To answer your question it is important to know which compound dissociate completely in water when in aqueous solution. For example, consider following reaction:

$$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3_{(aq)}} \tag1$$

When ionic compound dissolves in water, each molecule separates to corresponding ions, which influenced by water. All of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ are ionic compounds, which are soluble in water. $\ce{AgCl}$ is, onthe other hand, an ionic compound, but it is insoluble in water (refer to solubility rules).  Hence the reaction $(1)$ can be rewrite as follows:

$$\ce{Ag+_{(aq)} + NO3-_{(aq)} + Na+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} + Na+_{(aq)} + NO3-_{(aq)}} \tag2$$

Like in algebra, you can canceled out common ions from each side of the equation $(2)$, which leaves:

$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} } \tag3$$

We called the equation $(3)$ a net ionic equation.

Now, let's look at your equation:

$$\ce{HCl_{(aq)} + NaF_{(aq)} -> HF_{(aq)} + NaCl_{(aq)}} \tag4$$

Now, $\ce{NaF}$ and $\ce{NaCl}$ are ionic compounds, which are soluble in water, hence they separated into ions in aqueous medium. How about $\ce{HCl}$, and $\ce{HF}$? They are soluble in water, but are they ionic? Since both are acids, to answer this question, we should look into their respective $\mathrm{p}K_\mathrm{a}$ values (or $K_\mathrm{a}$ values). Ref.1 is listed followings:

$$
\begin{array}{c|ccc}
\text{acid/base (chemical formula)} & \mathrm{p}K_\mathrm{a} & K_\mathrm{a} & \\
\hline
\text{hydrogen iodide }(\ce{HI}) & -10.0 & 1.0 \times 10^{10} \\
\text{hydrogen bromide }(\ce{HBr}) & -9.0 & 1.0 \times 10^{9} \\
\text{hydrogen chloride }(\ce{HCl}) & -8.0 & 1.0 \times 10^{8} \\
\text{hydronium ion }(\ce{H3O+}) & -1.7 & 50.1 \\
\text{hydrogen fluoride }(\ce{HF}) & 3.2 & 6.3 \times 10^{-4} \\
\text{acetic acid }(\ce{CH3CO2H}) & 4.7 & 2.0 \times 10^{-5} \\
\hline
\end{array}
$$

Accordingly, $\ce{HCl}$ should be ionic in aqueous solutions and should dissolve in water according to following equation:

$$\ce{HCl_{(aq)} + H2O_{(l)} -> H3O+_{(aq)} + Cl-_{(aq)}} \tag5$$

I put forward arrow because it'd dissociate completely, since $K_\mathrm{a} = \frac{\ce{[H3O+][Cl-]}}{\ce{[HCl]}} = 1.0 \times 10^{8}$.

However, $\ce{HF}$ is a different story. Since its  $K_\mathrm{a} = 6.3 \times 10^{-4} \lt \lt 1$, its dissociation is not complete, and most of $\ce{HF_{(aq)}}$ molecules stay intact (keep its covalent nature):

$$\ce{HF_{(aq)} + H2O_{(l)} <=> H3O+_{(aq)} + F-_{(aq)}} \tag6$$

Note that $K_\mathrm{a}^\ce{HF} = \frac{\ce{[H3O+][F-]}}{\ce{[HF]}} = 6.3 \times 10^{-4}$, and hence $\ce{[HF]} \approx 1.6 \times 10^{3}\ce{[H3O+][F-]}$. Like sugar molecules, which are soluble in water but did not dissociate, $\ce{HF}$ stays as $\ce{HF_{(aq)}}$ in water. Thus, the equation $(4)$ can be rewritten as:

$$\ce{H+_{(aq)} + Cl-_{(aq)} + Na+_{(aq)} + F-_{(aq)} -> HF_{(aq)} + Na+_{(aq)} + Cl-_{(aq)}} \tag7$$

When you made common ions cancelled out, the net ionic equation renains as:

$$\ce{H+_{(aq)} + F-_{(aq)} -> HF_{(aq)} }$$

References:

K. Peter C. Vollhardt, Neil E. Schore, In Organic Chemistry: Structure and Function; Fifth Edition; W. H. Freeman & Co.: New York, NY, 2007 (ISBN: 0-7167-9949-9).


Current definition implies that $\mathrm{pH}$ is a function of relative activity.
Originally, the amount concentration of  $\ce{H+}$ in $\pu{mol L-1}$ was proposed, which is also often used these days as an approximation [1].

$\mathrm{pH}$ was originally defined by Sørensen in 1909 … in terms of the concentration of hydrogen ions (in modern nomenclature) as $\mathrm{pH} = −\lg(c_\ce{H}/c^\circ)$ where $c_\ce{H}$ is the hydrogen ion concentration in $\pu{mol dm–3},$ and $c^\circ = \pu{1 mol dm–3}$ is the standard amount concentration. Subsequently …, it has been accepted that it is more satisfactory to define $\mathrm{pH}$ in terms of the relative activity of hydrogen ions in solution
$$\mathrm{pH} = -\lg a_\ce{H} = -\lg (m_\ce{H}γ_\ce{H}/m^\circ)\tag{1}$$
where $a_\ce{H}$ is the relative (molality basis) activity and $γ_\ce{H}$ is the molal activity coefficient of the hydrogen ion $\ce{H+}$ at the molality $m_\ce{H},$ and $m^\circ$ is the standard molality. The quantity $\mathrm{pH}$ is intended to be a measure of the activity of hydrogen ions in solution.
However, since it is defined in terms of a quantity that cannot be measured by a thermodynamically valid method, eq. 1 can be only a notional definition of $\mathrm{pH}.$

So, if the numerical value for normality coincides with the one for molarity, and one can omit the activity in favor of molarity, i.e. when both equivalence factor $f_\mathrm{eq}$ and activity coefficient $γ$ are approx. 1, then yes, normality can also be used to estimate the $\mathrm{pH}.$
I'm not sure how to interpret "present by itself" part of your question, but if it means there is a single solute, then it seems to be incorrect.
The simplest counter example would be a solution of di-, tri- and polybasic acid/base, say $\ce{H2SO4},$ for which the equivalent concentration is double the value of amount concentration.
References

Buck, R. P.; Rondinini, S.; Covington, A. K.; Baucke, F. G. K.; Brett, C. M. A.; Camoes, M. F.; Milton, M. J. T.; Mussini, T.; Naumann, R.; Pratt, K. W.; et al. Measurement of pH. Definition, Standards, and Procedures (IUPAC Recommendations 2002). Pure and Applied Chemistry 2002, 74 (11), 2169–2200. DOI: 10.1351/pac200274112169. (Free Access)


Does freezing a solution with water always cause the water to separate and form the ice lattice?

The purification that occurs when an aqueous solution freezes is called Fractional Freezing and is based on the idea that impurities of a solvent (and this can be any solvent, not just water) have a much lower solubility in the solid phase of the solvent relative to the liquid phase.   Several techniques for performing fractional freezing for purification purposes can be found here.   As the solid is formed, the impurities are expelled into the liquid that hasn't yet solidified.  The purified crystals can then be removed by physical methods (e.g. filtration).  In the case of water, where ice is less dense than the liquid phase, the purified crystals float to the top of the impurity-laden liquid.  This is probably the reason why some people have observed a layer of impurities deep within a block of ice.  As mentioned above, the process is not limited to just water, and freeze crystallization is use to eliminate the purities of silicon wafers and has been known for many years as a way to make hard liquor.  This latter use of fractional freezing is close to the OPs question, which inquires about miscible substances (in this case, ethanol and water).  Despite their miscibility, ethanol and water can be separated by fractional freezing, just like they can be separated by fractional distillation.  Certain substances cannot be separated completely using distillation, and these substances are called azeotropes.  Water and ethanol make an azeotropic mixture that is approximately 95% water.  (So the moonshine you were thinking of making using the reference above will be limited to 190 proof.)  What I find interesting is the following figure,

which shows the freezing point of water/ethanol mixtures as a function of ethanol content.  At a point around 95% there is a significant dip in the freezing point.  At this point, called the eutectic point, the solution behaves as if it is a single component (much like an azeotrope) and will not likely be separable through 'simple' fractional freezing means.  Note I use the term simple because azeotropes can be separated using special distillation techniques.

So in summary: fractional freezing will work with any aqueous solution, regardless of the miscibility of the components, unless the component ratio is close the eutectic point.  We don't need to limit ourselves to aqueous solutions, as many species can be purified in this manner.  The purification mechanism is not so much a factor of the crystal lattice, but the relative solute solubilities in the solid and liquid phases of the solvent.  Finally, many water solutions will freeze solid even with impurities.  This and other types of separations work on equilibrium processes, and multiple iterations are necessary to "completely" remove impurities.  (or more correct, remove the impurities to a level that is acceptable for your application.)


Can you separate CO2 into carbon and oxygen using pyrolysis?

Standard enthalpy of formation of carbon dioxide is -393.5 kJ/mol. Standard enthalpy of formation of both oxygen and graphite is 0 J/mol. Standard entropy of carbon dioxide is 213.8 J/mol-K (NIST). Standard entropy of oxygen is 205 J/mol-K (Chem LibreTexts). Standard entropy of graphite is 5.69 J/mol-K (ChemLibreTexts).
Everything below is a lot of fairly inexact approximations/assumptions because I didn't want to use a better model or search literature for empirical adjustments and/or constants.
Assuming constant T,P (which I doubt would be true under reactor conditions) G = H - TS. Approximating $\Delta H, \Delta S$ as constant with T, $\Delta G = (0 - (-393500)) - (T \cdot (5.69 + 205 - 213.8)) = 393500 + T \cdot (3.11)$. Then $\Delta G < 0$ regardless of temperature. Catalysts also do not affect thermodynamics, so this process can never occur spontaneously. Increasing pressure of the reaction chamber would not help appreciably either, as the reaction has 1 mol gas on each side. While assuming $\Delta H, \Delta S$ are constant with T is quite incorrect as conditions become more extreme, the approximation should suffice to show that the reaction does not occur easily. Additionally, even if one were to account for the perturbations in $\Delta H, \Delta S$ with T and P, these conditions would be expensive to maintain in a reactor. (Note that this analysis does not even account for the activation energy/kinetics problem! I did not look up the activation energy of the problem, but it is entirely possible that the extreme conditions would still not allow the process to occur at a reasonable rate.)
As a result, this reaction is (probably) practically only driven forward by direct energy input (aka a coupled reaction/external energy). This often, if not always, involves some energy input, such as light or a coupled chemical reaction, that directly impacts the electron states of the participant compounds. These changes in electron states change the identity/properties and stability of compounds, and, when applied effectively, they can help drive the reaction forward. However, relevant changes to electron states using heat not only often involve very high temperatures, but the effect of heat on electron states is also a highly statistical process. Thus, it may be relatively inefficient. Thus, pyrolysis, while it may be theoretically possible, would be incredibly impractical at achieving separation of carbon dioxide. (Also, to compare to your methane example, methane combusts readily, carbon dioxide does not. Methane has a standard enthalpy of formation of -74.87 kJ/mol.)
Finally, if you still need more reasons to look at a different way to drive this reaction, I have not mentioned: energy/cost to continuously scrape/collect the produced carbon, purify/collect/pressurize an inlet carbon dioxide stream, (if you actually want to run the reaction, you'll probably want) catalytic surfaces to encourage carbon deposition, possible packed/fluidized bed catalysts, heat exchangers, and much, much more.
ELI5: Methane combusts/reacts readily and is an "unstable" compound. Carbon dioxide is very stable and does not react easily. Higher temperatures can make some reactions that do not happen at a lower temperatures occur, but this particular reaction receives no such benefit. Heat and catalysis can also drive forward the speed of a reaction, but the reaction has to be able to occur first. Pyrolysis, as you described it, involves catalysts and heat. As a result, pyrolysis cannot be used to separate carbon dioxide into carbon and oxygen.
tldr: Excepting conditions extreme enough to perturb $\Delta H_{rxn}, \Delta S_{rxn}$, conditions of constant T,P never favor breakdown of carbon dioxide into elemental carbon and diatomic oxygen. Pyrolysis cannot drive this reaction.


Heat of combustion per mole and per kilogram

Q is the heat of combustion per mole. Next, you would use the formula weight to determine the heat of combustion per g (or kg).


Which is most stable conformation of optically inactive butane-2,3-diol?

Conformation B is probably correct because of the hydrogen bonding, if we assume this substance is either pure or in a polar solvent. The Me-Me gauche interaction is on the order of +3.8 kJ/mol (see most textbooks, or it is one half of methyl's A value, which involves two gauche interactions), while the hydrogen bond can be on the order of −5 to −25 kJ/mol. According to the Wikipedia article on hydrogen boning, the typical $\ce{O-H-OH}$ hydrogen bond has an energy stabilization of −21 kJ/mol. Since these groups are not quite properly aligned, a safe assumption is that the hydrogen bond interaction would be about half of that ideal value.

However, in a nonpolar solvent, there might be a benefit to conformation C, which has a net-zero molecular dipole moment. The entropic repulsion between the nonpolar solvent and B would have to be more serious than the two Me-OH gauche interactions (probably about +1.8 kJ/mol each). However, without that information, conformation B should be correct.

The correct answer cannot be A or D since they are representing a chiral diastereromer.


For what is reaction enthalpy more relevant than reaction free energy?

While a detailed answer to this question could go on at length, the major practical distinction is the following:

The enthalpy change of a reaction/process indicates how much heat input or output is involved with the reaction, with a negative $\Delta H_\mathrm{rxn}$ indicating heat output.
As you note, the Gibbs free energy change of a reaction/process indicates whether a reaction is thermodynamically favored or disfavored, with a negative $\Delta G_\mathrm{rxn}$ indicating a favorable reaction.

Note that, e.g., a reaction/process can have $\Delta H_\mathrm{rxn} > 0$ (endothermic), but also $\Delta G_\mathrm{rxn} < 0$ (thermodynamically favored).  That is, despite the reaction requiring heat input, it is still favored to occur.  This is because the change in entropy of the process, $\Delta S_\mathrm{rxn}$, is sufficiently large that the $T\Delta S_\mathrm{rxn}$ term is able to overcome the positive $\Delta H_\mathrm{rxn}$.  A common example is the endothermic dissolution of various salts, such as ammonium nitrate $\left(\ce{NH4NO3}\right)$.

Thus, the paper probably used the enthalpy results because they were comparing their calculations to experimentally-measured enthalpies (heats) of their process(es) of interest.


The red line in your diagram corresponds to the redox potential of hydrogen, which depends on $\mathrm{pH}$:

$$\begin{alignat}{3}
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\
\ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14)
\end{alignat}$$

The blue line in your diagram corresponds to the redox potential of oxygen, which depends on $\mathrm{pH}$, too:

$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\
\ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad\quad &&E^\circ = +0.401\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14)
\end{alignat}$$

Therefore, only areas between these lines are stable in aqueous solutions.

Your initial species is elemental chromium ($\ce{Cr}$). It is shown below the red line; therefore, it is unstable with respect to oxidation to $\ce{Cr^3+}$ at $\mathrm{pH}=0$

$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\\
\ce{Cr^3+ + e- \;&<=> Cr^2+}\quad\quad &&E^\circ = -0.408\ \mathrm{V}\\
\ce{Cr^2+ + 2e- \;&<=> Cr}\quad\quad &&E^\circ = -0.913\ \mathrm{V}\\\\
\ce{2Cr + 6H+ &-> 2Cr^3+ + 3H2}
\end{alignat}$$

and to $\ce{Cr(OH)3}$ at $\mathrm{pH}=14$

$$\begin{alignat}{2}
\ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\\
\ce{Cr(OH)3 + 3e- \;&<=> Cr + 3OH-}\quad\quad &&E^\circ = -1.33\ \mathrm{V}\\\\
\ce{2Cr + 6H2O &-> 2Cr(OH)3 + 3H2}
\end{alignat}$$


Seems to me there is an error in the last line of this list:

  No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
  One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
  Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$
  This results in a final ratio of 1:8:18:8:1

Simultaneous J-couplings to two $\ce{^{195}Pt}$ should result in a triplet, not a singlet, given that the J-couplings are identical in magnitude.

Therefore the expected relative intensities in the quintet are as follows:

$$\begin{align} &0.34^2\times \frac14 &= 0.029      \tag{1}
\\&0.66\cdot0.34\times1 &=0.224    \tag{2} \\& 0.34^2\times \frac24 + 0.66^2 \times 1 &=0.493   \tag{3} \\& 0.66\times0.34\times1 &=0.224   \tag{4}  \\& 0.34^2\times \frac14 &=0.029\ \tag{5} \end{align}$$

Note there is an implicit assumption that molecules with two $\ce{^{195}Pt}$ result in a 1:2:1 triplet contributing to peaks (1) (3) and (5) above, and that molecules with one $\ce{^{195}Pt}$ account entirely for peaks (2) and (4).

Normalized so that the central peak (3) has an integral of 18 the expected integrals become

$$\begin{align} & 1.0      \\&8.2     \\& 18.0    \\& 8.2     \\&1.0\end{align}$$

Separately I attempted digitizing and integrating the provided image of the spectrum. I obtained the following integrals:

$$\begin{align} & 0.25      \\&7.0     \\& 18.0    \\& 6.2     \\&0.7\end{align}$$

Given the resolution perhaps the disagreement is acceptable.


Ingredients for cleaning optical elements

It might possibly be intended to dissolve an adherent organic film, such as that left by evaporation of plasticizers from foam packaging.

However, it also might well partially dissolve paints and cements used in the lens assembly, leaving yet harder-to-remove dirt on the lens. Unless you have a specific need for those solvents, use distilled water and/or isopropanol, according to B&H.


Would sodium carbonate from carbon dioxide plus hydroxide be formed in solution or as a solid?

This depends on a lot of things: the volume of solution, the time carbon dioxide has to get dissolved and the temperature are probably the most important.

The actual reaction taking place is happening in solution. Thus, carbon dioxide must first dissolve in the solution (1) before dissolved carbon dioxide can react with the hydroxide base (2). You should view these steps as distinct even though the second is much faster than the first.

$$\begin{align}\ce{CO2 (g) &-> CO2 (aq)}\tag{1}\\
\ce{CO2 (aq) + OH- (aq) & -> HCO3- (aq)}\tag{2}\end{align}$$

Since these happen in solution, the counterion — a spectator ion — is irrelevant and the same would happen with $\ce{KOH, Ca(OH)2}$ and others.

At some point you will reach saturation, because either too much carbon dioxide dissolved or (much more likely) too much water evaporated. Thenceforth, sodium hydrogencarbonate or sodium carbonate (I never performed a formal analysis) will precipiate as a white crust. Incidentally in Munich in the inorganic teaching labs, there were 5 litre reservoirs of $6~\mathrm{M}\ \ce{NaOH}$ solution from which smaller bottles could be filled. If there was a spill (and there always was — it was undergrads in that lab after all) it would show up days later as a white crust. When I myself was a TA, part of my job preparing the labs for the next student generation was to clean those areas. Diluted hydrochloric acid worked wonders.


Diffusion of CO2 and O2 in Water/Atmosphere System

Such questions are relatively standard in discussing transport phenomena, so I'll suggest Bird, Stewart, and Lightfoot's Transport Phenomena as a good reference.

We assume a flat plane of contact infinite in extent. We assume that the bulk atmosphere has a constant concentration $c_A$ of some species of interest, and that the bulk liquid has a constant concentration $c_B$ of that species. The species is only allowed to mix between the two bulk phases for times $t > 0$. Then the problem reduces to the one-dimensional diffusion equation (Fick's first and second laws) $$\partial_t c(z,t) = D\,\partial_z^2 c(z,t)$$ subject to the initial condition $$c(z,0) = \frac{c_A+c_B}{2} + \frac{c_B-c_A}{2}\text{sgn}(z),$$ and the boundary conditions $$c(-\infty, t) = c_A, \quad c(+\infty, t) = c_B,$$ where $D$ is the species' diffusion constant and $\text{sgn}(z)$ is the signum function.

For ease of solution, make the substitutions $$\bar{c}(z,t) := c(z,t) - \frac{c_A+c_B}{2}, \quad \Delta c := \frac{c_B-c_A}{2}, \quad \phi(z,t) := \frac{\bar{c}(z,t)}{\Delta c},$$ which yields $$\partial_t\phi(z,t) = D\,\partial_z^2 \phi(z,t), \quad \phi(z,0) = \text{sgn}(z), \quad \phi(\pm\infty, t) = \pm1.$$ By definition $\phi$ is a dimensionless quantity, so it must depend on the variables relevant in the problem in a dimensionless form, for otherwise $\phi$ would have units. The relevant variables in our problem are the diffusion constant $D$, the length $z$, and the time $t$.

The only dimensionless variable that arises from this set of quantities is $\eta := z/\sqrt{Dt}$, so we know $\phi(z,t) = \phi(\eta)$. Substituting this result into our diffusion equation and applying the chain rule yields the ordinary differential equation $$-\frac{\eta}{2}\phi'(\eta) = \phi''(\eta).$$ Integrating twice yields the result $$\phi(\eta) = \alpha + \beta\int_0^\eta\text{d}\eta'\,\exp\left(-\frac{\eta'^2}{4}\right),$$ and it remains to determine the constants $\alpha$ and $\beta$ from our boundary conditions. Actually, for standardization, we will first take $\eta' \to 2\eta'$ and $2\eta \to \eta$, so $$\phi(\eta) = \alpha + 2\beta\int_0^\eta\text{d}\eta'\,\exp(-\eta'^2).$$ By symmetry $\alpha = 0$, and by explicit evaluation $\beta = 1/\sqrt{\pi}$, and hence we have $$\phi(\eta) = \text{erf}(\eta), \quad \text{or} \quad \phi(z,t) = \text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ Going back and untangling all our substitutions, we end up with $$c(z,t) = \frac{c_A+c_B}{2}+\frac{c_B-c_A}{2}\text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ The flux then follows as $$J(z,t) = -D\,\partial_zc(z,t) = -D\,\frac{c_B-c_A}{\sqrt{4\pi Dt}}\exp\left(-\frac{z^2}{4Dt}\right),$$ and we are done.

Regarding your second question, I don't believe the movement of one phase relative to the other will have any effect. As we've seen above, the xy-plane essentially drops out of the picture, and you won't be introducing any further gradients in chemical potential by translating one phase relative to the other, since the concentration of each phase is a function of $z$ only. Of course, if you had a finite area of contact, then things would be different. Perhaps see BSL 18.5, in this case.

In addition, keep in mind that there's important chemistry that's being neglected in this simplistic model. For example, the dissolution of carbon dioxide in water follows not just from diffusion but also from reaction with water, forming bicarbonate ion and related species in solution. This model doesn't account for that.


It's a consequence of the triple product rule for partial derivatives of thermodynamic state variables.

In summary, if you have three interdependent (not independent) variables, then this is true:

$$
\left ( \frac{\partial{X}}{\partial{Y}} \right )_Z \left ( \frac{\partial{Z}}{\partial{X}} \right )_Y = - \left ( \frac{\partial{Z}}{\partial{Y}} \right )_X
$$

Substituting in your variables, if we let $X = T$, $Y = P$, and $Z = S$, we get:

$$
- \left ( \frac{\partial{S}}{\partial{T}} \right )_P \left ( \frac{\partial{T}}{\partial{P}} \right )_S =  \left ( \frac{\partial{S}}{\partial{P}} \right )_T
$$

Note that the order of multiplication of the two partial derivatives doesn't matter.

Now, why this rule holds for thermodynamic variables is a little bit more complicated - but I think the wikipedia article that I linked to has a pretty good derivation if you are interested.


Forces of attraction of hydrated potassium ion in aqueous solution

Dipole-dipole interaction, specifically, will probably not happen, or they at least won't contribute significantly to IMFs. The thing with these different chemical bonds is that they are mostly based in identical features, and sometimes calculated differently, taking different things in account.
An ionic bond will be stronger than a polar bond, so that is why dipole-dipole interaction seems irrelevant, in comparison to ion-dipole forces.

All the kinds of interaction mentioned are derivatives of electrostatic interaction, also know as Coulomb interaction. LDFs are a direct derivative of Van der Waals bonds, but if you look more physically at all those bonds you can think of them as electrostatic interaction.


Why does ammonia react with water to produce hydroxide?

The shortest answer is "because the change in free energy is negative," but maybe you want to know how the mechanism works?

Ammonia is a Lewis base, and hydroxide is a stronger Lewis base. However, both are stronger Lewis bases than water is. Pure water always has some amount of "free" $\ce{H+}$ (it's actually hydronium ion) and $\ce{OH-}$ due to self-ionization ($pK_W \approx 14$):

$$
K_W = \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2}
$$

When you add ammonia to the system, some of those hydronium ions (which are very strong acids) react with the ammonia to form ammonium. Since the concentration of hydronium decreases, the concentration of hydroxide must increase to maintain equilibrium. Those hydroxide ions are stronger bases than ammmonia is, and so they will "win" the tug-of-war over protons. However, in the meantime, some other hydronium reacts with some other ammonia, and the net result is a dynamic equilibrium in which there is more hydroxide then there would be in pure water.

The same mechanism is responsible for any other acid/base reaction where a base accepts protons from water and leaves hydroxide as a free species, but the net concentration of hydroxide will be different for different strengths of bases.


Contamination would occur if a sufficient quantity of carbon monoxide wouldn't just bubble through but

would react with water to form an unhealthy product or
is dissolved in water

Both isn't really true:

$\ce{CO}$ does not react under these conditions.
The solubility of $\ce{CO}$ in water at 20 °C is as low as $\mathrm{30\,mg\cdot L^{-1}}$ (GESTATIS database)


That's not viable at all.

There's no good reaction to go from alkanes to alcohols. This would be incredible expensive to do, produces a lot of waste and is potentially very dangerous too.
There's no good reaction to convert alcohols back to alkanes. It would be again very expensive. And you would convert more expensive products into less expensive ones, which isn't a very good thing to do.
We know how to handle liquefied petroleum gas very well, it's not that dangerous. There are also hundreds of thousands of km of such pipelines [1] already in use. The safety aspect definitely doesn't justify the cost, danger, waste, ressources and energy needed to convert the product forth and back again.

[1] http://www.j-pipe-eng.com/abstract.cfm?cat_no=2145s


Neutralize the smell of alcohol in liquid chalk

Chemically, chalk is CaCO3. But blackboard chalk is mostly CaSO4.2H2O. If you want to mark something, the aim is to lay a coating of a certain (dry) thickness so as to be visible.

One way to reduce the odor is to reduce the amount of alcohol used by 1) making the paste richer in the chalk by a) just adding more chalk (but that might increase the viscosity too much), so b) grind the paste with a mortar and pestle (this may thin out the paste by causing a better variation in particle size and packing), or c) add some of whichever "chalk" you are not using (i.e., if you are using CaCO3, add some blackboard chalk and grind).

Or, 2) dilute the alcohol with a little water. You probably use alcohol to get a fast dry time, so may not be able to use much water. There are other solvents that have less odor (like heptane), but have other objections. Isopropanol has a higher boiling point (less odor, slower drying, but not much slower) than ethanol (but you are probably not using ethanol).

Or, 3) add a little TiO2. This will dramatically increase the visibility of thinner films, allowing application of much less paste and reducing the amount of alcohol exposure.

Or, 4) add a dispersant to reduce viscosity. Assuming you do not have access to many chemicals, I would suggest a little trisodium phosphate. Or a tiny amount of dishwashing detergent (be careful not to stir too vigorously so as to incorporate air bubbles); this may help wet particles or displace tiny air in pockets between particles. Then you can add more solid.

5) Chemical "masks" are available that claim to cover up obnoxious odors. In my experience they do not cover, but only add extra odor. But adding some delightful fragrance could distract from the alcohol odor to a more pleasing experience. Commercial additives I have used are coconut and cherry - both quite nice. You don't need much. Some food flavorings have enough odor to be useful here. Some people find acetone to be more pleasant than alcohol - a little acetone could change the odor to be more acceptable.

6) Slowing the evaporation by adding wax or high molecular weight alcohols (octadecanol) might reduce the odor while extending the dry time, if that is acceptable. Octadecanol is used to reduce the evaporation rate of water, and may work with alcohol; wax is often used with solvents.

Other than that, I can't think of anything.


Organic Chemistry - Reaction scheme and structures of molecules

Q is pentan-1-ol and not pentan-3-ol as on dehydration, of pental-1-ol, only one product (alkene) is formed. Whereas on dehydration of pentan-3-ol, we get two alkenes, as the alkene can show geometrical isomerism.


Would standing next to a pool of mercury do you any harm?

It depends. But there is little short term risk.

Mercury metal isn't very toxic. You can even ingest it without significant short or long term effects. But mercury vapour is nasty and you don't want to be continuously exposed to any significant amount of that (see this answer for more detail). And some mercury compounds are pretty nasty too.

Of course mercury metal emits mercury vapour so there will always be some in any room with exposed mercury. But the reason why this isn't an immediate worry is that it takes time to absorb enough vapour for it to become notably toxic. Chemists are used to handling liquid mercury in the open lab but are a lot less comfortable if it has been spilled and not cleaned up. This didn't used to be so which is one of the reasons we know the dangers: chemists who were exposed to vapour over years did suffer toxic effects. Mercury miners suffered similar long term toxic effects from the vapour. In modern labs we make sure the vapour doesn't build up by good ventilation and making sure any spills are cleaned up quickly so exposed mercury doesn't hang around emitting vapour.

So would you be in danger standing next to a pool of mercury? No, if you didn't do it for long. Or, if the ventilation in the room were good and the vapour wasn't allowed to build up. Even then, though, modern safety standards wouldn't be comfortable having pools of mercury exposed to the atmosphere: you wouldn't want to regularly work in that room with the pool.

In short, standing next to a pool of mercury even in a badly ventilated room wouldn't cause immediate harm; working next to that pool for months or years would certainly cause you harm.


$\ce{-OH}$ and $\ce{-O-H}$ are the same.

The single bond between oxygen and hydrogen is implied in the first representation and this representation is the most commonly used because of it's brevity.


The substituents go alphabetically, and as 'chloro' is before 'methyl' in the alphabet, the correct name is 2-chloro-4-methylpentane.


Why does a hard-boiled egg have more entropy than a normal egg?

I’ve come across several claims that a (hard) boiled egg has greater entropy than an unboiled egg. None of them cite any peer-reviewed research to back up their claim. While it is possible that entropy increases when you boil an egg, it is also possible that it does not. It is even possible that it stays the same.

I did my own literature search to see if I could find any publication that answered the question. What I found left the question very much up in the air. So I contacted Greg Weiss, a scientist who has done research on a related topic: ‘unboiling’ an egg. Dr. Weiss was kind enough to discuss the topic with me via email.

tl:dr

Whether a hard-boiled egg has greater entropy all boils down to the question of how structured are the assemblies found in aggregated ovalbumin. Furthermore, we can't just consider the assemblies found in the ovalbumin aggregate since we're talking about assemblies composed of all the types of protein in the overall albumen. (BTW, it's important to note that heated egg white doesn't just form any type of aggregate, it forms a gel, which presumably implies additional structure.)

For those who are interested in the details, here is the full email thread.


Using PV = nRT to solve an electrolysis problem

Both methods yield $\pu{4.45E-3}$ mol $\ce{O2}$. But to get $\pu{1 mol}$ $\ce{O2}$, you need $\pu{4 mol}$ electrons, as the half-reaction at the anode is : $$\ce{2 H2O -> 4H+ + O2 + 4 e-}$$So to get $\pu{4.45E-3 mol}$ $\ce{O2}$, the needed amount of electrons is $4$ times the amount of $\ce{O2}$, or $4$· $\pu{4.45E-3}$ = $\pu{1.78E-2}$ electrons. Now $2$ electrons are needed to produce $\ce{1 Pb}$. So $\pu{1.78E-2}$ mol electrons will produce half of it, or $\pu{8.9E-3}$ mole $\ce{Pb}$. And this amount weighs $\pu{8.9E-3} ·\pu{207.2 g}$ = $\pu{1.84 g}$


Converting Disolved Oxygen from micromole/L to millimole/kg

Oxygen is barely soluble in water, so honestly, treating the saltwater as though it were "pure" saltwater with no dissolved oxygen is roughly equivalent to treating it as heavily oxygenated.  The difference is in the fourth or fifth decimal place.  Note that this would start to fall apart if you are concerned about the solubility of high pressure oxygen in water, but for small quantities of oxygen (such as the micromoles and millimoles you're using) this is likely not a factor.

The conversion is relatively straightforward.  For a solution with concentration $X$:

$$(X\ \mathrm{\mu mol/L}) \times (1\ \mathrm{mmol}/1000\ \mathrm{\mu mol})\times(1\ \mathrm L/1.027\ \mathrm{kg}) = 0.0009737\ X\ \mathrm{mmol/kg}$$


This question is based on relative rates of reaction of different $\ce{H}$ with halogens. In the free radical mechanism of chlorination of alkane, relative rate of reaction of $\ce{3^\circ : 2^\circ : 1^\circ}$ $\ce{H}$ is in order $5:3.8: 1$.

Using this, we get an idea about product amount. To get a rough estimate of relative amounts of different products, we multiply number of identical hydrogen and its rate of reaction with chlorine.

So there is only $1$ $3^\circ$ $\ce{H}$ and its rel. rate is $5$. So relative amount is $5$. For the $2^\circ$ $\ce{H}$, there are $2$ of them and their relative rate is $3.8$. So its relative amount is $2 \cdot 3.8 = 7.6$. Similar calculation for rest of the hydrogens will show you why given answer is correct.

Refer: Wikipedia


Why doesn't this structure represent a substituted pentane?

You need to find the longest continuous carbon chain of:

$$
\begin{align*}
\ce{CH3-CH2-CH2-&CH-CH3}\\
&|\\
\ce{&CH2-CH2-CH3}
\end{align*}
$$

Stretch that out to $\ce{CH3CH2CH2CH(CH3)CH2CH2CH3}$ (with the $\ce{-CH3}$ coming off of the $\ce{CH}$) and you'll see that the book is correct.  The "side chain" of $\ce{-CH2CH2CH3}$ is actually part of the main carbon chain.  As far as the numbering, counting from either side give the 4- position due to the symmetry of the molecule.


If radium has such a long half-life, how can radon possibly be a threat to us?

There are two factors here:

Radon is a gas and can thus enter the body very easily, through inhalation.
$\ce{^222Rn}$ and its first 4 decay products are extremely radioactive, with halflives of miliseconds to a few days, releasing alpha and beta radiation.

When the two are combined, we have a problem. Even if the quantity is low, inhaling radon places an extremely radioactive substance at very close proximity to the cells in one's lungs. Normally, alpha and beta radiation are relatively harmless as they cannot penetrate tissue very well, but now they have direct access. All it takes is one mutation in the wrong place to cause lung cancer.

To give you an idea of what a difference these make, the volumetric activity which is one guideline for needing to mitigate radon accumulation in a house is around 800 $\mathrm{Bq/m^3}$, which corresponds to only 141 fg of radon per cubic metre of air. This isn't like a uranium mine or something, but at this level, a certain increased risk of cancer is expected. Apparently some study found a 50% increased risk at a much lower level ~150 $\mathrm{Bq/m^3}$, too.

N.B. a becquerel (Bq) is a reciprocal second—1 Bq means one atom of the radioactive substance disintegrates every second.


Determination of a chemical compound in a non-homogeneous sample

If you analyze non-homogeneous samples, the results can't be interpreted as being representative of the whole organism (or population). So, for example, if histamine is unevenly distributed within the body of a fish, your results would vary widely depending on what part of the fish you analyzed.


Why are protons in the same position in 1-chloro-4-nitrobenzene magnetically different?

The hunch lies in the definition of magnetic equivalence:

  For two protons to be magnetically equivalent they not only have to have the same chemical shift, but they must also each have the same $J$ coupling to other magnetic nuclei in the molecule.

(Quoted from Hans Reich’s NMR course online syllabus; emphasis present in the original.)

What does that mean for our specific example? Compare the molecule we have in figure 1 below.

Figure 1: 1-chloro-4-nitrobenzene with labelled protons.

Needless to say as you already figured that $\ce{H^1}$ and $\ce{H^3}$ are chemically equivalent (one is transformed into the other by $C_2$‡) meaning that they will display the same chemical shift; likewise for $\ce{H^2}$ and $\ce{H^4}$.

However, for magnetic equivalence we need to consider their coupling constants to each other. If $\ce{H^1}$ and $\ce{H^3}$ are magnetically equivalent, then $J_{\ce{H^1 H^2}} \overset{!}{=} J_{\ce{H^3 H^2}}$. But, $J_{\ce{H^1 H^2}}$ is an ortho coupling with a value of $J \approx 7~\mathrm{Hz}$ while $J_{\ce{H^3 H^2}}$ is a para coupling with $J \approx 0~\mathrm{Hz}$.

Thus we can easily see that $J_{\ce{H^1 H^2}} \ne J_{\ce{H^3 H^2}}$ and therefore the nuclei $\ce{H^1}$ and $\ce{H^3}$ are magnetically non-equivalent.

The same argument can be made for $\ce{H^2}$ and $\ce{H^4}$.

‡: They are also transformed into each other by the $\sigma_{\mathrm{v}1}$ plane. However, planes of symmetry create enantiotopic protons while rotations create isotopic ones. In this case — a non-chiral environment and a non-chiral solvent — enantiotopic protons cannot be distinguished and give a common signal, and thus are sometimes incorrectly called chemically equivalent. The argument is not of relevance here though, since the $C_2$ axis renders the protons isotopic thus equivalent.


Addition of Br2 to Double Bond and Carboxylic Acid

The first intermediate is indeed Br+ atom attached to the double bond forming a cycle (see step one):

 On the next step, however, oxygen of carboxylic acid attacks the intermediate and forms the product you show on second figure. This happens because carboxylic acid group is in the right position for intramolecular reaction.

First step follows the diagram. In second step the attacking nucleopline is not Br$^-$, but R-COOH or RCOO$^-$ depending on the reaction condition. To obtain high yield of the cyclic product on your second figure you need to slowly add dilute solution of Br$_2$ to a solution of organic reagent. If you add Organic reagent to neat (pure) Br2, then you get a significant amount of the product you predicted on figure 1. Don't do it in real life because a reaction of neat bromine with alkenes is violent and dangerous.


Optical isomerism in Co-ordination Compounds

Any 3rd-year inorganic chemistry textbook worth its weight in salt will provide some information about isomers in coordination compounds.  An optically active compound contains no rotation-reflection symmetry element.  Typically, this means that if a molecule has an inversion center or a mirror plane then it is not chiral.

It takes a systematic brute force and ignorance approach to determining the number of isomers a coordination compound has, and then a bit more visual thinking to determine if any of the isomers are enantiomers.  The following figure is taken from Miessler, Fischer, Tarr 5th edition and is useful for identifying isomers of 6-coordinate complexes.

Note that a 6-coordinate complex with 6 unique ligands in an octahedral field would be enantiomers. In fact, there are 15 different isomers, each of which is chiral resulting in a total of 30 isomers!

The method of determining isomers this way is to formulate a structure of the type M where  indicates two ligands that are trans to each other.  Then, you start varying a, b, c, ... until you've come up with as many unique combinations as possible.  For your case, you have 2 instances of a bidentate ligand.  Bidentate ligands are typically designated with capital letters in this notation, so the formula for your molecule would be $\ce{M(AA)_2b_2}$.  Further, since bidentate ligands are too small to coordinate in a trans conformation, you will decrease the number of possible isomers.  One isomer, for example, would be M, where I use A and A' to designate the different $\ce{(en)}$ ligands.

The answer to your question, then, is that there are two isomers, and they are enantiomers, since you can't rearrange the A, A' and b in the formula above to come up with another unique configuration.


How does critical temperature affect adsorption?

Higher critical temperature = more adsorbed.

So, what is critical temperature? When gas is heated above it, it can't become liquid, even if you'll increase pressure. Why? Because it's large kinetic energy makes intermolecular forces kinda "irrelevant". And adsorption is all about those forces.

High critical temperature means that gas particles are good in intermolecular forces - "tenacious". This allows them to adsorb (cling to surfaces) more easily than those that won't hold on (low critical temperature).


How was pH measured back in the day if you had nothing to calibrate to?

"Hydronium" content can be easily determined by simple acid base titration. Keep in mind that there is nothing fundamental about pH. It is just a convenient scale for expressing hydrogen ion concentration. For example, a kilogram is a recent invention for expressing mass measurement, it does not mean that ancient people did not weigh anything before kilogram was known.
In the same way, acid-base titration is more than 300 years old, with litmus among the earliest indicators. pH is a relatively new concept. Of course, the titrations were not analytically precise and accurate 300 years ago.
Historically, the simplest way to do acid base titration was to use potash (naturally occurring mineral $\ce{K2CO3}$). One could continue titration until the evolution of bubbles stopped.* No indicator needed. This is the way vinegars were assessed. Note that in those time moles, molarity, did not exist. There was the concept of equivalent weights, i.e., this weight of X will react this that weight of Y completely. The concept of equivalents is still taught in some countries, although it has vanished from Western curricula.
Kolthoff (father of modern analysis) has written books on acid base indicators. Check internet archive. Your question is worth an entire essay.
*From Bishop, Indicators: International Series of Monographs in Analytical Chemistry


What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH?

Part A:  The conceptually part: Scientific concepts and conception

1-"I thought because 1 mmol of KOH reacts with 1mmol of CH3COOH this will form 1 mmol of CH3COOK"

yes ,the reaction reached at the equivalent point, the weak acetic acid in mixture has been fully neutralized by the stong($\ce{KOH}$) base (no unreacted acid or base present anymore), the solution will only contain $\ce{H2O}$ and the basic salt ($\ce{ CH3COOK}$),so the concentration of the salt :
 $$[\ce{CH3COOK}] =\frac{\text{moles of salt}}{\text{Total volume}} =\frac{1\text{m}mol}{(10+10)\text{ml}}=0.05\text{M}$$
 2-"which leads to no formation of OH−ions. Because there is no excess."

No, the result, at the equivalence point, will be the same as dissolving $(\ce{CH3COOK})$ in water at the same concentration(0.05${M}$) as you have at the equivalence point.
the salt ($\ce{ CH3COOK}$) is  electrically neutral substance formed by cation $\ce{K+}$ (an acid)and  anion$\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COOK   -> K^+_{(aq)}   + CH3COO^−{(aq)}}$$
a)  Since $\ce{K+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{K+}$  actually spectator ion.

b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions$\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :

$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$

Part B:: Calculating the  approximate pH of $\pu{0.05 mol L-1} \ce{CH3COOK}$ with neglecting water autoionization:

The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:

$$ K_\mathrm{b}=\frac{K_\mathrm{w}}{K_\mathrm{a}} =\frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}$$

Assuming :${[\ce{OH-}]=[\ce{CH3COOH}]}$, ${[\ce{CH3COO-}]=[\ce{CH3COOK}]_0}$

Substiute $K_\mathrm{a}\ and\ [\ce{CH3COO^-}]$ in the equilibrium equation and solve for $[\ce{OH-}]$.

$$[\ce{OH-}] = 5.27\times{10^{-6}} , \mathrm{pOH}=5.278 , \mathrm{pH}=8.72$$


Reasoning for redox reactions to be classified as such

$\ce{Zn + 2AgCN -> 2Ag + Zn(CN)2}$
Zinc changes its oxidation state from $0$ to $+2$, and silver changes its oxidation state from $+1$ to $0$. So, zinc is reducing agent and silver is oxidising agent.
$\ce{Zn^0 ->[-2e^-] Zn^{+2}}$ (oxidation);
$\ce{2Ag^{+1} ->[+2e^-] 2Ag^0}$ (reduction).
Other reactions are just equations of ion exchange without changing the oxidation states of chemical elements.


$\ce{Al2O3.nH2O}$ is just a general (and slightly incorrect, see below) way of writing the formulae of the aluminium oxy-hydroxides that occur in bauxites. These are solid minerals, so there is no "$\ce{Al2O3}$" molecule in them. Instead, it is an (almost) infinite array of Al and O ions in a three dimensional structure. Therefore, what it basically means is that you have three cations of aluminium for every two anions of oxygen. $\ce{Al2O3}$ means exactly the same as $\ce{AlO_{1.5}}$ or $\ce{Al4O6}$ in this context. We just use $\ce{Al2O3}$ because it's simpler: smallest numbers with no fractions.

In the case of the hydroxides, they are also crystalline minerals. You can see some nice pictures of gibbsite, böhmite and diaspore when they are well crystallised. Their formula are:

Gibbsite: $\ce{Al(OH)3}$ which is the same as $\ce{AlO_{1.5}.1.5H2O}$
or $\ce{Al2O3.3H2O}$.
Böhmite and diaspore: $\ce{AlO(OH)}$ which is the same as $\ce{AlO_{1.5}.0.5H2O}$ or $\ce{Al2O3.H2O}$

Bauxites usually contain mostly gibbsite, which is unfortunate economically because it contains the most water in it. When you dig the stuff out, you want the most aluminium per tonne of material, and dilution by water is undesirable.

It's also important to note that even though we write the formulae of the minerals as $\ce{Al2O3.nH2O}$, the water is not hosted in molecular form in the crystal. You do not have proper $\ce{H2O}$ molecules in the structure of the mineral. Instead, you have hydrogen cations attached to some of the oxygen anions. There are, however, minerals where the water is in molecular form. One example is gypsum: $\ce{CaSO4.2H2O}$.

Usually, as a rule of thumb, the way a formula is written can give you an idea on how the water is hosted. If the formula is written with $\ce{(OH)}$ in it, it's probably in the form of hydrogen bonded to oxygen. If it's $\ce{.H2O}$ then you probably have molecular water. This is also important for the physical properties of the crystals. Minerals with molecular water are usually very soft and will lose their molecular water with gentle heating (up to 150 °C or so). Minerals with hydroxide groups in them require much higher temperatures to de-volatilise, around several hundreds of °C.

So, to sum it up, writing the formula as $\ce{Al2O3.nH2O}$ gives the impression that the aluminium hydroxides have molecular water in them, while they actually do not. However, it's a convenient "catch-all" way to refer to all of them together.

Also related:
What is the meaning of the "dot" notation in chemical formulas?


What is the Si–N–C bond angle in silyl isothiocyanate?

The answer that you are supposed to give is presumably something along the lines of "dπ–pπ" bonding between silicon and nitrogen. This essentially means that the lone pair of nitrogen is involved with backbonding into a silicon 3d orbital, leading to a linear geometry at nitrogen (i.e. the bond angle is 180°). This can be represented with the following resonance structures:

To be very clear, this is very likely incorrect. The extent of d-orbital involvement in main group compounds is usually tiny. If there is any backbonding, the acceptor orbital is probably a Si–H σ* orbital rather than a d-orbital (cf. backbonding from transition metal to phosphine).

Electron diffraction studies of H3SiNCS show that the Si–N–C bond angle is actually 163.8°.1 A computational study (TZVP + MP2 level of theory) found 165.91°.2. In fact, the linear structure above is not an energy minimum, but rather a transition state for the interconversion of the two energy minima:3

I don't actually know how one can predict this a priori without access to a computer, but clearly this system is not amenable to bond angle prediction with VSEPR theory, and invoking the so-called "Si–N dπ–pπ bonding" doesn't give the correct answer either.

References

Glidewell, C.; Robiette, A. G.; Sheldrick, G. M. Electron diffraction study of large amplitude vibrations in H3SiNCO and H3SiNCS. Chem. Phys. Lett. 1972, 16 (3), 526–529. DOI: 10.1016/0009-2614(72)80416-0.
Palmer, M. H.; Nelson, A. D. The structures of the azido-, isocyanato- and isothiocyanato-derivatives of methane and silane and their derivatives. A comparison of ab initio with experimental results. J. Mol. Struct. 2004, 689 (1–2), 161–173. DOI: 10.1016/j.molstruc.2003.11.008.
Palmer, M. H.; Guest, M. F. The methyl, silyl and germyl esters of the pseudo-halogen acids. A comparison of structural data from experimental and ab initio studies. Chem. Phys. Lett. 1992, 196 (1–2), 183–190. DOI: 10.1016/0009-2614(92)85951-6.


Assuming you are given normality of sodium thiosulphate as 0.1 N (or sometimes called standard), you could use this formula to find the iodine value:

iodine number=$$\frac{(mL 0.1N Na_2S_2O_3 blank  – mL 0.1 N Na_2S_2O_3 test) (12.7)(100)}{(1000) (1.1)}$$

Where “mL 0.1 N Na2S2O3 Blank” = the volume of sodium thiosulphate required to titrate
the blank solution,
Where “mL 0.1 N Na2S2O3 Test” = the volume of sodium thiosulphate required to titrate the
test solution that contained either solid fat or oil as the sample,
Where “12.7” is the amount of grams of iodine contained in one litre of 0.1 N iodine,
Where “1.1” is the amount of fat taken in grams  and,
Where “100” converts the sample to per 100 grams of fat.


What is the chemical reaction between coins (copper and nickel) and polyvinyl chloride?

Far, far too broad a question. I don't know what is going on in your situation.
I can, perhaps, clear up some misconceptions you may have. PVC is NOT commonly used in sandwich bags. PVDC is a close relative and used to be common as "saran wrap". But most sandwich bags are made out of polyethylene (PE). Chemically both PVC and PVDC degrade due to weathering and other things. Dehydrohalogenation is the reaction of organic chlorine compounds (both PVC and PVDC) which can be expressed in the partial formula C-HC-Cl → C=C + HCl (this is a crude simplification). Note that this reaction generates HCl, also known as hydrochloric acid (or muriatic acid). HCl is a powerful acid, attacking most metals. If you live by the ocean or in a desert chloride levels in the air (dust or mist) can be quite high and can accelerate corrosion dramatically. Coins are designed to be inert. Placed clean and dry into a sealed PE bag, they should not corrode. But. Whenever two different metals (or metal alloys) come into contact, a electrical potential is created. A random bag of coins will be sorta like a bag of little batteries. That is, you shouldn't be surprised if they react (discharge), especially in the presence of moisture. Neither copper nor nickel form green oxides, but both do form green chlorides and green carbonates. You need a source of chlorine for the chlorides, but the carbonates might form when simply exposed to air and moisture when a electrical potential exists. Sweat, sweaty hands, are potentially one source of chloride/chlorine, but as I stated up front I am not familiar with your situation, I carry (mixed) coins in my pocket for days at a time without any green forming.


Total oxidation of toluene free enthalpy

Your problem is that you are trying to calculate a kinetic property (spontaneous versus nonspontaneous; on a compound level that’s inert versus reactive) with thermodynamic data (exergonic versus endergonic; on a compound level that’s stable versus unstable).

Thermodynamically, we should all be carbon dioxide and water, as should any other organic compound. However, the activation energy of combustion of organic compounds is rather large. Thus, while the reaction is thermodynamically favourable, it is not spontaneous in the absence of activation energy.


I see two possible answers. Not (A) and (B) but rather (A) and (D).

Chakravarthy Kalyan is responsible for reminding us that (D) can be an answer too (refer to comments).

For option (D):

$\ce{HC ≡ CH + 2AgNO3 + 2NH4OH→ AgC ≡ CAg↓ + 2NH4NO3 + 2H2O}$

The greyish precipitate of silver acetylide will readily form when ethyne is passed through freshly prepared tollen's reagent.

You may read more from this answer by @ron: Does acetylene give a positive Tollens' test? If not, why is this so?.

For option (A):

While $\ce{Br2}$ looks like a seemingly unsuitable candidate, it can actually be used as a crude way to differentiate them.

Take the same volumes of equal strength of bromine water in two vessels and allow equal amounts of ethyne in one and ethane in the other to pass through them (this shouldn't be difficult, just passing equal volumes at STP will do the trick).

Carefully observing which vessel has the more intense color will help you identify which one is ethene, which would consume half the amount of bromine consumed by ethyne.

If you're not convinced, put the final solutions in a cuvette each and measure the absorption or transmittance in a spectrophotometer at 530 nm.

P.S. Make sure that you have a sufficient amount of bromine water, or you'll end up comparing two colorless solutions.

P.P.S. You could also use titration to find which one reaches the end point first, which would be ethene.


I believe that you would have to know at least these things to start with:

All enthalpies of bond dissociation are endothermic
Complete dissociation of triple bonds is more endothermic than double bonds, which is more endothermic than single bonds (when compared for the same elements).
The difference between going from a triple bond to a double bond is smaller than double to single, and that is smaller than going from single to complete dissociation (you referenced this in your question)

We can summmarize these facts with the following figure:

Here D(O-O) represents the bond dissociation enthalpy of an oxygen-oxygen single bond. The numbers inside of circles represent "steps" between each successively stronger bonds, and are there just for easier reference.

With that in mind, let's look at the options again, and count how many of each type of bond are broken and formed:

$ a) \space O_3(g) \rightarrow O_2(g) + O(g) $

Ozone has a resonance structure, which means that we need to consider the bonds as being somewhere between a single and double bond in strength. We could imagine them being about halfway up step 2.

Broken: 2 x single/double resonance bonds

Formed: 1 x double bond

So we are looking at an overall reaction that would be slightly more endothermic than the breaking of a single bond. To make that easier to see, think about taking these energetic "steps" using the figure above:

(+1.5, +1.5) - breaking the two resonance bonds

( -1, -2) - forming the double bond

$ b) \space  2O_3(g) \rightarrow 3O_2(g) $

Here we are breaking two sets of resonance bonds, but forming 3 double bonds. So our steps look like:

(+1.5, +1.5, +1.5, +1.5) - breaking six resonance bonds

(-1, -2, -1, -2, -1, -2) - forming three double bonds

There are three -1, -2 steps, each of which is larger than a +1.5 step. However, they are not much larger, and the total of the difference may or may not be greater than the fourth 1.5 step. This means we can't say for sure whether it is exo- or endo-thermic, but we can at least say it would be less endothermic than breaking a single bond.

$ c) \space  3O_2(g) \rightarrow 2O_3(g) $

Now we are doing the opposite of (b). We still don't know the sign, but we know the magnitude is the same as (b), which was less than (a), so either way it can't be (b) or (c).

$ d) \space  O_2(g) \rightarrow 2O(g) $

Here we are breaking a double bond, and forming nothing. This is the clear winner so far, with steps +1, +2.

$ e) \space  O_3(g) + O(g) \rightarrow 2O_2(g) $

In the final case, we break two resonance bonds, and form two double bonds. Our steps are:

+1.5, +1.5, -1, -2, -1, -2

Since the formation of each double bond releases more energy than the dissociation of each resonance bond, this is clearly exothermic, as you stated in your question.

Taking all of this together, we have to choose (d) as the correct answer.


Why is aluminium(III) considered a hard acid if it forms covalently bonded compounds like aluminium oxide?

Short summary - HSAB theory has it's limits of applicability, and this is a good example.

More recent discussion of acid-base theory center on the frontier orbital concept, e.g. W. B. Jensen, The Lewis Acid-Base Concepts (1980) pp. 112-155.

In this concept, an "acid-base" reaction creates new HOMO-LUMO orbitals of the product. So it's about energy matching between the acid (empty LUMO) and base (filled HOMO) (and also symmetry and orbital overlap). It's still very much a Lewis picture, since you have the electron pair on the base and an electron acceptor on the acid.

For example, if we react $\ce{NH3 + H+ -> NH4+}$ we combine the HOMO of $\ce{NH3}$ (i.e., a lone pair) with an empty LUMO on $\ce{H+}$ to create the adduct.

IMHO, this is a nice concept, since the orbital energy match between Al and O is good, you get covalent character to the resulting product. (Now you also get ionic character, but that's a different question.)


Why does vibrational frequency influence quantum tunneling of a proton?

It is not the frequency per se but the mass that is important,; the probability of tunnelling an energy E is given by

$$p(E)=\exp \left(-\frac{4\pi\sqrt{m}}{h^2}\int_{x_a}^{x_b}\sqrt{V(x)-E}\;dx \right)$$

where $x_a$ and $x_b$ are two points either side of the potential barrier through which tunnelling occurs. The integral is in effect a measure of the area of the potential $V$ above position $x_a$ to $x_b$ through which tunnelling can occur.

As m increases, for a given potential barrier and energy, the tunnelling probability falls.


How is "F2/sig(F2)" the data quality in SXRD?

$F^2/\sigma(F^2)$ is basically the signal-to-noice ratio, a simple measure of how well your crystal diffracts. It is hardly related to $R_{int}$ at all. Indeed, to calculate $R_{int}$ you need to have the whole dataset, otherwise it just would not make sense. As for $F^2/\sigma(F^2)$, you have it as soon as you've collected a little bit of data.


Why does not aqua regia burn filter paper?

For "burning" of paper by sulphuric acid, the better term would be "aggressive dehydration".

Note that aqua regia is rather a diluted acidic mixture without significant dehydrating or oxygen based oxidation behaviour, but manifests aggressive chloration behaviour, that dissolves precise metals.

Cellulose has quite good endurance against chlorine, which is used in various solutions for cellulose whitening.


Can you explain how we get gold(II) sulfide?

Brauer's Handbook of Preparative Inorganic Chemistry ([1, p. 1062] and references therein) lists the following method for gold(II) sulfide synthesis:

  $$\ce{$\underset{1/10}{}$\qquad$\underset{242.7}{\ce{8 AuCl3}}$ + $\underset{20.2\,l.}{\ce{9 H2S}}$ + $\underset{7.2}{\ce{4 H2O}}$ = $\underset{183.3}{\ce{8 AuS}}$ + $\underset{87.5}{\ce{24 HCl}}$ + $\underset{9.8}{\ce{H2SO4}}$}$$

  A neutral 1-3% solution of $\ce{AuCl3}$ is precipitated in the cold (the temperature must not exceed $\pu{40 ^\circ C}$) with $\ce{H2S}$ or an alkali sulfide. The precipitate is filtered, thoroughly washed with water, and
  treated with alcohol, anhydrous ether, $\ce{CS2}$ and again with ether.
  The product is dried at $\pu{130 ^\circ C}$ over $\ce{P2O5}$.

  Alternate method: A solution of $\ce{AuCl3}$ is added in drops to an
  aqueous solution of sodium dithiosulfatoaurate(I) [...].

  Properties: Formula weight 229.07. Deep black. Insoluble in water and acids; soluble in aqua regia and solutions of potassium cyanide and alkali polysulfides. Resistant to KOH in the cold, decomposes after prolonged boiling, liberating gold. Thermal decomposition begins at $\pu{140 ^\circ C}$.

Reference

Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1965; Vol. 2.


Do thermodynamically favourable chemical reactions always proceed?

Because a reaction is thermodynamically favoured does not mean that it will be a fast reaction. So 'yes' they always proceed but possibly infinitesimally slowly. Temperature is the most common way of quickening a reaction. Of course a catalyst also does so but this generally by changing the way the reaction occurs, i.e. the mechanism is different using a catalyst but the product is the same. Enzymes are natures catalysts, in synthetic chemistry (and in you car exhaust) metal based compounds are often used.

All reactions have an activation energy $E_a$ between reactants and products and a small increase in the size of this can slow a reaction exponentially. Experimentally, the rate constant is generally found be of the form $k=k_0 exp(-E_a/RT) $, which is the Arrhenius equation, with R the gas constant and T the temperature in Kelvin.

The Boltzmann thermal distribution explains why this type of equation applies. This shows that as temperature increases molecules have more energy and so surmount the activation energy more frequently and so reaction rate increases,


The short way is to proceed as noted in the comments. The equivalent long way is to start from the total differential for $S_m$ in terms of $p$ and $T$:

$$dS_m = \left(\frac{\partial S_m}{\partial p}\right)_Tdp + \left(\frac{\partial S_m}{\partial T}\right)_pdT$$

The second term is given by

$$ \left(\frac{\partial S_m}{\partial T}\right)_p = \frac{C_p}{T}=\frac{5R}{2T}$$

The first can be determined using the identity

$$ \left(\frac{dS_m}{dp}\right)_T = -\left(\frac{\partial V_m}{\partial T}\right)_p=-\frac{R}{p}$$

Substituting,

$$dS_m = \left(-\frac{R}{p}\right)dp + \left(\frac{5R}{2T}\right)dT$$

It is not difficult to show that this is equivalent to the result obtained by substituting

$$dV=\frac{R.dT}{P}-\frac{RT.dP}{P^2}$$

into the expansion in terms of $V$ and $T$ shown in the OP.


Does electrolysis of water alter the electrodes?

It depends on the substance you have used for your electrodes. If they are in platinum, they will not be altered. If they are in carbon (charcoal, graphite), they will not be chemically tansformed, but after a while, they will be slightly corroded, and some powder of charcoal will progressively appear under the electrode on the ground of the flask used for the electrolysis. If they are in copper, there will be oxidized and a chemical reaction will occur at the anode. The copper metal will be oxidized and transformed into $\ce{Cu^{2+}}$ ions, which will produce a precipitate if the solution is basic or contains an anion like phosphate or carbonate. If the electrodes are in iron or zinc, the anode will be quickly corroded and transformed into the corresponding ions. In the beginning, the cathode will not be altered, but in the long run, they will also be corroded, specially if the solution is acidic.
Anyway you have not described the composition of the electrolyte. Common metals like iron and zinc are altered by acids, with or without electrolysis.


The two ions you are comparing to each other are very different. In $\ce{ClO3-}$ there are two more electrons than in $\ce{NO3-}$. This is also the reason why the former pyramidalizes, while the latter does not.

In nitrate the negative charge can be delocalised over the whole molecule, since there is one molecular orbital with $\pi$ symmetry. One basic requirement for these orbitals is, that they are antisymmetric with respect to a mirror plane, that includes the bonding axis.

This MO scheme would also be thinkable for the chlorate ion, but here you have two more electrons, so you have to fill an antibonding orbital. The charge can no longer be delocalised as well in this molecule. The conclusion is, that the symmetry of the whole molecule is lowered and the $\pi$ orbitals vanish. A lone pair is always more stable in an orbital with high $\ce{s}$ contribution, which leaves only the $\ce{p}$ orbitals of chlorine for bonding with oxygen.

Resonance is a model which is necessary to describe bonding situations, that cannot be described by one Lewis structure alone. It is therefore more a rule of thumb than any evidence. $\pi$ bonds (or orbitals) are not necessary for valid resonance structures. They are however necessary for ([negative] hyper-)conjugation.


Why doesn't this β-hydroxycarboxylic acid undergo dehydration upon heating?

In order to obtain an α-β-unsaturated carbon acid from a β-hydroxy acid, you need a secondary or tertiary carbon in α position, because otherwise you would have to break a carbon-carbon bond, which requires a lot more energy then simply heating.

The mechanism is basically self catalysed, which is why this may happen easily with a little bit of heat. The acid may protonate intramolecular (or intermolecular) the hydroxy group. You have water as a nice leaving group, creating a carbocation, and abstraction of a proton in α-position leads to the olefin.

If you do not have that α-hydrogen, you are stuck with the carbocation. You might be able to obtain a lactone,like Waylander suggests, but I doubt this would easily happen with a little bit of heating, as spiro-compounds with four-membered rings tend to be quite unstable.

In your attempt you choose the less acidic hydrogen to transfer and then break a carbon-carbon bond resulting in a very unstable molecule which would rearrange to cyclohexanecarboxylic acid. I'd suspect you would need insane temperatures to overcome the activation barrier.


Why does an electron lose energy so that it falls back to a lower energy state than just stay at the higher energy state?

Actually, this question is not as bad as the comments may make you think. It's actually quite a sensible question, because when we solve the usual, Schrödinger's equation-based, QM problem of energy states of hydrogen atom (or any other atom), we get eigenstates of the Hamiltonian we inserted into the equation. These solutions should thus be stable for infinite time.

In real world, however, an atom isn't isolated. It interacts with the electromagnetic field, and this interaction results in spontaneous emission, which is the reason why an atom can't be in excited state indefinitely.


Does using stonger soaps help me lose weight?

I'm afraid no: you are applying a chemical principle to a complex biological response.
The Le Chatelier principle state:

If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established.

We start with the definition of chemical system at equilibrium :

the temperature dependent condition in which chemical species are present at concentrations which have no further tendency to change with time.

So what you need are chemical species: molecules, ions. Although fatty acids could be a product "body" is not a compound or a molecule. So what you have is not a chemical system but a biological one. You could go a little more close if you think something like this:
$$\ce{Fatty acid <=> Chemical  \space Precursor_{Y} + Chemical \space Percursor_{X}}$$
However even this is a wrong oversimplification of many complex biochemistry reactions and equilibria.
Moreover you can't interact with these equilibria, perturbing them and so applying LC principle because sebum, which is a complex mixture of lipids, is stored inside sebaceous glands and is formed by the complete disintegration of the cell it self. So to from new sebum you have to from new sebaceous glands.
Soap simply remove sebum not seabaceous glands, so using stronger soap by the way don't affect sebum production. At this point is clear that is no more chemistry but physiology. In this case an equivalent of Le Chatelier's principle is the homeostasis. However from what I know the diet is one of the main factors that induce sebaceous glands to produce sebum so eat better seems to be the best way to lose weight and sebum!

References
Pub Med Sebaceous gland lipids


Can I get some pointers on what to do in calculating the hardness of water?

The hardness of water is defined as the sum of the dissolved $\ce{Ca^2+}$ and $\ce{Mg^2+}$ species and can be reported in different ways (moles/L, g/L, and so on). The one I have seen most commonly is mg/L, so that is the calculation presented below.

You need to determine how many milligrams of $\ce{Ca^2+}$ and $\ce{Mg^2+}$ are present and then divide that number by the solution volume (here, 10 L).

For $\ce{Ca^2+}$:

$$1.64\;\mathrm{g}\;\ce{Ca(NO3)2}\cdot{1\;\mathrm{mol}\;\ce{Ca^2+}\over 164.09\;\mathrm{g}}\cdot{40078\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Ca^2+}}
 = 400\;\mathrm{mg}$$

For $\ce{Mg^2+}$:

$$0.94\;\mathrm{g}\;\ce{Mg(Cl)2}\cdot{1\;\mathrm{mol}\;\ce{Mg^2+}\over 95.2\;\mathrm{g}}\cdot{24305\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Mg^2+}}
= 240\;\mathrm{mg}$$

Thus, you have 640 mg $\ce{Ca^2+}$ and $\ce{Mg^2+}$ in 10 L of solution so the hardness is:

$${640\;\mathrm{mg}\over 10\;\mathrm{L}} = 64\;\mathrm{mg}\cdot\mathrm{L}^{-1}$$


Why can't a nucleophile attack the intermediate of an electrophilic aromatic substitution?

The intermediate is non-aromatic arenium cation which is stabilized by resonance. Deprotonation of sigma complex is a fast process as it leads to much thermodynamically stable i.e., aromatic ring. Aromatization or deprotonation step is fast as Kinetic Isotopic effect is not observed. Hence even a weak base will extract the proton from the substituted site which is sp3 hybridized.

As for another nucleophile to attack the $\sigma$ complex, the positive charge is delocalized and deprotonation step is much faster. I think that it would be easier for the base to extract proton than to attack the Delocalized positive charge. The proton already has a $\delta$+ charge due to Hyperconjugation effect. I think that even if the base if a strong acid, it can dissociate to give H+ but the non-aromatic intermediate will not remain the solution as it is.


Which is the strongest base relative to ammonia?

Let's have a look at some of the definitions in the gold book and see how they relate to your problem.

  proton affinity
  The negative of the enthalpy change in the gas phase reaction (real or hypothetical) between a proton (more appropriately hydron) and the chemical species concerned, usually an electrically neutral species to give the conjugate acid of that species. Proton affinity is often, but unofficially, abbreviated as PA.

  gas-phase basicity
  The negative of the Gibbs energy, $\Delta G_\mathrm{r}^\circ$ change associated with the reaction:
  $$\ce{B + H+ -> B+-H}$$
  in the gas phase. Also called absolute or intrinsic basicity.

I was unable to find an exact definition for relative basicity, so I am going to assume based on the following reasoning. First we define the basicity for ammonia, $Bf(\ce{NH3})$, and an arbitrary amine $Bf(\ce{NR3})$.
\begin{align}
\ce{NH3 + H+ &~-> NH4+}  -\Delta G_\mathrm{r}^\circ(\ce{NH4+}) &
Bf(\ce{NH3}) &= -\Delta G_\mathrm{r}^\circ(\ce{NH4+})\tag{1} \\
\ce{NR3 + H+ &~-> NR3H+}  -\Delta G_\mathrm{r}^\circ(\ce{NR3H+}) &
Bf(\ce{NR3}) &= -\Delta G_\mathrm{r}^\circ(\ce{NR3H+})\tag{2} \\
\end{align}

We can follow up and define the relative basicity of an arbitrary amine  to ammonia $Bf_\mathrm{rel}(\ce{NR3})$ as the difference of the independent basicities.
$$Bf_\mathrm{rel}(\ce{NR3}) = Bf(\ce{NR3}) - Bf(\ce{NH3})\tag{3}$$

With (3) we would actually describe the following reaction:
$$\require{cancel}
\begin{align}
\ce{NR3 + \cancel{H^+} + NH4+ &~-> NR3H+ + \cancel{H^+} + NH3}\\
\ce{NR3 + NH4+ &~-> NR3H+ + NH3}\\
\end{align}$$

So the relative basicity describes in this scenario how likely it is that the arbitrary amine is abstracting a proton from the conjugate base of ammonia, i.e. ammonium. The higher the relative basicity, the stronger the base.
The conjugate acid $\ce{NR3H+}$ is the strongest if its parent compound has the smallest basicity, meaning most likely to loose the proton. The weakest conjugate acid is obviously the strongest base.

But what have we actually compared? In principle we have discussed with this comparison how likely it is for the amines to bind a proton and we have chosen to do that in comparison with ammonia. We could chose to do the comparison to water and with a couple of transformations and simplifications we would arrive at a $\mathrm{p}K_\mathrm{a}$ which we would then compare. To make things easier, we could also directly compare the values of $\Delta_\mathrm{r}^\circ$ in (1) and (2); the lower the value, the higher the basicity.

If we now choose to compare the enthalpies instead of the Gibbs energies, then we would be ultimately comparing (gas-phase) proton affinities. We would be still looking at the same reactions, just a different characteristic property. We would still be observing the same trend.

TL;DR It does not matter if you look at proton affinities or relative basicity, you are still comparing the same effect with different numbers, that are all linked.


Does reaction quotient Q take into account pure solids and liquids?

It isn't that solids don't appear in equilibrium constants and/or reaction quotients. They are simply assumed to have an activity equal to the value of 1! Hence people don't show them, but they should. I think it is important to understand and remember this assumption.

Also, equilibrium constants and reaction quotients are fundamentally a ratio of activities! The use of concentration and/or pressures in equilibrium constants/ reaction quotients is the results of approximations in themselves and this simplicity permeates chemistry and should be stopped or at least respected.

for the reaction you show, the equilibrium constant is

$$K_\mathrm{eq} = \frac{a_\ce{H2PO4-}a_\ce{H3O+}}{a_\ce{H3PO4}a_\ce{H2O}}$$

See the Chemistry LibreTexts for more details about solid and liquid activities and equilibrium constants.


Avogadro's number and percent error help?

When a problem requires calculations using values, always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.

Nevertheless, your approach is correct. You calculated the volume of the solution as
$$\begin{align}
V_\text{solution}&=\frac{23}{30\ \mathrm{ml^{-1}}}\tag{1}\\[6pt]
&=0.77\ \mathrm{ml}\tag{2}
\end{align}$$
the concentration as
$$\begin{align}
c&=\frac n{V_\text{solution}}=\frac m{M\cdot V_\text{solution}}\tag{3}\\[6pt]
&=\frac \gamma M\tag{4}\\[6pt]
&=\frac{1.2\times10^{-4}\ \mathrm{g\ ml^{-1}}}{284.48\ \mathrm{g\ mol^{-1}}}\tag{5}\\[6pt]
&=4.2\times10^{-7}\ \mathrm{mol\ ml^{-1}}\tag{6}
\end{align}$$
and the amount of stearic acid as
$$\begin{align}
n&=c\cdot V_\text{solution}\tag{7}\\[6pt]
&=\frac{1.2\times10^{-4}\ \mathrm{g\ ml^{-1}}}{284.48\ \mathrm{g\ mol^{-1}}}\cdot\frac{23}{30\ \mathrm{ml^{-1}}}\tag{8}\\[6pt]
&=3.2\times10^{-7}\ \mathrm{mol}\tag{9}
\end{align}$$

Furthermore, you can calculate the area of the surface as
$$\begin{align}
A&=\pi r^2\tag{10}\\[6pt]
&=\frac{\pi d^2}4\tag{11}\\[6pt]
&=\frac{\pi \left(15.5\ \mathrm{cm}\right)^2}4\tag{12}\\[6pt]
&=189\ \mathrm{cm^2}\tag{13}
\end{align}$$

For simplicity’s sake, you may assume that a single stearic acid molecule is a small block with a length $l$ and a width $w$.

Depending on the used model, you may find that the ratio is roughly

$$l=6w\tag{14}$$

Assuming that the surface is densely packed with stearic acid molecules, the number $N$ of molecules can be estimated as
$$N=\frac A{w^2}\tag{15}$$

The corresponding volume of stearic acid is

$$\begin{align}
V&=A\cdot l\tag{16}\\[6pt]
&=A\cdot 6w\tag{17}
\end{align}$$

You also know that
$$\begin{align}
V&=\frac m\rho\tag{18}\\[6pt]
&=\frac {n\cdot M}\rho\tag{19}
\end{align}$$

where $\rho$ is the density of stearic acid. Instead of using the density of solid stearic acid at room temperature, you may want to assume the density of liquid stearic acid close to its melting point, which is

$$\rho=0.847\ \mathrm{g\ cm^{-3}}\tag{20}$$

Solving $\text{(17)}$ for $w$ and inserting the result into $\text{(15)}$ yields

$$N=A\cdot\left(\frac{6A}{V}\right)^2\tag{21}$$

And inserting $\text{(19)}$ for $V$ yields

$$N=A\cdot\left(\frac{6A\cdot\rho}{n\cdot M}\right)^2\tag{22}$$

The Avogadro constant is defined as
$$N_\mathrm A=\frac Nn\tag{23}$$

Inserting $\text(22)$ for $N$ yields

$$\begin{align}
N_\mathrm A&=\frac{A\cdot\left(\frac{6A\cdot\rho}{n\cdot M}\right)^2}{n}\tag{24}\\[6pt]
&=\left(\frac{6\rho}{M}\right)^2\left(\frac An\right)^3\tag{25}\\[6pt]
&=\left(\frac{6\times0.847\ \mathrm{g\ cm^{-3}}}{284.48\ \mathrm{g\ mol^{-1}}}\right)^2\left(\frac{189\ \mathrm{cm^2}}{3.2\times10^{-7}\ \mathrm{mol}}\right)^3\tag{26}\\[6pt]
&=6.6\times10^{22}\ \mathrm{mol^{-1}}\tag{27}
\end{align}$$

Even when taking the uncertainty of the used values into account, this result is significantly smaller than the literature value of the Avogadro constant. Therefore, you may conclude that a large part of the difference may be attributed to experimental error. I guess, you did not use the floating device method to push the stearic acid molecules together, which makes sure that the molecules are densely packed and lined up on the surface of the water with polar heads on the surface and nonpolar tails sticking up away from the surface.


Determining the order of molar conductivity

I would agree with the answer key. The trick here is it's limiting molar conductivity, the molar conductivity at infinite dilution. For a neutral electrolyte compound, I was taught the notation $\Lambda ^0$, and $\lambda ^0$ was reserved for individual ions. (The $0$ superscript represents zero concentration, equivalent to $\infty$ dilution.)

Anyway, limiting molar conductivity is an interesting property because it does not represent a physically possibly scenario (a solute conducting electric charge when no solute is present). It can be extrapolated by the Kolhrausch Law, where it $\Lambda^0$ is the y-intercept of molar conductivity $\Lambda$ vs $\sqrt C$, concentration (note this relationship is only valid for strong electrolytes).

The degree of ionization of a weak electrolyte depends on the concentration, and tends to unity (100% dissociation) in the limit of concentration approaching zero. Limiting molar conductivity $\Lambda^0$ is defined (only) for this exact limiting scenario.

So when we consider limiting molar conductivity, we are doing so in a (abstract) condition where both strong and weak electrolyte fully dissociate. Thus for $\Lambda^0$ it makes no difference whether the electrolytes being compared are "strong" or "weak" at finite concentrations (for we are comparing them at "zero" concentration, i.e. infinite dilution). So we should have no problem accepting that a "weak" electrolyte might have a higher limiting molar conductivity value than a "strong" one, as in this case.

Why might $AcOH$ (organic chemist abbreviation), $HCl$, and $NaOH$ have higher $\Lambda^0$ values than $KCl$? Well, the $H^+$ ion is better at transporting its charge through an aqueous medium than $K^+$ because it can "water hop" via the
Grotthuss mechanism. This makes acidic compounds very effective electrolyte conductors. $OH^-$ is also an extremely effective charge carrier by an analogous "deprotonation-chain" mechanism. Because they use water itself to "tunnel"* through the bulk solution, protons and hydroxide have extremely high ion mobility in aqueous solutions and thus provide abnormally high molar conductivity.

So in general acids and bases would be expected to out-conduct a pH-neutral salt ($KCl$), and in the limit of zero concentration/infinite dilution, even "weak" acids and bases will have higher limiting molar conductivity values.

*Not in the quantum sense.


What form of zinc's found naturally in vegetables & meat?

The most likely form is $\ce {Zn^{2+}}$ bound inside a protein, or maybe chelated to some amino acids. It is not very likely that there is zinc oxide, or zinc sulfide, or zinc sulfate in organic matter.


Can the Bohr atomic model be applied to sodium and other single valence electron species?

One of the great successes of the Bohr model was the ability to explain the emission spectrum of hydrogen (and hydrogen-like monoelectronic ions), by allowing a simple theoretical derivation of the experimentally determined and remarkably accurate Rydberg formula:

$$\frac{1}{\lambda}=RZ^2 \bigg(\frac{1}{n_f^2}-\frac{1}{n_i^2}\bigg)$$

$\lambda$ is the wavelength of light emitted, $R$ is a proportionality constant known as the Rydberg constant (which actually varies slightly with changes in atomic mass), $Z$ is the amount of positive charge attracting the electron (the nuclear charge in multiples of $q=1.6022\times 10^{-19}\ \mathrm{C}$), and $n_f$ and $n_i$ are positive integers (called quantum numbers) which previously had no experimental interpretation, but which now arise from the postulates of the Bohr model, being attributed to the concept of electron shells.

After this great new insight into atomic physics, naturally one would like to investigate how generally applicable the model is. Given the success in describing the emission spectra of hydrogen, one of the first places to look would be the spectra for other elements. So here they are:

(Source - isn't it a beautiful image? Kudos to the designer! Open in a tab for gigantic resolution)

As you can see, most elements have rather elaborate spectra. After undoubtedly many frustrated attempts, it was found that the Bohr model, as accurate as it was for hydrogen, applied quite poorly to the rest of the elements, and no spectra could be reliably predicted.

However, the alkali metals stood out in that a modified version of the Bohr model had some success, producing a Rydberg-like formula, though salvaging it came at the cost of additional experimental parameter fitting, again with no theoretical background.

The formula for the energy of a electron in a hydrogen atom according to the Bohr model is:

$$E_n=-\frac{hcR}{n^2}=-\frac{13.6\ \mathrm{eV}}{n^2},\ \ \ \ n \in \mathbb{N} \ \ (n=1,2,3,4...)  $$

For the alkali metals, the formula had to be modified to:

$$E_n=-\frac{hcR}{n_{eff}^2}=-\frac{hcR}{(n-\delta_l)^2},\ \ \ \ n_{eff} \in \mathbb{R_+}  $$

To keep the same structure, the mismatch between theory and experiment is brushed into the quantum numbers $n$, which are replaced by effective quantum numbers $n_{eff}$ that are no longer positive integers. To recover integer quantum numbers, a value called the quantum defect ($\delta_l$) had to be introduced. This value is different for each value of $n$, and also varies with another parameter $l$. This may seem like an arbitrary mathematical attempt to salvage the equation, but while the values of $\delta_l$ had to be tabulated, it was found that they were relatively constant with changes in $n$, and that there were a limited number of different cases ($l$ was a parameter limited to only a few values). The resulting Rydberg-like equation is:

$$\frac{1}{\lambda}=R \bigg(\frac{1}{(n_f-\delta_{l_f})^2}-\frac{1}{(n_i-\delta_{l_i})^2}\bigg)$$

While not as tidy as the result from the Bohr model, at least the equation can be used to approximately explain the atomic emission spectra for the alkali metals. Details of an experimental procedure for chemistry undergrads to determine the emission spectrum of sodium and apply the modified Rydberg formula can be found here, along with theoretical background.

Nowadays, we know why the alkali metals seemed to be a better fit to the Bohr model than the rest of the elements. With finer knowledge about the details of the electronic structure of atoms, borne out of the full quantum mechanical model of the atom, we know that alkali metals share an important property with hydrogen; they all possess a single electron in their outermost shell, i.e. have a valence shell configuration of $(ns)^1$. This is the only electron which is important in electronic transitions in the near-ultraviolet, visible and infrared regions of the electromagnetic spectrum, as the core electrons of the alkali metals are held much more tightly to the nucleus and require high energy photons to excite. This creates a much simpler spectrum.

Furthermore, atoms with multiple electrons suffer from electron-electron repulsions, which are not taken into account at all in the Bohr model, and thus are a significant source of error. However, in the alkali metals the consequences of these repulsions are simplified, as all electrons are in core shells, except the single valence electron. For all other multielectronic elements, electron-electron repulsions in the valence shell preclude a simple application of the Bohr model.


How much gas is inside a pressurised gas cylinder?

The cylinder contains $\pu{19kg}$ of LPG, $\pu{60\%}$ of which is propane (presumably by mass) so we have $\pu{11.4kg}$ propane and the rest is butane.

The molar mass of propane $\ce{C3H8}$ is $\pu{3*12+8*1 = 44 g/mol}$, so we can work out how many moles of propane there are. Then do the same for butane.

Now we use the ideal gas law:

  But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.

You are not interested in the pressure inside the cylinder. You are interested in what the volume is when the LPG uncompresses. Therefore the pressure you need to use is atmospheric pressure.

If you apply the formula for the ideal gas law $PV=nRT$ you will find that $1$ mole of gas occupies about $\pu{24 L}$ at $\pu{25^oC}$ ($\pu{298K}$) and atmospheric pressure. That is a good shortcut to remember for cases like this.

Note that the ideal gas law only works when the conditions are in the range where the gas behaves like an ideal gas. You know the volume of the cylinder, but you cannot use the ideal gas law to find the pressure inside the cylinder, because the gas behaves non-ideally at high pressure (It liquifies!)


The radius ratio of $\ce{Li+}$ and $\ce{Cl-}$ is very small, this means the size of $\ce{Li+}$ is so small than $\ce{Cl-}$ that the crystal lattice of $\ce{LiCl}$ becomes unstable. So to make the size of $\ce{Li+}$ bigger $\ce{H2O}$ molecules surround the caption bigger and make the structure stable.


Difference between baking soda and baking powder

As LDC3 says, Baking soda is simply sodium bicarbonate, $\ce{NaHCO3}$. When one adds an acid, the following reaction occurs: $\ce{HCO3- + HA -> H2CO3 + A-}$ and the carbonic acid releases carbon dioxide: $\ce{H2CO3 -> CO2 + H2O}$ (these are both equilibrium reactions, but as you can observe when you add vinegar, a lot of gas is produced and the gas is generally lost, pushing the equilibrium towards making carbon dioxide).

Baking powder is a mixture of sodium bicarbonate and one or more weak acids. When dry, no reactions occur, but when wet, the components dissolve and can react. The observed differences when mixing vinegar with baking powder probably result from the fact that there is less sodium bicarbonate in a given volume of baking powder than baking soda, and baking powder often contains starch, which can influence the properties of the added liquid, giving the bubbles a different appearance.


As Waylander states, calcium and magnesium hydroxides do have some solubility in water.  In the case of magnesium hydroxide you can demonstrate this by simply buying some milk of magnesia and applying red litmus paper to it; the paper turns blue like there were an alkali.

Simply dissolving either calcium or magnesium hydroxide into water would raise the pH beyond the target (you get 10 for magnesium, 12 for calcium).  Additional calcium or magnesium salts may be introduced to shift the dissolution equilibrium for the sparingly soluble hydroxide, and thus control the pH.

Here and here the use of lime and magnesia for water treatment is mentioned.  Both work by making the water slightly alkaline so as to displace heavy metals as less soluble hydroxides.


Is a weak deactivator considered a stronger activator than a strong deactivator for the purposes of directing effects

I think you are reaching the limitations of directing group rules. These rules are not perfect and can often break.

In this example, chloride and nitro meta to each other, the group should add meta to the nitro group because the nitro group has a much more significant effect on the electronics of the molecule.


How did Joseph-Louis Proust know what gases weighed?

He knew because he weighed them

This might seem ridiculously obvious but the early pioneers of modern atomic theory based the idea on actual results they got from weighing things before and after reactions.

It is perfect possible to weigh a container full of gas if you have an accurate enough balance (the buoyancy of the air cancels out if you use the same vessel for different gases). Note that the idea of a vacuum was developed in the 17th century so was well-known by the time of Proust and he could also have measured the empty weight.

The idea is simple: fill a vessel with some gas and weigh the vessel very accurately. A litre of oxygen weighs about 1.5g which is large enough to measure and with good equipment and technique large enough to measure accurately. A liter of hydrogen would be a little under .1g, a big difference.

A series of careful experiments like this with a range of gases would reveal something about the weight-ratios and carefully counting weights and volumes before and after reactions would tell you more (2 volumes of hydrogen react with one of oxygen, for example).

This is what people like Proust did, though with equipment we would now regard as primitive. The reason why it isn't mentioned a lot in modern descriptions is probably because it seems too basic and simple. Nowadays we just teach the theory not so much about the careful experiments that came first and actually led to the theory.


Intermolecular Forces for Ionic Compound?

Ions in ionic compounds are held together by electrostatic attractions, i.e. the idea that "opposite charges attract".

The strength of an electrostatic attraction is given by Coulomb's law:

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}$$

where $q_1$ and $q_2$ are the charges on the two ions and $r$ is the distance between them. In a completely ionic bond, $q_1$ and $q_2$ are multiples of the elementary charge, $e = 1.602 \times 10^{-19} \text{ C}$; however, no bond is completely ionic.


In complex analytical chemistry problems like this, it's useful to consider a stoichiometric matrix of the reaction network.

                               | H2PO4- | H3O+ | H3PO4 | CH3COOH | HPO4-2 | PO4-3
------------------------------ | ------ | ---- | ----- | ------- | ------ | -----
1st phosporic aciddissociation | 1      | 1    | -1    | 0       | 0      | 0
2nd phosporic dissociation     | -1     | 1    | 0     | 0       | 1      | 0
3rd phosporic dissociation     | 0      | 1    | 0     | 0       | -1     | 1
Acetic acid ionization         | 0      | 1    | 0     | -1      | 0      | 0

Each score represent the stoichiometric coefficient of that species in the given reaction, with a negative sign when it's a reactant and a positive value otherwise. I am not using $\ce{KH2PO4}$ or $\ce{HCl}$ as species because they are always fully dissociated. It's better to think of them as $\ce{H2PO4-}$ and $\ce{H3O+}$ respectively in regards to their molar concentration.

We can approach the problem by consider a "mixing $\to$ reaction" two-step process. Because equilibrium is achieved regardless of the path chosen, we can do this freely, that is, consider the "initial" concentration as though we mixed all parts together with no reaction occurring.

The mixing step mass balance will be $C_{\textrm{initial}} \times V_{\textrm{pre-mixing}} / V_{\textrm{solution}}$. Here's the full table:

                               | Solvent | H2PO4-       | H3O+         | H3PO4 | CH3COOH      | HPO4-2 | PO4-3 | CH3COO- | Total
------------------------------ | ------- | ------------ | ------------ | ----- | ------------ | ------ | ----- | ------- | ------------
Initial Concentration (mol/L)  | 0       | 2,2044883383 | 4            | 0     | 1            | 0      | 0     | 0       |
Initial volume (L)             | 0,03    | 0,02         | 0,008        | 0     | 0,5614062187 | 0      | 0     | 0       | 0,6194062187
Solution concentration (mol/L) | 0       | 0,0711806976 | 0,0516623809 | 0     | 0,9063619346 | 0      | 0     | 0

Since we are basically solving a nonlinear problem, I took the liberty to assign a random value to the initial acetic acid volume, which we will later determine by minimizing an objective function.

Let's call the stoichimetric matrix $A$, and the solution concentration (post-mixing, pre-reaction) vector $C_{\textrm{post-mix}}$. Let's define the extent of reaction vector $\delta$ as the array of values representing the amount of mols (per unit volume) produced or consumed by each reaction.

Reaction                   | Extent (mol/L)
-------------------------- | --------------------
1st phosporic dissociation | 0,0034158968
2nd phosporic dissociation | 1,46105223300036E-05
3rd phosporic dissociation | -2,218252003941E-14
Acetic acid ionization     | -0,0547767774

Once again, we do not know these values a priori. They will be optimized later on. The amounts of each reactant that is produced or consumed ($C_{\textrm{reacted}}$) is then given by the matrix product

$C_{\textrm{reacted}}^T = A^T \delta$

The equilibrium concentration is then the post-mixing concentration plus the reacted concentration,

$C_{\mathrm{eq}} = C_{\text{post-mix}} + C_{\textrm{reacted}}$

Now consider the vector of potentials of the equilibrium constants fpr each reaction ($\mathrm pK = -\log(K)$),

Reaction                       | pK
------------------------------ | -------------
1st phosporic aciddissociation | 2,1611509093
2nd phosporic dissociation     | 7,2076083105
3rd phosporic dissociation     | 12,3187587626
Acetic acid ionization         | 4,7447274949

For the equilibrium step, equaling the equilibrium expression to the $K$ of each reaction is the same as imposing the following matrix product (think of why?)

$A (\mathrm pC_{\textrm{eq}})^T = \mathrm pK$

Where the vector $\mathrm pC$ is calculed by taking the negative log of the species concentrations vector $C_{\mathrm{eq}}$. One of these values is the $\mathrm pH$, which is given by the problem.

We are then left with the task of finding the initial acetic acid volume by solving:

Find $V_\ce{CH3COOH}$ and $\delta$ (remember, this is a vector) that minimize the residuals in $A (\mathrm pC_{\mathrm{eq}})^T = \mathrm pK$ and $\mathrm pC_{\mathrm{eq}} \Bigg|_{\text{species}=\ce{H3O+}} = (\mathrm pH)_{\text{given}}$.

I used libreoffice-nlpsolver to yield

$$V_{\ce{CH3COOH}} = 0.5614062187\ \mathrm L$$

But, wait, this is the volume of acetic acid we'd employ to react with the entire solution volume. But the problem asks about only $25\ \mathrm{ml}$, it is thus just a matter of scaling the proportion using the total volume ($V_\mathrm t = 0{,}6194062187$),

$$V = 0{,}5614062187 L \times \frac{0{,}025 L}{0{,}6194062187} = 0{,}022659048 L \ \mathrm L$$

Or about $22{,}7\ \mathrm{ml}$

I purposely did not take into account the self ionization of water, because of the low pH in question. But it could have been done nonetheless. The only changes would be the need to include $\ce{OH-}$ in the stoichiometric matrix and take $\mathrm pW$ into action.


What is the charge of the amino acid lysine at pH 2?

At $\mathrm{pH}=2$, the charge will be:

(charge near neutral pH, 2 groups positive, one negative) + (the fraction of the pKa = 2.18 group that is protonated)

=1 + (the fraction of the $\mathrm{p}K_\mathrm{a} = 2.18$ group that is protonated)

$=1 + 1/(1 + 10^{\mathrm{pH}-\mathrm{pK}}) =1 + 1/(1 + 10^{2-2.18})=1.60$

So the charge is +1.60, which is closer to 2 than 1, as you are saying.