DFT Calculations, Atomic Ionization Potentials -- Which Exchange-Correlation Functional to Use, to Preserve Koopmans Theorem?

I think the most widely-used approach at the moment comes from Roi Baer and Leeor Kronik and others, e.g.

Phys. Rev. Lett. (2010) 105, art. 266802
Phys. Rev. B (2011) 84, art. 075144

Basically the idea is to tune the range-separation parameter $\gamma$ between short-range and long-range electrostatic effects in a range-separated hybrid functional. You find a match such that Koopmans' theorem holds.

The result is an "optimally-tuned" range-separated hybrid DFT functional (OT-RSH).

It's not perfect, for example, it violates size consistency: J. Chem. Phys. (2013) 138, 204115


The time scale of calculations in computational chemistry

No, that is not possible. What is possible is to estimate how much more time you need to calculate the electronic energy if you increase the system size by some factor. For any method which uses LCAO-MO expansion, it is the number of basis functions $m$ which primarily determines the computational cost, so that it is usually used as a measure of the system size. For instance, HF (without any tricks like density fitting, or taking the spatial symmetry into account) scales approximately as $m^4$, i.e. if you double the number of basis functions by using a basis which is twice as large, the computational cost will be approximately $2^4=16$ times larger. But that is it: estimating the time for a calculation using the well-known scaling behavior is possible only if you already know some timings on the same setup. And besides, this is all about just the electronic energy, which is quite often not the end of the story.


Protecting elemental Li from the atmosphere

Industrially, this has been done by coating with Parylene.

The Rubber and Plastics Age, Volume 46 says:

  Parylene coated lithium metal granules on the left exhibit no reaction with water on which they float

and  New linear polymers says:

  Encapsulation of reactive chemicals in thin parylene C coatings provides protection. Parylene-coated lithium metal granules at left exhibit no reaction with water on which they float Unprotected granules at right react instantly and violently with water.

However, I don't know on what timescale this protection lasts.  As others are saying, lithium metal is an extreme fire hazard.  What if the coating is scratched or peels? Be careful!


What is this device? (Constant level device?)

Thanks Joseph. Knowing that it is a Fi-Streem product, I called/emailed them and got an answer from their friendly support tech. From his email response: [This] is a "Switch Level Device, front" that is used in conjunction with an older series Fi-Streem (Fisions) Still, which is no longer manufactured. The part number for this item is, 06756."


This question is related: Handling and storing sodium hydroxide, hydrochloric acid, hydrogen peroxide and isopropyl alcohol

In the industries the handling of chemicals maybe periodic and, with routine, one tends to forget the hazardous and risks related to this. There's a lot of information on the web (http://water.usgs.gov/admin/memo/policy/wrdpolicy94.006.html):

STORAGE: Hydrochloric acid should be kept in a tightly closed container and
stored in chemical area that is compatible with other chemicals.
Store in a secure, well-ventilated area, that is well marked, and
away from the general work population.  Do not store near oxidizing
materials.

This manual should be helpfull too: http://www.jsia.gr.jp/data/handling_02e.pdf

1. Types of Containers:
The shipping containers used for hydrochloric acid are those rail tanks be tightly sealed, such as acid-resistant bottles (including polyethylene containers, etc.), and steel drums tank trucks, tank cars, and tankers with corrosion resistant linings

If you keep reading the manual there is valuable information about handling, transportation, and rules to keep the human safety.


Calculate the lattice energy of ammonium carbonate

The lattice energy is not directly measurable, and it is difficult to calculate. In a paper about estimating lattice energies, the estimation for $\ce{Na2CO3}$ is 30% off from the reference value (which is from CRC Handbook of Chemistry and Physics). I mention sodium carbonate because ammonium carbonate is not one of their examples. The introduction explains why it is not possible to directly measure the lattice energy:

  Lattice  potential  energy  (UPOT)  is  a  dominant  term  in  the thermodynamic analysis of the existence and stability of ionic solids.  Direct  experimental  determination  is  generally  not possible since, in practice, the crystalline solid dissociates into atoms  and not into  gaseous  ions,  as  is  required  in  the  lattice energy  evaluation.  Therefore,  its  indirect  experimental  determination, computation, or estimation is of considerable interest in modern materials science; indeed, whenever the energetics of condensed-state materials are studied, the chemical processes under consideration may be rationalized if the appropriate lattice energy steps can be incorporated into the thermochemical cycle.

They go one to discuss how you can obtain the lattice energy from experimental or theoretical data:

  A  variety  of  estimation  methods  for  lattice  energies  is vavailable. These include the Born-Haber-Fajans thermochemical cycle (which requires ancillary thermodynamic data) and modern computational methods (which generally require knowledge of the lattice constants and the coordinates of the ions as well as needing an established force field). The computational methods  range  from  direct  energy  calculational  procedures through to programs that produce lattice energies in the course of their modeling of the solid. Quantum mechanical procedures are also available, but are highly computationally expensive and are generally applied only to the simpler systems.

Then they talk about cheaper methods that are less accurate. The paper is about a new quick estimation methods.

  By contrast,  a  few  exceedingly  rapid  estimation  methods based on ionic radii or volume per formula unit (“molecular”volume) have  been  developed.  Originally  only  applicable  to the simpler binary ionic solids (such as MX (with a 1:1 charge ratio)), the volume-based approach has recently been extended by  us to  include  MX2(2:1)  and  M2X  (1:2))  salts.

So check the CRC Handbook of Chemistry and Physics.


Is didehydrobenzene valid IUPAC nomenclature?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for the compound that is given in the question is 1,2-didehydrobenzene. The name cyclohexa-1,3-dien-5-yne is acceptable for general nomenclature. The formerly used name benzyne is no longer recommended.


How to draw wedged double bond in ChemDraw?

As Zhe pointed out in the comments, bolding both bonds looks 'weird', and if you look in textbooks or throughout the chemical literature, BINAP (and related systems) are commonly drawn with only the outer edge bolded (as shown on the left in the image below).

Although the feature you want isn't directly supported in ChemDraw,  one can get both lines bolded by simply selecting the bond in question, right-clicking, and changing the line width to match the bold width (this gives you the structure on the right hand side of the image below).

That said, just because you can do this, doesn't mean you should, it goes against the standard way of drawing these systems, and doesn't add anything in terms of clarity (the first diagram emphasises the stereochemistry perfectly well, the additional bold bond adds nothing).


Why do BH3 and NaBH4 have different selectivities?

Your assertions are correct, for the most part. $\ce{BH3}$ is a Lewis acid, and it does not reduce carbonyl groups by directly donating a hydride group like $\ce{NaBH4}$ does. The $\ce{NaBH4}$ reduction mechanism is fairly short and involves a direct transfer of the a hydride ion to an electron deficient carbonyl carbon:

$\hspace{2.4cm}$

$\ce{NaBH4}$ is incapable of reducing carboxylic acids only because the acidic hydrogen is more reactive towards the hydride ion than is the carbonyl carbon. It is, in fact, capable of reducing carboxylate anions to primary alcohols, though the reaction is very slow. Electron-rich carbonyls favor reactions with electrophilic reducing agents. To help in understanding what makes a carbonyl electron-deficient or rich, let us consider the contributing resonance structures of amides, esters, and acid chlorides:

$\hspace{4.7cm}$

Nitrogen atoms are much better than bearing a positive charge than are oxygen than is chlorine. In this way, you can think of nitrogen donating its lone pair into the carbonyl the most, making it electron-rich, and the chloride donating its lone pairs the least, making it electron-deficient.

The $\ce{BH3}$ reduction mechanism is quite a bit longer and has some interesting intermediates, but here it is:

$\hspace{2cm}$

Note that this mechanism does in fact include hydride transfers (it has to). Though you were right in thinking that $\ce{BH3}$ wouldn't immediately donate a hydride, coordinating to a lone pair weakens the $\ce{B-H}$ bonds and makes hydride transfers possible. $\ce{BH3}$ and $\ce{NaBH4}$ really aren't as different as you might first think.

Though $\ce{BH3}$ is more reactive towards electron-rich carbonyls, it is also capable of reducing electron-deficient carbonyls like acid chlorides and esters through similar pathways. This should show you that there isn't any contradiction here, and that just because two compounds have different reactivities does not mean that they cannot perform the same type of reactions—they just won't perform it in the same way.


Is it possible to calculate the momentum of an electron precisely?

To preface this answer I suggest reading a proper QM text which should address many of your issues. Griffiths' Introduction to Quantum Mechanics served me well but you may prefer any of the other excellent texts out there.

Option 4

This depends on your interpretation of quantum mechanics, but in the usual Copenhagen interpretation, the measurement of an observable causes the wavefunction of the system to collapse into the corresponding eigenstate. To give a more concrete example, if we have energy eigenstates $|1\rangle, |2\rangle, |3\rangle, \cdots$ with corresponding eigenvalues $E_1, E_2, E_3, \cdots$ then the wavefunction may be written as a linear combination of energy eigenstates. [This is guaranteed to be true because the Hamiltonian is a hermitian operator, and hermitian operators have complete sets of eigenstates.]

$$|\psi\rangle = c_1|1\rangle + c_2|2\rangle + c_3|3\rangle + \cdots$$

When you measure the energy of this system, the chance of getting the result $E_i$ is equal to $|c_i|^2$. Let's say you measure the energy to be $E_1$. Immediately after the measurement, the wavefunction of the system collapses to $|1\rangle$:

$$|\psi\rangle \overset{\text{collapse}}{\longrightarrow} |1\rangle$$

Now let's say you want to measure some other observable $A$. Again, you can express the original wavefunction $|\psi\rangle$ as a linear combination of eigenstates of $A$.

$$|\psi\rangle = a_1|A_1\rangle + a_2|A_2\rangle + a_3|A_3\rangle + \cdots$$

and the probability of measuring the eigenvalue $A_i$ would be $|a_i|^2$. However, the new wavefunction after collapse will have different coefficients in this expansion:

$$|1\rangle = a_1'|A_1\rangle + a_2'|A_2\rangle + a_3'|A_3\rangle + \cdots; \qquad a_i \neq a_i' \text{ (in general)}$$

and hence the probability of measuring the eigenvalue $A_i$ is now $|a_i'|^2$, which is in general, not the same as $|a_i|^2$.

In the absolute simplest case, we can consider measuring the energy again immediately after measuring the energy and obtaining the value $E_1$. Since the measurement caused the wavefunction of the system to collapse into $|1\rangle$, a second measurement done immediately after this will inevitably return the same value $E_1$ again.

Hence option 4 is true, which means it is not the answer you are looking for. Measurement of an observable causes wavefunction collapse, which in turn causes the probabilities of measuring other eigenvalues to change.

Option 2

First, we note that for a measurement of an observable to have no uncertainty, the system must be in an eigenstate of the relevant operator. This follows from the definition of uncertainty

$$\sigma_A^2 = \left\langle (A - \langle A\rangle)^2 \right\rangle$$

If the wavefunction of an electron is an eigenstate of the momentum operator $\hat{p}$, then a measurement of the momentum will always return the same value. In real life, if we carry out this measurement, will there be some uncertainty due to the precision of the machine? Yes, of course. Is it practical to set up an electron in a momentum eigenstate? Probably not. (By the way, your electron cannot really be constrained within your apparatus, because of quantum tunnelling. That is, unless the walls of your apparatus are an infinitely thick infinite potential energy barrier.)

But, that question is not asking you how you carry something out in the lab. The question is asking purely about theoretical considerations - and there is no fundamental theoretical problem with having an electron in a momentum eigenstate and measuring its momentum to arbitrary precision without changing its state. You are correct that its position will be indeterminate, but that is not a problem in the context of this question.

For interest, an "ideal measurement" of momentum is described in Shankar's Principles of Quantum Mechanics, 2nd ed. (p 123):

  Consider a particle in a momentum eigenstate $|p\rangle$. The postulate tells us that if the momentum in this state is measured we are assured a result $p$, and that the state will be the same after the measurement (since $|\psi\rangle = |p\rangle$ is already an eigenstate of the operator $P$ in question). One way to measure the momentum of the particle is by Compton scattering, in which a photon of definite momentum bounces off the particle.

  Let us assume the particle is forced to move along the $x$-axis and that we send in a right-moving photon of energy $\hbar\omega$ that bounces off the particle and returns as a left-moving photon of energy $\hbar\omega'$. (How do we know what the photon energies are? We assume we have atoms that are known to emit and absorb photons of any given energy.) Using momentum and energy conservation:

  $$\begin{align}
cp' &= cp + \hbar(\omega + \omega') \\
E' &= E + \hbar(\omega - \omega')
\end{align}$$

  it is now possible from this data to reconstruct the initial and final momenta of the particle:

  $$\begin{align}
cp &= -\frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\
cp' &= \frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\
\end{align}$$

  Since the photon always loses energy to the particle (as is clear in the particle rest frame) $\omega' < \omega$ and by sending $\omega \to 0$, we can make the change in momentum $p' - p$ arbitrarily small.


Simple alternative to UNIFAC to prove that a mixture containing 1ml of x and 5 litres of propylene glycol is not flammable?

UNIFAC is not a procedure, unless by chance in a similar field two things got the same name, which is unlikely.

UNIFAC is used by activity coefficient models for predicting liquid properties. It is often coupled with a vapor phase equation of state, and quite often simply the ideal gas law...

A flash point calculation is a procedure, and must be done in whole. Note: you can do either an experimental flash calculation, or a computational one!. UNIFAC is used in the latter. It combines the fact that at equilibrium a specie in both liquid and vapor must have the same chemical potential (gotten through via activity coefficients in the liquid phase and fugacity coefficients in the vapor typically) coupled with mass balance. You cannot do half of it and have an answer... like all calculations it must be done fully and correctly. It is not that difficult, but it should be done by someone who is a scientist or engineer... This is the type of thing that would be expected of a chemical engineer.

Most chemical engineering Thermodynamics textbooks have a decent section covering flash and bubble/dew point calculations.

There are many programs that can do flash calculations for you. VMGSim, Aspen/HYSIS to name just two of the many commercial/industrial simulators plus a plethora of small programs online made by people like me (mine isn't online ;) ).

I just did a quick google and this looks like it would work https://checalc.com/solved/multi_flash.html

Note, you do not need to use UNIQUAC in the liquid, but, it usually by default will use UNIFAC to fill in unknown interaction parameters, so it is basically a safe activity coefficient model. Depending on what is in the vapor, the peng-robinson or even ideal gas equations of state are probably good enough. If there is association or the molecules are polar, then I recommend the Haydon O Connell.

The link I sent looks like it uses the Peng-Robinson equation of state for liquid and vapor. It is probably okay for what you are doing. Actually the link I sent doesn't look like it has PG in its database... this one may... https://www.daoftx.com/usingflsc.htm

But you know what to look for now... These premade programs just need you to tell them the species, initial composition, T, P and then you just hit calculate. You just need to find one that has your molecules in its database.


Relative stability of cis and trans cycloalkenes

Usually trans-olefins are more stable than their cis isomers for steric reasons, like you suggested.  However in small and medium size rings this is not the case; here the cis-cycloalkene is more stable than the corresponding  trans isomer.

trans-Cyclooctene is the smallest trans-cycloalkene that is stable at room temperature (trans-cyclohexene and trans-cycloheptene have been detected as short-lived intermediates).  In your drawing up above of trans-cyclodecene (a) draw in the two hydrogens on the trans double bond.  One of them is "inside" the ring and causes steric problems with the carbons on the other side of the ring as pictured below.  Because of this destabilizing transannular interaction trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules.  They exist in a conformation as pictured below for trans-cyclooctene.

(image source)

If you can, build a model of trans-cyclooctene.  Notice how you have to pull and stretch the methylene chain in order to attach that last methylene back onto the other end of the double bond - there is a lot of strain in this molecule. Now build a model of cis-cyclooctene; it is easy to loop the methylene chain from one end of the double bond to the other end.

Not until you get to around trans-cyclododecene (12 carbons) do you have enough carbons to easily connect from one end of the trans double bond back to the other end.  trans-Cycloalkenes with less than 12 carbons have additional strain because of this difficulty in spanning the double bond. Once you get to around 12 carbons there is no difficulty spanning the double bond and so once again trans-cycloalkenes become more stable than their cis counterparts.

Since double bond carbons are $\mathrm{sp^2}$ hybridized with 120° angles, if we add more double bonds, the 120° angles make it more difficult to loop the more flexible $\mathrm{sp^3}$ methylene (with their 109.5° angles) chain around the double bonds.


All calcium chloride hydrates give away water of crystallization upon heating without forming a hydroxide, and the anhydrous $\ce{CaCl2}$ melts without decomposition [1, p. 162]:

  anhydrous salts melts at 772 °C, while the mono-, di-, tetra- and hexahydrates decompose at 260 °C, 175 °C, 45.5 °C and 30 °C, respectively; the anhydrous salt vaporizes at 1935 °C.

$$\ce{CaCl2 · x H2O ->[\pu{260 °C}] CaCl2 + x H2O}$$

Calcium chloride dihydrate would remain intact until its melting/decomposition point at 175 °C.
When heated quickly in the absence of desiccant, the powdered salt is dissolved in its own water of crystallization forming glass-alike solid upon cooling.

References

Patnaik, P. Handbook of Inorganic Chemicals; McGraw-Hill handbooks; McGraw-Hill: New York, 2003. ISBN 978-0-07-049439-8.


Why do negative oxidation states not extend to -8?

Why do the negative oxidation states not go all the way down to -8?

Let's think about what we would need for this to occur - i.e. an oxidation state of -8.

We would need a central atom that has considerable electronegativity - one that is at least more electronegative than its attached atoms. This already excludes numerous molecules; most molecules' central atoms are, on a relative basis, more electropositive rather than electronegative.

In addition, this atom would need to have one of the following configurations; "allocated" refers to the assignment of oxidation state - since oxidation states are assigned on the premise that all bonding is ionic - the more electronegative element in a bonding configuration is "allocated" all the bonding electrons in the calculation of oxidation states. I.e. in H2O, the oxygen would be allocated both pairs of bonding electrons in the H-O bonds, and oxygen would then "have" 8 electrons, and its oxidation state would be -2.

1 valence electron and be "allocated" 9 electrons

2 valence electrons and be "allocated" 10 electrons

3 valence electrons and be "allocated" 11 electrons,

and on the other extreme:

6 valence electrons and be "allocated" 14 electrons,

7 valence electrons and be "allocated" 15 electrons,

8 valence electrons and be "allocated" 16 electrons.

As we can see, the first set of criteria are basically incompatible. Elements only having 1 or 2 valence electrons are going to be small (on a relative basis). These elements would also much rather be oxidized than reduced. Oxidation allows these elements to easily become isoelectronic with a noble gas.

The other set of criteria for a negative 8 oxidation state is also hard to fulfill. None of the group 6 elements except the possibility of sulfur and elements below it are hypervalent - i.e. they can accommodate more than 8 valence electrons. Even if we consider sulfur as able to accommodate 12 valence electrons (as some depictions of sulfuric acid suggest) we must remember that sulfur is not very electronegative and unlikely to be allocated any electrons.


Structural isomerism in iso-butyl and tertiary butyl alcohol

I think the best way to describe this isomerism is constitutional isomerism because it takes into consideration of both position of functional group and structure at the same time.

  Constitutional isomers may have different carbon skeletons (as in
  isobutane and butane), different locations of a functional group along
  the chain (as in isobutyl alcohol and tertbutyl alcohol), or different
  functional groups (as in ethanol and dimethyl ether)

Positional isomers have the same carbon skeleton and same functional groups but differ in location of such functional groups (carbon skeleton remains unchanged). On the other hand, structural isomers the atoms are in completely different order , they arise because of the possibility of branching

I think the reason why these isomers are considered both structural and positional is that they both have the longest chain of max 3 carbons (hence same carbon skeleton), at the same time different location of -OH group  (a different chain - structural)

Hope this helps

References

Organic chemistry Mc Murry

Chemguide: Structural isomerism


What is the correctly diagrammed electron probability distribution for p orbital?

The probability distribution of an electron in the p sub-shell is determined from its wavefunction. The wavefunction of a system contains all the information about the quantum states of the system. Once of the most important properties of the wavefunction is that the squared modulus of the wave function, $|ψ|^2$, is a real number, interpreted as the probability density of the particle. Here is the wavefunction for the 2p orbital:

So if you square this function you will get the probabily distribution which looks like this:

As you can see, this corresponds with the p orbital drawn in your book, as there is a node in the middle, on either side is high probability density and at minus and positive infinity probability density is actually very, very small despite being shown as 0 on the wavefunction. The reason for this is that you need to remember that the orbital shows where an electron will be 90% of the time, not where it will be at all times. This means that the electron can be found outside the orbital.


What are all the structural isomers of a tertiary amine with the sum formula C6H15N?

Please note that this is just a visualization of the comment given by LDC3!

Here are the two missing isomers:


Oxymercuration demercuration, PCC, NaCN/HCl reaction

Check the mechanism of the last reaction here.
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch20/ch20-3-5-1.html

It is not the displacement of nitrile by carboxylic acid(which you do not have, btw). It is the combination of water addition and the displacement of ammonia by second equivalent of water.


What is the final temperature of each substance upon absorbing 2.25 kJ of heat?

You are almost correct, however, to calculate these values you need to put a bracket around he $t-29$ term (which is the $\Delta T$ term from $Q = mc \Delta T$). Hence, from your working out step:

$$2250 = 0.240 \times 28 \times (t - 29)$$

You'll need to expand the brackets.


Iron ring is a correct term often used for this piece of equipment, according to the Wikipedia page, also this lab equipment tutorial page and in a very comprehensive pictorial lab equipment guide from Iowa State University.

Additionally, it is also called an iron support ring and a retort ring.


Is the bond in HF ionic while it is covalent in HCl?

Bonds can be completely covalent as in $\ce{Cl-Cl}$, $\ce{H-H}$, etc. In these cases the electron density is shared equally between the two atoms.  Bonds can also be ionic as in $\ce{Na^{+} Cl^{-}}$ where much (~80%) of an electron has been transferred from the sodium atom to the chlorine atom.  Between these two extreme cases exist a continuum of bonds that are called polar, covalent bonds.  In such cases the electron density between the two atoms is not shared equally, but neither is the electron transferred extensively from one atom to the other.

Just as we can draw resonance structures for acetone and discuss how the real molecule is a weighted superposition of these two structures, we can draw resonance structures for inorganic compounds and, again, discuss how much the various structures count towards the "real" description of the molecule.  In the diagram below, two resonance structures are shown for sodium chloride.  The structure on the left represents a covalent bond and the structure on the right represents an ionic bond. Just as acetone is a mix of the two resonance structures, so is sodium chloride.  When the weighted mix of the resonance structures lies far to the left, we can say the compound or bond is covalent; when it lies far to the right we can say the compound or bond is ionic.  Everything in between represents a polar, covalent bond.

One can crudely estimate how polar a diatomic covalent bond is by using the following methodology.  The dipole moment of a molecule is given by
$$\ce{\mu~=~Q*d}$$
where the dipole moment is in Debye (D), Q is the partial charge on an atom measured in coulombs and d is the distance between the two atoms (the bond length) in meters.

In the case of $\ce{HF}$, the dipole moment is 1.86 D and the bond length is 91.7 x 10^(-12) m.  Noting that 1 D = 3.3356 × 10^(−30) C·m, solving for Q we find a partial charge on one of the atoms of
$$\mathrm{Q~=~\frac{[1.86 ~x~3.3356 ~x~ 10^{-30} ]}{[91.7~x~10^{-12}]}~=~6.77~x~10^{-20}~C}$$

Now since one electron equals 1.6022 × 10^(−19) C, by taking the ratio
$$\mathrm{\frac{[6.77~x~10^{-20}]}{[1.6022 × 10^{−19}]}~=~ 0.42}$$
we estimate that roughly 42% of an electron has been transferred from the hydrogen atom to the fluorine atom.  This gives us a rough idea of how ionic the polar, covalent bond is in $\ce{HF}$.  Repeating this calculation for $\ce{HCl}$ suggests that roughly only 18% of an electron has been transferred in this case.

Clearly the polar, covalent bond in $\ce{HF}$ has more ionic character than the analogous polar, covalent bond in $\ce{HCl}$.  The reason why the $\ce{HF}$ bond has more ionic character is the greater electronegativity of fluorine (3.98) compared to chlorine (3.16).


Why is there a difference between O2 and B2 sigma 2p molecular orbitals in diagrams?

This topic is really fun. You see in case of boron molecule $\sigma_{2s}$ and $\sigma_{2p_z}$ (being on the intermolecular axis) will have the same symmetry and hence the orbitals will overlap and result in the formation of 2 bonded $sp_z$ orbitals. So the $\sigma_{2p_z}$ orbital will obtain somewhat $s$ character and it will become unstable or in other words will be of higher energy. The $\sigma_{2s}$ MO will obtain somewhat $p$ character. So in a energy level MO diagram $\sigma_{2p}$ bonding orbital is placed above the $\pi_{ 2p}$ orbital.

However, in case of $\ce{O2}$, due to the higher nuclear charge of oxygen, the energy difference of $2s$ and $2p$ is large so there is no such kind of intermixing between the bonded MOs. Whereas in boron, the energy difference of $2s$ and $2p$ is less, facilitating intermixing.

You can find this topic in any higher level chem books. Hope you got my point; thank you.


Is the vapor pressure of a liquid dependent on the pressure of the atmosphere in which it is measured?

Yes, in principle it could be different, but the effect would be very small unless the increase in pressure were very large.  To put it in simplified terms:  The increase in ambient pressure compresses the liquid, bringing it to a higher-energy state, causing a higher proportion of the liquid molecules to want to escape from the liquid, and into the gas state, thus causing the vapor pressure to increase.  This assumes ideal gas behavior, i.e., that the increased ambient pressure affects only the liquid, not the gas.

DETAILS:

Let's call your liquid "$X$".  At equilibrium, the chemical potential of $X_{(l)}$ equals the chemical potential of  $X_{(g)}$.  The pressure of $X_{(g)}$ at which that occurs is called the vapor pressure.

Now suppose you increase the ambient pressure at constant temperature. Consider what happens to the chemical potential of the liquid: $d\mu = -S_mdT +V_m dP$, where $\mu$ is the chemical potential, and $S_m$ and $V_m$ are the molar entropy and volume, respectively.  Thus, at constant $T$, $d\mu = V_m dP$.  Hence the chemical potential of $X_{(l)}$ increases.  This effect, by itself, would cause the vapor pressure to increase (to get the chemical potential of $X_{(g)}$ to match the increased chemical potential of $X_{(l)}$).

But what about the chemical potential of $X_{(g)}$?  If the gas behaves ideally, then it is unaffected by the presence of other gases, and thus its chemical potential is unaffected as well.  I.e., as with the liquid, its chemical potential is determined by its pressure.  But unlike the case with the liquid, the pressure of $X_{(g)}$ is independent of the ambient pressure -- it depends only on the partial pressure of $X_{(g)}$ itself.  Thus the only effect to consider would be the increased chemical potential of the liquid, and the vapor pressure would increase.

However #1: At high concentrations, gases tend not to behave ideally.  Thus there are two possibilities: the increased partial pressure of the other gas increases the chemical potential of $X_{(g)}$, or decreases it.  If it's the former, and the effect is sufficiently strong, the vapor pressure will decrease.  If it's the latter, the vapor pressure will increase.

However #2: If the ambient gas is soluble in the liquid, increasing the ambient pressure will increase the concentration of the ambient gas in the liquid, which could reduce the chemical potential of the liquid, thus reducing the vapor pressure!

Again, let me emphasize that these are very small effects.


Photochemical ligand substitution of metal carbonyls - role of water cooled lamp-jacket?

The main point is to stop the heat flow from the lamp, while allowing UV through. You want to do photochemistry, not thermochemistry.


It looks like a vessel for sublimation. The tube going in the top would be filled with cold liquid making a tiny cold finger. The crude mixture is put in the cylindrical cup at the bottom. A vacuum is pulled from the bottom to lower the pressure. The side arm is for pressure release. After the material sublimates and deposits on the cold finger, it can be removed with the pure material on it in order to easily recover the material.


What causes passivity of specific metals in conc. nitric acid?

Aluminium

This link states that:

  Aluminium metal dissolves readily in dilute sulphuric acid to form solutions containing the hydrated Al(III) ion together with hydrogen gas, $\ce{H2}$. The corresponding reactions with dilute hydrochloric acid also give the hydrated Al(III) ion. Concentrated nitric acid passivates aluminium metal (due to formation of thin protective oxide layer, $\ce{Al2O3}$ on surface of metal which cuts off further reaction. )

$$\ce{2Al(s) + 3H2SO4(aq) → 2Al^3+(aq) + 2SO4^2-(aq) + 3H2(g)}$$
$$\ce{2Al(s) + 6HCl(aq) → 2Al^3+(aq) + 6Cl-(aq) + 3H2(g)}$$

Lead

This link states that:

  The surface of metallic lead is protected by a thin layer of lead oxide, $\ce{PbO}$. This renders the lead essentially insoluble in sulphuric acid. Lead reacts slowly with hydrochloric acid and nitric acid. In the latter case, nitrogen oxides are formed together with lead(II) nitrate, $\ce{Pb(NO3)2}$.

$$\ce{Pb + HCl -> H2[PbCl4] + H2}$$
$$\ce{Pb + 4 HNO3 → Pb(NO3)2 + 2 NO2 + 2 H2O}$$

Tin

This link states that:

  Tin is attacked only slowly by dilute acids such as hydrochloric acid and sulfuric acid. [...]Tin dissolves easily in concentrated acids.

$$\ce{Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O}$$
$$\ce{Sn + 3HCl → H[SnCl3] + H2}$$
$$\ce{Sn + 4HNO3 → H2SnO3 + 4NO2 + H2O}$$

Summarizing as a table,

  Do they have something in common or are there individual reasons?

The three metals do not have common properties and reacts with different acids differently as we can see in the table.


What considerations are made in determining whether to use a sand bath or oil bath?

The following is the comparison list from the top of my head:

Oil for the oil bath is often available commercially, every batch has the same properties and is arguably easier to store and transport; sand bath requires sifted and refined sand (quartz sand, ideally). It all depends on what's easier to find a good supplier and storage for.
Oil bath heats up relatively quickly, especially if the oils isn't very viscous; sand bath heats up very slowly due to slow heat transfer in powdered/granulated material.
As a consequence of the former statement, oil bath is uniformly heated, whereas sand bath always has a temperature gradient with the maximum temperature at the bottom and minimum on top layer.
This requires that the tip of the thermal sensor is located at the same depth as chemicals in the reactor.
Oil bath can usually only be heated up to approx. 250 °C max depending on the oil used; sand bath can be heated up to 400 °C (provided a ceramic pod is used).
When operated at high temperatures, mineral oil in the oil bath may start to decompose producing volatile health hazardous organics; sand bath, if the sand was prepared correctly, doesn't deteriorate even at high temperatures.
Silicone oil is much safer in this regard, but is also considerably more expensive than mineral oil.
In case of temperature control fail, oil bath may even ignite (again, depends on oil used); whereas sand bath will just overheat – however, the aftermath may be comparable.
Oil bath can be used with rotavap; sand bath can't.


At which temperature does carbon monoxide explode?

At a wide range of temperatures.

If mixed with air in a proportion between 12 and 75% it is explosive (assuming a there is an ignition source) - I think this is at standard temperature and pressure - 273K, 0.1 MPa.

Note that 1% of CO in air will cause death within 3 minutes due to its toxic effects.


When does Alcohol act as Solvent and as a Nucleophile?

The proposed solution to this roadmap problem has several errors. I am assuming that $\ce{K2Cr2O7}$, methanol and $\ce{LiAlH4}$ are employed in excess. In the oxidation of 1 (A), chromic acid, $\ce{H2Cr2O7}$, is formed from $\ce{K2Cr2O7}$. To obtain the purported hydroxy acid 6 suggested in the solution, exclusive oxidation of the primary alcohol would be required, a pathway that leads to lactone 5 via hemiacetal 4. One could argue that hydrolysis of the lactone occurs to form 6 but, if this were the case, then hydroxy acid 6 would oxidize to levulinic acid 2 in the presence of excess oxidant. Clearly, levulinic acid 2 fits the formula $\ce{C4H8O3}$ presented in the original problem. As the problem is set forth, oxidation of the secondary alcohol is faster than that of the primary alcohol of diol 1.
The Fischer esterification of levulinic acid 2 leads to methyl levulinate 3 ($\ce{C6H10O3}$). Such an esterification is invariably conducted with excess methanol. The reduction of 3 with $\ce{LiAlH4}$ is conducted in a solvent such as ether or tetrahydrofuran (step 1) and then decomposed cautiously with water (step 2) to reform diol 1 (A) by reduction of both the ketone and ester groups of 3. Clearly, this sequence of transformations is an academic exercise and not one to be conducted in a laboratory. Hopefully, the person who provided the solution was not the same person as the one who created the question.


Potassium carbonate vs. Potassium hydroxide in Wood Ash

You can never get $\ce{KOH}$ directly out of wood ash. One can recall that potash $\ce{=K2CO3}$ is indeed pot-ash! So you are right that historically calcium hydroxide was needed to obtained $\ce{KOH}$ out of potash. The rest of the stories are all false. $\ce{KOH}$ is unstable in air with respect to carbonate formation. It is hygroscopic as well. Wood ash remains bone dry forever. I don't think there is even a trace of potassium hydroxide in wood ash.

Until and unless you don't have a furnace, which can generate temperatures above 1300○ Celsius chances of getting $\ce{KOH}$ in useful amounts are almost negligible.

Secondly, a soap made with tons of calcium salts will be nothing but literally scum (=insoluble calcium soaps). Calcium should not be a part of soap.

Here is a nice abstract on the composition of wood ash as a function of temperature. [1]

  The elemental and molecular composition of mineral matter in five wood types and two barks was investigated as a function of temperature using thermal gravimetric analysis, differential thermal analysis, inductively coupled plasma emission spectroscopy, and X-ray diffraction. Low temperature ash was prepared at 500°C, and samples were heated in a tube furnace at temperature increments to 1400°C. The dissociation of carbonates and the volatilization of potassium, sulfur, and trace amounts of copper and boron were investigated as a function of temperature. Overall mass loss of the mineral ash ranged from 23–48% depending on wood type. The mass of K, S, B, Na, and Cu decreased, whereas Mg, P, Mn, Al, Fe, and Si did not change with temperature relative to Ca which was assumed to be constant. Sintering of the ash occurred, but fusion of the ash did not occur. In the 600°C ash CaCO3 and K2Ca(CO3)2 were identified, whereas in 1300°C ash CaO and MgO were the main compounds. The implications for ash deposition in furnaces is discussed.

References:

Misra, M. K.; Ragland, K. W.; Baker, A. J. Wood ash composition as a function of furnace temperature. Biomass Bioenergy 1993, 4 (2), 103–116 DOI: 10.1016/0961-9534(93)90032-Y.


Major product of the nitration of benz[cd]indol-2(1H)-one

According to this patent here, N-ethyl 1,8-naphtholactam nitrates in the 4 position i.e. structure a. It seems entirely likely the N-unsubstituted material will do likewise.


They occur because that's just how physics works. There isn't anything profound going on. Let's take water as an example. Water boils because enough energy has been introduced into the system such that molecules are able to break away from the interactions keeping them in the liquid state.


One of the major ingredients in modern printer toner cartridges (for laser printers) is magnetite, Fe3O4. It is used a charge control ingredient, known as a triboelectrification agent. 'Tribo' comes from the Greek meaning 'to rub' (carfeul if you google that one), and essentially refers to the creation of static charge through friction. This is required to have the printer particles stick to the paper. You can read about triboelectrification here. Magnetite, of course, is magnetic, and this is why your toner can produce a ferromagnetic fluid.

Other ingredients in toner cartridges include waxes, polymers and pigments. Many formulations are water or alcohol based and the toner is actually prepared as an emulsion which is filtered and dried. In order to get this material dissolved into kerosene, you probably need to add an emulsifying agent. Oleic acid is commonly used to achieve this, as it is a long chain acid that has both polar (hydrophilic) and non-polar (hydrophobic) properties.


It's because in ideal gases there are no intermolecular forces whereas in a real gas the force with which it collides is reduced due to intermolecular forces.Now for the volume occupying part ideal gas particles are point sizes but in a real gas the molecules do occupy some volume.That's why some corrections were made to volume and pressure in the ideal gas equation.
Note that the above were assumptions of the Kinetic theory of gases.


For sulfur we can start by identifying the number of electrons lost in the oxidation half reaction:

$$\ce{PbS^{2-} ->PbS^{6+}O4 +8e-}$$

Now we can balance the oxygens using water, which also leaves us with an excess of protons:

$$\ce{PbS + 4H2O -> PbSO4 + 8e- +8H+}\tag1\label{Oxidation}$$

Now consider reduction half reaction:

$$\ce{H2O^{1-}2 +2e- -> 2H2O^{2-} }$$

It will obviously require a proton source, so we balance with protons:

$$\ce{H2O2 + 2 H+ +2e- -> 2H2O }$$

Multiply by a factor of 4 to match the number of electrons needed by the oxidation reaction:

$$\ce{4H2O2 + 8 H+ +8e- -> 8H2O }\tag2\label{Reduction}$$

Combine the half reactions $(1)$ and $(2)$:

$$\ce{PbS + 4H2O2 -> PbSO4 + 4H2O}$$


Rate of EAS in chlorobenzene and fluorobenzene

We need some data to answer your question. As presented in this earlier answer (which is also relevant to this question), here are the relative rates for electrophilic aromatic nitration of the halobenzenes compared to benzene itself.

\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative Rate}  \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} & 0.11 \\ \hline
\ce{Ar-Cl} & 0.02 \\ \hline
\ce{Ar-Br} & 0.06  \\ \hline
\ce{Ar-I} & 0.13 \\ \hline
\end{array}

You are correct that inductive and resonance effects operate in opposite directions here. All of the halogens are more electronegative than hydrogen so they inductively remove electron density from the aromatic ring and tend to slow the reaction down. Whereas resonance effects would donate electron density into the aromatic ring and speed the reaction up.

However it is not correct to say

  On comparison of resonance effect and inductive effect, resonance
  effect dominates over inductive effect

Here is a series where inductive effects play a larger role than resonance effects.  This is why all of the halobenzenes react more slowly than benzene itself.

In fluorobenzene the overlap between carbon 2p and fluorine 2p orbitals is very effective, but the carbon 2p orbital cannot overlap as effectively with higher p orbitals (e.g. chlorine 3p, bromine 4p, etc.). It is the stronger resonance contribution from this effective electron donating resonance interaction with fluorine that makes fluorobenzene more reactive towards electrophilic aromatic substitution than chlorobenzene.

The electrophilic reactivity of the other halobenzenes is also controlled by a blend of resonance and inductive effects.  As we move down the series from chlorine to iodine the electronegativity decreases and the strength of the electron withdrawing inductive effect also decreases - this should speed up the rate of reaction. However the strength of the resonance effect also decreases due to less effective orbital overlap between halogen and carbon - this should slow the reaction down. Looking at the data we can see that the reaction rate slowly increases as we move from chlorine to iodine, suggesting that the inductive effect is still the primary effect with resonance, as expected due to poorer overlap, still playing a secondary role.


How can I make a better whiteboard cleaning solution?

Andselisk's recommendations are great, but you'll save yourself a lot of trouble if you just throw all the erasers out. They get crud on them quickly and after a while, they put more back on the board than they wipe off.

A microfiber cloth is much more effective, and works for a shockingly long time. I would bet on a filthy cloth over a brand new eraser, and unlike erasers, they can be thrown in the wash. You can get them specifically for whiteboards or just stop by an auto parts store.


Chlorine and Free Chlorine

Chlorine is the term commonly used to describe the addition of chlorine gas, sodium hypochlorite (i.e., bleach), or calcium hypochlorite (i.e., bleach powder) to water to form free chlorine. Free Chlorine is the sum of hypochlorous acid ($\ce{HOCl}$) and hypochlorite ($\ce{OCl-}$); a monoprotic acid with a $\ce{pK_a} = 7.54$.

Chlorine gas ($\ce{Cl2}$) does not exist in water (except highly acidic) as it hydrolyzes to $\ce{HOCl}$. Chloride ($\ce{Cl-}$) forms from the redox reaction of $\ce{HOCl}$/$\ce{OCl-}$, but it is not reactive.

Both $\ce{HOCl}$ and $\ce{OCl-}$ are volatile, with $\ce{HOCl}$ being $\approx\!5\!\times$ greater (Henry's solubility constant being $H = 0.076$ and $0.013$, respectively).$^{[\text{citation needed}]}$ When you open a pitcher of water containing free chlorine to the atmosphere, the free chlorine comes into equilibrium with the atmosphere and thus it volatilizes. As such, you do not lose "$100\%$ or all" of the free chlorine, but enough is removed that the concentrations drops below your taste threshold.

Side Note

If the water contains dissolved organic carbon and organic nitrogen (e.g., natural organic matter, pathogens), then the free chlorine could be reacting while it is sitting and thus you could be losing free chlorine this way as well. Further you can also add combined chlorine (e.g., monochloramine) to water in place of free chlorine. This also forms from free chlorine reaction with nitrogen compounds in water (e.g., urine) and is the common "pool smell" observed in swimming pools that are not properly chlorinated.


What's the name of this electrolysis device?

Based on this article, in which the authors describe a system that can do 25 cells simultaneously, the term 'Parallel Electrolyser' would seem appropriate.


Reaction of alkenes with bromine in presence of sodium chloride

The first step is $\ce{Br-Br}$ approaching the pi-cloud, being polarized, loosing a $\ce{Br^{-}}$ to form a 3-cyclic bromonium cation.  Given solubility and reactively inert solvent, $\ce{Br^{-}}$ (large poor nucleophile) competes with $\ce{Cl^{-}}$ (smaller, better nucleophile) for ring-opening capture.  Now factor in relative concentrations.  Now factor in thermodynamics (reversibility, $\ce{Cl^{-}}$ wins for the stronger $\ce{C-Cl}$ bond) versus kinetics (smaller $\ce{Cl^{-}}$ vs. initial tight ion pair rapidly collapsing to product).  Temperature probably makes a product ratio difference, too.

If your linear ion intermediate is controlling, the $\ce{Cl}$ comes in at the 2-position by default.  If my cyclic ion intermediate is controlling, the $\ce{Cl}$ comes in at the 1-position by steric hindrance. And if you do it in water, you get halohydrin byproducts.

Keep the reaction way from light or you will get radical abstraction and allylic substitution.


How can I make a graphene supercapacitor?

Chemistry World UK have made this YouTube Clip that has a professor from UCLA explaining what is needed and how to go about it - even though the video is about 2.5 minutes, but it is a good demonstration.  The associated article is DVD player burns graphene to disc

Additionally, according to this JensLabs page DIY Graphene, it states that the following readily available materials are needed:

  Graphite oxide
  Light scribe capable DVD burner
  Light scribe DVD
  Substrate
  Sonicator - the article suggests that a ultrasonic cleaner would suffice
  Pipette
  Ion-porous separator
  Electrolyte


Can some one please explain alkyl shifts for me

The reaction of the ketal 1 with aqueous H2SO4 is more likely to follow the pathway 1 --> 2 --> 3 to form α,β-unsaturated ketone 3. This is a typical ketal hydrolysis in acid medium. For the proposed reaction pathway 1 --> 4 --> 5 --> 6, species 6 is unstable. Indeed, if it were formed as an intermediate, it would afford ester 7a and ultimately the carboxylic acid 7b. Is there a possibility that the actual reactant doesn't bear methoxy groups but methyl groups? If this is the case, then the reaction 8 --> 9 is more reasonable.


Role of acetone in the Finkelstein reaction

Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide. $\ce{R-CH2-Cl + KI -> R-CH2-I + KCl}$. The driving force in this reaction is precipitation of $\ce{KCl}$. It is important to find a solvent that would dissolve $\ce{KI}$ better than $\ce{KCl}$. Then solution will always have $[\ce{I-}] > [\ce{Cl-}]$ (more iodide anions than chloride anions). As a result this equilibrium will be driven to $\ce{R-CH2-I + KCl}$. Acetone is exactly this type of a reagent.

For this case Le Chatelier's principle means that reaction will be driven to the direction in which one of the reagents gets outside of the solution. Since $\ce{KCl}$ keeps precipitating the reaction will be driven to the desired product.

This reaction is important because there are many ways to make $\ce{RCH2Cl}$, but making $\ce{R-CH2-I}$ is difficult. Finkelstein reaction allows to convert available $\ce{R-CH2-Cl}$ to $\ce{R-CH2-I}$.


The following answers might be vague or overtly simplified or basic, but the questions by nature are fairly open-ended, aren't they? feel free to inquire further if something isn’t clear enough.

Take $$\ce{2NO2 \to 2NO + O2}$$ You can (extremely basically) think of this reaction as two $\ce{NO2}$ particles colliding, (forming a complex), then coming out as $3$ molecules, $2$ being $\ce{NO}$ and $1$ being $\ce{O2}$. But you needed two $\ce{NO2}$ molecules to collide. Collisions are unlikely, first the two molecules have to find each other, then they have to have enough energy, and be in the right orientation.
I’m not sure about multiple activated complexes, but I would only imagine that the reaction becomes even less likely if more than one activated complex is necessary for it; that basically means you need a second reaction from the first activated complex, which means a second collision. (likely slower rates? Someone else feel free to correct if I’m wrong)
Three molecules colliding at once ($3$ rd order reaction): way less likely, and that’s why it’s not studied so much, it’s much harder getting $3$ molecules to collide (especially in the gas phase without) in the right orientation with enough energy.
By general rate law do you mean finding the constant $k$? Or deriving its form entirely? Finding the orders for each reactant and the constant can only be done experimentally by using different concentrations and seeing what comes out.
I think there is more than one way to derive the general rate law’s form. One way is to take the steady state approximation. It has a useful problem that might explain more clearly. The steady state approximation is a commonly made assumption in various situations. Essentially, it assumes that the intermediate is consumed as soon as it’s produced (so there’s no net accumulation of intermediate – which makes sense for many situations). Using the info gained from this approximation you can use the intermediates and the relations that result to form the general rate law. Of course, it turns out the $k$ is composed of multiple constants.
It’s true because they told you it was. This gives you details that you wouldn’t have necessarily been able to deduce just by having the reaction equation there. Importantly, it’s there to tell you straight up that the equation you have can’t be broken down to further reactions... Again you can think of it like this: $\ce{a A}$ molecules have to collide with $\ce{b B}$ molecules to produce a single activated complex. That’s one step. If they didn’t specify that, maybe there could have been intermediate steps that could change the rate law.
Maybe it'll be easier to see if you consider extremes. Say Reaction $R$ can be divided into $4$ reactions, $r1, r2, r3, r4$. Now, say $r1, r2$, and $r4$ happen in microseconds, while $r3$ takes $100$ seconds. Already you're at the point where you can basically ignore the times of $r1, r2$, and $r4,$ because they're so small compared to the time $r3$ took (comparing $1\mathrm E-6$ to $1\mathrm E+2$ here). So as you can see in this example, $r3$ was literally the rate determining step.

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html


2-Bromo-1-chloro-4-nitrobenzene reacts with sodium ethoxide

The reaction over here is a $\text{S}_{\text{N}}(\text{Ar})$ reaction. Normally, nucleophilic substitutions do not occur, but in presence of strong nucleophiles and electron withdrawing groups, it is possible.

Now, coming to your reaction, let's see what really happens:

Consider the resonation structures of 4-chloro-3-bromonitrobenzene:

You can see that the resonance structures produce a carbocation at the chlorine attached carbon, but not at the bromine attached one. Nucleophiles always tend to attack electrophilic centers, so it is the chlorine that will get replaced.

Continuing with the reaction, we will get:

So, the product would be 3-bromo-4-methoxynitrobenzene.


Of course, electrons do exist. We have a myriads of observations confirming their existence. What does not exist is working classical picture of an electron. Electrons can behave in certain circumstances like waves, in some other - like particles. That is already nonsense from the classical physics point of view with its universal dichotomy: something is either a particle or a wave.

But the trickiest part of the wave-particle duality is that despite the fact that electrons can behave as waves and as particles, their are, strictly speaking, neither of them. Our language developed on the basis of observations of macroscopic surroundings with its well-established classical notions of waves and particles is simply inappropriate for the description of the microscopic phenomena. Thus, whenever one tries to use any language except for that of mathematics to describe microscopic world he/she quickly runs into apparent paradoxes.

With respect to the desire of pictorial classical description of an electron I feel like the only thing I have to do is to quote Paul Dirac:

  […] the main object of physical science is not the provision of
  pictures, but is the formulation of laws governing phenomena and the
  application of these laws to the discovery of new phenomena. If a
  picture exists, so much the better; but whether a picture exists or
  not is a matter of only secondary importance. In the case of atomic
  phenomena no picture can be expected to exist in the usual sense of
  the word 'picture', by which is meant a model functioning essentially
  on classical lines. One may, however, extend the meaning of the word
  'picture' to include any way of looking at the fundamental laws which
  makes their self-consistency obvious. With this extension, one may
  gradually acquire a picture of atomic phenomena by becoming familiar
  with the laws of the quantum theory.


Strong or weak force between protons and neutrons?

I think it is a poor phrasing. The strong force is responsible for keeping quarks together inside a proton or a neutron, but also between quarks belonging to different proton/neutron (but I am not sure for proton-proton interaction because of the repulsive electromagnetic force).

By contrast, the weak force is responsible for turning one type of quarks into another type or quarks, thus transforming a proton into a neutron in the case of beta+ radioactive decay, with emission of a positron (the overall charge is always conserved).


Testing this on version 9 will have to wait until tomorrow, but on version 8 (which I have in front of me, but which has limited differences to version 9 by memory)

1)Rescale the chemical shift axis to focus only on the relevant sections of my spectra.

Do a manual zoom.

->View >Zoom >Manual Zoom (or just type M, or click the manual zoom button)

Type in the values you want

2)Set a threshold so that only peaks above a certain intensity will automatically label.

Do Pick Picking via a Manual Threshold

->Analysis >Peak Picking >Manual Threshold (or just type K)

Left click and hold to define a box within which you will have your peaks picked

3)how on earth do you get the parameter box to fit onto the page? Ive managed to get it up but it is about three times larger than my spectra and rescaling using the little green buttons leaves it the same size but just cuts loads out.

You need to resize the text to a smaller font size

First select the parameters box.
Left click the parameters (which should have the green selected boxes shown) and choose properties. Make sure under the geometry tab, the Auto box is selected for size.
Then from the text toolbar (by default is usually at the bottom of the screen, but if not shown go to >View>Toolbars>Text) change the font size to something small enough to fit on the screen.

Try that for your version - if no luck I'll check on one of my other processing stations tomorrow which has version 9 installed. Let me know how you go.


Peptide bond formation involving side chains of charged amino acids

Equilibrium in aqueous solution

In aqueous solution (such as our digestive tract), the equilibrium of the reaction as written is on the side of the reactants. Digestive enzymes catalyze the hydrolysis of peptide bonds, showing that they are not thermodynamically stable in water. However, in the absence of catalysts at neutral $\mathrm{pH}$, the peptide bond is very stable against hydrolysis, explaining how it is possible for proteins to fulfill their biological role.

Peptide formation in the cell

In the cell, amino acids form a mixed phosphoric acid anhydride by reacting with ATP, which then reacts with hydroxyl groups of tRNA to form an ester. This ester reacts with the growing peptide chain, catalyzed by ribosomes using mRNA as a template to make a specific polypeptide rather than a random one. All steps are enzyme-catalyzed, ensuring that only the main chain carboxylate reacts while the side chain carboxylate does not.

Peptide solid state synthesis

In a peptide synthesizer, amino acid building blocks start out as esters and amides with suitable protecting groups on the side chains. Only when the entire polypeptide has been synthesized are the side chain protecting groups removed. Below is a schematic of the process:

Source: https://www.biosyn.com/images/userfiles/image/general-scheme-solid-phase.jpg

  Why does that extra group not participate in peptide bond formation?

It depends on the context but it does not react because

it is not activated (carboxylic acid in water does not form amide)
it is protected
enzymes are selective for specific functional groups on a molecule

  It should be possible for the $\ce{-COOH}$ group attached to $\ce{C}$3 in aspartic acid and $\ce{C}$4 in Glutamic acid to form a peptide bond with $\ce{−NH2}$ of another amino acid. Do these bonds occur in any peptide linkage? If they do, what is the nomenclature associated with them and how do their chemical properties differ?

For side chains with amino groups, there are cases of peptide formation, for example in ubiquitination reactions where the protein ubiquitin is attached to a lysine via its carboxy-terminus. Other side chains are glycosylated, methylated, or attached to lipids. Together, these reactions are described as post-translational modifications. There is a report of glutamate methylation (https://www.ncbi.nlm.nih.gov/pubmed/16707700).

Some cyclic peptide antibiotics form peptide bonds with the glutamate side chain called gamma peptides (instead of alpha peptides). Here is an example:

Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1176672/?page=7


Ratios of produced gases in water electrolysis

We know from stoichiometry that you did produce twice as many moles of $\ce{H}$ at the negative electrode as $\ce{O}$ at the positive electrode. However, we don't know a few important things:

What were the actual volumes of the gases that were finally collected? Volume is notoriously hard to estimate by eye, unless you are comparing the levels in two identical containers. So, we'll take your word for it that it was "more", but we won't place any bets on the factor of 10.
What happened to the oxygen and hydrogen after the water split? Did it react?
What else was in the water that could have reacted? We know there is $\ce{NaCl}$, but maybe there are other dissolved compounds as well?

Question 1 can't really be answered here - the best we can do for now is assume that the volume of hydrogen was 2-10 times larger than that of oxygen. To find out for sure, you would need to conduct another experiment using graduated collection vessels.
Question 2 we can partially answer. If the electrode was corroded, then something reacted - either chlorine, oxygen, or some other more complicated set of reactions. We can get a rough estimate of the volume of oxygen this might consume by assuming it was oxidized by oxygen. If that 0.1 g of the electrode was oxidized, and that it was oxidized into $\ce{Al(OH)3}$, then at 1 atm and 298 K it would consume roughly 137 mL of $\ce{O2}$:
$\ce{Al^3+ + 3OH^- -> Al(OH)3}$
$0.1 \space \rm{g \space Al} * \frac{1\rm{\space mol \space Al}}{27\rm{\space g \space Al}} * \frac{3 \space mol \space O}{1 \space mole \space Al} * \frac{1 \space mol \space \ce{O2}}{2 \space mol \space O} = 0.0056 \space mol \space \ce{O2}$
$PV = nRT$ (oxygen is close enough to ideal under normal lab conditions)
$V = \frac{nRT}{P} = \frac{(0.0056 \space \rm{mol} \space \ce{O2})(0.08206 \space L \space atm \space mol^{-1}K^{-1})(298 \space K)}{1 \space \rm{atm}} = 137 \space \rm{mL}$
$\ce{Al(OH)3}$ is a likely candidate for a reaction product because of the white flakes you observed - aluminum hydroxide is insoluble in water. Aluminum oxide ($\ce{Al2O3}$) is also a potential candidate, and would consume 68 mL of $\ce{O2}$ under the same conditions.
So, assuming 0.1 g is a reasonable estimate for the mass of corroded aluminum, it seems very possible that what happened is the oxygen reacted with the aluminum electrode, reducing your yield of oxygen gas.
How can you get around this? Well, my five minutes of google research did not turn up any good candidates for inert electrodes. Usually graphite electrodes are recommended for other electrolysis setups, but they will react with oxygen as well. I have seen stainless steel recommended, but I would think that gold/platinum/silver are probably your best bets.
Question 3 is also hard to answer. We know that the $\ce{NaCl}$ will react and produce hydroxide ion (further supporting our hypothesis for aluminum hydroxide formation) at low concentrations, and chlorine gas at higher concentrations. If it were producing chlorine gas, you would know - it would be a pale yellow/green and it would make you sick as soon as you smelled it. So it is unlikely that your concentration of table salt was high enough for that to be the competing reaction. However, it would contribute to the aluminum oxidation. The easiest way to rule this out is to not use table salt. I have seen baking soda (sodium bicarbonate) recommended instead, and that would probably work. However, it might still produce hydroxide in solution. What about other unknown substances? If you used tap water, it is likely that there is additional chloride, possibly fluoride, and probably various magnesium and calcium salts. All of these would react during the hydrolysis, and would affect yields. The way to eliminate them is to buy or make distilled or deionized water.
Summary:

Why did I get less gas than expected at the electrode connected to the +?
Was the oxygen consumed by oxidising the electrode? (It appeared quite damaged at the end.)

The answer is most likely yes, the oxygen reacted with the aluminum. I would bet that it formed aluminum hydroxide, which then precipitated into the white flakes you observed.

How can I get a ratio of 1/2?

You might not be able to get it to exactly 1:2, but if you use sodium bicarbonate or a weak acid (vinegar would probably work) instead of salt as an electrolyte, and an inert electrode - like gold, or platinum, you can probably get close. Keep in mind - hydrogen gas is usually the desired product in hydrolysis. Oxygen is cheap if you don't need it to be pure! And even pure oxygen is relatively cheap compared to other gases.


As water evaporates, the vapor pressure above the more concentrated urea solution drops. Either a better vacuum pump is needed, or perhaps gentle warming, to extract more water.

You might also try chilling to (partially) separate the solution. At ~260 K, achieved in many home refrigerators, much of the urea will have crystallized from the remaining saturated solution... just don't mix this up with food in the freezer.


Ageing is a disease.  Just as treatments have been found for other diseases, ageing can be slowed or reversed as well.

Not that long ago, scientists noticed that at the end of each DNA strand there were thousands of nucleotides.  Nucleotides are used for genetic coding.  That is to say, certain sequences of nucleotides instruct the body to synthesize  certain proteins when the genetic code in the DNA is being read.  DNA is really like a massive software program.

In any case, when scientists noticed all of this code at the end of a DNA strand, at first they couldn't figure out what is was for, so they thought it must be gibberish.  Later they noticed that everytime a cell divided (e.g. everytime the DNA was read) a piece of this gibberish was lopped off.  Eventually it became understood that this "gibberish" was the basis of aging.

Each lopped off piece was termed a telomere.  Each telomere contains the same nucleotide sequence (code).  There are about 15,000 telomere nucleotides at the end of the DNA strands in an embryo.  This has been shortened to about 10,000 when you are born and when it reaches a length around 5,000, the body dies.  It turns out that the telomeres play a role in stabilizing DNA strands during the copying process, when their length gets down to around 5,000 nucleotides the telomere length becomes to short to provide adequate stabilization and errors (genetic mutations) occur in the copying process and/or the cells become senescent (become inactive, go to sleep).  Errors or sleep, either way, the body stops functioning efficiently.

This telomere shortening occurs during cell division in all human cells except for sperm and egg cells.  Sperm and egg cells are immortal.  There is an enzyme called telomerase, it is expressed only in the sperm and egg cells.  After these cells divide a telomere is lopped off, but telomerase re-adds a telomere to the DNA in sperm and egg cells.  Hence, their telomere length never shortens - they live forever.

All cells contain the code to make telomerase, but that code  (or signaling) is turned off in all cells except for sperm and egg cells.  A specific protein combines with a certain portion of DNA and that is all that is required to prevent telomerase expression in a cell.

If scientists could find a way to prevent that (repressor) protein from complexing with DNA, then telomerase expression could be turned on and that cell would then be immortal.

As you can imagine, this is a big area of research.  Here is a link to a company that has been active in this area for a long time, and here is a link to a basic tutorial on the subject of telomeres.  Poke around the web site, the videos and other tutorials are extremely interesting.  Even aspects such as the social consequences of curing aging are discussed.

Enough progress has been made that you can buy products in the over-the-counter market that reduce telomere shortening, but they're very expensive at the present time.


Why does the structure of haemoglobin change when the ferrous ion changes to ferric ion?

The $\ce{Fe^{II}}$ ion is held in the centre of the heme's porphyrin ring and is also attached to a nitrogen atom on a nearly histidine group which is part of the protein. When there is no oxygen present the Fe ion is in a high spin state. The Fe has two electrons in the $e_g$ orbital and is 5 coordinate. The other 4 bonds are to the porphyrin N atoms making the geometry square pyramidal. In this state the Fe is just a little too big to fit into the porphyrin ring which becomes a distorted, bending towards the histidine. The Fe is 0.07 - 0.08 nm out of the plane and the Fe-N distance 0.22 nm (to the imidazole N on histidine F8). When an oxygen molecule binds the Fe becomes low spin, (the $e_g$ orbitals are now empty) and 6 coordinate, and  has a slightly smaller radius and so fits into the porphyrin ring which becomes planar. As the Fe has moved on binding oxygen, the histidine is also moved and is pulled towards the porphyrin ring by $\approx$ 0.075 nm. This motion is transmitted and amplified through the protein to the other three binding sites which affects their ability to bind. This is called an 'allosteric' interaction as it is 'non-contact' between the sites.
The picture (b) is self explanatory. The structure (a) shows the protein backbone as cylinders and the porphyrins and Fe at atoms. The Fe is yellow.

(Picture from 'Protein Physics' A. Finkelstein & O. Ptitsyn, publ. Academic Press.)


Ionic compound having highest solubility in water

Caesium salts are unapologetically ionic, and they typically have quite high mass solubilities in many solvents, including water.

Assuming organic ions are allowed, caesium acetate ($\ce{H3CCO2^-Cs+}$) in particular has a remarkably high solubility of 9451 g/kg water at −2.5 °C, increasing to 13 455 g/L water at 88.5 °C.

Caesium formate ($\ce{HCO2^-Cs+}$) is also quite soluble, with a solubility of 4880 g/kg water at 20 °C, resulting in 2.56 L of solution with a density of 2.297 g/mL (reference, .docx file). However, its solubility increases much faster with increasing temperature (J. Chem. Soc., Trans., 1922, 121, 1837-1843). At 100 °C, it reaches an outstanding value of 20 071 g/kg water! That's 11.6 molal, or roughly 20-25 mol/L assuming the density doesn't change too much. This saturated solution is 67.7% cesium formate by number of moles, which means more than two caesium ions and two formate ions per molecule of water.

I believe I have read somewhere that caesium formate is the record holder for highest mass solubility in water (evidently only at high temperatures). If this is not true, then I can scarcely believe it will be topped by much.

Tangentially, Ivan mentions Clerici's solution, which is actually a mixture of thallium(I) formate and thallium(I) malonate. The mixture doesn't count (though the individual components are quite soluble themselves), but it interesting to analyse. Apparently 300 g of each compound will dissolve in 40 g of water without saturating it room temperature (ref), giving a lower bound to their combined solubility of 15 000 g/kg water. This value rises with heating, and is the only way I can see to beat the mass solubility of caesium formate.

For further entertainment, I recommend these two solubility tables with a large number of entries (1, 2). The second one can be ordered by solubility at different temperatures. It's interesting to see the variety of cations and anions which can be combined to display extreme mass solubility.

Edit: If room-temperature ionic liquids are allowed, then it is quite likely some of them are miscible with water in any proportion, which is effectively "infinite solubility in water". Something as simple as ethylammonium nitrate ($\ce{C2H5NH3+NO3-}$) may suffice.


Equivalent mass in case of disproportionation reaction

I like and dislike the question. I assume this is from your Indian university entrance exams. The good part is that it is tricky, the sad part is that the concept of equivalents is so outdated that any North American student will barely recognize it. They should abolish teaching equivalents exactly because this problem you are encountering.

Fact number no. 1, unlike molar equivalents of a given molecule equivalent weights are reaction dependent. Hence addition of equivalent weights is irrelevant. Equivalent weights are useful in analytical chemistry because they simplify calculations. The case of disproportionation reactions and their equivalents is rarely discussed because nobody uses disproportionation reactions in titrations, gravimetry etc.

In your case, you will have two equivalent weights one with respect to oxidation, and one respect to reduction. Solve the disporptionation reaction below and see what you get as an equivalent weight for $\ce{Br2}$ in each case.

$\ce{3Br2 + 6OH^- -> 5Br^- + BrO3^- + 3H2O}$


Why is a volumetric flask shaped the way it is?

A volumetric flask is designed to store a relatively great volume of liquid with great accuracy. If you want to reach a determinate volume of liquid in a large backer when you add liquid to it so the liquid height increases a little bit the volume increase a lot. What you measure with your eyes for determining the volume is the height of the liquid, so the error in the volume measure is directly linked with the error in measuring the liquid height. The relation between the increase of the height ($\Delta h$) and the volume for a cylinder is very simple:

$$ V= \pi r^{2} \Delta h  $$

So suppose that you make a systematic error $\Delta h$ of 0.5 mm, that is quite a good esteem of average max human accuracy in measuring the liquid height in a cylinder with naked eyes. If you make this error in glassware with different radius you will change the accuracy of your measurements, according to the previous equation:

So if you use ten times greater radius you will have a 100 times greater error!

Why the body has not the conical shape of the Erlenmeyer?

This actually is not so important. Generally, the spherical surface more evenly distributes stress across its surface and is less prone to fractures. This is why we use round-bottom flasks. It doesn't seem there are a lot of differences in the production of both Erlenmeyer and volumetric flask, so it's probably mainly due to a convention. In fact, you can find also volumetric flask with Erlenmeyer style. Like in the image below.


Chemical reaction of radicals with different multiplicity

With respect to the question "if a chemical reaction is possible or not", here is some background material on oxygen species.
In particular, Singlet oxygen apparently reacts with oxygen (also referred to as dioxygen) creating a very strong (but transient) oxidizing atomic oxygen.
Examples of possible reactions of interest, including atomic oxygen, can be found, per Equations (1) to (5), from a work discussing graphene patterning here: http://olab.physics.sjtu.edu.cn/papers/2017/29.Huan%20Yue_PCCP_2017.pdf , where apparently, O(3P) is created from severe collision quenching of O(1D) atom with air or oxygen and acts as the major oxidant in the work by Huan Yue and colleagues ‘Exploring the working mechanism of graphene patterning by magnetic-assisted UV ozonation’.
Note, O(3P) is also known as highly reactive ground-state 3P oxygen and a form of atomic oxygen.
Also be mindful, in the real world if any water vapor is present, Singlet oxygen reacts with H2O forming H2O2, where O(1D) has a half-life of around 45 minutes.


What is the hybridization of oxygen in furan?

Furan is aromatic because one of the lone pairs of electrons on the oxygen atom is delocalized into the ring, creating a $4n+2$ Hückel aromatic system.

The steric number is sometimes incorrectly described as the sum of the number of sigma bonds and the number of lone pairs. However, it is more accurately given by the number of sigma bonds plus the number of localised lone pairs.


Role of intermediate in determining the exothermic or endothermic nature of reaction

A typical method for determining whether a reaction is endo or exothermic is to compare the heats of formation ($\Delta_f H$) of the products to those of the reactants. If we do not have measured values for the enthalpies of formation, we can get a fairly good estimate by summing up typical values for the bond enthalpies of the bonds in the compounds.
When using bond enthalpies, we can simplify the calculation by considering only those bonds that are broken or formed during the reaction, since the other bonds appear in both the products and reactants and so cancel out when we look at the difference.
Thus,
$$\textrm{Enthalpy of reaction = }\Delta_r H = \textrm{[sum of enthalpies of bonds broken]} - \textrm{[sum of enthalpies of bonds formed]}$$
[Here I'm using the standard convention that bond enthalpies are reported for the breaking of bonds and so are positive values.]
Comparing the four reactions given, we can simplify even further by noting that all four reactions involve breaking of a $\ce{H3C-Cl}$ bond and formation of a $\ce{H3C-H}$ bond. Since those occur in all four reactions, we can ignore them for the purpose of comparison.
We now have reduced the problem to a simple comparison of the change of an $\ce{R-H}$ bond to an $\ce{R-Cl}$ bond.
You are correct that it does not matter what intermediates might occur on the reaction path. The radical stability is relevent, however, because the bond enthalpy is defined as the change in enthalpy needed for homolytic bond cleavage, ie
$\ce{A-B -> A. + B.}$
Since the products are radicals, the stability of the radicals gives us information about the bond enthalpy. It is for this reason that comparing the stability of radical "intermediates" is a way to tell which of your given reactions is not exothermic.


Why does this protein seem to have multiple chains?

Some parts a the protein are a bit more flexible, also in the protein crystal. These parts don't give clear electron density after x-ray diffraction, and are removed from the resulting model structure. Often the N- and C-terminal ends are removed, this is something you'll find in many .pdb structures, but sometimes also loops are removed, resulting in two extra loose ends.

The authors were nice to note in the comments section of the pdb file exactly which residues were removed from the model (#465).

There might be tools available to model loops into the structure so you'll have one continuous chain in the .pdb file, but I really don't know which one is best (of course they're all completely unreliable).


How to perform Mulliken population analysis by hand with GAMESS output?


No. The 6 membered ring is also a part of its aromaticity.
This can be better understood by considering the Zwitterion of Caffeine.

The entire molecule is planar, conjugated and has 10π electrons(4n+2)
Reference:Slides by Nevina Keskineva


How much ionising radiation is released during an atomic blast?

The amount of energy in an atomic blast releases as radiation (gamma rays, beta particles, alpha particles and neutrons) depends on how the bond is designed but isn't usually a big proportion of the total energy released. But that's not why we worry about the radiation from a bomb.

The reason why we worry about radiation is the radioactive products created in the blast (eg from  fission of the large nuclei in a fission bomb or from the potentially radioactive nuclei created when the initial radiation blast interacts with other materials). Typically many dangerous isotopes are generated that continue to release radiation for days or years. Worse the explosive blast scatters many of these over a large area.

But not all bombs are the same. Neutron bombs are designed to release far more of their energy as radiation (in this case a large blast of high energy neutrons) with the express purpose of killing people in the immediate area while having a much smaller blast. The original idea was to be able to use them on invading Russians in Western Europe without doing too much damage to the infrastructure. Neutron bombs can release ~40% of their energy as neutron radiation which is far more than the 5% released in a bomb tuned to create a large physical explosion.


Precise estimation of air properties as a function of temperature at 1 atm

Here are some links that look like they might be useful, there seems to be promising information on some engineering resource sites:

https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470516430.app2
This is the appendix from what appears to be an engineering textbook, it gives a chart of the information you're looking for at 1 atm and 15 C. It also walks you through the temperature dependent equations for some of the properties.
https://www.me.psu.edu/cimbala/me433/Links/Table_A_9_CC_Properties_of_Air.pdf
This is a table of the properties of air at 1 atm from -150C to 2000C. It doesn't provide equations but you could maybe enter the data into excel and run a regression to extract a an equation.
https://irc.wisc.edu/properties/ This is a handy little calculator that outputs a whole host of properties of fluids, including dry air.

The point that has been made about water vapor fluctuations is valid, but most calculations of air properties assume dry air with constant composition.


The standard for equilibrium constants of gas phase substances is to use $K_p$, so in this case $p^\circ$ would be more appropriate than $c^\circ$ for the organics as well as the hydrogen. If you use $c^\circ$, you should use it for all gas phase components and indicate your $K$ as $K_c$. Using different standards for gas phase components in the same reaction is unorthodox.
And if you use your equilibrium constant for calculation of $\Delta G$ for example, the simple form of the equation $\Delta G^\circ = -RT\ln K$ requires that the $K$ be $K_p$ for gas phase reactions. Using $K_c$ will give an incorrect result.


What causes the "Gd break" in the trend of lanthanide-EDTA formation constants?

If you consider the solvent extraction of lanthanides by reagents such as H-DEHPA, then to a first approximation the distribution ratio is given by the equation.

$$\ce{DLn}= k[\ce{H-DEHPA}]^3[\ce{H+}]^{-3}$$

If you plot k as a function of atomic number for DEHPA dissolved in most organic solvents (solvent extraction term is diluent) then the value goes up as a function of Z. But the line has bumps on it.

The reason is that the extraction reaction is

$$\ce{Ln(H2O)_n^3+(aq) + 3H-DEHPA(org) -> [Ln(DEHPA)3](org) + H^+(aq)}$$

As you go from one side of the lanthanide group to the other the binding of the lanthanide to the anionic DEHPA ligands is likely to become stronger, but the value of $n$ changes in several steps. This will give bumps on the graph of $K$ for both the extraction of the lanthanides and also for their binding to ligands such as EDTA.


What is the minimum mass of iron(III) chloride required to be added until the dark blue precipitate first appear?

Yes that is right.

However I would not round intermediate results to avoid round off errors. In fact I'd carry two extra significant figures for intermediate calculations and only round off final answer. (Easy now with a calculator, not so easy to do when I was doing these problems with a slide rule...)  So I'd get Fe as $1.3478\times10^{-8}$ molar. Using that number I get 4.3749893530170255598967635197789e-7 as the final answer which rounds to 4.37e-7.

The problem's answer is really wrong since the concentration of the hexacyanoferrate(II) is only given to two significant figures. So the answer shouldn't have more than two either. If the problem wanted three significant figures then the hexacyanoferrate(II) should have been specified as 0.00100 molar.

You have to realize that significant figures are a crude way to do error propagation. So two significant figures for 0.0010 molar means +/- 0.00005 molar, so a +/-5% specification. However if the answer to two significant figures is 9.9 then that means +/- 0.05 which only has an error of +/-0.5%. So although crude using significant figures keeps you from stupidly assuming that all the figures in the calculator's answer, like 4.3749893530170255598967635197789e-7, are significant.


Decomposition reaction of tin(II) nitrate

No wonder you cannot find any literature on $\ce{Sn(NO3)2}$, because I don't think it exists in solid form. This is supported by the fact that I even can't find its CAS number online. This book (Ref.1) and relevant paper (Ref.2) support my suggestion, both of which state that:

  Attempts to prepare a covalently bound tin(II) nitrate, by the reaction of tin(IV) tetranitrate with anhydrous nitric oxide, have produced only a white solid solid of formula $\ce{SnN2O6}$, which gives a tin(IV) Mössbauer resonance ($\delta = \pu{0.29 mm\:s-1}$, $\Delta = \pu{0.96 mm\:s-1}$)...

Nonetheless, a study on thermal decomposition of metal nitrates (Ref.3) states that:

  Due to a back-donation of electronic cloud from the nitrate to an unfilled $\mathrm{d}$-orbital of transition and noble metals, their nitrates generally exhibited lower decomposition temperatures ($T_d \lt \pu{700 K}$) than those of the base metals ($\gt \pu{850 K}$).

They suggest those metal nitrates with lower decomposition temperatures decompose to corresponding oxides together with $\ce{NO2}$ and $\ce{O2}$. Thus, it is safe to assume that if $\ce{Sn(NO3)2}$ exists, it would decompose as following reaction suggest:

$$\ce{2 Sn(NO3)2 (s) + heat -> 2 SnO (s) + 4 NO2 (g) + O2 (g)}$$

However, keep in mind that this is pure speculation.

References:

R. Greatrex, “Chapter 8: Mössbauer Spectroscopy,” In Specialist Periodical Reports: Spectroscopic Properties of Inorganic and Organometallic Compounds, Volume 6; N. N. Greenwood, Ed.; The Chemical Society Burlington House: London, United Kingdom, 1973, pp. 494-622.
P. G. Harrison, M. I. Khalil, N.Logan, “Concerning anhydrous tin(II) nitrate,” Inorganic and Nuclear Chemistry Letters 1972, 8(6), 551-553 (https://doi.org/10.1016/0020-1650(72)80139-9).
Shanmugam Yuvaraj, Lin Fan-Yuan, Chang Tsong-Huei, Yeh Chuin-Tih, “Thermal Decomposition of Metal Nitrates in Air and Hydrogen Environments,” J. Phys. Chem. B 2003, 107(4), 1044-1047 (https://doi.org/10.1021/jp026961c).


Diluting 4 % NaOCl to get a 1 % solution

There is a simple dilution formula: $\mathrm{C_iV_i=C_fV_f}$, which is valid for all concentration units. The "i" and "f" indicate initial and final concentrations or volumes.

Using this formula, you can see that you don't need to prepare buckets full of 1% bleach. All you need to decide is what should be your final volume and final concentration.

As you highlighted, the issue, you cannot prepare a 1% solution and expect it to last for a day. It is better to prepare on need basis. The stabilizer in bleach is sodium hydroxide itself. Upon dilution, of course the base is also diluted and the pH is reduced.

One can look at the distribution diagram of "bleach" as a function of pH (yes this is from a Handbook of Surfactant. It is a 4000 paged volume- books are still useful)


I cannot understand how can degree of hydrolysis be very small
  compared to 1 when $\ce{NH4+}$ is a strong acid.

That is wrong.

You also wrote:

  It is a salt of a strong acid and a weak base.

Well, the salt of a strong acid and a weak base, i.e. the conjugated acid of that weak base is a weak acid.

In the case of the ammonium ion, $pK_a=9.25$, very weak indeed.

Also, the hydolysis reaction equilibrium is (because free $\ce{H+}$ don't exist in water and $\ce{NH4OH}$ is old notation, long obsolete):

$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$

$h$ is negligibly small because this equilibrium leans very strongly to the left, so $[\ce{H3O+}]\approx 0$.


How do I call the functional group of this alkane?

Under the 2004 provisional recommendations by IUPAC, one possible systematic name is 4-(2-methylpropan-2-yl)octane, which indicates that the longest chain in the substituent is three carbons long, is connected to the main carbon chain via the second carbon (-2-yl), and has a methyl group on the second carbon (2-methyl).

4-(1,1-dimethylethyl)octane is also possible, and indicates that there is an ethyl substituent on the fourth carbon in octane, which is further substituted with two methyl groups on the first carbon where it branches off (see section P-29.4).

However, the preferred IUPAC name is 4-tert-butyloctane; the tert-butyl group is used to describe a $\ce{-C(CH3)3}$ group and is preferred over either of the two above systematic names (see section P-56.2.2.2).

I don't have access to the 2013 recommendations but I believe it should not be different.


What are the molecular orbitals of the hypothetical linear H₃⁺ molecule?

The lowest energy MO is : $\psi_1 = (1/2)(\phi_A+ \sqrt2\phi_B+\phi_C)$.

Then comes the MO: $\psi_2 = (1/\sqrt2)(\phi_A- \phi_C)$.

The highest energy MO is  $\psi_3 = (1/2)(-\phi_A+ \sqrt2\phi_B-\phi_C)$.

Where  $\phi$ denotes the atomic orbital $1s$ on each hydrogen atom. $A$, $B$ and  $C$ denote hydrogen atoms, where  $B$ is the central one.


Does gold always become gold chloride when dissolved with HCl and an oxidizer and can it always be precipitated with sodium metabisulfate?

You may find to your dismay that getting dissolved gold separate from dissolved copper and nickel is dicey (to give you some idea, I still have a dissolved solution of gold and copper sitting in my garage ca. 2 years ago because I hesitate to mess with it; also it looks kind of cool).

You should use urea to neutralize first, just going on experience. I've done the same procedure you describe here to try to recover gold from RAM cards etc. Add urea slowly and check pH regularly. I wouldn't try sodium bicarbonate only because I've never done that and it seems you ought to go with the most well-used method if you're risking a big loss.

Sodium metabisulfate is indeed a common reagent for precipitating the gold. It smells pretty terrible before and after adding it (because of sulfur content). I think $\ce{SO2}$ gas is released when you react with gold so you should have a well-ventilated area for this. You should get a brownish precipitate. I would keep adding it to the neutralized solution until no more precipitate is formed, then let it settle.

This point is probably your best hope for separation of the gold. Recover the precipitate by filtering and leave off as much of the fluid as possible (wash with DI $\ce{H2O}$, and melt the brown salt to remove metabisulfate and get the pure metal.

In the future I would not use any oxidizer when stripping the gold from $\ce{Cu}$/$\ce{Ni}$. I found that a much better method is to stir/air-bubble the material in $\ce{CuCl2}$ which can be made from pure copper, $\ce{HCl}$, and careful additions of $\ce{H2O2}$. It will take longer but it is much safer. You'll recover solid bits of gold instead of dissolving it undesirably. i.e.

$\ce{CuCl2}$ > $\ce{HCl} + \ce{H2O2}$ > aqua regia

A lot of this procedure is difficult to improve because (even academic) literature is nebulous on the topic. I once bought a ~$250 book on the topic. "The Properties of Electrodeposited Metals and Alloys: A Handbook" by Safranek. It's hard to get a hold of.

P.S. it does not form gold chloride when dissolved in aqua regia ($\ce{HCl}$ alone won't effectively dissolve it). It forms a complex with $\ce{Cl-}$, chloroauric acid $\ce{HAuCl4}$.


What are the differences between D and L glucopyranose?

They are mirror images of one another and therefore their chemical and physical properties are the same, except for the deflection of polarized light, which will be in opposite directions.

Mirror image compounds that are not the same are called enantiomers and it does not matter if they are in the open or cyclic form.
They will, however, have different properties if they are in a chiral medium.
For instance, if placed in a chiral solvent, they will have different solubilities.
Chiral means that the compound has a mirror image that can not be superimposed on the original structure.


What can be used as an electrolyte in water to perform electrolysis other than salt?

You can actually just use salt. It is because that the chemical reaction on the top is for one condition which is that your halide solution ($\ce{NaCl}$) has a very high concentration.

If you halide solution has lower concentration,the hydrogen ion and hydroxide ion in the water will discharged and the two half equation will be

$\ce{4H^+ + 4e^- -> 2H_2}$

$\ce{4OH^- -> 2H_2O + O_2 + 4e^-}$

So, the overall equation is

$\ce{4H^+ + 4OH^- -> 2H_2 + 2H_2O + O_2}$

Thus, is okay to use salt.

Besides Salt

You can also use hydrochloric acid besides salt. Just remember that don't use a high concentration hydrochloric acid as the anode will just discharge the chloride ion instead of hydroxide ion.


How many electrons are transferred in this redox reaction?

The carbon atom must have an oxidation state (O.S.) of -4 on the left hand side since each hydrogen bonded to it must have an O.S. of +1. On the right hand side, the carbon atom has an O.S. of +4 since each oxygen bonded to it has an O.S. of -2. The carbon's oxidation state increased by 8. This is a hint that 8 electrons were transferred. To verify, examine the oxygen atoms. On the left, we only have free oxygen, therefore the O.S. is 0. On the right, we have 4 total oxygen atoms, each with an O.S. of -2. Thus our net change in oxidation state is 0 - 8 = -8. We can almost conclude that all these electrons were transferred from carbon, but we must check the oxidation states of the hydrogen atoms. On both sides, each H atom has an O.S. of +1, so there is no net O.S. change. Thus we can confirm that 8 electrons were indeed transferred in this reaction.


There are lots, and I mean lots of methods and software programs to produce partial charges. See Wikipedia for a small (incomplete list)

Let's start with the basics. The idea of a partial atomic charge, while useful for concept, can not be defined uniquely. Quantum chemical methods (whether wavefunction or DFT) produce some sort of electron density. Different schemes divide up that electron density in different ways.

I like Cramer and Truhlar's categories:

Class I: Some sort of intuitive or empirical approach (i.e., non-QM). These include methods based on experimental dipole moments and atomic electronegativities, including Gasteiger-Marsili partial charges and other "electronegativity equalization methods" (EEM).
Class II: Partitioning using wave functions / orbital schemes (including Mulliken charges while very efficient to compute, are basis set dependent.
Class III: Partitioning the electron density or electrostatic potential (e.g., Hirshfeld, density-fitting schemes, etc.
Class IV: Semiempirical mapping from Class II or Class III (preferred) to match experimental data like dipole moments.

Almost any modern quantum package will include several schemes to fit partial charges. My personal preference is for something like Hirshfeld charges (i.e., from the electron density, so not basis-set dependent), or electrostatic potential fitting schemes like Merz-Kollman or CHelpG.

I think the latter (electrostatic potential fitting) will be better generally for molecular dynamics, since you're attempting to produce point charges that will best represent the quantum mechanical electrostatic potential.


What is exposed in the surface tension of water?

Surface tension may be described regardless if all / some / none of the water molecules expose one or both hydrogen atoms or the oxygen atom toward air, even if there are studies in this field (e.g., example, example).
Key points however are

the movement (translation, vibration) of the water molecules in the bulk of water droplet in restained, because of intermolecular attractive forces (e.g., hydrogen bonds), as well as intermolecular repulsion (molecules normally do not come closer to each other than the sum of the corresponding van der Waals radii).  Like with humans, there is an optimal intermolecular distance where overall the energy passes a minimum.
In extension of your drawing of the situation in the water droplet, I add the following figure where molecules are reduced to a black sphere; these are either in the inner of the droplet (light blue background), or at the interface (dark blue background) exposed to air.  The black arrows indicate intermolecular attractive forces (responsible for cohesion).

(derived from wikipedia)

For a molecule in the inner of the droplet, surrounded completely by molecules of the same type, the net force on this molecule is zero.

For a molecule at the interface water / air, the attractive forces to molecules of water in the inner of the water droplet exceeds the attractive forces between this water molecule and air molecules.  Thus, there is a net force pointing to the inner of the water droplet; this is symbolized by the red arrows.
Because of this inward directed force, the droplet thus tends to contract, building an inner pressure until compensated by the earlier mentioned intermolecular repulsion.

Overall, the lowest energy of the water droplet would be achieved as a sphere.  Sometimes, this is shown in microgravity (Scott Kelly at 8:28min).  On earth, however, things like gravity, contact with a surface (adhesion), relative air speed (e.g., if a drop falls down the rain gutter) etc. lead to deviations from this ideal.  However no, because the molecules in the inner of the water droplet are the same as the ones at the interface exposed to the air, there is no difference in the intermolecular, pairwise attractive forces between molecules of same type.