Why does 2,4 DNP give orange colour crystals?

Generally, you will notice that compounds containing chromophore and auxochrome groups have a colour. Chromophore groups are groups that are responsible for colour of the molecule once they are “attached” to it: benzene is colourless, but nitrobenzene is pale yellow. Apart from a chromophore group there are groups called auxochrome groups which alter this colour and so the intensity of absorption ( example: nitrobenzene is pale yellow, but p-nitroaniline is a bright yellow to red, the auxochrome enhanced the colour ). Among the chromophore structures, there are groups such as nitro, azo, nitrile, thioketone, conjugated double bonds ( like in retinal, carotene ). Some auxochrome structures are: the aldehyde group, the methyl mercaptan group, hidroxyl group, amine group. To sum it up, functional groups such as these will account for the colour.
Now, onto analysing DNP. This compound has 2 nitro groups and a hydrazo group ( 2,4-dinitrophenylhydrazine ). From the start, we have a colour: red ( if it is pure ). When it reacts with an aldehyde or a ketone in a condensation reaction, a double bond forms between the carbon on which oxygen was attached and the unsubstituted nitrogen of DNP. The formation of this double bond will alter the colour, shifting it to an orange. It also depends on the carbonylic compound you use. If the carbonylic compound contained conjugated double bonds ( like propenaldehyde ), the condensation product will contain conjugated double bonds which are chromophoric ( so in this case I can only guess what the resulting colour would be  because we have more chromophoric groups ). So, DNP is red in its purest form. Its colour is the result of the existing functional groups. A reaction with an aldehyde or ketones will shift its colour to orange.


What is the difference between physical and chemical bonds?

In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is the sharing of electrons between two or more nuclei. In contrast, other interactions are often said to be intermolecular (which is somewhat more specific than the term “physical”.

In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).

Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In The Nature of the Chemical Bond (1960), he gives the following definition:

  The Chemical Bond Defined.—We shall say that there is a chemical bond between two atoms or groups of atoms in case that the forces acting between them are such as to lead to the formation of an aggregate with sufficient stability to make it convenient for the chemist to consider it as an independent molecular species.

A bond is what links atoms into molecules, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…
The complete opposite: consider all interactions as chemical bonds, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like this one:

  A chemical bond is the physical phenomenon of chemical species being held together by attraction of atoms to each other through sharing, as well as exchanging, of electrons and is a phenomenon that is fully described by the laws of quantum electrodynamics. In general, strong chemical bonds are found in molecules, crystals or in solid metal and they organize the atoms in ordered structures. Weak chemical bonds are classically explained to be effects of polarity between molecules which contain strong polar bonds. Some very weak bond-like interactions also result from induced polarity London forces between the electron clouds of atoms, or molecules. Such forces allow the liquification and solidification of inert gases. At the very lowest strengths of such interactions, there is no good operational definition of what constitutes a proper definitional "bond".

This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).
You can classify interactions by energy: decide that chemical bonds are those interactions that have an energy higher than a certain threshold, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.
Finally, what I believe is the most common description is to look at the nature of the interaction, and classify it following a certain convention. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.
I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the New Oxford American Dictionary as well, which I quite like:

  chemical bond a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.

Short and powerful. What I like in that is that it gives a general prescription, allowing one to argue individual cases and not based too much on convention. What are the features of a chemical bond? Well, it has to be an attractive force between atoms, sure… but I think the most relevant criterion of all is sharing (or transfer) of electrons. That is, after all, what chemistry is about: description of electronic clouds around two or more atoms. And I think when this criterion is applied to the list of interaction types commonly classified, it works quite well (whithout being dogmatic).

Also, what I like in it is that a given interaction type can be considered one way or another depending on its strength. The best example for that may be the hydrogen bond, which is the archetype of the almost-chemical-bond…


IUPAC name of a molecule with a cyanide group and a double bond

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

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds

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

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)

You have correctly identified the suffix (“nitrile”) for the principal characteristic group. Thus, according to the first criterion, the parent structure could be pentanenitrile or but-3-enenitrile since both include the nitrile group.
The next criterion for the principal chain is the greater number of skeletal atoms (i.e. the longest chain). Thus, the principal chain corresponds to the pentanenitrile part and not to the but-3-enenitrile part since pentanenitrile has a longer chain than but-3-enenitrile. Therefore, the correct name is 2-ethenylpentanenitrile.
Many people would probably still call it 2-vinylpentanenitrile; according to current IUPAC recommendations, however, the prefix “vinyl” is retained only for general nomenclature, “ethenyl” is recommended as the preferred prefix.


Most useful reagent for gravimetric analysis of silver

You are missing something: before you even put the silver in, NH4Cl is soluble, while BaCO3 has already precipitated (insoluble)!

This means that Cl- will precipitate more completely with NH4Cl.  Therefore, D.


How does this brominated intermediate lose its methyl group?

A methyl cation definitely won’t leave by itself. However, you can consider bromide ions floating around in solution. These bromide, being nucleophilic can attack the methyl group in an $\mathrm{S_N2}$ manner, because any positively charged oxygen is a good leaving group. Therefore, the side product would be bromomethane.

Other than that, your mechanism is looking good.


Is it possible to build a non-degrading battery?

But is it possible to have a battery that will never degrade and always keep its performance without upkeep/maintenance?

Well, consider the dry pile powering the Oxford Electric Bell, which has been running since 1840. However, that is an example of a nonrechargeable pile.

The question is, can a rechargeable battery be designed that retains it's charging/discharging current profile, reflecting a conserved structural integrity, during an unlimited number of cycles?

While in practice there are many reasons this is difficult to do (loss of electrolyte, occurrence of irreversible passivating reactions, etc) in principle there seems to be no reason why this should not be the case, at least over time on the order of a human lifetime. We manufacture many things which last an awful long time, and batteries would last even longer if designed or treated in an exceptional way, shielded from the environment and charged/discharged in a highly regulated manner. Some battery designs such as thin film lithium-ion allow an exceptional amount of cycling.

  I just didn't know if there were some types of laws in chemistry that prevents this from happening.

The second law of thermodynamics places limits on what you can expect to achieve given the constraints of a particular choice of battery chemistry. This reveals itself for instance in Peukert's law, particularly applicable to lead-acid batteries operating at constant temperature, which says that the capacity of a battery is restricted by its discharge rate. Discharging quickly reduces the total work you can do with the battery.

The second law however does not say we cannot use work to recharge a battery to the same original state a very large number of times. But it does make it very hard! It begs the question, "is such a  battery useful?" Probably not, except theoretically. The design and operation of a battery involves tradeoffs between charge density, discharge and recharging rate, and preservation of structural integrity during charging/discharging cycles.

By the way, you could generalize the question to include other portable high-capacity sources of electrical power. Think of probes that are sent into outer space and continue to operate after, well, a long time. Or consider the quantum battery.


How much carbonation is in one liter of commercial fountain Coca-Cola

If the soda is not distributed, the rate of loss of $\ce{CO2}$ would be an exponential decay. So changing from 5 volumes to 3 volumes is faster than from 3 to 1 volumes.

A can of soda with 3 volumes of $\ce{CO2}$ has about 15 psi of extra pressure inside the can. If it was increased to 25 psi (5 volumes, like the fountain sodas), the can would need to be thicker to ensure it doesn't rupture, which would increase the cost of the can. There is a balance of the amount of $\ce{CO2}$ and the cost for the drink.


Just to sum up what has already been pointed out in the comments: Methanol can have infinite conformations in free space due to free rotation of the C-H and C-OH sigma bonds. Hence, for any particular spatial conformation you can think of for the molecule, it can always develop a perfect mirror image for the same almost instantly(in case you are wondering about some time lag between them) as conformers are in chemical equilibrium.

So what this essentially means is that, suppose some conformation A of methanol has polarized light in some direction (assuming the right-handed coordinate system, let's say the positive X direction) in some fashion(let's say circularly polarized, clockwise). Then, this particular interference pattern of light will be internally compensated by a conformation B of the molecule which will circularly polarize the light in the anticlockwise direction in the negative X direction. So, the emergent light ray will appear to be unpolarized for the observer


Turbidity is measured by the examining the light passing through a sample.  A sample's turbidity is formally measured using standardized methods (e.g. ISO 7027 or "Standard Methods," No. 2130B. source).  Standardized methods require the instrumentation called for by the standard.

An informal measure of visible differences in turbidity can be obtained by comparing the 20mL sample to a set of calibrated standards.  An example of a set of calibration standards for turbidity are shown in the following link http://www.water.ncsu.edu/watershedss/info/turbid.html.


Carbonic Acid Titration with Sodium Hydroxide

The two pKa values for carbonic acid are given as 6.37 and 10.25, so you may want to pick the midpoint between these (pH 8.31) as your equivalence point. At that pH, a tiny amount of base will change the pH a lot, so the amount of base needed should not depend strongly on the exact value you select for the equivalence point. But the pH range is rather narrow, so I would recommend that you use a pH meter instead of a visual indicator for the titration.


How to draw chair-form for trans-decalin (with slight variation)?

Use of 3D models
3D version of Figure 1:

Viewed upside down (rotate about z by 180 degrees):

Now it looks like we are viewing from the bottom, so we will turn the molecule a bit around the x axis:

Use of 2D projections
If you want to use the 2D projections, consider whether hydrogens are pointing up or down on the bridge head. First, we will interpret the chair as viewing the top of the ring from above (left panel) rather than viewing the bottom of the ring from below (right panel) to avoid ambiguity.

No matter whether you use two fused chairs as drawn by the OP (Fig 1 bottom panel) or its mirror image, the front hydrogen will point down and the back hydrogen will point up. This works well for Fig 1.
In order to use the same chair building block but show the front hydrogen pointing up, you could rotate the flat model, and connect two chairs accordingly, as shown below. In effect, you choose the location of the two bridging position to match the stereochemistry you want.


Rusting of iron is an electrochemical phenomenon.
A certain site on the iron object acts as an anode.
$$\ce{(Fe(s) -> Fe^2+(aq) + 2e ) × 2}$$
Electrons released at this spot go to another site on the object and reduce oxygen in presence of $\ce{H+}$ ions which are either obtained from water out from acidic substances in water:
$$\ce{H2O -> H+ + OH-}$$
$$\ce{CO + H2O -> H2CO3}$$
$$\ce{H2CO3 -> 2H+ + CO3^2- }$$
This site behaves as cathode and reaction is
$$\ce{4H+ + O2 + 4e -> 2H2O}$$
Therefore the overall reaction is :
$$\ce{2Fe(s) + O2 + 4e -> 2Fe^2+(aq) + 2H2O}$$
Now the ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in form of hydrated ferric oxides.
$$\ce{Fe^2+(aq) + 2H2O + 1/2O2 -> Fe2O3 + 4H+(aq)}$$


what is mechanism for reaction phenol + dimethylsulphate -> anisole

The methylation of Phenol using Dimethyl Sulfate is an Electrophilic Methylation and happens through a SN2 nucleophilic attack. Phenol is first deprotonated by the sodium hydroxide as you already thought. The deprotonation of Phenol is favored because the pKa of Phenol is lower than the pKa of water meaning $\ce{OH-}$ is a stronger conjugated base than $\ce{PhO-}$:

The formed Phenolate is a good nucleophile, and will attack the methyl groups of the Dimethyl Sulfate. Those methyl groups are readily attacked by nucleophiles because of the great induction effect caused by the neighbor atoms which are much more electronegative than the carbon atom, which makes those more positively charged. This makes the methyl groups good electrophiles, hence the reaction name "Electrophilic Methylation": If we focus on the methylating reactant, it's performing an Electrophilic Attack.

Below is the reaction mechanism:

EDIT: Thanks @user55119 for pointing out that only one methyl group of the Dimethyl Sulfate reacts under normal circumstances, requiring 1 equivalent of Dimethyl Sulfate per Hydroxyl Group and not 1:2 as I previously thought


Why does the nature and oxidation state of a central metal affect field split?

Let me start this answer — as usually for an answer I give about coordination complexes — by showing the molecular orbital scheme of a typical octahedral complex ignoring $\ce{L\bond{->}M}$ π interactions.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

It is easy to see where field splitting comes from: two of the ligand group orbitals transform as $\mathrm{e_g}$ as do two of the metal’s $\mathrm{d}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). These interact in a bonding/antibonding manner to give $\mathrm{e_g}$ (considered a ligand-centred orbital due to the electronegativity difference) and $\mathrm{e_g^*}$ (a metal-centred orbital).

It is important to note that this scheme depends strongly on the relative positions of the orbitals. The closer the energies of the ligand and metal group orbitals, the larget the split will be.

If you consider different aquacomplexes, you can always be sure that the ligand orbitals start off from the same energy (they are the same water, after all). However, the metal orbitals will be differently positioned. The higher the first ionisation enthalpy (typically corresponding to the removal of an electron from $\mathrm{4s}$ — here $\mathrm{a_{1g}}$) the lower these orbitals will be in this scheme. Loosely, the ionisation enthalpy depends on the number of protons in the nucleus, so we might expect nickel to have lower-lying orbitals than titanium. Whatever the actual difference is (there are many more effects), no two different metals will ever have the same orbital energies and therefore any two different metals will always have different field splits corresponding to a different colour. You know this: hexaaquachromium(III) is green while hexaaquairon(III) is yellow-orange.

The oxidation state influences the orbital scheme in a similar way. Each oxidation corresponds to the removal of an electron meaning less electronic repulsion meaning lower-lying orbitals (in slightly larger contexts, this corresponds to the contraction of cations). Thus, if you change the oxidation state of a metal, e.g. from $\mathrm{+II}$ to $\mathrm{+III}$, the metal orbitals will be lowered in energy and the field split will most likely increase since there is better overlap with the ligand orbitals. Therefore, iron(II) and iron(III) have different colours in solution.


What is the meaning of the word root "aceto"?

The root "acet-", according to wiki, is derived from latin acētum, meaning vinegar, i.e. diluted acetic acid. Consequently, direct derivatives of acetic acid, like acetamide and acetonitrile are called so. Also, acetone called so because it can be produced from acetic acid (directly or by dry distillation of calcium acetate)

What isn't obvious is why acetonitrile  is considered a derivative of acetic acid and why nitriles in general are considered derivatives of carboxilic acids. Partially it's tradition and partially it's because compounds with carbon skeleton and oxidation state of carbon atoms unchanged are often rather easy to convert into each other.


How can ionic substances have an infinitely-repeating lattice structure?

To be technically accurate, it's "practically" infinite.

It's not infinite since as you mention a grain of salt (or even a large crystal) has some finite size.

You have to remember the actual atomic dimensions $10^{-10} \:\mathrm{m}$. So while it's not truly infinite, there are an incredibly large number of atoms in a crystal.


Visualizing atoms in XYZ file with colors representing partial charges

As far as I know, no software can do this right away but with some work, this should be possible.

Use CYLview to create a povray file, change the atoms to the desired color and render it with Povray. The problem here is that the coordinates of the xyz file and the povray are not the same. But as far as I can tell the numbering stays the same. So if the 3rd atom in your xyz is this:

N       -1.227679000      0.558365000      0.000000000

the 3rd sphere in the povray file would look like that:

sphere {<-1.9762436, 0.9491377, 0.0000000> 0.2812500
   pigment{color rgbt <0.4,0.4,1.0,0.00>}
   finish{F_normal}

All you need to do is change the color here. It should also be pretty easy to automate this color change with some lines of code.

So you can go from this

to for example this

by simply changing the color value shown above to

pigment{color rgbt <0.2,0.8,1.0,0.00>}


Determination of hardness of water using EDTA

There really is no need to add EBT in a fixed amount. Calcium EDTA complex is far more stable at $\mathrm{pH} \approx 10$ (ammonia buffer) and will be the dominant Ca-compound in the solution even if there is a huge excess of EBT.

However, the amount of EBT is important to track the color change. If there is too much EBT, the color will be too dark, and one can easily miss the red-blue transition. If there are only trace amounts of EBT, the colors will be pale and again the transition is going to be barely visible.

The amounts of EBT given in the lab manuals are typically approximate and are estimated for the convenience of the personnel and to reduce the error in determining equivalence point due to unclear reading of the solution's color.


How to derive the molar fundamental relation

You have  $$ds=\frac{du}{T}+\frac{P}{T}dv+\left(\frac{u}{T}-s+\frac{Pv}{T}-\frac{\mu}{T}\right)\frac{dN}{N}$$
But, $$\mu=u+Pv-Ts$$
That is, $\mu$ is the Gibbs free energy per mole.  So the term in parenthesis drops out.


Simple carboxylic acid or carboxylate ions (like CH3CH2CH2COO–) do not undergo decarboxylation to yield corresponding alkanes because in this case the leaving group would be a carbanion (in this case CH3CH2CH2–) which is very strong base and hence a very poor leaving group (the reason behind this is that strong base wants to give electrons, but leaving group has to take electrons and leave the compound, strong base is poor leaving group and weak base is good leaving group)

For decarboxylation to happen a keto group must exist to stabilise the negative charge, which is not present in salt of butanoic acid CH3CH2CH2COO–Na+

Yield of propane is very less if you chose decarboxylation, that is why Corey-House synthesis is best for preparation of propane.

I want to know why do you think that Option D is correct, mention in comment.


Why aren't these two resonance structure same?

There are called equivalent resonance structures since the number of pi bonds, lone pairs and sigma bonds are the same. On the other hand, resonance is the delocalization of pi electrons. These structures show how pi electrons are delocalised from ortho position to para position and then back to ortho(on the other side of the molecule). So both the structures you mentioned are different in this context and do have a purpose in explaining the resonance phenomenon.0

Note: This rule only applies to hydrocarbons. In other cases, the two resonance structures need to have the same structural formula to be equivalent.


Using group theory to make molecular orbital diagrams

Group theory really just formalises the process that you're going through when you construct molecular orbital diagrams by inspection (though as the molecules get bigger it gets significantly harder to intuitively guess at what will mix and in what combinations).

In order to construct an MO diagram for water, we'll take a stepwise approach:

first, determine the way in which the hydrogen atoms can combine (in phase or out of phase, as per the standard dihydrogen MO diagram) which will give us a set of symmetry adapted combinations
second, we mix these symmetry adapted orbitals with the atomic orbitals on the oxygen, bearing in mind the 'rules' of the game

only orbitals of similar energy are able to combine
only orbitals of correct geometry/orientation are able to combine (i.e. head on or side to side)
only orbital of correct symmetry are able to combine. This is where group theory becomes useful

1. Symmetry of the central oxygen's orbitals

The symmetry of the oxygen orbitals (2s and 2p, the 1s isn't relevant to the bonding) can easily be read off from a character table. In this case, water has C2v symmetry, the character table for which is given below:

$$\begin{array}{c|cccc|cc} \hline
C_\mathrm{2v} & E & C_2 & \sigma_\mathrm{v}(xz) & \sigma_\mathrm{v}'(yz) & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & z & x^2, y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_1} & 1 & -1 & 1 & -1 & x, R_y & xz \\
\mathrm{B_2} & 1 & -1 & -1 & 1 & y, R_x & yz \\ \hline
\end{array}$$

To summarise the information obtained:

the s-orbitals have a1 symmetry
the px orbital has b1 symmetry
the py orbital has b2 symmetry
the pz orbital has a1 symmetry

(as @Orthocresol noted in the comments, there is no convention on axis in the C2v point group,meaning x and y can switch depending on what book you get the character table from. The one given is the same as the commonly used set from Atkins' Physical chemistry, but overall it doesn't really matter as long as you're consistent)

It's worth noting that you can derive these symmetries above by considering how the atomic orbitals moves under the various elements within that point group, this is shown below for the hydrogens, and not really needed to be repeated here.

2. Symmetry of the hydrogen group orbitals

To construct the group orbitals for hydrogen (which will go on to be mixed with the oxygen atomic orbitals), we simply have to consider the possible combinations and then work out the symmetries of those combinations.

For the case of two hydrogens interacting, only two cases are possible: in phase, and out of phase.

If we consider these two cases pictorially, we can then determine what happens to them under the various symmetry elements within the C2v group. If the orbitals stay the same, we assign a character of 1, if they switch, we assign -1.

Once we have the full set of characters for both possible combinations, we can assign symmetries, again, using the character table for the relevant point group:

the in-phase combination has a1 symmetry
the out of phase combination has b2 symmetry

3. Mixing

Mixing is the part of this process that you're likely more comfortable with, except now rather than considering whether interaction (on symmetry grounds) is likely by inspection, we can do it by mixing orbitals with the same symmetry labels:

The diagram above is the one that you would likely have arrived at without any consideration of group theory (you can actually derive a more complex diagram in which the molecular orbitals also mix), some key points to note:

the oxygens orbitals have all (qualitatively) been placed lower than those of hydrogen. Electronegative atoms have more contracted orbitals and are lower in energy
two of the p-orbitals end up non-bonding, these are the lone pairs on oxygen
the s-orbital on oxygen interacts with the in-phase combination of hydrogen, this is the lowest energy combination rather than using the p-orbital (also with a1 symmetry)

Further reading

The general approach taken here was a mix between that taken in Group Theory for Chemists by Molloy, and Cottons' Chemical Applications of Group Theory. Most (if not all) books on group theory will provide examples of how to construct MO diagrams based on formal symmetry arguments, and many approaches exist.

I've followed a more qualitative method, relying on the character table to gain the required information, but there is a more formal approach in which irreducible representations are used... possibly someone else will provide an answer with this more formal method (the amount of MathJax required put me off doing this, especially given the ease at which the information can be accessed and the fact that given you're asking the question, you're probably not at the point at which this has been formally taught).


Is there any book where specific heat capacities are given?

There are several databases of thermochemical properties, a number of them being listed on this Wikipedia page (at the bottom). Most of them are not freely available online, but require subscription. I've searched online but couldn't find that information freely available.

Another possible way to get the information would be through a search of the published literature (scientific papers). You can, for example, find this information on magnesium nitrate in a 1944 paper in J. Am. Chem. Soc.. However, I had no such luck with Al(NO3)3 · 9 H2O .

Clarification: the answer above is for the crystalline compound aluminium nitrate nonahydrate, which as @Ivan stated, may not be what you actually need for your calculation if the reaction is in water.


Cheap silocone casting resin (1 or 2 component resin)?

The problem with attempting this is the nature of the cheap silicone.  Most store bought silicone is already polymerized before use.  For example, the silicone caulk which can be purchased in home improvement areas is a finished product which  has been infiltrated with solvent.  These caulks contain no catalysts to facilitate reactions.  Any curing is the result of solvent drying from the caulk and solidifying the resin.

If you were to attempt to further dissolve one of these materials with something like ethyl acetate (see here for a simplified list of solvents), it probably would not go well.  All you would be doing is further impregnating the silicone with solvent.  While this would decrease the viscosity, it would increase your drying time.  Shrinking would occur during your casting, and I expect that you would also experience a very messy result, no matter the quantity of release agent you applied.

Consider a very simple silicone (aka polysiloxane).  Two silanol molecules react by hydrolysis to form a disiloxane and a water molecule.  In order to reverse the process and obtain your starting material, you would need to put the water back into the disiloxane to cleave it.  This is energetically expensive.

Even the full polymerization process is not a spontaneous one.  When you purchase a silicone casting resin from a supplier, you are actually buying a Part A composed of n-siloxane (where n is a small enough number to be a liquid) and a Part B composed of crosslinker and catalyst (this is often silicic acid and tin because they are inexpensive).  So what you are actually doing is taking a short chain polymer and crosslinking it into a solid mass.

Adding silicic acid and tin to a store-bought silicone product such as caulk would probably cause a reaction.  The result would probably be a very hard rubber with incompatible properties for casting.  However, I have not attempted this.  Furthermore, I am not sure silicic acid would be easily accessible to a home chemist.

Despite how fun it would be to synthesize your own silicone casting resin at home from everyday chemicals, it probably isn't feasible.  There are some inexpensive options out there for resin casting, and I encourage you to go with those.


Rate of hydrolysis of ester in presence of an acidic medium

We need to consider 4 quantities for the equilibrium reaction:

$$\ce{R1COOR2 + H2O <=> R1COOH + R2OH}$$

The equilibrium constant
The reaction quotient
The forward reaction rate
The reverse reaction rate

The forward reaction above is of the pseudo first order, as the professor says.

But its rate constant depends on acidic  catalytic factors and the net reaction rate is affected by the reverse reaction rate

The concentration of products- the organic acid and the alcohol - affects the net reaction rate by increased rate of the reverse reaction
and by shifting of the reaction quotient.

The content of a mineral acid  catalyzes the reaction, what means speeding up both reactions, therefore keeping the same equilibrium constant.(required as consequence of the energy conservation law).

The reaction is still the pseudo first order, but the rate constant depends on concentration of the catalysing acid.

The alternative approach is considering the mineral acid as self-regenerating reagent. But it must be then evaluated as reaction chain.

The content of a mineral base decreases reverse reaction rate and shifts the reaction quotient by reacting with the organic acid. It also directly reacts with the ester.

$$\ce{R1COOR2 + OH- -> R1COO- + R2OH}$$

This reaction is of the 2nd order.


Reactivities: interhalogen compounds versus halogens

Here is the answer for which bond (ionic or covalent) is stronger.
There is no direct answer to this question as it depends on atoms that take part in bond formation.For example you might be knowing that a Fluorine- Fluorine bond is weaker than a chlorine-chlorine bond due to high inter electronic repulsion.In another example iodine-iodine bond is weaker than chlorine-chlorine bond.Similarly there is certainly this sort of variation in ionic bond as well.
So if you compare the weakest ionic bond and strongest covalent bond then covalent bond would be stronger.And if you compare stongest ionic bond and weakest covalent bond, then of course ionic bond would be stronger.Also in many substances its not only the bond strength that matters but also the crystal structure.
An example where covalent bond is stronger:Carbon-Carbon bond is stronger than NaCl bond.
Whereas NaCl ionic bond is more stronger than Br-Br covalent bond.

So it depends on the atoms that participate in bond formation.

visit https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf for bond enthalpy.


It should be named as 2-bromo-4-methylpentan-3-one

The alphabatical order should be followed while naming this compound. As b in bromo comes before m in methyl, it should be named as above.

See here and here for more information.


If SOCl2 reacts with alcohols via SNi, why doesn't POCl3?

The step in the red box is wrong. You probably get dissociation to form an ion pair. The pair quickly collapses, leading to chloride attack on the cation before any dissociation or diffusion occurs so that you get the retention of stereochemistry.

For phosphorus based chlorination, this ion-pair mechanism is probably not favorable, though it's not clear to me why this is the way it is.


Most negative and most positive value for proton chemical shifts

I don't know if the following two examples present the "largest" upfield and downfield proton-nmr chemical shifts, but I suspect they're in the running.  The dihydropyrene dianion example has 16 pi electrons

delocalized around the periphery of the pyrene frame. It fits the 4n rule with n=4, so it is antiaromatic.  The [16]-annulene dianion has 18 pi electrons and fits the 4n+2 rule with n=4, so it is aromatic.  Note how the direction of the ring current reverses between antiaromatic (paramagnetic current) and aromatic (diamagnetic current) systems.  More intesting proton chemical shifts can be found in this compilation.

Edit

Guilty of thinking organic.  If we're including inorganic proton shifts, then how about IrHCl2(PMe(t-Bu)2)2 which has a chemical shift of -50.5!


Does variations in air pressure affect the melting point of PLA and ABS while 3D Printing?

Boyle's law refers to properties of an idealized gas, where the product of pressure and volume remains constant.  Taking into account your description, as well from filament-based 3D printers seen so far, my answer is no, Boyle's law does not apply here.

If running by extrusion, polylactic acid (PLA) is heated to overpass melt, however this liquid is not heated to become a gas. Temperature control clearly is of importance, because the viscosity of the polymer depends on it.  And viscosity co-influences the precision of details you may print, actually down to the scale of less as a tenth of a millimetre.  Without experience to run one 3D print in at sea level, and the same again during an intercontinental flight, I assume it is more likely, machine control takes into account the pressure difference between ambient air and polymer melt, and not (yet) the (in comparison) lesser variations of air pressure.

The chemical structure of PLA shown in Wikipedia's article shows a relatively low number of functional groups susceptible to water, relatively low to the inner backbone of the polymer (in the structural scheme, this is written in the square brackets).  The article refers to typical molecular weights of 130 kDa when PLA is produced.  These two let me say, humidity does not play an important role.


Why is Cubic closed packing called "cubic"?

Start with the hexagonal A-B-A-B structure.  It is hexagonal because, choosing one atom in a layer, it is surrounded by 6 equivalent atoms. So, you have a 6-fold symmetry axis, hence the 'hex-'. The fact that it is close-packed can be gotten at mathematically (or at least shown that it is closer-packed than other crystal structures).

Cubic close packing is, at least in most materials science and metallurgy circles, much more commonly called face centered cubic. The 'standard' unit cell is a cube of atoms with additional atoms at the center of each face.  It is somewhat harder to visualize how this is related to close-packed planes, although there are many pictures available on-line (one, chosen somewhat randomly, is from Florida State). The point is that the stacking direction of the close-packed planes are along the 111 body diagonal of the cube.

Rotate the cube so that it is standing on one point with the opposite point on the body diagonal pointing up. Your close packed planes are:
(1) the bottom point, (2) the top point, (3) the three corner atoms closest to the bottom point as well as the face atoms closest to the bottom atom, (4) the three corner atoms closest to the top atom as well as the face atoms closest to the top atom. It may take some squinting...


Should moderately soluble compounds like calcium hydroxide be denoted (aq) or (s)?

In principle, you cannot know without conducting the experiment. It depends on the concentration of a moderately soluble compound such as $\ce{Ca(OH)2}$ whether it is dissolved or whether there is stuff still sitting at the bottom of the vessel undissolved.

However, you are performing a reaction which is in aqueous solution. This reaction will typically only work if all the reactants are already in solution. Moreover, this is a precipitation reaction. Thus, it wouldn’t make sense to have a non-dissolved reactant to react it with a dissolved one. Given this context, you should assume both reactants are in solution.

$$\ce{Al2(SO4)3 (aq) + 3Ca(OH)2 (aq) -> 3CaSO4 v (s) + 2 Al(OH)3}$$

Note that whether or not aluminium hydroxide precipitates depends on the resulting $\mathrm{pH}$ value. If you started off with excess hydroxide, it may well stay in solution.

The exact answer for a given compound can only be known if you know both concentration and solubility product.


Change in spin angular momentum during transition in one-electron atoms

In general, transitions in which the "direction" of spin changes are forbidden. More formally, the spin conservation rule (also referred to as the Wigner rule) states that

  For both radiative and radiationless transitions,
  transitions between terms of the same multiplicity are spin-allowed,
  while transitions between terms of different multiplicity are
  spin-forbidden.

But spin-orbit coupling relaxes this rule, so in general, lines corresponding to spin-"forbidden" transition are just much weaker and not totally absent from the spectrum. And the bigger the atomic number, the stronger is the spin-orbit coupling, and consequently, the bigger are the intensities of spin-"forbidden" transitions.

$
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\ket}[1]{\left|\,{#1}\right\rangle}
\newcommand{\abs}[1]{\left|\,{#1}\right|}
$
With respect to your question on addition of spin angular momentum you raised in comment section: you formulated it incorrectly which confused me, so my comments (now deleted) were wrong. Thus, I decided to write a short answer here.

So, for angular momentum $J$ the simultaneous eigenstates of $\linop{J}^{2}$ and $\linop{J}_{z}$ operators are given as follows,
$$
\begin{align}
        \linop{J}^{2} \ket{j(j+1),m} &= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\
        \linop{J}_{z} \ket{j(j+1),m} &= m \hbar \ket{j(j+1),m} \, ,
\end{align}
$$
where $j$ could take non-negative integer and half-integer values,
$$
    j = 0, 1/2, 1, 3/2, \cdots \, ,
$$
and $m$ ranges from $-j$ to $j$ in integer steps
$$
    m = -j, -j+1, \cdots, j-1, j \cdots \, .
$$

When there are two sources of angular momentum (like two electrons with their spins) we first we have to make a choice when specifying the state of the system: either we use the uncoupled picture $\ket{j_{1}, m_{1}, j_{2}, m_{2}}$, which leaves the total angular momentum unspecified, or we use the coupled picture $\ket{j_{1}, j_{2}, j, m}$, which leaves the individual components unspecified.

The very language OP uses (i.e. talking about multiplicities of states) suggest that we choose the second picture for further discussion. In the coupled picture the question arises as to the permissible values of total angular momentum quantum numbers $j$ and $m$. It is known from quantum theory of angular momentum that the allowed values of $m$ follow immediately from the relation $m = m_{1} + m_{2}$, while the situation with the allowed values of $j$ is more complicated: its permited values are given by
$$
    j = j_{1} + j_{2}, j_{1} + j_{2} - 1, \dotsc, \abs{j_{1} - j_{2}} \, .
$$

All this is true for any kind of angular momentum (orbital or spin) with the only one exception: with respect to half-integer values of $j$ we note that for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible.

Now for each and every electron we have a source of spin angular momentum characterised by quantum numbers $j = 1/2$ and $m = \pm 1/2$. Thus, in accordance with what is written above for a system of two electrons we have 4 possible state which in the coupled picture are written as follows: $\ket{1/2, 1/2, 1, -1}$, $\ket{1/2, 1/2, 1, 0}$, $\ket{1/2, 1/2, 1, 1}$ and $\ket{1/2, 1/2, 0, 0}$. The first three states are three components of a triplet, the last one is the only component of a singlet.

Note, however, that projection quantum numbers for individual components ($m_1$ and $m_2$) are not specified in the coupled picture, so while it might be tempting to think that for $\ket{1/2, 1/2, 1, -1}$ state they are $m_1 = m_2 = -1/2$ this is wrong. And that what was essentially wrong with the question raised by OP in the comments section: he used the coupled picture terminology ("multiplicities") and at the same time specified the values of projection quantum numbers for individual components ("two electrons both have spin of -1/2"). This could not be dome simultaneously.


Why does it yield iron(II) chloride instead of iron when copper meets iron(III) chloride?

Take a look at standard reduction potentials at $\pu{25 ^\circ C}$:

$$
\begin{array}{clr}
\hline
\text{Reaction} & E^\circ~(\pu{V}) \\
\hline
  \begin{align}
    \ce{Fe^3+ + e- &-> Fe^2+} \\
    \ce{Cu^2+ + 2e- &-> Cu} \\
    \ce{Fe^3+ + 3e- &-> Fe} \\
    \ce{Fe^2+ + 2e- &-> Fe}
  \end{align}
&
  \begin{array}{r}
    +0.77 \\
    +0.34 \\
    -0.04 \\
    -0.41
  \end{array} \\
\hline
\end{array}
$$

A more positive potential means a greater tendency to be reduced, so $\ce{Cu(II)}$ is easier to reduce to copper metal than either $\ce{Fe(II)}$ or $\ce{Fe(III)}$ being reduced to iron metal. So, copper metal can't force the formation of iron metal.

But $\ce{Fe(III)}$ can be reduced just to $\ce{Fe(II)}$ at a relatively high potential, high enough to allow the copper to be in its oxidized form as $\ce{Cu(II)}$. So, we can have $\ce{Cu(II)}$ and $\ce{Fe(II)}$ at equilibrium and thus we can get the corresponding chlorides $\ce{CuCl2}$, $\ce{FeCl2}$.


The molar mass $M$ of $\ce{NaCl}$ is $58.44\ \mathrm{\frac{g}{mol}}$ indeed:

$\displaystyle\text{molar mass}(\ce{NaCl}) = \frac{\text{mass}}{\text{amount of substance}} = 58.44\ \mathrm{\frac{g}{mol}}$

$\displaystyle\Rightarrow \text{amount of substance} = \frac{\text{mass}}{\text{molar mass}}=\frac{10.0\ \mathrm g}{58.44\ \mathrm{\frac{g}{mol}}} \approx 0.171\ \mathrm{mol}$

$\displaystyle\text{concentration} = \frac{\text{amount of substance}}{\text{volume}}=\frac{0.171\ \mathrm{mol}}{0.75\ \mathrm L}=0.228\ \mathrm{\frac{mol}{L}}$

Assumptions: Adding $\ce{NaCl}$ to these $750\ \mathrm{ml}$ does NOT change the volume noteworthy.


Why is the boiling point of alkyl iodide is more than alkyl fluoride?

The main factor affecting the boiling point is the molecular weight, as can be pictured from the noble gases boiling point in the periodic table (the noble gases are those which have the weakest interactions due to their complete outer shell). But note that if you want to consider all the effects altogether, you must use Kelvins instead of Celsius. So going from 44°C to 88°C is not twice as much but a mere 14% increase.

The other factor affecting boiling point is the nature and strength of intermolecular interactions, usually in the order ionic > hydrogen bond > dipole-dipole > Van der Waals.

In the comparison of RF vs RI boiling point, molecular weight is definitely in favor of higher boiling points for RI, at least for small molecules. It is true that the C-F bond is much more ionized than C-I so it will create stronger dipole-dipole interactions but not sufficiently to compensate for the effect of the mass increase.

HTH


Nomenclature of a cyclohexyl ether and locant position enumeration

The smallest sum of locants rule does not exist in the IUPAC recommendations. The application of this ‘rule’ can lead to wrong results in many instances.
The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually reads as follows:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
(…)

With regard to numbering of locants, simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl and ethoxy) are considered together with equal seniority:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, the example is named as 2-ethoxy-1,1-dimethylcyclohexane rather than 1-ethoxy-2,2-dimethylcyclohexane since the locant set ‘1,1,2’ is lower than ‘1,2,2’.


What is the formula of potassium ammonium sulfate?

$\ce{K(NH_4)SO_4}$ is the correct formula. You have two unipositive cations and a single dinegative anion. The total charge must be zero.


Production of Succinic Acid from Crude Glycerol using Fermentation

Glycerol has 3 carbon atoms and succinic acid has 4. So you need (apart from some redox reactions) a way to add another carbon atom.

Also, succinic acid appears in the citric acid cycle. What you have to research are anaplerotic reactions. When you remove any of the metabolites (like succinic acid) from the cycle, you have to replenish them. One way is to turn pyruvate (3 carbons) into oxaloacetate (4 carbons), as shown in the wikipedia article on anaplerotic reactions:

$$\ce{pyruvate + HCO3− + ATP -> oxaloacetate + ADP + P_i + H2O}$$

Then, you have to find a pathway from glycerol to pyruvate. Here is a schematic of glycerol metabolism:

Source: https://www.researchgate.net/publication/224950286_Effect_of_impurities_in_biodiesel-derived_waste_glycerol_on_the_performance_and_feasibility_of_biotechnological_processes/figures?lo=1

As you can see, it is possible to turn glycerol into pyruvate fairly directly (using reactions of glycolysis).

To put everything together, you just need an organism that secretes succinic acid, and make sure that these pathways are turned on sufficiently that most of the provided glycerol gets metabolized to succinic acid. As this involved oxidation, the organism has to have access to oxygen or some other oxidation agent.

UPDATE

Prompted by Andrew's comment, I found a paper where they optimize succinate production from glucose. As you can see from the scheme above, glycerol can be funneled into glycolysis by two enzymes (using one ATP for phosphorylation and one NAD+ for oxidation). The enzymes used for the remainder of the reaction (and to provide NADPH via the pentosephosphate pathway) are shown in the figure from the paper below. You would need to provide some glucose to feed in the pentosephosphate pathway to have sufficient NADPH, or invoke some other pathways:

Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4983090/figure/Fig1/?report=objectonly


What is Drago's rule? Does it really exist?

Going through what you posted, I think ‘Drago’s rule’ (which I never encountered, either at school or at university) gives good predictions but uses a largely terrible set of arguments.

For any atomic system with more than one electron, quantum chemistry predicts the energies of the $\mathrm{s}$ and $\mathrm{p}$ subshells of a shell to differ — while for hydrogen-like systems (one-electron systems) all subshells of a given shell have the same energy. The $\mathrm{s}$ subshell of a certain shell in multi-electron systems has a lower energy than the $\mathrm{p}$ subshell, which in turn has a (far) lower energy than the $\mathrm{d}$ subshell and so on and so forth. Since $\mathrm{s}$ subshell is determined to be more stable by this method, it is not surprising that the first two electrons of a given shell are added to said $\mathrm{s}$ subshell; the next six electrons are added to $\mathrm{s}$ subshells whereafter a new shell is used.

This immediately explains that it is most stable for any atom to treat the $\mathrm{s}$ electrons of its valence shell as ‘extended core electrons’; i.e. we can predict any configuration that does not touch the $\mathrm{s}$ electrons to be energetically favourable with respect to any configuration that does a priori. (You can consider that part of the quoted statement to be correct.) Therefore, the a priori predicted most stable configurations of certain coordination numbers are:

mono-, di- and tri-coordination: use 1, 2 or 3 $\mathrm{p}$ orbitals to form bonds; keep one lone pair in an $\mathrm{s}$ type orbital. (Predicted bond angle: $90^\circ$.)
tetracoordination (no additional lone pairs): since the former is not possible, hybridise $\mathrm{s}$ and $\mathrm{s}$ orbitals to form $\mathrm{sp^3}$ hybrid orbitals. (Predicted bond angle: $109.5^\circ$.) Actually, in MO terms you should consider the four ligands to supply four group orbitals whose symmetry always fits one of the three central atom’s orbitals, resulting in one lower bond energy and one higher bond energy corresponding to triple-degenerate bonds. The $\mathrm{sp^3}$ description is mathematically equivalent, though.
penta- and hexacoordination; tetracoordination with additioal lone pairs and related: attempt to form as many normal bonds with p orbitals as possible; keep one lone pair in an $\mathrm{s}$ orbital if possible. Use remaining $\mathrm{p}$ orbitals to construct four-electron-3-centre bonds to the remaining atoms. (Predicted bond angles: diverse. $90^\circ$ going from 2e2c bonds to 4e3c bonds; $180^\circ$ between a pair of coordinating atoms contributing to the same 4e3c bond.)

Using the aforementioned 4e3c bonds nicely explains why electronegative atoms prefer to occupy these positions. You should always remember that 4e3c bonds can be described by the following resonance structures (drawn for $\ce{ClF3}$):
$$\ce{F-Cl^+\bond{...}F- <-> F^-\bond{...}Cl^+-F}$$
In this resonance, each outer atom taking part in a 4e3c bond has a formal average charge of $-0.5$; it is generally better to assign charges to electronegative atoms.

The theories I have outlined so far do not adequately predict the bond angles of water, ammonia and related compounds. Indeed, these compounds are remarkably complicated although they seem simple. The gist, however, is that a bond angle of $90^\circ$, while being most favourable on a pure orbital basis, introduces steric strain between the outer atoms which approach each other too closely if the central atom is relatively small. To combat this destabilisation, the bond angle is extended by mixing $\mathrm{s}$-contribution into the bonding orbitals — which in turn means that the lone pair will be less $\mathrm{s}$-like. The ideal bond angle for dicoordinated systems based on sterics is $180^\circ$, showing that the electronic contribution ($90^\circ$) is more important than the steric one ($180^\circ$). Larger central atoms than oxygen and nitrogen, such as phosphorus, antimony, selenium, etc., allow for smaller bond angles since the bond lengths are larger increasing the spacing between the ligands.

The reason for the special behaviour of nitrogen and oxygen is not their high electronegativity but their small size. Chlorine, which has an electronegativity similar to nitrogen, behaves as predicted for large atoms.

It has been commented that you should not invoke $\mathrm{d}$ orbital participation for main group elements. Remember what I said in the second paragraph: a shell’s $\mathrm{d}$ subshell has a higher energy than the next shell’s s subshell. Thus, before any same-shell $\mathrm{d}$ subshell is added to hybridisation, the next shell’s $\mathrm{s}$ subshell should be. Since hybridisations involving two different shells’ $\mathrm{s}$ subshells are not observed, we should reject the possibility of using $\mathrm{d}$ orbitals for hybridisation. Therefore, we can also flat-out reject the central two paragraphs of your quotation.


Temperature and pressure for three phase equilibrium of aqueous sodium chloride solutions

But how can we find this range of temperature or pressures? IS there an experimental methods?

The range of temperatures is fairly easy. Mix saturated aqueous NaCl (liquid solution, concentration about 5 mol/L) with ice (pure water, solid state), and measure the temperature. It will be about equal to the temperature of a household freezer (no coincident, freezers were using salt/water mixtures in the old days), i.e. -20 $^\circ$C. If you keep measuring while the ice melts and the NaCl concentration drops, you will get the temperature ranges, which will have the normal freezing point of water at the high end.

Partial pressure of water measured at atmospheric pressure

Using the same setup, you could measure the humidity above the ice/solution slush. In the experiment, water exists in the solid, liquid, and gas phase, just like pure water at the triple point. The gas phase contains mostly nitrogen and oxygen, so you would have to measure the partial pressure of water rather than the total pressure (which is about 1 atm and depends on the weather). The partial pressure will be a bit lower than that of pure water at the same temperature (the activity of water is less than one in a solution, so the partial pressure will be lower relative to pure water, Roult's law).

Pressure of water measured with the aid of a vacuum pump

In a slightly more involved setup, you would place the slush into a vessel that you evacuate with a vacuum pump. This removes all gas components (nitrogen, oxygen) from the gas phase, and replaces them with evaporating water until the water boils and there is pure water in the gas phase. At that point, you can close off the system and measure the pressure. The result should be the same as the first experiment.


How did Noe et al. derive $k_{AB}$ in their PNAS 2009 paper?

Let us start by defining all the terms in Eq. 1 carefully, as the OP does not do so. I have always found it easier to interpret such models in the context of particles flowing from state to state, and that is the picture that I will try to present here.

$T_{ij}$ measures the probability of going from state $i$ to state $j$ within a time $\tau$.
$\pi_i$ is the equilibrium probability of being in state $i$.
$q_i^+$ is the probability that, starting at state $i$, a particle will reach the set of states $B$ before the set of states $A$.
$q_i^-$ is the probability that, ending at state $i$, a particle has more recently encountered $A$ than $B$. (At equilibrium, all trajectories are reversible, so we can alternately think of $q_i^- = 1 - q_i^+$ as the probability that, starting at state $i$, a particle will reach $A$ before $B$, but that is irrelevant for this problem.)
$k_{AB}$ is the normalized rate of particle flow from $A$ to $B$. In other words, given particles that start out in $A$, $k_{AB}$ is the fraction of particles per unit time that end up in $B$.

Now we simply need to interpret the numerator and denominator of Eq. 1. To think about these quantities in the particle picture, let us multiply top and bottom by the number of particles $N$ evolving in the system, defining $N_i = N\pi_i$, so that $$k_{AB} = \frac{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+}{\tau\sum_{i\in\text{all states}}N_iq_i^-}.$$ In the numerator $$\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+,$$ the quantity $N_iT_{ij}$ is the number of particles that leave $A$ (going from $i$ in $A$ to $j$ not in $A$) in a time $\tau$. A fraction $q_j^+$ of these particles will end up in $B$. Now, in steady state, this must also be the number of particles that are arriving at $B$ from $A$, or otherwise we would have an accumulation of particles in the system.

In the denominator $$\tau\sum_{i\in\text{all states}}N_iq_i^-,$$ the quantity $\sum_{i\in\text{all states}}N_iq_i^-$ is the number of particles in the system that started out in $A$, which serves to normalize the number of particles in the numerator, and $\tau$ is the time interval in which this all occurs.

Putting all this together, then, $$k_{AB} = \underbrace{\frac{1}{\sum_{i\in\text{all states}}N_iq_i^-}}_\text{normalized}\underbrace{\frac{1}{\tau}}_\text{rate of}\underbrace{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}}_\text{particle flow from A}\underbrace{q_j^+}_\text{to B.},$$ as desired.


Corrosion of a galvanic couple made with silver and gold

As you state, though silver is not very reactive, silver jewelry would corrode more rapidly when in contact with gold and a bit of sweat as electrolyte. Though the difference in electronegativity of pure gold and pure silver is 0.15V, and is considered acceptably corrosion resistant, commercial jewelry contains copper and other metals in the alloys. There is clearly evident corrosion in photos of copper/silver jewelry. According to the Victoria and Albert Museum staff, "All metals, with the exception of 24 carat gold, suffer the effects of corrosion."

However, in a bangle, physical abrasion is likely to cause more damage and loss of material than galvanic corrosion. Wear the jewelry and enjoy it, but it will not last forever.


Does 1,3-butadiene in bromine/water do addition twice?

If reaction is carried out in 1:1 ratio of Br2 and 1,3 Butadiene, then the above two products will be formed. If more Br2 is taken, then surely 4 positions will be brominated.


What are the major factors that influence LogP

Partition Coefficient, represented as P and defined as: P = [Organic phase] / [Aqueous phase]

Where "[]" represents concentration in the  phase and $\mathrm{\log (P)= \log_{10}(Partition Coefficient)}$ if log P value is higher then its more lipophilic ;those drugs were more absorbed by cells and elimination of those drugs from body will be less .The partition coefficient is a ratio of concentrations of un-ionized compound between the two solutions so for a chemical which is having both hydrophobic and hydrophilic ends also have log P values.

Eg: Oleic acid: $\mathrm{\log (P)= 7.64}$

chemicals which are hydrophopic(lipophilic) tends to have more Log P values and hydrophilic(lipophobic) chemicals have less log P.

eg: pentane($\mathrm{\log (P)= 3.45}$) [lipohilic], 1-pentanol ($\mathrm{\log (P)= 1.51}$) [lipophobic]


Is it harmful to mix Hydrogen Peroxide with oven cleaning spray?

Chemically, we are talking about mixing hydrogen peroxide with sodium hydroxide. They will react to form sodium hydroperoxide, but the latter is not as good an oxidizing agent as hydrogen peroxide, and a weaker base than sodium hydroxide. Probably the added base helps to dissolve some components of the stain. I don't see this as risky, especially if it is on a small scale (spraying laundry), and uses 3% peroxide, rather than mixing them in bulk. Note: With hydrogen peroxide, you have to watch out because some metal ions catalyze its decomposition to oxygen and water.


A correct standard notation for the galvanic cell would also include the physical state of each species involved. As such, the correct notation for the given cell is:

$$\ce{Cu(s)}~|~\ce{CuSO4(aq)}~(10^{-3}~\mathrm{M})~||~\ce{CH3COONa~(aq)}~(10^{-2}~\mathrm{M})~|~\ce{Pt(s)}~(\ce{H2(g)})~(1~ \mathrm{atm})$$

Recall that aqueous solutions of salts are generally accompanied by their molarity in solution. Also, metals are usually solid, while gaseous species are accompanied by a reference electrode (platinum in this case) and their pressure value.

In your case, notice that the ionic salt $\ce{CuSO4}$ is assumed to be a $100\%$ dissociated ionic salt, hence, its aqueous solution will produce $[\ce{Cu^{2+}}]=[\ce{SO4^{2-}}]=10^{-3}\mathrm{M}$. That should answer your final query.

Apart from that, your expression for $\mathrm{Q_c}$ is wrong. It should be $\displaystyle\mathrm{Q}=\frac{p_{\ce{H2}}}{[\ce{Cu^{+2}}]}$. Since $\ce{H2}$ is present in gaseous form, use its pressure value in the expression for $\mathrm{Q}$  (and not molarity which doesn't exist)

PS: The concentration ($\pu{10^−3M}$) that's given to you is the concentration of copper ions after equilibrium has been achieved for the interconversion of $\ce{Cu+2(aq)}$ and $\ce{Cu(s)}$.


Preparation of sulfonyl chlorydes and mechanism of preparation of tosyl chloride

The way to understand this is to think of the autoprotolysis of water:

$$\ce{H2O + H2O <=> H3O+ + OH-} \tag{1}$$

In general autoprotolysis can be represented in the following form (you can recover eq. (1) by setting $\ce{A} = \ce{OH}$):

$$\ce{HA + HA <=> H2A+ + A-} \tag{2}$$

Now set $\ce{HA}$ as $\ce{ClSO3H}$ (or $\ce{(HO)SO2Cl}$, for clarity) and you have

$$\ce{(HO)SO2Cl + (HO)SO2Cl <=> (H2O+)SO2Cl + ^-OSO2Cl} \tag{3}$$

The cation produced in eq. (3) fragments to give water and the reactive chlorosulfonyl cation:

$$\ce{(H2O+)SO2Cl -> ^+SO2Cl + H2O} \tag{4}$$

The water in eq. (4) reacts with a third molecule of acid in an ordinary acid-base reaction:

$$\ce{H2O + (HO)SO2Cl -> H3O+ + ^-OSO2Cl} \tag{5}$$

The combined equation you have is simply the sum of eqs. (3), (4), and (5).


How Does One Create A Cellulose Ingredient For Use In The Creation Of Glucose?

Cotton wool, also called "purified cotton", is inexpensive and available at pharmacies ("chemists" in the UK). I could not find the percentage of impurities (probably lignin and minerals), but it is considered one of the purest forms of cellulose commercially available, and should not need pretreatment to make glucose.


This is unfortunately not a real answer to your question as such, but I'd like to provide some pointers why this approach is a lot more complicated than it seems at first. It might very well be (currently) impossible.
That is probably also a reason why you don't find a lot of publications on the theory of it and so far no black-box method has been developed. Solvation is still one of the major problems in computational chemistry, hence the use of implicit solvents and counter-charges. Such methods naturally come with the implicit disclaimer that these are approximate corrections only.
The most elaborate method I know so far is COSMO-RS, which still has large error margins, at least that is the my current state of knowledge.

I think for calculations you would at least need to run molecular dynamics simulations with many explicit solvent molecules to describe the equilibrium. This is important, because the proton which dissociates will be caught in the bulk of water molecules and a highly complex network of hydrogen bonds. This is very well demonstrated by the Grotthuss mechanism, which shows how easily bond lengths can change. I would say it is impossible to describe this complex system with a single quantum chemical calculation.

Furthermore, DFT is notoriously bad at describing non-covalent interactions, and including dispersion corrections will only get you so far. I have not checked recently, but I believe there are quite a few publications discussing various sizes of water clusters on this level. I have neither used nor heard of AFPD before, so it would be wise to thoroughly search for literature (benchmarks) about its performance on these clusters. Suffice to say, that one particular functional will not give you a reliable result as you have nothing to compare it to. Especially in such complicated cases you have to calibrate your results.

Another problem here is that you are the describing heterolytic bond dissociation. More often than not this may lead to additional problems. I don't always have confidence in my results when I treat simple donor-acceptor complexes where there is no charge separation and all singlet species. A supermolecular approach would probably a better choice, but could heavily suffer from a BSSE. However, it might be worth considering $$\ce{H2O\bond{...}HA <=> H2OH\bond{...}A}.$$

I think the key for calculating acid dissociation constants is treating the whole system as one ensemble.

Recall the common definition of the equilibrium constant for the reaction
\begin{align}\ce{AH (aq) &<=> A- (aq) + H+ (aq)},&
K &= \frac{a(\ce{A-})\,a(\ce{H+})}{a(\ce{HA})}.\end{align}
We conveniently ignore water here, because it is abundant and we assume ideal solution. However, when your system is too small, you will get a large error for $K$. This is most likely due to your solvated proton, which will be described worst with solvation models. Explicitly you are writing
$$\ce{HA (aq) + H2O (aq) -> A− (aq) + H3O+ (aq)},$$
which is not really what you are calculating. Instead you are (more ore less) using already approximations to treat thermal and solvent corrections
$$\ce{HA (g, th, solv) + H2O (g, th, solv) -> A− (g, th, solv) + H3O+ (g, th, solv)}.$$
And this doesn't even consider that you are calculating two states in an equilibrium and you will get the equilibrium constant (using activities, not concentrations) for only these two states (which might not even be the ground sates),
$$K = \exp\left\{-\frac{\Delta G}{RT}\right\}.$$
You have also not accounted for the possible conformational space of $\ce{HA}$/$\ce{A-}$, which might become very important as soon as you have ethyl moieties, or even cyclohexyl moieties.

These are just some issues that immediately come to mind, there might very well be more underlying insufficiencies. I have looked into this a couple of years ago and came to the conclusion not to bother with it any more.

What can you do instead?

Assessing acid strengths does not need to be at the level of what experimental chemists understand. If they really need the $\ce{pKa}$ they should run the appropriate measurements and obtain reliable results instead of educated guessing. If they want you to classify whether an acid is stronger or weaker, use a rigorous approach to obtain the conformational space for all species and compare them with isodesmic equations. This will have the advantage that insufficiencies of the methodology will likely cancel each other.

Another purely theoretical approach would be calculating gas-phase proton affinities. From this you can judge if a compound behaves stronger or weaker. Keep in mind that this approach is better suited for bases or nucleophiles. If you have the resources, you can also evaluate Lewis acidities,
$$\ce{Base + BX3 <=> Base-BX3}.$$

This should give you at least a good idea how on a relative level your compounds behave.

You can also use a first-principle derived value for your proton, which will likely cause the largest error. A quick check turned up a few promising results. (For example: Chang-Guo Zhan and David A. Dixon. J. Phys. Chem. A 2001, 105 (51), 11534–11540.) You can then go ahead and calculate know compounds and try to find a correlation between $\ce{pKa}$ and $\Delta G$. This way of "calibration" might lead to some significant error compensation.

I would also recommend to look at COSMO-RS in more detail. There might be a lot of development that I have missed and they might have found a way to obtain more reliable results. As far as I remember, the foundation of this was solid and a good step into easier capturing solvent effects without the need to resort to full blown MD simulations.


Does dissolving something in water change the specific heat of the solution?

If you were adding two liquids ideally, that is, assuming that the mixing occurs without any change in enthalpy - which occurs for chemically similar substances, then yes, the heat capacity of the mixture is a weighed average of the heat capacity of the components ($A$, $B$):

$$c_p = x_Ac_{pA}+(1-x_A)c_{pB}$$

However, when we are talking about dissolving solids, the situation is a little different. That is because infinite-dilution properties of a solute that derives from a solid source do not mimic the properties of the (crystalline) solid. For example, table salt ($\ce{NaCl}$) has a certain density as a powder, but this value has nothing to do with the density it acquires in solution once the ions dissociate and live freely in solution. The same goes for heat capacity, which is also a thermodynamic property.

So the answer is yes, the heat capacity of the solution does change by adding a solute and, in fact, it depends on the amount that is added. Only experimentally can you measure the exact function of concentration that defines the heat capacity of a particular solution. From a molecular perspective, it will be highly dependent on how the phenomenon of solvation develops.

As always, you can always look up the scientific literature to get the data of the thermodynamic functions of certain solutions including heat capacity.


Yes, by definition chemical potential is related to Gibbs potential in the following way:

$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_{j\neq i}} = g_i$

Those two notation are synonyms. You will find very detailed and precise explanation of this in: Atkins, Physical Chemistry.


What is the oxidation state of Calcium in teeth?

As a group II metal, you will be hard pressed to find calcium in the 1+ oxidation state (a quick google search yields an isolated report of calcium(I) chloride). Calcium's willingness to oxidise is evinced by its vigorous reaction with water. As you have noted, the calcium in tooth enamel is hydroxyapatite, often with varying degrees of fluorination. Which brings me to your next question, which hopefully you or a mod can slipstream into the main body -

I would definitely warn against eating chalk without good reason as it will evolve $\ce{CO2}$ in your tummy and raise your stomach pH, and is associated with a variety of negative health outcomes:

Milk-Alkali Syndrome
Hypercalcaemia
Antacid side effects

As always, all things in moderation. On a personal note eating chalk sounds like exactly the thing certain sectors of the 'alternative health' crowd would be into (namely, those who assert that the body is 'too acidic' (there's another crowd who claim the opposite and will recommend, say, drinking $\ce{HCl}$ (through a glass straw, natch)), ignoring for a moment the fact that the human body is pretty good at maintaining its pH and will up and have a cardiac arrest or something if the pH drifts more than a bit)


Enolate formation and tosylate displacement in cis/trans decalin

I went looking for this question again today after talking to someone about a similar reaction, then realised I never got round to trying to actually answer it.

Leading reference: Tetrahedron 1969, 25 (21), 5267–5274.

In the first case, with the trans-fused ring system, we first get tosylation as expected. Using the bulky potassium tert-butoxide, we get enolate formation at the least hindered alpha position. This enolate is then setup, as can be seen in the conformational diagram below, to displace the tosylate forming the new ring. This is confirmed in the cited literature.

In the second case, with the cis-fused ring system, things are a little more complicated. Once again we get tosylation, however the the cis-fused product can be drawn in multiple 3D conformations, only one of which is productive (shown in orange below).

In this case, the steric argument for enolate formation doesn't hold, as the thermodynamic enolate is unproductive, and as such only the enolate formed at the most hindered position goes forward to form the product.

As you mentioned in your initial post, "the obtained structure would be too strained", which is what prevented me from thinking this pathway happened in the first place. However, looking at the original Tetrahedron paper, they do have evidence to back up the formation of these ring systems. One possible conclusion is that due to the conformational preferences of the 6,6-ring systems, the bonds are closer together than you'd maybe expect, meaning there isn't such a penalty to formation of the cyclobutane.


Hybridisation of carbon atom in cyclopentadienyl anion

It entirely depends on what that electron pair is doing.

If the lone pair is hanging out by itself in its own orbital, then the carbon is probably sp3 hybridized, as greater hybridization is typically more stable. This is what you'll see on an alcohol or an amine lone pair. Those lone pairs aren't really interacting with anything else in the molecule, hence the "standard" logic of the rule you invoke.

However, in certain compounds you're able to stabilize the system further by delocalizing the electron pair. In this sort of case, the sp3 hybridized lone pair orbital is not able to participate in the delocalization, and thus will be higher in energy than an electron pair that is in the remaining p orbital of an sp2 hybridized carbon. This is the sort of thing you see with amide groups, where the "lone pair" on the nitrogen is sp2-like, which allows the electron pair to delocalize into the carbonyl, allowing for resonance stabilization.

Note that delocalization is not always going to happen. You can potentially draw up examples where delocalization would decrease the stability of the compound. For example, 1,4-Dihydropyrazines would be anti-aromatic if both lone pairs from each nitrogen were to participate in the ring system. This is highly disfavorable, and the molecule contorts itself to avoid it, putting the nitrogens into sp3 (pyramidal) hybridization, except where it needs to be planar to conjugate with an exocyclic double bond.

In this particular case - the one of the cyclopentadienyl anion - you would be able to form an aromatic system if the lone pair were to delocalize into the ring - much like pyrrole does. The lowered energy obtained from this aromatic delocalization more than makes up for the small amount of energy gained by going from sp2 to sp3, so the carbon should be sp2 hybridized.

The important thing here is not simply to invoke rote-memorized rules to determine hybridization states, but think about what the electrons are doing and what the comparative stability of the various options are. The hybridization will change to be the most stable (lowest energy) state. -- To be perfectly honest, neither sp2 nor sp3 is the "true" answer. The actual   answer to these sorts of questions lies more toward Molecular Orbital Theory, where the molecule is taken as a whole. Per-atom hybridization is a convenient shorthand, but still one which ultimately rests on the basis of the true molecular stability.


What is the mechanism behind the decomposition of enediol phosphate intermediate to methyl glyoxal?

This reaction can occur only when the $\ce{C-O}$ bond to the phosphate group
  lies in a plane that is nearly perpendicular to that of the enediol so
  as to permit the formation of a double bond in the intermediate enol
  product.

In solution, the enediol intermediate readily breaks down with the elimination of the phosphate at C3 to form the toxic compound methylglyoxal:

The spontaneous decomposition of the enediol intermediate in the triose phosphate isomerase (TIM) reaction to form methylglyoxal through the elimination of a phosphate group.

In order for this elimination to occur, the bond to the phosphate group must lie, as shown above (a), in the plane perpendicular to that of the enediol.

This is because, if the phosphate group were to be eliminated while this bond was in the plane of the enediol as shown in (b), the $\ce{CH2}$ group of the resulting enol product would be twisted 90° out of the plane of the rest of the molecule. Such a conformation is energetically prohibitive because it prevents the formation of the enol’s double bond by eliminating the overlap between its component $\mathrm{p}$ orbitals.

Reference

Voet and Voet Biochemistry


So you have the reaction :

$$\ce{LiAlH4 + 4H2O <=> LiOH + Al(OH)3 + 4H2}$$

You want to get the "energy" given by this reaction, which is nothing but the reaction enthalpy.

$\Delta_rH^0(T) = \sum_{i}\nu_i.\Delta H_{i,f}^0(T)$

Now you get this, just look for the theorical values of each $\Delta H_f^0$ you need for the 5 compounds you have (they are tabulated) and sum them up pondered with the stoechiometric coefficients.


Black Smoke Produced from Regular Lighter but Not Torch Lighter Applied to Copper

Common copper wire, especially used in electronic components like inductors and transformers, is usually covered with insulating varnish. This is the common source of black smoke, sometimes it may even burn on its own. I doubt, however, that kitchenware is coated with such varnish, there is no point in it.

There are two types of devices with flames: ones where the fuel is premixed with air, that usually produce blueish flame, and ones where fuel is not pre-mixed with air, but burns and mixes with it in same time, that may produce bright yellow flame.

For many organic compounds, especially ones that require large amount of oxygen per molecule to fully burn, in case the fuel is not premixed with enough air, the flame usually contains oxygen-deficient region where tiny particles of carbon and molecular hydrogen are formed. These carbon particles, since they are heated, give thermal light emission and are responsible for bright non-transparent orange-yellow organic flames. A typical example of such flame is a flame of a candle. It is a flame of heavy hydrocarbons or fatty acids (c20+), that require tens of molecules of oxygen per molecule of the fuel to fully burn.

In case gases from this yellow part of the flame are cooled without mixing with more air, this carbon particles may be obtained in form of deposits or black smoke. For this, a cold effective conductor of heat (i.e. metal, especially copper and silver) or a solid surface with high heat capacity introduced into yellow part of the flame may be used. In some cases, when too many of such particles are produced and the flame is not particularly hot, there is no need even to cool the flame. This is the usual case of hydrocarbons with big amount of carbon, like aromatics, and chlorinated hydrocarbons, like PVC (as long as they burn at all, which often is not the case even if there is no much chlorine in the compound).

Propane torches usually have a mean to regulate amount of air premixed with fuel. Torch lighters are small propane torches, usually without such means. When propane in such torch is premixed with excess of air, a very hot blueish short flame is formed. However, when air is not premixed with propane, the flame is organized in layers with more oxidized outer layers and more reduced inner layers. In inner layers, the dominating reactions are thermal decomposition

$\ce{C3H8 = 3C + 4H2}$

and incomplete combustion

$\ce{C3H8 + 2O2= 3C + 4H2O}$

(this is an oversimplification, as numerous other intermediates are formed, but that's advanced stuff)

In outer layers, propane combusts completely, just as if it was premixed with air

$\ce{C3H8 + 5O_2= 3CO2 + 4H2O}$

and carbon particles (and other reduced products) from inner layers also combust here.

Now, to get carbon particles from the flame in form of smoke the key is to cool them before they mix with more air. An ideal solution would be to suck them with tiny pipe and cool with water. However, it seems, that skein of copper mesh, at least until it is heated enough, works good for this purpose, both cooling the inner layers and hindering their mixing with air.

Since flames of fuels that are not pre-mixed with air have very efficient emitter of visible and and infrared light in it (many carbon particles), they tend to be cooler, especially the inner part. The flames with fuel pre-mixed with air (like in torch lighters with blueish flame like this), the flame is thin, quite short and much hotter.

Copper wire, when heated in air, also oxidizes. If done properly, it is a good way to produce pure black copper (II) oxide. However, it usually stay on the surface of the wire and indeed, can be washed with hydrochloric acid. If there are black particles that are not dissolved in hydrochloric acid, they are carbon particles.


Spectrochemical Series - Sigma Donor Capability

The spectrochemical series is an ordering of relative ligand interaction strength with metal ions. It can be used to infer relative $\sigma$-donor strength, if $\pi$-interactions are not possible for those particular ligands. Examples of this type ligand with no $\pi$-interactions include $\ce{NH3}$, $\ce{CH3-}$, and $\ce{H-}$.

Most ligands will exhibit $\pi$-interactions, operating as $\pi$-donors or $\pi$-acceptors, and these interactions dominate the ordering of the spectrochemical series. The $\sigma$-donor interactions are obscured and heavily affected by the $\pi$-interactions. Experimentally all of these effects are important, and it is difficult to separate these interactions from one another.

For example, the $\sigma$-donor interaction of a $\pi$-acceptor ligand is drastically increased from some initial level. $\pi$-backbonding reduces electron density at the metal center, leading to increased $\sigma$-donation (which increases $\pi$-backbonding, and so on). This will make a given $\pi$-acceptor ligand act as a better $\sigma$-donor than would be expected. The converse effect is observed with $\pi$-donor ligands reducing $\sigma$-donor strength.

The complexity of these interactions tempers the question of gauging $\sigma$-donor interactions only. The combined bonding effects influence one another.

Theoretically separating out $\sigma$-donor interactions from all $\pi$-interactions could be possible computationally, but I am unaware of any research on this specific topic. From a very general standpoint, improved metal-ligand orbital overlap will increase $\sigma$-donor strength, which may help estimate these characteristics qualitatively.

Most inorganic textbooks should be able to provide further detail on this topic under sections on ligand field theory.


Is "ice point" the same as "freezing point"?

Ice point is the freezing point of water.

Freezing point is the point where liquid turns to solid.

The difference is that in the case of ice point is substance is always water.

(Note that there is a slight discrepancy of 0.01 K between the triple point of water, melting point, freezing point and/or ice point. And technically, Ice point is the point where water and Ice are in equillibrium.)

See e.g. http://www.thefreedictionary.com/freezing+point, http://www.thefreedictionary.com/ice+point.


Confusion in calculating $\Delta U$ from a bomb calorimeter

The formula in the book is correct.  They are trying to get the change in internal energy per mole of sample.  From the first law, for this constant volume system (no work), $$\Delta U_{\textrm{total}}=q=C\Delta T$$where C is the heat capacity of the calorimeter.  This equation assumes that the heat capacity of the water in the bath is lumped into C, and that the temperature change of other parts of the calorimeter is the same as that of the water.
The number of moles of sample is m/M.  So, $$\Delta U_{\textrm{per mole}}=\Delta U_{\textrm{total}}\frac{M}{m}=C\Delta T\frac{M}{m}$$In their notation, they use the symbol Q to represent the heat capacity of the calorimeter C.


It is unlike other acyl chlorides as formyl chloride dissolves in water. At the same time, it reacts vigorously with it, and no other acyl chlorides produce carbon monoxide and hydrochloric acid.


How does dimethylsulfoxide serve as a cryoprotectant?

This question has a simple, but non-obvious answer. DMSO and water are miscible. Mixtures, and in particular homogeneous ones like that formed by water and DMSO, usually have lower melting (or more relevant in this case, freezing) points than either substance does alone. This is the reason salt melts ice: the salt solution has a lower freezing point than pure water. The physics behind this melting point depression are complex, but a simple enough approximation involves disruption of the intermolecular forces holding the solid crystal lattice together.

Water freezes at 0 °C, and DMSO freezes at 19 °C. Mixtures of the two have lower freezing points. More importantly, mixtures of the two likely form less disruptive crystals when frozen. Water is well know for having a solid state that is less dense than its liquid state. Ice crystals take up more room than the liquid water they formed from. Single water crystals can grow quite large, especially as smaller crystals merge. Additionally, water crystals tend to form this hexagonal fractal pattern (snowflakes) which have "pointy bits". DMSO, even at the 10% by weight customarily used, must prevent such large disruptive crystals from forming, especially when the everything is cooled to -80 °C.

DMSO is used for this purpose instead of other water-miscible liquids such as ethanol because it has very low toxicity.


Can a chemical reaction have two balanced equations?

You can expand the second equation so it has the same amount of educt on the left side as the first one:
$$\ce{6XeF4 + 12H2O → 3Xe + 3XeO3 +  24HF + 1.5 O2}$$
Difference on the product side, compared to first equation, is
$\ce{-1 Xe + 1 XeO3 - 1.5 O2}$.
This excess of one $\ce{XeO3}$ can go
$$\ce{XeO3 → Xe + 1.5 O2}$$
, exactly the amount of xenon and oxygen that was missing compared to the first equation.
Your equations describe not one reaction, but (at least) two, which don't have to occur in a predefined proportion. It's not directly clear, and would involve an in-depth investigation, what exactly the elementar reactions are.
So you could make an infinite number of chemically correct, but quantiatively different equations for the hydrolysis of xenon flouride. All depends on the reaction conditions, probably.


Electrophilic addition of HX to 2,3-dimethylbut-2-ene

Electrophilic addition would likely still occur. Remember that Markovnikov's rule is an observational heuristic for "in electrophilic addition reactions of alkenes in which carbocations form, the more/most stable carbocation forms preferentially, and thus the product originating from that cation is the major product".

In this case, a carbocation can form. Nothing about the tetrasubstituted pattern prevents addition. The molecule is still planar and can be aproached by the electrophile. The methyl groups are electron donating by induction. The methyl groups 1) increase the electron density of the alkene and make it more nucleophilic, and 2) stabilize the carbocation intermediate.

Thus, 2,3-dimethyl-2-butene is likely more reactive than other alkenes toward electrophiles. For example:

$$\ce{(CH3)2C=C(CH3)2 + HBr -> (CH3)2BrC-CH(CH3)2}$$


Why do alpha-hydroxy ketones respond positively to Tollens', Fehling's, Benedict's, Schiff's and HgCl2 tests?

Its because of two tautomerizations happening, the ketone becomes en-ol, and then another tautomerization happens with the original hydroxy group (which is terminal), making it an aldehyde group. All three forms exist in equilibrium, so it responds to Tollens' test.


Calculating standard combustion enthalpy of dodecane from that of heptane & octane

Heptane is $\ce{CH3-CH2(5)-CH3}$

Octane is $\ce{CH3-CH2(6)-CH3}$

Given combustion enthalpies for each of $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$  respectively, we now have a simple set of algeraic equations:

5A + 2B = $-4816~\mathrm{kJ~mol^{-1}}$

6A + 2B = $-5470~\mathrm{kJ~mol^{-1}}$

First we can solve for A as the combustion enthaply of a CH2 in a linear hydrocarbon and we get the value $-654~\mathrm{kJ~mol^{-1}})$.

We can now solve for B,

In the first equation we get 2B = ($-4816 - (654 * 5)~\mathrm{kJ~mol^{-1}}$, A = $-773~\mathrm{kJ~mol^{-1}}$

In the second equation we get 2B = ($-5470 - (654 * 6)~\mathrm{kJ~mol^{-1}}$ , A = $-773~\mathrm{kJ~mol^{-1}}$

dodecane  has CH3-CH2(10)-CH3 , for 10A + 2B.

10 * A  + 2 * B  = ($-6540$ + $-1546$)$~\mathrm{kJ~mol^{-1}}$ = $-8086~\mathrm{kJ~mol^{-1}}$


If I remember correctly, we (an analytical laboratory) would generally get all copper minerals into solution using a nitric/perchloric digestion, depending on the sample matrix.  Unless the sample matrix as a whole were highly silicious in nature an additional HF digestion was not required. As commented above though, $\ce{HNO3/HF}$ may work for you also.


Why are inert gas (especially Xenon) compounds powerful oxidizing agents?

Let's be clear here. Xenon by itself is not an oxidizing agent. Instead, certain compounds of xenon tend to be strong oxidizing agents.

First, let's consider an analogous case - the nitro group. Molecules containing nitro groups ($\ce{-NO2}$) tend to be explosive - the nitro groups contain oxygen which facilitates combustion. Remember, combustion is just a form of oxidation.

Back to xenon!

Xenon, being a noble gas, shouldn't like bonding too much. It already has a complete octet of outer shell (valence) electrons.

So it is likely the case that compounds of xenon are only weakly held together. Compounds of xenon can therefore said to be relatively thermodynamically unstable. Therefore, it is also likely that compounds of xenon tend to react in the presence of other molecules - especially ones containing atoms to which fluorine can form stronger bonds.

We verify the above by examining bond enthalpy data: the Xe-F bond has a bond energy of about 130 kJ/mol. Compare this to a typical C-F bond with a bond energy of 485 kJ/mol.

It is no wonder therefore that compounds such as xenon difluoride are used in fluorinating (oxidizing) organic molecules.

Here are some examples from Wikipedia. Why the nitro group in the second photo has been reduced to an amino group - I have no clue - and I suspect it's just an error. Further updates soon ...


Comparison of Resonance Energies of some Carbocations

It is not like π bond between fluorine and carbon is very stable and that is where your line of thinking is wrong. While it is true that $\ce{{2p}-{2p}}$ π bonds are more stable in general than their $\ce{{2p}-{3p}}$ counterparts, one should not take that alone as a decisive factor.

The question is comparing the electron-donating capabilities of fluorine and chlorine, and that italicised word there alone should ring a bell. Fluorine is much more electronegative than chlorine which means that a carbon bound to fluorine has a much smaller electron density which destabilises cations much more than could be offset by the overall weak π donating capacity of fluorine.

Remember that for electrophilic aromatic substitutions, halide substituents are classified as deactivating — mainly due to their $-I$ effect — and only weakly ortho/para-directing — due to the smaller $+M$ effect. This should serve to show which effect predominates and it only gets more pronounced if instead of a large aromatic system you only consider the atom directly adjacent.


By definition (in Woodward-Fieser rule), an exocyclic double bond is a double bond where one of the $sp^2$ carbon atoms of the bond is a part of a ring, while the other $sp^2$ carbon of the bond is not a part of same ring (it could be a part of another ring).

If both the $sp^2$ carbon atoms of the double bond is part of the same ring, then it is called endocyclic double bond. From these definitions, you can understand that exocyclic would stand for a double bond outside the ring and endocyclic would stand for a double bond within the ring. Following examples would illustrate the definition:


There’s a simple way to calculate how many bonds you need to form a Lewis structure, and how many lone pairs remain. It is four simple steps:

How many valence electrons do we have? (add up)

For bromate, the calculation is:

$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$
How many valence electrons are required for full octets/hydrogen dublets?

This equates to $2m + 8n$, where $m$ is the number of hydrogens and $n$ the number of other elements.

$$4\times 8 = 32\tag{2}$$
How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$

$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$
How many electrons are remaining? These will form lone pairs. $(1) - (3)$

$$26 - 6 = 20 = 10~\text{pairs}\tag{4}$$

Thankfully, knowing the general structure is ‘halogen in the centre, oxygen atoms around it’ and knowing that there are three oxygen atoms, this clearly shows us that all $\ce{Br-O}$ bonds are single bonds. The halogen would then need one more lone pair to be satisfied, the remaining nine lone pairs go to the three oxygen atoms (3 each). This leads us to the following, final Lewis structure:

(Unfortunately, the $2+$ on iodine and its remaining lone pair intersect in the image. Originally taken from an answer about iodate hence the central iodine. Replace it with bromine to get bromate.)


Reaction of ethylene glycol with PI₃ (P + I₂)

User @tato_mazo already pointed out the reason that due to two iodine atoms being adjacent to each other, ethylene diiodide is sterically crowded and decomposes to produce ethylene and iodine. My answer basically points out the actual reaction involved and some properties of ethylene diiodide.

$$\ce{OHCH2CH2OH + PI3 -> [CH2I-ICH2] -> CH2=CH2 + I2}$$

Ethylene diiodide is an unstable compound which is made by mixing ethene and iodide (damp) under light or slight heating crystallizing into solid, colorless needle like structure (m.p. 75 °C). It again decomposes into corresponding reactants on stronger heating and occasionally at room temperature, though slow. It is only be sublimed or preserved colorless under ethene gas.

References

Ahluwalia, V. K. Textbook Of Organic Chemistry; Ane Books Pvt. Ltd, 2010; Vol. 2. ISBN 978-93-80156-02-6.
Strecker, A. Short Text Book of Organic Chemistry, 1881.