Identity of a green substance formed during electrolysis reaction?

Without further details of your equipment, this might be a possible answer. I assume the electrolyte you added was table salt (NaCl).

What you saw as a yellowish cloud was gaseous chlorine which discharged on the clip you connected to the positive electrode (cathode) of your battery. The reason why you did see this, instead of seeing just colorless bubbles of oxygen, is because of something called "overpotential". It's a joint property of the species in your cell and the material and surface termination of the electrode; in your case, the clip (made of wire steel) has a lower overpotential for the discharge of chloride anions (from the salt you added) than oxygen.

Elemental chlorine is a very reactive oxidant, and you had a thin chlorine film around your clip electrode. The reaction between chlorine and iron in aqueous solution most probably yielded iron (II) chloride tetrahydrate which is the reason you saw the greenish colour of the previously submerged fraction of the clip.


Why Isodecane (B.P.: 327 *C) has a higher boiling point than decane (B.P.: 174.1 *C)?

The only problem here is the units - the boiling point of isodecane is given in Fahrenheit and the boiling point of decane in Celsius.


Transition from ground state to excited state

First you need to have the ground state electron configuration, for Neon this is 1s$^2$2p$^6$ or what ever it is for your atom. This will give you the term symbol for the ground state. The excited state has one electron promoted, say p to a d orbital, then 1s$^2$2p$^5$3d$^1$.

A term symbol has the form $^{2S+1}L_J$, where S is the total spin ($2S+1$ is the multiplicity) , L the orbital angular momentum and J the total angular momentum.

Because angular momentum is added vectorially it is necessary to follow some rules to get all the values. Each type of angular momentum forms a series of values (the Clebsch-Gordon series). If total spin angular momentum is S and for each electron the spin quantum number is s1 and s2 then the series is $S= |s1+s2|..|s1-s2|$ with steps of 1 between values. Note that all values are positive and there may only be one value in the series. The same type of series applies to L and J total spin and orbital angular momentum.

A p$^5$ can be considered as the same as p$1$ as only 1 electron is unpaired. Thus using the series if there are two electrons each with spin quantum numbers s1, s2 = 1/2 each the the total spin for two electrons is S = 0 or 1 and the spin multiplicity is 1 (a singlet) or 3 a triplet. The series is |s1+s2| = 1, and |s1-s2| =0.

The L values is calculated with l1=0 (s orbital) and l2=1 (p orbital) thus L only has 1 value = 1. Thus far the possible term symbols are $^1P$ and $^3P$.

The total angular momentum  J is found in the same way as S and L values were using the Clebsch-Gordon series, $$J= |L+S|...|L-S| $$ Again this means find the maximum value, then the minimum make it positive, and separate any other values by 1. Sometimes there is only 1 value so be careful. There are $2J+1$ degenerate J levels, these are split in an external magnetic field into separate states.

This for the triplet state $S=1$ and so $J=|1+1|..|1-1|$, or 2, 1 and 0. (note three values from the series). Thus the term symbols are $^3P_2$ , $^3P_1$, $^3P_0$. But there is also a singlet term so the J values for this are $J=|1+0|..|1-0| = 1$ (one term only). Thus the last term symbol is $^1P_1$. Although this looks rather complicated it becomes quite quick with a bit of practice.

Once you have all the term symbols you can then apply selection rules to see which transitions are allowed.

Finally if the configuration contains, say, p$^2$ then one has also to consider the Pauli principle and it is more tricky. The way to do this is to list all s1, s2, l1, l2 values and remove any with the same set of quantum numbers. Finally when you have the term symbol Hund's rules are used to determine which levels and states are lowest in energy. (Many text books work through the p$^2$ case, See for example Steinfeld, 'Molecules & radiation')


Estimating the absolute entropy of water at room temperature

To evaluate $S=Nk\ln(\Omega)$ it is necessary to use statistical mechanics to calculate $\Omega$.

In quantum mechanics, for any independent system in a potential there is generally a set of permissible energy levels $\epsilon_1,\,\epsilon_2,\cdots$. The energy of the whole assembly of $N$ similar systems is $E$ and the volume $V$. The number of microstates or configurations (complexions) is $\Omega$ and this depends on $E,\,V,\, N$. It is assumed that all of the different microstates with the same values of $E,\,V,\,N$ are equally probable.

There are $n_1$ systems with energy $\epsilon_1$, $n_2$ systems with energy $\epsilon_2$ and so on. The total number of different states of the assembly is the number of ways of dividing $N$ things into groups with $n_1, n_2\cdots$ in each which is $N!/(n_1!n_2!\cdots)$. The total number of different states is then the  total of all of these and is

$$\sum_{\text{all possible sets }n_i} \frac{N!}{\prod_i n_i}$$

As the total energy $E$ is fixed as is $N$, then

$$\Omega = \sum \frac{N!}{\prod_i n_i}  \quad \text{with constraints} \quad \sum_i n_i =N, \sum_i \epsilon_i n_i=E$$

This is a difficult expression to evaluate but it happens that the largest term makes such an overwhelming contribution that only this term need be considered. The mathematical method to find the maximum is an interesting use of Lagrange's method of undetermined multipliers and is given in texts on stat mech but covers several pages so is too long to include here.

The result is that

$$S=k\left(N\ln\left(\sum_i e^{-\epsilon_i /(kT)}  \right) +\frac{E}{kT}\right) $$

where $\displaystyle Z=\sum_i e^{-\epsilon_i /(kT)}$ is the partition function. The entropy thus comes down to calculating the partition function. In fact all other thermodynamic properties can be found once $Z$ is known.

In a molecular system the $\epsilon_i$ are the energies of translation, vibration and rotation and electronic excited states (if any). Knowing the energy levels, say from spectroscopy, then allows each of these terms to be calculated. The translational partition function is calculated using the Sakur-Tetrode equation and a value $\ln(Z)\approx 70 $ is typical. For rotational motion $\ln(Z)\approx 10$ and for vibrations $\ln(Z)\approx 1.5$ for each vibrational mode. The vibrational contribution is small because at room temperature $kT\approx 210$ wavenumbers and vibrations are typically several hundred to a few thousand wavenumbers.  In the case of liquid water you have a particular problem which may be mitigated somewhat by adding H bonds to a molecule as an extra vibration.


Osmium tetroxide and meta-chloroperoxybenzoic acid

With osmium tetroxide, no it does not appear that dihydroxylation and acetalization can be done simultaneously. Usually dihydroxylation is carried out with catalytic osmium tetroxide with a stoichiometric amount of N-methylmorpholine N-oxide (NMO). The latter regenerates the Os(VIII) species necessary to carry out the dihydroxylation and results in the formation of N-methylmorpholine, which is a weak base. Without an acid available for catalysis, the acetalization does not occur. The sequence can be accomplished in "one-pot" however. In practice, you can monitor the progress of the dihydroxylation reaction. When it is deemed complete, add catalytic acid directly to the same reaction mixture, which allows the acetal formation.

Reference: 10.1016/j.tetlet.2012.02.045

Similarly, the overall transformation has been accomplished with mCPBA, but it's not clear if happens without any further intervention. In the referenced paper below, it's not entirely clear to me when the epoxide opens, because the procedure specifies that the epoxidation step is carried out with exclusion of water. It's likely that the epoxide is stable until addition of water, which opens the epoxide to the diol and then the acetalization can occur, catalyzed by the meta-chlorobenzoic acid byproduct from the epoxidation. In some cases, they simply isolate the epoxyketone as well.

Reference: 10.1002/ejoc.201101087


How does the relative amount of oxidation of different alcohols affect the amount of energy released?

It doesn't have to do with incomplete combustion, as the enthalpy values you've seen reported are for the enthalpy of complete combustion.

It has to do with the alcohol already being partially oxidized, so there is less energy available for release upon the oxidative process of combustion.  In other words, if a particular carbon atom is already partially oxidized, like in an alcohol, then you can't get as much energy from combustion as compared to a more reduced carbon, like that in a methyl group for example.

This effect is most pronounced for the low-molecular weight alcohols.  Methanol has 1 oxygen per carbon atom, while ethanol has 0.5 oxygens per carbon atom.  That is a much larger relative difference than there is between a $\ce{C10}$ alcohol and a $\ce{C11}$ alcohol.  In other words, the larger a mono-hydroxy alcohol is, the more it resembles a pure alkane with respect to the enthalpy of combustion.


If TLC is the only analysis we have, it may be difficult to determine exactly what is happening (or not happening). Other things that might be happening include homocoupling (biaryl formation), proteo-deborylation, and debromination. These may be hard to discern by TLC alone.

If there is no reaction of starting materials this might be due to:

Failure to reduce Pd(II) precatalyst to active Pd(0) species: your catalytic cycle never initiates. Consult chapter 1 of Negishi's Handbook of Organopalladium Chemistry for Organic Synthesis. Consider adding a catalytic amount of exogenous reducing agent, changing the solvent to EtOH/PhMe/H2O, increasing the temperature, switching to a Pd(0) catalyst system, or a system that is more easily reduced to Pd(0) like Pd(OAc)2 + xs phosphine ligand.
Stalling in the catalytic cycle: there is insufficient energy to get over the activation barrier in the oxidative addition, transmetallation, or reductive elimination step. Try increasing the temperature.
Catalyst destruction by oxidation: did you degas your system with N2 or Ar?

Alkyne poisoning of the homogeneous Pd catalyst seems unlikely since there are plenty of examples with Sonogashira coupling of terminal alkynes involving dppfPdCl2 that work just fine.


Integration of the first order rate equation

Klaus and Binary Geek have already made the main points but I think it would be helpful to see a full explanation.

The differential equation we are trying to solve is: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}]$$

Separating the variables: $$\int{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int{-k\,\mathrm{d}t}$$

Substituting in the limits; initial concentration $[\ce{A}]_0$ at time $t=0$ and final concentration $[\ce{A}]$ at time $t$: $$\int_{[\ce{A}]_0}^{[\ce{A}]}{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int_0^t{-k\,\mathrm{d}t}$$

Integrating both sides: $$\bigg[{\ln[\ce{A}]}\bigg]_{[\ce{A}]_0}^{[\ce{A}]} = -k\bigg[{t}\bigg]_0^t$$

Evaluating: $$\ln[\ce{A}] - \ln[\ce{A}]_0 = -k(t-0)$$

Condensing the logarithms: $$\ln{\left(\frac{[\ce{A}]}{[\ce{A}]_0}\right)} = -kt$$

Solving for $[\ce{A}]$: $$[\ce{A}] = [\ce{A}]_0 \mathrm{e}^{-kt}$$


What route to take in converting ethanol to crotonic acid?

Allylic bromination and subsequent hydrolysis might be replaced by an allylic oxidation of the butene using selenium dioxide - if that's inorganic enough for you ;)

But I'm wondering whether there's a shorter way. What about the Grignard reaction you described, followed by oxidation (PCC on silica or manganese dioxide) to 2-butenone. Subsequent bromination with bromine in hydrobromic acid should yield an $\alpha,\alpha'$-dibromoketone which, upon treatment with base (Favorskii rearrangement) and, eventually acidic workup should yield crotonic acid.


How can an insoluble compound be a strong electrolyte

It sounds contradictory, but soluble and insoluble are relative terms. Silver chloride's $\mathrm{K_{sp}}$ is $1.77\times 10^{-10}$, so one can generally think of it as insoluble, but actually about a milligram will dissolve in a litre of water.

Much like the terms strong acid and weak acid, strong electrolyte and weak electrolyte refer to the dissociation of a substance in a solvent, though they include all electrolytes, not just acids. In the case of silver chloride, though little of it dissolves, what does is present only as $\ce{Ag+}$ and $\ce{Cl-}$, not solvated  $\ce{AgCl}$. In water, all salts are strong electrolytes, but in other solvents, things can be different. For example, while perchloric acid is a strong electrolyte in water, it doesn't dissociate completely in acetic so it's a weak electrolyte.

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


How do I determine the mass percent of Barium nitrate and Sodium acetate in an original unknown solution

It occurred to me that you might have been asking how to complete this separation experimentally rather than conceptually, which is what the comments and previous answer all were pointing toward.  If that is the case, then once you've precipitated out the barium sulfate, you have to isolate it from all other substances in order to find its true mass.

The precipitate is a solid, so you could separate it from most of its contaminants by pouring the solution through a pre-weighed filter paper, making sure to wash the walls and bottom of the reaction vessel with distilled water and pouring this wash through the filter as well in order to collect all of the precipitate.

The filtrate would contain the bulk of the sodium acetate and sodium sulfate in the reaction mixture.  What remains in the filter paper would be barium sulfate and a small amount of reaction mixture, including the sodium salts.  To eliminate the sodium salts, add additional distilled water to the filter to wash the precipitate (and filter paper) clean.

Once you have a filter paper with clean barium sulfate, dry it thoroughly and weigh.  I don't believe that barium sulfate forms a hydrate, so the mass of the precipitate should all be barium sulfate.

Note: Barium salts are quite toxic, particularly highly soluble ones like barium nitrate.  Be careful using them and dispose of the product appropriately.


Acidity of cyclopentadiene when fused to benzene rings

Clar's rule states that (Solà, 2013):

  The Kekulé resonance structure with the largest number of disjoint aromatic π-sextets, i.e., benzene-like moieties, is the most important for characterization of properties of polycyclic aromatic hydrocarbons (PAHs).

Applying Clar's rule to the the question, we would realise that in the indenyl and fluorenyl anions, the resonance structures where there are 6 π-electrons delocalised exclusively within the fused benzene rings would contribute more to the actual structure than the resonance structures where the π-electrons are delocalised along the 5-membered ring. In the major resonance structure for the fluorenyl anion, we would have the negative charge concentrated on the carbon atom, without delocalisation.

In the cyclopentadienyl anion, the negative charge is completely delocalised, as shown below in the resonance structures. Thus, it is the most stable among them. In the indenyl and fluorenyl anions, the negative charge is not as delocalised and thus, these are less stable.

Due to this, the π-electrons in the 5-membered ring experiences the most delocalisation and achieves most aromatic stabilisation in the cyclopentadienyl anion, and this gradually decreases going to the indenyl and fluorenyl anions.

Reference

Solà, M. (2013). Forty years of Clar's aromatic π-sextet rule. Frontiers in Chemistry, 1(22). doi:10.3389/fchem.2013.00022


Can rocket candy explode during the preparation?

See James Yawn's website ${\it Recrystalized\; Rocketry}$, which is all about solid propellant rocketry of the type you're interested in. It's really an exceptionally comprehensive site.

Can this mixture explode during the heating somehow?

It appears that it's more likely to catch fire, and do so rather vigorously, as in cases where it's exposed to a spark or an open flame. Because the system (pan and mixture of $\ce{KNO3}$ and sugar) would be open to the atmosphere (and not confined to a fixed volume), there's more of a risk of fire than of an explosion. As reported by James Yawn in the subsection on "Skillet Rcandy":

This stuff looks innocent enough in the pan.  But if you have ever seen a quantity this size burn you will have more respect for it.  This batch could send a tongue of flame 20 feet in the air and engulf the person stirring it, if only briefly.  Burns to the body would likely be second degree, and I am willing to tolerate that level of risk.  But eyeballs don't recover from this kind of trauma easily.

See this safety page from James Yawn's site.

Or it can just burn, not explode?

Yes, it can just burn: that's essentially what's going on in the rocket engine. The exhaust from the burn is channeled in one direction (${\it via}$ the hollow core in the engine itself), resulting in thrust in the other direction.

How to prevent it, and do it the most secure way?

From James Yawn's site:

Basic Safety Precautions
At present, I do NOT recommend making this propellant in a household kitchen.
Make small batches only!  No more than one standard recipe, which yields about 160 grams of propellant.
Wear protective gear. You should have a full-face shield, long sleeved heavy cotton shirt, heavy apron and heat-resistant gloves.  These must be worn at appropriate points in the process.
Do not allow children or rambunctious pets in the room where you are working.
Do not allow distracting, inattentive, inebriated or otherwise incompetent adults in the room, especially if they smoke.
Be vigilant of possible sources of sparks, flame, or high temperature that could ignite propellant on contact.
If you smoke, place your cigarettes and lighter in another room so you will not light up unconsciously.
Clear your work-area of any flammable materials.
Have a good fire extinguisher available. Also a 5-gallon bucket of water nearby and a garden hose outside.
Make sure there is good ventilation, and a quick escape route to fresh air.
Avoid stupid rocket tricks.
Avoid excessive exposure to Potassium Nitrate.

All of the above points are elaborated upon at the linked page.


Why does dimethyl sulfoxide have higher boiling point than acetone?

Firstly, sulfur has a larger atomic radius than carbon, so we would expect DMSO (Ddimethyl sulfoxide) to have a much larger and thus polarisable electron cloud than acetone. The London dispersion forces between DMSO are thus stronger, causing DMSO to have a higher boiling point.
Secondly, if we compare the dipole moments of the $\ce{S=O}$ and $\ce{C=O}$ bonds, the $\ce{S=O}$ bond is much more polarised. This is because the $\mathrm{3p-2p}$ overlap between $\ce{S}$ and $\ce{O}$ is much less favourable than the $\mathrm{2p-2p}$ overlap between $\ce{C}$ and $\ce{O}$. As such, the resonance structure on the right is actually the more significant resonance structure. This makes the $\ce{S=O}$ bond much more polarised as compared to the $\ce{C=O}$ bond, causing the permanent dipole-permanent dipole interactions between DMSO to be much stronger than the same interactions between acetone molecules.

Lastly, we would realise that DMSO has a trigonal pyramidal shape while acetone is planar. Due to this, the net dipole moment in DMSO is likely to be stronger due to it being a more asymmetric molecule.


Oxidation of acetophenone to benzoic acid

From a SciFinder search, there are ~10 examples of oxidations of aryl methyl ketones resulting in a carboxylic acid using a chromium based reagent (usually dichromate but also chromium trioxide). This suggests the reaction is at least possible, although I think including it in a textbook is unwarranted.

One reaction slipped through my search that is informative (below). Notice that in the presence of a benzylic position that does contain a hydrogen, the methyl ketone is unchanged, while the other group is oxidized.

Taken together, it seems that chromium oxidation of an acetophenone to a carboxylic acid is slower than alkyl benzylic oxidation.


Does a certain "Badecker reaction" actually exist?

The reaction in question is actually called the Boedeker reaction. How the book came from Boedeker to Badeker I do not know, but maybe there was one instance of Boedeker being written Bödeker which was later misparsed by a human or a robot and mistook for an a. (Replacing oe for ö is a common mistake with German names since the latter is typically used in orthography but names don’t adhere to orthography.)

The reaction is well-known and often used as a qualitative analysis for sulphite ions in solution, much like the same nitroprusside ion can also be used to detect sulphide ions. In chemical formulae, it is:

$$\ce{[Fe(CN)5(NO)]^2- + SO3^2- <=> [Fe(CN)5(NOSO3)]^4-}$$

The reaction in itself is also rather interesting since for over 100 years the structure of the addition complex was unknown. Bottomley et al. were finally able to obtain crystal structures of the probably structurally similar ruthenium complex $\ce{[RuCl(bpy)2(NOSO3)]}$.[1] These show that the addition product is — very much like in the sulphide case — formed by nucleophilic attack of the sulphite sulphur’s lone pair to nitrosyl nitrogen creating a nitrosylsulphite anion $\ce{O3SNO-}$. Thus, the complex formed should be named pentacyanidonitrosylsulphito-κN-ferrate(II).

Reference:
[1]: F. Bottomley, W. V. F. Brooks, D. E. Paéz, P. S. White, M. Mukida, J Chem. Soc., Dalton Trans. 1983, 2465. DOI: 10.1039/DT9830002465.


Phase diagram of water using Gibbs free energy instead of pressure

Phase diagrams are not limited to temperature vs pressure. They can be made with any set of thermodynamic quantities. The diagram in your question is for a particular pressure at which the liquid phase exists. If you changed the pressure, the lines would shift. A triple point will be represented by all three lines intersecting through the same point.


So which one is it?

Both! Proactinium-234 exhibits nuclear isomerism. The metastable form $\ce{^{234m}Pa}$ has a half-life time in the range of 1 min, whereas $\ce{^{234}Pa}$ shows a half-life time around 6 hrs.


Why is benzoic acid a stronger acid than acetic acid?

It is all about the electron withdrawing/ donating nature of the acid's functional groups.  The more electron withdrawing the group, the stronger the acid will be.  The more electron donating the weaker the acid will be.  A phenyl ring is electron withdrawing, while a methyl group is electron donating.  I don't know why resonance would destabilize an organic molecule or ion.  if you are referring to the carboxylate donating electrons to the phenyl ring, I don't see how this could be possible and not violate the octet rule.


Why Benzene Cannot be chlorinated using HOCl?

One possibility:  When you chlorinate with, let us say, $\ce{Cl2 + FeCl3}$, you generate a "leaving group", $\ce{FeCl4^-}$, that comes off easily because it is highly polarizable and only weakly basic.  The $\ce{HOCl +H^+}$ combination would have to displace a more strongly basic and less polarizable water molecule.

Chlorination can be effected with hypochlorous acid by adding silver ion as another catalyst/co-reagent.  See here.


Testing thermite for chemical properties

Typical military-grade thermite mixtures (e.g. used in incendiary hand grenades) contain a large amount of barium nitrate (e.g. 29.0 % in Thermate-TH3, see Patent No. US 6766744). It should be possible to detect the barium oxide that is left over after the fire, maybe first with a simple flame test on site followed by AAS (atomic absorption spectroscopy) in a laboratory.


Why is $\ce{OH-}$ more basic than $\ce{F-}$?

To answer this question, let us consider the $\mathrm{p}K_\mathrm{a}$'s of the conjugate acids of these anions: $\ce{H2O}$ and $\ce{HF}$. $$\mathrm{p}K_\mathrm{a}(\ce{H2O})=15.7\ \\  \mathrm{p}K_\mathrm{a}(\ce{HF})=3.17$$

You didn't need me to tell you that $\ce{HF}$ is a stronger acid than water though, but it is -- $10^{12.53}$ times stronger to be exact. This means that this is also how many times stronger $\ce{OH-}$ is than $\ce{F-}$

  In a polar aprotic solution, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

Small, electron-dense ions are always great nucleophiles, except when the solvent crowds the nucleophile with hydrogen bonds, as in the case of $\ce{F-}$. It is so heavily surrounded by solvent molecules in polar protic solvents that it can't make an attack very well. $\ce{I-}$ is not as heavily solvated in these solvents, and it is very easily polarizable, making it a better nucleophile in these solvents.

In polar aprotic solvents, the small $\ce{F-}$ ion can easily slip between solvent molecules and make attacks. Now in comparison, $\ce{I-}$ is a much poorer nucleophile.

  In a lecture, my teacher said $\ce{RO-}$ is relatively stable because of the high electronegativity of $\ce{O}$.

Again, we can answer this by looking at $\mathrm{p}K_\mathrm{a}$'s. The $\mathrm{p}K_\mathrm{a}$ of ethanol, for example, is $15.9$! Only $10^{0.2}=1.58$ times weaker an acid than water! That means that ethoxide will deprotonate everything $\ce{OH-}$ will and then a little more, so it's not really that much more unstable.

  When people say something is stable, are they referring to its acid/base characteristics, or can stability also be referring to nucleophilicity/electrophilicity?

The two are pretty well related. You have exceptions like how $\ce{F-}$ is a poor nucleophile in water despite being a good base, but as a rule of thumb, I would say people mean acid/base reactivity when they talk about "stability."


So if that is the case, shouldn't period 3 have more elements, since
  it can hold up to 18 electrons, and therefore it can have up to 18
  more protons from the largest atomic number element in period 2?

Indeed, elements of the same period have the same number of electron shells, but the "problem" is that in accordance with the Madelung/Janet/Klechkowski  rule, the $\mathrm{4s}$ orbital is occupied before the $\mathrm{3d}$ orbital. As a result, the $\mathrm{3d}$ orbital could be filled only when the $\mathrm{4s}$ orbital is already filled, i.e. only for elements of the 4th period.

Similarly, the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are occupied before the $\mathrm{4f}$ orbital. As a result, the $\mathrm{4f}$ orbital could be filled only when the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are already filled, i.e. only for elements of the 6th period.


Is cyclohexane saturated or unsaturated?

Cyclo comes from the Greek word 'kyklos' which means circle.

The cyclohexane molecule is in the form of a circle therefore it can have the formula $\ce{C_nH_{2n}}$ and still be saturated.


What is the problem with forming 2,3-dimethylcyclohex-2-en-1-one?

Your starting product for A is symmetrical, but your starting product for B will be asymmetrical - one carbon less on one side.  This is going to mean you may get 2 or more different products; B may not be your major product, and you'll have to separate them afterwards.  I don't believe there's inherently a tricky difficultly in the mechanism for B.


You can buy barium hydroxide. See the listing from Flinn Scientific although Flinn is by no means the only seller. However, Flinn's safety information is written at a very accessible level.


Understanding secular equation derivation

It doesn't matter how you obtain an expression for the partial derivative, as long as you obtain one, and set the partial derivative to zero. The method you described above is probably easier than differentiating the fraction. All the steps that you did are mathematically valid, so your final result is valid. (The "differentiate both sides" step is valid because the equality holds for all values of $c_a$ and $c_b$.)

If you want to, you can differentiate the fraction and you will end up with the same result (i.e. the secular equations). I actually typed the maths out before, so here you go: https://chemistry.stackexchange.com/a/42636/16683 The notation is a tiny bit different, but you should be able to understand it.

As for the linear algebra parallel, the secular equations are only analogous to $Ax = \lambda x$ if you make the additional assumption that $S_{ab} = 0$, in which case $H-ES$ is directly analogous to $A - \lambda I$ (since the overlap matrix is now equivalent to the identity matrix). In this case, the secular equations reduce to $Hc = Ec$.

If you do not make that assumption, then the secular equations take the form of $Hc = ESc$ and the parallel you should draw is to the generalised eigenvalue equation $Ax = \lambda Bx$. The eigenvectors $x$ are said to be the "eigenvectors of $A$ with respect to $B$". There is a lot more information online if you are interested - simply google "generalised eigenvalue".


Is it possible for this Iron-Carbon Compound to exist?

Short answer: No

Long answer: Assuming you had hands tiny enough to manipulate individual atoms, you could make this molecule. The amount of time it would stay together would be as close to 0 as it gets. To start iron can't accommodate that amount of electrons, and the formal charges of this molecule are extreme to say the least.


Carbon decrease from 1-octyne to 1-hexene

In principle, this can be realized in two steps:

Oxidative cleavage using catalytic amounts of $\ce{RuO2}$ and excess of oxone converts 1-octyne to heptanoic acid. (DOI)
Heating a mixture of acetic anhydride and a carboxylic acid in the presence of $\ce{Pd}$ catalyst turns the acid to a terminal alkene lacking one carbon atom (DOI). Admittedly, the high temperatures needed might be a problem here and may force you to perform the reaction in a sealed tube in an autoclave.


Sodium borohydride is a weaker reducing agent than, say, lithium aluminium hydride, so, although it will react with ketones and aldehydes, it is not strong enough to react with carboxylic acids or esters. This generalization is quite broadly applicable, so you already know your borohydride will reduce the ketone carbonyl on that molecule. $\ce{NaBH4}$ is a weaker reducing agent than $\ce{LiAlH4}$ because boron is less electropositive than aluminium. Boron has a Pauling electronegativity of 2.04 while aluminium's pauling electronegativity is only 1.61, therefore $\ce{NaBH4}$'s hydrides are significantly less nucleophilic and they won't be able to react with a weaker electrophile.

To answer your question directly, you have to think about which carbonyl is more electrophilic. Generally, resonance effects are more important than inductive effects when considering something's nucleophilicity/electrophilicity.

The ketone has two resonance forms, while the ester has three, therefore the partial positive charge on the carbon is more delocalized on the ester, making the ketone more electrophilic. The weak $\ce{H-}$ nucleophile provided by $\ce{NaBH4}$ will only be able to react with the stronger electrophile, the ketone.


Does crystal field splitting energy Δ increase going down a group of transition metals if the ligands are weak field?

I see where you are coming from, but you forgot that even with π-donor ligands, there is still a σ-bonding present in the complex. In the 4d/5d metals, the $\mathrm{e_g}^*$ orbital is pushed up by a lot due to the σ-type overlap.

The $\mathrm{e_g}$ overlap is always stronger than $\mathrm{t_{2g}}$ overlap because the metal $\mathrm{e_g}$ orbitals point directly at the ligands ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$), whereas $\mathrm{t_{2g}}$ orbitals point in between the ligands ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$).

So, yes, the $\mathrm{t_{2g}^*}$ orbital is pushed up higher by the π-donor, but the $\mathrm{e_g^*}$ orbital is pushed up even higher. Thus, for example, the complex $\ce{[RuCl6]^2-}$ has a $(\mathrm{t_{2g}})^4$ configuration (i.e. low spin) even though $\ce{Cl-}$ is a π-donor ligand.


I very much doubt that all diatomaceous earth is food grade.  For some reasons why, consider the Wikipedia page for diatomaceous earth: "Kieselgur" was discovered when folks in the 1800s were trying to dig a well.  Since knowledge of hygiene was some decades away, it's not implausible that those people may have been trying to dig a well that happened to be near an outhouse.  I'd doubt that diatomaceous earth recovered from a site near an outhouse would be food grade.

Also, Wikipedia lists "salt-water-derived" diatomaceous earth and specifically mentions that it is:

not suitable for human consumption
not effective as an insecticide
often calcined, which converts the safer amorphous silica present in most natural diatomaceous earths into a far more dangerous crystalline silica.  This can lead to silicosis if long-term exposure to airborne microparticles of crystalline silica occurs.

Thus, any diatomaceous earth that contains lots of salts or which has been calcined is unlikely to be food grade.


What is the difference between carboxylate ions and groups?

Carboxyl

The carboxyl group $\ce{-COOH}$ is an organic functional group consisting of a carbon atom double bonded to an oxygen atom and single bonded to a hydroxyl group.

Compounds with a carboxyl group are called carboxylic acids (e.g. acetic acid, $\ce{CH2COOH}$).

Carboxylate

A carboxylate $\ce{-COO^-}$ is a salt or ester of a carboxylic acid. For example, acetic acid reacts with sodium bicarbonate to form sodium acetate.

Compounds with this group have the carboxy- prefix or -ate suffix (e.g. sodium acetate $\ce{CH3COONa}$). When these salts are ionised (e.g. dissolved in water), the anion with the carboxylate group is called a carboxylate ion.

The carboxylate ion is the negative ion (anion) formed as the conjugate base when a carboxylic acid loses a proton.

$$\ce{R-COOH -> R-COO^- + H+}$$

Being its conjugate base, sodium acetate mixed with acetic acid makes a useful buffer solution for many biochemical reactions where a 'mildly acidic' pH of 4-6 is desired.


The dipole moment of $\ce{HF}$ is greater than $\ce{HI}$:

$\ce{HF} = \pu{1.91 D}$
$\ce{HI} = \pu{0.42 D}$

This is due to high electronegativity of fluorine. Hence from $\mu = \vec{q} \cdot \vec{d}$, the charge of fluorine is larger than iodine, but the bond length changes only a small amount.


N-heterocyclic carbene metal complexes. Question about the delocalization of the nitrogen lone pairs

Some parts of this answer probably border a bit on personal opinion, but I will at least try to justify having such an opinion.

The first thing to do is to actually examine the heterocycle. What kind of ligand is it?

This is a carbene. It donates two electrons via the lone pair on the carbon in plane with the ring. Notice that there is a smear of positive charge from the delocalization of the $\pi$ system between the N-C-N system. That is balanced by the negative chage on the carbon, however. Overall, this ligand is neutral, so it is an L-type ligand.

What's the overall charge on this metal complex? This is a ruthenium(II) ion (2 chlorides, everything else neutral), so this is a $d^{8}$ metal ion. You have 2 electrons from the NHC, 2 from each chloride (4 total), 2 from the phosphine ligand, and 2 from the carbene (the double bond is back-bonding). That gives 10 electrons from ligands, so this is a fine-looking neutral 18 electron complex.

For a complex that's overall charge zero, why is there a plus charge hanging out by itself? Well, it turns out that we're extremely sloppy with charges. For example, the chlorides are X-type ligands but the phosphine is not. Do we stipulate this difference in the structure? No we do not, though we use it during electron counting. The authors here felt that the smear of positive charge of the carbene, which is how you would draw the structure in isolation, should be shown but chose not to show the negative charge of the carbene, which is the case by convention. That seems somehow inconsistent to me because the overall charge of this ligand is neutral.

Look at these other representations:

Mendeleev Communications, 2019, 29 (1), pp 38-40. (10.1021/om3009298)

Organometallics, 2012, 31 (23), pp 8355–8359. (doi: 10.1016/j.mencom.2019.01.011)

Both representations of the NHC are used: without explicit $\pi$ system and with.

Note that neither of them say anything about the positive charge. This seems more consistent to me, and these authors, at least, agree with me.


Why do halogen substituents make molecules more lipophilic?

Halogens, except for fluorine, aren’t all that electronegative. Chlorine is just slightly less electronegative than nitrogen and bromine is even less again. Iodine almost has carbon’s electronegativity. Thus, there won’t be so much of a dipole moment created by substituting a hydrogen with a halogen. This applies even more if a hydroxy group is replaced with a halogen.

To illustrate this, take dichloromethane. Of all the chlorinated methanes it has the highest permanent dipole moment. Yet it is still almost immiscible with water but fully miscible with hexane. Note by the way that your data somewhat reinforces that: the molecule with the highest dipole moment of all your halocarbons, fluoromethane, has the lowest water solubility.

A second point is that halogens come with free lone pairs. Lone pairs are not polarised per se but can easily be polarised in London interactions. Thus, halogens actually increase molecule’s ability to interact with lipophilic, unpolar media by allowing stronger London forces.

Note also that while nitrogen is capable of forming hydrogen bonds with surrounding aquaeous media, chlorine — in spite of its similar electronegativity — cannot since its lone pairs are much more diffuse. Only fluorine is in theory able to accept hydrogen bonds. In principle, this is another consequence of the larger size of higher-period halogens.

In all this, fluorine is somewhat separate and special. But fluorine is special in itself since perfluorinated molecules are typically insoluble in anything but fluorocarbons. Fluorous extraction is actually a technique used by some green chem labs.

To sum it up, just looking at the simple variable polarity cannot explain the solubility of halocarbons adequately. Their size needs to be considered.


What is the meaning of the slope in a graded dose-response curve?

Since the drug concentration axis is logarithmic, even a small change in the nominal drug concentration can have a large effect if the slope is steep, so getting the correct dosage could be problematic. Accordingly, this requires care in drug concentration preparation and, in particular, it might make it harder to avoid overdoses.
For a shallow slope, a relatively large change in drug concentration may be needed in order to achieve the desired result. But side effects of the drug do not necessarily follow the same dose response relationship. So increasing dosage by a factor of 10, for example, might pose a much higher than factor of 10 risk of side effects.


Alkylation of conjugated nitriles - regioselectivity

I found this article that sheds some light on deprotonation mechanism - it differs significantly from presented in the question and although it's for acetonitrile not crotonitrile it still provides some useful information.

Lithium amide exists as $\ce{(LiNH2)2}$ dimers which tend to coordinate to nitrile groups creating Li-N bonds. In acetonitrile deprotonation occurs intramolecularly with opening of $\ce{Li2N2}$ rings, not pericyclicly, but for crotonic nitrile analogous mechanism isn't possible what suggests intermolecular reaction.

  On the basis of our MP2(full)/6-31+G*//MP2(full)/6-31+G* calculations (Scheme 2 and Figure 1), the adduct, which is formed from CH3CN (7) and the dimeric base 8, is stabilized by about 12 kcal/mol relative to the separated reactants. It is likely that in solution this coordination proceeds by replacement of a solvent molecule from the dimeric lithium amide. The existence of a complex comparable to 9 and the dimeric character of solvated lithium amides 6 are confirmed by NMR spectroscopy and structural studies. To allow hydrogen transfer from the nitrile to the amide moiety, the central four-membered ring of 9 must open.  The resulting intermediate 10, an “open dimer”, is also supported by various investigations of related reaction pathways. 7 The cleavage of an Li-N bond requires about 11 kcal/mol, while the transition state for proton transfer, 11, lies only just above slightly more than 1 kcal/mol higher in energy than 10. This low barrier is indicated in a lenghtened C-H bond of the open dimer (1.094 Å, compared with 1.082 Å in acetonitrile). After the formation of another “open dimer”, 12, which is stabilized by 14 kcal/mol relative to the transition state, and extrusion of ammonia, the final intermediate, 14, is a mixed (dianionic) aggregate of LiNH2 and LiCH2CN, comparable to Boche’s QUADAC.

Another article provides some info about anions of nitriles:

  The  exceptional nucleophilicity stems primarily from the powerful inductive
  stabilization  of  metallated  nitriles  rather  than  resonance
  stabilization as is the case with enolates.

and influence of nitrile group on carboanion:

  Essentially, the  powerful  electron  withdrawal  of  the  CN  group stabilizes the adjacent negative charge in a manner akin to an ylide. Consistent with this inductive stabilization is the range of geometries for the metallated carbon which spans a continuum varying from planar in 12 to pyramidal in the cyclopropanecarbonitrile 13.

Mainly inductive stabilisation suggests that in metallated anion of crotonitrile negative charge is concentrated on alpha carbon and thus it's more nucleophilic then gamma carbon and in low temperature, under kinetic control, attack via alpha carbon should be favoured.


1H NMR of p-methoxy phenol, 32 possible lines?

I think your theory is correct, but my tests were inconclusive. You can try your own hand at simulating the spectrum, for instance with Hans Reich's program DNMR. It resulted in the following:

This is the downfield half of the spectrum. I did not attempt to replicate the experimental spectrum exactly, it's only to illustrate. I used the following parameters:

$\sigma_A = \pu{7.5 ppm}$
$\sigma_X = \pu{8.19 ppm}$
$J_{ortho} = \pu{8 Hz}$
$J_{meta} = \pu{3 Hz}$
$J_{para} = \pu{0.5 Hz}$

On to your question. If you increase the J sufficiently, while retaining symmetry among the J, you get...

... 20 peaks! (only one set of multiplets is shown, so the total number of peajs is twice the number visible above) . I continued tinkering with the values, and breaking the symmetry among the couplings led to ...

... 26 peaks!

Presumably breaking the symmetry in the shifts will increase the number of peaks further, but surprisingly this had just the opposite effect.

But this is not the most illuminating approach. Better would be better to dig into the QM.


Niobium has a density of 8.57 g/cm3 and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom

This is how I would do the calculation...
Given data:
$$\text {Weight of crystal} = \pu{92.90638 g/mole} \ce{->[Round] \pu{92.906} g/mole}$$
$$\text {Density} = \pu{8.57 g/cm^3}$$
Ok, the density is good to only 3 significant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures throughout all the intermediate calculations to try avoid rounding errors within  the multiple calculations. I'll round the final result to 3 significant figures.
$$\text{Volume of mole} =\text{V}_{mole} = \frac {92.906}{8.57} = \pu{10.841 cm^3/mol}$$
In BCC, there are 2 atoms in one unit cell. Also note that the accepted figure for Avagadro's constant to five significant figures is now $6.0221\times10^{23}$
$$\text{Volume of unit cell} = \text{V}_{cell} = 10.841 \times \frac {2}{6.0221 \times 10^{23}} = 3.6004 \times 10^{-23}\text{ cm}^3$$
A unit cell is a cube with each side being $a$
$$\text{V}_{cell} = a^3$$
$$\therefore a = \sqrt[3]{3.6004 \times 10^{-23}} = 3.3021\times 10^{-8}\text{ cm}$$
But the BCC crystal, spheres of radius, $r$, packed inside a cube with side, $a$, the geometric relationship between $a$ and $r$ is: $$a = \dfrac{4}{\sqrt{3}}r$$ so
$$r = \dfrac{\sqrt{3}\times(3.3021\times 10^{-8})}{4} = 1.4299\times 10^{-8}\text{ cm}$$
Now rounding $r$ to 3 significant figures gives
$$r = 1.43\times 10^{-8}\text{ cm} = 143\text{ pm}$$
You shouldn't chase values through websites and literature. You can only calculate given whatever data you're suppose to take as given. For instance no temperature was given for the density which certainly changes with temperature. Also, to make the point again, with a calculator you should carry extra significant figures in the intermediate calculations to try to avoid round off errors in the final result(s).


This is essentially the same question as to why carbon concatenates through up to four bonds with binding energies low enough to be broken again yet stable enough to last given typical terrestrial environmental conditions. Great for entities like us.
On the other hand, the binding energies of Silicon, Germanium, from the same group IV, for instance, are higher. The binding energies are directly related to the bond length/bond distance, which is great for determining the energies through crystallography, along with proper model fitting.
Sulfur is one of few elements amenable to catenation, - forming long molecules with each other.
So whichever configuration releases the greatest enthalpy is typically the most stable one, but depends upon the initial energies thus giving rise to allotropes - elemental structures of different molecular configurations. The bound system typically has lower potential energy.
If a double bond lowers the overall energy of the bound system more it will typically form over a single bonded one.
Update: I found this (see also this table of dissociation energies):
Update 2: I'd like to add that from a quantum mechanical perspective, which is a viewpoint much closer to nature, the bonds arise from 3D space where electron densities are highest - which means most energetically favorable.
Since for whatever weird reason, nature is really quantized, and I cannot stress that "really" part enough, one gets standing waves for the electron equations around a given nucleus - rather than a continuum - as is the case of a sphere, like say a falling waterdrop to add a few exterior forces into the mix as well. The classical and the quantum object have only in common the "energetically favorable" state.
Somehwat simplified then, those standing waves become "orbitals" when doing the calculations for 3D space. The orbitals change in shape with increasing interference with other electrons and higher potential energies due to larger nuclei , meaning higher positive proton counts.
The energies involved result in very different packing i.e. different lattices. A necessity for life is breaking bonds or "bond accessibility". There simply isn't "enough energy" in today's protein-scaffold based catalyists on earth, to use anything else than the constituents of the periodic table with few protons in case of homogenous bonds (i.e. atoms of the same species like say Oxygen O2).
It does not have to be this way. We from earth are biased by our own biochemistry we observe. Every human generation breaks down the biases of the previous ones and encounters new ones.
Bond energies

1 S8: 1808 kJ/mol
1 O8: 1168 kJ/mol
4 S2: 1408 kJ/mol
4 O2: 1976 kJ/mol


Well, if you're talking about melting points, the simple argument that more compact molecules should have higher mp because of increased dispersion forces can easily fail.  The problem is that melting is a much more complicated phase transition than boiling, with significant entropic effects coming from considerations of packing and entanglement -- especially relevant for floppy large molecules -- so that arguments based purely on the enthalpic changes due to intermolecular forces can easily be insufficient.


Is it possible for one specific atom in a molecule to have a non-integer oxidation state?

It depends. Consider various radicals such as the superoxide anion $\ce{O2^{.-}}$ or $\ce{NO2^{.}}$. For both of these, we can draw simple Lewis representations:

In these structures, the oxygen atoms would have different oxidation states ($\mathrm{-I}$ and $\pm 0$ for superoxide, $\mathrm{-II}$ and $\mathrm{-I}$ for $\ce{NO2}$). That is the strict, theoretical IUPAC answer to the question.

However, we also see that the oxygens are symmetry-equivalent (homotopic) and should thus be identical. Different oxidation states violate the identity rule. For each compound, we can imagine an additional resonance structure that puts the radical on the other oxygen. (For $\ce{NO2}$, we can also draw resonance structures that locate a radical on both oxygens and another one that expands nitrogen’s octet and localises the radical there.) To better explain this physical reality theoretically, we can calculate a ‘resonance-derived average oxidation state’ which would be $-\frac{1}{2}$ for superoxide and $-\frac{3}{2}$ for $\ce{NO2}$. This is not in agreement with IUPAC’s formal definition but closer to the physical reality.


The correct answer is B in both the case the final concentrations are not changing. Therefore same will be the heat of rise. But heat released will be less but volume of the final solution also got halved. So temp. raise will be same.


What's the source of electrical energy in galvanic cells?

Judging by your above comment to beginner's answer, you seem to be confused about how reactions proceed in general.

Take a look at this graph : http://en.wikipedia.org/wiki/Activation_energy#/media/File:Activation_energy.svg

The oxidation in the beginning certainly requires energy, but at the end of the reaction, the system ends up at a lower energy state overall. This causes the reaction to be "spontaneous".


Most of what I would be saying now is stuff I already highlighted in another answer of mine to a related question.

$\ce{H2SO3}$ exists in two forms, one of which is equivalent to the structure you drew, the other is the structure you proposed. Taking the monoanions (because that was the focus of the other question and I am too lazy to redraw the picture) will give you these two structures:

(All horizontal bars are formal $-$ signs, never lone pairs.)

The crystal structure of $\ce{CsHSO3}$ was solved in 1980 by Johansson, Lindqvist and Vannerberg; they conclude that $\ce{CsHSO3}$ is not cesium hydrogen sulphite but rather cesium sulphonate.[1] See the other answer for details.

In cis-bisbipyridylbishydrogensulphito-κS-ruthenium(II) $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$, a crystal structure solved by Allen, Jeter, Cordes and Durham, the sulphur atom directly coordinates the ruthenium centre which is only possible if it has a lone pair; thus we must be dealing with hydrogensulphite.[2]

Therefore, depending on the conditions both forms are possible and observed.

In case of the free acid, it is likely more stable in one of the two forms. I do not have analytical or thermodynamic data at hand to tell you which tautomer is preferred.

References:

[1]: L.-G. Johansson, O. Lindqvist, N.-G. Vannerberg, Acta Cryst. 1980, B36, 2523. DOI: 10.1107/S0567740880009351.
[2]: L. R. Allen, D. Y. Jeter, A. W. Cordes, B. Durham, Inorg. Chem. 1988, 27, 3880. DOI: 10.1021/ic00295a003.


How do I find the theoretical pH of a buffer solution after HCl and NaOH were added, separately?

First I want to point out a couple things.

When you add more liquid to a solution, the concentration of the solute is going to decrease. Therefore, after the addition of 10 mL of HCl, the concentration of sodium acetate and acetic acid will no longer be 0.5 M since you diluted the solution a little bit. When calculating the concentration after dilutions, keep in mind the amount of moles is still the same, so just use the formula $\ce{M1V1 = M2V2}$.
Also, sodium acetate is a soluble salt (see solubility rules), so in the net ionic equation, don't write out the sodium as it is a spectator ion. The same applies for HCl. It is a strong acid, so it completely dissociates in water.
For this problem, using the Henderson-Hasselbalch equation is not really the right way to do it. When you are using this formula, keep in mind that it is only good for calculating pH when you already know the equilibrium concentrations of the acid/conjugate base. Here's a derivation of the formula if you're interested that will probably help clear things up a bit.

I think you mostly understand this point last point since you did made an ICE table before you applied the equation. However, in the context of this problem, making an ICE table will actually lead you to the pH value without requiring the HH equation. You'll see why.

Alright; now we can actually get to how to approach the problem. First, you're going to want to look out for a reaction. When solving these problems, being able to identify a reaction will make it 10 times easier.

How do you identify a reaction? Well, look at the reagants in this problem: $\ce{HCl}$, which is essentially just hydronium and chloride, sodium acetate, which as mentioned earlier is essentially just sodium and acetate, and acetic acid. Acids react with bases, so using that, it isn't too hard to determine what exactly happens in the solution.

You actually got the net ionic equation for the reaction correct (without the Na, however), but you didn't solve the ICE table correctly.

Actually, we're going to start with simple stoichiometry as a reaction is occurring. To simplify the process, we're going to assume the reaction reacts to completion, and then reverses until it reaches equilibrium.

For this, just use limiting reactants and such to calculate the final concentrations of $\ce{CH3COOH}$ and $\ce{CH3COO-}$.

As I said earlier, we're going to first assume that the reaction reacts to completion, and then goes back on itself and reaches equilibrium. We're at the second part, the equilibrium part. This is where an ICE table comes in.

First, write out the equilibrium reaction. This is pretty simple; it is just the acid dissociation reaction. It's going to look something like this: $$\ce{HA<->H+ +A-}$$

For your ICE table, you didn't really set it up properly. Remember these three things:

You usually won't be writing any unknowns for your initial concentrations. Everything should be known.
You usually won't be writing any numbers for your change. It's an equilibrium. You don't know the exact amount it is going to react to reach equilibrium. That's why we write out the ICE table, so we can calculate how much needs to react for the system to reach equilibrium.
Don't use the Henderson-Hasselbalch equation. Use the $K_a$ and its definition to calculate the concentration of $\ce{H+}$, which leads to your pH.

I know I pretty much just wrote a wall of text, but hopefully that helps nonetheless.


What is the modern hardware/supermarket term for 'laundry bluing'?

The active ingredient is prussian blue which you can make yourself from potassium ferrocyanide and an Fe(III) salt.


How to calculate the energy of a single proton?

An isolated proton will have energy given by the Schrodinger equation (just like any molecular system):

$$
-\frac{\hbar^2}{2\mu}\nabla^2\psi + V\psi = E\psi
$$

($\mu$ is reduced mass; $\hbar$ is $h/2\pi$; $V$ is potential energy; $\psi$ is the wavefunction)

The potential of this single-proton system is zero; the kinetic energy will depend on the momentum of the particle.

If you're using a generic quantum-chemistry software, the Born-Oppenheimer approximation is probably being applied, i.e. the proton ($\ce{H+}$) will be fixed in space with zero momentum. Thus its total energy will be zero, unless you deactivate BO-approximation or introduce another particle.

It's not typical to consider the energy of an isolated proton for obvious reasons (try finding one in nature!). Nonetheless it is calculable if the wavefunction is known.

Another way to consider its kinetic energy is by the classical equation $K = \frac{1}{2} mv^2$; if you consider an approximation of the lonely proton's speed as roughly that of atoms in liquid water, 1 angstrom per picosecond, you obtain ~ $5.03$ kJ/mol. This is similar to the thermal energy available at room temperature, $k_B T$, ~ 2.48 kJ/mol.

You're better off calculating energy of $\ce{H+}$ by getting energy for a molecule, and subtracting the energy of its deprotonated form. In that way you can evaluate proton affinity for a given molecule.


Why is AlCl₃ is known as aluminium chloride while PCl₃ is known as phosphorus trichloride?

Names of binary molecular compounds (which contain only two different elements, normally nonmetals, and do not contain positive and negative ions) can contain the prefixes mono, di, tri, tetra, penta, etc.
Normally, names of ionic compounds do not use these prefixes to indicate the number of positive or negative ions. $\ce{CaCl2}$ is calcium chloride, not "calcium dichloride".

However, aluminum chloride, $\ce{AlCl3}$, is sometimes called aluminum trichloride which is not incorrect in this case because it actually is a molecular compound (it has very polar aluminum-chlorine covalent bonds) even though it looks like it should be ionic since it contains metal and nonmetal elements typical of ionic compounds.
 $\ce{MnO2}$ is often called manganese dioxide rather than manganese(IV) oxide for the same reason. Usually, it is best to play it safe with compounds like these and use the ionic names:

$\ce{FeCl3}$ is ferric chloride or iron(III) chloride rather than "iron trichloride." $\ce{PCl3}$, a molecular compound (expected since only nonmetallic elements are present), is commonly called phosphorus trichloride, but is sometimes called "phosphorus(III) chloride."


Can electrons be present outside orbitals. If yes how does this affect chemical reactions

There are more levels of abstraction involved. At first, orbitals are simplification and they are best described as "one electron wavefunction", ie. area of space where one electron can reside (or two, if you think of spatial orbitals, as opposed to spin-orbitals).

Being expressed as a wavefunction, the orbital (and therefore electron residing in it) has nonzero probability in whole space (I mean, whole universe). But the probability decays very fast.

If you plot a probability density, see image below, you can clearly see the behaviour. If, on the other hand, want to plot the 3d surface, which are often shown in textbooks, you have to state some cutoff value. Often it is probability density of 0.9, ie. 90%. By this you obtain the typical pictures of orbitals.

I would deliberately leave your question without answer, as they stem from severe confusion caused by your textbook.


Is propanone responsible for photochemical smog?

Most of the information is taken from Wikipedia or is mentioned therein

First some definations.

What is a VOC?

Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air, a trait known as volatility. For example, formaldehyde, which evaporates from paint and releases from materials has a boiling point of only –19 °C (–2 °F).

What is a photochemical smog?

Noxious haze of minute chemical droplets and ozone, resulting from sunlight induced chemical reaction between industrial and automobile pollutants (mainly nitrogen dioxide, hydrocarbons and aldehydes).

Indeed, propanone(acetone) is a VOC. Majority of acetone is released from industries origins. It indeed evaporates rapidly, from water and soil and has a 22-day half-life and degraded by UV light via photolysis primarily into methane and ethane or other alkyl, alkoxy or acyl radical.

As said, the main components of photochemical smog is nitrogen dioxide, hydrocarbons and aldehydes. The reaction include breakdown of hydrocarbons and aldehydes into alkyl and alkoxy radicals. These radicals react with nitrogen dioxide to form PAN(peroxyacetyl nitrate) which is the main cause of photochemical smog. The reaction mechanism is given here:

Another mechanism is given here:

What is observed is that the active component is acyl radical which is necessary in the PAN formation. Both aldehyde and acetone necessitate in acyl radical through their decomposition. From here:

  The primary photochemical decomposition of aldehydes produces organic
  free radicals. For ex: acetaldehyde decomposes into methyl and formyl
  radicals.

  $$\ce{CH3CHO + hv → •CH3 + HCO•}$$

  Similarly, acetone decomposes into methyl and acetyl radicals.

  $$\ce{(CH3)2CO + hv → •CH3 + CH3CO•}$$

But acetone contribution to photochemical smog is negligible as acetone is quite stable to irradiation and photolysis. From here:

  [...]Acetone has shown little reactivity to irradiation. Its
  contribution to produce significant amount of PAN or ozone in ambient
  air is negligible.


What is the mechanism of photosynthesis?

I'm definitely not going to outline light harvesting in the light harvesting complexes (LHC), the role of the quinones in electron transfer, etc.

But let me give a short answer on your last question:-

The molecular aggregates, such as the LHCs, involved in the different steps of photosynthesis are not optimized for duration. In contrast to man-made machines, they don't have to! Remember that all these fragile molecular assemblies are part of a living system. If parts break, they will resynthesized and replaced.

While it is partly possible to mimic photosynthesis in the lab, it is the self-repair mechanisms that we are unable to replicate in artificial systems.


How would I write an input Lammps command for continuous change in temperature?

(Disclosure: I'm not a LAMMPS user, but I've used other MD packages.)

You'll probably want to use one of the thermostat settings such as the Nose-Hoover type described here in the LAMMPS documentation, but there's a more general HowTo guide on temperature control and calculation in the docs as well that looks pretty useful.

I would strongly recommend reading the sections of the manual concerning any given thermostat type you're going to use fully before running serious calculations with it, though, since there are some things to look out for, such as the tendency for setting the thermostat interval too low to cause large oscillations in effective temperature.

The Nose-Hoover type (and some of the others, but I only took a quick look) also lets you step up or down the temperature with a given interval. Depending on exactly what type of thermal expansion you're looking at, I'd just like to note: you really need to set the stepping interval appropriately to let the structure equilibrate at each temperature you want to look at, otherwise your results won't be meaningful. (Or, at least, won't have the meaning that you might want.) Relatedly, I'd also suggest that if you weren't planning to already, you run each simulation both ways: one stepping the temperature up, and one bringing the temperature down. If the behaviour matches at the temperature steps in each, you can be relatively confident that you've given the structure enough time at each temperature.


Software for viewing a structure and altering bond angles

Have a look at Avogadro!

build your own structures or import files in various formats.
optimize the geometry using different force fields
change and measure bond length, bond angles and dihedral angles

Once you've used it, you don't want to miss it again.

Geoff Hutchison, who cannot be praised enough for his child Avogadro, has given the right directions in his comment. I'm just adding some images to illustrate how easy angle adjustment is:

Draw a molecule in Avogadro
From the main menu, select Build -> Properties -> Angles or whatever the localized menu of your system shows (on my German machine, it is Ansicht -> Eigenschaften -> Winkeleigenschaften)
In the table that pops ups, select the angle you want to change. Note that the atoms of the selected angle are highlighted

Change the value for the angle. Note that the geometry of the molecule is immediately updated!

So again, I think that Avogadro is one of the best tools available and it is free and open source! Use it in education, use it in research and whenever you do, cite the Avogadro paper:
Hanwell, M. D.; Curtis, D. E.; Lonie, D. C.; Vandermeersch, T.; Zurek, E.; Hutchison, G. R.; Avogadro: an advanced semantic chemical editor, visualization, and analysis platform. J. Cheminf. 2012, 4 (1), No. 17. DOI: 10.1186/1758-2946-4-17.


Determining the concentration after half-life period

I'm afraid you are mistaken. The DNPA essentially turns into DNP in the course of the reaction, so the answer is $84\ \mu M$, assuming the initial DNP concentration is zero. Formally, the stoichometric coefficient of DNPA is -1, which illustrates nicely that the sum of the amounts of DNPA and DNP is constant in this system.


What happens to a molecule when it is unstable?

Think of a reaction-energy diagram.  Specifically, think of a two-step reaction energy diagram linked like this one:

.

The carbocation is the intermediate $\mathbf{I}$.  The relative energy for alkyl-substituted carbocations will be lower.  This also practically means that $\mathrm{TS1}$ will have a lower relative energy as well.  The fact that $\mathrm{TS1}$ and $\mathbf{I}$ are of lower energy for a substituted carbocation (vs. a primary or methyl carbocations) means that it will be formed more often under the quasi-equilibrium assumption of transition state theory.  Basically, if the barrier is lower, then then the intermediate can form more often.  And thus the reaction can happen more often.

  I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation?

I think you are correct as far as your thought goes.  If you could somehow instantaneously create a beaker full of primary carbocations, they might well react more quickly than other carbocations.  (And that's not only via the SN1 reaction we are considering but also via other pathways such as elimination as well.)  But instantaneously creating a beaker full of primary carbocations is way more difficult than creating a corresponding beaker full of secondary or tertiary carbocations.  So (referring to the diagram again) yes, iff you have lots of $\mathbf{I}$, for $\mathbf{I}=$ primary carbocations, reactions would probably be quicker than the case where $\mathbf{I}=$ higher carbocations.  But in reality in the primary carbocation case essentially no $\mathbf{I}$ is ever generated.  So the overall reaction rate is slower.


Iodometric titration of copper – standard electrode potentials


How can magnesium hydroxide be a strong base and be sparingly soluble at the same time?

I think we should not conflate "solubility" with "dissociation". They are different phenomena. The former has to do with a molecule breaking away from the other solute molecules and forming interactions with the solvent. The latter is a process which occurs after dissolution, whereby the solute molecules break apart to form ions, separately solvated by the solvent molecules. For a simple covalent substance, the two phemomena are clearly different.
For an ionic solid, the two may seem to occur together since when a soluble ionic salt is placed in water, its ions are in fact pulled apart by water molecules. In this sense, it is dissolving and dissociating at the same time. Hence, by the Arrhenius definition of acids and bases, insoluble hydroxides such as $\ce {Mg(OH)_2}$ may not even be considered to be strong bases. Arrhenius theory would tell us that the strong base is the hydroxide ion, that is produced from the dissolultion and dissociation of the salt.
Actually, in particular for some metal oxides such as $\ce {MgO}$, the first reaction it undergoes upon contact with water is $\ce {MgO + H2O -> Mg(OH)2}$. Hence, we can rightfully say $\ce {MgO}$ is not an Arrhenius base, but it is the precursor to the weak Arrhenius base $\ce {Mg(OH)2}$.
I would like to disagree with the part of the answer you have referenced in your post because as aforementioned, dissolution and dissociation are the same phenomena for ionic salts. It is thus not correct to say that "what does dissolve is still $\ce {100}$% ionised into [ions]" and hence, say that insoluble (or sparingly soluble) ionic salts are strong bases.
To generalise, Arrhenius theory would consider soluble ionic hydroxides such as $\ce {NaOH (s)}$ to be strong bases while it would consider sparingly soluble ionic hydroxides such as $\ce {Ca(OH)2}$ to be weak bases. In other words, the strength of these metal hydroxides as bases is directly linked to their solubility and this makes perfect sense since dissolution and dissociation for an ionic salt are the same process.


Does almond extract made from bitter almonds contain any cyanide?

Crude extracts from bitter almonds or apricot seeds do contain amygdalin, which is hydrolyzed in the human body and releases cyanide.

Food-grade bitter almond extract does not contain amygdalin, because the crude extract is typically treated with a with a mixture of calcium hydroxide and $\ce{FeSO4}$. In this process, the amygdalin is decomposed and the cyanide is extracted in the form of $\ce{Ca2[Fe(CN)6]}$.

The typical flavour of the resulting product stems from benzaldehyde.


Facial selectivity of iodonium formation in an iodolactonisation

Here, it does not matter which face is preferred. The formation of the iodonium ion is reversible and it can be formed on both faces. However, the iodonium ion can only be attacked by the carboxylate if it is formed on the opposite face from the carboxylate, which is the bottom face in your diagram. Attack on the top face is said to be unproductive because it cannot lead to iodolactonisation.


Reaction mechanism for oxidation of primary alcohol to carboxylic acid

A dichromate salt and aqueous sulfuric acid are used to prepare chromic acid in situ, which is among the most powerful oxidizing agents commonly available. A primary alcohol is first oxidized to an aldehyde (via a chromate ester intermediate). Under acidic conditions, the aldehyde's oxygen acts as a base and abstracts a proton, the carbonyl carbon is attacked by water (whose oxygen acts as a nucleophile) and simultaneously the carbonyl π-bond breaks, and then finally the resulting molecule is deprotonated by another water molecule, yielding a diol. That diol is then subject to further oxidation by the remaining chromic acid, which occurs by the same mechanism as the initial oxidation to an aldehyde. The ultimate result is a carboxylic acid.


Does hyperconjugation stabilize a carbocation more than a free radical?

A carbonation is deficient in 2 electrons whereas a free radical is only 1 electron deficient. Thus any type of stabilisation working on a carbonation will be much more effective as it is more electron deficient. Hence, hyperconjugation or inductive electron donation stabilises a carbocation more than it does a free radical.


Why is 1-bromotriptycene inert to nucleophilic substitution?

Sn1 reaction proceed through carbocation intermediate which has planar structure. The structure of 1-bromo triptycene  is cage like structure so the bridgehead carbon cannot assume planarity , hence,the formation of a carbocation at the bridgehead position does not take place. Consequently 1-bromotriptycene is inert to sn1reaction.


What is the function of PhLi in this reaction?

I was wrong, this is not a Wittig reaction. Just posting the answer to let anybody who comes in hunt for this in the near future.

On googling for about an hour, I discovered that this is indeed a preparation of phosphonium ylide which is the witting reagent.

On adding $ \ce{Ph_3P} $ to 1,2-dichloroethane , $\ce{PhLi}$ acts as an catalyst and eliminates $\ce{HCl}$ from both the carbons resulting in a phosphonium di ylide.


How to calculate what temperature change in entropy for the reaction equal to the change in entropy for the surroundings given Enthalpy and entropy

You're looking for the equation of Gibbs free energy:
$$\Delta G^\circ =\Delta H^\circ - T\Delta S^\circ$$
Per Wikipedia's Gibbs free energy page:

The equation can be also seen from the perspective of the system taken
together with its surroundings (the rest of the universe). First
assume that the given reaction at constant temperature and pressure is
the only one that is occurring. Then the entropy released or absorbed
by the system equals the entropy that the environment must absorb or
release, respectively. The reaction will only be allowed if the total
entropy change of the universe is zero or positive. This is reflected
in a negative ΔG, and the reaction is called exergonic.

So effectively you're trying to deduce the $T$ that produces $\Delta G = 0$ which is the "tipping point" (then the reaction will occur for $\Delta G < 0$)
So you solve for $T$ in:
$0 = \Delta H - T \Delta S$
Thereafter as $T \rightarrow \infty$ you see that $\Delta G$ keeps decreasing, making the reaction more favorable.
Thus $S_\textrm{system} = S_\textrm{surroundings}$ is a fancy prerequisite for reaction occurrence, which we commonly associate with $\Delta G \le 0$.


Determining stoichiometry in ICP-OES data analysis

Let R be the ratio of number of Na atoms to number of Fe atoms. Then, for the first compound, we have

$$\pu{R = }\frac{\pu{112300 ppm } /\ \pu{22.98977 g mol^{-1} } }{\pu{357200 ppm } / \ \pu{55.845 g mol^{-1} } } = \pu{0.7637  }$$

Since there are two Fe atoms in the compound, the Na number for the first compound is simply 1.527, i.e., twice the value of R calculated above.

Following the same procedure for the remaining three compounds, I get Na numbers of 1.662, 1.731 and 1.635. The last one is furthest from the value of 1.68 the authors published, but maybe 1.63 was read as 1.68.


How many gauche butane interactions are present in cis-1,2-dimethylcyclohexane?

While cyclohexane does have gauche interactions, they are irrelevant because chair cyclohexane is taken as the reference point. The axial methyl group in cis isomer 1 has two gauche interactions with the ring as illustrated in red in 1a and 1b. A third gauche interaction is between the two methyl groups in 2. By comparison the trans isomer 3 only has gauche interactions between the methyl groups.

The difference in the number of gauche interactions - namely, two -  between these epimers is approximated by the difference in their heats of combustion[1] which is $\pu{1.54 kcal/mol}$. [1,2 cis: $\Delta H^\circ_\mathrm{comb} = \pu{-1248.31 kcal/mol}$; 1,2 trans: $\Delta H^\circ_\mathrm{comb} = \pu{-1246.77 kcal/mol}$.] A typical value for a gauche interaction is $\pu{0.9 kcal/mol/interaction}$.
References:

Walter H. Johnson, Edward Prosen, and Frederick D. Rossini, Heats of Combustion and Isomerization of the Eight $\ce{C8H16}$ Alkylcyclohexanes. Research Paper RP1S12, Volume 39, July 1947. PDF via NIST


Relation between electron gain enthalpy and electron affinity

You are correct, at absolute zero $Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e}$ for the gas-phase act of gaining an electron:

$$\ce{X(g) + e-(g) → X-(g)}$$

$$A_\mathrm{e} = E(\ce{X(g)}) - E(\ce{X-(g)})$$

The term $5/2RT$ arises from the so-called "electron convention" when electron is treated as ideal gas with corresponding heat capacity $C_p$ from Boltzmann statistics and the element of enthalpy

$$H_T(\ce{e-}) - H_0(\ce{e-}) = \frac{5}{2}RT,$$

resulting in exact relation

$$Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e} - \frac{5}{2}RT.$$


You could try the mechanism behind mist fountains or other "cold fog" machines:

These use an ultrasonic oscillating diaphragm to produce a fine mist of room temp water, no heating performed. The resulting vaping machine would be larger and require more power, but wouldn't have the harmful aldehydes inherent in current vaping liquids. Nicotine itself is hygroscopic but not really water soluble at room temp, but it can form salts that are water-soluble so it could be introduced to the body this way. Various flavors currently used might be more or less water-soluble. Finally, the fog, being simple water vapor, wouldn't be visible when exhaled which is one reason the glycerol compounds are used; they produce a "smoke" that dissipates more quickly than tobacco smoke.


Can a first order reaction change to a second order reaction under specific temperatures?

It's theoretically possible, though I don't know if there exists specific examples.

The first thing to recognize is that reactions don't really "switch" in an on/off fashion (at least not based on temperature). If there's two physically possible reactions, they both occur under all conditions ... theoretically. The main reason you see one and not the other is due to the comparative rates of reaction - one reaction might be so slow that no appreciable amount of product forms in reasonable timeframes.

So while the rates of reactions change based on temperature, as Zhe mentions, the way that rate changes in most cases (the Arrhenius equation) means that the two reactions aren't going to swap positions based on those considerations alone.

However, kinetics is not the only knob we're playing with here, when we change temperatures. We're also messing with the thermodynamics. Specifically, temperature is a factor in the free energy equation

$$\ce{\Delta G = \Delta H - T \Delta S }$$

If the two reactions have different $\Delta H$ (enthalpy) and $\Delta S$ (entropy) values, then we can change the spontaneity of the reactions. At low temperatures, something with a very favorable enthalpy will be spontaneous, even with a disfavorable entropy. Conversely, at high temperatures, something with a very favorable entropy will be spontaneous, even with a disfavorable enthalpy.

So if you have a system which can undergo first-order and second order reactions (for example, a molecule which has ) and choose your reactant such such that the first order reaction has a more favorable enthalpy and a less favorable entropy than the second order reaction (or vice versa), there will be a temperature where the thermodynamically favored product will go from that of the first-order reaction to that of the second order reaction. Whether this would be enough to manifest itself as an appreciable difference in product depends on the particular numbers.

While we're on the discussion, there is another technique related to controlling product outcome, that exploits the balance between the two factors already discussed. That's using thermodynamic versus kinetic reaction control. In this setup, you have one reaction which provides a more stable product, but due to things like steric hindrance the reaction rate is lower.  The other product forms faster, but the product is less stable. If you set up reaction conditions such as short reaction times at high temperature where the speed of the reaction is the dominant factor, the faster-made "kinetically controlled" product dominates. However, if you use conditions such as long reaction times at low temperatures, where the stability of the products is the dominant factor, the more stable "thermodynamicly controlled" product dominates.


How do you calculate the norm of a function describing a molecular orbital?

I'm unsure if you're meant to solve this via the Hückel molecular orbital theory (see Chemistry LibreTexts), however, if you need to calculate the norm of a wave function $\Psi$, this is $|\Psi |$. Given what we know in your post about both $\Psi_{1, 2}$ then
\begin{align}
  |\Psi_1|^2
    &= \langle\Psi_1 | \Psi_1\rangle \\
    &= 2\cdot(0.602^2)+2 \cdot (0.372^2) \\
    &= 1.001576 \\
  \to |\Psi_1| &= 1.000788 > 1\\
\end{align}
I write that this is greater than one, as, on closer inspection,  is down to rounding from the coefficients of each $p_z$ orbital.

The repeat process can be found for the overlap part, which I am interpreting as $\langle\Psi_1 | \Psi_2\rangle$.


Your teacher is correct, that

  a carbocation connected to 3 cyclopropane rings is among the most stable carbocations.

This because of various electrical effects, such as:

The Inductive Effect
Resonance
The Mesomeric Effect
Electromeric Effect
Hyperconjugation

Here's a link for you, to study them: Click Here.
Please make sure that you learn this correctly and from standard resources, in order to avoid confusion.

There may be a little confusion between mesomeric effect and resonance effect, so here's another link just to avoid confusion. : refer to the answer by Maria Dar

Please, do ask your teacher to help you understand this topic better :-)


How to find the mass of the solute given a concentration as a mass fraction?

You have been given a mass fraction, $w_{\ce{CH2O}}=37\%~\mathrm{w/w}$, and the mass of the solute, $m_{\ce{CH2O}}=250~\mathrm{g}$. You calculated the mass of the solution wrong.
The mass fraction is defined as
$$w_i = \frac{m_i}{m}\tag1$$
with $m$ being the total mass and therefore
$$m = \sum_i m_i.$$
The latter is also the mass of your solution, the sum of the solute and the solvent.
$$m=m_{\ce{H2O}}+m_{\ce{CH2O}} = m_{\ce{H2O}} + 250~\mathrm{g}$$
Substitute what you know into $(1)$ and transform:
\begin{align}
w_{\ce{CH2O}} &= \frac{m_{\ce{CH2O}}}{m_{\ce{H2O}}+m_{\ce{CH2O}}}\\
0.37 &= \frac{250~\mathrm{g}}{m_{\ce{H2O}} + 250~\mathrm{g}}\\
m_{\ce{H2O}} &= \frac{250~\mathrm{g}}{0.37}-250~\mathrm{g} = 426~\mathrm{g}
\end{align}


Which carbon(s) in cyclohexene have the highest electron density?

:-( The book gives an answer, not necessarily the right one.

Looking at the $\ce{^13C}$ NMR of cyclohexene the carbons at positions 4 and 5, in your diagram, are shifted the least and therefore have the most shielding. Thus those carbon atoms also have the greatest electron density around them.


In a galvanic cell,do charges accumulate on electrodes 1st (then redox occur) which produce E in the wire when we connect between electrodes?

You can imagine the cell electrodes as capacitors, that are charged by chemical way.

But the capacitance of these capacitors

$$C=\frac {\mathrm{d}q}{\mathrm{d}E}$$

is very small, so does the accumulated charge $q$.

The charge loses due pushing electrons  through the wires of a closed circuit are balanced by simultaneous recharging by ongoing redox reactions.

If we consider the zinc electrode of the Daniell cell, the reaction :

$$\ce{ Zn -> Zn^2+ + 2 e-}$$

is charging the electrode negatively, while the opposite reaction:

$$\ce{  Zn^2+ + 2 e- -> Zn}$$

is charging the electrode positively.

The electrode gains such a potential, where both reactions are ongoing at the same rate.


How to calculate the frequency of cyclohexane conformational interconversion?

Look up application of the rate of reaction equation given barrier height and energy available at room temp, $kT$ (Boltzmann's constant).

$k = A \mathrm e^{ - \frac{E_\mathrm a}{RT} }$

http://research.cm.utexas.edu/nbauld/teach/cyhexconf.jpeg

Twist-boat leads to mathematically perfect geometric chirality,
Symmetry: Culture and Science 19(4) 307-316 (2008).


You could calculate an equilibrium constant at a given temperature for a reaction from the standard free energy change ($\Delta G^{\circ}$) for any reaction using:

$\Delta G^{\circ} = -RT \ln K$

$\Delta G^{\circ}$ can be calculated for reactions using tabulated standard free energies of formation ($\Delta G_{f}^{\circ}$) using the following equation:

$\Delta G^{\circ} = \sum{\Delta G_{f}^{\circ} \mbox{(Products)}} - \sum{\Delta G_{f}^{\circ} \mbox{(Reactants)}}$

If the reaction only contains solids and liquids, then K will still depend on the ratio between the product and reactant concentrations.  However, these concentrations will not change as a reaction moves towards equilibrium, which implies that the reaction quotient ($Q$) remains constant


How does ferrous iron coordinate nitrogen in a porphyrin ring?

This is something I have asked myself for a long time until I finally realized by myself. I have superficial knowledge on this type of chemistry but I believe this will be enough, so i will explain it in a very simple fashion. Well you have to look at Nitrogen first, nitrogen doesn’t change it’s charge just like you said, which seems weird at first since it forms four bonds. Now think of nitrogen in the lewis formula, what does it look like? Nitrogen has a free pair of electrons available. Think of the d-orbitals in Fe, their character and shape. Think of metals in general, having their electrons in a type of diffuse cloud. Now to the bond, the Iron is capable of grabbing these free electron pairs and use them in a bond, pulling them into it’s direction. This way the charge doesn’t change. The nitrogen “donates” it’s electrons which do not participate in regular bonding like you would see it in ammonia, but in complex bonds. Hope this is enough to clarify :)
Greetings from Germany


When temperature is decreased, why do reactions occur at all?

The activation energy varies a lot for various reactions. As long as T > 0 K, there is still disordered kinetic energy in the molecules. But the reason gasoline doesn't spontaneously combust, is that the temperature isn't high enough to overcome the activation energy - a spark is needed. So you are onto something, just remember that the activation energy varies with each reaction.


How can I emulsify Beeswax with Aloe Vera Gel?

Aloe vera gel is water based, whereas beeswax is, well, wax-based. As they say, "oil and water don't mix" -- at least, not on their own.

So, the trick is to make them willing to mix. To do this, you need what's called an emulsifier. These are chemicals (natural or synthetic) that have both oily and watery parts to them, and thus are able to make oil and water play nice(r) with one another.  Soap is one well-known, everyday example of an emulsifier; in particular, it's an example of a surfactant.

Anyways, you need an emulsifier. If you're looking to stick with natural products, probably the most readily available option is lecithin, commonly sourced from soy but also available derived from other plants such as sunflower.  While I don't have detailed recommendations, I would melt the beeswax together with lecithin and the aloe vera gel over very low heat, mixing thoroughly, then allowing the blend to cool.  You'll have to experiment until you find a ratio among these that gives the consistency you want in the final product.


How does absence of nd orbital in Oxygen affect its valency?

Whether or not oxygen can reach oxidation states higher than $\mathrm{+II}$ is completely independent of the type of orbitals the second shell has. If the orbitals would matter, nitrogen would not be able to reach every single oxidation state between $\mathrm{-III}$ and $\mathrm{+V}$ but would also be restricted to <insert arbitrary number here>.

The actual reason why oxygen’s accessible oxidation states are limited is its very high electronegativity, second only to fluorine among the non-noble gas elements. There is simply no partner that can draw away electrons strongly enough to access high oxidation states. Fluorine itself, while typically being able to access high coordination numbers (higher than oxygen, compare $\ce{XeF6, NF3}$ or $\ce{SF6}$ to $\ce{XeO4, NO3-}$ or $\ce{SO3}$) is not the element of choice for accessing high oxidation states — that is oxygen. But oxygen cannot oxidise itself since it has the same electronegativity. Fluorine itself, as is evident by $\ce{NF5}$ being unknown as of today, would not be able to stabilise an oxidation state of more than $\mathrm{+II}$ on oxygen given our current knowledge.

Furthermore, the statement’s claim that group 16 elements except oxygen only exhibit positive oxidation states is bogus. $\ce{Na2S}$, $\ce{Na2Se}$, $\ce{Na2Te}$ and $\ce{Na2Po}$ are all known and in all of these compounds the chalcogen has an oxidation state of $\mathrm{-II}$.


Determination of metal state from protein X-ray structure

X-ray diffraction experiments, regardless if about small molecules or proteins / RNA / DNA, are usually one element characterizing a sample.  On occasion, you may infer the oxidation state from the transition metals environment: how many closest neighbors, their distance, their number, their spatial arrangement.
You are free to complement findings by any other technique either using the same crystal sample as mounted on the diffractometer (e.g., UV-Vis), or with sample material from the same batch characterized at the same temperature as the crystal mounted on the diffractometer: spin characterization by EPR, Moessbauer, NMR (your are not constrained to CHN) and SQUID  to mention a selection. Because PubMed includes records about the structure characterization of large biomolecules, this database may serve as a source of inspiration how this was done, e.g., for heme


Why does the dipole moment order of group 15 hydrides increase down the group?

The query you have is a common mistake, and this is aksed as a very common question in examinations, well, I wouldn't blame you as the reason is simple but not easy to think of.

What you said about E.N. of phosphorous being more than arsenic and antimony is absolutely correct. But what you missed, is that lower down the group the direction of dipole itself reverses.

E.N. of $\ce{P}$ is $2.19$, while that of $\ce{As}$ is $2.18$ and of $\ce{Sb}$ is $2.05$

Whereas H as an E.N. of $2.20$. So hence now you can see clearly that your order is justified, as in actuality, H becomes the negative end of the dipole.

Another factor that further widens this gap is that the dipole vector in chemistry notations is from + to-, which means all vectors add up in case of $\ce{As}$ and Sb


Can glucose and insulin exist in the same container without reacting?

Glycoproteins come in O-linked and N-linked forms. For O-linked sugars, a glycosidic bond forms between the sugar and the hydroxyl of a serine or threonine side chain. For N-linked sugars, a glycosidic bond forms between the sugar and the amide of asparagine. The equilibrium for all these reactions lies on the side of hydrolysis in aqueous solution.

  Is it possible to make a reasonable prediction what would/not happen in the mixed glucose/insulin solution in the container?

Yes, in aqueous solution there is no expectation that a glycoprotein forms between insulin and glucose. In vivo, glycoproteins form in a coupled reaction involving the carbohydrate (usually an oligosaccharide rather than a single glucose), the protein and a nucleotide triphosphate.

However, if an individual has high blood sugar levels (i.e. high concentrations of glucose in the blood) for extended periods of time, hemoglobin (specifically, the N-terminal amino group and the sidechain amino group of lysine) will react with the sugar and become glycalated. This is used in the HbA1c diagnostic test for checking if someone has diabetes. After the sugar binds (reversibly), there is a further reaction (Amadori rearrangement) that leads to a long-lasting product.


Quantitative MO theory- how is the overlap integral and the weighting coefficients of the molecular orbital actually evaluated?

In essence, one picks a computational level, chooses a system to run the calculation on (typically a molecule) and lets a computer and an appropriate program do the heavy numeric and algebraic lifting. No really, that's it.

In more detail: One would want to solve the multi-electron, multi-nucleus Schrödinger equation (that's already an approximation, because we ignore relativistic effects here). That's really difficult. So we make more approximations, such as fixed nuclei, a product ansatz for the wave-function (made up of one-electron wave-functions) and often a finite-size basis set. A basis set are functions that can be linearly combined into molecular orbitals; a large one usually gives better results, but make the computation longer. This choice is part of the computational level. Other, more mathematical and technical choices make up the other part (and this is a science in itself).

For a molecule like $\ce{C2H6}$, the computational time can vary between several days to weeks and less than a second, depending on sophistication of the computational level and its accuracy. In the end, we obtain an approximate wave-function (containing molecular orbitals), from which approximate properties such as energy or NMR shifts can be obtained. The computer is useful in this by calculating integrals (partially analytic, mostly numeric) and keeping track of many very large matrices.

At times, the problems can be worked out using pen and paper, making use of high symmetry and tiny basis sets. This leads to easily interpreted, but not very accurate wave-functions.


What does it mean when "a gas is dilute in CO₂"?

In this case, it's comparing post-air-combustion $\ce{CO2}$ capture with pre-combustion capture, and with post oxy-combustion capture.

In the two latter cases, the  $\ce{CO2}$ stream is very highly concentrated.  However, with air-combustion, the flue gases contain a lot of nitrogen and NOx; hence the $\ce{CO2}$ is diluted in the flue gas, compared to the other two methods, where it's concentrated.

And as to Leonardo's supplementary question in the comments, about why the pre-combustion  stream has higher concentrations of $\ce{CO2}$ than the post-air combustion stream:

the comparison here isn't between two points in the same process. It's between analagous points in two completely different processes.

With post-combustion capture, the $\ce{CO2}$ is stripped out after the hydrocarbons (coal, gas, oil) have been burnt in air.

Whereas with pre-combustion capture, the hydrocarbon goes through a gasifier/shift process, resulting in a stream of $\ce{H2}$ mixed with $\ce{CO2}$. And at this point, the $\ce{CO2}$ is stripped out, and the $\ce{H2}$ is burnt.

And that's why the $\ce{CO2}$ concentration is higher in the pre-combustion capture process, than in the post air-combustion capture process.


Friedel-Craft reaction with amide substituent

-NHCOCH3 is a slightly activating group due to the lone pair of N which shows +M effect. Hence it is Ortho para directing.


C=O bond length comparison among cyclohex‐2‐en‐1‐one and cyclohex‐3‐en‐1‐one

The case 1 has partial double bond character (due to resonance) while case 2 has total double bond character of the C-O bond. So, relative to case 2, the C-O bond length in case 1 is longer.


What is the difference between strontium-86 and strontium-90?

For all practical purposes, Sr90 is chemically identical to Sr86. However, the extra neutrons crammed into Sr90 make it radioactive.

As mentioned in the comments above, alkaline earth metals (e.g. Ca, Sr, Ba) behave chemically in a similar fashion, and tend to deposit in bone like calcium does. This keeps the radioactive Sr90 in contact with the rapidly-growing hematopoietic material in bone, increasing the risk of cancers such as leukemia.


Is enthalpy of combustion the 'heat released' or 'enthalpy change'?

The enthalpy of combustion (aka heat of combustion) is defined as the enthalpy change in going from pure reactants (each at 1 bar pressure) to pure products (again, each at 1 bar pressure) at 298 K.

In this problem, it is first of all necessary to assume ideal gas behavior to proceed.  For ideal gases, the change in enthalpy in going from pure reactants each at 1 bar pressure to the starting gas mixture in the calorimeter, there is no change in enthalpy at constant temperature.  Then, application of the first law of thermodynamics to the mixture before and after the reaction leads to the conclusion that $\Delta U=Q$, since the reaction volume is constant.  Since the temperature change between the beginning and end of the combustion in the calorimeter is essentially zero, this means the the calorimeter has measured the internal energy change of the reaction.  As Buck Thorn has pointed out, the change in enthalpy for the gases in the calorimeter is given for an ideal gas by:
$$\Delta H=\Delta U+\Delta (PV)=\Delta U+RT\Delta n$$where $\Delta n$ is the stoichiometric change in the number of moles of gas.  He also correctly pointed out that there is nothing said in the problem statement about the stoichometry of the reaction or the change in the number of moles of gas.  So, as he indicated, the problem statement is flawed.  However, if we arbitrarily assume that the change in the number of moles of gas is zero, we would have to conclude that the change in enthalpy for the process in the calorimeter is equal to the change in internal energy.  Then, since, again, the products form an ideal gas mixture, there is no change in enthalpy in going from the final gas mixture in the calorimeter to the pure products at 1 bar pressure.  So, based on these assumptions, the enthalpy of combustion is equal to the internal energy change in the calorimeter (also Q) per mole of species being burned.


Can i use deionized water instead of distilled water

Water from a dehumidifier is similar to distilled, in that it contains little in the way of dissolved materials.   It condensed on (typically) anodized aluminum,
and dripped through polymer collection trays into a bucket.

How clean is the bucket?  There may be residue from dust/lint etc.

It's water.   You can moisten plants, fish can swim in it, and it's pure enough
to drink or wash your windows (in fact, it's softer than most tap water, it
makes a good final rinse for windows).  I'd stick to deionized or distilled
for lab use, though, because those didn't come from a dirty bucket.


Removing Heavy Rust from Very Old Carbon Steel Carving Knife on the Cheap

First, read my answer here where I cite the electrochemistry background. The latter include source links to galvanic corrosion and, in particular, in the current instance, a classic case of pit corrosion.
Stop long-term soaking, as this in the presence of salts and an H+ together with dissolved oxygen will just further promote the slowly evolving electrochemistry.
I, personally, ignored stake-knives left in a sink (water and oxygen exposure) that were used to cut a mango. The sticky mango juice, rich in Ascorbic acid (Vitamin C) is apparently particularly notorious for promoting pit corrosion.
Never have completely removed the stains, so do not have high expectation here,