How do sodium-potassium pumps differentiate between sodium and potassium?

$\ce{Na^+-K^+}$ ATPase (more commonly called the$\ce{Na^+-K^+}$ pump) is a transmembrane protein that consists of two types of subunits: The 1000-residue  $\alpha$ subunit contains the enzyme’s ATP and ion binding sites, and the 300-residue $\beta$ subunit facilitates the correct insertion of the  subunit into the plasma membrane.

The  subunit of this $160 \ Å$ long protein consists of a transmembrane domain (M) composed of 10 helices of varied lengths and, and three cytoplasmic domains, namely: the nucleotide- binding domain (N), which binds ATP; the actuator domain (A), so named because it participates in the transmission of major conformational changes; and the phosphorylation domain (P), which contains the protein’s phosphorylatable Asp residue.

The $\ce{Na^+-K^+}$–ATPase is an antiport that generates a charge separation across the membrane, because for every three positive charges exit the cell for every two that enter.

The key to how it functions lies in the precise phosphorylation of a specific Asp residue of the transport protein. ATP phosphorylates the transporter only in the presence of $\ce{Na^+}$, whereas the resulting aspartyl phosphate residue is subject to hydrolysis only in the presence of $\ce{K^+}$.

Thus, the "pump" has two conformational states (called E1 and E2) with different structures, different catalytic activities, and different ligand specificities.

Outline of the entire process

E1.ATP, which acquired its ATP inside the cell, binds three sodium ions to yield the ternary complex $\ce{E1.ATP.3Na^+}$.
The ternary complex reacts to form the “high-energy” aspartyl phosphate intermediate $\ce{E1-P.3Na^+}$.
This “high-energy” intermediate relaxes to one lower in energy, i.e $\ce{E2-P.3Na^+}$, and releases its bound sodium ions outside the cell.
$\ce{E2~P}$ binds two potassium ions from outside the cell to form an $\ce{E2-P.2K^+}$ complex. (structure pictured earlier)
The phosphate group is hydrolyzed, yielding $\ce{E2.2K^+}$.
$\ce{E2.2K^+}$ changes conformation to E1, binds ATP, and releases its two potassium ions inside the cell, thereby completing the transport cycle.

This enzyme appears to have only one set of cation binding sites, but they changes both their orientation and their specificity during the course of the transport cycle. The difference in specificity the two conformations have arises principally because the two ions differ in size.

The illustration given below, shows the ion binding site in this ATPase and one can easily note the difference in positioning of residues between the two conformations.

Particularly, I would like to call your attention to figure(s) a, b and c. The purple spheres represent sodium ions, and the green ones correspond to potassium ions. Fig a and b represent the same view of the ion binding site, and one can clearly see the slight positional differences of the residues which account for the difference in specificity.

Image Source(s):
Voet and Voet, Biochemistry
and http://www.nature.com/nature/journal/v502/n7470/full/nature12578.html


Calculating solubility of iron(III) hydroxide in water

In my version of the book the original assumption (step 7a) is shown to lead to a contradiction when later checked (logically, this form of proof is called reductio ad absurdum or proof by contradiction). The point is that you should 1) learn not to be afraid to apply assumptions, but that you should then 2) verify that the original assumption was indeed acceptable. This is a strategy repeatedly encouraged in the book.

To show how this leads to a contradiction,

make the assumption, so that $ \ce{3[Fe^{3+}]} = \ce{[OH-]}$
solve for $S = \ce{[Fe^{3+}]}$  based on the $K_{sp}$ and the assumption
Compute $\ce{[OH-] = \ce{3[Fe^{3+}]}  }$, following the original assumption
Compute $\ce{[H+]}$ from $K_w$ and $\ce{[OH-]}$
Check the accuracy of the assumption

It is by the way not difficult to solve the problem exactly, by computing the $\ce{[OH-]}$ as follows:

$$ \ce{[OH-]}=\mathrm{\sqrt\frac{K_w+\sqrt{K_w^2 +12K_{sp}}}{2}}$$

and then applying the expression for the solubility constant to compute the solubility.
However the point of the problem is to illustrate how to follow this line of reasoning when solving problems (the use and verification of assumptions).


How to recover 15g gold from 50 gallons of 30% Acid?

You could distill the waste easily enough, but the distillation gases will be rather nasty. You will also need a condenser to cool everything down again or the environmental penalties (and corrosion damage to the exhaust system) will be a lot more than the gold value.

Once you have the waste boiled down (literally) into a small flask you could use any of the known gold-extraction methods to get it out in elemental form.

At current market prices you have about US$580 of gold in the drum. No idea how much you will spend in fuel and cooling water to recover it but it will be a substantial percentage. Why don't you just tell "Someone" that they can either pay for the gold they dumped or mind the still themselves over the weekend (and pay for the propane). This will also encourage others to pay more attention to what is in the beaker.


Understanding Standard Hydrogen Electrode

This is a concentration cell question where 2 electrodes of same stuff are dipped in their own ions and if there is non equal concentrations of anything a "potential" develops called electrode potential (to be understood in qualitative way)(kind of more concentration of reactant forces product formation)
as you can understand this would be a redox reaction where reduction would occur in one and oxidation in another, Now, by standard it means that all the components of the reaction (participants, i.e. reactants and products) are in standard state i.e. $1M, 1$ $bar$ etc, so, we know that concentration of $\ce{H+}$ in standard one is $1M$, we need to create biggest gap between them, as the concentration is really high and we cant compete with it, just reduce our own $\ce{H+}$ to increase their difference and force ours to undergo Oxidation to gas and standard undergoes reduction

Just for extra information,
$E=E_0-{RT\over nF}\ln Q$ and can be derived by $G=G_0+RT\ln Q,\;G=-nFE$ {$n$ is number of electrons being exchanged ex- in $\ce{2H+->H2},\;n=2$; $F$ is the charge on 1 mol electrons} and Standard Hydrogen is the standard bearer so, we assign its standard value as $0$, $E_0=0$.

EDIT : I can't comment due to reputation
, basically, assume complete dissociation and and concentration of $\ce{H+}$ in $\ce{HCl}$ would be $0.1M$ , $\ce{H2SO4}$'s second ionization is not complete like others but can be assumed for a good approximation, so concentration becomes $0.2 M$ , and basic thing to remember is

IF ONE SIDE  UNDERGOES OXIDATION THEN OTHER MUST UNDERGO REDUCTION

and if we strengthen either reaction, potential increases, concentration is only in a solvent, how can a solid would have active $\ce{H+}$ ions (everything is activity not concentration or partial pressure flat out, they are approximated, solids are assumed to have activity as unity) and log is defined as the power to $10$ that would give us that number


About the nomenclature: manganate or manganite?

Let's go down the ladder.  $\ce{KMnO4}$ is permanganate, Mn(VII).  $\ce{K2MnO4}$ is manganate, Mn(VI).  $\ce{K4MnO4}$ is manganite, Mn(IV).  $\ce{LaMnO3}$ is called both manganate and manganite, but Mn(III). In a proper world it should be lower than that manganite.  They are pedagogically both wrong if it is La(III).

http://dictionary.sensagent.com/potassium%20manganite/en-en/

Manganese oxidation state nomenclature is a real world mess. Drawing nomenclature parallels to Cl(VII) and downward is not valid.  Drawing nomenclature parallels to titanium is poor, for Ti(IV) is the highest common oxidation state.  $\ce{SrO + TiO2}$ gives $\ce{SrTiO3}$, properly called titanate, for it is Ti(IV) on both sides.

Organic had the same problem. The highest valence carbocation was tricoordinate, hence a carbonium ion.  Magic Acid gave $\ce{CH5^+}$.  Carbonium ions were renamed carbenium ions to general confusion.  Then, $\ce{CH6^{2+}}$.  Organic dropped the ball with direction of optical rotation versus geometric left and right.  It was flat out stupid designating E ("trans") and Z ("cis") olefins.  The designation should have been abstract rather than language-originated.


What's the reaction between H2SO4 and NaH2PO4?

Actually you have two reactions, since sulfuric acid can give off two protons:

$\ce{H2SO4 + NaH2PO4 -> NaHSO4 + H3PO4}$

$\ce{NaHSO4 + NaH2PO4 -> Na2SO4 + H3PO4}$

The $\ce{NaHSO4}$ is called sodium bisulfate.  Bisulfate is the thing you form first when sulfuric acid acts as an acid, then you get the sulfate ion if you have a strong enough base to react with the almost but not quite strongly acidic bisulfate ion.  In this case, actually, you can look up some acid dissociation constants and discover that phosphoric acid has almost exactly the same amount of strength as the bisulfate ion acting as the "second stage" of sulfuric acid.  So the second reaction is really reversible:

$\ce{NaHSO4 + NaH2PO4 <=> Na2SO4 + H3PO4}$

For more on what makes acid-base reactions reversible, see here.


Sulfuric acid is  unreactive with carbon- indeed only hot concentrated sulfuric acid reacts with carbon to produce carbon dioxide.


What happens when hydrogen bonds stretch?

Yeah this question is actually surprisingly difficult to to answer.  But let me explain it this way.   In thermodynamics you can't have negative temperature.  Thus anytime you are getting heat, you are actually gaining electrons.  When you gain electrons these chemical bonds change and rearrange themselves, which is how they can store energy.
We do not think about this normally because we look at hydrogen and we just see a proton and electron.  But in reality, unless hydrogen is in an extreme environment, it likely has absorbed energy in some manner.
For instance,  if you put a glass of water under concentrated sunlight it will absorb photons, these photons will be absorbed by the moving electrons inside the atom. Which we understand as "heat", moreover If there are more electrons than water molecules than it will start to evaporate because they run out of room.


How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?

Explanations of bonding and hybridization in non-metals that involve d-orbitals have generally fallen out of favor for two reasons:

Usually d-orbitals are considerably higher in energy than the s- and p-orbitals they would interact with leading to poor mixing, and
d-orbitals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak

A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved.  Such bonding involves p-orbitals with overlap geometry and molecular orbitals as pictured in the following diagram.

In the case of $\ce{IF5}$, the molecule has the following geometry.

Using the hypercoordinated bonding approach, one would say that the 4 identical "equatorial" fluorines (they're not quite equatorial, we'll return to that later) and 2 p-orbitals on iodine would participate in this type of bonding.  Each of the two iodine p-orbitals contains two electrons from iodine and bonds to two fluorines which each contribute one electron.  We wind up with two linear hypercoordinate bonds at right angles to each other that lie in a plane.  Each of these bonds is comprised of 3 atoms ($\ce{F-I-F}$) and contains 4 electrons as described in the MO diagram above.  The only orbitals remaining on iodine are an s- and a p-orbital.  These two orbitals mix in traditional fashion (like the carbon in acetylene) to produce two $\ce{sp}$ orbitals.  these two orbitals are arranged in a linear fashion (again, like acetylene) and are perpendicular to the plane containing the two 3 atom - 4 electron hypercoordinate bonds.  One of these $\ce{sp}$ orbitals contains an iodine lone pair of electrons and the other forms a bond to fluorine using one iodine electron and one fluorine electron.

Because of the asymmetry introduced by having a lone pair on one side of the molecule and a fluorine on the other side, our 4 "equatorial" $\ce{I-F}$ bonds distort from planarity tilting away from the electrostatically repulsive lone pair as illustrated in our figure (above) showing the geometry of $\ce{IF5}$.

In this bonding scenario the central iodine is sharing 4 electrons in each of the two hypercoordiante bonds (total=8) and two electrons in the apical $\ce{I-F}$ bond; plus there is a lone pair on the iodine.  Therefor the iodine is neutral (7-[(8+2)/2]-2 = 0), but is it surrounded by an octet of electrons?  The answer is, yes.  Look again at the MO diagram for the hypercoordinate 3 center - 4 electron bond.  Notice that the nonbonding orbital ($\Psi_2$) contains 2 electrons, but these electrons are centered entirely on the fluorine atoms, not the central iodine.  Therefore, around iodine we have 2 (lone pair) plus 2 (apical $\ce{C-F}$ $\ce{sp}$ hybridized bond) plus 2 times the two $\Psi_1$ electrons in our hypercoordinate bonds, yielding a total of 8 electrons around the iodine.


PLP-Dependent Cyclization and Decarboxylation

A PLP-dependent enzyme could certainly accomplish a reaction like this. An example of something similar is the PLP-dependent enzyme KAPA synthase in the biotin biosynthesis pathway. In that reaction, alanine is decarboxylated and a new bond is formed between the alpha carbon and an electrophilic thioester. Interestingly, the proposed mechanism involves deprotonation of the alpha carbon rather than decarboxylation as the initial step. Only after the new C-C bond is formed is the carboxylate lost:

1: Source: Webster, S.P, et al. (1998) Characterisation of S-amino-7-oxononanoate synthase: A bacterial PLP-dependent, acyl CoA condensing enzyme. Biochemical Society Transactions 26(3):S268. DOI: 10.1042/bst026s268
Although this is an intermolecular reaction, your proposed intramolecular reaction should be even easier to accomplish. An example of a PLP-dependent enzyme catalyzed reaction that results in intramolecular cyclization (but not decarboxylation) is that of aminocyclopropane carboxylate synthase.


Good question.  Let's examine your system in more detail.

$$ \frac{dA}{dt} = - k_1 A $$
$$ \frac{dB}{dt} = k_1 A - k_2 B $$
$$ \frac{dC}{dt} = k_2 B $$

The third equation is redundant because $C$ can always be found from the conservation equation $C = A_0 - A - B$.
This system can be solved exactly without any further assumptions.  So thank you for picking a system that has this nice feature.  An exact solution will let us compare to any approximate solutions obtained via simplifying assumptions.  Assuming that the initial conditions are $A(0) = A_0$ and $B(0) = 0$, then the exact solution is

$$ A(t) = A_0 e^{-k_1 t} $$
$$ B(t) = \frac{k_1}{k_2-k_1} A_0 \left (e^{-k_1 t} - e^{-k_2 t} \right )$$

What about the quasi-steady state approximation (QSSA)?  The QSSA is an assumption that $B$ does not really vary with $t$ as far as $C$ is concerned.  That is, predicting the rate of formation of $C$ can be done by assuming therate of formation of $B$ is zero.  So if $k_1 \gg k_2$, then $A$ is more or less "instantly" and quantitively converted to $B$, after which $B$ converts to $C$ via the usual first order kinetics with rate constant $k_2$. So this case, $k_1 \gg k_2$, is already a clue that the QSSA is invalid for this mechanism.
Why is the QSSA invalid for this case of $k_1 \gg k_2$?  Usually, the QSSA is invoked on catalytic reaction schemes.  In these schemes, it is assumed that the concentration of catalyst is very small relative to the concentration of substrate.  Thus there is an upper bound on the possible concentration of the intermediate, a bimolecular complex of reactant and catalyst: the total amount of catalyst added limits how much bimolecular complex there can be at any time.    This assumption that substrate is in excess relative to catalyst must be satisfied for the QSSA to make sense.
What about the other case?If $k_2 \gg k_1$, then at time scales where $k_2 t \gg 1$, then $B \approx \frac{k_1}{k_2}A_0 e^{-k_1 t}$.  The maximum concentration of $B$  in this case is the ratio of rate constants, a ratio that we've already assumed to be much less than one.  And since the maximum concentration of $C$ is not similarly bounded, this is a case where the QSSA makes sense.  This implies that the conservation equation can be written as $A_0 = A + B + C \approx A + C$.  Proceeding with the QSSA assumption to find the rate of formation of $C$:

$$\frac{dB}{dt} \approx 0$$
$$B = \frac{k_1}{k_2}A$$
$$A_0 = A + B + C \approx A + C$$
$$\frac{dC}{dt} = +k_2 B \approx k_1 A = k_1 (A_0 - C) $$

That last equation is now a single-variable first-order linear ODE and could be solved for $C(t)$ if desired.  But the main thing is that with the QSSA, $B$ was eliminated from the expression for $C$

I can't think of a physical reason why a system where $k_1 = k_2$ exactly couldn't exist.  But someone else will have to come along and explain how to get the math to work out.

If $k_1 = k_2 = k$, then the system of equations is :
$$ \frac{dA}{dt} = - kA$$
$$ \frac{dB}{dt} = k(A - B)$$

The solution is:
$$A = A_0 e^{-k t}$$
$$B = A_0\; k t\; e^{-k t}$$
In this case, there is still no generally valid way to apply the QSSA.  The derivative of $B$ is $B'(t) = k A_0 e^{-k t} \left(1 - kt \right)$ and is equal to zero only when at the precise time where $kt=1$.  It isn't approximately equal to zero in any other time range, except as $t \rightarrow \infty$.  So, the QSSA isn't valid to the case when $k_1 $ is exactly equal to $k_2$, and it isn't valid when $k_1 \neq k_2$ either.
Bottom line: using the QSSA in this problem is equivalent to assuming $k_2 \gg k_1$.  You might not have thought so without going through this whole analysis.  The key fact is that the QSSA requires that variation in the concentration of the intermediate be small relative to the variation in concentration of the products or substrates that you care about.  In mechanisms involving catalysts, this happens any time that catalyst levels are small relative to substrate(s).  However, in other mechanisms, this happens only with arbitrary restrictions on the kinetic parameters that may not be obvious a priori.


Reason for triphenyl methane being less acidic than acetylene

As mentioned by @orthocresol, just comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units and the amount of resonance required to overcome this difference is very high.
Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.


I think this is due to the disordered atomic arrangement at the grain boundary which implies higher  internal energy of the constituting atoms and consequently more reactivity (attack by acid, corrosion,...).


Starch is a weak acid, but a stronger acid than water.

If you make a slurry of starch and add sodium hydroxide the starch will react with the sodium hydroxide, the starch slurry will then change in viscosity. It normally becomes a lot more viscous in my experience. This is part of the method by which starch is alkylated. I have made moderate sized batches of benzyl starch from corn starch by first stirring the starch with sodium hydroxide and then treating the resulting paste with benzyl chloride and then heating it up.

If you were to get a very senstive pH meter you would see that a slurry of starch in water is slightly more acidic than water would be. An alternative would be to use a dilute sodium hydroxide solution to make the measurement, I will warn you that it can get messy as the starch can be transformed into a paste / slime.

MaxW is right in part, a pure starch will give no ash but zinc sulfate will dehydrate to anhydrous zinc sulfate unless you heat it very very hard.

Another method is to add iodine, many starches will form a deep blue / black colour with a small amount of iodine. Zinc salts will not.


I do not think that hydrates will serve as good moisturizing agents, because they usually do not loose their crystal water unless when being heated. Magnesium sulfate heptahydrate, for example, starts to loose water at temperatures above 50 °C, and complete dehydration of the salt is only achieved at temperatures above 310 °C. Some hydrates can even have a dehydrating effect when they react with moisture to hydrates with higher $\ce{H2O}$ content. Calcium sulfate hemihydrate, for example, reacts with water to form the dihydrate in an exothermic reaction. The stability of hydrates is the reason why many anhydrous salts which are easily hydrated are used as drying agents, for example, sodium sulfate and magnesium sulfate.


Which of the resonance structures is more stable?

There are few things you should keep in mind while deciding comparative stability of resonance structures:

Non-polar structures are more stable than dipolar.

Resonance structures with a greater number of covalent bonds are more stable than those with lesser number. Thus, nonpolar structure of buta-1,3-diene is more stable than any other resonating structures.

Resonance structures with similar charges on adjacent atoms are insignificant due to electrostatic repulsion and thus are unstable.

Structures with positive charge on electropositive element and negative charge  on electronegative element are more stable.

Structures in which charge is delocalised are more stable than those in which there is charge separation.

Structures with complete octet are more stable.

All the atoms in a molecule taking part in resonance should be co-planar. because this enables effective overlap of p-orbitals and delocalization of electrons.

But when you have to decide among these properties itself, you can use the following observational facts:

Although during comaprison, there may be a positive charge on the more electronegative atom(O in this case) with complete octet, but it will be preferred over a structure with an electropositive atom ( C in this case ) having positive charge and incomplete octet.

But when number of covalent bonds are same, positive charge on C atom with incomplete octet is more preferred than positive charge on more electronegative O atom.

A stable neutral structure is a major contributor of a resonance hybrid.

An important note: when you are comparing the stability of the resonance structures , you are actually trying to figure out the major contributor to the resonance hybrid (that which best describes it). Technically, all the resonance structures are contributors to the same true structure of the ion. Since all of your mentioned structures represent the same species, they cannot have have different energies, and therefore they cannot have different stabilities.
Your question is actually asking "which contributor is more important in describing the structure and behavior of the hybrid?"
So fact number 1 and 2 makes it very clear that structure 4 in this case will be the most stable.


Effect of increase in volume of container on concentration of chlorine at equilibrium conditions

According Le Chatelier's Principle, when a change in concentration or pressure is applied to an equilibrium, the equilibrium changes to counteract the change.

In your equilibrium, the volume of the container is increased, causing a decrease in pressure. To counteract this, the side of the equilibrium with the highest number of moles is favoured. I.o.w. the equilibrium will move somewhat to the left ($\ce{4HCl + O2}$).


On adding sugar to water, does the entropy increase or decrease?

While it may seem that randomness always increases when a crystal is dissolved into a liquid phase, it does not have to be that way.

Concerning sugar, the molecule has a large number of hydroxyl groups and is generally rather large when compared with the water molecule around it — much larger than your average sodium or chloride ions. Every hydroxyl will act as both a hydrogen bond donor and an acceptor, likely to two different water molecules, creating a cage of water that is rather ordered and large again when comparing to the sodium or chloride ion’s solvation cell. This will decrease the disorder (and hence entropy) of the free water so much, that it counteracts the increase in entropy of the crystal dissolving. (The effect is more pronounced with sucrose when compared to glucose due to the molecule’s size and the number of hydroxyls.)

On the other hand, the molecules in a sugar crystal are likely less ordered than the ions of a salt crystal. They will form extensive hydrogen-bonding networks, yes; but overall if you shift one row of molecules relative to its neighbour you won’t necessarily break the crystal as you won’t suddenly have ions of the same charge next to each other. And finally, the single salt crystal will break down into much smaller fragments than the sugar molecules.

So summed up, the effects are:

large solvation cells
crystal breaks down into larger (less disordered) fragments upon solvation
sugar crystals themselves are less ordered than salt crystals

Meaning that there is a lot less entropy to be gained.


Why are triple covalent carbon to carbon bonds drawn linearly in skeletal structure?

Triple bond in carbon means that the carbon atom is sp hybridized. sp hybridized carbon has the two bonded groups at 180 to each other. Therefore it is displayed linearly which is closest to the actual structure. Single and double bonds are at an angle of nearly 109.5 and 120 degree therefore are not drawn linearly.


Why does attraction always cause release of energy making the system more stable?

Attraction is a shorthand way of saying "potential energy gets lower as the distance gets lower".  The opposite is true of repulsion.  The potential energy of two like-charged particles, e.g. two $\ce{Na+}$ atoms, will increase as they are brought closer together.

Attraction (and its opposite, repulsion), are forces with units (in SI) of Newtons, or $\frac{\rm kg\cdot m}{\rm s^2}= \rm N$.

Energy has units of Joules, or $\frac{\rm kg\cdot m^2}{\rm s^2} = {\rm N\cdot m} $.

The relationship between a force $F$ between two particles separated by a distance $r$ and the potential energy $E$ of those particles is $F = -\frac{dE}{dr}$.

If the energy gets lower ($dE$ is negative) when two particles get closer together ($dr$ is negative), then the force $F$ is attractive. If the energy gets higher ($dE$ is positive) as the distance gets lower ($dr$ is negative), then the force is repulsive.

Force and energy are two different mathematical ways of looking at the same thing.  That's why they are always related.$%edit$


My guess is that the acid is just there to protonate the carboxylate salt and allow decarboxylation. Hydrolysis of the ester with $\ce{KOH}$ generates the potassium carboxylate salt:

Since the mechanism for thermal decarboxylation involves a cyclic transition state including the carboxyl $\ce{-OH}$ group, I expect that the carboxylate salt must be protonated before decarboxylation can take place, hence the addition of aqueous acid.


At equilibrium (that is, if volume/temperature/mass is stable and not changing over time), the external pressure applied to the gas will equal the internal pressure of the gas.

This is (more or less) a consequence of Newton's third law. The external environment pushes on the gas with a certain force, and the gas pushes back with an equal and opposite force. At equilibrium, no work is being done on the gas or on the environment, so the forces have to balance. (This is very crude and hand-wavy - a more thorough description would require much more math.)

So (at equilibrium), there really isn't any difference between the two. In the Boyle's law case, the value is described as the external pressure, because the typical experimental setup in that case involves using something like a piston to control and vary the external pressure applied to the system. In the ideal gas case, you may not explicitly be controlling the external pressure, instead you're controlling temperature, volume, or particle numbers. In those case, you're normally measuring the internal pressure of the system, rather than (directly) controlling it or the external pressure. But (at equilibrium) the pressure of the gas pressing out is the same as the pressure of the container pushing in.

It's all just a matter of perspective, and how the system is set up.


Why does alcohol crack acrylic (plexiglass)?

If a solvent causes an amorphous polymer to craze, this is because the polymer is principally soluble in it. It diffuses into the surface, but without instantly dissolving it, as the polymer chains are still stuck in the glassy network of the substrate. The polymer swells (like a fruit gum in water), but the volume increase can only happen perpendicular to the surface.

The polymer chains thereby end up oriented towards the surface, weakening the structure. When the solvent dries off again, the chains adhere to each other again, but cannot reentangle in time. The surface has become thicker now, but as it is still the same amount of polymer, the additional volume is made up by the cracks and crazes appearing in it.


Is it 'Phenyl' or 'Phenol' that is used as a cleaning agent?

It's definitely not phenyl in there. This isn't stable and would immediately react to something else. Phenyle seems to be a name for a certain disinfecting agent out of pine oil.

It says "phenolic type" on the bottom, so my best guess, without seeing the actual ingredient list, is that it is a derivative of phenol. Derivative means that phenol is the base structure and it's somehow modified. Phenol itself is toxic but in the past it was used as a disinfectant, especially by doctors. It actually turns your skin white and that color goes away after some hours. Pine oil seems to have phenols in it, but I cannot find information on the structure of those.


How do i create a significantly correct world with large crystal formations in my novel?

Do you mind a scenario somewhat less dramatic than exploding suns? ;)

Just let selenite ($\ce{CaSO4*2H2O}$) cristallize very slowly und undisturbed in a system of caves for a long time.
Let the inhabitants of your world discover the cave system or let parts of the cave system collape to make it more accessible.

A model for this already exists: the Cave of the Crystals, connected to the Naica mine in Mexico.

Update

Note that I'm not a geoscientist, but to me, large hexagonal columns that can exist under earth's atmosphere cry basalt, which in turn indicates vulcanic activities.

The English wikipedia has a list columnar structures resulting from vulcanism. Some examples are

Giant's Causeway in Northern Ireland
Fingal's Cave in Scotland
Devil's Postpile in California


Relation between First Law of Thermodynamics and Ideal Gas Law

Why thermodynamical equations are just for gases?

They are not. The equation $\Delta U = q + P\Delta V$  applies to any phase (gas, liquid, solid...) when only pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work.

Gases are (1) an easy way to introduce thermodynamics concepts because they allow for a simplified analysis and may exhibit dramatic behavior, and (2) they provide a link to understanding the behavior of other phases. They also happen to be inherently important for practical and historical reasons during the development in the science.

  I want to know my mistakes in above consideration.

There are no mistakes. Consider the ideal gas law $$V=\frac{nRT}{p}$$ If you assume that $p$, $n$ and $T$ are constant, then the dependent variable $V$ will also be constant.

What's probably causing the confusion is that the water is undergoing a phase change. We assume that the liquid does no work, that the change in volume is only due to the formation of vapor. In practice $\pu{\Delta n=+ 1 mol}$ for the gas, $\pu{\Delta n= - 1 mol}$ for the liquid, and $\pu{\pu{\Delta n= 0}}$ for all of the water. We ignore the work done when reducing the amount of liquid because it is small compared to that done when the gas is formed (the change in volume of gas is much greater).

  How is Δn=1 ?

See the answer to the previous question. You are converting $\pu{18 g}$ of liquid water into vapor. Since the molecular weight of water is $\pu{18 g/mol}$, you are converting $\pu{1 mol}$ of water.

  Why temperature is taken as 373 K and not 298 K, since the process starts at 298 K we should use it?

I agree, if the work is actually performed at a lower temperature than $\pu{373 K}$, so this is an estimate. It is assumed that water is "boiling off" the skin at $\pu{373 K}$ and $\pu{1 atm}$ vapor pressure (the boiling point of water at $\pu{1 atm}$ of pressure is $\pu{373 K}$). Maybe not an accurate portrayal of what is going on, but it gets you to practice the theory.

  I have lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).

The problems more generally are that (1) we have an intuition about the way the world works, based on everyday observations, and this intuition sometimes misinforms us; some of the effort of education is to develop a more accurate intuition; and (2) when teaching science, practice problems are sometimes abstract and don't reflect real life situations except approximately. For instance, when you dry yourself after swimming, you are far from a thermodynamic equilibrium, with air currents, sun heating your skin, and a low water vapor pressure all playing a potential role. Modeling all this is beyond the scope of an introductory course. Probably the most important approximations here (assuming a closed system free of mass flow) are that the enthalpy of vaporization is constant over a broad temperature range and/or (as you rightly pointed out) that the water evaporates at $373 K$.


Volume of sucrose (solute) contributing to volume of solution? (solvent = water)

The English wikipedia page for sucrose notes a solubility of 2000 g/L at 25 degrees Celsius, corresponding to an approximately 5.8 M solution. 3 M shouldn't be a problem in that respect, but to make dissolving easier you could warm up your solution slightly if that is appropriate in your protocol.

Now, molar is a concentration equivalent to mole per liter and this volume is the total volume including solvent and the dissolved material. So 25 ml of a 3 M solution of anything, say x, should have a final volume of 25 ml, including the solvent and 75 millimole of x.

This means your second approach is correct (being 25.6725 g of sucrose, to which you add water until the total volume is 25 ml).


Reaction between RCH2=CH2 and CCl4 -- Why does it form RCHClCH2CH(R)CH2CCl3 along with RCHClCH2CCl3?

The propagation series starts with the addition of the trichloromethyl radical to the alkene to give the more stable secondary radical. This radical can abstract a chloride from carbon tetrachloride to give the expected product. Alternatively, this radical can add to the alkene to give an a new carbon-centered radical with an extended chain. Abstraction of chloride gives the 'unexpected' product.

The key realization here, as permeakra suggested in the comments, is that any radical in solution can theoretically add to the alkene. This means that the penultimate radical intermediate could add to the alkene as well extending the chain. This is the basis of a radical polymerization.


Is corovin Polypropylene or flat polyester?

To my understanding, Corovin is a technology using to manufacture Nonwoven fabric. These fabric could either be polypropylens or polyesters, yet technique is first develop for polypropylene (Ref.1). Following is the phrase I borrowed from the book (p. 134):

  Companies such as Du Pont, Freudenberg and Corovin began their research and development work on generating spunbonded webs. In Europe, Freudenberg, Terram, and Corovin started production of polypropylene (PP) spunbonded webs at the beginning of the 1970s. Within these technologies, filaments are spun directly out of the polymer melt through small die heads arranged side by side, stretched down to a diameter of $\pu{10-30 \mu m}$ and laid down on a porous lay down and transport belt.

A recent patent has been assigned to Fiberweb Corovin GmbH for manufacture of polyethylene-based, soft nonwoven fabrics as well (Ref.2). The abstract states:

  The present invention concerns a nonwoven fabric the fibers of which have polyethylene on at least part of their surface, where the fibers are thermally bonded and the non-woven fabric exhibits an abrasion of less than $\pu{0.5 mg/cm2}$, in particular of less than $\pu{0.4 mg/cm2}$, and an embossed area fraction of less than 35%, in particular of less than 28%. Furthermore a device is made available for the manufacture of a non-woven fabric, using a polyethylene-containing polymer, with a take-down system underneath a spinning plate that causes the take-down of the polyethylene, where the spinning plate has an L/D ratio of between 4 and 9.

References:

H.-G. Geus, "Chapter 5: Developments in manufacturing techniques for technical nonwovens," In Woodhead Publishing Series in Textiles, Number 181: Advances in Technical Nonwovens; George Kellie, Ed.; Woodhead Publishing: Amesterdam, The Netherlands, 2016, pp. 133-153 (ISBN: 978-0-08-100575-0).
Markus Haberer, Henning Röttger, "Polyethylene-based, soft nonwoven fabric," US Patent: 8420557, 2013 (Filed: 2006).


Colorful flame
To create a colorful flame you could just buy some packages of Colorful flame which is not toxic.
To buy it on amazon click here.

Gender part
For the gender part you could treat some sawdust with aqeous Copper(I)-chloride ($\ce{CuCl}$). Let that sawdust dry and throw it in the fire (after some time passes so they can be amused by the "Colorful fire"). It'll create a blue colour which would be great if its a boy.
However, if its a girl you could do the same thing, but instead of the Copper(I)-chloride you would use Strontium(II)-chloride ($\ce{SrCl_2}$) or Strontium(II)-nitrate ($\ce{Sr(NO_3)_2}$).
I don't know how long it'll  last but it depends on how much you put in the fire. (More stuff lasts longer)

Precautions

Be careful! You could be burned or something could catch on fire.

Definitely don't breathe in the fumes from the fire.

Don't use methanol because its a poison if ingested.

And the general advice: Wear safety gear!

Sources linked here and here.


Relationship between heat and standard enthalpy for a reversible reaction

In this description, the reactants in the initial state and the products in the final state are all pure.  Now, if you only let the system go partially to completion, and then separated all the remaining reactants and products into pure substances at the standard conditions (as part of your process), you could determine the heat of reaction from that also.


Let me reword the question:

Does $\Delta^9$-tetrahydrocannabinol (THC) significantly degrade when its solution in propylene glycol is heated to 115-123 °C?

This was somewhat answered in a study by Michael A. Repka et al. (DOI) on the stability of THC embedded in a  hydroxypropylcellulose polymer matrix. When heating this system to 120 °C for 75-90 min, about 2.7% of the initial THC degraded. With other words, 97.20% remained unchanged.

Heating to higher temperatures did not lead to significantly higher degradation, but among the degradation products, the amount of cannabinol (CBN), the oxidation product of THC, was much higher here.

Technically, heating to ~120 °C seems to be no problem.

You might want to check the reference given, have a look at the articles cited, compare toxicological data of THC and CBN and decide for yourself.


How to tell the difference between lysine and the N-terminal amino acid in DNP labeling?

Actually DNP-labeling of amino groups was first introduced in 1923 when 2,4-dinitrochlorobenzene (DNCB) was used for the identification of the terminal groups of a partial hydrolysate of silk fibroin (Ref.1). Then, Sanger has introduced 1-fluoro-2,4-dinitrobenzene (DNFB) in 1945, which bears his name, the Sanger’s reagent. DNFB was first used to detect free amino acids of insulin (Ref.2). In a nutshell, DNFB undergoes nucleophilic aromatic substitution with the $\ce{N}$-terminal amino group of a peptide or protein. DNP-group is stable to acid hydrolysis. After the DNP-labeled peptide or protein is hydrolyzed, the individual amino acids can be separated and only the DNP-labeled $\ce{N}$-terminal amino acid can be detected by a colorimetric detection at specific wavelength. Hence, DNFB can be used in protein sequencing to determine $\ce{N}$-terminal amino acids.
Not only the $\ce{N}$-terminal amino acid of the protein is labeled by Sanger’s reagent, but also the backbone amino acids with an active group to DNFB such as $\epsilon$-$\ce{NH2}$ in lysine, $\ce{SH}$ in cysteine, $\ce{OH}$ in tyrosine, and $\ce{NH}$ in histidine would be labeled as well. The labeled protein is then hydrolyzed (DNP derivatives are stable to acid hydrolysis), and extracted with ether. If lysine, cysteine, tyrosine, histidine are terminal amino acids, they should be di-DNP-labeled amino acids. If any of them are in the backbone (within $\alpha$-helix), they should have free amino group after hydrolysis. These mono-DNP-labelled amino acids with free amine group are soluble in acid phase and won't come to ether layer when extracted (Ref.2).
The concentrated ether layer is loaded on silica gel chromatography column and the elution profile is matched with the standard profile obtained from individual DNP derivative of all the amino acids. The reagents produce yellow colored derivatives which can be easily detected by absorbance (Ref.2). This method can be applied to any protein (best works with less than 50-70 amino acids). For larger proteins, before DNFB treatment, the protein should be hydrolyzed to small fragments by enzymes such as trypsin or pepsin, or by chemicals like cyanogen bromide.
Note: Extraction and separation methods are detaied in Ref.2, hence I attached the PDF file. For instance, in insulin (MW: ~$\pu{12000 Da}$), there are four terminals (two glycines and two phenylalanines), and two $\epsilon$-DNP-lysines.
References:

E. Abderhalden, W. Stix, Hoppe-Seyl. Z. 1923, 129, 143 (Reference 2 of Ref.2).
F. Sanger, "The free amino groups of insulin," The Biochemical journal 1945, 39(5), 507–515 (doi:10.1042/j0390507)(PDF).


The clue is the moles in the dilute solution is the same as the moles of the concentrated stock solution - it has to be constant.

The method I have used is (therefore) correct.


What's the mechanism of the reaction of (E)-methyl 3-(furan-2-yl)acrylate in acidic methanol?

Unfortunately I cannot agree with the currently accepted answer. This 1,5-hydride shift that was invoked seems to have a very strained transition state (the cyclic, planar system means that the "1" and "5" atoms are quite far apart) and with all due respect, I am quite skeptical as to whether it would really work. Furthermore furan does not enjoy a particularly large aromatic stabilisation and "rearomatisation" as a driving force is not too convincing. Furan, in all its reactions, has a very pronounced tendency to lose its aromaticity. For example even nitration with the mild reagent acetyl nitrate ($\ce{AcONO2}$) leads to a 2,5-addition product and you actually need to add a second reagent (pyridine) to eliminate $\ce{AcOH}$ and rearomatise. For all intents and purposes furan reactivity is much closer to that of an enol rather than that of benzene.

Instead of a 1,5-hydride shift, proton transfers should suffice, especially since there are carbonyl groups all over the molecule (i.e. the proton transfer is simply a tautomerism and doesn't go through any particularly unstable intermediates).

One way of thinking about this question would be to try and do it in reverse. 1,4-Dicarbonyl compounds are known to cyclise to form furans in acidic solution, so this should simply be the microscopic reverse of that reaction. The key cyclisation step involves the attack of an enol oxygen onto a protonated carbonyl group. So, in this reaction, the fragmentation step should be the reverse: the ejection of an enol oxygen from a (hemi)acetal.

I would propose this mechanism instead.

So far, this is the same as in the currently accepted answer. However the usage of the furan lone pair to push electrons saves time in drawing all the resonance forms (this is what was meant by K_P's comment also).

Protonation of the furan oxygen (not drawn) brings us to the start of the next step:

As described previously this fragmentation is the reverse of the cyclisation reaction used to generate furans. Following tautomerisation of two of the enols, most of the molecule is already in place and the remaining question is how to generate the second ester functionality. The issue is that the carbon that should become the ester has an extra hydrogen on it. Instead of having a hydride shift it makes much more sense to move it around by tautomerisation!

The ketene acetal can be protonated and attacked by methanol at the carbonyl carbon, but that would be unproductive as there is no way of getting rid of one of the methyl groups on oxygen (unless we invoke a $\mathrm{S_N2}$ reaction on the methyl group with methanol acting as the nucleophile). So, it probably needs to wait until water is added during the reaction workup, at which point it is hydrolysed to give the desired product.


Difference between backbonding and hyperconjugation and conjugation

OK, let's step through these terms and let's add "cross-conjugation" and "homoconjugation" for good measure to round out the series.
Backbonding:
This term is most often used to describe a type of bonding that occurs in inorganic compounds.  Metals bonding with carbon monoxide provide good examples of backbonding. The carbon atom in carbon monoxide has a lone pair of electrons that can be used to form a sigma bond with a metal.  Because carbon monoxide has low-lying $\pi^{\ast}$ orbitals, it can accept electrons back from the metal and further strengthen the bonding between the metal and the carbon monoxide ligand.  This process of "accepting electrons back from the metal" is termed backbonding. Here is a good picture of the $\ce{\sigma}$ and $\mathrm{d}-\pi^{\ast}$ (backbonding) bonding interactions.

It happens most often when 1) the metal is electron rich, 2) the ligand has low-lying $\pi^{\ast}$ orbitals and 3) the metal has available a $\mathrm{d}$ orbital with the proper symmetry to overlap constructively with the carbon monoxide $\pi^{\ast}$ orbital.  Here is a link to a great, one page article on backbonding if you'd like to read more.
Hyperconjugation:
This term is most commonly used in discussions about resonance.  Hyperconjugation is a stabilizing interaction that results from the interaction of the electrons in a $\sigma$-bond (usually $\ce{C-H}$ or $\ce{C-C}$) with an adjacent (empty or filled) $\mathrm{p}$-orbital. This interaction creates an extended molecular orbital that increases the stability of the system.  Most typically it involves the breaking of a carbon-hydrogen bond and then using what were the $\ce{C-H}$ bonding electrons to stabilize an adjacent charge. In resonance structures, we can't move nuclei, but we can move electrons. For example, below is a diagram showing resonance structures that involve hyperconjugation that serve to stabilize the n-propyl carbocation. Hyperconjugation has been used as part of the explanation for the carbocation order of stability being tertiary > secondary > primary.

Recently it has been suggested that hyperconjugation, rather than steric effects, is what stabilizes the staggered conformation of ethane (very interesting!).    Here is a link to a fairly extensive, but very well written article on hyperconjugation, it covers the effects of hyperconjugation on resonance structures and conformational stability.
Conjugation:
This term is used to describe a system of continuous, overlapping p-orbitals where electron delocalization can occur.  Examples of conjugated systems include the 1,3-butadiene, acrolein, benzene, beta-carotene, etc.

Cross-conjugation:
Sometimes you'll hear the term "cross-conjugation" used.  In this case, the continuity of the overlap between the p-orbitals does not follow a simple linear pattern as seen in the examples of conjugation pictured above.  Benzophenone and 3-methylene-1,4-pentadiene are examples of molecules with cross-conjugation.

Homoconjugation:
Sometimes $\mathrm{p}$-orbitals that are not directly conjugated (e.g., that are not directly connected, there is a "spacer" in between them) can still interact or overlap in a "through space" fashion.  The solvolysis of the exo- and endo-norbornenyl chlorides shown below provides a striking example. The exo-isomer generates the corresponding carbocation thousands of times faster than its endo-isomer. This difference is attributed to homoconjugation. The $\mathrm{p}$-orbitals on the remote norbornenyl double bond can interact through-space with the developing $\mathrm{p}$-orbital that is being formed on the carbon where the exo-chloride is leaving from, dramatically stabilizing the formation of the resultant carbocation. This interaction is geometrically impossible for the endo-isomer.  Again note, there is no direct conjugation between the remote double bond and the carbon atom where the carbocation will form, it is entirely a through space interaction that generates a pseudo-allyl system.


More ions form pairs when concentration of acid increases, does that decrease the speed of reaction?

In general, the conductivity of any liquid increases with the concentrations of its ions, and decreases with their dimensions. In pure sulfuric acid, the only ions existing are $\ce{H3SO4+}$ and $\ce{HSO4-}$. They are big and their concentration is low. So the conductivity of pure sulfuric acid is low. But when you add some water, the $\ce{H2O}$ molecules are totally transformed into $\ce{H3O+}$ and $\ce{HSO4-}$ ions which are numerous and sometimes small (for $\ce{H3O+}$). The conductivity must increase.


Which of the following are true based on this comparison?

A is not true - no such connection exists.
B is true (and given the same temperature indicating the phase transition it is in fact likely that K and L are the same matter).
C is true and can be explained as follows: if K and L have the same mass, they must have different specific heat capacities, defined as:

$$
c = \frac{\text{the amount of energy to be added to raise the temperature by 1°C}}{\text{mass}}
$$

This is a property that is unique for each matter - same as the melting and boiling point. It can be said to be a function of the matter. Note that the $T(t)$ graphs appear to start at the same temperature, but it takes longer to heat L than K to 85°C although the same heating is applied.


Synthesis of Benzaldehyde and Aniline Schiff Bases

The reaction needs acid catalysis but ethanol is fine as solvent. There will always be some starting materials left, but since the product is a solid and the reagents are liquid, the best (and cheapest) is to purify it through recrystallization.

For this, dissolve the product in the smallest quantity of hot ethanol and then add water until precipitation occurs.


Why does decreasing the distance between the electrodes in an electrolytic cell increase the rate of electroplating?

It's a bit of an oversimplification to explain this with a simple invocation of Ohm's law. The thing to remember about electrochemistry is that electron transfer only happens at every close distances to the surface (<10 nm) and it's the potential difference between the electrode and this tiny region at the surface that is important. You might be applying 10 V between the two electrodes, but if you have a lot of solution resistance between them, the voltage near the surface may only be 1 V as the other 9 V are dissipated as heat in the electrolyte. Thus, by moving the electrodes closer, the resistance goes down and the voltage the electrode "sees" is higher.

The trouble with this type of two-electrode cell is that many factors are being conflated. The current that is measured (the rate of electroplating) is dependant on what is happening at both electrodes as well as the resistance of the bulk solution. As the electrode voltage increases, the current usually increases too as increasing the voltage makes it easier for electrons to transfer between the electrode and the redox-active species in solution. Eventually, however, the rate will not be limited by electron transfer, but rather by how fast copper ions can move from the bulk of the solution to the active region near the electrode and a "limiting current" is reached. You can see this below as the voltage at an electrode is swept to higher values:
 (Source: Own work)

At low voltage, there isn't enough voltage for any electron transfer to occur, but as it increases along the x-axis, we see an increase in current until it reaches a peak and the current is limited by diffusion to the electrode. (the reason it goes down after this point rather than levelling out has to do with the time-scale of the experiment, but it's not important for this)

Because the system is so complicated, when we're interested in examining the reactions that are happening at just one electrode without worrying about the other electrode or the solution resistance, an alternate setup is used. If we use a feedback amplifier (known as a potentiostat) to control the voltage applied between the two active electrodes using a third (reference) electrode to measure the voltage close to the surface of the electrode of interest, we can decouple the effects of the solution resistance and other electrode.

A sidenote:

  My initial thought was that less distance to travel=ions arriving at cathode quicker=greater current. That doesn't seem right though, because different ions are depositing and liberating electrons and hence not really flowing through (right?).

You were actually onto something interesting here. You're right in that it takes a while for copper ions from one electrode to make their way to the other so most of the copper being deposited is coming from what was already in the solution (at least on relatively short time scales), but what would happen if we moved the electrodes very close to one another? Eventually we would reach a situation where copper could dissolve from one electrode and be almost immediately deposited on the other and we would not see the limiting current behaviour as above.


Could you suggest me a good coupling method for amide synthesis between tetrahydroisoquinoline and a carboxylic acid?

A SciFinder search gave 200 or so examples of coupling tetrahydroisoquinoline with an unsaturated carboxylic acid (including benzoic acids and the like). Surprisingly dicyclohexylcarbodiimide (DCC) was not one of the reagents used, although N-(3-dimethylaminopropyl)-N′-ethylcarbodiimide (EDC) was used in more than 50 cases. Other common amide coupling agents including PyBOP, BOP, HBTU, and HATU were reported. Another related strategy through a mixed carbonate/anhydride by activating the acid with isobutylchloroformate.

I'm not sure if you have to isolate the free base of your amine first, but you should certainly include a base to liberate the active nucleophile. If you haven't tried it yet, add an equivalent of triethylamine or diisopropylethylamine.

Aapptec, a supplier of reagents and building blocks for peptide synthesis, has a nice compilation of the different amide coupling reagents that are used:
Aapptec coupling reagents.


How do I balance the combustion reaction of butene?

The balanced equation for the combustion of butene is
$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}.$$
Nitrogen is included only to indicate that it is present in the medium with some specific ratio to the oxygen amount – well technically it is included to make question look complicated.

$\ce{C4H8}$ in the products side is included to indicate that not all of it can be combusted but some escapes. And it really is arbitrary, depends on the the engineering of combustion room, etc. – which is why you have 4 equations, 5 unknowns: d is free to vary.  But you are given the dry analysis to fix its value.

Looking at the equations given, you have

\begin{align}
  b &= c = 4(1-d)\\
  a &= e = b + c/2 = 6(1-d) = 3b/2
\end{align}

And dry analysis says ratio of $\ce{CO2}$:$\ce{C4H8}$ is $14.95:0.75 = 19.93 (b/d)$ after the combustion. And this is the last equation you need.


In Chapter P-4 "Rules for Name Construction", section P-44 "Seniority Order for Parent Structures", of Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, it is stated that (p 490):

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

The Blue Book lists several helpful examples, one of which is that the principal chain "octane" is senior to "hept-1-ene". Therefore, the following compound

is named 3-methylideneoctane instead of 2-ethylhept-1-ene.

The authors mention that this preference is in contrast to previous recommendations (e.g. 1979):

  In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.


Overall effective activation energy for parallel reactions

The activation energy $E_\mathrm a$ should, in general, be viewed as an empirical quantity that characterises the sensitivity of the rate to temperature. From this perspective, it may be obtained via the equation

$$E_\mathrm a = RT^2 \left(\frac{\mathrm d \ln k}{\mathrm dT}\right) \tag{1}$$

This relation can be "derived" from the Arrhenius equation. (Quotation marks because the Arrhenius equation itself is only an empirical relation.)

$$\begin{align}
k &= A\exp\left(-\frac{E_\mathrm a}{RT}\right) \\
\ln k &= \ln A - \frac{E_\mathrm a}{RT} \\
\frac{\mathrm d \ln k}{\mathrm dT} &= \frac{E_\mathrm a}{RT^2}
\end{align}$$

If a reaction rate has Arrhenius-type behaviour, i.e. $k = A\exp(-E_\mathrm a/RT)$, then equation $(1)$ simply returns the usual interpretation of the activation energy. However, equation $(1)$ is useful in that it allows us to define an "activation energy" in the case where the rate constant does not have the mathematical form $A\exp(-E/RT)$. The parallel reactions that you have described fall exactly into this category. The sum of two exponential functions generally does not return an exponential function, so the overall rate constant $k = k_1 + k_2$ should not be expected to conform to an Arrhenius description.

If we substitute

$$k = k_1 + k_2 = A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)$$

into equation $(1)$, then we get:

$$\begin{align}
E_\mathrm a &= RT^2 \cdot \frac{\mathrm d}{\mathrm dT}\left\{ \ln\left[A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)\right] \right\} \\[8pt]
&= RT^2 \cdot \frac{A_1(E_1/RT^2)\exp(-E_1/RT) + A_2(E_2/RT^2)\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&= \frac{A_1E_1\exp(-E_1/RT) + A_2E_2\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&= \frac{E_1k_1 + E_2k_2}{k_1 + k_2}
\end{align}$$

as desired.


Are the electrons also distributed for the alpha decay?

The equation for the $\alpha$-decay omits the charges.

Since the $\alpha$-particle actually is $\ce{^4_2He^{2+}}$, the resulting nucleus is initially left with a surplus of two electrons. These are emitted too, probably on the same timescale.

So, the answer is yes: the $\alpha$-decay of a neutral nucleus yields another neutral nucleus, a double-positively charged $\alpha$-particle and two electrons.


How to synthesize 1,3-pentadiene from butanal?

I believe one way would be to start with butanal participating in a Wittig reaction to form 1-pentene.

$\ce{CH3CH2CH2CHO + H2C=PPh3 -> CH3CH2CH2CH=CH2}$

Then allylic bromination, as you said.

$\ce{CH3CH2CH2CH=CH2 ->[{NBS}][{hv}] CH3CH2CHBrCH=CH2}$

Dehydrohalogenation with a hindered base like t-butoxide (to favor the elimination reaction) should give the diene. The only thing I'm not sure about is whether it would be E1 or E2. It isn't very hindered, but it is a secondary halide. With those factors, plus the alkoxide, I think it will be an E2 mechanism.

Let me know if I made a mistake somewhere.


Need something that is dishwasher safe that breaks down paraffin

This may be doable, per a source, to quote:

Paraffin wax is mostly found as a white, odorless, tasteless, waxy solid, with a typical melting point between about 46 and 68 °C (115 and 154 °F).

So as a pre-treatment, try applying a jet of steam to facilitate the removal of the paraffin wax.


What's the meaning of wavefunction being finite?

A broader point regarding functions that are finite:
A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $\lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number.
This is a generalisation of the comments written above, not specifically quantum related, but a more general point on mathematical functions.


The following link from a very old text (pages 279-280) suggests boiling the wax in an alkaline solution. This method may be specific to that wax described as a new vegatable wax from a laurel like shrub from the forests of Para and Bahia. Please explain the application. Why do you want the wax to be alkaline? What are you going to do with the wax?


Is there any substance that's a 4-4-4 on the NFPA diamond?

Answering my own question based on the comments, tert-butyl-hydroperoxide is at least one such chemical. As stated on this MSDS from a government website, it's a 4-4-4, with additional special warning of being a strong oxidizer. The only thing that it does not do that could make the 704 diamond any worse is react strongly with water. It is in fact water soluble, though marginally, preferring to float on top (and therefore traditional water-based fire suppression is ineffective, but foam/CO2 will work).

If anyone else can find a chemical that, in a form that is used in the lab or industrially, is a 4-4-4 that is a strong oxidizer and reacts strongly with water, that's pretty much "as bad as it gets" and they'll get the check.


Does rate of Evaporation change with surface area?

Why does vapour pressure decreases with decrease in surface area.

Well, it does not. Vapour pressure is a function of temperature only. It is in fact the equilibrium constant for the reaction $\ce{H2O_{(l)}->H2O_{(g)}}$, and like all other equilibrium constants, is independent of surface area, amount of the species but depends only on the temperature. At a given T, only for one specific value of the partial pressure is water and its vapour in equilibrium.

That said, water is not actually in equilibrium when in an open vessel evaporating into the atmosphere. The vapour pressure above the water filled vessel is far less than what is required for equilibrium. Water constantly evaporates as a desperate attempt to increase the vapour pressure to its equilibrium value, but since the vessel is open and the vapours not confined, the equilibrium value is seldom reached. If the surface area is larger, it means that the liquid phase can evaporate a larger amount of water while attempting to attain equilibrium. Therefore, larger surface area increases the amount of water that is lost as vapour while attaining equilibrium, since it is that part of the liquid exposed to the vapour phase which tries to maintain equilibrium by evaporating or condensing. Evaporation is essentially a surface phenomenon, as opposed to boiling which becomes considerably a bulk phenomenon.

Since it is practically the surface layer which participates actively in the phase equilibrium, larger the surface, greater the attempt to attain equilibrium and quicker is the evaporation.


Why has the mole been redefined for 2019?

You have already mentioned the last definition of the mole before the proposed change:

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is “mol”.
(…)

According to this definition, the mole still depends on the definition of the kilogram. The main purpose of the proposed change is to find an independent definition. Another purpose is the possibility of linking the definition to an exact value (similar to the speed of light.)
The formal decision for the change was made during the 26th meeting (13–16 November 2018) of the General Conference on Weights and Measures (CGPM).
The motivation behind the change of the definition of the mole can, for example, be found in the resolutions of the 24th meeting (17–21 October 2011) of the General Conference on Weights and Measures (CGPM); however, the discussions and preparations started already much earlier.

On the possible future revision of the International System of Units, the SI
(…)
considering
(…)
that of the seven base units of the SI, only the kilogram is still defined in terms of a material artefact, namely, the international prototype of the kilogram (1st meeting of the CGPM, 1889, 3rd meeting of the CGPM, 1901), and that the definitions of the ampere, mole and candela depend on the kilogram,
(…)
that it is also possible to redefine the mole so that it is linked to an exact numerical value of the Avogadro constant $N_\mathrm A$, and is thus no longer dependent on the definition of the kilogram even when the kilogram is defined so that it is linked to an exact numerical value of $h$, thereby emphasizing the distinction between amount of substance and mass,
(…)


Difference between calcination and pyrolysis

Essentially calcination is decomposition at high temperature with inorganic reactants such as calcium carbonate in making cement.
$$\ce{CaCO3 ->[\Delta ] CaO + CO2}$$
Pyrolysis is decomposition or oxidation at high temperatures of organic materials such as cellulose in making charcoal.
$$\ce{(C6H12O6)_{n} ->[\Delta] C7H4O + C_{x}H_{y} + H2O}$$
where $\ce{ C_{x}H_{y}}$ is volatilized byproducts.


How do electrolytes allow electrolysis to occur?

In the electrolysis of $\ce{H2O}$, a small amount of sulphuric acid is added. Why? Because pure $\ce{H2O}$ doesn't ionize much on its own ($K_w=10^{-14}$ approximately). So traces of an electrolyte ($\ce{H2SO4}$ in this case) is added so that it can help break one $\ce{O-H}$ bond and $\ce{H}^{+}$ released forms a coordinate bond with one $\ce{H2O}$ to form $\ce{H3O}^{+}$ ion. So now the ions present in the solution are: $\ce{H3O}^{+}$, $\ce{OH}^{-}$ and $\ce{SO4}^{2-}$. Of these, the oxidation potential of $\ce{OH}^{-}$ is higher than that of $\ce{SO4}^{2-}$ so at the anode $\ce{OH}^{-}$ gets oxidized to $\ce{O2}$. In this case only one cation goes to the cathode so it gets reduced to $\ce{H2}$. In case you use an electrolyte that gives a cation $\ce{M}^{n+}$ other than$\ce{H}^{+}$ then $\ce{H2}$ would be formed only if the reduction potential of $\ce{M}^{n+}$ is lower than that of $\ce{H3O}^{+}$. So, if the electrolyte used is $\ce{ZnSO4}$ then $\ce{H2}$ will be formed but if $\ce{CuSO4}$ is used then $\ce{H2}$ is not formed, but $\ce{Cu}$ is deposited at the cathode.

So the ionization of water doesn't have anything to do with the electrons used in the redox process. The ionization depends simply on the electrolyte used.


May I introduce you to the Barton-McCombie reaction for the reduction of alcohols to alkanes.

image and further information here


I need Iron powder. What's the best way to produce it?

Sigma-Aldrich has a product that is very pure but rather expensive. I suspect you can have high purity or low cost but not both.


The difference does not lie in stereochemistry. Rather, N$^\alpha$-acetylornithine and N$^\delta$-acetylornithine are regioisomers of one another.

The superscript Greek letter denotes which nitrogen is substituted with an acetyl group. The lettering is relative to the carboxyl group as shown in the image below:

Note that:

The carboxyl group’s carbon does not get the $\alpha$, because it is considered part as the functional group from with the labelling begins
The $\alpha$ carbon is asymmetric. Both L- and D-compounds (or (S) and (R) according to CIP nomenclature) are possible for both N$^\alpha$- and N$^\delta$-acetylornithine.


Calculating the number of neutrons in a sample

The formula

$$\text{amount of substance} = \frac{\text{mass}}{\text{molar mass}}$$

tells you that you have 1/12 moles of magnesium, since you are taking the mass of magnesium (2 g) and dividing by the molecular mass of magnesium (24 g/mol). However, the question is not asking about the number of moles of magnesium.

Each magnesium atom has 24 - 12 = 12 neutrons. Consequently, 1/12 moles of magnesium has (1/12)(12) = 1 mole of neutrons.


Calculating the Ionic strength of sodium acetylsalicylate

Aspirin has a solubility of ~3 g/L. In addition from a patent by Galat [1]:

  Sodium aspirin is almost 1000 times as soluble in water as aspirin itself.

All of the $\ce{NaOH}$  can be assumed to react with acetylsalicylic acid (aspirin) to form the sodium salt.  This means the amount of solubilized sodium aspirin is roughly equal to the amount of added $\ce{NaOH}$ (up to the equivalence point).

When computing the ionic strength you can therefore assume that the concentration of solubilized sodium aspirin is roughly equal to the amount of added $\ce{NaOH}$.

References

Galat, A. Stable Sodium Acetylsalicylate and Method for Its Manufacture. US3985792A, October 12, 1976. (PDF)


This is a Peterson olefination, specifically this one is from the synthesis of Roseophilin published in J. Am. Chem. Soc. (2001) 123 8509. The LDA deprotonates next to the silicon as both it and the adjacent imine stabilise the anion.

The Wikipedia page gives a good summary of the mechanism and why an acidic work up is used in this case to give the required double bond geometry (Wikipedia).

(Image from www.chem.wisc.edu)


You might have success with amorphous carbon sheets[1]. It's not cheap, but it's effective. It's the material used in the windows of the AMPIX cell[2] developed at Argonne. It does react at low potentials, though, so might not work for anode studies.

Have you considered aluminum? It's crystalline, but cubic so there aren't very many peaks to worry about.

This paper[3] might be helpful as you work on your experiments. You're right to think that not having a current collector for part of the electrode would affect the degree of lithiation; it's actually worse than that. Those kapton window cells are convenient, but always put an asterisk next to your results.

I'm doing a fair amount of in-situ characterization for my PhD. If you can share more about you goals, I'd be happy to chime in.

[1] http://www.htw-germany.com/products.php5?lang=en&nav0=3&nav1=6

[2] http://scripts.iucr.org/cgi-bin/paper?S0021889812042720

[3] https://pubs.acs.org/doi/abs/10.1021/acs.jpclett.5b00891


Does an acetal need to have one hydrogen attached to its carbon?

According to the IUPAC definition, an acetal is any compound with a formula $\ce{R2C(OR')2}$ (R' ≠ H). In the past, "acetal" referred only to compounds of this type, which were derived from aldehydes (one R = H), however, it now broadly encompasses all compounds of this type, including those derived from ketones, which were formerly only referred to as ketals. Interestingly, there was a period of time in which use of the term "ketal" was discouraged (IUPAC says 'abandoned', but many chemists kept using it anyway), but it has now been re-embraced as a term, for the subset of acetals that are derived from ketones.

One last note - the molecule shown in this question is a "spiroacetal" - a type of acetal where the acetal carbon is the only atom joining two rings.


Hoffman's Isocyanide test: How does addition of concentrated HCl degrade the isocyanide formed?

Isocyanide is not as toxic as its isomer cyanide and can be tolerated. The purpose of its removal is due to its disagreeable odor which is not at all tolerable.

  While some isocyanides (e.g., cyclohexyl isocyanide) are toxic, others
  "exhibit no appreciable toxicity for mammals". Referring to ethyl
  isocyanide, toxicological studies in the 1960s at Bayer showed that
  "oral and subcutaneous doses of 500-5000 mg/kg can be tolerated by
  mice".(Wikipedia)

Isocyanide is removed on the basis of its property of acidic hydrolysis.

  Isocyanides are stable to strong base (they are often made under
  strongly basic conditions), but they are sensitive to acid. In the
  presence of aqueous acid, isocyanides hydrolyse to the corresponding
  formamides:

  $$\ce{RNC + H2O → RN(H)C(O)H}$$

  This reaction is used to destroy odorous isocyanide mixtures. Some
  isocyanides can polymerize in the presence of Lewis and Bronsted
  acids.

But the story does not here. The formamides are still pretty unstable and further degrades to form amines and formic acid in the presence of acidic medium.(Here)

  Isocyanides are hydrolyzed by dilute mineral acids to give primary
  amine and formic acid (Isocyanides are not hydrolyzed by bases).
  Isocyanides are hydrolysed only by acids and not alkalies because the
  negative charge present on carbon atom in isocyanides initially
  attracts electrophiles ($\ce{H+}$) but repels nucleophiles ($\ce{OH–}$). When a
  proton gets attracted to negatively charged carbon atom, the tendency
  of this carbon atom to attract a nucleophile increases due to presence
  of positive charge on the N atom and hydrolysis is facilitated as
  shown below:

  Mechanism:

Here is a research paper for a reference.


Do all conversion factors have an infinite number of significant figures?

Your intuition is correct. There are two types of conversion factor.
One kind of conversion factor is based on definitions. These conversions are exact (infinite significant figures).
In addition to your examples of the specific heat of water (in calories), the kilogram/gram conversion, and the quart/gallon conversion, other conversions are exact, including:

The speed of light - $c=299,792,458\ \mathrm{m/s}$ the length of the meter is then derived from this constant.
The price of gasoline/petrol.
The inch is defined as exactly $25.4\ \mathrm{mm}$, and thus the conversions from all imperial to metric lengths (including your  are exact as well.
The specific heat of water in joules - the relationship between the calorie and the joule is defined exactly as $\mathrm{1\ cal = 4.184\ J}$

Note that most conversions of this type are unit conversions. Many units have been defined exactly in terms of SI-units.
Other conversion factors are based on measurements. One common example is atomic mass. Atomic masses are valuable conversion factors that are based on measurements. However, some unit conversions live in this category.
The electronvolt is based on the measurement of the elementary charge of the electron (which is not exact). Thus the electronvolt is $\mathrm{1\ eV=1.6021766208(98)\times 10^{−19}\ J}$.


Why is pH used as a measure of acidity if not only hydronium ions can act as a acid when in solution?

HCl being a strong acid, it is quantitatively dissociated in water, meaning that it fully dissociates into hydronium and chlorides. The hydronium are therefore the main contributors to the acidity of the solution. In a solution of ethanoic acid, the majority of acid molecules are undissociated. This is what "weak acid" means.

The amount of base necessary to neutralise a given acidic substance can be used as a scale of acid "strength" in chemical risk assessment.


Three parts to this answer:

You have two electro-philes, the tertiary carbo-cation formed by protonating the alcohol and breaking off a water molecule and then the bromine.  Looks like you got this part.

Assume the two electro-philes attack consecutively.  The methoxy group is more strongly activating and ortho/para directing than the alk-yl group for the second attack.  So the bromine and the newly formed carbon-carbon bond will be meta to each other.

The sterically favored positions for each are as shown in the given answer.  The carbo-cation favors the position para to the methoxy group.  The bromine picks the position ortho to the methoxy group that's away from the alk-yl function.

Please excuse my use of hyphens.  Device auto-correct was putting in wrong words and I had to defeat it.


Proton transfer equilibrium in bisulphite adducts

I think your confusion arose because of an error in the translation of the text you linked (the discussion on p. 359) and that this text was unclear about which equilibrium it was referring to. In that figure, the equilibrium that the authors are presumably referring to is the (planar) carbonyl- (tetrahedral) hemiacetal equilibrium (as described on page 358). So the equilibrium that the text refers to isn't the "proton transfer equilibrium," but the planar-tetrahedral equilibrium.

In contrast, the scheme you posted with your question assumed that the formation of a hemiacetal (or hemiketal) is irreversible, as indicated by a unidirectional arrow. This reaction should be treated with an bidirectional equilibrium arrow as shown in p. 359 of the document that you linked.

In this context, the steric argument is reasonable and the reaction as a whole analogous to the more well-known relative hydration equilibria of aldehydes and ketones (which is related the general notion of aldehydes being more electrophilic than ketones).


Nucleophilic acetyl substitution is common among carboxylic acid, acid chlorides, anhydrides, esters, and amides.

Two very common examples are anhydride decomposition and esterification of carboxylic acids. For example, if you look at esterification, carboxylic acid and alcohol first make an addition, giving tetrahedral intermediates 1 and 2. Intermediate 2 then collapses and displaces the water molecule as a leaving group to give the corresponding ester (see the scheme).

Your substrate is an amide. Like all other carboxylic acid derivatives, this molecule also undergoes a nucleophilic acetyl substitution with Grignard reagent, which is a strong nucleophile. The polarized carbonyl group has partially positive carbon, which is a good electrophile.

The nucleophilic carbon on Grignard reagent attaches to carbonyl carbon and gives tetrahedral intermediate as a metal alkoxide complex. But, this alkoxide can collapse to reform strong C=O bond, displacing the amino portion of the amide as a leaving group, in the form of the amide ion, $\ce{R1R2N-}$. This produces the ketone as a final product, assuming you have used only one equivalent of Grignard reagent (Channel reactions of Grignard reagent with esters).

  Why does the reaction continue after the tetrahedral carbon forms?

After nucleophilic attack on a carbonyl, the resulting tetrahedral species either gets protonated to form the final addition product, or it is an intermediate on the way to the substitution product. This depends on the type of reaction. For example, a geminal diol will usually dehydrate to form a ketone. There is a nice overview of the different scenarios, including some nitrogen chemistry, in this document.


Alphabetic order dictates which order the substituents are written in, but not the numbering of their positions.  Numbering is determined by what minimizes the numbers.  E.g. 3-ethyl-2-methylhexane is correct, but 4-ethyl-5-methylhexane is not.


First, human sweat contains urea, so it doesn't require incontinent commuters. Despite constant cleaning, this effluvia accumulates in the rail cars.
Second, air conditioners precipitate water, entrained dust and volatile organic compounds, making an ideal environment for bacteria, fungi and protists to swim around, excreting their wastes. Microbial volatile organic compounds are none too pleasant, and may have the smell of urine or ammonia (though the waste products of yeast in baking bread can be quite appealing).
Third, there are probably birds in and around the train stations. Birds produce uric acid, which is rapidly converted to ammonia by fecal flora.
Fourth, in older drainage systems, rain water and waste water may be mixed. Given a high water level, wastes may be closer to the surface... or overflow.


Is it okay to heat a test tube for several minutes for evaporation of solvent inside?

There are two issues here:

Heat from a Bunsen burner is pretty strong. You better make sure whatever is left after the evaporation will survive.
If the solvent is organic (and not super halogenated), this is a terrible idea because you run a serious risk of igniting your solvent vapors.


How are van der Waals parameters determined experimentally?

At the critical point, the first and second derivatives of pressure with respect to volume are zero.

This provides two equations to solve for the two unknowns (a and b) in terms of the critical temperature and critical volume (or pressure).

See Critical Constants of the van der Waals Gas


The picture you showed does have an unequal number of sodium cations and chloride anions.  However, the picture shows only part of a crystal.  Every atom that is on a boundary of the shown cube, whether on a face, edge, or vertex of the cube, is shared with other "cubes" in the crystal that aren't shown in the picture.

Each of the 8 corner Cl atoms in your picture is shared with 8 cubes (7 not shown). The 6 face-centered Cl atoms are shared with 2 cubes. Each of the 12 edge Na atoms is shared with 4 cubes (3 not shown). The center sodium atom is not shared. Thus there are 8/8 + 6/2 = 4 Cl atoms per unit "cube" in your picture, and 12/4 + 1/1 = 4 Na atoms per unit "cube" in your picture. 4=4, so the charge balances.

You might be thinking that this math checks out only insofar as a crystal is actually infinitely sized.  And you may have noticed that no salt crystals are infinitely big in the real world.  These things are both true.  But even tiny specks of salt crystals are gigantic relative to atoms.  The surface of a salt crystal could involve imperfections that mean that the count of sodium atoms and chloride atoms aren't precisely equal.  But instead of 14 vs. 13, the difference is more like 100,000,000,000,000,000 vs 99,999,999,999,999,999.  And since the imperfections are at the surface, on the outside of the crystal, any charge imbalance can be corrected if an opposite-charged particle from outside the crystals floats by and neutralizes the extra charge from the extra atom.


Historically, the relative masses of the isotopes were measured relative to O=16.

There is a good historical explanation in A New Mass-Spectrograph and the Whole Number Rule (1927).

At this time (1927), though it was understood that many elements had more than one isotope, only one isotope of oxygen was known.


How can I search molecules by constituent atoms

You can use ChemSpider's full search:

http://www.chemspider.com/FullSearch.aspx

and click on the 'Elements' section. Alternatively, there is a direct link here:

http://www.chemspider.com/ElementsSearch.aspx

This gives you two periodic tables that you can select elements from for your search. I think the upper one is 'allowed' and the lower one is 'forbidden', but I haven't tested this.

Of course, as pointed out in the comments, the set {C, H} will give you most of the database! However, for rarer elements this can be a better filter.


There are a few reasons, but the most direct answer is that the wavelength of X-ray photons is on the order of the distance between atomic nuclei in solids, e.g. ~ 4 ångströms (bonds are roughly 1.5-2.5 Å). You can think of it like the waves fit nice and snugly between the atoms and "fill" the crystal and thereby give us information about where the "cavities" are, and thus by reciprocal analysis (inverse Fourier transform) we can determine where the "rigid" stuff is (the atoms).

Other kinds of incident radiation are used e.g. neutron beams (for crystallography) and even electron beams (for surface spectroscopy), but X-rays are a more readily available source of radiation. In other words, it's easier/cheaper to produce X-rays than a beam of pure neutrons.

Gamma rays are about 100 times smaller in wavelength than X-rays (so instead of 4 Å, something like 0.04 Å, and so they generally penetrate through any solid material and don't produce a clear diffraction pattern like X-rays. Also if I'm not mistaken, it is much more difficult to produce gamma rays.

It's important to realize that diffraction is not the same as reflection and refraction. The latter two are surface phenomena - diffraction is a bulk phenomena completely unrelated to surfaces.

Electrons interact strongly with the surface of solids not only because of their charge but their energetic wavelength - gamma rays interact very weakly with a bulk crystal and go right through it. X-rays and neutrons (bearing no charge) are in the "happy medium" and produce a sufficient interaction with crystals as to give information about the arrangement of particles.

Picture taken from wikipedia's X-ray article:

More useful info in Economou's book "The Physics of Solids" can be found.


Oxidation number of phosphorus in phosphine

Here is a way to get the answer. If you place a reagent with oxidation number of H = -1 (hydride) it would produce dihydrogen gas. For example: $\ce{KH + H2O-> KOH + H2}$.

If you place a molecule with oxidation number of $\ce{H} = +1$ dihydrogen gas is NOT formed. $\ce{NH3 + H2O -> NH4+ + OH-}$. Phosphine doesn't form dihydrogen gas on contact with water so $\ce{H}$ is in oxidation state +1 and $\ce{P}$ is in oxidation state -3. Reaction with strong acids leads to $\ce{PH4+}$ which has similar structure to the ammonium ion. For these reasons even for $\ce{As}$ in $\ce{AsH3}$ the oxidation state of $\ce{As = -3}$.

The problem with electronegativity is that we are trying to assign a single number, while we actually need a function. We take average of electronegativity of $\ce{H}$ in $\ce{HF, NaH, H2O, CH4, NH4+}$ and come up with a single number. This works well if you need to guesstimate the behavior, but it is not accurate. It doesn't even account for softness/hardness of a partner.

Analysis "will $\ce{PH3 + H2O}$ form $\ce{H2}$?" (no) vs "will $\ce{PH3 + P2O3}$ form $\ce{P + H2O}$?" (yes) is more accurate way to assign formal charges.


Is the 5d orbital involved in triiodide ion?

Professor Klüfers presents a structure of the $\ce{I3-}$ ion in his general and inorganic chemistry introductory course at the LMU Munich. The ion is linear and the $\ce{I\bond{...}I}$ distances are identical, measured experimentally to be $293~\mathrm{pm}$.

In neutral iodine, the $\ce{I-I}$ distance is measured to be $272~\mathrm{pm}$.

We see a notable increase in bond length which indicates a decrease in bond order. A reduction of bond order — say from a bond order of $1$ (single bond) to one of $0.5$ (four-electron three-centre bond) — means that we cannot add an additional bonding orbital to the mixture. Otherwise we would have something along the lines of two single bonds (bond order $1$), which is how the ‘octet expansion’ traditionally was explained.

The reduction in bond order is best explained — as mentioned — by a four-electron three-centre bond and thus without d-orbital participation.


Hypothetical: What happens to water as pressure increases to infinity?

Ron's answer gives us a good idea of what might happen in terms of "normal" chemistry, but if you really mean "increases to infinity", some very strange stuff happens. Suffice it to say, it doesn't stay water after a certain point. The intense temperatures created by the compression will cause the water to break apart, eventually no longer even having oxygen atoms due to nuclear reactions. Because we're talking about an externally applied pressure, the Chandrasekhar limit doesn't apply, so there is a point at which electrons and protons combine (when the electron degeneracy pressure is overcome) and a mass of neutrons remains. Neutrons themselves also have a degeneracy pressure (though we don't have good models to predict the exact pressure that has to be overcome). From here, we don't know what happens with as much certainty, but the formation of quark matter has been predicted.

Eventually, we reach a singularity. We can think of this as all the matter we had before being compressed into an infinitesimal volume with infinite density and our applied pressure ceases to mean anything. If we started with enough water, this would behave much like any other black hole, though micro black holes are hypothesized to have some special properties.


Why protons combined with electrons will release heat?

If the process of uniting protons and electrons is assumed to occur isothermically in a vessel of constant volume, then the heat released will be equal to the change in internal energy of the system:

$$q_v=(\Delta U)_v$$

Now if the mixture were to behave as an ideal gas, the change in internal energy would be zero and therefore so would the heat. But we can expect a mixture of oppositely charged species such as protons and electrons to behave as an extremely non-ideal mixture, consistently with the large Coulombic potential between the particles. Since $\Delta U$ for the process of uniting protons and electrons (and thereby also minimizing repulsions) is negative, the process is then expected to be exothermic ($q<0$).


For ionic compounds there is really no such thing as a "molecule" of the compound. In the solid state the ions exist in a 3D crystalline structure and there is no unique way to associate a particular $\ce{Mn^{4+}}$ ion with a particular pair of $\ce{O^{2-}}$ ions. Therefore the simplest integer ratio is used as the "molecular formula."


Relative atmospheric lifetimes of carbon monoxide and methane

The dominant oxidation (via hydroxyl radical) pathways for CO don't have any intermediate species as unstable as the methyl radical, which is step one for methane oxidation via the hydroxyl radical.


Viability of proposed carbon-carbon bond-forming reaction

With some some minor modifications, your two stage reaction

Kolbe nitrile synthesis
alkaline hydrolysis of the nitrile to corresponding carboxylic acid

will work in principle.
In the Organikum, 21. edition, 2001, p. 256ff, a general procedure for the reaction of alkyl bromides with sodium cyanide is provided.
The procedure distinguishes between reactive and less reactive halides.
Reactive halides are benzyl halides (unsubstituted, methoxy- or methyl-substituted.
Less reactive halides are halogen-substituted benzyl halides and plain alkyl halides.
For the latter, the following procedure is described:

In einem 1-l-Dreihalskolben mit Rührer, Rückflußkühler und Innenthermometer werden 250 ml Triethylenglycol, 1,25 mol gut gepulvertes, trockenes Natriumcyanid und l mol Alkylbromid oder Alkylchlorid unter gutem Rühren vorsichtig erhitzt. Der Beginn der stark exothermen Reaktion ist bei den niederen Alkylhalogeniden daran zu erkennen, daß die Lösung aufsiedet. Man steigert die Temperatur langsam auf 14O °C, bei der Umsetzung von Benzylhalogeniden nur auf 100 °C, und rührt bei dieser Temperatur noch 30 Minuten.

(Translation: In a 1-liter three-necked flask equipped with stirrer, reflux condenser and internal thermometer, 250 ml of triethylene glycol, 1.25 mol of well-powdered and dry sodium cyanide and 1 mol of alkyl bromide or alkyl chloride are carefully heated. The beginning of the strongly exothermic reaction in the lower alkyl halides is evident from the fact that the solution settles. The temperature is slowly raised to 140 °C, while the reaction of benzyl halides is only to 100 °C, and stirring is continued at this temperature for 30 minutes.)
The solvent is triethylene glycol with a boiling point of 291 °C, and the reaction is performed under exclusion of water using dry sodium cyanide.
As I have mentioned in my comment, the low boiling point of bromoethane is a disadvantage!
In fact, the Organikum recommends the use of diethyl sulfate (bp 209 °C) for the synthesis of propionitrile (bp 118 °C) and suggests to isolate the product by distilling it off the reaction mixture.
For further purification of the more volatile nitriles with an alkyl chain shorter than $\ce{C5}$, it is mentioned to

wash then with brine
dry over calcium chloride
re-distill over a 30 cm Vigreux column

For this reaction (to propionitrile), the yield is given as 90%. (For valeronitrile, hexanenitrile and heptanenitrile, which all can be prepared from the corresponding bromides, the given yields are around 80%).
According to the Organikum (p. 500) the hydrolysis of primary nitriles is achieved by heating 1 mol of the nitrile with 2 mol of 25% aqueous sodium hydroxide solution under reflux until no more ammonia evaporates (2 to 1o hours).
Whether you really want to use this two stage procedure of perform a Grignard reaction of the bromides with $\ce{CO2}$ is up to you. As mentioned in the comments, the advantage of the Grignard pathway is that it's pretty much a one-pot reaction all the way down to the carboxylic acid. The downside is the requirement of a dry solvent.

Update
In a comment, you asked for a recommendation on which carboxylic acid to synthesize.
I would synthesize a compound if

I need it (for further studies)
it is not commercially available
it costs a fortune. (Make sure that the halide that you need as a starting material is not more expensive than the final carboxylic acid.)

If it is just for practise, I'd go for phenylacetic acid (or a donor-substituted derivative), which has the following advantages:

benzyl bromide is a (weak) lacrymator, but not a CMR compound
the reaction with sodium cyanide can be carried out in refluxing acetone (instead of triethylene glycol)
phenylacetic acid, the final product, has a melting point of 77 °C and a rather low solubility in water (< 2 g/L). The product can be purified by crystallization.


Although the work of Wilds and Werth1 does not align perfectly with the conditions of the OP's question, it does illustrate that Robinson annulation and decarbomethoxylation can occur under alkaline conditions. Catalytic base was found to be optimal to effect the Michael addition with methyl vinyl ketone (MVK). The aldol condensation was conducted in aqueous KOH at reflux.
Whether decarbomethoxylation occurs prior to aldolization (2-->3-->5) or subsequently (2-->4-->5) was not addressed. α'α-Dialkylated β-ketoesters generally undergo attack at the keto group rather the carboxylate by base. This generality may be tempered by the aromatic nature of the keto group in 2. Initial aldolization of 2 to 4 affords a structure that is a vinylogous β-ketoester; it is susceptible to base initiated decarboxylation.

1) A. L. Wilds and R. G. Werth, J. Org. Chem. 1952, 17, 1154.


Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?

The simple explanation is that weak intermolecular forces (the forces that make something condense to a liquid when things are cold enough) depend on the surface area (as well as many other things). But in the case of relatively similar non-polar isomers (where weak intermolecular forces are the dominant forces), the larger surface area will lead to the large force and hence the highest boilling point.

2,2 dimethyl propane is a compact, almost spherical molecule. Pentane is long and "floppier" so will experience more forces between molecules.

The underlying, but slightly simplified, explanation of this is that intermolecular forces (often called van der Waal's forces) depend on attractions caused by quantum fluctuations in the surface electrons of the molecule. These lead to short-lived dipole moments that can also induce dipoles in neighboring molecules to which the original dipole is attracted. The larger the surface area, the more opportunity for such dipoles to exist and therefore a stronger force.


Viewing and capturing real time values of Electronic balance measuring instrument on television set and computer monitor larger screens

Three possibilities come to mind:
(1) Purchase a scale with a digital output e.g, https://www.arlynscales.com/digital-indicator/industrial-scales-with-analog-and-digital-outputs/ , send it to your computer, and output it to a monitor (depending on the scale's output format, it may be possible to send it directly from the scale to a monitor)
(2) Point a USB webcam towards the scale's digital readout, plug that into to your computer, and then display on a monitor.
(3) Same as no. 2, but use a digital camcorder that has a digital output that can be displayed directly on a monitor, without the need for a computer.


What is the proper way to prepare a solution aiming to minimize errors?

Methylene blue isn't a primary standard so I'm not sure how much accuracy/precision you need.

Typically when wanting the best accuracy the solid would be dried in
a hot oven, then cooled in a desiccator.
Do not to handle the beaker with your hands. Even if it has been
"left out" in humid air that amount of water would be stable. So use
tongs to pull in and out of balance.
Another option is to weigh more and do a serial dilution. So 1 gram
in liter, then 10 ml of that solution into 1 liter.


Separation can be done in two ways:

Natural:

Natural separation term is not a good one but i'm gonna use it here. If at equilibrium (When the free energy of all the chemical constituents are same), those two specific liquid like to separate themselves out from the rest of the mixture, than you just need to give it enough time to reach equilibrium to separate them out. And this is the only method i can think of which will have no significant effect on the liquid in that container.
Forced:

Forced separation: Now if the above mentioned equilibrium condition needs different temp. and pressure than room temp. and pressure. Then by simply changing temp. and pressure you can achieve the above separation. Or you can utilize the different in various properties to separate them, e.g. Viscosity, boiling point, melting point etc.
Distillation: If they have separate boiling point than you can use distillation column and collecting the liquid at various height will give you separated solvent.
Solvent extraction, Filtration, mixer settlers, centrifugal force, ion exchange, membrane filtration, using bio separation like algae etc you can also achieve liquid liquid extraction.