber, M. Zerara. 'Chemical Pressure'. CHIMIA International Journal for Chemistry, 2002, 56, 12, pp 685–689. DOI: 10.2533/000942902777679858
[6] R. S. Liu, C. H. Shen, T. S. Chan, R. Gundakaram, S. F. Hu, J. G. Lin, C. Y. Huang. 'Chemical Pressure Induced Phase Transition in Single and  Double Perovskites with Magnetoresistance Effect'. Tamkang Journal of Science and Engineering, 2002, 5, 1, pp 59–61. Archived link: bit.ly/2fvxrsv
[7] A. Huon, D. Lee, A. Herklotz, M. R. Fitzsimmons, H. N. Lee, S. J. May. 'Effect of chemical pressure on the electronic phase transition in $\ce{Ca_{1-x}Sr_xMn7O12}$ films'. APL Materials, 2017, 096105, pp 096105-1–096105-7. DOI: 10.1063/1.4994089
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How to determine the purity of lactic acid with titration?

At the end point, you have a solution of sodium lactate (weak base). The concentration of sodium lactate is given by the equation:
$$
C' =
\frac{C_\mathrm{b}\, V_\mathrm{b,eq}}{V_\mathrm{a} + V_\mathrm{b, eq}},
$$
where $C_\mathrm{b}$ is the concentration of the base, $V_\mathrm{b, eq}$ is the volume of the base at the end point and $V_\mathrm{a}$ is the volume of lactic acid.
$$
C' =
\frac{\pu{0.1010 mol L-1} \times \pu{13.1 mL}}{\pu{25 mL} + \pu{13.1 mL}}
 = \pu{0.03473 mol L-1}
$$
The $\ce{pH}$ of the solution at the end point is given by the equation:
\begin{align}
  \mathrm{pH}
    &= \frac{1}{2}(\mathrm{p}K_\mathrm{w} + \mathrm{p}K_\mathrm{a}
      - \mathrm{p}C')\\
    &= \frac{1}{2}(14 + 3.86 + \log 0.03473) = 8.20
\end{align}

To calculate the purity percentage of lactic acid, we calculate the amount of substance of pure lactic acid: At the end point
$$n_\mathrm{a} = n_\mathrm{b}.$$
This means:
$$
n_\mathrm{a}
  = C_\mathrm{b} V_\mathrm{b}
  = \pu{0.1010 mol L-1}\times \pu{13.1E-3 L}
  = \pu{0.001323 mol}
$$
The mass of pure lactic acid is:
$$
m
  = \pu{0.001323 mol} \times \pu{90.1 g mol-1}
  = \pu{0.1192 g}
$$
Or the mass of $\pu{0.5 \%(w/v)}$ lactic acid is:
$$
m'
  = \frac{\pu{0.5 \%(w/v)} \times \pu{25 mL}}{100\%} = \pu{0.125 g}
$$
The purity percentage of lactic acid $P\%$:
$$P\% = \frac{m\times 100\%}{m'}=95.36\%$$


Enthalpy of formation and reaction equations

As far as I can tell, your reasoning is sound, though the question is actually not written spectacularly well. By "enthalpy of the reaction" it seems to be implicitly assuming that the stoichiometric numbers are referring to moles (making all the answers have units of kJ instead of kJ/mol). The question doesn't specify standard enthalpy change of reaction, but even if it did, the most common convention for balancing equations is to use the smallest whole numbers for stoichiometric coefficients, which is what is written there and would give the answer you got. You can calculate an enthalpy change of reaction for any reaction conditions you like and the value you get will change depending on the quantities of the substances involved, but if the question is implicitly assuming moles for coefficients, you still get the same answer.

Incidentally, there's no reason you couldn't use some other definition of "standard" in which you set the coefficients such that the smallest ones are equal to 1, though I can't remember seeing anyone do it this way. IUPAC convention is that whenever a quantity referring to some standard state is used, the conditions of the standard should be specified (though chemists like to ignore IUPAC).


Why is the para product major in the nitrosation of phenol?

Ortho and para positions are activated due to the +R effect of the $\ce{-OH}$ but the preference of the para product can be explained by the mechanism.

The mechanism proceeds through a dienone. The ortho dienone intermediate will experience strain as  $\ce{=O}$ annd$\ce{=NOH}$ lie in the same plane. Para nitrosated intermediate on the other hand is free from strain.

Reference: González-Mancebo S., Lacadena J., García-Alonso Y., Hernández-Benito J., Calle E., Casado J. Monatsh. Chem. 2002, 133 (2), 157-166. DOI: 10.1007/s706-002-8245-0.