Reconcile the Beer Lambert Law with rate of absorption

To answer your second question first, just use the Beer-lambert law. The first question needs us to go back a few steps and re-define the Einstein B coefficients in terms of the intensity which means that the 'new' B or $B^I=B^{\rho}/c$ and from now on for simplicity we just use B instead of $B^I$.

The intensity I is defined (per unit frequency interval) as flux across unit area per unit time thus has units  $\pu{W/m^2.s^{-1}}$ or mass time$^{-2}$ and thus is not a rate. The energy density $\rho$ means density of energy per unit volume and is related to I as $I=c\rho$ where c is the speed of light. (The rate of absorption is unchanged but written as $B^II$ rather than $B^{\rho}\rho$.)

The energy density $\rho$ has units $\pu{J.m^{-3}/s^{-1}}$ which is mass length$^{-1}$ time$^{-1}$. In these units the Einstein B coefficient has dimensions time mass$^{-1}$ or $\pu{s kg^{-1}}$ in SI units.

The Beer Lambert law derives from the differential form
$$-dI_{\nu} =I_{\nu}k_{\nu}dx$$ where for clarity  x is used to indicate length rather than l. The constant $k_{\nu} = \epsilon [c]$ at frequency $\nu$ and has dimensions of length$^{-1}$.

If the absorption band is wide we take a small element about frequency $\nu$, which is $\delta\nu$,  and calculate the rate of energy removal from the beam during absorption from state 1 to state 2 with populations $N_1$ and $N_2$ respectively, with Einstein coefficients for (stimulated) absorption $B_{12}$ and stimulated emission $B_{21}$
$$ -dI_{\nu}\delta\nu =  h\nu I_{\nu} [N_1B_{12}dx -N_2B_{21}dx  ]$$
rearranging gives
$$-\frac{1}{I_{\nu}}\frac{dI_{\nu}}{dx}\delta\nu=h\nu[N_1B_{12} -N_2B_{21}  ]=k_{\nu}\delta\nu $$

If the light intensity is low, the population of the upper level is only a minute fraction of that of the lower level and can be set to zero. In this case B can be obtained directly as $$B_{12}=\frac{k_{\nu}\delta\nu}{h\nu N_1}$$

Then integration over $\delta\nu$ gives the B value for the absorption band. The value of $N_1$ is effectively the total population.

This shows that the B values are related to the extinction coefficients as expected and that time does not come into the calculation. However, this is only part of the story.

A quantum calculation starting with the Schroedinger equation and using the dipole approximation and first order perturbation theory is necessary to explain the electric dipole interaction. The result of this calculation (see Atkins & Friedman Molecular Quantum Mechanics chapter 6 for gory details) is that the probability of absorption from state a to b is
$$P_{ab}= \frac{|V_{ba}|^2}{\hbar^2}\frac{\sin^2((\omega-\omega_{ab})t/2)}{(\omega-\omega_{ab})^2}$$
where $V_{ab}= <a|H|b>$ is the interaction energy, $\omega$ the radiation frequency and $\omega_{ab} = (E_a-E_b)\hbar$. This oscillating behaviour as a function of frequency is not ordinarily observed in thermal samples but has been observed in a cold molecular beam of HCN molecules. Time was made constant by passing the beam at constant speed through the fixed illuminated region (see Dyke et al. J. Chem. Phys. vol 57, 2277, 1972).

The rate of absorption is the time derivative of the probability which is
$$ R=\frac{|V_{ba}|^2}{2\hbar^2}\frac{\sin((\omega-\omega_{ab})t)}{\omega-\omega_{ab}}$$

and as the the absorption coefficient is directly proportional to the rate of absorption  $\gamma(\omega,t)=\beta R$ (where $\beta$ only contains constants) which indicates that near resonance absorption should increase in time. However, this is not what is usually observed for molecules at room temperature as a vapour or in solution. The reason for the generally observed behaviour is that the absorption is continually interrupted by first order processes, for example by collisions.

Suppose that the system is initially in state a and then the radiation field is turned on. The molecule is then in a superposition state involving a and b, for example $\Psi=c_a(t)\psi_a+c_b(t)\psi_b$ where the c are time dependent coefficients that move the system from a to b. A collision will destroy the superposition and return the molecule to state a or b each with a certain probability, then the radiation causes the superposition to grow again.

If we assume that the number of superposition states is n and that a first order process destroys them with lifetime $\tau$ , then $n(t)=n(0)\exp(-t/\tau)$. There is now a competition between the superposition state growing and being destroyed. The average absorption coefficient is obtained by averaging over the probability distribution of it being interrupted thus

$$\gamma(\omega)=\frac{\int_0^{\infty}\gamma(\omega,t)\exp(-t/\tau)dt}{\int_0^{\infty}\exp(-t/\tau)dt}$$
evaluating gives
$$\gamma(\omega) \propto  L(\omega_{ab}-\omega)$$
where L is the Lorenzian function
$$L(\omega_{ab}-\omega)=\frac{1}{\pi }\frac{\tau^{-1}}{\tau^{-2}+(\omega_{ab}-\omega)^2 }  $$
and examining the function the half-width  at half-height $\Delta \omega = 1/\tau$. The first order process thus explains why the absorption does not increase in time.  However, the experimentally line shapes on many molecules do not have a Lorenzian shape and the reasons for this are not that the idea of interrupting the superposition is wrong but that other effects dominate, doppler broadening for example, or in solution there is always inhomogeneous broadening. This is caused by interactions with solvent molecules which shift energy levels up and down and thus the solution effectively contains molecules with a mixture of energy levels. That this is so can be observed at very low temperatures (say 10K) where it is possible to burn a spectral hole in a sample and so remove a small part of the inhomogeneous population. The spectral hole produced is very narrow as expected from the calculation above.


Wavelengths of colors emitted by alkali metals during flame test

The NCERT textbook is partially wrong for elements after sodium. The flame colors and wavelengths do not match. Keep in mind that for alkali metals, the Bunsen burner flame shows multiple wavelengths for potassium, rubidium and cesium.  It is very easy to excite their outermost electron in such a low temperature flame to higher energy levels. My point is that the table should mention multiple lines for K, Rb and Cs.
What you textbook is showing is showing the wavelength of the strongest intensity for all elements except Cs.
The strongest lines or the so-called resonance lines for the alkali metals are
Li = 670 nm (falls in the red region)
Na = 589 nm (orange-yellow)
K  = 769 nm (deep deep red)
Rb = 780 nm and 794 nm  (still deep red)
Cs = 852 and 894 nm (near infrared, not visible to human eye)
Now you are right, K flame appears violet to us and Cs is indeed blue. As stated above, potassium salts in Bunsen flame also show
K = 404 nm (deep violet)
Now violet and red wavelengths appear somewhat lilac to us.
The same goes for Cs, the strongest emission is near infrared, but the visible spectrum has a faint line of 455 nm - this is bluish. Hence the flame appears blue. The name Cesium also comes from bluish gray. Rubidium comes from red from its red lines in the spectrum.
The source of wavelengths is a very old book by Grotrian.
The Bunsen flame spectrum of Lithium and Sodium is pure Li and Na spectrum. The sodium line is a doublet. I could not find the experimentally measured emission spectrum of K, Rb, and Cs.


Yes, this is called distance-of-flight mass spectrometry. It offers some advantages for large (e.g. biological) molecules and because of its high speed for identifying multiple targets, in chemical and biological warfare.


What do you call the number of atoms in a molecule?

The "number of atoms"'s name is... the number of atoms. I am not aware of any special word for that. Databases may not let you search/filter that kind of criterion, i.e. the total number of atoms, but usually let you search by "chemical formula" which is, in your case, C2H2O2.

I have just tried with Chemspider by typing "C2H2O2" in the search box and it disappointingly gave me just 3 results, i.e. glyoxal (OHC-CHO, not in your list), 42879-41-4 and 16005-17-7.

DID YOU KNOW? Just to expand the scope, did you know that the last digit in CAS# is actually a check digit? It helps any software checking that you did not mistyped the number.


Range of distance for Van der Waals force

Van der Waals force is due to interaction of two dipoles. It is never strictly zero, but it diminish very fast with distance.

It is common to look at VdW interaction at Van der Waals radius. The largest radius should be around 2.60 for Fr making the total distance between nuclei more than 4 A. Heavier elements will have larger radius, but heavier elements are not stable. Are we really interested in VdW force between two atoms that will decay in 10^-15 sec?

Also, by "4-12A" do you actually mean distance between nuclei or the surfaces of atoms (VdW radius A + VdW radius B + "4-12 Angstrom")? In the first case iodine has VdW = 2.15A, so nuclei interact on distances more than 4A. This is the distance between atoms of neighboring I2 molecules in the crystal. In the second case you will see some interaction, but it will be quite small.


What is the difference between ψ, |ψ|², radial probability, and radial distribution of electrons?

All the four terms are different and they represent different concepts in quantum mechanics.Firstly, the term $\Psi$ represents wave function of a particle which is distributed in a three dimensional space. This wave function is a function of four coordinates ($x$, $y$, $z$, and $t$), and it gives the values which are in complex-space. For a typical example, $$\Psi(x,y,z,t) = \sqrt{\frac{8}{abc}} \sin\left(\frac{n_x\pi x}{a}\right) \sin\left(\frac{n_y\pi y}{b}\right) \sin\left(\frac{n_z\pi z}{c}\right) e^{-2\pi iEt/h}$$ is an example of a wave-function.

But the $|\Psi|^2$ is mathematically defined as $\Psi\cdot\Psi^*$. Max Born interpreted the value of this real valued function as the probability of finding the particle in 3 dimensional space. If you consider the previous example then $$|\Psi(x,y,z,t)|^2 = \frac{8}{abc} \sin^2\left(\frac{n_x\pi x}{a}\right) \sin^2\left(\frac{n_y\pi y}{b}\right) \sin^2\left(\frac{n_z\pi z}{c}\right)$$ Here the complex part will not appear as in the previous example because $\Psi$ is multiplied with its complex conjugate.

The probability of finding the particle in a unit volume element $\mathrm dV$ is $|\Psi|^2 \mathrm dV$. In spherical polar coordinates, it is $|\Psi|^2 r^2 \, \mathrm dr \, \sin\theta \, \mathrm d\theta \, \mathrm d\phi$. When you are only concerned about the radial part, the polar angular integral and azimuthal angular integral are replaced by $4\pi$ as, $\int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta \, \mathrm{d}\theta \, \mathrm{d}\phi =4\pi$. thus we are left with only the radial part which is your radial probability distribution function. $$\Pr(r) = |\Psi|^2 4 \pi r^2 \, \mathrm dr$$ The radial probability can be thought as the probability of finding the particle within an interval of length $\text{d}r$ at $r=r_0$. So, the radial distribution is a function but the radial probability as described can be calculated by integrating that function from $0$ to $r_0$.


Why was the Deville process of Alumina production given up in favor of the Bayer process?

The product of the Bayer process is purer than that of the Deville process.

Trevor I Williams, Aluminium: latecomer to the metal industry, Endeavour, Volume 17, Issue 2, 1993, Pages 89-93


Entropy change of isothermal irreversible expansion of ideal gas

Here is a cookbook recipe for determining the change in entropy for a system that has suffered an irreversible process:

THE RECIPE

Apply the First Law of Thermodynamics to the irreversible process to determine the final thermodynamic equilibrium state of the system
Totally forget about the actual irreversible process (entirely), and focus instead exclusively on the initial and final thermodynamic equilibrium states.  This is the most important step.
Devise a reversible (alternative) path between the same two thermodynamic equilibrium states (end points).  This reversible path does not have to bear any resemblance whatsoever to the actual irreversible process path.   For example, even if the actual irreversible process is adiabatic, the reversible path you devise does not have to be adiabatic.  You can even separate various parts of the system from one another, and subject each of them to a different reversible path, as long as they all end up in their correct final states. Plus, there are an infinite number of reversible process paths that can take you from the initial state to the final state, and they will all give exactly the same value for the change in entropy.  So try to devise a path that is simple to work with (i.e., for which it is easy to apply step 4).
For the selected reversible path, evaluate the integral of dq/T from the initial state to the final state, where dq is the incremental amount of heat added to the system along the sequence of changes comprising the reversible path.  This will be your change of entropy S.  That is, $\Delta S=\int{\frac{dq_{rev}}{T}}$ , where the subscript rev refers to the reversible path.

For further details and worked examples, see:
Reference https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/


Why isn't the hydroxyl group of citric acid deprotonated while the carboxyl groups are?

You are correct in saying that the carboxyl groups will have resonance structures after being deprotonated. Thus, the carboxyl groups are fairly acidic and will deprotonate in water. On the other hand, deprotonation of the hydroxyl group will not lead to any resonance stabilization, such that the hydroxyl group will be much less acidic than the carboxyl groups. In fact, the $\mathrm{p}K_\mathrm{a}$ of alcohols is usually in the range 16-20, which means that the alcoholic hydrogen will not deprotonate in water. It can deprotonate, however, in other solvents.


What is going on when pen stops working on a section of paper?

If in doubt, try it out.

I just spilled some olive oil on a sheet of paper and wiped it off again.

Result: The ballpoint pen does not write on the oily spot.

Did you fat-finger the paper?

EDIT

The experiment above confirmed the finding in the question but didn't give an explanation.

Actually, two hours later it is still not possible to write on the oil-soaked and more transparent parts of the paper with a ballpoint pen, while a felt-tip pen (here: an Edding, ther German equivalent to a Sharpie) leaves perfect lines on every part of the paper.

Paper is fibrous material with a porous structure - without that and the capillary effects coming with it, fountain pens wouldn't be able to write on it.

For ballpoint pens, this probably isn't so important - the ink is very viscous.

But paper also has a rough surface - and this might be crucial, as already suggested by Lord Stryker in his comment. The rough paper surface provides friction, allows the tungsten carbide ball to roll, take up ink and spread it on the paper.

On a smooth or slippery surface - and this seems to be the case for the oily regions on the paper - the tip of a ballpoint pen only slides but does not roll. As a result, no ink is transfered to the paper.


Naming ionic compounds with multiple cations and anions

I propose you consult the recommendations of the International Union of Applied Chemistry (IUPAC), in particular the Red Book as reference.  Obviously, you deviated for a reason from the Hill formula of $\ce{NaKCl2}$ (that were $\ce{Cl2KNa}$) some data bases adhere to.

As already presented in rules IR 4.4.2.1, and IR 4.4.2.2, the order of the elements within the formula may be arranged either by their relative electronegativity (starting by the element of least electronegativity), or by alphanumeric order, respectively.  So rules in section IR 5.4 subsequently discriminate between electropositive constituents to be named prior to electronegative ones in the "overall name".  In addition (quote):

"The order of citation is alphabetical within each class of constituents (multiplicative prefixes being ignored), except that hydrogen is always cited last among electropositive constituents if actually classified as
an electropositive constituent."

The same reference provides numerous examples of application useful to look at, too.  Page 76 (of the book, not the pdf) is a good place to start.  As in nomenclature of organic compounds, there are instances you want to point the attention not only to the stoichiometric composition of the compound, rather than to different constitutional isomers.


Dimerisation of HOOC-COOH in non polar solvents

I believe your answer key is wrong on this one. Carboxylic acids like acetic acid (a) are well known to form doubly hydrogen bonded dimers in which each the carbonyl group on one acid is hydrogen bonded to the OH of the other.

For more explanation and structures, see https://www.chemguide.co.uk/organicprops/acids/background.html.

Oxalic acid, being a carboxylic acid, can participate in this kind of intermolecular association, but the result would be a polymeric structure, rather than a simple dimer.

The alcohols can hydrogen bond to each other only singly - too weakly to form stable dimers.

Kudos to you for having the right instincts to challenge your answer key an probe more deeply!


The problem with method 1 is that you discard the protonation process, which effectively would increase pH as it produces $\ce{OH-}$ ions.

The criteria of $K_1 > 1000K_2$ is applicable when you have a polyprotic acid in its completely (de)protonated state. The ionisation of H2A produces only a small amount of HA- and K2 is small as well, so ionisation of HA- would be negligible and we can ignore it.

However with an ampholytic species, while K2 is smaller than K1, the initial reactant is the same (HA-) and it is in high concentration, so both processes occur. The formula in method 2 is the result of this consideration.


Does the main washing effect of soap come from foam?

Foam is a side effect of using a tensio-active agents (though some tensio-active agents are engineered to produce as little foam as possible, e.g. in washing machines). In fact, you could consider that, if soap has nothing better to do, it will form foam. This answers some of your questions: foam does not (really) carry away dirt and its percentage of efficiency in the whole process is low.

Molecules of soap are made of 2 parts: a hydrophobic tail (which loves fat, grease, lipids, whatever you call them) and a hydrophilic head which loves water, and hate lipids.

When the hydrophobic tails are busy linking to a greasy stain or grease on your hands, they no longer participate into forming bubbles/foam.

The easiest way to verify that is to pour a little bit of oil in your hands and rub them. Wash your hands with soap a first time: there will be almost no foam. Now rinse your hands and wash them again: there should be much more foam!

Note that it does not mean that your hands are cleaner, regarding bacteria, but just that they are less greasy!

If you want to remove bacteria, rub your hands with soap for at least 30 seconds. So-called "anti-bacterial" soaps do not achieve better and promote bacterial resistance.


Cyclopentanone is treated with piperidine

As per the reaction sequence what you said about A is absolutely correct and in B I think you didn't consider hydrolysis of the product. The product what you have written is correct before hydrolysis, and after hydrolysis the carbonyl reforms and the other component becomes piperidine again. So,the final product B is 2-(phenylmethyl)cyclopentan-1-one. Hope that helps!


Note that I can only comment on the English translations!

Cyclohexanol is correct, cyclohexol is just wrong. This is not the only error in the abstract of the thesis.

  Cyclohexone is the most [...]

isn't correct either. The name is cyclohexanone!


Time-dependent DFT can be used to predict excitation energies through a linear-response formulation.

In this Gaussian result, beyond the first line, you are looking at the largest coefficients in the configuration-interaction (CI) style expansion.

(It's not strictly CI, but the implementation of time-dependent HF or RPA is essentially the same for TDDFT or Tamm-Dancoff Approximation TDDFT.)

Anyway, since you're performing a calculation on an open-shell system, the formulation for the state will include all one-electron excitations AND all one-electron de-excitations. That state has a de-excitation from 454B into 453B (i.e., the SOMO will fill the hole left by the excitation). Now you may think "but that doesn't accomplish anything." Remember that Gaussian only prints the dominant coefficients.

For this reason, Rich Martin created the Natural Transition Orbitals method as a nice way to express the excitation, rather than a set of coefficients and the ground-state orbitals.

  Oftentimes, there is no dominant configuration in the list of excitation amplitudes, thereby making a straightforward interpretation of the excited state difficult. This is particularly unsatisfactory when attempting to determine the qualitative nature of an excited state. An additional complication in the DFT case is that in principle all orbitals in DFT but the HOMO are devoid of physical significance. However, chemical intuition is built on the orbital construct, and a simple orbital interpretation of "what got excited to where" is important.

This is available in Gaussian from the Population keyword and in other codes.

Incidentally, there's a nice review of TDDFT in: "Progress in Time-Dependent Density-Functional Theory" Annual Review of Physical Chemistry
Vol. 63: 287-323.


I googled it for a while and this was the best thing I found (Source):

  This is a tough question, that is probably not known with any certainty.
  Even Jerry March (Adv Organic Chem, 4th ed, page 1219) doesn't commit himself to a mechanism.

  You'll have to speculate.   Add a hydride on the N next to the alkyl group, then
  grab a proton on the terminal N.    Add a hydride on the terminal N, then grab a proton on the middle N.  After that, it gets a little sketchy for breaking the N-N bond.

  Try a SN2 attack by hydride on the middle N, giving hydrazine and  RNH-, which is a lousy SN2 leaving group, but we're stuck with it.
  Protonation of the RNH- gives the amine and hydrazine.   Or, you could do a SN2 on the N next to the alkyl group, and kick out a NH2NH- anion, which is also a lousy leaving group for the SN2.
  You can see why most books conveniently leave out a mechanism for this reduction.


You're exactly right: the phenomenon is called boiling-point elevation:

  Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. [...] It is an effect of the dilution of the solvent in the presence of a solute. It is a phenomenon that happens for all solutes in all solutions, even in ideal solutions, and does not depend on any specific solute-solvent interactions. The boiling point elevation happens both when the solute is an electrolyte, such as various salts, and a nonelectrolyte. In thermodynamic terms, the origin of the boiling point elevation is entropy and can be explained in terms of the vapor pressure or chemical potential of the solvent. In both cases, the explanation depends on the fact that many solutes are only present in the liquid phase and do not enter into the gas phase (except at extremely high temperatures).

The empirical method for computing boiling point elevation for a given solute is described here. In short:

  $\Delta T_b = \sum_i{n_iK_{b,i}m_i}$

Here, $\Delta T_b$ is the observed increase in boiling point, and $K_{b,i}$ and $m_i$ are the boiling point elevation constant and the molality (moles of solute per kilogram of solvent), respectively, of solute $i$.  The $n_i$ factor is the number of ions into which solute $i$ dissociates upon dissolution (e.g., $n_{sucrose} = 1$ and $n_{\ce{NaCl}} = 2$).

Your best bet for a full thermodynamic description is a college or graduate level textbook on the topic. I don't know offhand of any that specifically address boiling-point elevation, though.  The Wikipedia page on colligative properties actually has a bit of thermodynamic detail, providing the expression for computing the boiling point elevation constant from the enthalpy of vaporization $(\Delta H_{\mathrm{vap}})$ and other physical parameters:

  $K_{b,i} = \frac{RM_iT_b^2}{\Delta H_{\mathrm{vap}}}$

where $R$, $M$, and $T_b$ are the ideal gas constant, the formula weight of solute $i$, and the boiling point of pure solvent, respectively.


However, my problem is that I'm not getting clear enough peaks. The obvious thing I can think of to do is increase the number of scans or lengthen the dwell time.
But I was wondering if there are other solutions other people usually
do that I might be missing. I do see some papers that try to
deconvolute Ru and C in the same region but my professor says that
this is an incorrect method because it is impossible to consider the
correct charge correction that way.

Do you have an Atlas of XPS spectrum of elements? Check the instrument's manual. I do recall CASA used to have element by element spectrum.
Thermo Fisher has one periodic XPS spectrum here. If gold is causing problems, choose another metal whose peaks do not overlap with your sample.
The only solution for weak signals is that you collect the data for a longer time (ensemble averaging).
Don't try deconvolution in XPS especially when it is peak fitting. Your mentor is right. It is not reliable. You can fit as many peaks as you like. There must be a rationale for fitting peaks in an analytical signal (you exactly know peak shapes, peak model, baseline etc.)


How many d-electrons are in the outer shell of metals like Cadmium (this is counting the 4d electrons)?

Without doing some sort of at least half-serious calculations, you cannot really answer the question. For all we know, a cadmium(II) ion could be $[\ce{Kr}]\,\mathrm{4d^{10}}$, $[\ce{Kr}]\,\mathrm{4d^9\,5s^1}$ or $[\ce{Kr}]\,\mathrm{4d^8\,5s^2}$.

However, given the analogies to other transition metals and the general structure of practically all transition metal ions, the first configuration, $[\ce{Kr}]\,\mathrm{4d^{10}}$ is most likely. I can’t recall a case where a transition metal ion had any s electrons in the valence shell before the d electrons were completely populated; i.e. even if you have two s electrons in the neutral atom and ionise only once you would end up with only d electrons in the ion.


Will all thermodynamic equations have a linear dependence with moles?

Extensive properties actually have a linear dependence with the common magnitudes used for amount of substance.

Mathematical speaking, you work in some representation in which one magnitude is represented with a function of some variables, and the later represent other physical variables. For example $S(U,V,N)$. When a magnitude represents a extensive property, the function is a homogeneous function of of order 1 of the extensive variables, that is:
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$

We choose $\lambda = 1/N$, so
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$, using lower case symbols for (say molar) properties,

$ N \,S(e, v, 1) = S(E,V,N)$

So, taking in mind that your equations are formulated for finding values for your thermodynamic functions, they must be stated in the form (for the previous example):

$S(E,V,N) =$ $whatever$

If you work with molar properties you will find just:

$S(e, v, 1) =$ $whatever$ $for$ $a$ $mol$, but using the third equation:

$S(E,V,N) = N\times$ $whatever$ $for$ $a$ $mol$

So the homogeneous (order 1) property will assure you that there is linear dependence with $N$.

Note: Thermodynamic functions that also depend on intensive variables have a zero order dependence in the homogeneous sense with these variables.

EDITED:

After reading the @Curt F. answer, to avoid confusions I decided made a comment: The explaining above works in the scope of 'traditional' thermodynamics, where no surface effects are involved. There are another reasons that can leads to draw wrong conclusions about this linearity, for example, if external fields are involved. But these cases are not in the scope of any course on thermodynamics that I aware of.


It is said that atoms cannot be created. If so, then how did atoms get created after the Big Bang?

Atoms can be created and destroyed.  For example, in nuclear fusion two or more atoms make one atom.  In nuclear fission, one atom makes two or more atoms.

  how did the atoms form after the Big Bang?

The theory of this process in known as Big Bang Nucleosynthesis

Starting from protons and neutrons, light elements formed.  Later, in stars, nuclei up to iron formed through nuclear fusion.  Nuclei heavier than iron formed from supernovae by the r-process and asymptotic branch giant stars by the s-process.

  Also, does this mean that the number of atoms in our universe has remained the same since then?

No.  The number constantly changes due to fusion, fission, supernovae, creation of neutron stars, black holes and possibily quark stars.

  Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?

Currently, this process does not create or destroy atoms.  However, according to the theory of accelerating expansion, there is a senario referred to as the "big rip" in which all atoms would be destroyed.


Fluorescin(left), a non-fluorescent compound, reacts with iron in residual blood samples to produce highly fluorescent fluorescein(right).

The reference$\ce{^{[2]}}$ from which the following excerpt is taken describes the chemistry, preparation of reagents, and application to forensics.

  Fluorescin is the oxidized colorless form of fluorescein. Fluorescin reduced with hydrogen peroxide is immediately oxidized to fluorescein in the presence of blood-associated proteins and iron ion found in hemoglobin. Fluorescein fluoresces when viewed using blue light and orange or yellow filters, whereas fluorescin will not fluoresce under the same lighting conditions.

Literature Resource

An excellent starting point to consider for searching for relevant literature is PubMed.

References

Image sources - PubChem
Google Books - Complete Crime Scene Investigation Handbook, Everett Baxter, Jr.
Latent-Prints.com - Fluorescein Detection of Latent Bloodstains
The Presumptive Reagent Fluorescein for Detection of Dilute Bloodstains and Subsequent STR Typing of Recovered DNA
A Comparison of the Bluestar and Luminol Effectiveness in Bloodstain Detection


Can atomic number change the type of electromagnetic radiation being emitted when a electron drop from a energy level to another?

Absolutely, the differences in potential energies will be different for every element. This is because the potential energies will be different for every element. Notice that the potential energy for each energy level can be determined from the ground state. Let $E_g$ be the potential energy of the ground state (with the sign). Then for energy level $n$ the potential energy is $\frac{E_g}{n^2}$. Thus Differences between energy levels $n$ and $m$ can be written as $E_g\left|\frac{1}{n^2}-\frac{1}{m^2}\right|$.

Clearly it will be different for a different ground state energy which is precisely what the number of protons affects.

Note that the only important part of the mathematics is that all differences rely on the ground state potentials.


Different products using different indicators

An indicator does not affect a particular reaction. But in your solution, you have two successive reactions when adding HCl to this solution : First $$\ce{CO_3^{2-} + H^+ -> HCO_3^-}$$ and a given indicator must be added to determine the end of this reaction. If you don't, all you see is a colorless solution being transformed into another colorless solution, even though its composition has changed. This added indicator may be phenolphthalein, which goes from red to colorless precisely at the end of this first titration. Then, by adding more HCl, you carry out the following reaction : $$\ce{HCO_3^- + H^+ -> H_2O + CO_2}$$ And you need is another indicator, maybe methyl orange, to determine the end of this second reaction. The color of the methyl orange in solution changes from yellow to reddish exactly after the end of this second reaction.

Phenolphtalein and Methyl orange are molecules which reacts with $\ce{H^+}$ exactly like $\ce{CO_3^{2-}}$ and $\ce{HCO_3^-}$. But they react before or after these two ions. To summarize, $\ce{H^+}$ reacts first with  $\ce{CO_3^{2-}}$ and nothing else, and it takes a couple of milliliters to do it. Then it reacts with Phenolphtalein and nothing else, and it takes less than one drop to do it. Then it reacts with $\ce{HCO_3^-}$ and nothing else, and it also takes a couple of milliters to do it. And last but not least, it will react with methyl orange.


What would happen if the reaction of reduction of carboxylic acids by LiAlH4 is carried out in alcohol?

Alcohols are not an ideal solvent for LiAlH4, but the main issue is that there is no carboxylic center to where the hydride transfer from LiAlH4 can take place.

So in this case, the side-reaction between alcohol and LiAlH4 can be regarded to be minimal and the major product will be the alcohol produced from the carboxylic acid.

P.S. : A diagrammatic representation of the reactions for LiAlH4


White phosphorus has P4 molecules packed into a crystal,these dissolve readily in Carbon Disulfide.

Whereas Red phosphorus is polymeric in nature. It is a derivative of white phosphorus where one P-P bond is broken and an addtional bond is formed with neighboring tetrahedron molecule resulting in a chain like structure .

Like all long chain polymers this will lower its ability to dissolve and its solubility in solvents


Why are two three-membered rings preferred to a single five-membered ring in the reaction of a dihalide with alcoholic KOH?

I think the product may be formed by the following pathway.
First, the carbanion will form when the strong base abstracts the proton from the α-position. The lone pair generated would obviously have two points of attack, either at the nearer carbon to form the cyclopropyl ring or at the farther carbon to form a five membered rkng where the whole controversy lies.
While apparently, the position five membered ring appears to be more stable than a small highly stained three membered ring, a very interesting phenomenon occurs if such a three membered, strained ring forms.
This is similar the basic resonace observed in cyclopropane.
The carbonyl functional group also triggers the reaction foreward towards formation of the product.
Let us begin by observing the resonance structures of the carbonyl functional group.
Clearly, out of the two contributing structures, positive charge may be seen residing on the carbonyl carbon while negative charge may be seen on the carbonyl oxygen.

A combination of these two resonance hybrids leads to generation of partial positive charge on carbon and partial negative charge on the net resonance hybrid.
Next, we have to observe the structure of the 1-methylcycloropyl carbocation.
Due to ring strain the bonds are bent and a special kind of resonance involving the σ C-C single bond is observed.

In fact the bonds are actually sp5 hybridized i.e. having large p character to ease the ring strain.
The enormous stablity can be alluded by the following resonce diagram.
Moreover, the immense stablity of tris(cyclopropyl)methyl carbocation, is actually due to this σ-resonace. It is actually the one of the most stable carbocations known.

As the number of cyclopropyl groups increase, the stablity increases.

Now to observe a combination of both.
In the final product the M- effect of the carbonyl group will favour the extended delocalisation of electrons of the σ-bond over the oxygen.
Thus, the enourmous stablity concept of cyclopropane may be attributed to the compound under question by the following diagram:

To draw the resonance structures in a simplified way we first show complete charge seperation between the carbon and oxygen of the carbonyl group. Then the structure becomes similar to a cyclopropylmethyl carbocation showing the required stablising resoance of σ-bond.
Due to this unique resonance effect, also known sometimes as dancing resonance, the product becomes exceptionally stable, more stable than the 5 membered ring obtained in the other case.
Hope this helps.
References:
What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?


Family is an informal term. Only a group and a period is an accepted terminology for the periodic table by the International Union of Pure and Applied Chemistry, a widely respected authority on chemical nomenclature and terminology.

The Wiki page of periodic table should be modified because a dubious and layman type reference (www.shmoop.com) for group (= also known as family) is being cited.


How to predict the reason for color in a transition metal complex?

Nobody can do it just by "looking" at the color of any compound. Spectroscopists have to do calculations to figure that out. There are at least 15 reasons for the "color" of chemical compounds.
Now qualitatively one can "predict" whether it is a d-d vs. it is charge transfer transition for transition metal compounds.
Keep in mind that d-d transitions are "forbidden" (Laporte selection rule). Therefore, majority of the d-d transitions correspond to low molar absorptivity. For example, Cu(II) solution is light blue, manganese (II) is almost invisible pink, chromium (III) is light green.
Charge transfer transitions are very intense! Add two three crystals of potassium permanganate to water, and it is intensely purple.


Technique to measure the volume of a crystal

Potash alum is soluble in water (~14g/100mL at 20C) but not in acetone. I would submerge the sample in a known volume of acetone and measure displacement. If you want to be super thorough, cool the acetone first.

I wouldn't use reported density because you don't necessarily know if you made a pure product etc.


How does a space group determine crystal structure?

To build lattice you need the space group, the three unit cell dimensions, the three unit cell angles, each atom's fractional coordinates (e.g., a *.cif file overall),

Acta Cryst. B38, 598 (1982)  doi:10.1107/S056774088200346X
Ni3Sn4, C2/m, a=12.214, b=4.060, c=5.219 A, beta=105.0 degrees (others 90 degrees by default), V=250 A^3 (verify), Z=2 (formula units/unit cell)

plus

http://www.chem.gla.ac.uk/~louis/software/ortep3/ or

http://www.shapesoftware.com/00_Website_Homepage/ or

http://www.iucr.org/resources/other-directories/software?result_42405_result_page=A
the big kahuna.

For 3D visualization, if you are not practiced,
http://www.berezin.com/3d/3dprism.htm


Why is sulfur the central atom in the Lewis structure for SO2?

I think you messed up something. Because it is exactly opposite.

"Central part of atom is normally least(not the most) electronegative atom i.e. S in this case.


Why is BCl3 a monomer whereas AlCl3 exists as a dimer?

Introduction

The bonding situation in $\ce{(AlCl3)2}$ and $\ce{(BCl3)2}$ is nothing trivial and the reason why aluminium chloride forms dimers, while boron trichloride does not, cannot only be attributed to size.

In order to understand this phenomenon we need to look at both, the monomers and the dimers, and compare them to each other.
Understanding the respective bonding situation of the monomers, is key to understand which deficiencies lead to dimerisations.

Computational details

Since I was unable to find any compelling literature on the subject, I ran some calculations of my own. I used the DF-M06L/def2-TZVPP for geometry optimisations. Each structure has been optimised to a local minimum in their respective symmetry restrictions, i.e. $D_\mathrm{3h}$ for the monomers and $C_\mathrm{2v}$ for the dimers.
Analyses with the Natural Bond Orbital model (NBO6 program) and the Quantum Theory of Atoms in Molecules (QTAIM, MultiWFN) have been run on single point energy calculations at the M06/def2-QZVPP//DF-M06-L/def2-TZVPP level of theory.
A rudimentary energy decomposition analysis has been done on that level, too.

Energy decomposition analysis

The dissociation energy of the dimers $\ce{(XY3)2}$ to the monomers $\ce{XY3}$ is defined as the difference of the energy of the dimer $E_\mathrm{opt}[\ce{(XY3)2}]$ and double the energy of the monomer $E_\mathrm{opt}[\ce{XY3}]$ at their optimised (relaxed) geometries $\eqref{e-diss-def}$.
The interaction energy is defined as the difference of energy of the relaxed dimer and double the energy of the monomers in the geometry of the dimer $E_\mathrm{frag}[\ce{(XY3)^{\neq}}]$ $\eqref{e-int-def}$. That basically means breaking the molecule in two parts, but keeping these fragments in the same geometry.
The deformation energy (or preparation energy) is defined as the difference of the energy of the optimised and the non-optimised monomer $\eqref{e-def-def}$. This is the energy required to distort the monomer (in its ground state) to the configuration it will have in the dimer.
$$\begin{align}
E_\mathrm{diss} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{opt}[\ce{XY3}]
  \tag1\label{e-diss-def}\\
E_\mathrm{int} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{frag}[\ce{(XY3)^{\neq}}]
  %\ddag not implemented
  \tag2\label{e-int-def}\\
E_\mathrm{def} &=
  E_\mathrm{frag}[\ce{(XY3)^{\neq}}] - E_\mathrm{opt}[\ce{XY3}]
  \tag3\label{e-def-def}\\
E_\mathrm{diss} &=
  E_\mathrm{int} + 2E_\mathrm{def}\tag{1'}
\end{align}$$

Results & Discussion

The Monomers $\ce{XCl3; X{=}\{B,Al\}}$.

Let's just get the obvious out of the way: Boron is (vdW-radius 205 pm) smaller than aluminium (vdW-radius 240 pm). For comparison chlorine has a vdW-radius of 205 pm, too. That is pretty much reflected in the bond lengths and the chlorine-chlorine distance.
\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.0 &173.6&--\\
\mathbf{d}(\ce{Cl\bond{~}Cl'})&/\pu{pm} & 356.8 & 300.6 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}

From this data we can draw certain conclusions without further looking. The boron monomer is much more compact than the aluminium monomer. When we compare the bond lengths to the covalent radii (Pyykkö and Atsumi) we find that the boron chloride bond is about the length that we would expect from a double bond ($\mathbf{r}_\mathrm{doub}(\ce{B}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 173~\pu{pm}$). While the aluminium chloride bond is still significantly shorter than a single bond ($\mathbf{r}_\mathrm{sing}(\ce{Al}) + \mathbf{r}_\mathrm{sing}(\ce{Cl}) = 225~\pu{pm}$), it is still also much longer than a double bond ($\mathbf{r}_\mathrm{doub}(\ce{Al}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 191~\pu{pm}$).

This itself offers compelling evidence, that there is more π-backbonding in $\ce{BCl3}$ than in $\ce{AlCl3}$. Molecular orbital theory offers more evidence for this. In both compounds is a doubly occupied π orbital. The following pictures are for a contour value of 0.05; aluminium (left/top) and boron (right/bottom)

In numbers, the main contributions are as follows (this is just a representation, not the actual formula):
$$\begin{align}
\pi(\ce{BCl3}) &=
   21\%~\ce{p_{$z$}-B} + \sum_{i=1}^3 26\%~\ce{p_{$z$}-Cl^{$(i)$}}\\
\pi(\ce{AlCl3}) &=
   13\%~\ce{p_{$z$}-Al} + \sum_{i=1}^3 29\%~\ce{p_{$z$}-Cl^{$(i)$}}
\end{align}$$

There is still some more evidence. The natural atomic charges (NPA of NBO6) fairly well agree with that assesment; aluminium is far more positive than boron.
$$\begin{array}{lrr}
                    & \ce{AlCl3} & \ce{BCl3}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.4       & +0.3     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
%\mathbf{q}(\ce{X})~\text{[QTAIM]}  & +2.4       & +2.0    \\
%\mathbf{q}(\ce{Cl})~\text{[QTAIM]} & -0.8       & -0.7     \\\hline
\end{array}$$

The analysis in terms of QTAIM also shows that the bonds in $\ce{AlCl3}$ they are predominantly ionic (left/top) while in $\ce{BCl3}$ are predominantly covalent (right/bottom).

One final thought on the bonding can be supplied with a natural resonance theory analysis (NBO6). I have chosen the following starting configurations and let the program calculate their contribution.

The overall structures in terms of resonance are the same for both cases, that is if you force resonance treatment of the aluminium monomer. Structure A does not contribute, while the others contribute to about 31%. However, when not forced into resonance, structure A is the best approximation of the bonding situation for $\ce{AlCl3}$. In the case of $\ce{BCl3}$ the algorithm finds a hyperbond between the chlorine atoms, a strongly delocalised bond between multiple centres. In this case these are 3-centre-4-electron bonds between the chlorine atoms, resulting from the higher lying degenerated π orbitals.

This all is quite good evidence that the monomer of boron chloride should be more stable towards dimerisation than the monomer of aluminium.

The Dimers $\ce{(XCl3)2; X{=}\{B,Al\}}$.

The obvious change is that the co-ordination of the central elements goes from trigonal planar to distorted tertrahedral. A look at the geometries will give us something to talk about.

\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.7 &175.9&--\\
\mathbf{d}(\ce{X-{\mu}Cl})&/\pu{pm} & 226.1 &198.7&--\\
\mathbf{d}(\ce{Cl\bond{~}{\mu}Cl})&/\pu{pm} & 354.1 & 308.0 & --\\
\mathbf{d}(\ce{{\mu}Cl\bond{~}{\mu}Cl'})&/\pu{pm} & 323.6 & 287.3 & --\\
\mathbf{d}(\ce{B\bond{~}B'})&/\pu{pm} & 315.7 & 274.7 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}

In principle nothing much changes other than the expected elongation of the bonds that are now bridging. In case of aluminium the stretch is just below 10% and for boron it is slightly above 14%, having a bit more impact. In the boron dimer also the terminal bonds are slightly (> +1%) affected, while for aluminium there is almost no change.

The charges are not really a reliable tool, especially when they are that close to zero as they are for boron. In both cases one can see that charge density is transferred from the bridging chlorine to the central $\ce{X}$.

$$\begin{array}{lrr}
                    & \ce{(AlCl3)2} & \ce{(BCl3)2}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.3       & +0.2     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
\mathbf{q}(\ce{{\mu}Cl})~\text{[NPA]} & -0.4       & +0.1     \\\hline
\end{array}$$

A look at the central four-membered-ring of in terms of QTAIM offers that the overall bonding does not change. In aluminium they get a little more ionic, while in boron they stay largely covalent.

The NBO analysis offers a maybe quite surprising result. There are no hyperbonds in any of the dimers. While a description in these terms is certainly possible, after all it is just an interpretation tool, it is completely unnecessary. So after all we have two kinds of bonds in the dimers four terminal $\ce{X-Cl}$ and four bridging $\ce{X-{\mu}Cl}$ bonds. Therefore the most accurate description is with formal charges (also the simplest). The notation with the arrows is not wrong, but it does not represent the fact that the bonds are equal for symmetry reasons alone.

To make this straight: There are no hyperbonds in $\ce{(XCl3)2; X{=}\{B,Al\}}$; this includes three-centre-two-electron bonds, and three-centre-four-electron bonds. And deeper insight to those will be offered on another day.

The differentiation between a dative bond and some other for of bond does not make sense, as the bonds are equal and only introduced by a deficiency of the used description model.

A natural resonance theory for $\ce{(BCl3)2}$ gives us a overall contribution of the main (all single bonds) structure of 46%; while all other structure do contribute, there are too many and their contribution is too little (> 5%). I did not run this analysis for the aluminium case as I did not expect any more insight and I did not want to waste calculation time.

Dimerisation - yes or no

The energies offer us a clear trend. Aluminium likes to dimerise, boron not. However, there are still some things to discuss. I am going to argue for the reaction
$$\begin{align}
\ce{2XCl3 &-> (XCl3)2}&
\Delta E-\mathrm{diss}/E_\mathrm{o}/H/G&.
\end{align};$$
therefore if reaction energies are negative the dimerisation is favoured.

The following table includes all calculated energies, including the energy decomposition analysis mentioned at the beginning. All energies are given in $\pu{kJ mol^-1}$.
\begin{array}{lrcrcrcrr}
 \Delta & E_\mathrm{diss} &(& E_\mathrm{int} &+2\times& E_\mathrm{def}&)& E_\mathrm{o} &H &G\\\hline
\ce{Al} & -113.5 &(& -224.2 &+2\times& 55.4&)& -114.7 & -60.4 & -230.4\\
\ce{B}  &   76.4 &(& -111.2 &+2\times& 93.8&)&   82.6 & -47.1 &  152.5\\\hline
\end{array}

The result is fairly obvious at first. The association for aluminium is strongly exergonic, while for boron it is strongly endergonic. While both reactions should be exothermic, stronger for aluminium, the trend for the observed electronic energies ($E_\mathrm{o}$ including the zero-point energy correction) and the (electronic) dissociation energies reflect the overall trend for the Gibbs enthalpies.

While it is fairly surprising how strongly entropy favours association of $\ce{AlCl3}$, it is also surprising how it strongly disfavours it for $\ce{BCl3}$.
A look at the decomposed electronic energy offers great insight into the reasons why one dimer is stable and the other not (at room temperature).
The interaction energy of the fragments is double for aluminium than it is for boron. This can be traced back to the very large difference in the atomic partial charges. One could expect that the electrostatic energy is a lot more attractive for aluminium than it is for boron.
The deformation energy on the other hand clearly reflects the changes in the geometry discussed above. For aluminium there is a smaller penalty resulting from the elongation of the $\ce{Al-Cl}$ bond and pyramidalisation. For boron on the other hand this has a 1.5 times larger effect. The distortion also weakens the π-backbonding, which the additional bonding would need to compensate.
The four-membered-ring is certainly not an ideal geometry and the bridging chlorine atoms come dangerously close.

Conclusion, Summary and TL;DR:

The distortion of the geometry of the monomer $\ce{BCl3}$ cannot be compensated by the additional bonding between the two fragments. Therefore the monomers are more stable than the dimer. Additionally entropy considerations at room temperature favour the monomer, too.

On the other hand, the distortion of the molecular geometry in $\ce{AlCl3}$ is less severe. The gain in interaction energy of the two fragments well overcompensates for the change. Entropy also favours the dimerisation.

While size of the central atom is certainly the distinguishing factor, its impact is only severe on  the electronic structure. Steric crowding would not be a problem when the interaction energy would compensate for that. This is quite evident because $\ce{BCl3}$ is still a very good Lewis acid and forms stable compounds with much larger moieties than itself.

References

The used van der Waals radii were taken from S. S. Batsanov Inorg. Mat. 2001, 37 (9), 871-885. And the covalent radii have been taken from P. Pyykkö and M. Atsumi Chem. Eur. J. 2009, 15, 12770-12779.

Computations have been carried out using Gaussian 09 rev D.01 with NBO 6.0. Additional analyses have been performed with MultiWFN 3.3.8. Orbital pictures were generated with the incredible ChemCraft.


What's the difference between ΔG° and ΔG°'?

The prime usually denotes a standard free energy that corresponds to an apparent equilibrium constant where the concentration (or activity) of one or more constituents is held constant.

For example, for $\ce{HA <=>A- + H+}$ the equilibrium constant is $K=\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$ and the corresponding standard free energy change is $\Delta G^\circ=-RT \ln(K)$. If you know the value of $K$ and the start concentration of $\ce{HA}$ then you can compute the equilibrium concentrations of $\ce{HA}$, $\ce{A-}$, and $\ce{H+}$.

However, if the pH is held constant then $[\ce{H+}]$ is no longer a free variable and the apparent equilibrium constant is $K'=\frac{[\ce{A-}]}{[\ce{HA}]}$ and the corresponding standard free energy change is $\Delta G'^\circ=-RT \ln(K')$. So $\Delta G'^\circ=-RT \ln(K/[\ce{H+}])=\Delta G^\circ+RT\ln[\ce{H+}]$

In biochemistry there is often several important constituents in addition to  $\ce{H+}$ that are held constant such as $\ce{Mg++}$, phosphate, etc.


Physical implications of the simple harmonic oscillator quantum mechanical solutions

However, to reach this state from, for example, x=0 in the n=2 energy
  level it must pass through a node.

To start, ask yourself: "What should pass through a node?" The answer is: it is the probability that "passes through the nodes", not atoms (or, I should better even say nuclei). Probability is just a mathematical quantity ant the zero value is allowed for it, so there is nothing to worry about. Yes, quantum harmonic oscillator has some forbidden regions of displacement from the equilibrium. But it can also penetrate into classically forbidden regions. Quantum world is weird.

The root of this (and most of the other) apparent quantum paradoxes is the classical way of thinking: you think that you have some funky quantum balls on a funky quantum spring out there, but you don't. Quantum harmonic oscillator is a quantum system that essentially has the Hamiltonian similar in its form to the Hamiltonian of a classical harmonic oscillator. And this is it. No balls, no springs, just a similar Hamiltonian.

It should be mentioned as well that atoms (again, more precisely, nuclei) are not vibrating in molecules even in a classical sense: their motion can be only approximately separated into vibrational and rotational ones. Besides, the vibrational motion can only approximately be treated as being harmonic.


Why are there no hydrogen atoms in the crystal structure of a biomolecule?

Typically crystal structures are determined by X-ray diffraction off of electrons within a bond and around the nuclei of each atom. Since H only has 1 electron, and its often involved in a polar type bond (hence doesn't spend much time by the H nucleus when bonded) it is notoriously hard to "see" with X-rays.

In order to have X-rays diffract accurately off H atoms the sample crystal needs to be large and very regular; this is not usually the case with proteins. As a workaround information about the H position can be calculated by the other atoms hybridization. This is often unreliable especially with all the hydrogen bonding in proteins which pulls them out of their idealized positions.


Why are metal oxides generally more stable than their corresponding elemental forms?

The oxidation of gold is not favorable.  For many other metals the metallic bonding is weaker than the bonding with oxygen.

Let's look at the steps in oxidation:

Breaking metallic bonds (endothermic).

Ionizing the metal atoms (endothermic).

Electron affinity of oxygen (exo for the first, endo for the second).

Lattice energy of the oxide (exothermic).

The lattice energy is so high that the process is favorable.


What is the mechanism of the reduction of organic acids with borane?

Source:
Henry Rzepa. Chemistry with a twist. Mechanism of the reduction of a carboxylic acid by borane: revisited and revised. October 16th, 2011. http://www.ch.imperial.ac.uk/rzepa/blog/?p=5114
DOI: 10.15200/winn.142795.57029 provided by The Winnower, a DIY scholarly publishing platform.


What can be used to thionate a monosubstituted urea?

I believe Lawesson's Reagent could do this. It is useful to thionate a large number of carbonyl compounds including amides, so I think it can do for urea as well. I don't see any effect on a monosubstituted urea that would negatively affect the reaction.


In chemistry it has often been explained that entropy increase had something to do with increasing disorder. The word disorder stems from Helmholtz (1882). But this is outdated, it was only a first attempt at theoretical interpretation.

Entropy = Spreading of energy or of particles of molecular (or sub-molecular) scale against constraints on their free motion. These constraints may be outer (volume) or inner (aggregative) constraints. This spreading leads to quantum mechanical stabilization effects (which may be explained with the “particle in a box”).[1] — This is the reason why entropy increase leads to more stability.


Ambigious nature of aldehydic group in glucose

The distal hydroxy groups are in a perfect distance to form pyranose (six-membered oxane) or furanose (five-membered oxolane) rings. Any equilibrium reaction that leaves the carbonyl group liberated for extended periods is prone to lose free aldehydes to the (unreactive) hemiacetal forms. Any reaction whose equilibrium is strongly product-sided or which proceeds somewhat irreversibly due to precipitation etc. will not be affected by that effect.

This explains why Fehling’s and Tollens’ solutions react with glucose: in both cases one of the products precipitates which drives the reaction to completion. None of the aldehyde that is used up in the test reaction will be regenerated. The same is true for the osazone formation which involves a redox step or the reaction with bromine which is also a redox reaction.

The reaction with sulphite to give sulphonic acids is an equilibrium: sulphite can attach but is equally quickly liberated again. However, sulphite is a rather soft nucleophile and an acceptable leaving group. The reaction is reversible and due to the formation of hemiacetals not much of the sulphonic acid will be observed. In general sulphite testing, the precipitation of the strongly hydrophilic sulphonate drives the reaction to completion but sugars themselves are already strongly hydrophilic so this driving force cannot exist.

A cyanohydrin, however, is a much more stable intermediate because cyanide is a much worse leaving group. Thus, here again we are driving a reaction to completion due to a practically irreversible step — and contrary to sulphite which is based on the same mechanism but is much more reversible.

The same difficulty occurs when attempting to generate imines: these are also prone to attack by the distal hydroxy functions giving O,N-acetals. While these may be stable products, they do not have the same features as the imines usually generated by the reaction of carbonyl compounds with amines and thus are not captured by the tests. This also prevents productive reaction of glucose with Schiff’s reagent — and additionally, Schiff’s test requires the addition of a sulphite which we learnt already is unfavourable.

The big question remains why some hydrazines react with glucose according to your list but some don’t. I have no satisfactory explanation why that would be the case. The only difference between 2,4-DNPH and phenyl hydrazine is the former’s electron-withdrawing groups that could lead to an overall slower reaction or less tendency to undergo the required redox step to form an oxazone. It is not a great explanation but maybe a starting point.


Type of reaction between (+)-bis salt of (+)-phenylsuccinic acid and HCl

The reaction between the bis salt of proline and phenylsuccinic acid will give free phenyl succinic acid and proline hydrochloride as HCl is a much stronger acid than phenylsuccinic acid. All you are really doing is exchanging one acid for another so it is hardly an acid/base reaction.


Will the bond order of CO be greater in its existence as carbon monoxide or in metal carbonyl?

In figure 1 you can see the molecular orbital scheme of a $\ce{[ML6]}$ complex including σ and π interactions. The class of ligands is not important; however, care should be taken when correctly assigning orbitals.

Figure 1: Octahedral $\ce{[ML6]}$ complex including σ and π ligand-metal interactions.

The first thing we should be doing is assigning the original, pre-bonding orbitals. The metal orbitals on the left are obvious. The ligand group orbitals of lower energy, $\mathrm{a_{1g} + e_g + t_{1u}}$ are the six ligand orbitals that interact with the metal in a σ fashion; in carbon monoxide, these represent the lone pair on carbon (the HOMO). The twelve p type orbitals, of which nine are omitted and the remaining transform as $\mathrm{t_{2g}}$ are π symmetric with respect to the $\ce{L\bond{->}M}$ bond and represent the $\pi^*_{\ce{C=O}}$ bonds of carbon monoxide; the LUMOs.

In a metal carbonyl complex, the left-hand side is usually higher in energy due to the metal’s lower oxidation state. Thus, the lower an orbital in the centre of the scheme is, the more ligand-centred it is.

While filling in the final complex’ electrons, we start from the bottom; first the twelve ligand electrons that represent the carbon monoxide lone pairs. Once we are done with this, the next orbitals happen to be $\mathrm{t_{2g}}$ — the bonding ones which have more ligand contribution. These can not only be seen as bonding with respect to the $\ce{M\bond{<-}L}$ bond but also antibonding with respect to the $\ce{C=O}$ bond. The maximum we can fill in here is 6 electrons. These six electrons reduce the $\ce{C#O}$ bond order as can be seen in the formula:

$$\text{B.O.} = \frac{(12 + 2\times12) - 6}{2\times6} = \frac{30}{12} = 2.5$$

The additional factor 6 in the denominator is because I am analysing six $\ce{C#O}$ bonds at the same time.

The result ($2.5$) clearly shows that the bond order is reduced with respect to noncoordinating $\ce{CO}$.

The discussion I did with an octahedral complex can be done with any coordination geometry; octahedra are just a very typical case.


What is created when "Iron-Out®" is mixed with household bleach?

Good that you left for fresh air.

Assuming it is the same product I know, it calls here for caution:

The below mentioned data safety sheet available (SDS) here mentions for the very product as composition

and recommends as first aid (same source):

Upon the action of bleach, you likely generated sulfur dioxide, gas with pungent smell. Refraining from a remote diagnosis, but if you accident happened in a less ventilated space and you inhaled a lot, consultation of a physician may be adequate as sulfur dioxide ($\ce{SO2}$) is harmful and maybe you equally inhaled vapours of bleach, too.  Likely, mixing "iron out" and bleach together yielded a considerable heat, too.


Converting 26.62 NOx µg/L into mM (millimolars)?

I would make it easy for yourself and convert your µg/L to g/L. You can then divide it by the formula weight to obtain the molarity (in M). After this it is just a simple conversion to mM.

I suppose in Excel you can create a new column and use a formula something like this:

=((CELL/106)/Formula weight)*1000

So for your example of phosphate:

=((0.00000773)/94.97)*1000

=0.00008139 mM // 8.139x10-5 mM


How to calculate the equilibrium composition after a change in partial pressure of one component?

As follows:

Just to simplify the math a little bit, assume pressure of $\ce{CO2_{(g)}}$ is p. Thus, pressure of $\ce{NH3_{(g)}}$ is 2p at equilibrium. Total pressure in this case is 3p.

Let 3p equal new pressure of $\ce{NH3_{(g)}}$.

Added pressure of $\ce{NH3_{(g)}}$ would shift the equilibrium to the left, so say we lose 2x pressure of $\ce{NH3_{(g)}}$ and x pressure of $\ce{CO2_{(g)}}$. New equilibrium pressures will be 3p-2x for $\ce{NH3_{(g)}}$ and p-x for $\ce{CO2_{(g)}}$. Thus, ratio of new to old is $\frac{4p-3x}{3p}$. You can solve for x in terms of p using $K_p$.

Solve for x and find the ratio, which should be $\frac{31}{27}$


There are a few method to synthesize $\ce{SrO2}$ from $\ce{SrO}$ and $\ce{O2}$. One of them uses $\ce{KClO3}$ as the source of $\ce{O2}$ [Ref. 1]. The paper is in German but has English abstract which states:

  Abstract. Single crystals of $\ce{SrO2}$ have been oblained after high
  pressure/high temperature reaction of a $\ce{SrO}$/$\ce{KClO3}$ mtxture at $\pu{20 kbar}$, $\pu{1400 ^\circ C}$. ...  From a refinement of the site occupation factor for oxygen a composition $\ce{SrO_{1.95(2)}}$ has been found for the crystal investigated.

Note that $\pu{20 kbar}$ pressure was inserted by the $\ce{O2}$ released by the following reaction at high temperature (Ref. 2): $\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}$.

Experimental information on the $\ce{Sr-O}$ has been reviewed in Ref. 2 and stability of $\ce{SrO2}$ and phase-diagram of $\ce{Sr-O}$ is graphically presented (see Figure 1 & 2, respectively). Accordingly, $\ce{SrO}$ under $\pu{0.5 atm}$ of $\ce{O2}$ pressure at $\pu{800 ^\circ C}$ would not produce $\ce{SrO2}$ (see Figure 1: Only $\ce{SrO}$ exists under the conditions, $\log_{10} p = \log_{10} 0.5 = -0.30$ and $\frac{1000}{\mathrm{T}} = \frac{1000}{1073.15} = 0.93$).

References:

K.-J. Kange, F. Rau, U. SchieBl, U. Klement, Verfeinerung der Kristallstruktur von $\ce{SrO2}$ (Refinement of the Crystal Structure of $\ce{SrO2}$), Z. anorg. allg. Chem., 1994, 620, 879-881 (https://onlinelibrary.wiley.com/doi/pdf/10.1002/zaac.19946200521).
D. Risold, B. Hallstedt, L. J. Gauckler, The strontium-oxygen system, Calphad, 1996, 20(3), 353-361 (https://doi.org/10.1016/S0364-5916(96)00037-5).


What is purpose of sulfuric acid in iodine extraction?

You need a strong acid to provide hydrogen (technically hydronium) ions, which is true for most aqueous oxidation reactions:

$\ce{2I- +H2O2 +2H+ -> I2 +2H2O}$

Unfortunately, bisulfate ion appears to be a weaker acid than water $(K_{\mathrm{a}} =  10^{-2})$, so it may not work unless you use a very concentrated amount to obtain enough hydrogen ions. Try it!

A fairly cheap alternative could be hydrochloric acid, which may be available in hardware stores as muriatic acid, and is stronger than sulfuric acid. But read the MSDS, as hydrochloric acid fumes and is nasty stuff.

Best of luck, it’s a cool experiment!


As you heat up a substance will it's bonds begin to dissipate?

Remember the energy of the bonds never increase or decrease. They remain the same. When things are heated what actually happens is the molecules begin to move around (in solids they just oscillate while in liquids and gases the movement becomes much more prominent) and their kinetic energy increases and the molecules bump into each other harder and quite often which makes it increasingly difficult for the bonds to hold the molecules together (intermolecular forces). Once a certain threshold is reached the bonds become inefficient at their job and hence small lumps of molecules move apart enough for us to perceive the action as a phase change.


Slight FTIR band intensity changes across several different copper salt spectra: relevant for characterization?

(Answered after your edit of your question:)

Some variation in the location of the absorption bands (vs. 1/cm) and their intensities (vs. remnant transmittance) is unavoidable due to the interaction of your analyte (here: your copper salt/salts) with the matrix (here: pellet of KBr) compared to recording a neat film of your sample.  The spectra depicted by the trace in magenta and blue seem to be a little bit different from the others, perhaps due to some humidity (KCl exposed to air, or a sample not completely dried, or both, ...).  On first view of the current representation, the other look very similar, compared with each other.

As your data represent infrared spectroscopy, figure out how to change the representation such that your ordinate is in remnant %transmittance instead of absorption.  Let the spectrometer's software do this for you.  For a visual inspection of the data, a split in the abscissa, too, for example around 2000 1/cm eases the visual inspection a lot -- equally something typically offered by the software of the spectrometer. (On good instruments, both changes do not alter the raw data collected by the spectrometer.) For example:

(source, the picture was edited by addition of horizontal lines)

Determine the barycentre of your absorption bands. Likely, as you start with IR spectroscopy, a deviation of about 1/cm is not an issue  (ask your instructor).  In other words, if the peak pic by the software suggests 1837.54 1/cm, a report like 1838 1/cm likely is good enough (ask your instructor).
For each important absorption band, have a look about the shape, note if it is a particular broad absorption band -- in your example, around 3350 1/cm, for example.
Beside location, an other important information provided by IR spectroscopy is the intensity of the absorption band.  In the representation of %T, imagine three equal parts between the baseline on top, and the most intense absorption band on the bottom.  A bit like the three horizontal blue dotted lines added to the picture above. The absorbtion bands in the first strip are then (semiquantitatively) classified as of weak, the ones reaching till the second strip of medium, and the ones reaching down to the third strip of strong intensity.  (As an example, the edited spectrum above contains an absorption band of medium intensity at about 3500 1/cm, and two strong ones at about 1700 and 1500 1/cm.)
Compare localisation and absorption band intensity of the spectral data recorded by you, with data found in a reference (textbook, journal article, data base).  Are all important absorption bands found in your spectrum mentioned in the literature reference?  Do you find all important absorption bands described in the literature reference in your spectrum, too?  Are there some important bands in addition or missing?
Of course you may compare sample #1 with sample #2, with sample #3, etc.,
too.

Only by growing experience you will learn what important absorption bands in infrared spectroscopy are.  Because -- in contrast to other spectroscopic techniques -- it is rather the exception than the general case that you may correlate each absorption band with a structural element in the structure of your compound.

For future reference I suggest to record the IR data either already in the %T mode, or (second best) to convert them into this form.  This allows you to check rapidly if even the most intense absorption band does not hit a transmittance of 0% (such an occurence were a rapid indication of a sample with to high concentration of X in your pellet of KBr.)


a C=O double bond becomes two C-O single bonds, and those C-O single bonds are slightly more stable than the C=O double bond resulting in ΔH being slightly positive.

The tables I found (e.g. OpenStax Chemistry, Wikipedia) have the C=O bond with roughly twice the bond dissociation energy compared to the C-O bond, and in all cases the C=O bond is slightly more stable than the two single bonds. So the statement in the textbook is surprising. As these values are aggregates derived from large sets of molecules, their values will vary depending on which set of molecules they are based on.

If you go from less stable bonds (easier to break, smaller bond dissociation energy) to more stable bonds (harder to break, larger bond dissociation energy), the enthalpy of reaction should be negative (exothermic), not positive as claimed in the statement above.

For completeness, these are not the only changes. In addition, the O-H bond of the hydroxyl group is broken, and the O-H bond of the hemiacetal is formed. If you are estimating bond dissociation energies from a simple table, however, this will just cancel out because it is treated as the same kind of bond.

Estimating enthalpy of reaction with bond energies

Bond energies give you a rough estimate of reaction enthalpy, but will not pick up subtle differences between cases that are treated equally using this estimation. Here are two other examples where a C=O double bond is replaced by two C-O bonds, and the counts of all the other bonds are equal:

Hydration of a ketone, e.g. acetone and water reacting to form 2,2 propanediol. According to this web site, the reaction is slightly endothermic (about +6 kcal/mol). On the other hand, formaldehyde is mostly hydrated in water, so ketones and aldehydes behave slightly differently.
Butanone isomerizing to tetrahydrofurane. This is a hypothetical reaction, but NIST data suggests that it is also endothermic, again different than the estimate from bond dissociation energies.

So first, you have to look at the experimental evidence, and then you can try to rationalize it with a simple model, which sometimes will give the expected answer and sometimes will not.

"energetically favorable" vs major species

Just because a reaction is exothermic does not mean the product will be the major species. Not all monosaccharides form rings, and mixtures of ketones and alcohols typically do not form hemiacetals as major species. There are other energetic factors  (e.g. ring strain), and entropic factors (e.g. intramolecular vs intermolecular, conformational degrees of freedom) that also contribute to making one or the other species more prevalent.

  Why are two C-O single bonds more stable than a C=O double bond (hemiacetal)?

This is what we observe. A simple model like adding up bond energies seems to support it, but maybe that is just luck. Computational methods are sufficiently sophisticated to predict that certain monosaccharides form cyclic hemiacetals. I'm not sure this answers the question of "why?", though.


Why do molecules having a higher relative molecular mass have stronger inter-molecular forces?

There are a lot of reasons for why some molecules have stronger IMF's than others, but the trend of increasing IMF for increasing relative molecular mass ($M_\mathrm r$) is due to an increase in London Dispersion Forces, part of the greater set of van der Waals forces. As you may or may not know, these alkanes are non-polar molecules (that is, they don't have a real internal dipole--all $\ce{C-C}$ bonds are non-polar and the electronegativity of the $\ce{C-H}$ bond is close enough that those bonds are also considered non-polar). This matters because polar interactions are a relatively strong IMF.

Let's take a detour into e- distributions. Typically, electrons have the highest probability of being evenly distributed about the atoms in a molecule, proportional to the electronegativity of those atoms. That being said, that is only the most probable state, and so, of course, there are times when the e- are NOT evenly distributed. Remember how alkanes normally aren't polar? Well, when the electrons get unbalanced in the molecule, surprise! An Instantaneous Dipole is temporarily introduced.

What does this mean? Well, when an alkane is next to another alkane that is subject to an instantaneous dipole, the electrostatic forces from the quasi-polarized alkane cause the other alkane to also attain a lopsided distribution of electrons, causing it to gain a temporary polarity, and this effect both attracts these molecules while also propagating throughout the surrounding molecules.

So now that you have a little understanding of what LDF's or instantaneous dipoles are, why the trend with molecular weight? The short answer is that larger molecules have more e-, and are affected by LDF's more than smaller molecules with less e-. With more electrons and more atoms for those electrons to be distributed over, there is a greater chance for frequent or large instantaneous dipoles, increasing the overall Van der Waals forces.


In case of nitration in aniline, the Nitric Acid protonates the aniline to form the Anilinium ion.

Now since the nitrogen atom has no lone pair to conjugate, it has no mesomeric effect on the ring, the but as nitrogen is now protonated, it has a high negative inductive effect.

Which simply leads to the fact that the position farthest from the nitrogen substituted carbon will have the highest electron density (as inductive effect decreases as we move away from the atom causing it).

So in this case the best site for attack for the electrophile was para>meta>ortho.

The Hydrogen bonds would stabilise then ortho product once its formed, but the Rate Determining step of an electrophilic substitution on benzene is the sigma complex intermediate wherein the most stable intermediate is formed as major product. The positive charge in sigma complex stays the farthest in para position from the anilinium ion hence(most stable) is the major product.


Why are there quantised energy levels in the vibrational energy of a molecule?

"Why" is a good question, and one that science has yet to fully answer. We generally have a good understanding of "how" things work at the subatomic level, based on over a century of observation followed by theory and math backed up by experimentation. However, the really basic, naively simple questions, like "where do particles get the charges, spins and masses that define them" are the questions scientists are asking now.

As for the discrete quantum energy states theorized by quantum theory, those are the result of a lot of math that explains what we see much to well to be coincidence. Basically, an electron can be thought of as orbiting its nucleus, but it's really more like a vibration. Similar to a vibration, there are frequencies above the "fundamental" that produce "harmonics"; standing waves with wavelengths of a perfect integer fraction of the fundamental, and perfect integer multiples of frequency, which are self-stabilizing. Electrons can only exist in these discrete energy states because if they didn't, they'd "break orbit" and go zooming off as free radicals, or have an unstable or unenergetic path that sends them crashing into the nucleus.

What's weird is that that doesn't ever happen as far as we know; not only do electrons exist in stable states, they only exist in these states, and we can't get them to exist in any energy state between two discrete states.

Max Planck first posited these quantum energy states existed while attempting to solve a conundrum of classical physics. Back to the vibration analogy, consider a guitar string. By damping the string at precise points on the string, we can induce clear harmonics. however, the higher the harmonic, the harder you have to pluck to produce a wave of the same amplitude. As the wavelengths get smaller, the energy you have to put into them to obtain a constant amplitude (and thus the energy inherent in the wave; Newton's third law) increases exponentially, diverging to infinity as the wavelengths approach zero. Now, this was obviously wrong; observed visible spectra of a glowing piece of metal as it is heated clearly shows a "peak" in the frequency of light produced, and at frequencies above this peak, amplitude of emitted light drops off sharply.

Planck solved this "ultraviolet catastrophe" by describing what was going on as similar to this string having a number of springs of different length attached to it, with weights on the other end of each spring. As the string vibrated, it in turn induced vibrations in the springs (known technically as "harmonic oscillators"), and as the frequency increases, it approaches, matches, and then falls away from the harmonic frequencies of the springs, increasing their relative amplitudes and then making them fall off as the vibrations begin to cancel out, while simultaneously increasing the energy imparted to a different string whose harmonic was being excited. The relative harmonic frequencies of the springs are what determine the colors of light seen, and the total energy of the string is divided among the springs, keeping the same total energy as frequency increases instead of diverging.

To make the math work, Planck had to describe the springs as each receiving an integer portion of the energy of the entire system, and described these energy levels as proportional to the length of the spring. The proportionality is defined by a slope known today as Planck's Constant. Thus, Planck asserted that all energy existed in discrete amounts, which he called "quanta", hence "quantum theory".

While this worked, it seemed to do so by convenience, until Einstein then further developed photon theory: light was made of of photons, "packets" of energy that could be quantified in proportion to the photon's frequency, based on Planck's Constant. This in turn explained puzzling properties of the "photoelectric effect", namely that the threshold where electrons began to be ejected from metal when exposed to light (forming an electrical charge) was dependent not on intensity of light, but on the frequency of the light. Einstein's photons explained that, because an electron was likely to only be hit by a single photon of light, no matter how many there were, the energy imparted to the electron could only come from that single photon, and as a photon's energy increases with frequency, there is a minimum frequency required to impart enough energy with a single photon to eject an electron.

Planck's constant, and the highly definite nature of how atoms interact with photons, continues to show up in math that accurately describes things we observe about matter, such as how materials emit very specific wavelenghs of light while absorbing all others. That was explained by the Bohr model of the atom, which was the first model to introduce quantum theory to atomic structure. The Bohr model was based partially on a mathematical model first proposed by Johann Balmer and then improved on by Johannes Rydberg, showing that the emission spectra of hydrogen was predictable by an equation with integer solutions. Bohr showed that the energy states of electrons, which can be altered by absorption of photons, must exist in certain integral states, defined by an equation derived from Rydberg's but that also includes, you guessed it, Planck's Constant. The famous "double-slit experiment" showing that light is both wave and particle, and the resulting mathematical models by Louis de Broglie showing the movement of electrons as a series of stable standing waves of vibration in the electromagnetic field of the nucleus (electrons, in fact, are similar to photons in that they are as much a wave as a particle), reproduced the hypothetical energy states of Bohr's model by a different method, and in math, when you get independent confirmation of your results by a completely different mathematical model, you know you're on to something.

Now, that's the theory to date (well, actually I stopped about 100 years' worth of research ago, but them's the basics of the discovery of discrete quanta). What we are still trying to explain is why these quanta actually exist. A spaceship can have an infinite number of stable orbits around the Earth, and there are countless stable orbits of planets and planetoids around the Sun; why then does an electron stabilize so deterministically in these discrete states? Our best answer is that the electron's wave motion is only stable in these discrete states; otherwise the wave would interfere with itself or others and cancel out. However, we have never seen this "turbulence" happen and can't seem to force it to happen; it seems to be a fundamental quality of matter.


Formulations for wax thermostatic element for effect in alternative temperature range

Although typical applications for wax elements are centered around 40 °C, it is possible to create formulations for wax elements that change volume in the lower range of 5 °C to 20 °C.  The example below, although not linear in the lowest temperature extreme of the range of interest, behaves roughly linearly in the majority of the range.


Why Gold and Silver react minimally with atmosphere?

Copper, silver, gold. What is the energy of the highest "s" orbital versus the highest "d" orbital?  Gold is a very heavy nucleus.  An "s" orbital has non-zero probability at its nucleus.  Special relativity effects make its "s" electrons more massive, bringing them into the atom.

A silver atom is a soft Lewis acid.  It reacts well with large soft Lewis bases.  Silver quickly tarnishes with sulfide.  Gold is not chemically inert at all to even mild oxidation if the product cation is coordinated by ligands.  Salt water plus air will tarnish gold.  Dilute cyanide plus air will dissolve it (e.g., heap leaching of gold ore).

Thermodynamics proposes, kinetics disposes.  Aluminum is tremendously reactive toward air and water.  Anything that permanently breaches its adherent oxide coating (e.g., abrasion plus amalgamation) sets it off.


Cannizaro Reaction, Hydride Shift Reason?

Hydride shift occurs because in the intermediate dianion there is no other suitable leaving group. It is either $\ce{H-}$, $\ce{R-}$ or $\ce{O^{2-}}$. Of those $\ce{H-}$ is the best leaving group. Note though that even under these conditions $\ce{H-}$ will of course not be solvated on its own (it is way too unstable for that). It is directly transferred from the dianion to the aldehyde.

This reaction will go "to the right" since the hydride shift is irreversible. There is simply no mechanism to form a dianion from the alcohol and carboxylic acid.


Regarding logic behind Organic Chemistry

Sometimes it was a lucky accident, sometimes there was a lot of math and logic going in the designing experiments.  The organic chemistry of that era, though, was a lot more applied and empirical than the physics of today.  They usually weren't setting out to find the absolute laws-of-nature governing chemistry, they were just trying to make better dyes, medicines, rubber, etc, and stumbled upon new phenomena along the way.
Further Explanation
Almost all these guys worked in Germany in the late 1800s, when chemical industry was the hot new thing (think of biotechnology in today's world).  There was a lot of money to be made in making better dyes, medicines, rubber, etc, so people would hack/engineer the established methods for making that stuff, and see if the resulting products were any better.
For example:
Chemical A cures headaches really well, but causes nausea.  Chemical B gives a little relief from headaches, but doesn't have any side effects.  I bet I could make chemical C (which no one had ever made before), which has a structure that's sort of a combination of chemicals A and B, and maybe that will cure headaches with no side effects.
==> {made the chemical and did some tests to find out the structure}.
What the heck!  This has a completely different structure than what I was expecting, it must have undergone a second reaction to rearrange itself.
==> {publish paper},{get famous},{people 100 years later question the logic of your experiments} :)


What is p defined as for the expansion and compression of an ideal gas system in the equation ΔU = ΔH - pΔV?

In the following I limit discussion to the particular expression $$ΔU = ΔH - pΔV$$
which follows (spoiler alert) from $$H\equiv U+pV$$ when the pressure $p$ of the system (the container) is  identical at the beginning and end of the process.

  Is the pressure outside the container assumed to always be constant?

For an isobaric process such as the stated process involving zinc, by definition yes. Generally speaking the expansion work in the 1st Law is written  $$dw = -p_{ext}dV$$ where $p_{ext}$ is the pressure of the surroundings against which the system does work. If $p_{ext}$ is constant work is still against or by the pressure of the surroundings and the integral can be evaluated as $$w=-p_{ext}\Delta V $$ For a reversible process the pressure of the system and surroundings are equal and we evaluate $$w=-\int_{Vi}^{Vf} pdV$$ where $p$ is the pressure of the system.

  Is the pressure inside the container assumed to always be constant? I assume no looking at the graph of Boyle's Law.

In the case of a reversible process against constant external pressure, yes.

BTW Boyle's law ($p \propto V^{-1}$) does not consider changes in the number of particles during a reaction.

  For expansion of the gas, work is done by the system (i.e. the gas in the container) on the surroundings, so is p the pressure of the surroundings as work is done against the pressure of the surroundings?

The pressure of the system (container) and surroundings are assumed to be equal (balanced) for a reversible process. However, if the pressure of the system is greater than that of the surroundings (the latter assumed constant) then the work is still $$w=-p_{ext}\Delta V$$

  This doesn't seem to be the case in several examples in my lecture notes: e.g. calculate the work done against pressure of 1 bar at 298 K when 1 mol of zinc is dissolved in aq. HCl

$pV$ work involves a change in volume. Here the volume change is due to formation of gas which causes the pressure in the container to increase by an infinitesimal amount $\mathrm{d}p$. The gas pushes against the piston, increasing the volume and returning the system to mechanical equilibrium when the internal pressure equals the external pressure $p$.

  For compression of the gas, work is done by the surroundings on the system, so is p the pressure of the gas in the container as work is done against this pressure?

Assuming no chemical processes, compression requires the surrounding pressure to be higher than that of the system. It suffices that the difference be infinitesimal in magnitude, resulting in a reversible process. During such a process, the pressure $p$  of the system may vary during the entire process, remaining always equal (to within an infinitesimal amount) to the applied pressure.


Finding initial amount of ammonia using equilibrium constant and amount of nitrogen gas

Since the volume is given and it's a constant, the initial amount of ammonia $n_0(\ce{NH3})$ can be found from its initial concentration $c_0(\ce{NH3}):$

$$n_0(\ce{NH3}) = c_0(\ce{NH3})\times V\tag{1}$$

To find $c_0(\ce{NH3}),$ an ICE table might indeed come in handy; however, yours needs corrections.
First, I suggest to rewrite it according to the process occurring in the system, namely dissociation of ammonia:

$$
\begin{array}{lccc}
\ce{&2 NH3(g) &<=> &N2(g) &+ &3 H2(g)} \\
\text{I} & c_0 && 0 && 0 \\
\text{C} & -2x && +x && +3x \\
\text{E} & c_0 - 2x && x && 3x \\
\end{array}
$$

Second, note that the provided equilibrium constant $K_c$ is given for the synthesis of ammonia, hence in our case a reciprocal value is relevant:

$$
\begin{align}
\frac{1}{K_c} &= \frac{[\ce{N2}][\ce{H2}]^3}{[\ce{NH3}]^2}\tag{2.1}\\
\frac{1}{K_c} &= \frac{x (3x)^3}{(c_o - 2x)^2}\tag{2.2}\\
\frac{1}{K_c} &= \frac{27x^4}{(c_o - 2x)^2}\tag{2.3}\\
\frac{1}{\sqrt{K_c}} &= \frac{3\sqrt{3}x^2}{c_o - 2x}\tag{2.4}
\end{align}
$$

$$c_0 = 3\sqrt{3K_c}x^2 + 2x\tag{3}$$

At the equilibrium there is $\pu{0.200 mol}$ of nitrogen in the $\pu{1 L}$ vessel, so $x = \pu{0.200 mol L-1},$ and the initial amount of ammonia can be found as follows:

$$
\begin{align}
n_0(\ce{NH3}) &= (3\sqrt{3K_c}x^2 + 2x) × V\\
  &= (3\sqrt{3\times 4.20}\times (\pu{0.200 mol L-1})^2 + 2\times \pu{0.200 mol L-1})\times \pu{1 L} \\
  &= \pu{0.826 mol}\tag{4}
\end{align}
$$


Are the collisions between the real gas particles perfectly elastic?

Well, what is an inelastic collision, really? Suppose you have two balls made of steel; they collide, then fly away with some lasting deformation, so some energy is lost. With molecules, it is not quite like that. You can't leave a dent on a molecule. It has certain discrete energy levels, and that's it. You either excite the molecule to one of these levels, or you don't excite it at all.

To sum it up, some collisions of molecules are perfectly elastic, and others are combined with excitation of some rotational or (more probable at higher temperatures) vibrational mode in one of the molecules, or maybe in both.

Noble gases which have no molecules and hence neither rotational nor vibrational modes may enjoy perfectly elastic collisions up to pretty high temperatures.


nomenclature of some organoselenide moieties

Based on usage:

Selenenylate is a verb. It means to attach a selenium group to an organic compound - See this passage from Comprehensive Organic Synthesis: Selectivity, Strategy, and Efficiency in Modern Organic Chemistry, Volume VII

  The analogous morphinoselenamide has been shown to selenenylate the $\alpha$-ketoester...

Selenolate is a noun. It is the $\ce{RSe-}$ anion, especially in ionic or coordination compounds. See various usage at the links in this Google search


Calculating Delta H for Non Ideal Gasses

Are you familiar with the following equation?:  $$dH=C_pdT+\left(V_m-T\left(\frac{\partial V_m}{\partial T}\right)_P\right)dP$$


Titration curve of a diprotic acid with a strong base titrant

The pKa values for sulfuric acid from Wikipedia are −3, and 1.99. So for all practical purposes the first ionization happens completely and would be unobservable with a pH electrode in aqueous solutions. So if you had a 0.1  molar solution of sulfuric acid it would almost look like you were titrating a 0.2N of $\ce{H^+X^-}$ where that acid had a pKa of 1.69 (pH at half NaOH used). This would be a strong acid too and the inflection point would be that for water at pH 7.

Note that the pKa 1.69 is assuming $\ce{H^+X^-}$ and is taken at 1/2 of the $\ce{Na^+OH^-}$ used. Knowing that $\ce{H2SO4}$ has two protons, the second pKa should be taken at 150% on the plot which would give the correct pKa value of 1.99.

If you had an acid with two lower pKa values then you'd see two pH transitions in the titrations. Look at the figure below. With two widely separated pKa values maleic acid shows two nice transitions. With two pkA values closer together the two transitions for malonic acid aren't as sharp. For succinic acid the knee at a volume of about 5 ml is a hint that there are two dissociations but you don't see two nice complete transitions like you do for maleic acid.

To go back to the sulfuric acid tritration, we'd really expect to see a knee at a low pH something like the titration curve we got for succinic acid. Such a knee isn't seen because a pH electrode just doesn't work at such high acid levels (pH range from -3 to 0).

PS - Thanks to the commenters for help to vastly improve the answer.


Probably not so much calcium hydroxide, but alkali metal hydroxides mixing with each other in the liquid phase.  The pairs $\ce{LiOH-NaOH}$ and $\ce{KOH-NaOH}$ are not only fully miscible, they have been studied as fuel cell electrolytes.  The picture below, from the referenced article, shows results for the a range of electrolyte compositions that imply the extensive miscibility.  The fuel cell itself is based on carbon oxidation to carbon dioxide.


Will ice cream freeze faster with frozen brine or adding salt to ice?

If the brine melts at the same temperature you reach with the ice and salt, then it should take the same time to freeze the ice cream since the freezing time depends on heat transfer. The benefit of frozen brine is that there is more heat capacity in the frozen brine than in ice-salt cooled water, so the ice-salt water may become warmer than the frozen brine.

It may be that there is poor heat transfer from the cream to the ice so it takes longer than you think. Also, if you are the one turning the crank, time seems to stretch out since your muscles get tired and yet the ice cream is still not frozen.


Change in Entropy and Enthalpy found over a temperature range and asked to compute the change in Gibb's Energy

This should address your confusion about constant T:

$$G = H - TS \Rightarrow dG = dH -TdS - SdT$$

$$ \Rightarrow \text{(at constant T) } dG = dH -TdS \Rightarrow \Delta G = \Delta H - T \Delta S$$

So, yes, you need $\Delta T = 0$ for $\Delta G = \Delta H - T \Delta S$ to hold.

However, this holds at any T.

E.g., suppose you do the reaction at $300 K$. $\Delta T = 0$, so $\Delta G = \Delta H - T \Delta S$ holds.

But suppose you repeat reaction at $350 K$. It's still the case that $\Delta T = 0$, so $\Delta G = \Delta H - T \Delta S$ still holds.

As far as the question itself goes, I agree with Chet Miller that it appears to be badly worded.  It seems what your tutor meant to say was:

"Between 300 ºC and 350 ºC, assume that $\Delta H$ and $\Delta S$ are constant, with values of 1700 kcal and 15kcal/K, respectively. Calculate the range of temperatures over which $\Delta G <=0$, i.e., the range the textbook labels as 'spontaneous'".