This is simply how the chemist chose to draw the molecule in that particular instance, and cetyl alcohol is formally regarded as a straight-chain molecule. Cetyl alcohol has all single bonds and no rings, so it is not structurally locked into any particular conformation or geometry, and every bond is free to rotate.

For very simple saturated molecules like this, where the conformation is not in any way fixed (under typical conditions), the chemist is free to represent the molecule as one of any of its possible conformers.

In reality, the most stable (i.e., lowest energy) conformer should statistically predominate, and this is the one with all groups staggered and the largest groups antiperiplanar (q.v., the IUPAC Gold Book entry on torsion angles for details), the conventional explanation being that this minimizes steric hindrance/strain. The skeletal formula corresponding to this minimum-energy conformer is the simple straight-chain representation (seen here, for example).


How do you make sense of acid base reactions in organic chemistry?

The answer to both questions lies in the fact that a reaction can be directed to a certain extent by controlling the concentration of the reagents.  The first reaction is perfectly fine since an alkoxide ion is more basic than a hydroxide ion. As for the second reaction, a carbon attached to an electronegative element is quite acidic. Hence is capable of possessing a negative charge. This along with the excess amount of water we supply for this reaction and the fact that water is neutral in nature adds to the reaction progressing in the forward direction.

Hope this helps


What happens to minerals once you boil off liquid in a water and fat mixture?

Because dietary minerals are predominantly water soluble and not significantly soluble in fats, your first statement is correct; the minerals will not dissolve in the fats as you boil away the water.  This is true even after all the water has boiled off, so the answer to your second question is no. As the water boils away, the minerals will precipitate out until no water remains. At that point, you will predominantly have the minerals at the bottom of the pan, covered in the fats. It is unlikely any of the minerals will have a low enough density to float on top of the fats. With mixing, you may be able to keep a suspension of the minerals within the fats even though they don't technically dissolve.

It will probably be difficult to tell exactly when all of the water is gone. There should be a difference in the appearance of the way the stuff boils once the water is gone and it's just fats boiling, but this could be a subtle change. You might be able to tell when the water is gone by monitoring the temperature of the mixture. It should stay close to the boiling point of water until the water is gone and then increase when you just have the fats boiling.

I'm far from a chef, so I hope I haven't made any improper assumptions on that front. Don't hesitate to ask for clarifications in the comments.


Likely the green solution contains copper(II) phosphate, $\ce{Cu3(PO4)2}$. "It is commonly encountered as the hydrated species Cu2(PO4)OH, which is green."

Try crystallizing it!

Libethenite, $\ce{Cu3(PO4)2}$, crystals.

You have $\ce{H+}$, $\ce{Na+}$, $\ce{Cu+}$ and/or $\ce{Cu++}$, $\ce{OH-}$, $\ce{Cl-}$ and $\ce{PO4{^-}{^3}}$ in solution. You could identify the white crystals by color (white, so probably no Cu in them) shape, birefringence, and solubility; they might be $\ce{NaCl}$ or $\ce{Na3PO4}$. The less soluble should crystallize first, of course.


Ultrasound absoprtion by polypropylene beaker for HF acid

Polymers, like other materials as well, attenuate / absorb acoustic waves.  Beside reflection (echo, similar to active sonar), this renders ultrasound suitable for non-destructive inspections.  The extent of absorbing ultrasound depends on several factors (e.g., temperature, density/crystallinity of the sample, deployed frequency) and may be expressed as attenuation and by acoustic impedance.

The following table is part of a lecture about medical diagnostic applications of ultrasound:

(source, slide 5.)

A more detailed listing, specifically addressing polymers, may be found here. From data like

|---------------------------+----------+---------+-----------+-------------|
| Plastic material          | Velocity | Density | Impedance | Attenuation |
|                           |    [m/s] | [g/qcm] |           |     [dB/cm] |
|---------------------------+----------+---------+-----------+-------------|
| Polyethylene, low density |     1950 |    0.92 |      1.79 |         2.4 |
| Polypropylene, Profax     |     2740 |    0.88 |      2.40 |         5.1 |
| Polypropylene, white      |     2660 |    0.89 |      2.36 |        18.2 |
|---------------------------+----------+---------+-----------+-------------|

some differences in polymers of same name, but different brand (and degree of polymerisation, etc.) become apparent. (Interestingly, Teflon/PTFE is not mentioned in their listing, yet according to this study possess a much higher acoustical impedance than PP.)

Hence, beside chemical resistance against HF, you likely have to test the efficiency of the ultrasound waves passing through the walls of your beaker anyway.


Why are pi antibonding orbitals lower in energy than sigma antibonding orbitals?

When two atoms form a bond, if two atomic orbitals mix, then two molecular orbitals are created, one of lower and one of higher energy than that of the atomic orbitals.

Now, just how much the energy changes is dictated by how well the atomic orbitals overlap. Overlapping depends on three things:

Orbital energy
Distance between atoms
Orbital symmetry (they must be "aligned")

Now, assuming that the atomic orbitals are similar in energy and the symmetries are correct, which is the case for C and O.

The sigma bond is a result of two orbitals very close together, directly facing each other frontally. The overlap is great. The pi bond emerges from two orbitals overlapping side-by-side, which isn't as efficient because it happens at a greater distance.

Here is a pic to exemplify:

Thus, not only is the bonding combination of a sigma bond lower in energy, but also the antibonding combination is higher in energy.


Neutrons bind with protons and one another in the nucleus through the strong force, effectively moderating the repulsive forces between the protons and stabilizing the nucleus.$^{[1]}$

$\ce{^2He}$ (2 protons, 0 neutrons) is extremely unstable, though according to theoretical calculations would be much more stable if the strong force were 2% stronger. Its instability is due to spin–spin interactions in the strong force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the $\ce{^2He}$ nucleus a negative binding energy. $\ce{^3He}$ (2 protons, 1 neutron), on the other hand, is stable, and is also the only stable isotope other than $\ce{^1H}$ with more protons than neutrons.$^{[2]}$

$^{[1]}$ Wikipedia, Neutron, Beta Decay and the Stability of the Nucleus
$^{[2]}$ Wikipedia, Isotopes of Helium


What are dipole moments in a molecule supposed to act upon?

Your confusion appears to arise from trying to consider dipole vectors as representing forces, which is incorrect.

Dipole vectors don't "act on" anything; they describe the displacement of the "center-of-charge"‡ of a given region of space (including both the electron cloud and the nuclei) from a particular reference point. In particular, I believe you are confusing dipole moment vectors with electric field vectors:

The electric field vector at a point describes the force experienced there by a point charge
The dipole moment vector at a point describes the displacement of the "center of charge" of the system from that point

Thus, whereas the electric field contributes to the forces present in a system, the dipole is simply a post facto descriptor of a system.

Further, in this case the law you quoted is not applicable:

  Two equal and opposite vectors do not cancel each other out when acting on different bodies.

It only makes sense to calculate an overall dipole moment if you use the same reference point for all of the 'subsystem' dipole vectors. Thus, in this example of $\ce{CO2}$, the two $\ce{C-O}$ 'subsystem' vectors point in exact opposite directions from the same reference point, since $\ce{CO2}$ is linear.  Thus, since the outside atoms are both oxygens, the magnitudes of both 'subsystem' vectors are equal and their sum is the zero vector.

‡
For a non-electroneutral region of space, by center of charge I mean the charge-weighted centroid of the system: (1) Choose a reference point $\vec r_0$. (2) At each point $\vec r$ in the region of interest, multiply the charge at that point, $\rho\!\left(\vec r\right)$, by the distance from the reference point, $\vec r - \vec r_0$. (3) Integrate the resulting product, $\rho\!\left(\vec r\right)\cdot\left(\vec r - \vec r_0\right)$, over all space. (4) Divide by the net charge of the region of interest. Dipole moment calculations are more complicated for electroneutral systems, but the principle is similar.


If I assume you refer to 1,2-cylohexanediols, the two isomers to consider were (cis), and (trans)

than there are three conformers to consider.  For the (cis)-configuration, there is only the (axial, equatorial) orientation of the two O-substitutents:

This contrasts to the case of the (trans)-configuration, where the two substitutents may be either in axial, or (likely preferred by thermodynamics) equatorial orientation:

From the later picture, taking into account i) the relative orientation of the two hydroxyl groups towards each other as well in respect to the cyclohexane moiety and ii) the (assumed) conformational preference for this form over the conformer with two axial oriented hydroxyl groups, I would assume the two individual vectorial contributions along $\ce{C -> O}$ are better lined up than in the instance of the (cis)-configuration to yield a larger (global) dipolar moment.


Mechanism of 5‐(iodomethyl)oxolan‐2‐one synthesis from 4‐pentenoic acid and iodine in acetonitrile

https://en.wikipedia.org/wiki/Iodolactonization
I+ reacts with the CC-π electron donor. A cyclic iodonium three-membered ring intermediate is formed. The carboxylate group opens the electrophilic intermediate to form the product lactone. 5-membered rings (gamma-lactones) https://en.wikipedia.org/wiki/Lactone are prefered due to optimum entropic and enthalpic parameters of activation.


Markovnikov's rule is used to predict how (for example) HX, where X is a halide, adds to an unsymmetrical olefin.  As the diagram below shows, Markovnikov addition follows the rule that the "H" will add to the least substituted carbon while the "X" will add to the most substituted carbon.  This is due to the fact that these reactions proceed by protonaton of the olefin to yield a carbocation.  Protonation at the least substituted carbon yields the most stable carbocation (in this case a secondary carbocation).  Finally the carbocation reacts with $\ce{X^-}$ to yield the final product.

In some cases the Markovnikov rule is not followed.  Such additions to olefins are termed "anti-Markovnikov".  These reactions typically proceed through a radical intermediate rather than a carbocation.  For example, addition to the olefin might be initiated by the addition of a bromine radical to generate the most stable radical intermediate, which would place the radical center at the most highly substituted carbon. Addition of a hydrogen radical completes the process and generates the final product which has the reverse addition pattern as compared to the Markovnikov addition product.


Hydrogen chloride in its standard state is a gas. We usually do not make solutions of $\ce{HCl}$ by the method of dissolving a known mass of the the solute in enough solvent to make the appropriate volume of the specific concentration since gaseous substances are difficult to weigh (and weight is one of the most direct ways we have to determine mass in the lab).

Hydrochloric acid (aqueous solutions of $\ce{HCl}$) is commercially available in several concentrations, usually $1\ \text{N}$, $0.1\ \text{N}$. and "concentrated".

"Concentrated" simply means that you have in that bottle the highest concentration of $\ce{HCl}$ water that is stable. For $\ce{HCl}$, a concentrated solution is 36-38% by mass. If you know the density of the solution (should be on the bottle), you can convert % by mass into moles of $\ce{HCl}$ per liter of solution (you also need the molar mass of $\ce{HCl}$).

For example, if you had a solution that was 20% $\ce{HCl}$ by mass with a density of $1.098 \ \text{g/mL}$, you can determine molarity by converting mass of $\ce{HCl}$ into moles and mass of solution into liters:

$$20\% \ \ce{HCl} \ \text{by mass}=\frac{20 \text{ g }\ce{HCl}}{100\ \text{g solution}}$$
$$20 \text{ g }\ce{HCl} \times \frac{1 \text{ mol }\ce{HCl}}{36.46 \ \text{g }\ce{HCl}}=0.5485\ \text{mol }\ce{HCl}$$
$$100\ \text{g solution}\times \frac{1\ \text{mL solution}}{1.098\ \text{g solution}}\times \frac{1\text{ L}}{1000\ \text{mL}}=0.09107\ \text{L solution}$$
$$\frac{0.5485\ \text{mol }\ce{HCl}}{0.09107\ \text{L solution}}=6.023 \ \text{M}$$
However, other folks have the done the math for you. The Wikipedia article of hydrochloric acid lists the concentrations (and densities) for several $\ce{HCl}$ solutions as a function of % by mass. This listing of concentrated reagents lists $37.2\% \ \ce{HCl}$.

With a concentration for $37.2\% \ \ce{HCl}$, you can proceed to work through the dilution process using $c_1 v_1 = c_2 v_2$.


Is it possible to get energy from frozen methane without oxygen?

The planet will have gaseous oxygen if it does have an atmosphere, and if that is the case then with enough energy(like a spark) atomized hydrocarbons could be lit for energy. What is atomizing you say? Well its just breaking a liquid apart into little particles, making it easier for the oxygen to react with said methane.

As for recycling the oxygen from the combustion reaction, it will not be possible to do in an energy efficient way, CO2 is too stable and the bonds are too strong.


Mass number of isotopes of copernicium denoted with 'm'

The 'm' does indeed stand for metastable. The word 'isomer' in this sense refers to nuclear isomers, in which at least one of the nucleons is in an excited state. Think of it as somewhat analogous to an electron in an excited state, but occurring within the nucleus. Just like electrons release photons when an excited state decays to the ground state, when nuclear isomers decay to the ground state (which is different than 'normal' nuclear decay like $\alpha$ or $\beta$ decay), a gamma ray is release.


How long does it take to run an MS/MS experiment

The MSMS part is the detection of the LC in this case (nb. you don't need to do LC-MSMS, you can do direct infusion quite happily if you don't need to separate anything), so the actual experiment wouldn't need to take much longer than any other LC (as CHM says).

However, calibration + set-up of these things, and extensive cleaning afterwards, can make it a day's job to run just a few samples (even direct infusion). If you don't have much information on the system already, you may also find that you need to do a 'sighting' run, to work out which ions you need to examine more closely (MSMS). And then there is data processing.....

All in all: quick, if you know exactly what you want, and someone is running everything for you. If you want to argue that it is an easy job, you can, but be aware that this is not always gong to be the case.


How is it possible to safely use carbon monoxide as a fuel?

During WWII many cars were powered by a gasogene engine, which transformed wood or charcoal into a synthetic gas composed by $\ce{CO}$, $\ce{N2}$, $\ce{H2}$ and $\ce{CO2}$. Thus the use of $\ce{CO}$ (albeit not as a pure gas) as a fuel is not something new (the former use of piped coal gas for house heating and cooking has already been mentioned above). I am not sure however about how safe those gasogene engines were.


How does solution which electrodes are immersed in effect electrode potential?

The short answer is that we want a cathodic reaction so we can use the metals with a positive standard reduction potential (because 0 is for $\ce{H2} $ reduction).

In reality however to really find a suitable electrolyte cyclic voltammetry is typically used and electrolytes with suitable electrochemical windows for the cell reaction are chosen.

The electrochemical window is the voltages where the current is zero in the picture above. If the wrong electrolyte is chosen then a full redox reaction happens(with the electrolyte participating) rather than half reactions at each electrode negating the need for the external circuit and creating heat


How much CO2 gets free when combusting 6 kg n-heptane (C7H16)

For these types of problem, concept of mole is very handy. Here, you know the reaction stoichiometry, that means you have at least three information from that reaction equation. How?

** For every mole heptane, you need 11 mole oxygen for complete combustion
** For every mole of heptane combustion, you will get 7 mol carbon di oxide and 8 mol water.

So, now if you know the mole corresponding to 6 Kg heptane you can easily find out the amount of carbon di oxide produced in mole unit.
The relation is
$$n_{CO_2}=7 \times n_{Heptane}$$
$$n_{Heptane}=\frac{m_{Heptane}}{molecular~weight~of~heptane}$$
Now if you want to also calculate amount of carbon di oxide produced in mass unit or volume unit, you need to use the second equation or ideal gas law equation.


To sum it up, the adhesive property of an epoxy is the result of inter-molecular forces and the formation of mechanical bonding.

Adhesives cure when the small resin molecules (mers) join together to form extremely large molecules known as polymers.
Some adhesives are used only with specific substrates because of the cure time needed to allow for good bonding (both with the substrate and for polymer formation.)
I had trouble finding information saying that all epoxies use ionic bonding, and it  seems to me that most epoxies will actually just utilize intermolecular forces such as

Ion-induced dipole forces
Ion-dipole forces
Van der Waals forces
Hydrogen bonding

An important first step in applying an epoxy is the wetting of the surface and then the polymerization of the epoxy. The ability to wet is related to viscosity and the inter-molecular forces. If there is a low viscosity and strong attraction between the substrate and the adhesive then we can witness the formation of mechanical bonding. Below is an explanation:

  In order for an adhesive to bond  two surfaces, there must be several types of interaction between the adhesive and both substrates. The first type of interaction is that the adhesive must wet the substrate, meaning that the adhesive must spread itself out into a film that covers the substrate surface. In order for this to happen, the adhesive must have a low enough viscosity so that it will flow. Because viscosity is temperature dependent, the application of a cold adhesive to a substrate, or the application of an adhesive to a cold substrate, may result in poor wetting. Another factor that affects wetting is the relative strengths of cohesive forces (between like molecules, such as two adhesive molecules) and those of adhesive forces (between unlike molecules, such as an adhesive molecule and a substrate molecule). If the cohesive forces among adhesive molecules are weaker than the adhesive forces between the adhesive molecules and the substrate surface, then the adhesive molecules will spread out over the substrate and wet its surface. An adhesive that has a relatively low viscosity and is able to wet the substrate surface will flow into any tiny cracks or pores on the substrate surface, thus promoting what is known as mechanical bonding. Mechanical bonding increases the strength of an adhesive bond and, as a result, a forced separation of the two substrate surfaces is more apt to tear the substrate surfaces.

  Mechanical bonding is one of several ways that an adhesive bonds substrates. All surfaces, except those that are highly polished, have pores. If the adhesive flows into these pores and then polymerizes, a mechanical bond is formed. It is similar to placing a wick into liquid candle wax. Once the wax solidifies the wick can not be easily removed. A mechanical bond has formed.

Quotation and information from this answer were sourced here.


Can electrostatic interaction affect nucleophilic attack by thiolate anion

It would since the S atom is now negatively charged. However, S is a very poor hydrogen-bond acceptor as evidenced by its weak basicity (thiols are more acidic than alcohols). So this electrostatic effect will be greatly diminished if you compare with more basic nucleophiles like F- and O-. This is because sulfur larger and will be more polorizable making the localization of the negative charge more disperse around the atom and much weaker to form ion-ion interactions. It will still remain as a relatively powerful nucleophile, so you can expect side reactions.


Why don't dicarboxylic acids show hydrogen bonding in spite of the fact that carboxylic acids exist usually as dimers in polar aprotic solvent?

I would like to object the statement dicarboxylic acids would not engage hydrogen bonding.  Terephthalic acids ($\ce{HO2C-C6H4-CO2H}$) are a prominent example since early attempts to formalize the description of intra- and intermolecular hydrogen bonding in the solid state. Consider the pairing of benzoïc acid and $p$-terephthalic acid, sharing the same motif of hydrogen donors and acceptors, described as $R_2^2(8)$ in Etter's notation:

(Etter, Acc. Chem. Res. 1990, 23, 120-126,  doi 10.1021/ar00172a005)
But there is no limit to $p$-substitution to find this motif again, e.g.:

(Holy et al., Collect. Czech. Chem. Commun. 2006, 71, 139-154, doi 10.1135/cccc20060139, researchgate.net)


Are there substances people can put on themselves to prevent catching fire?

It depends how long one is exposed to heat, and how hot the source is, but yes, there are "fire retardants" for skin:

A mix of alcohol (ethanol or isopropanol) and water or butane in soap sud can be set afire on your palm without burning you, if done very carefully. The water soaks up heat energy both because of its high specific heat and through evaporation.
Insulation applied to the skin or between skin and flame can protect (momentarily) from a torch.
A wet hand is temporarily protected from molten lead by the leidenfrost effect. As water on the skin boils, it acts, briefly, barrier to the heat.

NB: Do not try these experiments yourself (or on others). You get no points for being a crispy critter.


Solubility of silver acetate in the presence of nitric acid

The process you are trying to investigate is:

$$\ce{AgOOCCH3(s) + HNO3(aq) <=> AgNO3 (aq) + CH3COOH (Aq)}$$

with the net ionic equation:

$$\ce{AgOOCCH3(s) + H+(aq) <=> Ag+(aq) + CH3COOH(aq)}$$

The equilibrium constant is

$$K_c = \dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]}$$

Note that this is law of mass action can be derived from the $K_a$ and the K_{sp}$ you have:

$$\begin{aligned}
K_{a}&=\dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}\\
K_{sp}&=[\ce{Ag+}][\ce{CH3COO-}]\\
K_c &= \dfrac{K_{sp}}{K_a}=\dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]}
\end{aligned}$$

The problem is now an ICE (Initial-change-equilibrium) problem.

$$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{H+}] & [\ce{Ag+}] & [\ce{CH3COOH}]\\ \hline
I & 0.1\ \mathrm{M} & 0\ \mathrm{M} & 0\ \mathrm{M}\\
C & -x & +x & +x \\
E & 0.1 -x \ \mathrm{M} & x \ \mathrm{M} & x \ \mathrm{M}\\ \hline
\end{array}$$

Plug values into the law of mass action, and solve for $x$, the concentration of $\ce{Ag+}$, and then you can determine the solubility.


Is there a function to approximate atomic mass from the atomic number?

The short answer is that you can find a power-law fit ($1.61Z^{1.1}$) with low average error.

I'd never really thought about it much, but after downloading the IUPAC Atomic Weights, I decided to do some curve fitting.

Here's a linear fit between atomic number and atomic mass:

As you say, the fit isn't very good for small $Z$, but the overall fit isn't bad - the mean absolute error (MAE) is $2.821 \:\mathrm{u}$, and taken as a whole, the data is surprisingly linear. (Well, surprising to me.)

So I thought of a quadratic fit, requiring the intercept to be 0,0 to ensure the best fit for small $Z$:

Looks better, right? Certainly the fit is much better for first and second row elements, but the MAE only reduces to $2.749 \:\mathrm{u}$.

So I went up to a cubic fit, again requiring 0,0 for the intercepts:

Aha, now we're talking! We get the subtle nonlinearities, and the MAE is down to $1.36 \:\mathrm{u}$.

Thanks to the comment by Nicolau below, I performed a power-law fit.

So that gives a power-law fit with MAE of $0.01 \:\mathrm{u}$ and a fairly easy-to remember function:

$m_a \approx 1.61Z^{1.1}$


Calculating the density of a saturated salt solution

Often, solubility is given as mass per volume of solution (not solute). Nevertheless, unless you have the mass percentage (w) too, you are not able to calculate the density of the solution.


Estimating protein binding and disassociation

If we call the drug L, and the binding sites S, the equilibrium is the following:

$$\ce{LS <=> L + S}$$

The equilibrium constant expression is:

$$K_d = \frac{[\ce{L}] [\ce{S}]}{[\ce{LS}]}$$

With $[\ce{L}]_\mathrm{tot}$ as the total concentration of ligand and $[\ce{S}]_\mathrm{tot}$ the total concentration of binding sites, we can solve of $[\ce{L}]$, the free ligand concentration to arrive at a quadratic expression like the one in the question.

If the ligand concentration is small compared to the concentration of available binding sites, you can estimate the ratio of free to bound ligand:

$$ \frac{[\ce{L}]}{[\ce{LS}]} \approx \frac{K_d}{[\ce{S}]_\mathrm{tot}}\tag{1}$$

Now, we can look at the statements 1 through 5 and see what information we can extract to determine $K_d$ and $[\ce{S}]_\mathrm{tot}$:

1.The drug has a molecular weight of 150

This information is not directly relevant. I'll get back to it later.

2. Drug binds only to serum albumin, whose concentration is 4.4% and whose molecular weight is 67,000

This allows us to calculate the concentration of protein capable of binding to the drug. 4.4% refers to 0.044 kg of albumin in a solution with a volume of one liter. With a molar mass of M = 67000 g/mol, we get the following concentation:

$$c(\mathrm{albumin}) = \frac{\pu{44 g / L}}{\pu{67000 g/mol}} = \pu{0.66 mM} $$

If we knew the number of binding sites for each molecule of albumin, this would also give us the concentration of sites.

3. At sufficiently low drug concentrations, the drug is 89% bound

If 89% are bound, 11% are free, and the ratio of free to bound is 0.12. "At sufficiently low drug concentration" means the drug concentration is low with respect to the concentration of binding sites, so we can use equation (1):

$$\frac{K_d}{[\ce{S}]_\mathrm{tot}} \approx \frac{[\ce{L}]}{[\ce{LS}]} = 0.12$$

So we can estimate the ratio of dissociation constant and total concentration of sites.

4. The distribution volume is 50 ml/kg

This means for every kg of body weight, the apparent volume in which the drug dissolves is 50 mL. Picture a solvent of unknown volume. You dissolve a solute and then remove some solution to test the solute solution. You should be able to calculate the unknown volume from the amount of solute added and the concentration obtained. If, however, the solute absorbs on the container wall or diffuses through it, your answer is inaccurate.

The human body has an average of 70 mL of blood for each kilogram of body mass. A substance that remains in the blood stream should have a distribution volume of 70 ml/kg if it does not bind to anything.

I'm not sure what to do with this piece of information.

5. Free drug has a half-life of 30 min.

This ensures that the binding reaction has a chance to almost attain equilibrium. If this were not the case, we could not use equilibrium calculations, and would have to resort to non-equilibrium methods such as the ones described in the paper cited by the OP (which is behind a paywall, so I just read the abstract).

  Using these details is there a way to estimate the parameters of P and Kd and if so please explain how they can be derived.

The missing piece of information is the concentration of sites. If we knew the number of sites on each albumin molecule, we could calculate this concentration because we know the concentration of albumin.

Experimentally, what we are missing is data for saturating conditions. As it is, we have one equation for two unknowns, so we only know their ratio. If you assume one binding site per albumin molecule, you would be done. However, looking at the molar masses, it is entirely possible for more than one drug molecule to bind to a single albumin molecule.


How do orbitals fill, what is the difference between them, and why is this stable?

What is stable?

In this context, stability is that it neither gains electron to form negative anions nor loses electron to form positive cations. I will only discuss noble gases because they are so exceptionally stable.

It does not lose electrons

This is not unique to noble gases. Elements in group VII such as fluorine and chlorine also do not tend to lose electrons.

This is also not entirely true. Compounds like $\ce{XeF4}$ and $\ce{Cl2O7}$ do exist in which $\ce{Xe}$ and $\ce{Cl}$ bear partial positive charges respectively, although $\ce{Xe}$ is a noble gas.

$\ce{Xe}$, being a noble gas, indeed has a high ionization energy (energy taken to remove an electron) but it is not infinite. This means, as long as we have enough energy, we can make them form compounds.

The ionization energy increases from Group III to Group 0 because there are more protons in the nucleus which exerts a higher pull to the orbitals, which lowers their energy. For example, the ionization energy of helium is twice that of hydrogen, and helium has exactly twice the number of protons than hydrogen.

It does not gain electron

This is because the next orbital has a positive energy. Orbitals with negative energy means that it takes work to free the electrons from the orbitals. Orbitals with positive energy means that no electrons can reside in the orbitals: they are already free.

For example, the electronic arrangement of argon is $1s^2 2s^2 2p^6 3s^2 3p^6$. The next orbital, if it existed, would be $4s$, which has a positive energy. Therefore, its electron affinity (energy released by gaining electrons) is 0. This is evidenced by the ionization energy of the next element, potassium, which is small, suggesting that the extra proton pulls the $4s$ orbital just enough so that it has a slightly negative energy.

Conclusion

Noble gases are stable because they have a full $p$ orbital, so the next electron would need to go to the next orbital which has a positive energy, and the extra protons pulling on the orbital lower the energy of the electrons in the orbital, making it not tend to lose electrons.


Flame Ionization Detector versus Mass Spectrometer

FID requires separation as it is not specific and cannot distinguish 2 non separated compounds.

MS is specific and can determine the particular compound even qualitatively.

For complex matrices, it is always good to have saparation even for MS, as MS spectrum of mixture would be confusing.


Polymerization Isomerism In Transition-Metal Complexes Not True Isomerism?

I’m pretty sure that polymerisation isomerism should mainly be considered a historical term nowadays. Going back in the history of science, all people knew at some point was which elements a specific compound is made out of. That worked well for main group metal salts but not too well for transition metal salts. Later, quantitative analysis established the law of multiple proportions, which basically says that if two compounds are made up of the same set of elements, the ratio of element mass ratios will always be an integer ratio. Sounds hard to understand, here’s an example:

$\ce{CuCl}$ has one chlorine per copper, $\ce{CuCl2}$ has two. So for every $x~\mathrm{g}$ of copper, one of the two will have $1~\mathrm{g}$ of chlorine, the other will have $2~\mathrm{g}$, and $1$ and $2$ form a simple integer ratio.

And all of a sudden there were compounds that were clearly distinct but gave the same ratio of elements, a simple example being $\ce{NO2}$ and $\ce{N2O4}$. To understand this, the term polymerisation isomer must have been introduced.

Nowadays, isomers are defined as having the same elemental composition on a per-fragment basis. So examples such as $\ce{N2O4}$ (which is a different molecule than $\ce{NO2}$) or the platinum complexes you give would no longer be considered isomers — neither constitutional ones nor stereoisomers. Hence why so many people said ‘your second book is right.’


Computational Chemistry (Ab Initio), what should I study?

Computational chemistry is not a new field, but yes it can be difficult to find the right approach to it, because it is so diversified.  And that means to some degree what you need depends strongly on what you want to study.  Most of the time, you're basically solving the Schrodinger equation all kinds of different ways.

It will be better if you can specify what level of knowledge you do have at this time. Assume that you've just got into college, then there's a list of things you should study:

Basic theory

Math.

You need linear algebra and mathematical analysis.  Any good old textbook will do.

What you use the most everyday would be linear algebra, so you better be very very good at it, along with a good expertise in vectors, matrices and in some cases tensors.  There may also be some functional analysis involved but for most people it really just boil down to linear algebra.

Mathematical analysis is basically just differentiation, integration, and all kinds of tricks related to it.

Quantum mechanics.

You some of it from P-Chem II, but if you take it from the physics department it will be more helpful.  I recommend the Cohen-Tannoudji textbook.  Make sure you start from the beginning, go through every chapter to the very end.

Quantum chemistry.

With a background you can now proceed to understand the conventional methods of quantum chemistry.  For this I would recommend Modern Quantum Chemistry by Szabo and Ostlund.

Courses to take:

In addition to what was already mentioned above, many universities will also have quantum chemistry and computational chemistry courses. I also recommend you to take group theory and calculus of variation.

Programming skills

And that prepares you for the theoretical part.  Whether you need comprehensive programming skill depends on what you want to do.  Some may make do with existing programs, some may need to write there own.  But even if you use an existing program, it requires no less amount of understanding of the quantum chemical theories you're using.

For programming skills, you will certainly benefit by taking numerical methods course, and mastery over one programming language. Take statistics and machine learning if you have time. Any one will do, it doesn't matter. For what we do, its straight forward to switch to a different language whenever you want.  Computational chemists are not usually very big on the algorithms themselves. Most of the times, you just take something right out of a numerical analysis text book or program package and adapt a little bit.  It is much more about the underlying physics, i.e., what form of wave function you choose, what approximation you make, etc.

The most important thing about programming, I feel, is hands on experience.  If you can understand quantum mechanics, programming is not gonna be hard.

Chemistry

As for other branches of chemistry, you don't have to be the expert, but you definitely need a good understanding because those are the problems you're trying to solve.  You need to understand very well the experimental procedure you're trying to study.


Why does the C–C bond have extremely weak absorptions?

For vibrational spectra the primary transition under investigation is the $v = 1 \leftarrow 0$ excitation (because $\hbar\omega >> k_\mathrm{B}T$, so excited states have negligible thermal population). The intensity of the transition depends on the transition dipole moment:

$$R_{10} = \langle 1 | \hat{\mu} | 0 \rangle$$

in that the intensity $I \propto |R_{10}|^2$. Taking a Taylor expansion of the operator $\hat{\mu}$ with respect to $q$, the vibrational coordinate, at $q = 0$, we have:

$$\hat{\mu} = \mu_{q = 0} + \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} + \cdots$$

and plugging this in we get

$$\begin{align}
R_{10} &= \langle 1 | \mu_{q = 0} | 0\rangle + \left< 1 \middle| \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} \middle| 0 \right> + \cdots \\
&= \mu_{q=0}\langle 1 | 0\rangle + \left( \frac{\mathrm d \mu}{\mathrm d q}\right)_{q=0} \langle 1 | \hat{q} | 0 \rangle + \cdots
\end{align}$$

where we took out the values of $\mu$ and $\mathrm d\mu/\mathrm d q$ evaluated at $q = 0$ because those are just ordinary numbers. The first term in the Taylor expansion vanishes because $|1\rangle$ and $|0\rangle$ are orthogonal (a key property of the harmonic oscillator, which is a good model for the low-lying vibrational states at the bottom of the well), and if we neglect the second- and higher-order terms, we have†

$$R_{10} \propto \left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0}$$

The RHS represents how much the dipole moment $\mu$ changes upon vibration of the bond (which is measured by $q$). As such:

for something like a C–C bond, where vibration leads to nearly no change in the dipole moment, this quantity is small, the transition dipole moment is small, and the peak is weak.
On the other hand, the stretching of the C=O unit leads to a large change in dipole moment and the corresponding peak is therefore stronger.
The C=C stretch of alkenes tends to be weak as well, unless it is conjugated to an electron-withdrawing group like a carbonyl group. This can also clearly be linked to the magnitude of $(\mathrm d \mu/\mathrm dq)_{q=0}$.
And incidentally, this is also why molecules like $\ce{H2}$ or $\ce{O2}$ cannot be studied by IR spectroscopy (no matter how much you vibrate, there is no dipole moment), and also why the symmetric stretch of $\ce{CO2}$ is not IR active. In all of these cases $(\mathrm d \mu/\mathrm dq)_{q=0}$ is identically zero.

Unfortunately most of what you've written is kind of off the mark.

It is indeed "reluctant" to absorb the photon, but that is not because of the stiffness of the bond. If the bond is stiffer ($k$ is larger), then it simply needs a photon of shorter wavelength (as the vibrational frequency $\omega = \sqrt{k/\mu}$ will be larger). Instead it is related to the poor interaction of the C–C vibrational mode with the oscillating electric field of the photon. This is going into stuff that I don't know about, so can't explain much more (search "Fermi golden rule" in Atkins MQM or some other quantum book).
$D_\mathrm{e} = kr^2/2$ is not a valid equality here as it simply makes no physical sense. If the energy of the oscillator is at $D_\mathrm{e}$, then physically speaking, the corresponding bond length $r$ should be infinity, as the bond is dissociated. Furthermore, the equilibrium bond length which you used in the calculation ($r_\mathrm{e} = \pu{154 pm}$) refers to the middle of the potential well. If you really wanted to model the C–C bond as a classical oscillator, then your equation should read
$$U = \frac{1}{2}k(r - r_\mathrm{e})^2$$
because the thing that gets squared is the displacement from equilibrium.
In that calculation you used the dissociation energy for one mole of carbon–carbon bonds, and that doesn't make sense if you're trying to calculate the force constant of one carbon–carbon bond.
Even changing no. 3 wouldn't make any difference, because the harmonic oscillator model simply does not work near dissociation (in fact if a bond were to be a harmonic oscillator, it would never dissociate). Therefore bond dissociation energies cannot be treated within a harmonic model; you have to use something else such as a Morse potential.
A typical value of a C–C force constant is on the order of $\pu{400 N m-1}$. A better way of calculating the force constant is with
$$\bar{\nu} = \frac{\omega}{2\pi c} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} \implies k = \mu(2\pi c\bar{\nu})^2,$$
where $\bar{\nu}$ is the IR wavenumber and $\mu$ is the reduced mass of the oscillator. A very rough calculation is as follows. Plugging in $\mu = \pu{10^-26 kg}$ (for a $\ce{C-C}$ bond), $\bar{\nu} = \pu{1000 cm^-1} = \pu{10^5 m^-1}$, and $c = \pu{3 x 10^8 m s^-1}$, we have $k = \pu{355.3 N m^-1}$. Correct order of magnitude, at least.
It's also nearly impossible that $k_\ce{C=O} < k_\ce{C-C}$, as force constants tend to be correlated with bond order. A quick Google search shows that $k_\ce{C=O}$ is on the order of $\pu{1000-2000 N m^-1}$. (If you go searching for this, you may wish to know that $\pu{1 mdyne/Å} = \pu{100 N m^-1}$, as CGS units are widely used.)

† Of course, there's an additional point that must be made, which is that $\langle 1 | \hat{q} | 0 \rangle \neq 0$, but you'll have to take my word for it as I'm not keen on proving it here. It is most readily shown using ladder operators, and it should be covered in any QM text under the harmonic oscillator.


How do you Benzylate (add a benzene group to) cyclic compounds?

Brenton, the term "benzene group" is a bit misleading; it seems that you'd like to add a benzyl group, i.e. $\ce{C6H5-CH2}$.

Typically, these reaction are carried out using freshly distilled benzyl halides. In the case of the diamines, monoalkylation is usually achieved using the bis-hydrochloride of the amine and a stochiometric amount of base, thus keeping one nitrogen atom protected.

Note however, that the Australian regulations, kindly mentioned by Richard Terrett, are likewise true in the EC: 1-benzylpiperazine, an infamous psychoactive party drug, is a controlled substance!


Yes, uranium ore is a mixture.

Uranium compounds such as pitchblende (uraninite) are only present as a small part (usually $<1$%) of the ore.

Extensive purification is needed. Uranium ore purification would be a good choice for your project.

See for example:

http://www.areva.com/EN/operations-677/uranium-processing-from-ore-to-yellow-cake.html

http://www.iaea.org/sites/default/files/publications/magazines/bulletin/bull23-2/23204882428.pdf

image source: http://chemwiki.ucdavis.edu/Development_Details/Approaches/ChemCases/Nuclear_Chemistry_and_the_Community/08._Uranium_Production


Do different oxidation states of the same element have the same emission spectrum?

Interesting question. Keep in mind that the elemental emission spectrum in a flame or plasma and even a discharge does not remember its history in solution or a solid phase.

The punchline is that the emission spectrum is dependent on the elements gas phase chemistry in the flame/plasma/discharge. Prof. Ed has explained you the example of sodium. Let us say we have the following

(a) a block of element sodium
(b) a block of sodium chloride crystal
(c) solution of sodium chloride in water

If you introduce (a), (b) and (c) in the flame, the flame will be colored yellow in each case, which means that the emission is coming from a common emitter. That emitter is a elemental sodium atom excited by high temperature in the gas phase. Thus atomic emission spectrum is a fingerprint of the element.

You may ask that you introduced Na(+) in the flame in the case of b and c. Flames can easily reduce an ion to the elemental state.

How to see the sodium ion spectrum: As we just said, the emission spectrum is independent of this original state. You can only cause ionization by increasing the temperature. This is the way to see the spectrum of an ion. If we were using a high temperature flame/electrical discharge, we would start seeing Na(+) spectrum along with elemental sodium spectrum.

Coming to your particular example: You introduced copper (I) and copper (II) into a flame and they all colored it beautiful blue-green. The reason is that if the flame temperature is low, compounds cannot fully dissociate into atoms (not enough energy to break the bonds). In such cases, very simple diatomic or triatomic molecules are formed in the flame which emit their characteristic colors. In the case of copper, CuCl is formed in the flame. CuOH may be formed as well. Whether you introduced Cu(I) or Cu(II), as a chloride, it does not remember its solution phase or solid phase history.

If you were indeed using a high temperature flame, you will never ever such a blue green coloration, because this time, the emission is from Cu atoms (in UV). Hope that clarifies your confusion.


Differentiation between zinc, aluminium, and magnesium ions in solution

Apart from the methods, Ringo already described, you can do a few other tests.

Aluminium

This is loosely translated from the German chemgapedia.de. Look at the pretty pictures.

Probably the easiest test you can do is reacting it with Morin in ethanoic acidic medium. It forms a yellow-green chelate complex, which has strong fluorescence under UV light. The colour and UV activity should be persistent when adding glacial acidic acid (water free), but vanish when adding concentrated hydrochloric acid.

Unfortunately sodium is also fluorescent, so keep your equipment clean and if you use a base, use potassium hydroxide instead. Also perform blind tests with all the chemicals involved, to see if they are clean enough.

Other elements that can show fluorescence are $\ce{In^3+, Ga^3+, Sc^3+, Zr^4+, Be^2+}$, usually very pH dependent.

From my own experience in first year of university, this is a very efficient test. Another test, that can be performed from the salt, is reacting it to Cobalt blue. If you have a solution, you have to precipitate aluminium first: $$\ce{Al^3+ + 3 {}^{-}OH ->C[NH3/NH4Cl] Al(OH)3 v}$$
Put the precipitate on a magnesia groove and add some cobalt nitrate. Glowing the mixture will result in the characteristic cobalt blue:
\begin{align}
\ce{2 Al(OH)3 &->[\Delta] Al2O3 + 3 H2O ^}\\
\ce{Co(NO3)2 &->[\Delta] CoO + 2NO2 ^ + 1/2 O2 ^}\\
\ce{CoO + Al2O3 &->[\Delta] \color{\navy}{CoAl2O4}}\\
\end{align}

This is an almost fool proof method for aluminium and is not influenced by other elements. And it looks quite nice.

Zinc

Zinc is probably one of the only elements that form a white precipitate with sulfide ions.
$$\ce{Zn^2+ + H2S -> ZnS v + 2 H+}$$
This precipitate can be used to form Cobalt green. The procedure is the same as for the above Cobalt blue.
$$\ce{ZnS + Co(NO3)2 + O2 ->[\Delta] \color{\green}{ZnCo2O4} + 4 NO2 ^ + SO2 ^ }$$

The only interference with this test is, that using too much cobalt nitrate will produce the black cobalt oxide, which can make it hard to see the green colour.
$$\ce{3Co(NO3)2 ->[\Delta] \mathbf{Co3O4} + 6 NO2 ^ + O2 ^}$$

Again you can have a look at the pretty pictures at chemgapedia.de.

Magnesium

A common problem with magnesium is, that all tests only work well and specific if you have a pure solution. Most heavier elements react more readily with the chemicals, than magnesium itself does. Another problem is, that salts from magnesium have no flame colouration at all.

I am again loosely translating from chemgapedia.de.*

You can precipitate magnesium with 8-hydroxyquinoline:

It forms a yellow-green complex. This is only a specific test, if you know you have no heavier elements in your test tube. For example, aluminium forms the complex Tris(8-hydroxyquinolinato)aluminium: $$\ce{Al^3+ + 3 C9H6NOH -> Al(C9H6NO)3 + 3 H+}$$

Another possibility is precipitation as magnesium ammonium phosphate (struvite). And yes, you guessed it, this happens with all other alkali and earth alkali elements (except lithium). But if you are certain you only have got magnesium, you can grow nice crystals. The pH for this reaction should be buffered at around 8-9 with ammonia/ ammonium chloride.
$$\ce{Mg^2+ + NH4+ + {}^{-}OH + HPO4^2- ->C[H2O][NH3/NH4Cl] Mg(NH4)PO4⋅6H2O v}$$

There are two more tests, that I would consider best suitable, because of the pretty colour.

The reaction of magnesium hydroxide with titan yellow in presence of sodium hydroxide will produce a red precipitate. It is important to do a control/ blind experiment, as older solutions of titan yellow will form a precipitate with sodium hydroxide alone.

The reaction of magnesium hydroxide with azo violet (Magneson) or Magneson II (Magnezon II) in presence of sodium hydroxide will produce a blue precipitate. Again test with sodium hydroxide to observe the violet solution. This reaction also occurs with $\ce{Ca^2+, Al^3+}$.

Testing for magnesium is as far as I remember is some kind of a lottery when in combination with other elements (qualitative analysis). Good luck.

* Please note that the first reaction equation has depicted hydroxyquinoline with two circles for the aromatic system, which is wrong. In benzene one circle stands for 6 $\pi$ electrons. So that would lead to hydroxyquinoline having 12 $\pi$ electrons, two too many.


Effect of substituents on Ullmann reaction

I think that the answer given in your textbook is wrong as they are trying to conduct ULLMANN reaction on a chloride a strong EWD needs to be present at the ortho/para positions for the reaction to be feasible.

There Are generally 2 Mechanisms Proposed for the reaction :

Free Radical Mechanism
Ionic Mechanism

  FREE RADICAL MECHANISM
  The Copper Metal (used as catalyst) Generate a Aryl Radical By transferring a electron, and then 2 Aryl Radical combine to form the final product.

  IONIC MECHANISM
  Acording to this a Organometalic intermideate of Aryl copper halide is fromed. The copper is added to the Aryl halide by Oxidative addition. then a electron is transfered to the intermidiate to for a organocuppric radical which again perform the oxidative addition on another aryl halide which result in formation of a biaryl copper complex which in turn undergo reductive elimination and make the final product.

You can easily see that we need a EWG to make the Reagent more reactive thus making the reaction feasible thus your answer must be either wrong or you might have checked it wrong. nevethless I also find this paper affirming mine and your belief.

Source:

https://link.springer.com/article/10.1007/s40828-016-0036-2
www.name-reaction.com/ullmann-reaction


What's the relationship between chemical polarity and vapor pressure？

The vapor pressure of a liquid is the equilibrium pressure of vapor above the liquid at a given temperature. Liquids with higher vapor pressures must have less strong forces holding the molecules together in the liquid phase, thus more of the molecules are able to escape the liquid and be in the vapor phase. This is the connection between inter-molecular forces and vapor pressure.

Polar molecules can attract each other through dipole-dipole and dipole-induced dipole interactions in addition to the induced dipole-induced dipole intermolecular interactions that all molecules experience. Thus, all else (molar mass, for example) equal, a polar molecule will generally have a lower vapor pressure than a non-polar molecule.


What can we learn from the energies of molecular orbitals?

Not a full answer, but some general thoughts:

In general, what we can learn from individual orbitals (and their energies) is very limited, since orbitals are just an abstract tool to represent the many particle electronic wave function. This is manifested by the need for methods beyond Hartree-Fock (DFT, post-HF, etc) and the fact that the orbitals in these methods become even more abstract: Kohn-Sham DFT considers the orbitals to be non-interacting and moves the electron-electron into an external potential. In post-HF the ansatz for wave functions is extended in different (yet similar) ways, but ultimately still based on orbitals. Thus, an exact theory does not consider individual orbitals, but always combines them. In the context of such a theory, it is thus very hard to relate a molecular property to orbital energies only.

However, there still seems to be some rough correspondence, as MO diagrams are widely used to explain things like stability and reactivities. But I think this is all just empirical. Also, the fact that orbitals are an approximation means it does work on some scale. For example, we can distinguish between core and valence electrons based on the orbital energies, as those are typically well separated. (the cutoff energy may however depend on the system). But arguing which valence electron is more strongly bound is more difficult, as their orbital energies are more similar. But this is not really helpful, since we can separate core and valence electrons without applying quantum mechanics.

But we can extend that idea a little bit: Looking at differences in orbital energies, we can estimate the distribution of the electrons and predict a high-spin or low-spin situation of unpaired electrons (e.g. in transition metal compounds). Or the stability of radicals. Or the stability of bonds, e.g. in $\ce{C=C}$ bonds, the $\sigma$ bond is more stable than the $\pi$ bonds. We can even explain the more stable bonding $\ce{N2}$ compared to $\ce{O2}$, although we are comparing two systems with different nuclear charges and number of electrons here. But these arguments mostly work without looking at the actual numbers for orbital energies. It is mostly just qualitative.

Ultimately, we still know something about the orbital energies that we can use. The values are just very fuzzy (high uncertainty, large error bars), therefore we have to be careful with any conclusion. This may actually be the reason for how a large part of chemistry actually works: Often there are concepts which kind of work, but are not applicable in every situation, e.g. the HSAB concept. There may be other competing theories which cover other situations, but a full exact theory (other than elaborate quantum mechanics) rarely exists. In chemistry everything is a bit fuzzy and often it comes down to just know or test the properties of a certain compound. Physics on the other hand can be much more systematic.


How to calculate the percentage yield for a reaction with excess reagent?

The yield is typically calculated according to equation $(1)$.

$$\text{yield} = \frac{\text{actual amount of substance obtained}}{\text{theoretical amount of substance to be obtained}}\tag{1}$$

(Whether it is given in percent or as a fractional number is not important, but percentages are the usual value.) So for any reaction, you need to figure out:

the reaction equation
the amount of each reactant substance involved
the amount of desired product theoretically possible by stoichiometric coefficients
the actual amount of desired product.

The first is rather easy, especially since the equation $(2)$ has been given previously.

$$\ce{2 P + 3 Cl2 -> 2 PCl3}\tag{2}$$

Thus, for every mole of phosphorus, we expect one mole of product and for each mole of chlorine gas, we expect two-thirds of a mole of product. The amounts of the reactants are easily calculated using molar masses:

$$\begin{align}M &= \frac mn\tag{3}\\
n (\ce{P}) &= \frac{12~\mathrm{g}}{30.97~\mathrm{g/mol}}\\
&= 0.39~\mathrm{mol}\tag{4}\\[1em]
n(\ce{Cl2}) &= \frac{35.5~\mathrm{g}}{70.9~\mathrm{g/mol}}\\
&= 0.50~\mathrm{mol}\tag{5}\end{align}$$

For each of these two we would need to check the amount of product we would expect using the stoichiometric coefficients.

$$\begin{align}n_\text{theor}(\ce{PCl3}) &= \frac 23 n(\ce{Cl2})\\
&= 0.33~\mathrm{mol}\tag{6}\\[1em]
n_\text{theor}(\ce{PCl3}) &= n(\ce{P})\\
&= 0.39~\mathrm{mol}\tag{7}\end{align}$$

Since $0.33 < 0.39$, chlorine is the limiting reagent and our theoretically possible yield is calculated from the amount of chlorine added to the reaction. Therefore, we now know our denominator which is $0.33~\mathrm{mol}$. What is the numerator? Again, it is easily calculated:

$$\begin{align}n_\text{exp}(\ce{PCl3}) &= \frac{42.4~\mathrm{g}}{137.32~\mathrm{g/mol}}\\
&= 0.31~\mathrm{mol}\end{align}\tag{8}$$

And now all we need to do is calculate the yield:

$$\frac{0.31~\mathrm{mol}}{0.33~\mathrm{mol}} = 0.93 = 93~\%\tag{9}$$

This is consistent with your result.


The $2n^2$ rule is nothing but a mathematical consequence of the mathematical description of electrons around an atom – what chemists call quantum theory and is currently the most accepted description of atomic matter by a mile. Thus, until quantum theory is significantly amended, the fifth shell does, in theory, have the capacity to hold $2\times 5^2=50$ electrons.

I say in theory, because shells aren’t filled all at once bottom to top. In fact, the $n^\text{th}$ shell has $n$ subshells denoted by the letters s, p, d, f, g, h, … which in neutral atoms are (usually; there are exceptions) filled in a specific order known as the aufbau principle. For the first two shells this doesn’t have any effect, but the 4s shell is filled before the 3d shell and the 6s shell before the 4f shell, to name just two examples.

To completely fill the fifth shell (i.e. fill the 5g shell), we would need to also fill the 6s, 6p, 6d, 7s, 7p and 8s shells. The heaviest known (i.e. synthesised according to the standards laid out by IUPAC and IUPAP) element as of 2019 is oganesson (Og) which also happens to be the last 7p element. Two heavier elements (the 8s elements) must be discovered before the third in line will most likely be the first 5g element. The g subshell can hold 18 electrons, so until we have one with a completely filled 5g shell some time will pass.


Would a salt bridge work through which the ions could not diffuse into the half cells?

It is a very good thought experiment as to why we don't cap the salt bridge ends with gold plugs. By capping, you are essentially providing an interface which can source and sink for electrons in the cells, thus each end will act as an electrode in the cell. One end of the salt bridge will act as a cathode and one end will act as an anode (strange as it may sound!)

As to your comment above "I want them to be non porous and able to let the electric field through that the accumulating charges have. I think a closed glass tube would also work". No a close tube will not work. You need the ions or electrons to move in order to observe current flow. A closed glass tube will have an infinite resistance and hence no current can flow!


Draw the cycle as if it were planar (in fact it is not, but just imagine for a moment), with one hydrogen at each carbon pointing upwards with respect to the plane, and the other downwards. Now, if we introduce two substituents to different carbons on the same side of the plane, that would be the cis isomer, otherwise trans.

As for the stability of equatorial conformers, that's purely for sterical reasons. Draw the cycle, this time in its real geometry, and see what happens if we have two bulky groups in equatorial positions on the same side. Clearly they will be unpleasantly close to each other.


Effect of substituents on aniline C-N bond length

You think right, but for the wrong reason. The inductive effect is not that great and does not quite reach that far. It is the effect of conjugated $\pi$-system that really matters. See, we may draw an alternative resonance structure for I with minus on O, plus on N, and rearranged double bonds (quinone-like ring instead of aromatic). This means the C-N bond in I is kind of "partially double" (i.e., has a noticeable $\pi$ component), hence it is shorter.

Indeed, there is some conjugation between the aromatic ring and nitrogen's lone pair in unsubstituted aniline as well. That's what makes aniline a weaker base than ammonia, and its aromatic ring more reactive towards electrophiles than benzene. It is just that the $\ce{-CH=O}$ group in para position makes that conjugation stronger yet, as we may see qualitatively by drawing the resonance structures, or (supposedly) quantitatively by means of quantum chemistry.


Enzyme Kinetics two enzymes in complex with one substrate

As this seems like a homework question (and may be closed), as an example consider the reaction  $\ce{H2 + I2 = 2HI}$. The rate, r is $\displaystyle r=\frac{1}{2}\frac{d\mathrm{[HI]}}{dt}=k_1\mathrm{[H_2][I_2]}-k_2\mathrm{[HI]}^2$ where the rate constants $k_1,\;k_2$ are for the forward and reverse reactions respectively. (This assumes that the reaction occurs as written.) As   steady state the intermediate or transient species has a rate of change of zero and you note in your question.

In the general case if the reaction is $aA+bB= dD+eE$ then $\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{d}\frac{d[D]}{dt}=-\frac{1}{e}\frac{d[E]}{dt}$


If you placed silver metal into silver nitrate, would a reaction occur?

For OP: As I have realized from James Gaidis' answer, you were very minimalistic in describing context of your question, what is very bad practice, as much more burden is put on shoulders of responders via the question interpretation and clarification. As I have just now realized, you may, or may not speak about a solution.

Yes, the reaction will occur, and no, the reaction will not occur.

There is ongoing equilibrium reaction $\ce{Ag <=> Ag+ + e-}$ with zero net result.

Initially, when metallic silver is inserted, very tiny net result of the above reaction happens, until there is achieved the equilibrium potential. It is determined by the initial potential before inserting into the solution and by the effective capacitance of the electrode.

If you charged positively ( e.g. electrostatically ) a piece of metallic silver before inserting into the solution, a very small amount of silver would dissolve. If negatively, a very small amount would deposit itself.

As @MaxW has mentioned in his answer, isotope tracing is frequently used to track the reaction mechanisms of complex processes and also kinetics of systems in equilibrium, that seem macroscopically static.


Is it possible to measure light absorbance for analysis of lead concentration without photospectrophotmeter

Another possibility is the grease spot photometer (aka Bunsen photometer) German Wiki page. This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light. Also the darker the room, the better.

The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.

If you move a light source to achieve this condition, you can conclude relative intensities from the distance (intensity goes 1 / d² with d the distance from the respective light source).

Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48

How to measure photometrically

so schematically we have:

o--U----|----o

+---d---+

with light sources o, cuvette U and the spotted paper |. Measure distance $d$ between equal intensity paper position and the light source with the cuvette.

We have $I_{transmitted} \sim \frac{1}{d^2}$

Next, a calibration series is acquired as usual: blank (which serves as $I_0$) and concentrations over the calibration range. Almost as usual:

$$E = -\lg \frac{I}{I_0} \sim \varepsilon l c$$
$$-\lg \frac{d_0^2}{d^2} = b c$$
$$\lg d = a + b' c$$

$a$ and $b$ are determined by calibration - I use here $a$ instead of $2 \lg d_0$ because several choices are possible and in general the fitted value should be used rather than the theoretical one since it includes some effects in which practice differs from theory (e.g. differences between % transmission in the grease spot from % reflection outside).

Standard addition instead of calibration would be possible as well.


Ionic liquids are not magical liquids. They are just typical organic compounds with certain properties. These compounds generated a storm in the tea cup in the late 1990s and 2000s, but they are as toxic as other solvents. Certainly they have niche applications especially as a gas chromatography stationary phase, where they can be used at high temperature (up to 350 centigrade!).
Now you can guess based on the GC application, that ionic liquids cannot be distilled or vaporized. They start to decompose by complex path ways above these temperatures. We can easily see that in a gas chromatograph. Surprisingly, not much has been studied as what happens to ILs at high temperatures.
Only one recent study exists Dicationic ionic liquid thermal decomposition pathways. There are complex decomposition products in the absence of oxygen. Like any organic compound, they will burn in the presence of air. NaCl is another story. It is not an organic compound. In short do not compare NaCl with ionic liquids.


Decrease of Bicarbonate to compensate Respiratory alkalosis

Bicarbonate ion is amphiprotic, as you've recognized. It can react as both a Brønsted base and as a Brønsted acid in water solution:

As an acid:

$K_a = 4.8 * 10^{-11}$

$\ce{HOCO_2^- + H_2O <=> OH_3^+ + CO_3^{2-}}$

As a base:

$K_b = 2.1 * 10^{-4}$

$\ce{HOCO_2^- + H_2O <=> HO^- + H_2CO_3}$ (not really true carbonic acid).

Through K value inspection, we can see that the bicarbonate ion preferentially reacts as a base in water solution. The magnitude of this difference is massive; $K_b$ is more than a million times greater than $K_a$.

So by decreasing the concentration of bicarbonate ion, we can combat respiratory alkalosis.


Why do things want to lower their energy?

Maybe my shortest answer ever:

Energy is loosely a measure of how disturbed a system it is from its natural, grounded, non-excited state, and it is from here that the "less energy, more stable" thing comes from. Your question reverse-engineers that.


Based on their respective number of degrees of freedom, what's the most efficient fuel; hydrogen or gasoline?

It sounds like most of your confusion is coming from the fact that you are assuming that the vibrational degrees of freedom are completely frozen out at moderate temp. This is a common misconception. To see that this assumption is not true, we can look at the simple example of carbon dioxide. It is a linear molecule, so it has three degrees of translational freedom and two of rotation. If vibrational modes do not contribute at all, we expect that $\overline{C}_v=\frac52 R$ and $\overline{C}_p=\frac72 R$, so $\gamma=\dfrac{C_p}{C_v}=1.4$. If we include the 4 vibrational degrees of freedom, each of which contributes 1R to $\overline{C}_v$, we would have $\dfrac{C_p}{C_v}=1.15$. The experimental value at $20$ $^\circ C$ is about 1.3, indicating that the vibrational modes are already contributing to the heat capacity at this temperature. A plot of the heat capacity as a function of temp shows a gradual increase across a very wide range of temperature.

More rigorously, the contribution of a vibrational degree of freedom to the molar heat capacity is

$$R\cdot\left(\frac{\Theta}{T}\right)^2\cdot\frac{e^{\Theta/T}}{(e^{\Theta/T}-1)^2},$$

where the "vibrational temperature" $\Theta=\dfrac{hcv}{k_B}$. For the lowest energy vibrational normal modes of $\ce{CO2}$, $v=667 cm^{-1}$, so $\Theta = 961 K$. Although this is quite a bit higher than typical temperatures of interest, the above expression tells us that these vibrations contribute $\sim 0.45 R$ to $\overline{C}_v$ at 300 K. Since there are two modes at this frequency in $\ce{CO2}$, we expect that $\overline{C}_v$ and $\overline{C}_p$ are both increased by 0.9R at 300 K relative to the values excluding vibrations. This correction gives a value of $\dfrac{C_p}{C_v}$ very close to observed experimental value.

Going back to your case of octane, there are a number of normal modes with frequencies around 720 cm$^{-1}$, certainly low enough to contribute at 300 K. We have 18 + 8 = 26 total atoms, so we expect a maximum of 3*26=78 degrees of freedom. Since octane is not linear (in the strict sense of all atoms in a single line), there are 6 non-vibrational DOF. Thus, we expect that the number of degrees of freedom calculated using your method based on the experimental specific heat should be somewhere between these numbers.

Your calculation, however, needs to be corrected to account for the fact that vibrational degrees of freedom contribute 1R rather than R/2, so the equation becomes

$$\overline{C}_v = 3 R + (f-6)R.$$

From that, I get $f=33.8$, well within our expected range. You can repeat the rigorous calculation using the exact frequency of each normal mode of octane and see how close the result is to the experimental value.

UPDATE: To address the efficiency calculation, note that including the vibrational modes changes your equation for $\gamma$. Now we have that $\overline{C}_v=R(3+f-6)$ for (nonlinear) octane and $\overline{C}_v=R(\frac52)$ for hydrogen (ignoring its one vibrational mode). If we use the apparent degrees of freedom of octane of 33.8, $\gamma_{octane}=1.0325$ and $\gamma_{H2}=1.4$. [Note also that there is really no reason to calculate those apparent degrees of freedom except for curiosity, since we can just add R to the given $\overline{C}_v$ to get $\overline{C}_p$ and divide to get $\gamma_{octane}$.] From that I get an efficiency of 6.5% for the octane and 56% for the hydrogen.


Why is vapour pressure not dependent on shape and size of the container?

Vapor pressure does not depend on surface area because it is derived from the thermodynamic equilibrium between the liquid and gas phases of a substance. The closer the vapor pressure is to atmospheric pressure, the closer that substance is to boiling. At its boiling point, vapor pressure equals atmospheric pressure and the molecules that comprise the liquid now have enough thermal energy to overcome the intermolecular forces that are maintaining it in a condensed phase. So it might help to think about how container geometry does not affect the boiling point of a liquid; the same reasons are true for why it does not affect the vapor pressure.

The evaporation rate in terms of mass/time is absolutely a function of the container geometry, but equilibrium vapor pressure is an intrinsic molecular property.


Where can we buy chlorinated paraffin for lab application?

Sigma-Aldrich is a global supplier of purified chemicals for research users.

They have a whole page of the haloalkane compounds they sell (in small quantities such as you desire).  You might be interested in compounds such as trichloropropane, pentachloroethane, etc.

A reason it might be difficult to get the higher-molecular weight compounds is that they are toxic and difficult to dispose of properly.  In the US, their use is subject to increasing regulatory requirements.


Finding equilibrium constant for the reaction between non-stoichiometric mix of gaseous components

As usual, I suggest to start with a RICE table to get an overview, denoting initial partial pressures with $p_0$ and the change in partial pressure with $x$:
$$
\begin{array}{lccccc}
&\text{R} &\ce{&CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)}\\
&\text{I} && p_0(\ce{CH3OH}) && p_0(\ce{NOCl}) && 0 && 0 \\
&\text{C} && -x && -x && x && x \\
&\text{E} && p_0(\ce{CH3OH}) - x && p_0(\ce{NOCl}) - x && x && x
\end{array}
$$
You are interpreting given conditions wrong.
The problem explicitly provides you with the initial partial pressure of methanol $p_0(\ce{CH3OH}) = \pu{50.1 mbar}.$
As for the initial partial pressure of nitrosyl chloride, it can be found applying the ideal gas law:
$$
\begin{align}
p_0(\ce{NOCl}) &= \frac{m(\ce{NOCl})}{M(\ce{NOCl})}\frac{RT}{V}\\
  &= \frac{\pu{0.059 g}}{\pu{65.46 g mol^-1}}\frac{\pu{8.314 J mol^-1 K^-1}\times\pu{323 K}}{\pu{4.33E-4 m^3}} \\
  &\approx \pu{5590 Pa} \\
  &\approx \pu{55.9 mbar} \tag{1}
\end{align}
$$
Note that methanol and nitrosyl chloride reaction mix is apparently not equimolar.
Also, the total pressure is not relevant for obtaining the solution.
Do keep in mind though it is not the initial partial pressure of the first component because the second gaseous reagent has been added afterwards.
The unknown $x$ can be found trivially since you are given the equilibrium partial pressure of nitrosyl chloride $p(\ce{NOCl}) = \pu{27.6 mbar}$:
$$
\begin{align}
p_0(\ce{NOCl}) - x = p(\ce{NOCl}) \quad\implies\quad x &= p_0(\ce{NOCl}) - p(\ce{NOCl}) \\
  &= \pu{55.9 mbar} - \pu{27.6 mbar} \\
  &= \pu{28.3 mbar} \tag{2}
\end{align}
$$
Finally we can write an expression for $K_p$ (by the way, here $K_p = K_c$ since $\Delta n = 0)$:
$$
\begin{align}
K_p &= \frac{p(\ce{CH3ONO})\cdot p(\ce{HCl})}{p(\ce{CH3OH})\cdot p(\ce{NOCl})}\\
  &= \frac{x^2}{(p_0(\ce{CH3OH}) - x)\cdot p(\ce{NOCl})}\\
  &= \frac{(\pu{28.3 mbar})^2}{(\pu{50.1 mbar} - \pu{28.3 mbar})\times\pu{27.6 mbar}} \\
  &\approx 1.33 \tag{3}
\end{align}
$$
Don't forget to do dimensional analysis to check yourself and never omit units in your calculation.
Also, keep an eye on significant figures: the number for the molar mass you used was too sloppy; instead of two figures $(\pu{65 g mol^-1})$ you had to go with four $(\pu{65.46 g mol^-1}).$


The way I justify it relates to the main plane of the complex:
referring to your compounds, the plane I am talking about it the one which includes Co, the two NH3 and ethylenediamine.
If you want to "join together" with a line the two chlorides, you have to "cross that plane", therefore your compound is trans.

In the other case (the compound on the right), the plane includes Co, ethylenediamine, chloride and NH3. If you want to do the same thing, i.e. linking the two chlorides or the two ammonia together, you don't cross that plane. In the end, your compound is cis.

Also, take a look at this website as well.
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Coordination_Chemistry/Properties_of_Coordination_Compounds/Isomers/Stereoisomers%3A_Geometric_Isomers_in_Transition_Metal_Complexes_II

Feel free to add more questions about this!


what is a difference between outer electrons and unpaired electrons

Outer electrons and unpaired electrons are not the same. Outer electrons are, as you said, $(n-1)d$ and all $n$'s. But unpaired electrons are different. To understand what they are, you must know what is electron spin (which, I hope you do). If you don't, read on.
Every electron in an atom has a quantum spin state (either clockwise or anticlockwise, nothing else) denoted by $+\frac{1}{2}$ and $-\frac{1}{2}$. These spin quantum numbers have no classical analogue, i.e., you can't relate them to any phenomenon in real life situations that you can visually perceive.
So, you see, every electron is paired with another electron with opposite spin. Together, the two of them fill one orbital, as for example, $2p_x$, $3d_{xy}$ etc. These orbitals are found out by solving the Schrodinger Equation.
Now, whenever an electron in an orbital does not have another electron with opposite spin, it is called an unpaired electron.

The electronic configuration that is given is $4s^2\: 3d^{10}\: 4p^3$. As, you can see, the $4p$ is filled with $3$ electrons whereas its maximum capacity is $6$. $4p$ has $3$ orbitals and so, each orbital is filled with one electron each, as given (according to Hund's Rule). So, these three are unpaired electrons. All other shells are fully filled and so, there are no more unpaired electrons.
So, number of unpaired electrons is 3 not 15.


Calculating the pH of a dilluted aqueous KOH solution

Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.

$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.


It is good question given the fact that this idea is not commonly discussed. The main source of your confusion is the following idea

Why doesn't water dominate any electrolysis reaction since even if a small percentage of the collisions of water can accomplish the reaction? The sheer number of collisions on its side will ensure it wins.

Unlike chemical reactions, electrode reaction do not solely require collisions. It requires an additional process: electron transfer. No matter how much collisions are occurring, the electrode is unable to transfer or withdraw electron from water in high salt concentration. Thermodynamics tells us (from the Nernst equation) how easy it would be to remove or add an electron to a water molecule at a given salt concentration. It is silent as to why this happens and what is the mechanism.
Electrode processes are very complicated. Marcus got a Nobel Prize on electron transfer.


"chromic acid" is also a nickname for chromic trioxide.  It is stable enough to be sold commercially.  Chromic acid in the strict sense ($\ce{H2CrO4}$) is not sold commercially, I think.  For most purposes what you're looking for is Cr(VI) plus acidity, so the usual practice of adding concentrated sulfuric acid to sodium or potassium dichromate (which have multiple uses and can be found in every lab) is preferable to having to buy another reagent.


Typically, you will want to make sure that your solution is saturated. Don’t add a specific, certain amount; rather, add the solid to hot distilled water until it actually won’t dissolve any more. Then quickly filtrate the hot solution through a hot funnel (don’t burn yourself though, please). Let this filtered solution cool down and you should obtain acceptable seeding crystals in your beaker.

By using this supersaturation technique, you can be sure that

you have a saturated solution so crystallisation can occur immediately and
since you cool down the solution rather quickly and since solubility of aluminium potassium sulphate increases with increasing temperature, it becomes very supersaturated very quickly.

Once you have a seeding crystal, e.g. by this method, you can use the hanging technique mentioned by Carlos with the remaining solution (which is still supersaturated). If you do so, you can use an almost infinitely thin string; a human hair typically works perfectly and is even less visible. Another option is to filtrate the solution again so that the bottom of the flask is clean and place your crystals there to grow. You will need to turn them around every now and again to allow for even growth.

And you can make your aluminium potassium sulphate crystals even nicer by cocrystallising it with chromium potassium sulphate. They crystallise in exactly the same structure but the chromium(III) ion is coloured (aluminium(III) is not).


O-carboxylation in the Kolbe-Schmitt reaction

The resulting product would be a carbonate hemiester monoanion, specifically monophenyl carbonate.  Similar species have been observed in carbonated alcoholic beverages.  (Instead of phenol, the alcohol involved is ethanol so the carbonate hemiester is monoethyl carbonate.)  There is an equilibrium between bicarbonate and the hemiester:

$\ce{HCO3- + ROH <=> RCO3- + H2O}$

According to the paper I linked the equilibrium constant is about 2.0, i.e. $K_{eq} = \frac{[\ce{H2O}][\ce{RCO3-}]}{[\ce{HCO3-}][\ce{ROH}]} \approx 2$.  Under aqueous conditions, the concentration of water is very large and the concentration of bicarbonate is much lower, so usually the equilibrium concentration of the carbonate hemiesters is very low, on the order of a few mM.

In contrast, under the conditions of the Kolbe-Schmitt reaction, the carbon-carbon bond formation is effectively irreversible.  There is no equilibrium, and the desired product can be obtained in high yield.


Why does sulfur, but not oxygen, catenate?

First, a note: while oxygen has fewer allotropes than sulfur, it sure has more than two! These include $\ce{O}$, $\ce{O_2}$, $\ce{O_3}$, $\ce{O_4}$, $\ce{O_8}$, metallic $\ce{O}$ and four other solid phases. Many of these actually have a corresponding sulfur variant. However, you are right in a sense that sulfur has more tendency to catenate… let's try to see why!

Here are the values of the single and double bond enthalpies:
$$\begin{array}{ccc} \hline
\text{Bond} & \text{Dissociation energy / }\mathrm{kJ~mol^{-1}} \\ \hline
\ce {O-O} & 142 \\
\ce {S–S} & 268 \\
\ce {O=O} & 499 \\
\ce {S=S} & 352 \\ \hline
\end{array}$$
This means that $\ce{O=O}$ is stronger than $\ce{S=S}$, while $\ce{O–O}$ is weaker than $\ce{S–S}$. So, in sulfur, single bonds are favoured and catenation is easier than in oxygen compounds.

It seems that the reason for the weaker $\ce{S=S}$ double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the $\mathrm{3p}$ orbitals overlap is small and the $\pi$ bond is weak. This is attested by looking down the periodic table: $\ce{Se=Se}$ has an even weaker bond enthalpy of $\ce{272 kJ/mol}$. There is more in-depth discussion of the relative bond strengths in this question.

While not particularly stable, it's actually also possible for oxygen to form discrete molecules with the general formula $\ce{H-O_n-H}$; water and hydrogen peroxide are the first two members of this class, but $n$ goes up to at least $5$. These "hydrogen polyoxides" are described further in this question.


Expansion under isobaric adiabatic condition

If, at time = 0, we suddenly drop the external force per unit area to $P_{\text{ext}}$ and hold it at this value until our adiabatic system equilibriates, the first law of thermodynamics for this situation reads:

$$nC_\mathrm v(T_2-T_1)=-P_{\text{ext}}(V_2-V_1)=-\frac{nRT_2}{V_2}(V_2-V_1)=-nRT_2\left[1-\frac{1}{8}\right]$$

Solving this equation for $T_2$ yields: $$T_2=\frac{T_1}{\left[1+\frac{7}{8}\frac{R}{C_\mathrm v}\right]}$$

But, for $C_v=3R$, this gives:$$T_2=0.774T_1=\pu{232 K}$$

So, $$\Delta H=3(4R)(232-300)=\pu{-6.78 kJ}$$

So the final temperature is higher than in the adiabatic reversible case, and the decrease in enthalpy is less.


Why and how does lycopene absorbance increase with heating?

"Heating will result in an increase in amount of lycopene absorbed by
  the body."
  Is the statement scientifically correct?

It is a poorly phrased statement, but it is probably generally true.  It would have been better to say something like, "heating tomatoes will increase the amount of all-trans-lycopene available for ingestion."

Here's a link to a full paper on the subject (click the "View" button near the top-right of the page).  Basically two things are going on at the same time when lycopene containg food is heated.  It is the naturally occurring, all-trans isomer of lycopene that is thought to be beneficial to the human body. Heating lycopene causes it to isomerize to a variety of cis-isomers.  Heating in the presence of oxygen can also cause oxidation.  Both of these processes (isomerization and oxidation) reduce the potency of the lycopene in the broth.

However, in parallel with these deleterious effects, heating also makes more lycopene available in the heated broth or extract.  This is because heating disrupts cell walls and releases lycopene from the cells.  The above link shows that, at least at moderate temperatures, the latter effect predominates and so more active all-trans-lycopene is made available upon heating tomatoes, despite the loss of some lycopene to isomerization and oxidation.


If Electronegativity of Cl is greater than H2 then why is the bond angle of Cl2O greater than H2O?

Bond angle is largely governed by the location of electron cloud with regard to atoms in a molecule.
In dichlorine monoxide, the electron cloud is more spread out between oxygen and chlorine while in the water molecule it’s much closer to the oxygen atom. Therefore, the net bond paid-bond pair repulsion is higher in dichlorine monoxide than in water, while the lone pair-lone pair repulsion is almost the same.

Also, chlorine atoms are much bigger than hydrogen atoms, because of which they also undergo steric repulsion causing the bond angle to further increase.
That’s why the bond angle in dichlorine monoxide is greater than in the water molecule.


Very elementary question about valence electrons of ionic compound

You get 24 if you say Mg2+ has 8 valence electrons (its electron configuration is 1s22s22p6) instead of zero.

In that case, there are 8 in Mg2+, and 8 in OH-  (6 from O, 1 from H, and and extra 1 because of the -1 charge).  Therefore there are $8 + 2 \times 8 = 24$ per formula unit made of one Mg2+ and two OH-.

Some people do argue ionized Group 1A and 2A metals (above Period 2) have 8 valence electrons instead of zero, because technically there are indeed 8 electrons in the outermost electron shell and they are isoelectronic with the noble gas at the end of the previous row (e.g. in your case Mg2+ has the same electron configuration as Ne).

But I think most chemists would prefer to say Group 1A and 2A cations have zero valence electrons, that is, that the valence shell in these ions is the shell corresponding to their Period that used to have the outermost electrons.  In that case, the answer to your question is that there are zero valence electrons in Mg2+ and 8 in OH-, for a total of 16 in Mg(OH)2.  Note that this is slightly different from your reasoning, however.

An argument for saying Mg2+ has zero valence electrons is that if we draw the Lewis dot structure for the formula unit, I don't know anyone who would put any dots around the Mg.  On the other hand, an argument for saying that it has 8 is that we frequently argue that Mg forms the Mg2+ ion because that gives it a complete octet in the outer shell.  We rarely present the octet rule as requiring 0, 2 or 8 electrons in the valence shell.

I don't think the correct answer can be found in any mathematically rigorous sense, it depends on your preference in labeling, what you want to call "valence" electrons, and there are nontrivial arguments both ways.


On the process of heating a gas at constant pressure

OK, let's go one question at a time.

  does having constant pressure automatically mean there will be a temperature change and if so, why?

Not quite.  Adding energy means there will likely be a temperature change. What
constant pressure means is that it will be easy to calculate this change, because you can use the value of the constant-pressure molar heat capacity, $C_p$, which is the same for all diatomic ideal gases.  It is also independent of temperature, volume and pressure.  You will find it in your book.

The molar heat capacity tells you how much energy you need to transfer to one mole of the gas to raise the temperature by one degree.  You now have everything you need to know by how much the temperature changes, and therefore the final temperature.

  What about volume? Why do we not need to know the volume for this equation?

Actually you do know the volume, or can easily figure it out.  And though the volume is not important for finding the temperature change, for the reason explained above, it's helpful in finding the change in internal energy. Why?
The change in internal energy is the energy you put in (500 J) minus whatever energy the gas spends expanding.  Since the pressure is constant the energy spent expanding is simply the product of the pressure by the increase in volume.  You can calculate the increase in volume because you know the pressure, number of moles and initial/final temperature.

Now a bit of meta advice.

  I am trying to understand how to approach this question without simply plugging the numbers into an equation. Can someone please explain their thought process while solving this?

You don't make clear what, if anything, you know (either conceptually or equation-wise) and what, if anything, you've tried.
You'll get help faster if you show you've done your part.

  Basically, I want to understand what's going on conceptually rather than memorize a few rules.

Understanding what's going on conceptually goes hand-in-hand with learning the rules.  You can't really do one without the other.  (Well, you can memorize the rules without understanding the concepts, but you won't know which ones to apply.)


Finding out the value of rate constant on titration

From a mole balance, we find $C = n_x(t) + n_y(t) + n_z(t) = n_x(t) + 2n_y(t)$, where the second equality follows from the stoichiometry of the reaction. In particular, our boundary conditions are $C = n_x(0) = 2n_y(\infty).$

We assume that the reaction is first-order, so $n_x(t) = n_x(0)\exp(-kt) = C\exp(-kt).$ The two measurements give us the set of equations
\begin{align*}
80D &= n_x(5) + 5n_y(5) &&= C\exp(-5k) + \frac{5}{2}C(1-\exp(-5k)) \\
100D &= 5n_y(\infty) &&= \frac{5}{2}C,
\end{align*}
where $D$ is an undetermined proportionality constant for the titration, and the coefficient of 5 in $n_y$ accounts for the fact that 2 equivalents of $A$ react with $n_y$ and 3 equivalents with $n_z$. Dividing one equation by the other produces the final result $$\frac{4}{5} = \frac{2}{5}\exp(-5k)+1-\exp(-5k),$$ which can be solved for $k$ and hence $t_{1/2}$.


The origin of the word is historical, and has to do with the observed properties of the class of hydrocarbons we know refer to as alkanes. The name originated in the Latin for "little affinity."

From M. P. Crosland's text Historical Studies in the Language of Chemistry:

  The term paraffin was coined by Reichenbach (from Lat. parum + affinis = little affinity) to denote a particular substance, a mixture of hydrocarbons (Journal fur Chemie und Physik, 59 (1830), 455). The term was extended to cover any hydrocarbons of the series $\ce{C_{\rm n}H_{\rm 2n + 2}}$ by Watts (Dictionary of Chemistry, 1868, vol. iv, p. 344 and Supplement, 1872, p. 705).

Reichenbach, referred to above, is Carl Reichenbach, a multifaceted scientist who did pioneering work in isolating (and thus naming) hydrocarbons in the early 1830's.


Anhydrous calcium sulfate as a drying agent

I think your problem is with the idea of "irreversible" process.

The term irreversible here means that, at $\pu{393 K}$ and in the conditions where this is performed, the reaction

$$\ce{CaSO4.2H2O <=>  CaSO4 + 2 H2O}$$

is pushed towards products so much, that it can be considered "one-way" only:

$$\ce{CaSO4.2H2O -> CaSO4 + 2 H2O}$$

This is dependent on the temperature, and likely other conditions: for example, water exists as a gas at this temperature at a pressure of $\pu{1 atm}$ or lower. If you remove some of the water while heating (such as vacuum, or by circulating dry air around to get the water out), then you further push the equilibrium to the right (and make the backwards reaction impossible since water is no longer present, thus making the process irreversible).

Once you have your dry $\ce{CaSO4}$, it can be cooled down to room temperature (in a dessicator, to avoid it absorbing water). At temperatures below $\pu{393 K}$, the reaction will work backwards:

$$\ce{CaSO4 + 2 H2O -> CaSO4.2H2O}$$

Thus, your dry $\ce{CaSO4}$ will absorb water from the environment, becoming a drying agent. You may regenerate it by reheating it to $\pu{393 K}$, and so on and so forth.


Does coffee have the ability to dissolve less sugar into it than water?

I will agree with Ivan Neretin. That being said, just because other components are dissolved in solution does not mean that solubility of other components will be affected. (i.e. if you dissolve sodium chloride in water, the solubility of sucrose may or may not be affected). If you have some spare sugar, a scale, coffee, and a little extra time, I'd recommend playing with it a little bit. Look at the literature solubility of sucrose in water, make a known volume of coffee, mass a fair amount of sugar, and see how much you can get to dissolve. Do a few unit conversions and compare the value you get to literature. Could be a good weekend experiment.


What additives could make a masonry unit water-repellent but vapor-permeable?

The answer turned out to be waterglass. That's the carrier for silicate mineral paints.


Rank the following protons in order of acidity

the c hydrogen is more acidic because it is stacked between two electron withdrawing carbonly groups, and after the loss of hydrogen the carbanion is more stabilized, where as the same thing is not observed in case of d hydrogen the carbanion is in conjugation but only with one of carbonly group followed by b and c hydrogen.

now in case of second example, e is more acidic because it is allylic as well as next to the carbonyl group which means the anion after loss of hydrogen is more stabilized as come to f and g.


Given its heat resistance to thermal decomposition and the high chemical inertness of soapstone, as noted in this source, for example, to quote:

Soapstone is chemically inert, resistant to heat and durable, yet also malleable...As a metamorphic rock, it is very resistant to heat. In fact, soapstone is often used for casting metals, such as lead, pewter and silver...Because soapstone is primarily talc, which is chemically inert, it is highly resistant to etching and chemical reactions that may mar it’s beauty.

I would suggest placing a small amount of finely powdered soapstone in a vessel with highly alkaline aqueous mix of Sodium hydroxide (NaOH).  Apply conventional heating or, given an iron oxide presence, microwave heating. This is an attempted relatively safe synthesis of, for example, Sodium aluminate.
The proposed reaction is a hot alkaline attack of any of the several metal oxides present, including, for example, aluminum oxide, which could proceed as follows:
$$\ce{Al2O3 (s) + 2 NaOH (s) + 3 H2O (l) -> 2 Na[Al(OH)4] (aq) }$$
Support is provided per this reference, to quote:

All the solutions removed some $\ce{SiO2, Al2O3, and Fe2O3}$ from the samples, but the quantities were small. Sodium hydroxide attacked the kaolin group minerals more strongly...

I would then neutralize the soluble salts with $\ce{CO2}$, creating corresponding carbonate (or oxide, in the case of Aluminum ions) precipitates.
Please read the cited reference for more information on procedures (albeit, as alluded to per this 1971 source).


Why do power requirement figures for water electrolysis on the web seem to be wrong?

They said it right, "... requires at least 241.8 kJ ..." . You just found that one can simply calculate the molar formation enthalpy from the cell voltage, and vice versa, those two informations are redundant.

Of course you cannot run the electrolysis at the minimum voltage, because it will be extremely slow! You would just balance the electrochemical potential of the cell (created by the reverse reaction as soon as you make a small amount of hydogen and oxygen).

To make any actual electrolysis, you need a higher voltage, and the difference is converted into heat at the internal ohmic resistance of your cell. The minimum additional voltage is called the cell overvoltage, below that you don't get electrolysis. It is necessary to be able to break through the polarised electrolyte layers at the electrodes. So you cannot get towards 100% efficiency by running very slow.


Why can we disregard stoichiometric coefficients when computing reduction potentials?

You have to be careful of the difference between $E^\circ$ and $E$. $E^\circ$ is the standard electrode potential at defined conditions, some of which are mentioned in GODPARTICLE's answer. Standard thermodynamic conditions are usually

A temperature of 298 K
A pressure of gaseous components of 1 atm (or 1 bar)
A concentration of 1 M

The reference reduction potentials you quote are all standard potentials. These are the contributions to the cell potential when the half-cells meet the criteria above.

$E$ is the cell potential at all other conditions. The correction of $E^\circ$ to $E$ involves temperature, pressure, concentration, and stoichiometry. Behold the Nernst equation:

$$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$

$R$ is the ideal gas constant (this is thermo after all)
$n$ is the number of electrons transferred - here is the stoichiometry!
$F$ is the Faraday constant
$Q$ is the reaction quotient - which accounts for changes in pressure and concentration


What does under-stoichiometric or stoichiometric mean?

Yes, the convention is to refer to the oxygen atoms when talking about over- and under-stoichiometric in metaloxides. Normally they also indicate this explicitly by writing $\ce {SiO}_x$ (like in the article here for example) or $\ce {Al_2O}_x$. Being explicit about it is a good habit because it avoids ambiguity.


Software that determines whether a molecule can exist and draw it from a formula?

I believe this website will be of use to you

http://molview.org/

It appears to take names, formulas, smiles etc. If the name/formula/smile ID pops up in the search bar, it will draw it for you. Its database seems quite large, I use it for some pretty big drug molecules.

It draws 2D and 3D images, and it also does single/double/triple bonds.


Is the definition of anode and cathode failing?

entropid's answer is correct, but I'll try to express it differently. As Bard et al. state in Electrochemical Dictionary (1), the concept of anode and cathode is related only to the direction of electron flow: The anode is the electrode at which oxidation occurs (in an electrochemical cell) and electrons flow from electrolyte to electrode and the cathode is where reduction occurs and electrons flow from electrode to electrolyte. In a galvanic cell with a zinc anode, the electrons are flowing away from the interface in the electrode. Similarly, in a phototube, electrons emitted by the cathode are collected by the anode and again, electrons are moving away from the interface in the anode.

Where the electrons are actually coming from isn't important, just what direction they flow with respect to the electrode interface. Electrons moving from interface towards electrode = anode, electrons moving from electrode towards interface = cathode.

(1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008, pp 31.


What is the correct structure for sulfur dioxide?

See this.
It's because it can be thought as two interchangeable resonant structures. What happens in reality is that the electron is delocalized over the three atoms.