UPDATE:
Today I got a chance to read the abstract the original paper which made news everywhere that moon has rust on the side which faces the Earth.
The authors propose that upper atmosphere oxygen from the Earth reaches the moon. In their words
"Oxygen delivered from Earth’s upper atmosphere could be the major oxidant that forms lunar hematite. Hematite at craters of different ages may have preserved the oxygen isotopes of Earth’s atmosphere in the past billions of years. Future oxygen isotope measurements can test our hypothesis and may help reveal the evolution of Earth’s atmosphere."
Widespread hematite at high latitudes of the Moon in Science Advances (open access

Let us split your query into three parts:
a) Will iron oxidize without water and form oxides?
Iron would happily burn in oxygen to form iron oxides. But we cannot call it iron rust. So iron can oxidize without moisture. In fact, iron powder may spontaneously burn in air.
b) Will iron form rust without water?
Rust is more of a semantic issue.  Iron rust may not have constant composition. As per the dictionary definition (OED) "

A red, orange, or yellowish-brown substance which forms progressively
as a flaking, permeable coating on the surface of iron and its alloys
as a result of oxidation, esp. through exposure to air and moisture.

Therefore, rust is an product of ambient environmental factors that affect iron at (approx.) typical room temperature. Rusting in sea water might different than rusting on a mountain. Acidic gases such as carbon dioxide, sulfur dioxide, nitrogen oxides certainly accelerate this process. Iron rust can be green and it can be common brown version. Since, chemically rust is a hydrated oxide of iron plus its hydroxide, water's presence is a must. More importantly water is a necessary ingredient which works as mediator of the electrochemical cell that forms between oxygen and iron surface. If you search Google Scholar you would find tons of articles on the electrochemical mechanism of iron rusting.
Coming to the more interesting question of rust on the moon:

nasaspaceflight.com: Rust on the Moon. How is that possible without
oxygen and liquid water? "But how can rust form far from water ice
deposits on a barren oasis devoid of oxygen?" Axios: Researchers find
rust on the Moon "Scientists were surprised by the findings because
rust requires oxygen and water to form on Earth."

Since you would know more astronomy than I do, wasn't moon a part of the Earth? I recall this from a documentary. I do remember seeing layers of "rust" in moutainous paths, so the oceans had a lot of iron which settled as a layer in mountains. If moon were a part of Earth, why is that surprising?
a) The question is what was the age of this rust?
Secondly, Mars is also "rusty", its surface is highly oxidizing with a huge amount of perchlorate in the soil (there is a long story how that was discovered by the folly of analytical chemists). Who knows what is the moon soil like. What about the shower of radiation on the moon. That may promote oxidation as well.


Formulae for energy of photon (E=hf and E=hv)

I understand your confusion. They all are actually the same formulas. It is because lets start with the first equation,

$$E=h \cdot f$$

But we know that frequency can be written in terms of speed formula,
$$ v = f \cdot \lambda$$

Which can be rearranged into the following,
$$ \frac{v}{\lambda} = f $$

If you put that expression into the above equation, then that becomes

$$ E = f \cdot \frac{v}{\lambda}$$

Hence the same formula can be written in many ways.

Your first two equations are just the same. $f$ usually stands for frequency and in the second equation you have just used $\nu$ for it. That is the frequency of the photon that is coming in or being released.

Whereas your third equation actually comes from the speed formula as I have written above,
$$ v = f \cdot \lambda$$

Rearranging for frequency, you get:
$$ \frac{v}{\lambda} = f $$


What happens when ferric chloride is dissolved in water?

Iron(III) chloride indeed hydrolyses in water, or to be precise, $\ce{Fe^3+}$ does assuring acidic medium.
Hydrolysis isn't completely suppressed even in strongly acidic solutions.
In the absence of other factors, what happens exactly depends on the concentration of chloride and temperature.

In diluted cold solution a typical formation of metal aqua ions takes place:

$$
\begin{align}
\ce{FeCl3 + 4 H2O &<=> [Fe(H2O)4Cl2]+ + Cl-}\tag{R1.1}\\
\ce{[Fe(H2O)4Cl2]+ + H2O &<=> [Fe(H2O)5Cl]^2+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 0.30\tag{R1.2}\\
\ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)6]^3+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 1.48\tag{R1.3}\\
\ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)4(OH)Cl]+ + H3O+} &\quad \mathrm{p}K_\mathrm{a} &= 3.05\tag{R1.4}\\
\end{align}
$$

In concentrated cold solution aquacomplex is also accompanied by  tetrachloroferrate(III):

$$\ce{2 FeCl3 + 4 H2O <=> [Fe(H2O)4Cl2)]+ + [FeCl4]-}\tag{R2}$$

In boiling-hot solution iron(III) chloride decomposes to hydroxochloride

$$\ce{FeCl3 + 2 H2O ->[Δ] Fe(OH)2Cl(s) + 2 HCl}\tag{R3}$$

which is partially hydrolyzed further to colloidal iron(III) hydroxide ($\ce{Fe(OH)2Cl}$ is still more soluble than $\ce{Fe(OH)3}$).

Finally treatment with steam results in iron(III) oxide:

$$\ce{2 FeCl3 + 3 H2O ->[\pu{350 - 500 °C}] Fe2O3 + 6 HCl}\tag{R4}$$

Reactions were adopted from [1, pp. 420–421].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Solubility increases as Temperature increases.. Ks increases with temperature meaning dissolving is endothermic... But what about NaOH?

Let me address this statement first:

  We all know that the solubility of ionic substances increases as increases in temperature. That is a fact, we don't have to argue with that.

While this is frequently the case, there are plenty exceptions, and only one exception is required to prove the statement false. Lets consider the solubility of $\ce{Ca(OH)2}$.

$\begin{array}{rr}
\pu{0^oC}: & \pu{1.89g/L}\\
\pu{20^oC}: & \pu{1.73g/L}\\
\pu{100^oC}: & \pu{0.66g/L}
\end{array}$

$\ce{Li2SO4}$ is another example that follows this trend of decreasing solubility with increasing temperature, and there are plenty others.

Although the heat of a chemical reaction often follows Le Chatelier's principle, there are exceptions.  Based on the facts as laid out above and in the OP, the dissolution of $\ce{NaOH}$ is a case in which Le Chatelier's principle does not apply to the heat of the reaction; the dissolution of $\ce{NaOH}$ is exothermic, yet it's solubility increases with temperature.


Frequency, the only real characteristics of electromagnetic radiation ??

Recall that frequency $f$, wavelength $\lambda$ and the speed $v$ of a wave are all related as follows
$$
    v = f \lambda \, .
$$
So presumably, your book is talking about a situation when an electromagnetic wave crosses boundaries between different media. In such cases frequency indeed remains constant, while the speed of the wave, and consequently, its wavelength change.


Do high-energy chemical bonds indicate strong chemical bonds?

There are two factors here to consider.
First, the energy released by a given reaction is (partly) due to the difference between the bond strength in the product vs reactant, not the absolute bond strength, unless the reaction involves just breaking a bond and not forming a new one.
Second, even if your reaction is just a bond breaking, there are multiple ways to break a bond. Reported bond enthalpies are usually defined in terms of the energy required for homolytic bond breaking, ie one electron stays on each component. For example,
$$\ce{HCl -> H. + Cl.}$$
By that measure, the H-Cl bond is quite strong, but we know that in a polar solvent that mitigates the negative effect of charge separation, the H-Cl will readily undergo heterolytic bond cleavage, where both electrons stay on one atom:
$$\ce{HCl -> H+ + Cl-}$$
In that context, we might consider the H-Cl bond to be very weak. (Note, though, that I'm also overlooking the amount of "bonding" between solvent and ions in this case, which means that it's not an example of a pure bond breaking reaction.)
All of which is to say that before you can make a statement about bond energy and bond "strength" you need to define both terms more specifically.


Do different vibrational modes correspond to vibrational energy levels?

yes and no: all vibrational normal modes have a different energy (with the proviso that some may be accidentally degenerate) and hence different absorption frequency. (Fundamental frequencies have vibrational transitions that occur with selection rule $\Delta n=\pm 1$.) All normal modes exist in their ground state with zero point energy (quantum number $n=0$) as well as in other levels based on the same mode ($n>0$) which are populated according the Boltzmann distribution at temperature T. Thus each basic vibrational mode i has its own stack of energy levels, with quantum numbers $n= 0, 1, 2 \cdot\cdot$ and (harmonic oscillator) energy $E_i=\nu_i(n_i+1/2)$ with quantum numbers $n_i$ and frequency $\nu_i$ in  $\pu {cm^{-1}}$. The total zero point energy of the molecule is the sum of all the individual zero point energies above the zero of energy.

The number of vibrational levels populated at a given temperature depends on the vibrational level spacing, thus low frequency modes, say, $500 \pu{cm^{-1}}$ have far more levels populated than a high frequency, say $1500 \pu{cm^{-1}}$ mode.

As the energy increases the number of vibrational energy levels / unit energy can become very large indeed in polyatomic molecules. There are overtones where only one vibrational mode is excited, but with more than one quantum of energy, and combinations where two or more different vibrations of different frequency are excited and so each have non zero quantum numbers.

In water molecules there are 3 vibrations (of frequency approx $3650,1595, 3756 \pu{cm^{-1}}$) if we label them in order with quantum numbers then [000] is the zero point energy $(3650+1595+3756)/2=4500 \pu{cm^{-1}}$,  [010] a fundamental in the second frequency and [020] an overtone in this mode and [110] a combination of modes 1 and 2 with energy above zero of $4500+3650+1595 \pu{cm^{-1}}$


There should be no issue with putting carbonated water in plastic bags and freezing them (considering they're clean and sterile, of course). I imagine the benefit of freezing carbonated water instead of regular clean water has to do with the fact that there is much more dissolved gas in carbonated water. As the water freezes, the gas is expelled from the ice (only the water molecules will solidify), creating many pockets inside the solid ice which would weaken its structural integrity. This would likely make it much easier to shave or break into small pieces for a snow cone. I've never frozen carbonated water myself though. You should be a bit careful because as the carbonated water freezes, the carbon dioxide it releases will create pressure in a closed container, so you may end up with torn plastic bags, ruptured bottles, or a damaged freezer in the worst case.


Are glucose and galactose cis-trans isomers of each other?

Glucose and Galactose are best described as epimers of one another.
The three structures below all have the same molecular formula and the same connectivity, but differ in the absolute configuration at a single stereocentre (this is the definition of an epimer).
You can see that glucose and galactose are epimeric at the $\mathrm{C4}$ position, whilst glucose and mannose are epimeric at the $\mathrm{C2}$ position.

Image taken from Principles of Biochemistry, Lehninger et. al. (5 ed.)

To comment on the use of cis and trans.
The cis/trans nomenclature is used to describe the geometric isomerism of alkenes - it is not correct nor appropriate to use it in this context.  What you may be referring to is whether two adjacent stereo centres are syn or anti to one another.


Why dishwashers are corrosive to aluminum?

I think it is just alkalinity. Unlike many other metals, aluminum is amophoteric, so it dissolves in both strong base and in strong acid. Most dishwashing detergents are fairly alkaline, probably at pH 11 or 12 or so. That and the heat may be enough to corrode Al.  Most other common metals are attacked by strong acids but not by alkali.


None of the comparisons you might make suggests any sort of deatabilization by conjugation.

Let's look at the four-carbon compounds.  Depending on which source you believe for 1,3-butadiene, you could release up to 27.8 kcal/mol by hydrogenating one double bond (the figures given for this compound refer to hydrogenating both double bonds).  But then you release 30.3 kcal/mol hydrogenating the remaining double bond in 1-butene.  The combination of two double bonds in 1,3-butadiene has less energy to be released by hydrogenation than two independent bonds like the bond in 1-butene has.

Similarly for the five-carbon compounds.  When you hydrogenated both double bonds in the pentadiene isomers, a pair of conjugated double bonds in 1,3-cyclopentadiene has less energy to release than the non-conjugated double bonds in 1,4-cyclopentadiene.

Finally, something a little more subtle is going on between 1-butene and 2-butene.  There can no conjugation like that in 1,3-butadiene when only one double bond is present, yet 2-butene has clearly less energy to release by hydrogenation than 1-butene.  Thinking of carbocation nomenclature, the former is a like a "secondary carbication by having an alkyl group on only one end of the double bonds, while the latter is "tertiary" with an alkyl group on both ends.  Hyperconjugation, like that in carbocations, is at work here.


Unit of the equilibrium constant: contradiction of Bridgman's theorem?

The problem is that people are often sloppy with the definition of quantities. The equilibrium constant $K$ in your first equation is indeed a dimensionless quantity while the equilibrium constant $K_c$ that is usually used to describe an equilibrium in a solution is not. I will take some detour to show where they come from and how they are connected.

From thermodynamics it is known that the Gibbs free energy of reation is given by
\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ ,
\end{equation}

where $\xi$ is the extent of reaction and $\nu_{i}$ and $\mu_{i}$ are the stochiometric coefficient and the chemical potential of the $i^{\text{th}}$ component in the reaction, respectively.
Now, imagine the situation for a ideal system consisting of two phases, one purely consisting of component $i$ and the other being a mixed phase comprised of components $1, 2, \dots, k$, in equilibrium.
Since the system is in equilibrium and shows ideal behavior we know that the chemical potential of component $i$ in the mixed phase (having the temperature $T$ and the total pressure $p$), $\mu_{i}(p, T)$, must be equal to the chemical potential, $\mu^{*}_{i}(p_{i}, T)$, of the pure phase having the same temperature but a different pressure $p_{i}$, whereby $p_{i}$ is equal to the partial pressure of component $i$ in the mixed phase, namely
\begin{equation}
 \mu_{i}(p, T)= \mu^{*}_{i}(p_{i}, T).
\end{equation}
From Maxwell's relations it is known that
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = \left( \frac{\partial}{\partial p} \biggl(\frac{\partial G^{*}}{\partial n_{i}}\biggr) \right)_{T} = \Biggl( \frac{\partial}{\partial n_{i}} \underbrace{\biggl(\frac{\partial G^{*}_{i}}{\partial p}\biggr)}_{=\, V_{i}} \Biggr)_{T} = \left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}
\end{equation}

but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to
\begin{equation}
 \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}
\end{equation}

and one gets
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = v_{i} \ ,
\end{equation}

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase.
Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get
\begin{equation}
 \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p   \ ,
\end{equation}
so that, introducing the mole fraction $x_{i}$,
\begin{equation}
 \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .
\end{equation}
Please, note that there is a dimensionless quantity inside the logarithm.
Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$
\begin{equation}
a_{i} = \frac{\varphi_{i} p_{i}}{p^0}
\end{equation}
where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition.
When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises
\begin{equation}
 \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .
\end{equation}
Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets
\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}
where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure.
Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that
\begin{equation}
\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}
So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant
\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}
which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via
\begin{equation}
 K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
it is more or less proportional to $K$ and thus gives qualitatively the same information.

Edit: If the solution at hand behaves like an ideal solution then by definition it's activity/fugacity coefficient is equal to one. Furthermore the state of ideality is defined with respect to standard states: for an ideal solution this is $c^{\ominus} = 1 \, \text{mol}/\text{L}$. Using this together with the ideal gas law on the relation between $K$ and $K_{c}$
\begin{equation}
 K = \prod_i [\underbrace{\varphi_{i}}_{= \, 1}]^{\nu_{i}} \Bigl(\underbrace{\frac{RT}{p^{0}}}_{\substack{= \, 1/c^{\ominus} \, = \, 1 \, \text{L} / \text{mol} \\ \text{per definition}}}\Bigr)^{\sum_i \nu_{i}} K_{c} \qquad \Rightarrow \qquad K = \left(\frac{L}{\text{mol}} \right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
one sees that for an ideal solution $K_{c}$ is identical to $K$ scaled by a dimensional prefactor.

Edit: I forgot to mention there are also "versions" of the equilibrium constants that are defined in terms of partial pressures or mole fractions which provide a more suitable description for gas equlibria but all those "versions" can be traced back to the original equilibrium constant.


Why can't some salts be used in the salt bridge of specific voltaic cells?

The main reason would be that sulphates and chlorides can form precipitates in solution. Potassium nitrate essentially never forms in precipitates in solution by going off the solubility rules. The lead in the cell would form precipitates with the sulphate and chloride, and thus it would be ineffective as a salt bridge.


Magnetic moment of trivalent lanthanide cations

Your question, Why can't we use normal formula to find effective magnetic moment of tripositive rare earth elements (REs)? is answered excellently by  porphyrin' comment (vide supra). Thus, I won't attempt to answer it again except for to mention the calculated values from the equation, $\mu_\mathrm{eff} = g_J \sqrt{J(J+1)}\mu_\mathrm{B}$ (where $\mu_\mathrm{B}$ is Bohr magneton and $g_J$ is Landé factor), have good agreement with experimental values.

Table 1: Theoretical magnetic moment of tripositive rare-earth ions
$$
\begin{array}{c|ccc}
\text{Ion} & 4f^n & S^c & L^c & J^c & g^c & \mu_\mathrm{eff} = g_J \sqrt{J(J+1)}\mu_\mathrm{B} & \text{Observed }\mu^a \\
\hline
\ce{La} & 0  & 0 & 0 & 0 & - & 0.00 & -\\
\ce{Ce} & 1 & \frac{1}{2} & 3 & \frac{5}{2} & \frac{6}{7} & 2.54 & 2.4\\
\ce{Pr} & 2 & 1 & 5 & 4 & \frac{4}{5} & 3.58 & 3.5\\
\ce{Nd} & 3 & \frac{3}{2} & 6 & \frac{9}{2} & \frac{8}{11} & 3.62 & 3.5\\
\ce{Pm} & 4 & 2 & 6 & 4 & \frac{3}{5} & 2.68 & -\\
\ce{Sm} & 5 & \frac{5}{2} & 5 & \frac{5}{2} & \frac{2}{7} & 0.85 & 1.5\\
\ce{Eu} & 6 & 3 & 3 & 0 & - & 0.00 & 3.4\\
\ce{Gd} & 7 & \frac{7}{2} & 0 & \frac{7}{2} & 2 & 7.94 & 8.0 \ (7.23^b)\\
\ce{Tb} & 8 & 3 & 3 &6 & \frac{3}{2} & 9.72 & 9.5 \ (9.58^b)\\
\ce{Dy} & 9 & \frac{5}{2} & 5 & \frac{15}{2} & \frac{4}{3} & 10.65 & 10.6\\
\ce{Ho} & 10 & 2 & 6 & 8 & \frac{5}{4} & 10.61 & 10.4\\
\ce{Er} & 11 & \frac{3}{2} & 6 & \frac{15}{2} & \frac{6}{5} & 9.58 & 9.5\\
\ce{Tm} & 12 & 1 & 5 & 6 & \frac{7}{6} & 7.56 & 7.3\\
\ce{Yb} & 13 & \frac{1}{2} & 3 & \frac{7}{2} & \frac{8}{7} & 4.54 & 4.5\\
\ce{Lu} & 14 & 0 & 0 & 0 & - & 0.00 & -\\
\end{array}
$$
$^a$ Rare-earth and metals really show "Spin"?; $^b$ Reference 1; and $^c$ Reference 3.

The two experimental values from Ref.1 are in $\pu{emu mol−1 Oe−1}$ and have been converted appropriately to be able to compare with the values in hand. The reference also point out that crystals of mixed salt have approximately average of the pure monometal salts. The abstract of Ref.1 states that:

  Magnetic moment and susceptibility measurements of single crystals of pure and mixed rare earth fumarates of gadolinium and terbium were carried out at room temperature. The experimental values of molar susceptibilities for $\ce{Gd2(C4H2O4)3.7H2O}$, $\ce{Tb2(C4H2O4)3.7H2O}$, and $\ce{GdTb(C4H2O4)3.7H2O}$ are $2.68 \times 10^{−2}$, $3.89 \times 10^{−2}$, and $3.18 \times 10^{−2}$ (in $\pu{emu mol−1 Oe−1}$), respectively. The calculated effective magnetic moments are in good agreement with the theoretical predictions on rare earth ions.

The magnetic properties of rare earth elements (REs) are due primarily to the electrons in the incomplete shell ($\mathrm{4f}$). In fact the magnetic moment per atom is determined by the number of electrons in the $\mathrm{4f}$ shell according to the Russell-Saunders coupling scheme. The derivation of the equation $(1)$ is partially explained in Ref.2:

$$\mu_\mathrm{eff} = g_{J} \sqrt{J(J+1)}\mu_\mathrm{B} \tag{1}$$

According to Ref.2:

  The Russell-Saunders coupling scheme is normally adopted to describe the $\mathrm{4f}$ magnetism of the REs. The spins of the individual electrons are coupled to form a total spin $S$ by the exchange interaction and the individual orbital angular momenta are coupled to form a total orbital momentum $L$ by the Coulomb interaction. The state with the lowest energy is found from Hund’s rules by maximizing $S$ and thereafter $L$. The total angular momentum $J$ is given by $J = |L ± S|$, where the minus sign is used for less than half filled shells and the plus sign for more than half filled shells. The total magnetic moment due to the spin angular momentum is $µ_S = 2µ_B\sqrt{S(S + 1)}$, where $µ_B$ is the Bohr magneton. The total moment due to the orbital angular momentum is $µ_L = µ_B\sqrt{L(L + 1)}$. These moments process around the direction of $J$, therefore the magnetic moment due to the total angular momentum looks slightly more complicated: $µ_J = g_J µ_B\sqrt{J(J + 1)}$, where $g_J$ is the Landé factor [12].

The sighted Reference ([12]) is actually Ref.3 here. According to Ref.3:
$$
J =
\begin{cases}
L-S,  & \text{if $4F^n \le 7$} \\
L+S, & \text{if $4F^n \gt 7$}
\end{cases}
$$

Note: The equation $(1)$ does not obey experimental values of $\mu$ found for $\mathrm{3d}$ elements (Rare-earth and metals really show "Spin"?).

References:

B. Want, M. D. Shah, “Magnetic susceptibility measurements of pure and mixed gadolinium–terbium fumarate heptahydrate crystals,” Journal of Magnetism and Magnetic Materials 2016, 401, 391-393 (https://doi.org/10.1016/j.jmmm.2015.10.065).
I. L. M. Locht, Y. O. Kvashnin, D. C. M. Rodrigues, M. Pereiro, A. Bergman, L. Bergqvist, A. I. Lichtenstein, M. I. Katsnelson, A. Delin, A. B. Klautau, B. Johansson, I. Di Marco, O. Eriksson, “Standard model of the rare earths, analyzed from the Hubbard I approximation,” Phys. Rev. B 2016, 94, 085137 (https://doi.org/10.1103/PhysRevB.94.085137)(PDF).
Jens Jensen, Allan R. Mackintosh, In Rare earth magnetism: Structures and Excitations; The International Series of Monographs on Physics, J. Birman S. F. Edwards, C. H. Llewellyn, & Smith M. Rees, General Eds.; Clarendon Press: Oxford, United Kingdom, 1991.


How do we know that we have isolated a substance?

While modern tools make this easier, there are plenty of physical characteristics that help identify that a substance is pure

Modern tools like NMR, X-ray structure determination and others make identifying pure substances and their structures relatively easy, there are plenty of other characteristics than can support the identification of a pure substance.

Pure compounds tend to have sharply defined boiling points or melting points, for example. So if you can generate crystals, that is a good start. Crystallisation from solution, for example, tends to yield relatively pure compounds and this can be checked by measuring the melting point. Repeated fractional crystallisation can confirm this if the melting point doesn't change. For liquids you can distill them and observe the boiling point. Fractional distillation can separate mixtures and you can check that the results boil at a single temperature. Gases also have physical properties that can be checked after fractional distillation or other separation techniques.

Here is an example where attention to detail yielded the discovery of a new element. Rayleigh and Ramsay were experimenting with the separation of gases and noticed that nitrogen isolated from air (by removing all the other known gases) seemed to have different properties to nitrogen formed from chemical reactions (it was about 0.5% denser, for example). By careful fractional distillation, Ramsay was able to isolate a much denser gas from the nitrogen which we now call Argon. This was verified to be an element by observing its discharge spectrum which had new lines in it not present for other elements. So careful separation techniques and carful observation of physical properties could yield a new substance with different properties to nitrogen.

Before we had modern instrumentation this careful application of known chemical and physical techniques was the basis for the development of the laws of chemistry and the identification of new pure substances. It takes more time, but it works.


Temperature in the Maxwell-Boltzmann Distribution

If you are assuming a pseudo-steady state condition (where the bulk of the liquid is in thermal equilibrium with itself for the timescale of interest), and you want to study the bulk molecules of the liquid, then you want to use the temperature of the liquid.

Now, if you are going to be studying the surface of the liquid, using a Maxwell-Boltzmann distribution is much more questionable - it definitely isn't strictly correct, at least on time scales of any substantial length, but you might be able to get away with it for rough approximations.

If you gave some more details as to the specific application, I could probably give you a better answer.


Extracting "Crystal Radii" from a CIF file

The CIF file won't contain information on atomic radii, only the positions (in fractional space).

Consider that the atom might be ionic, covalent, etc. so you'll have to choose the atomic radius definition you feel is most appropriate. (Or run some sort of quantum calculation and use the resulting [electron density to define an atomic radius].)

For what it's worth, Open Babel does have a data table of covalent and van der Waals atomic radii, but it sounds like you want something else.


Overall equilibrium expression for competitive equilibria

OK, so essentially we have $[A],[B],[C],[AB],\text{ and }[AC]$, and a few equations: three conservation laws and two equilibrium constants.
$$\left\{
\begin{align}
[A]+[AB]+[AC] & = A_0\\
[B]+[AB] & = B_0\\
[C]+[AC] & = C_0\\
{[AB]\over[A][B]}& =K_1 \\
{[AC]\over[A][C]}& =K_2
\end{align}
\right.$$
With 5 unknowns and 5 equations, this should be solvable (because the nature manages to solve it somehow, if not for other reason). There is no guarantee, however, that the solution will be nice. Indeed, if we deal with the system exactly as it stands now, it is equivalent to a certain 4th degree algebraic equation which is better solved numerically.

Things get somewhat more manageable if we may disregard something. For example, if A is a relatively minor component, so that $[A]<<[B]$, then we may assume $[B]\approx B_0$ (and the same for C). All of a sudden, the problem becomes linear:
$$[AB]=K_1[A]B_0 \\
[AC]=K_2[A]C_0 \\
[A]+K_1[A]B_0+K_2[A]C_0=A_0
$$
which yields
$$
[AB]={K_1B_0\over1+K_1B_0+K_2C_0} \\
[AC]={K_2C_0\over1+K_1B_0+K_2C_0}
$$

Whether or not this approximation is realistic in your particular case is up to you.


What is the potential to use for non-dispersion interactions?

Because all three of these interactions are of the same order in the interelectronic distance ($1/r^{6}$), they are all captured within the Lennard-Jones potential to the extent of the accuracy of the potential. From the "Limits" section:

The term Van der Waals (vdW) forces roughly includes the following intermolecular forces:

  the permanent dipole-permanent dipole force (Keesom force), which is an electrostatic term,
  the permanent dipole-induced dipole force (Debye force), which is an induction or polarization term,
  the induced dipole-induced dipole force (London dispersion force), which is the dispersion term.

the forms of which are

$$
\begin{align}
V_{\text{Keesom}} &= \frac{-m_{1}^{2}m_{2}^{2}}{24\pi^{2}\epsilon_{0}^{2}\epsilon_{r}^{2}k_{b}Tr^{6}} \\
V_{\text{Debye}} &= \frac{-m_{1}^{2}\alpha_{2}}{16\pi^{2}\epsilon_{0}^{2}\epsilon_{r}^{2}r^{6}} \\
V_{\text{LJ}} &= 4\epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right]
\end{align}
$$

Note that the LJ potential contains both a repulsive ($1/r^{12}$) and attractive ($-1/r^{6}$) component. This is not the form of the dispersion interaction, but is both a suitable mathematical form that encapsulates a few noncovalent (vdW) interactions together and is easy to evaluate on a computer.

You may also be interested in the form of a more complicated potential, such as the AMOEBA polarizable force field, which uses a slightly different form for the pairwise vdW interaction ($i,j$ are atoms)

$$
\begin{align}
U_{\text{vdw}}(ij) &= \epsilon_{ij} \left( \frac{1.07}{\rho_{ij} + 0.07} \right)^{7} \left( \frac{1.12}{\rho_{ij}^{7} + 0.12} - 2 \right) \\
\rho_{ij} &= \frac{R_{ij}}{R_{ij}^{0}} \\
\epsilon_{ij} &= \frac{4\epsilon_{ii}\epsilon_{jj}}{(\epsilon_{ii}^{1/2}+\epsilon_{jj}^{1/2})^{2}} \\
R_{ij}^{0} &= \frac{ (R_{ii}^{0})^{3} + (R_{jj}^{0})^{3} }{ (R_{ii}^{0})^{2} + (R_{jj}^{0})^{2} }
\end{align}
$$

and also considers up to permanent and induced quadrupole moments

$$
\begin{align}
U_{\text{elec}}^{\text{perm}}(r_{ij}) &= M_{i}^{T} T_{ij} M_{j} \\
T_{ij} &= \begin{bmatrix}
1 & \frac{\partial}{\partial x_j} & \frac{\partial}{\partial y_j} & \frac{\partial}{\partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial x_j} & \frac{\partial^2}{\partial x_i \partial x_j} & \frac{\partial^2}{\partial x_i \partial y_j} & \frac{\partial^2}{\partial x_i \partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial y_j} & \frac{\partial^2}{\partial y_i \partial x_j} & \frac{\partial^2}{\partial y_i \partial y_j} & \frac{\partial^2}{\partial y_i \partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial z_j} & \frac{\partial^2}{\partial z_i \partial x_j} & \frac{\partial^2}{\partial z_i \partial y_j} & \frac{\partial^2}{\partial z_i \partial z_j} & \mathrm{L} \\
\mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{O}
\end{bmatrix} \left( \frac{1}{r_{ji}} \right)
\end{align}
$$

along with the other typical bonded terms found in modern force fields.

Here is a nice set of notes that describes the interaction between electric multipoles to different orders.


Why are more electronegative atoms better leaving groups?

You can interpret the term "stabilizing" to mean that some effect leads to a  more energetically favorable state (i.e. lower in energy) relative to your original state. So to answer your first question, if atom A is better at stabilizing negative charge than atom B, then it means that it is more energetically favorable to place a negative charge on atom A and have B remain neutral than the other way around. This could be because placing additional electron density around A is more energetically favorable than placing it on B, or because it is less energetically unfavorable to place more electron density on A than it would be to place it on B.

To answer your second question, more electronegative atoms are generally better at stabilizing negative charge because, by definition, electronegative atoms have a greater tendency to attract electron density. However, it is not only the electronegativity of an atom which determines its leaving group ability, but also its size. If an atom is larger, then it means the electrons will be spread out over a larger volume, which is more energetically favorable. This is one reason why iodide is a better leaving group than bromide, even though bromine is more electronegative than iodine. In addition, iodine is larger and it forms weaker bonds than bromine with other atoms due to poorer overlap of atomic orbitals, meaning it is easier to break iodine bonds (e.g. the bond dissociation energy is 284 kJ/mol for Br$-$CH$_3$ versus 232 kJ/mol for I$-$CH$_3$ at 298 K). Because iodine bonds are easier to break, it is easier for iodine to "leave" the molecule and is thus a better leaving group.

Along this same line of reasoning, fluorine, being very electronegative, can form a very stable anion, but it forms very strong bonds which are difficult to break, and fluoride cannot leave if the fluorine bond doesn't break. That is why fluoride is a poorer leaving group than the other halogens.


Selectivity in aldol condensation between pivaldehyde and acetone

Ketones have two alkyl groups attached to the carbonyl moiety, which are electron-donating by hyperconjugation. That is, the overlap of $\ce{C-C}$ and $\ce{C-H}$ $\sigma$-bonds with the $\pi$-type molecular orbitals of the carbonyl increases the electron density around the carbon of the carbonyl group, thus making it less electrophilic.

Alternatively, I think it's also equally (if not more) valid to consider the reactivity from the perspective of the frontier orbitals. Alkyl substituents tend to slightly lower the energy of the HOMO of the carbonyl (by comparison to hydrogens), but concomitantly they raise the energy of the $\pi^{*}$ LUMO of the carbonyl, into which the electrons of the enolate would flow, resulting in lower electrophilicity (since the energy of the tetrahedral intermediate and transition state resulting from attack by the enolate are going to be correspondingly higher).


The exponential must be dimensionless.  That means if $E_a$ is in units of $\mathrm{cal\cdot mol^{-1}}$, then $R$ must be units of $\mathrm {cal\cdot mol^{-1}\cdot K^{-1}}$.  Make sure you use the right value of $R$ for these units; $R=8.314~ \mathrm{J \cdot mol^{-1}\cdot K^{-1}}$, but $R = 1.987~ \mathrm{cal \cdot mol^{-1}\cdot K^{-1}}$.  Temperature must obviously be in Kelvins.
The example reaction you gave is bimolecular, so I would think $k_f$ is meant to have units of $\mathrm{cm^3 \cdot s^{-1}\cdot mol^{-1}}$.  That way, when you multiply $k_f$ by the molar concentration of both reactants, e.g. ethanol and hydroxyl, you get a reaction rate that is in units of $\mathrm{mol \cdot s^{-1}\cdot cm^{-3}}$.
The remaining terms must therefore combine to give the right units for $k_f$. Since the units of $T^b$ will have units of $\mathrm K^b$, then the units for $A$ will be the units for $k_f$ divided by $\mathrm K^b$, i.e. the units of $A$ are $\mathrm{{cm}^3 \cdot s^{-1}\cdot mol^{-1} \cdot K^{\it-b}}$.


Why does Azo coupling of β-naphthol takes place at alpha position and not at gamma position

Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.

If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 , with aromatic rings are possible.

If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.

Therefore attack at alpha position is the major product.

References:

https://pubs.rsc.org/en/content/articlehtml/2015/gc/c4gc02381a

https://en.wikipedia.org/wiki/Azo_coupling

Mechanism of formation of 2-naphthol red dye (aka Sudan 1)


Here is the reaction scheme for the Mozingo reaction (or Mozingo reduction) taken from this Wikipedia link.

The reduction works for both aldehydes and ketones and involves two steps. First, the carbonyl compound is converted into a thioketal (or thioacetal). The mechanism for this step is analogous to the mechanism for ketal or acetal formation (see here) except sulfur replaces oxygen as the nucleophile attacking the carbonyl.

In a second step, the thioketal is subsequently reduced to the corresponding methylene compound.  The reduction can be a standard Raney nickel hydrogenolysis or hydrazine can be used at higher temperatures. The mechanism of reactions on metal surfaces or at higher temperatures are usually not well understood.  Note however that a typical $\ce{C-S}$ bond is around 20 kcal/m weaker than a $\ce{C-C}$ bond.  It is also known that sulfur is readily bound at metal surfaces (catalyst poisoning by sulfur being an example).  The complexation of sulfur and the metal likely weakens the $\ce{C-S}$ bond further.  Whether a $\ce{C-S.}$ radical is formed that reacts with a hydrogen radical complexed at the metal surface, or whether the hydride anion is generated at the metal surface and displaces the sulfur at the weakened $\ce{C-S}$ bond is not known.

The Mozingo reaction is complementary to the Clemmensen (run under acidic conditions) and Wolff-Kishner (run under basic conditions) reductions in that it can be performed under neutral conditions.  So if acid- or base-sentive functional groups are present in the starting carbonyl compound, the Mozingo reaction is likely the pathway of choice to reduce the carbonyl.


Sodium thiosulfate and Hydrochloric acid

There's a nice demonstration of the experiment performed by Annis Hapkiewicz.

She uses a 0.15 M $\ce{Na2S2O3}$ stock solution.  She prepares 5 beakers with 50 40, 30, 20 and 10 mL of the stock solution filled up to a total volume of 50 mL with destilled water.

This corresponds to thiosulfate concentrations of 0.15, 0.12, 0.09, 0.06, and 0.03 M.

I couldn't figure out the concentration of the hydrochloric acid used, but the times for the appearance of the sulfur cloud (disapperance of the cross) were given.

The range is between 20 s and 5 min.

I'd probably go with the 2 M HCl for safety reasons and perform the experiment with the thiosulfate concentrations given above. Check the times and adjust the concentrations if necessary.

You don't want it to to fast to be measured by the kids, but you don't want the highest concentration to take half an hour.


How do I overlay spectra in an ACD/NMR Processor, version 12.01?

If anyone else is running into the same issue, after selecting which spectra you would like to overlay go to series -> collect spectra and then select "manual offset" if you want to adjust the height.


Fairly pure iron (or iron/nickel alloys) are found in meteoric iron. Some of these are surprisingly corrosion resistant... close to stainless steels. One well-known example is a dagger from Tutankhamun's tomb. See JSTOR for some more uses of meteoric iron.

As for iron smelting, there is much information on the web, and I suggest you try some research.


Let's take a sample reaction $\ce{\it aA + bB-> pP + qQ}$ with stoichiometric coefficients $a$, $b$, $p$, and $q$ for reactants $A$ and $B$ and products $P$ and $Q$.

The rate of reaction is the rate of change of the extent of reaction.  Extent of reaction is usually given the symbol $\xi$, so reaction rate is $\frac{d\xi}{dt} = \dot{\xi}=r$.

If the reaction rate is $\dot{\xi}$, then the rate of change of concentrations is:

$$\frac{d[A]}{dt}=-a \dot{\xi}$$
$$\frac{d[B]}{dt}=-b \dot{\xi}$$
$$\frac{d[P]}{dt}= p \dot{\xi}$$
$$\frac{d[Q]}{dt}= q \dot{\xi}$$

That is the stoichiometric coefficients link the rate of formation/consumption of the products/reactants, sort of by the definition of stoiciometry.

Now, when talking about the order of a reaction, what we are doing in effect is giving formulas (or partial formulas) for $\dot{\xi}$.  If the reaction is "elementary", then it will be $a$th order with respect to $A$ and $b$th order with respect to $B$.  This doesn't have to be true but often is.  This means that if you change the concentration of A while holding everything else constant, the change in rate will be proportional to the concentration of $A$ to the $a$th power, etc.  In math form that looks like

$$\dot{\xi} \propto [A]^a$$
$$\dot{\xi} \propto [B]^b$$

where $\propto$ means "proportional to", which is saying that the things are equal except for some one unknown constant.  Thus those two $\propto$ "equations" are equivalent to

$$\dot{\xi} = \kappa_1 [A]^a$$
$$\dot{\xi} = \kappa_2 [B]^b$$

It's important to note that $\kappa_1$ could depend on everything except $[A]$, i.e. it could be a function of $[B]$ etc.  Similarly $\kappa_2$ could depend on $[A]$ (but not on $[B]$).  The only way for both equations above to be true is if

$$\dot{\xi} = k [A]^a [B]^b$$

This means that $\kappa_1 = (k [B]^b)$ and $\kappa_2 =  (k [A]^a)$.  Now that we know the reaction order with respect to each reactant, and have written it down in equation form, we can relate it back to the change in concentration of each of the reactants/products:

$$\frac{d[A]}{dt}=-a k [A]^a [B]^b$$
$$\frac{d[B]}{dt}=-b k [A]^a [B]^b$$
$$\frac{d[P]}{dt}= p k [A]^a [B]^b$$
$$\frac{d[Q]}{dt}= q k [A]^a [B]^b$$

These equations clearly show the relationship between the order of a reaction and the rate of change of reagent/product concentrations.

  I have learnt for many years that rate of reaction is the rate of change of concentration of a reactant or product with respect to time.

By looking at the four equations above, the rate of reaction will only be equal to the product or reactant concentration change when the stoiciometric coefficient of that product or reactant is 1.  And that's true regardless of what the "order" of the reaction is.  The reaction $\ce{\it aA + bB-> pP + qQ}$ doesn't have to be $a$th order with respect to $a$.  It could have been that $\dot{\xi} \propto [A]^{1.7}$ or even that $\dot{\xi} \propto \frac{[A]^{0.2}}{K_m+[A]}$ or whatever.  But in chemistry, it does often happen that reactions do have orders that depend straightforwardly on stoichiometry, which is why those types of reactions are studied so often.


Can you make napalm out of gasoline and orange juice concentrate?

Here's an excerpt from an interview with the author of Fight Club, Chuck Palhniuk.

  How to make Napalm with Frozen Orange Juice and Gasoline?

  Well, Ed Norton changed one ingredient in every one to make them useless. So, that really pissed me off because I really research those really well. Actually its styrofoam and gasoline - it make the most incredible explosive.

http://www.dvdtalk.com/interviews/chuck_palahniuk.html

So no, Fight Club does not accurately convey any of the recipes. All were clearly modified. And most likely for safety reasons - i.e. you wouldn't want someone to walk out the theatre with a "bright" idea. On the other hand I would still expect a mixture of gasoline and orange juice concentrate to be rather flammable ...


How are reactivity and dipole moment related?

The reactivity of amines is generally based on how nucleophilic they are.  In turn, their nucleophilicity is dependent upon how much of the nitrogen lone pair of electrons really exists on the nitrogen atom (e.g. the lone pair electron density on nitrogen). In ammonia ($\ce{NH3}$), the lone pair electron density on nitrogen is very high making ammonia quite nucleophilic, quite reactive.  In the case of nitrogen trifluoride ($\ce{NF3}$), the three extremely electronegative fluorine atoms inductively remove much of the electron density from nitrogen; therefore $\ce{NF3}$ is less reactive than ammonia in a nucleophilic sense.


Charge density is the amount of charge per unit volume but what is charge?

An electron has a charge (lets say $e [C]$). If you have a certain amount of electrons in a certain volume ($V [m^3]$), you have a certain charge per volume $\rho = e/V [\frac{C}{m^3}]$. Which is a density.


Hydrochloric acid solutions can only readily reach about $\pu{40\%}$ concentration. So say you add $\ce{NaCl}$ to $\pu{98\%}$ sulfuric acid. You quickly end up supersaturated with respect to dissolved $\ce{HCl}$ and it goes into the gas phase which thus limits the ability for that reaction to go in the reverse direction.

The situation is similar for the second reaction, except in this case rather than removing carbonate from the system via volatilization, you are forming the insoluble $\ce{CaCO3}$, preventing the reaction from happening in the reverse direction.


Work done in an isothermal irreversible process

The key word here is irreversible. Irreversible processes are more like sudden jerks or a quick change in the system and therefore the ${P_{\text{ext}}}$ is not same as the ${P_{\text{int}}}$ throughout the process.

  Why do we NOT use the internal pressure (${P_{\text{int}}}$) to find the work done?
  It is very difficult to find the ${P_{\text{int}}}$ exerted by the gas in the container since ${P_{\text{int}}}$ is varying with time. For this, we would have to use integration. Something like this:$${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$$
  (where $P(V)_{\text{int}}$ is a function of $V$)
  But we know the value of ${P_{\text{ext}}}$ and this is constant throughout the process. This is why we use ${P_{\text{ext}}}$ instead of troubling ourselves with integration. Something like this: $$W_{ext}= P_{\text{ext}} \mathrm{d}V$$

So if a system undergoes a decrement in volume suddenly, it is the external pressure that does the positive work throughout this decrement. The equation would be: $$W_{\text{ext}}= P_{\text{ext}}\mathrm{d}V$$ (However we can use ${P_{int}}$ as well, please read the NOTE at the bottom)

And if a system undergoes an increment in volume suddenly, ${P_{\text{int}}}$ is doing the positive work. However, to find the $W_{int}$ we would have to use integration since ${P_{\text{int}}}$ is varying with time. It is much easier if we just find the negative work done by ${P_{\text{ext}}}$ and since we know that $${W_{\text{int}} = - W_{\text{ext}}}$$ we can find the work done. This is why we consider the ${P_{\text{ext}}}$ and according to the situation in the question, we apply the signs.

NOTE: Magnitude of work done by the internal pressure and external pressure is always same. We only use ${P_{\text{ext}}}$ because it is constant whereas ${P_{\text{int}}}$ varies with time. If we use integration to find ${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$, where $P(V)_{\text{int}}$ is a function of $V$, we will get the same magnitude as we would from $W_{\text{ext}}= P_{\text{ext}} \mathrm{d}V$. Only signs are opposite.


In an experiment testing catalyst influence on reaction, is NaOH a reactant or a catalyst?

In basic solution the half reactions are:

Reduction:

$\ce{H2O2 + 2 e- -> 2 OH-}$

Oxidation:

$\ce{I- + 6 OH- -> IO3- + 3 H2O + 6 e-}$

So the overall reaction is

$\ce{3 H2O2 + I- -> IO3- + 3 H2O}$

thus $\ce{OH-}$ is overall neither a reactant nor a product. Hence it is either a catalyst, an inhibitor, or a spectator ion.


Why do hydroxyl groups only produce a singlet in proton NMR?

You are both right and wrong.

At low temperatures, the NMR spectrum will indeed be as you predicted: the $\ce{H}$ from $\ce{OH}$ will produce a quartet and likewise the $\ce{H}$ from $\ce{CH_3}$ will produce a doublet. Note that at low temperatures the formation of $\ce{H}$ bonds is favored, hence leading to more stable components and significantly decreasing proton exchange.

At higher temperatures however, the following spectrum will be observed:

Due to the increase in temperature, the $\ce{H}$ bonds are weakened, and hence shields its signal. Most importantly, the intermolecular proton exchange, non negligible at those temperatures, will render the interactions between the two groups of $\ce{H}$ undetectable via NMR. Hence why you get two singlets.


pH of aqueous solution of HCl of low concentration

It is known that the equilibrium constant for the following reaction is $10^{-14}$.

$$\ce{2H2O(l) <=> H3O+(aq) + OH-(aq)}$$

That means,

$$\frac {[\ce{H3O+(aq)}][\ce{OH-(aq)}]} {[\ce{H2O(l)}][\ce{H2O(l)}]} = 10^{-14}$$

where the concentration of water $[\ce{H2O(l)}]$ is assumed to be $1$.

Initially, we have $[\ce{H3O+(aq)}] = 10^{-8}$.

Then, assuming that $x$ moles of $\ce{H3O+(aq)}$ is generated. An equal amount of $\ce{OH-(aq)}$ must also be generated:

$$(10^{-8}+x)(x) = 10^{-14}$$

Solving for $x$ gives $x = 9.51 \times 10^{-8}$.

So, the total concentration of $\ce{H3O+(aq)}$ is $1.051\times10^{-7}$.

Therefore, the pH is $6.978$.

The problem with your approach

Let the initial molarity of the hydronium ion be $a$.

Solving $(a+x)(x) = 10^{-14}$, if $a$ is well above $10^{-5}$, gives $x \approx 0$.

The approximation that the pH is equal to $-\log{{\left([\ce{HCl(aq)}]\right)}}$ is only accurate for normal concentrations.


Liquid floating on top of water due to density or insolubility?

Both are contributing factors.

The insolubility ensures that it does not mix with water.

The lower density of the liquid on top ensures that it stays on top of the water instead of below the water.


The question itself is poorly written as given.

The diagram should have "regions" and "points."

So what seems to be labeled "Region 2" and "Region 4" I would name as "Point 2" and  "Point 3" respectively.

So the innermost conical area would be "Region 1", the next conical area would be "Region 2" and the third conical area would be "Region 3."

Region 1 is where the mostly unburned gas and oxygen mixture are pushing above the lip of the Bunsen Burner. Region 1 exists because the gas coming out of the tube is cool. If the air port is open and the gas flow is too low then the gas will start to burn down the tube and you'll get a "strike back" where the flame is either (1) blown out or (2) burns at the jet. If the gas flow is too great you can blow the burning region off contact with the upper tube. If you increase the gas flow even more then you can in fact blow the flame out.

Region 2 would be a reducing region within the flame. This region is hot and burns the fuel and oxygen coming out of the tube.

Region 3 would be a oxidizing region of the flame. Here oxygen from "outside air" (oxygen which didn't come up the tube) is migrating into the flame to burn the excess fuel which is not bunt in region 2.

You'd use the reducing and oxidizing regions when doing bead tests for identification.

Point 2 would be the hottest part of the flame as shown by the composite image of a paperclip in the flame from the YouTube video.


What are simple applications of Monte Carlo simulations in chemistry, that can be done in Matlab?

I've done a little simulation as a senior year elective, though not in Matlab. Can you be a little more specific about how/what you're trying to model with nanoparticle growth? Perhaps direct us to an abstract or two.

The first Monte Carlo simulation you should attempt is the drunken sailor problem. The basic idea is that you have a drunken sailor who has equal probability of taking a step forwards or backwards, and you let him stagger around on a one dimensional line for some number of steps (1000) and repeat the experiment a large number of times. You then get some summary statistics and plots around your experiment.

After you're comfortable with that concept, you can look at increasing complexity into 2 or 3 dimensional random walks, changing the weighting function (not 50-50 for each direction), or some specific algorithm such as Metropolis-Hastings Monte Carlo.

I hope this helps. If you're looking for specific programming help, posting on stack exchange would be a good idea.


Is PCl5 really trigonal bipyramidal in shape?

The trigonal bipyramidal shape of $\ce{PCl5}$ is indeed the lowest energy conformation.

I have had a very hard time figuring out what your description of the structure meant and I am still not convinced I have actually understood correctly. However I was trying to model it in my head I either arrived at a variety of different angles, or at least two of them were equal to $72^\circ$, or both of these cases.

The only structure $\ce{ML5}$, where all $\ce{L-M-L}$ bond angles are equal is a regular pentagon, i.e. all ligands are in one plane, with $\angle(\ce{L-M-L})=72^\circ$, e.g. $\ce{[XeF5]-}$.

Another possibility for (almost all) equal bond angles is a square pyramid, where $\angle(\ce{L-M-L})=90^\circ$, e.g. $\ce{ClF5}$. Note however, that there are also other angles, so not all bond angles are equal, $\angle(\ce{L-M-L})=180^\circ$.

There is no better solution to arrange 5 point 'uniformly' on a sphere than the trigonal bipyramid. This is also known as the Thompson Problem, which is addressed in the linked question on Mathematics.se, provided by orthocresol: 5 Points uniformly placed on a sphere.
Another point of view would be to ask the question whether there is a way to put 5 points on the surface of the sphere so that they are indistinguishable, which is also answered there. Spoiler warning: there isn't.

You can dive deeper into the matter. For example for $\mathbb{R}^n$ there can only be $n+1$ points equally equidistant to each other. In our perceived reality $n=3$, which makes the tetrahedron the most complex molecule with this property. See on mathematics.se: Proof maximum number of equidistant points for a given dimension (duplicate), Prove that $n+2$ points in $\mathbb{R}^n$ cannot all be at a unit distance from each other (duplicate), $n$ points can be equidistant from each other only in dimensions $\ge n-1$? (Which are all duplicates, but still have answers attached to them.)

And then there are only five Platonic solids, none of which has five vertices.

Furthermore, I would like to add to all of that another word of caution: Molecules are flexible. The trigonal bipyramidal arrangement of ligands just happens to be the low energy conformation for a variety of reasons (considering the necessary and useful approximations, e.g. Born-Oppenheimer approximation).
The molecule (and other analogues) are special though, as the undergo pseudorotation according to the Berry Mechanism (see Wikipedia).

Side note:

  We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

This is only half the truth. It is certainly true that protons repel each other, more importantly though is that electrons repel each other, too. The most important part of which is also built into the very simple VSEPR theory. While this theory has its uses, there is a lot of criticism attached to it and one should be aware of that. I shall stress again at this point that VSEPR theory is unrelated to orbital hybridisation; there are quite a few textbooks which do make this post-hoc rationalisation, which can be misleading and dangerous.


If I understand what you are saying, then the issue is that the carbon and oxygen have already reacted. Carbon dioxide is thus not flammable because it is the fully oxidized product of the combustion between oxygen and carbon.

"Made of" has multiple meanings. You're confusing the meaning of physical mixture (flammable) with the chemical composition of a compound (not flammable).


What is the correct chemical formula for monosodium glutamate?

It should, but in organic molecules salts, often the counterion is written latter, to indicate its position on the molecule acting like so, for example, in sodium acetate, $\ce{CH3CO2Na}$, this would be something like a "semicondensated" formula.


In the second row of the periodic table, elements have relatively small differences between the size their $\mathrm s$- and $\mathrm p$-orbitals. Therefore, the orbitals of $\ce{NR3}$ can go from $\mathrm {sp}^3$ to $\mathrm {sp}^2$ with relatively little energy increase, so an amine can become planar and then reorient with the inverted stereochemistry. The same occurs with carbanions. I'm not certain, but I'm guessing oxonium cations do this, as well, and I would even guess that it would happen more quickly due to oxygen's smaller size.
The reason we don't see this with $\ce{NF3}$ is because orbitals with more $\mathrm p$ character are more affected by induction, so the highly electronegative fluorine atoms would lose electron density during inversion, which would be energetically unfavorable (to the point that it doesn't occur readily).
Chiral phosphines do not invert because the size difference between $\mathrm s$- and $\mathrm p$-orbitals for larger atoms is greater, so converting $\ce{PR3}$ $\mathrm{sp}^3$ orbitals to $\mathrm {sp}^2$ would require much contraction and, therefore, more energy than for the analogous amine.


Breaking degeneracy in phosphorus pentachloride

I would guess he means a spontaneous symmetry breaking, in which case the "perturbation" is any old force from the environment.  This can happen in physical systems: it's possible for a system of high symmetry to spontaneously fall to a lower-energy state of lower symmetry.  Only the tiniest of perturbations is required, if you wait long enough.  A familiar classical example is crystallization, in which below a certain temperature a system of atoms spontaneously loses complete translational symmetry even in the absence of an external potential.  A quantum example is magnetization when a ferromagnet cools below the Curie temperature -- the spins lose rotational symmetry and pick some essentially random preferred direction, in response to the tiniest of perturbations from the environment.  In the case of PCl5, one would assume, very roughly (and it would take detailed calculation to verify) that if 3 of the Cls pull in closer to the nucleus and 2 move further away, you get a better balance of nuclear-electron attraction and nuclear-nuclear and electron-electron repulsion than if all 5 remain at the same distance.  So the system spontaneously breaks rotational symmetry, in response to any old tiny perturbation in the environment.

That said, in the absence of an external potential, the rotational quantum state of the molecule will be entirely symmetric, and you will have an equal chance of finding the long Cl-P-Cl axis pointing in any direction.


Why is only bromine used in the degradation reaction?

It isn't.

Under alkaline conditions, the Hofmann rearrangement of primary amides to amines can be performed with chlorine too. But why work with chlorine when bromine is much more convenient.

Choosing bromine over chlorine when possible is an obvious decision for everyone with practical lab experience!

Under standard conditions, bromine is a liquid and can be added using a dropping funnel, while chlorine is a gas, which makes adding stoichiometric amounts much more tedious.

Hypohalites have been used too: in the presence of alkanols, rearranged carbamates are formed. Under acidic conditions, hypervalent iodine compounds have been shown to be effective.


The information is confusing precisely because your question itself is confusing. There are different kinds of dissociation, and different degrees thereof. Moreover, there are different areas of study, each with its own concepts of what is significant and what is not. For example, if one molecule in a trillion splits into ions, this is nothing for chemistry, but probably quite a something for atmospheric physics.

As for the question you supposedly intended to ask: no, electrolytic dissociation of dissolved substances never happens in gases the way it happens in water.

Also, welcome to Chem.SE.


The electron configuration and electron density distribution in singlet oxygen

Probably not a full answer, but hoping to point readers in a useful direction.

If we denote the two $\pi^*$ orbitals as $\pi_x^*$ and $\pi_y^*$ respectively, then the spatial wavefunctions for each state can be written as follows (all lower orbitals are ignored):

$$\begin{align}
{}^1\Sigma_\mathrm g^+ &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) + \pi_x^*(2) \pi_y^*(1)] \\
{}^3\Sigma_\mathrm g^- &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) - \pi_x^*(2) \pi_y^*(1)] \\
{}^1\!\Delta_\mathrm g &: \begin{cases}\pi_x^*(1) \pi_x^*(2) \\ \pi_y^*(1) \pi_y^*(2) \end{cases}
\end{align}$$

For the ${}^1\!\Delta_\mathrm g$ case, one would expect a spatial degeneracy of 2, because the letter $\Delta$ in the term symbol indicates that the quantum number $|\Lambda| = 2$. Therefore, there is one state with $\Lambda = +2$, and one state with $\Lambda = -2$. On the other hand, the $\Sigma$ term has $|\Lambda| = 0$ and hence is spatially non-degenerate (the $^3\Sigma$ term is triply degenerate due to spin).

(Digression: $\Lambda$ represents the projection of the angular momentum along the internuclear axis. As such, there is no further projection quantum number ("$M_\Lambda$") that can take values $-2, -1, 0, +1, +2$. In comparison, in atoms, $\ell$ indicates the total angular momentum and $m_\ell$ the projection of this angular momentum onto the $z$-axis. For more information about term symbols of diatomic molecules, I suggest reading this link. There are many other sources on the Internet, but many are sloppy with notation, which only leads to confusion down the road.)

So, the MO diagram above - which ties the ${}^1\!\Delta_\mathrm g$ state to only one wavefunction - is incomplete. The ${}^1\!\Delta_\mathrm g$ state corresponds to two possible wavefunctions, which are symmetry-equivalent; there is no "preference" for the $x$-axis over the $y$-axis.

If you want to go further than what I've written, then I'd point you to the case of the boron atom which I asked about earlier, which is exactly analogous to this. The point about boron is that it has one electron in the 2p subshell. Does this go into the $\mathrm p_x$, $\mathrm p_y$, or $\mathrm p_z$ orbital? In the case of dioxygen, do the paired electrons go into the $\pi_x^*$ or $\pi_y^*$ orbital?

As you said, it does not make sense for us to assign the electron(s) to any of the orbitals in particular, as that leads to an asymmetric electron density distribution. However, I must admit that I don't fully understand the CASSCF calculation there, and I never quite got round to following up on that question, so you'll have to ask somebody else about that.


If the gaseous molecules didn't have any kinetic energy would the surface of a filled balloon still be firm?

If ideal gas molecules had had no kinetic energy, there would have been no gas pressure. Inside nor outside the balloon. The outer gas would have been laying on the ground, the inner gas on the bottom of the balloon.

In such a case, the only remaining force, aside of gravity, would have been the elastic forces of the balloon, pulling the balloon rubber together to its original shape and volume.

For real gases compressed to volume about equal to tight arrangement of molecules, a sudden repulsive force and therefore pressure would have progressively appeared, due repulsion of electrons.


How does the position of hydroxyl group in a nucleotide monomer affect the dehydration synthesis of nucleotides?

The phosphate ion is of vital importance in biological systems. Its molecular geometry is tetrahedral. Thus, based on the $\mathrm{pH}$ of the system, hydroxy groups are inter-changeable with $\ce{P=O}$ form and hence all three remaining $\ce{O}$s are equivalent (see Waylander's comment). Therefore no impact on the dehydration.
However, the stereochemistry of the hydroxy group on the 3'-carbon on the deoxyribose nucleus of the nucleotide has an impact on the enzymatic dehydration process. For instance, if you change the original $(S)$-orientation of 3'-hydroxy group to $(R)$-orientation, DNA-synthase would not pick that nucleotide to synthesize intended DNA sequence.


find the pH where is maximum the concentration of a amphiprotic substance

Let's take an aqueous solution of a salt $\ce{NaHA}$ with the initial concentration $C$ when added to water. It will completely dissociate  according to the eaquation: $\ce{NaHA(s) \rightarrow Na^+ +HA^-}$.

$\ce{HA^-}$ will participate in three equilibria:

$\ce{2HA^- \leftrightarrows H2A +A^{2-}\quad \quad \quad  }$ ${K_1^0=K_{A2}/K_{A1}}$

$\ce{HA^- +H2O\leftrightarrows H3O^+ +A^{2-}\quad }$ $K_2^0=K_{A2}$

$\ce{HA^- +H2O\leftrightarrows OH^- +H_2A^\quad }$ $K_3^0=K_{B1}={K_w}/{K_{A1}}$

In most cases, $K_1^0$ is far bigger than $K_2^0$ and $K_3^0$. So, the first equilibrium is the preponderant reaction, and this reaction will impose the pH of the solution.

Let's now calculate the product $K_{A2}\times K_{A1}$:

$K_{A2}\times K_{A1}= \frac{[\ce{}A^{2-}].[\ce{H3O+}]}{\ce{[HA^-]}}.\frac{[\ce{}HA^{-}].[\ce{H3O+}]}{\ce{[H_2A]}}$

According to the stoichiometry  of the preponderant reaction, we have $\ce{[A^{2-}]=[H_2A^]}$. So the product
$K_{A2}\times K_{A1}= \ce{[H3O+}]^2}$ .
i.e. $\ce{pH}=0.5(\ce{p}K_{A2} +\ce{p} K_{A1})$


Your approach seems correct, however you seemed to have made a few computation errors. First of Sodium's molar mass is $ 23 \ce{g/mol}$ and not 11.

So assuming a total mass of 100g, we can re-interpret the given percentages as follows:

mass of sodium = 14.3 g

mass of sulphur = 10 g

mass of oxygen = 19.9 g

(note: the percentage you provided seems incorrect, the percentages add up to 91, not a 100. I've changed this data point to get the correct result.)

mass of water = 55.8 g

Dividing these by their respective molar masses to get no. of moles:

no. of moles of sodium  = 0.62

no. of moles of sulphur = 0.31

no. of moles of oxygen  = 1.2

no. of moles of water   = 3.1

The ratio we get is 2:1:4:10. Thus, the empirical formula is $\ce{Na_2SO_4.10H_2O}$


Are there any trends for electrical conductivity?

I am unsure of how to approach this problem.

No, you are not! You have already reduced the number of candidates by good reasoning!

At this point, one

could have a look at conductivity tables or remember in which period the metal blocks start for every main group or
remember typical electricity-related uses of $\ce{As}$ and $\ce{Sn}$

Did you ever form a conducting connection between electronic components (or just some wires) by soldering? $\Leftarrow$

(Do you recall that $\ce{GaAs}$ is used as a semiconductor material in thin film solar cells?)


Electrochemistry - reduction potentials - an IB Chemistry Question

First put your redox potentials on a horizontal line by order of increasing standard redox potentials. You will see that it starts on the left-hand side by $\ce{Zn^{2+}/Zn}$ at $\pu{-0.76 V}$. Then, going to the right, you find $\ce{V^{III}/V^{II}}$ at $\pu{-0.26 V}$. Now you find $\ce{Pb^{2+}/Pb}$, and so on.
Now try to learn how to use this scale. The reduced species of any couple can react with any oxidized species of any couple situated to its right. For example, $\ce{Zn}$ is the reduced species of the couple $\ce{Zn^{2+}/Zn}$. So $\ce{Zn}$  can reduce all oxidized species, and for example all oxidized forms of Vanadium ions. But $\ce{Zn^{2+}}$ cannot be reduced, because there is no reduced species at the left hand side of $\pu{- 0.76 V}$.
On the other end of the scale, you find $\ce{Cl2/Cl-}$ at $\pu{+1.36 V}$. Hopefully you admit that $\ce{Cl2}$ can oxidize all reduced species, and $\ce{Cl-}$ cannot reduce anything.
Now let's consider $\ce{Pb^{2+}/Pb}$, at $\pu{-0.13 V}$. $\ce{Pb}$ can reduce all oxidized species situated on the positive side of the axis, for example $\ce{VO^{2+}}$ which is at $\pu{+0.34 V}$. But it cannot reduce $\ce{V^{3+}}$ to $\ce{V^{2+}}$, because $\ce{V^{3+}/V^{2+}}$ is on tis left-hand side. And this is exactly your first question.
Now think a little bit about how to solve the second question. It is exactly the same approach.


Density of saturated n-octane in air at 300 K?

Thermodynamic parameter values for many substances (not just water and steam) can be found using so-called steam tables. In the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). For a simplified estimate, however, we only need the value of the vapour pressure of octane.

For simplicity’s sake, we may assume that the gas phase (containing air and octane) is an ideal mixture of non-reacting gases and that the liquid phase contains only pure octane. Thus, for the calculation of parameter values for octane, we may ignore the air and only consider the equilibrium in a system of pure liquid octane and pure gaseous octane.

Using tabulated data, we find that the saturation point for octane at a temperature of $T = 300\ \mathrm K$ corresponds to a pressure of $p_{\ce{C8H18(g)}} = 2060\ \mathrm{Pa}$.

The corresponding density can be estimated using the ideal gas law:
$$\begin{align}
p\cdot V &=n\cdot R\cdot T \\[6pt]
\frac nV &= \frac p{R\cdot T} \\[6pt]
\frac m{M\cdot V} &= \frac p{R\cdot T} \\[6pt]
\frac mV &= \frac {M\cdot p}{R\cdot T} \\[6pt]
\rho &= \frac {M\cdot p}{R\cdot T} \\[6pt]
&= \frac {114.23\ \mathrm{g\ mol^{-1}}\times 2060\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&= \frac {0.11423\ \mathrm{kg\ mol^{-1}}\times 2060\ \mathrm{kg\ m^{-1}\ s^{-2}}}{8.314462618\ \mathrm{kg\ m^2\ s^{-2}\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&= 0.09434\ \mathrm{kg/m^3}
\end{align}$$

By way of comparison, the corresponding density given by REFPROP is $\rho_{\ce{C8H18(g)}} = 0.094649\ \mathrm{kg/m^3}$.

Assuming that the total pressure is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, we may calculate the partial pressure for the remaining air:
$$\begin{align}
p_\text{tot} &= p_\text{air} + p_{\ce{C8H18(g)}} \\
p_\text{air} &= p_\text{tot} - p_{\ce{C8H18(g)}} \\
&= 101325\ \mathrm{Pa} - 2060\ \mathrm{Pa} \\
&= 99265\ \mathrm{Pa}
\end{align}$$

The density of dry air at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1770\ \mathrm{kg/m^3}$. The corresponding density at a partial pressure of $p_\text{air} = 99265\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1531\ \mathrm{kg/m^3}$.

Thus, the total density of the gas mixture is
$$\begin{align}
\rho_\mathrm g &= \rho_{\ce{C8H18(g)}} + \rho_\text{air} \\
&= 0.094649\ \mathrm{kg/m^3} + 1.1531\ \mathrm{kg/m^3} \\
&= 1.2477\ \mathrm{kg/m^3}
\end{align}$$

By way of comparison, I made a quick simulation using REFPROP assuming that air is a mixture with the mole fractions $x_{\ce{N2}}=78.12\,\%$, $x_{\ce{O2}}=20.96\,\%$, and $x_{\ce{Ar}}=0.92\,\%$. I added an excess amount of octane $x_{\ce{C8H18}}=10\,\%$ (in order to obtain a liquid phase) and renormalized the composition to $100\,\%$.
The resulting density of the vapour phase at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{g} = 1.2483\ \mathrm{kg/m^3}$.


Why is the effect of competitive antagonist less than the effect of non-competetive antagonists?

Personally, I think the statement "the effect of a competitive antagonist is less than the effect of a non-competitive antagonist" is too general to have much meaning. There's a large number of systems, each with a potentially large number of competitive and non-competitive antagonists, each with a potentially large range of activities. Saying a priori that a non-competitive antagonist is more effective than a competitive one is presumptuous.

However, there is a bit of sense in there, for a particular defined sense of circumstances.

By definition, a competitive antagonist is only active against the non-liganded state of the receptor. The native ligand (agonist) and the competitive antagonist compete for the receptor. In contrast, a non-competitive antagonist can generally work against both the liganded (with agonist bound) and ligand-free versions of the receptor. (How active it is against each depends on the particulars about the antagonist and receptor.)

So in a situation where the antagonist isn't at saturating concentrations (say at a concentration near it's IC50), saturating amounts of native ligand (agonist) will completely reverse the effects of a competitive antagonist, as the agonist will compete off the antagonist. However, the non-competitive antagonist will still have some remaining inhibitory function, even at saturating concentrations of native ligand, as it can bind to and inhibit the ligand-bound (agonist-bound) form of the receptor.

However, there's an apples-to-oranges comparison embedded in there. That comparison assumes that the competitive and non-competitive are each near their respective IC50s. This value can change greatly between drugs, so you can have a competitive inhibitor with a really low IC50, and a non-competitive inhibitor with a really high one, such that if you used identical molar concentrations of each antagonist, the competitive antagonist would be successful in almost completely shutting off receptor activity, whereas the non-competitive antagonist would show barely noticeable activity.


C4v point group complexes and the group orbitals (MO Theory)

I don't have time for a full answer or derivation right now, but here is a rough explanation. We can just consider s-type orbitals on each of the five ligands.

If the group orbital is to transform under the TSIR, then each of the four "equatorial" orbitals must have the same phase as each other. This is because these orbitals can be interconverted by symmetry operations, and if the character under every operation is to be $1$, then every orbital has to have exactly the same contribution to the group orbital. However, the equatorial ones don't need to have the same phase/magnitude as the "axial" orbital, because the equatorial/axial positions are not symmetry-related.

Therefore, you end up with two group orbitals that look somewhat like this:

Obviously, the exact relative contribution will depend on the system, but this should be the correct form.

For nearly any question of this nature, an excellent reference is Albright, Burdett, and Whangbo's Orbital Interactions in Chemistry, 2nd ed. Unfortunately, I can't find a diagram for C4v MOs, but in general most things can be found there.


Do catalysts undergo physical changes during a reaction?

I'll just give an example of a catalysis to show you that catalysts do really temporarily react with the starting product and eventually will be regenerated after the catalytic reaction.

An example is the catalytic decomposition of formic acid to carbon monoxide and water by using sulfuric acid as a catalyst. The sulfuric acid donates one proton ($\ce{H+}$ ion) which is then bonded to the central carbon atom in the formic acid molecule. Then next another hydrogen atom from the sulfuric acid reacts with the hydroxyl group present on the formic acid molecule to yield water. In the end the sulphate ion snatches the last two present hydrogen atoms from the formic acid molecule and yields carbon monoxide. The sulfuric acid is regenerated.

$\ce{HCOOH + H2SO4 -> H2COOH + HSO4- -> H2CO + SO4^{2-} + H2O -> CO + H2SO4}$

So catalysts do undergo chemical changes in the catalytic reaction but are regenerated after the catalytic reaction is finished.

This means that if you are using a catalyst, you can normally reuse it after your catalytic reaction because it has the same chemical (and physical) properties as before the catalytic reaction.

Also : when you use manganese dioxide as a catalyst to decompose hydrogen peroxide, you will see that the black solid compound is the same after the catalytic reaction.

There are also so called 'sacrificial catalysts' (which aren't regenerated and do physically/chemically change, but aren't real catalysts) -> see http://en.wikipedia.org/wiki/Catalytic_cycle


Which of the five main branches of chemistry has the most labwork?

All the five so-called branches of chemistry require heavy use of the brain. Modern chemistry is very interdisciplinary. At the research level, the distinction among these branches becomes blurry. Everyone is doing some part of each branch. Traditional inorganic and organic chemistry both require a lot of wet work and synthesis and characterization.
If you ask which branch of chemistry has the highest level of automation, and most of the work is done by instruments-it is analytical chemistry. Many modern physical chemists or theoretical chemists rely on in silico techniques (=computers, modelling, simulations etc). You may not be doing a single real experiment (if your not lucky). Computer is your virtual lab. As a hardcore theoretical chemist, finding a job afterwards in the field, may be difficult. My acquaintance switched to computer programming after a PhD, which was a wise decision after remaining jobless for a year. He learned programming languages in the mean time. In any case, all fields are good. It all depends on your passion.


Solids can't have a charge (neither can liquids or gases). The electrostatic forces are immensely great, way greater than you seem to think right now. Just imagine the forces that would be tearing apart a droplet of water (say, 36 microliters) if 0.001 of its molecules (not all of them, God forbid) would suddenly become positive ions. If you are aware of the Coulomb's law, you can help your imagination by actually calculating these forces.

(By now you may wonder how electrical phenomena are possible at all, but that's another story.)

Just in case if you don't like that answer, here is another:

Solid zinc, like any other metal, does have a charge. It consists of $\mathrm{Zn}^{2+}$ ions. These ions are immersed in a sea of electrons, which balances their charge, holds them together, and also is responsible for such properties typical of all metals as electrical conductivity, metallic lustre, etc.


How do I find change in Gibbs free energy of water formation?

You're using the correct equation, $\Delta G^0_r(T;P)=\sum \nu_i \mu_i^0(T;P)$,
where $\Delta G_r^0(T;P)$ is the standard free energy change for the reaction at $T$ and the pressure $P$.  The task is to calculate each $\mu_i^0$.  Provided the same convention for the standard state pressure $P^0$ and for the zero value of $\mu_i$ is used, then $\Delta G_r^0$ for any reaction will be the same.  Here are some ways to calculate $\mu_i^0(T;P)$ for each species to use in the equation and the data required to calculate them (available in tables). In the following, $P^0$ is the standard state pressure used in the tables and subscript $fi$ denotes the reaction forming the species from the elements in their standard states.  If the tabular values are available at a different pressure than the one you want ($P$), you must add $RT\ln(P/P^0)$ to the values of $\mu_i$ indicated below.

$\mu_i^0 = \Delta G^0_{fi}(T;P^0)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - T\Delta S^0_{fi}$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T) dT -Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T)dT -T\left[s_i^0(298.15) + \int_{298.15}^T \frac{c^0_{pi}}{T}dT\right]$

There are many other ways.  It's an interesting exercise to prove why all the above routes give the same value of $\Delta G^0_r$ for any reaction.


Hysteresis occurs when the state of the system depends on its past states---that is, the state of the system is memory-dependent.

The canonical example is a ferromagnet: if you apply and slowly ramp up an external magnetic field to a ferromagnet for some time, and then slowly ramp the magnetic field back down for the same amount of time, you might expect the ferromagnet to return to its original state, but it doesn't. Instead, if you track the magnetization of the magnet as a function of the applied magnetic field, you see these hysteresis loops. Each applied magnetic field corresponds, therefore, to two possible system states, and which state the system exists in will depend on its past history.

Desorption hysteresis is hysteresis applied not to magnetization but to adsorption and desorption, and your excerpt then goes on to describe the physical interactions that drive this phenomenon.


Wavelength-dependence of refractive index in internal reflection spectroscopy?

Yes it should, but it doesn't matter much.

In physics, roughly speaking, everything depends on everything. We wouldn't be able to do a thing if we couldn't tell what is important and what isn't. Now let's see how much the refractive index actually changes as $\lambda$ goes, say, from 400 to 800nm (thus covering the entire visible range).

(source)

Looks like $n$ of a heavy flint glass goes from 1.85 to 1.75 (that's a change of 5%, in most other materials even less), while $\lambda$ changes by... how many percent?


Is there a way to detect bleach residue on a surface?

What immediately comes to mind is something rather obvious, don't use ammonia on any surface that could possibly be contaminated with bleach.

A search revealed a video showing red cabbage juice used as an indicator.  I do not believe the limit of detection would be anything to write home about.  However, this got me thinking about a homemade pH paper.  This would allow you to dampen the surface with a very small amount of neutral water, apply the paper, and possibly get an accurate reading on whether something acidic or basic is present on the spot you tested.

Fortunately, someone else has already put together a procedure on how to make these strips, found here.  With another source showing the expected colors using this indicator.  Since bleach aka sodium hypochlorite is a base, I would expect the color of the test strips to be in the blue to green range.

This could also be a cool home science project.  The strips will be more effective at detection for lower concentrations of bleach than a liquid solution of the indicator. (pH is concentration dependent).

Good luck!

P.S. I linked many articles to fill in some relevant information but I was limited to 2 links.  If you have questions please feel free to ask!


Why is the equilibrium constant given in litre per mole?

$K_\mathrm c$ is defined as (for more information see here)

\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} = \frac{\prod \limits_{i \in \text{products}} [c_{i}]^{|\nu_{i}|}}{\prod \limits_{j \in \text{reactants}} [c_{j}]^{|\nu_{j}|}}
\end{equation}

where $\nu_{i}$ and $c_{i}$ are the stochiometric coefficient and the concentration of the $i^{\text{th}}$ component in the reaction, respectively.

If you use this definition for the reaction

\begin{equation}
  \ce{2SO2 + O2 -> 2SO3}
\end{equation}

you get

\begin{equation}
 K_\mathrm c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^{2} [\ce{O2}]}
\end{equation}

where the notation $[i] = c_{i}$ was used.
This equation tells you that $K_\mathrm c$ has the unit

\begin{equation}
 \text{unit of }K_\mathrm c = \frac{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2}}{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2} \left(\frac{\mathrm{mol}}{\mathrm{L}} \right)} = \frac{\mathrm{L}}{\mathrm{mol}} \ .
\end{equation}


When it comes to bond energy or molecular binding energy, do people mean enthalpy?

When in doubt, consult IUPAC Gold Book for definitions (italics below is mine).

bond-dissociation energy, $D$

The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> CH3^{.} + H^{.}}$, symbolized as  $D(\ce{CH3−H})$.

bond energy (mean bond energy)

The average value of the gas-phase bond dissociation energies (usually at a temperature of  298 K) for all bonds of the same type within the same chemical species. The mean bond energy for methane, for example, is one-fourth the enthalpy of reaction for: $$ \ce{CH4_{(g)} -> C_{(g)} + 4H_{(g)}} \, .$$ Tabulated bond energies are generally values of bond energies averaged over a number of selected typical chemical species containing that type of bond.

So, yes, conceptually you should use enthalpy. Note, however, that the enthalpy of a molecular specie in gas phase at, say 298K, is calculated as follows
$$H_{298} = E_{\mathrm{e}} + E_\mathrm{ZPE} + H_{\mathrm{trans}} + H_{\mathrm{rot}} + H_{\mathrm{vib}} + RT \, ,$$
where

$E_{\mathrm{e}}$ is the electronic energy;
$E_\mathrm{ZPE}$ is the zero point energy;
$H_{\mathrm{trans}}$, $H_{\mathrm{rot}}$ and $H_{\mathrm{vib}}$ are the temperature-dependent contributions to enthalpy from translation, rotational and vibrational motion, respectively.

And the problem (as Martin mentioned in comments) is that basically only the first term out of this five can be calculated accurately enough when the ultimate goal is to calculate energies (bond energies in our case) with the so-called "chemical accuracy" which is about $\pm 1$ kcal/mol.
For relatively large molecules harmonic frequencies and consequently $E_\mathrm{ZPE}$, as well as a rigid-rotor harmonic-oscillator approximation used to calculate thermal corrections, become inaccurate, and thus, the change of electronic energy should always be used as a reference.

$%edit$


Won't the net effect of a catalyst be zero if it creates a new path with lower activation energy?

Your realisation is correct and something chemistry teachers try to hammer into their students’ heads time and time again (and yet, the point is still often lost):

Catalysts will never change the thermodynamics of a reaction. They only ease the path of the reaction. Forward and backward reactions will be accelerated equivalently.

So what is the benefit of a catalyst? There are multiple ones.

Speed

Take for example the Haber-Bosch process to synthesise ammonia from nitrogen and hydrogen.

$$\ce{N2 + 3 H2 <=> 2 NH3}\tag{1}$$
$$\Delta_\mathrm{r} H^0_\mathrm{298~K} = -45.8~\mathrm{\frac{kJ}{mol}}$$

This reaction is exothermic and thus should, theoretically or thermodynamicly, proceed spontaneously, e.g. if you mixed nitrogen and hydrogen in the appropriate ratio and added a spark. It does not, however. Significant activation energy is required to cleave the $\ce{N#N}$ triple bond.

Typical methods to add activation energy include heating. In the Haber-Bosch process, the mixture is heated to $400$ to $500~\mathrm{^\circ C}$ to supply the required activation energy. However, since the reaction is exothermic, heating will favour the reactant side. Increasing the pressure improves the entropic term of the Gibbs free energy equation, hence why pressures of $15$ to $25~\mathrm{MPa}$ are used.

Catalysts, based on iron with different promotors, are used to accelerate the reaction. By using catalysts, one can lower the temperature required in a trade-off between speed of reaction and favouring the product side of the equilibrium. With the conditions and catalysts used, one achieves a yield of $\approx 15~\%$ of ammonia within a reasonable timeframe. Not employing a catalyst would give much lower yields at much longer timeframes — economically much less feasible.

Direct reaction path not accessable

This is mainly true for many transition-metal catalysed carbon-carbon bond formation reactions, but is also true for some inorganic processes like the disproportionation of hydrogen peroxide as in equation $(2)$.

$$\ce{2 H2O2 -> 2 H2O + O2}\tag{2}$$

Hydrogen peroxide is a reactive chemical that cannot be stored forever, but the direct disproportionation path is not typically what degrades it. However, you can add $\ce{MnO2}$ to it. Upon addition, oxygen gas vigourously bubbles out of the solution. In this case, there was a kinetic barrier impeding the direct transformation due to reactants and products having different multiplicities (oxygen gas’ ground state is a triplet, all others are singlets). The $\mathrm{d^3}$ ion manganese(IV) is a radical itself that can partake in different radical reactions, allowing the diradical oxygen to be liberated.

Selectivity

This is exceptionally true for transition-metal catalysed organic carbon-carbon bond formation reactions. Note first that the action of a catalyst is frequently depicted as a catalytic cycle: A reactant reacts with the catalyst to some intermediate species, this rearranges or reacts with other reactants/additives/solvents in a set of specific steps until finally the products are liberated and the catalytic species is regenerated.

Many such reactions require organic halides as one of the reacting species. And the first step is typically an oxidative addition as shown in equation $(3)$, where $\ce{X}$ is a halide ($\ce{Cl, Br, I}$).

$$\ce{R-X + Pd^0 -> R-Pd^{+II}-X}\tag{3}$$

Palladium typically prefers oxidatively adding to bromides or iodides and tends to leave chlorides alone. I myself have performed a reaction with near-quatitative yield in which a reactant contained both a $\ce{C-Br}$ and a $\ce{C-Cl}$ bond — selectively, only the $\ce{C-Br}$ bond took part in the palladium(0) catalysed Sonogashira reaction.

Although I did not try it myself, I am pretty sure that switching to a nickel(0) catalyst species would shift the reaction in favour of reacting with the carbon-chlorine bond rather than the carbon-bromine one.

Mildness

This is basically a reiteration of the first point albeit with different intentions. Many a time in organic synthesis, one has a rather sensitive reactant that would degrade or undergo side-reactions if subjected to standard reaction conditions, such as high pH-value or elevated temperatures. As an example, consider a transesterification as shown in equation $(4)$.

$$\ce{R-COO-Et + Me-OH <=> R-COO-Me + EtOH}\tag{4}$$

This reaction is, of course, an equilibrium and by using methanol as the solvent we can shift it to the product side. For the reaction to happen, one would need a base strong enough to deprotonate methanol, giving the methanolate anion, which can then attack the ester functionality. However, methanolate being a strong (and nucleophilic) base itself can introduce undesired side-reactions, including epimerisation of the α-carbon.

One can catalyse this reaction by using $\ce{Bu2SnO}$, which will activate the carbonyl group, making it more susceptible to a nucleophilic attack. The reaction speed is the same but the conditions are milder (no additional base required) and the number of side-reactions is those strongly limited. In particular, I noticed no epimerisation of the α-carbon in the tin(IV) catalysed method.


Why is the deprotonated form of dopamine more reactive than the protonated/neutral form of dopamine?

In general the free amine will react as a nucleophile i.e. by "donating" an available lone pair of electrons. In its protonated form, this lone pair is "blocked" and thus it has to lose a "blocking" proton to react (I use the words in "quotes" more freely - although I'm a chemist and not biochemist but hopefully it will help).

However re oxidation of dopamine, the reactive site is not the nitrogen (at least not the primary one - but will come back later) but the catechol part. Because of the electron donating effect of the 2 neighboring OH's, the aromatic ring is electron rich and more prone to oxidation to give o-quinone products. This oxidation reaction depends to an extend to the pH but mainly to the presence of (obviously enough) reactive oxygen species (ROS) and also enzymes (oxidases) or even Fe ions that can catalyse this oxidation.

Coming back to the nitrogen, after oxidation occurs with creation of reactive sites on the quinone part, it can act now as a nucleophile and attack, creating indolines that will further oxidize to the more stable but toxic indole products.


Smelly compounds can be destroyed by free radicals wich are formed by a burning candle. But you need a lot of candles for your apartment.
Candles were also used in bathrooms (after toilet visits) to destroy smelly compounds like $\ce{H2S, CH3-SH, CH3-S-CH3}$ ...

A better and safer way is it too use cyclodextrin. This compound absorbs molecules and destroys them as well. Cyclodextrin is the working stuff inside the air-improvement sprays.


(Note that I'm a rather bad cook, except in the lab!)

  Is there any sense to extracting the pears from the liquid?

Yes! Yomen Atassi has already explained that you actually have to consider two (distinct) processes in the recipe:

Cooking the pears for a short period of will yield the right consistency. Additionally, some transformations happening during the cooking will possible break up some cells, convert some ingredients to even more tasty ones, and maybe even facilitate the later take up of other of other compounds from the liquid. If you cook for too long, the pears will be sludge.
Marinating means that the compounds dissolved in the liquid will permeate the pear and add to the taste. Imagine that the pear acts like a spongue and sucks all the good stuff in from the marinade. This needs time.

With a two-step procedure given in the recipe, you should get the best of both worlds and I hope that the dessert was really tasty.


Atomic emission spectra: classical vs. quantum mechanics

First of all, classical mechanics does not explain the emission spectrum at all, since there is no energy quantization out there. Bohr introduced energy quantization in his model in an ad hoc way by requiring the angular momentum to be an integer multiple of a fixed unit $\hbar = h/2\pi$ (which in turn resulted in energy also being quantized). So, Bohr model is not a classical mechanical model, since there is quantization of angular momentum and energy. But it is also not a quantum mechanical one since quantization of angular momentum is not justified, but rather asserted.

I'm saying all this just to correct OP: it was not a fight of classical vs. quantum mechanics; classical mechanics was already knockouted and quantum mechanics didn't yet fully emerged. It was a battle between semiclassical-semiquantum theories and reality which today could be viewed as well as a fight between semiclassical-semiquantum theories and the real quantum theory invented later.

Anyway, there are in general many possible electron transitions for each atom (or molecule) that contribute to its emission spectrum and each transition has a specific energy difference and probability. The Bohr semiclassical model had shortcomings with respect to both these characteristics of electron transitions:

Multi-electron atoms do not have energy levels as predicted by the Bohr model, and consequently, the model predicts wrong positions of the spectral lines. In fact, it fails already for helium atom.
There is no general way of calculating probabilities of transitions, thus, it is not possible to provide relative intensities of spectral lines.


How much energy is there in 1 mol of diamond and how can you get it?

Diamond is just one of the allotropes of carbon. There is nothing unique about diamond that allows it to store significantly more energy than any other compound. It is true, however, that much energy is required to separate carbon atoms in a diamond lattice. This is due to the high bond dissociation energy of a carbon-carbon covalent bond.

Although bond dissociation energy in a C—C bond is relatively high, it is dwarfed by the energy predicted to be released by the mass-energy equivalence theorem.

$E = mc^2$

The nuclear mass of 1 mole diamond is about 12 grams. So the theoretical energy released in a fusion reaction involving one mole of diamond is around 1*$10^9$GJ.

I don't think this is the answer you were expecting. But this is the amount of energy in 1 mole of carbon. The method of releasing all this energy is a different story. Nuclear fusion would release a significant portion of this energy but not all of it.


Depending on your first calculations before dilution:
\begin{align}
[\ce{CH3COO-}]&=[\ce{CH3COONa}]=\pu{6.4\times{10^{-5} M}}=[\ce{Na+}]\\
[\ce{Na+}]_\text{(After dilution)}&=\pu{6.4\times{10^{-6} M}}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-4} M}}~\text{ (Before dilution)}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-5} M}}~\text {(After dilution)}
\end{align}
So,you have the the following $5$ independent equations in order to specify the unknown concentrations of $5$ species as present in an aqueous solution consisting of $\pu{1\times{10^{-5} M}}$  buffer solution made from acetic acid and sodium acetate :

\begin{align}
6.4 \times 10^{−6} &= [\ce{Na+}] \tag{1}\\
K_\mathrm{a} &= \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}
     = 1 \times 10^{−4.75} \tag{2}\\
K_\mathrm{w}
    &= [\ce{H+}][\ce{OH-}]
     = 1 \times 10^{−14} \tag{3}\\
1\times{10^{-5}}
    &=[\ce{CH3COOH}] + [\ce{CH3COO-}]\tag{4}\\
[\ce{H+}]+[\ce{Na+}]
    &= [\ce{OH-}] + [\ce{CH3COO-}] \tag{5}
\end{align}