How can you tell when a molecule forms a lattice

There is no easy way to guess or predict the answer in general, but experience will give you some clues

The way to tell whether something is a molecule or a covalent solid (Made from a lattice) is to look up the structure. When you first learn chemistry, some things have just have to be learned. But when you learn a few things, some patterns will emerge and you will get better at predicting the properties of other compounds.

The structure will usually tell you the answer. Other physical properties of the compound can hint at the structure. Carbon dioxide is a gas so you should be able to predict it is a small molecule; silicon dioxide is a refractory solid: this is a big clue that it is not a small molecule but is, in fact, a network solid mad from a lattice of Si-O bonds. Nitrogen and Oxygen are gases, so most likely small molecules; graphite and diamond (both forms of pure carbon) are solids so  out likely networks made from large lattices.

There isn't really a simple way to predict this, but once you have learned a few examples, patterns will emerge.


How can I clean solid sulfur and its oxides from test tube?

Since you stated:

  I'm not a chemist, so I don't have any strange chemicals at home.

you will neither have facilities for their proper storage and  waste management.

Consequently, the only reasonable solution is: Get some new test tubes (about 0.20 Euro for a 160 x 16 mm tube), but don't risk your own health or pollute the environment by messing around with carcenogenic ($\ce{CrO3}$ in $\ce{H2SO4}$) or neurotoxic ($\ce{CS2}$) agents.


The $K_\mathrm{sp}$ of $\ce{Fe(OH)3}$ is varied from source to source, but a reliable source gives the value of $2.79 \times 10^{-39}$ at $\pu{25 ^\circ C}$. We'll use this value throughout the calculations.
Suppose, $s$ amount of $\ce{Fe(OH)3}$ dissolves some in water according to its $K_\mathrm{sp}$, but assume water is not ionized:
$$\ce{Fe(OH)3(s) <=>[H2O] Fe^3+(aq) + 3OH-(aq)}$$
$$K_\mathrm{sp} = s \times (3s)^3 = 27s^4 \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{27}\right)^{\frac 14} = 1.008 \times 10^{-10}$$
This is $[\ce{Fe^3+}]$ in water. Thus, $[\ce{OH-}]$ would be $3s = 3.024 \times 10^{-10}$.  This value is much smaller than autoionization values of water. Therefore we have to consider $[\ce{OH-}] = 1.00 \times 10^{-7}$ for this calculation. Hence, if revise the calculation accordingly:
$$K_\mathrm{sp} = s \times [\ce{OH-}]^3 = (1.00 \times 10^{-7})^3s \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{1.00 \times 10^{-21}}\right) = 2.79 \times 10^{-18}$$
Thus, when autoionization of water is considered, $[\ce{Fe^3+}]$ in water at $\mathrm{pH} \ 7$ is much smaller (the actual value). I did these calculations to show the effect of autoionization of water. Similarly, when real $[\ce{Fe^3+}]$ is high in the solution like the case (a) here, $K_\mathrm{sp}$ plays a role.
Let's re do the calculation again considering $[\ce{Fe^3+}] = 0.037 M$. If all $\ce{Fe(OH)3}$ stay dissolved, them we can find the $[\ce{OH-}]$ in the solution using $K_\mathrm{sp}$ calculations:
$$K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3 =  0.037 \times [\ce{OH-}]^3 \ \Rightarrow \  \therefore \ [\ce{OH-}] = \left(\frac{2.79 \times 10^{-39}}{0.037}\right)^{\frac 13} \\ = 4.22 \times 10^{-13}$$
Thus, you need to keep $[\ce{OH-}] = 4.22 \times 10^{-13}$ to keep the solution away from precipitating back to $\ce{Fe(OH)3}$. Thus, $[\ce{H+}]$ should be at:
$$ [\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} = \frac{1.00 \times 10^{-14}}{4.22 \times 10^{-13}} = 0.024$$
Thus, after completely dissolving rust, you need to add additional $\ce{H2SO4}$ to keep the $[\ce{H+}]$ of the $\pu{100 mL}$ solution at $\pu{0.024 M}$. I think you can calculate that amount in $\pu{mmol}$ easily now (Keep in mind that you already added $\pu{5.6 mmol}$ of $\ce{H2SO4}$ to dissolve $\pu{0.4 g}$ of rust).
Note: Any deviation from the given answer should be due to the $K_\mathrm{sp}$ value used here (The given answer didn't give that numerical value).


Equivalencies and catalytic hydrogenation

Using the equipment that I have and working on the scale that I tend to work on (<1 mol typically), I can't deliver only one equivalent of hydrogen gas. We can measure the amount of hydrogen consumed as the reaction proceeds. The problem you have shown could be interpreted that way, i.e. 1 mol is consumed but more may be present.

On industrial scale it may be possible to deliver a more accurate amount of hydrogen. I'll leave it to others that might have more experience on large scale to confirm or deny.

There are alternative sources of hydrogen that can more easily be delivered accurately on small scale. Either ammonium formate or cyclohexadiene can be used to generate hydrogen in the presence of palladium. The former liberates ammonia and carbon dioxide as byproducts, while the latter forms benzene as byproduct.


How many hydrogen atoms are in a sample of NH3?

Before you use log you have to multiply the answer of $8.66~\mathrm{mol}$ by $6.02 \cdot 10^{23}$ by 3 because the answer of $8.66~\mathrm{mol}$ by $6.02 \cdot 10^{23}$ gives you the number of ammonia molecules and in each molecules there are 3 hydrogen atoms present.

So it is:

$$8.66 \cdot 6.02 \cdot 10^{23} = 5.21332 \cdot 10^{24}$$

$$5.21332 \cdot 10^{24} \cdot 3 = 1.563996 \cdot 10^{25}$$

$$\log(1.563996 \cdot 10^{25}) = 25.19$$


The Comprehensive Handbook of Chemical Bond Energies might be able to help.  It walks through bonds in different functional groups.  It can be found at http://www.ebah.com.br/content/ABAAABH2cAD/comprehensive-handbook-of-chemical-bond-energies-0849373662


is it possible to make ethanol from ethane through a chlorination / hydrolysis two-step procedure?

Yes, it is possible.

Although ,commercially(and ironically!) alkyl halides are produced from the corresponding alcohols.
The main reason why this is so because free-radical chlorination requires high temperatures and catalysts to promote the reaction.

Theoretically, the first reaction will give good yields(and not a mixture unlike most free radical halogenations because of the simple reactant involved)

Edit:As pointed out by bon, I forgot to take into account that because its a free radical reaction, the number of possible products is a lot.So no, the first reaction will not give you good yields.


Hybridisation of complex compound having octahedral geometry

If you calculate the oxidation state of the central metal atom in both of these cases, you'll find it to be $+3$. With $\ce{Co^3+}$, all ligands behave as strong field ligands except in the cases of $\ce{[CoF6]^3-}$ and $\ce{[Co(H2O)3F3]}$. Thus, in both of the cases you've mentioned, the ligands will cause pairing of electrons, the complex will be of a diamagnetic nature and the hybridization will be $\ce{d^2sp^3}$.


Any possible additives to a drain cleaner I use as a source of NaOH?

The product you've described contains the following ingredients (per the ingredient data sheet on the manufacturing company website):

Sodium hydroxide
Sodium chloride
Aluminium
Mineral oil

One possibility is that the aluminium is there to provide a foaming action, as its oxidation in the presence of aqueous sodium hydroxide liberates $\ce{H2}$ gas, per the following reaction (source):

$$\ce{2Al + 2NaOH + 2H2O -> 2NaAlO2 + 3H2}$$

The aluminium and sodium hydroxide would need to be kept separate and relatively anhydrous in the bottle to avoid reaction, so I wonder if the purpose of the mineral oil (which is hydrophobic) is to act as a moisture barrier and prevent mixing. If this is indeed the reaction that occurs, then the hydrated aluminium salts present are going to be at least somewhat soluble in solution, so I think it's improbable that what you have is pure sodium hydroxide. That said, if the quantity of aluminium is indeed small, this might not be a problem for your purposes.


How are the electrons of benzene actually configurated?

So, as I told you in chat what you have here is the orbitals of something which looks like, but which is not, benzene. It is the orbitals of one the all its two mesomeric form, it is so a static view.

Now let me show you the orbitals which are in the benzene. I only draw with the p-orbitals, I won't do the hybridation with the hydrogen it will be too difficult for me two draw and it will not be more effective.

How to draw a molecular diagram when the molecule has a lot of atoms?

First draw the diagram like if in the molecule there is just its squeleton (atoms which interact to create the shape of the molecule, for example with benzene only for the carbon atoms) and do like if these atoms were hydrogen atoms.

Second draw the correct orbitals with the same symetry of what you got with hydrogen just before.

You are obviously not obliged to do it but it is easier to see which orbitals in the fragments have the same symetry or not.

First step
Here for the benzene I will draw the MO's diagram of benzene with $\ce{H_6}$ first, with a molecule which is a rectangle of hydrogen atoms and the other fragment is a longer dihydrogen molecule than the original one. After seeing which of them have the good symetries I get this

Second step

Here we want the same but with the p-orbital of the carbons so keeping the same symetries or antisymetries you get this

And then here you have the correct orbitals.

Be careful

The color on the orbitals represent the sign of the wave function which has no meaning in the reality. So you can make the black and the white anywhere you want if you obviously respect the molecule symetries.
In this example I draw exactly the same orbital.

I say that because the orbitals I drawn for benzene are not exactly in the same color logic of the MO's diagram of $\ce{H_6}$ but as you can see each one have the same symetries or antisymetries.

And then the answer of your question

On the MO's diagram you show they show the interaction of the p-orbitals in one of the mesomeric structure, like this

And then the symetries are not the same as in the benzene but at one of its mesomeric form which is a static view like this

In fact you need to imagine the same but with an interaction of all p-orbitals and the get a ring orbital for $\psi_1$, and the it explain the real shape of the benzene like this

Remember that orbital diagrams are done to think about what you observe in the nature and not to predict it. Well you can use quantum theory to predict things but it is an other subject.

I hope it can help !


How to determine relative polarity (basic procedure)?

Determine which molecules are polar out of the given options.
Determine the polar bond (This will be the bond causing the molecule to be asymmetrical)
The molecule with the polar bond that has the greatest difference in electronegativity is the most polar. For example a carbon-oxygen bond is more polar than an oxygen-fluorine bond because the difference in electronegativity for oxygen and carbon is greater than the difference between fluorine and oxygen.

You can find electronegativity values in the following table or a table like it, I am not sure if you are provided with one. If not you can normally judge electronegativity by the how fare an atom is from fluorine, as seen in the table below. The size of the molecule also affects polarity, for example polarity decreases with the size of alcohols, but large complex molecules should not come out at your level.


Since this is pretty well covered in countless number of books and other sources I will try to just emphasise the key points.

The emission spectrum of atomic hydrogen was well known at that time, the only problem was that no one was able to explain it. :D Where does all this lines come from, i.e. what physical process gives rise to such discrete spectrum?
The wavelengths of spectral lines were known to be very well predicted by the Rydberg formula, but again no one was able to explain why this formula works, i.e. what are its physical justifications.
Bohr model explained discrete spectrum as follows: spectral lines are due to electron transitions between stationary orbits. Bohr was also able to determine the actual energy spacing between levels in hydrogen atom and explain the reason for the structure of the Rydberg formula.
However, Bohr failed to predict the lines in the emission spectrum even for helium atom. It happened simply because Bohr model is physically  wrong. Electrons do not orbit around the nucleus.


How to predict the solubility of an organic compound in different kinds of solvents?

There are some general guesses one can make from looking at the structure but the Abraham solvation equation is commonly used to estimate the solubility of a compound in a given organic solvent:
$$\log P_s = c + e E + s S + a A + b B + v V$$
This equation relies on a set of descriptors to characterize the properties of solvents and solutes to give the water/solvent partition coefficient $P_s$ (the ratio of the solubilities in water and solvent, under certain assumptions):

$p,e,s,a,b,v$ are coefficients that describe the solvent. Abraham and coworkers have collected coefficients for 85 different solvents.

The other variables are descriptors of the solute:

$E$ is the excess molar refractivity, which is a measure of the polarizability of a molecule.

$S$ is the solute dipolarity/polarizability.

$A$ is the overall (summation) hydrogen bond acidity.

$B$ is the overall (summation) hydrogen bond basicity.

$V$ is the McGowan characteristic volume.

You can look at the paper if you want to know exactly what these descriptors are, but their values can be obtained through various experimental and computational means, and combined with the solubility in water, used to estimate the solubility of a given compound in an organic solvent.

Fortunately, you don't have to do all this by hand as several websites, like this, that offer lookup and computation where you only have to specify a compound and pick a solvent from a list.


How does one tell (or conclude) if a substance is carcinogenic?

How do we tell or suspect one compound to be carcinogenic?

As written in the comments to the question, this the result of large studies on the human population, correlating blood or urine levels of some substance to instances of cancer. Alternatively or concurrently, studies on animal models, such as mice, rats or even in-vitro studies may be carried out. As a result, regulatory bodies may include a substance in lists, which may preclude them from use in consumer products, manufacturing procedures etc.

Are all carcinogens organic compounds?

No, this does not have to be the case: plain arsenic and basically all inorganic cadmium compounds have been classified as suspected or definitively carcinogenic.


Is mixing ammonia and bleach spray in the same parcel of air considerably dangerous to health?

This information is only for academic and educational purpose only. Please never ever try to mix ammonia and bleach.

Mixing bleach and ammonia is extremely dangerous, since many toxic vapors (nitrogen-chlorine species) will be produced. The primary toxic chemical formed by the reaction is chloramine vapor and hydrazine. (very, very toxic).

The following is the list of chemical species produced by the reaction:

$\ce{NH3}$ = ammonia
$\ce{HCl}$ = hydrochloric acid
$\ce{NaOCl}$ = sodium hypochlorite
$\ce{Cl2}$ = chlorine gas
$\ce{NH2Cl}$ = chloramine
$\ce{N2H4}$ = hydrazine
$\ce{NaCl}$ = sodium chloride or salt (non-toxic)
$\ce{H2O}$ = water (non-toxic)

The bleach decomposes to form hydrochloric acid:

$$\ce{NaOCl → NaOH + HOCl}$$

$$\ce{HOCl → HCl + [O]}$$

And then the ammonia and chlorine gas react to form chloramine, which is released as a vapor:

$$\ce{NaOCl + 2HCl → Cl2 + NaCl + H2O}$$

$$\ce{2NH3 + Cl2 → 2NH2Cl}$$

If ammonia is present in excess, toxic and potentially explosive liquid hydrazine may be formed.

$$\ce{2NH3 + NaOCl → N2H4 + NaCl + H2O}$$

References:

http://chemistry.about.com/od/toxicchemicals/a/Mixing-Bleach-And-Ammonia.htm
http://home.onehowto.com/article/what-happens-when-you-mix-bleach-and-ammonia-7512.html


Find the boiling point of a compound given a pressure

You could use the Clausius Clapeyron equation:

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}}{\Delta V \cdot T}$$

Since the change in pressure is rather small, it is fair to assume the molar enthalpy of evaporation to be constant and use the value given for standard conditions, which should be easy to obtain.

The other simplification you can make is to assume that the volume of the liquid is very small compared to the volume of the gas:

$$ V_\mathrm{l} << V_\mathrm{g}$$
so that
$$\Delta V = V_\mathrm{g}-V_\mathrm{l} \sim V_\mathrm{g}$$

Now by using the ideal gas law and expressing volume as molar quantity,

$$v=\frac{p}{RT}$$

you get the ODE

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}\cdot p}{T^2}$$

from whose solution you get a nice equation to solve your problem.


Stability of F- Cl- Br- I-? In aqueous state

If we consider $\ce{HX}$ as the common formula for hydrogen halogenides, then there is an equilibrium
$$\ce{HX(aq) + H2O <=> X-(aq) + H3O+(aq)}$$
that can be expressed also as:
$$\ce{acid1 + base2 <=> base1 + acid2}$$
The equilibrium is competition of 2 acids, $\ce{HX}$ ansd $\ce{H3O+}$. $\ce{H3O+}$ is the strongest acid stable in aquaeous environment. Acids we consider as strong acids ($\ce{HCl}$, $\ce{H2SO4}$, $\ce{HNO3}$ ) are much stronger acids than $\ce{H3O+}$. Therefore for $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, the equilibrium is practicaly fully pushed toward right:
$$\ce{HX(aq) + H2O -> X-(aq) + H3O+(aq)}$$
On the other hand, $\ce{HF(aq)}$ is a weak acid, weaker than $\ce{H3O+}$, with dissociation constant $\mathrm{p}K_\ce{a}=3.17$, for which the equilibrium is rather on the left side:
$$\ce{HF(aq) + H2O <<=> F-(aq) + H3O+(aq)}$$
Highly concentrated $\ce{HF(aq)}$ is rather a strong acid:
$$\ce{2 HF(aq) + H2O <=>> HF2-(aq) + H3O+(aq)}$$
and liquid HF is very strong acid ( H0 = −11 ) :
$$\ce{3 HF <=>> HF2- + H2F+}$$


No, you can't.

If you observe the mechanism of Kharash effect, an $\ce{HBr}$ molecule first undergoes homolytic bond cleavage to produce a $\ce{Br}$ free radical. To undergo that homolytic bond fission, the electronegativities of the two elements, attached by that bond, have to be very close, which is possible for $\ce{H}$ and $\ce{Br}$ ($\chi _ {\ce{H}} = 2.1; \chi_{\ce{Br}} = 2.96$). The benzoyl peroxide is just a free radical initiator.

Now if you come to water, electronegativity difference between $\ce{H}$ and $\ce{O}$ is huge $(\chi _ {\ce{H}} = 2.1; \chi_{\ce{O}} = 3.44)$. So, it is not possible at all to undergo homolytic bond cleavage of $\ce{H-O}$ bond even in presence of peroxides.

To do hydration according to anti-Markownikoff's rule (without any rearrangement), the most widely used method is Hydroboration-Oxidation. This is an alternative and better than doing first bromination in presence of peroxides and then hydrolysis, because in many other cases chances of rearrangement also remains there.


The mechanism of the addition of hydrogen bromide to ethene is called electrophilic addition. The region of highest electron density in ethene is the double bond, where the π electrons are. Hydrogen bromide acts as an electrophile toward the π electrons of the double bond. The proton of hydrogen bromide is positively polarized and thus electrophilic. (Obviously, the negatively charged bromide ion is not electrophilic.) Hence, the first step is an acid-base reaction in which hydrogen bromide donates a proton to ethene, forming a carbocation.


Why are so many wave functions associated with hydrogen?

1.Would an unperturbed electron ever go outside the lowest orbital?

First, there is no such thing as "outside the lowest orbital" because the lowest orbital has probability density at all points in space (no nodes).

But yes, at a given temperature electrons will be distributed amoung the energy levels in accordance with a Boltzmann distribution.

  2.At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?

There are infinite solutions none of which exceed the ionization energy (the ionization energy is only approached as n approaches infinity).

Think of a series like 0, 3/4, 8/9, 15/16, etc. where there are infinite members without reaching 1 (1 is approached in the limit of infinity).


Benzylic and allylic carbocation stability order

Yes tertiary allyllic carbocation is more stable than secondary because no.  of electrone donating methyl groups are greater than secondary (in tertiary 3 methyl groups  while in secondary  are 2) so they decrease  electrophilicity carbocation but it still it remain electrophillic.


Output of the Sugar & Sulfuric Acid Reaction?

The product is not pure elemental carbon, but instead an amorphous solid that consists of mostly carbon atoms, with at least 5-10% of the atoms being other elements.  For example, one academic paper from 2001 found that the "carbon" formed by sulfuric acid catalyzed dehydration of sugar had a $\ce{C}:\ce{H}$ molar ratio of 14:1.

Such materials are sometimes called hydrogenated amorphous carbon.


How can you detect iron hydroxide or iron sulfide in quarried rocks?

Rocks that turn green and then red seems to me like you might have some copper sulfides, like chalcopyrite (CuFeS2). At first, the copper reacts with the atmosphere to form something like malachite (similar to the green patina observed on things like the Statue of Liberty). Next, the iron oxidises to form iron oxy-hydroxides (as you correctly identified).

In order to detect these things rapidly, you have several options:

Use expensive methods like XRD or XRF. I would recommend against, because these only test the very thing you are analysing, and if you would like to test a whole rock face, this would be extremely expensive. It would however be beneficial to give one of the stained samples to a nearby research institute to test what the material is, either using SEM-EDS or XRD. I might be wrong and it would not actually be copper.
Just look at the rock very close, preferably with a hand-lens. Chalcopyrite (and other copper/iron bearing sulfides) and shiny, usually golden or slightly green-purple when tarnished. You can look at a gallery of chalcopyrite to have a rough idea of how it looks like. Note that the gallery has nice specimens, which yo clearly don't have, but try to see how the lustre and colour look like.
Spray bleach (sodium hypochlorite). This would speed up the reaction, better than acid. Note that this could affect other things in the rock and cause staining or dissolution of things that wouldn't have without the bleach.
Try to smell the rock, especially around the places where you see incipient greening. Can you smell a sulfurous smell? Our noses are surprisingly extremely sensitive to sulfur, and can detect it even when some instruments can't.


Both methods are used in practice (and a a lot more exist). Diazomethane is a last resort method when everything else fails. For example you might have a molecule that is acid sensitive or heat sensitive and then $\ce {MeOH/H^+}$ is out of the question. Although diazomethane is an excellent reagent to do the job, as it works selectively with almost every substrate, to give good yields under mild conditions, generally no extensive work-up or purification are required, it has some major disadvantages: It is not commercially available, it has to be prepared fresh before the reaction using a (somewhat) specialised equipment and a toxic starting material (Diazald). It is explosive and very toxic and obviously can not be stored. An alternative to using diazomethane is TMS diazomethane that is stable enough to be stored as a solution and it is commercially available however it is not innocent at all as this article  shows.


You might consider writing a correctly balanced equation, and then convert grams to moles.


Why do properties of colloidal solution depend only on dispersion medium's physical state?

The assumption that properties of colloidal dispersion only depends on the medium is completely incorrect. The vague part of the question is "properties" in the question, because the examiner is not clear about what type of properties he/she is talking about. It is not uncommon to find open ended questions in public exams. The properties of colloidal solutions heavily depend on the temperature (since their viscosity is temperature dependent as well as flow rate dependent). Similarly, the size of colloidal particle controls the optical properties of the dispersion. Similarly the surface charge on the particles determines the sign/magnitude of the zeta potential. If you want to read further, read DLVO theory (named after four persons) on the stability of the colloids from Wikipedia. One should not worry about an incorrect question. If it were a true-false question, one can simply mark it false.


It means that there are two phosphates connected together under loss of water, i.e.:

If you have any questions about the definitions of chemical nomenclature, have a look at the IUPAC gold book. See here. It basically says, that a pyro compound is one that is formed by heating another compound, usually under the loss of simple molecules like $\mathrm{H_2O}$ or $\mathrm{CO_2}$.


Following up on Permeakra's comment, the relative importance of resonance structures is typically determined by:

1) Maximize atoms that fulfill the octet rule.
2) Electronegative atoms have negative formal charges and avoid positive charges.
3) Minimize charge separation.

Carboxylic acid derivatives (esters, amides, etc.) have resonance structures that break #2, but since all atoms fulfill the octet rule, they are important contributors to the overall electronic picture of the compounds. Hopefully the image below is legible.

In general, resonance (mesomeric) structures that contain an electronegative atom that both breaks the octet rule AND bears a formal positive charge are not reasonable resonance structures and do not contribute to the overall electronic picture of the compound. The resonance structure of the carboxylic acid drawn in the original question falls into this category as the 'top' oxygen both breaks the octet rule and contains a positive charge.


Why can alkoxides leave in the E1cb mechanism?

The fact that the elimination of $\ce{EtO-}$ leads to the generation of a carbonyl bond makes this a feasible reaction. Also, the $\ce{EtO-}$ ion generated can then take $\ce{H+}$ from water taking the equilibrium further in the direction of the forward reaction ( the net constant would be ~ $10^2$).


Adsorption and distribution chromatography difference?

Adsorption Chromatography = Utilizes a mobile liquid or gaseous phase that is adsorbed onto the surface of a stationary solid phase. The equilibriation between the mobile and stationary phase accounts for the separation of different solutes.

Distribution Chromatography (aka Liquid-Liquid extraction) = Due to its liquid stationary phase, the ingredients are only separated by their respective polarity, i.e. according to their distribution coefficient KD. Distribution chromatography is also ideal for the concentration of trace quantities. This is more closely related to Flash, Column, and Gas Chromatography.


Why we can't see 3-d structures of certain compounds on PubChem?

Since it's been a couple of days, I guess I convert my comment into an answer.

MMFF94s is Merck Molecular Force Field.
From what I understand, the issue is that this FF has certain limitations when it comes to electrostatic interaction, estimating polarizability and intramolecular interactions, which confuses the method trying to predict 3D conformers for ionic compounds, metal complexes or molecules with twisted geometry.

From the Concluding Discussion section of the original publication [1, p. 515]:

  Despite encouraging success, certain limitations are evident.
  One of particular importance arises from the fact that MMFF94 uses static atom-centered charges. As such, it neglects both higher order multipoles and electrostatic effects that arise from molecular polarizability.
  Because of these simplifications, MMFF94, like a number of other force fields, employs “enhanced” charge distributions that emulate the effect of polarizability in amplifying electrostatic interactions for favorable contacts in a high-dielectric medium. Unfortunately, these enhanced charge distributions also amplify electrostatically unfavorable iteractions, whereas proper account of polarizability would
  diminish them.
  They also improperly enhance electrostatic interactions in gas-phase or low-dielectric environments. Furthermore, they may not be optimal for describing intramolecular interactions, and may thereby limit the ability of the force field to account for differences in conformational energies.
  Indeed, compounds containing two or more strongly polar functional groups in close
  proximity have proven to be the most problematic in this respect, though good results have been obtained in most cases to date. …  Other significant limitations include: the overly simplistic nature of the bond-charge-increment scheme used to assemble the partial atomic charges …; the lack of conformational dependence of the resultant charges …;
  and the omission of bond-torsion (and certain other) cross terms needed to account for significant geometrical changes that can occur when a torsion angle varies,… an example being the elongation of an amide partial C−N double bond by up to 0.1 Å when conjugation is broken.
  A further significant limitation is that no account is taken of metal-ligand interactions beyond that afforded by a relatively simplistic model that includes only electrostatic and van der Waals nonbonded interactions.

References

Halgren, T. A. Merck Molecular Force Field. I. Basis, Form, Scope, Parameterization, and Performance of MMFF94. Journal of Computational Chemistry 1996, 17 (5–6), 490–519. DOI: 10.1002/(SICI)1096-987X(199604)17:5/6<490::AID-JCC1>3.0.CO;2-P.


If the diazonium ion were formed in the way you described, it would be formed by N2 acting as a nucleophile and attacking the ring with a lone pair, displacing a leaving group - a nucleophilic aromatic substituion. In this case, the "one electron of carbon" would have left with the leaving group that was on the ring before. The carbon would now be neutral as it had lost an electron and essentially gained one from the dative bond from the N.

However, this isn't the usual way to form a diazonium ion. In fact, we normally start with a primary aromatic amine, and react it with a diazotizing mixture of NaNO2 and HCl. In this case, the positive charge on the N ends up, not from the N forming a dative bond whilst acting as a nucleophile towards the ring but in the final elimination of water to form the triple bond in the diazonium ion:

Image: Wikipedia


How do I print property matrices in Avogadro?

I am not aware of a menu button to print these tables.

There is however a different way to get your hands on the numbers.

Avogadro features objects: molecules, bonds and atoms are instances of classes, whose properties can be read and processed by programming.
The easiest way is the build-in Python-Terminal.

(Note that I'm using Avogadro 1.1.1 on Linux in German, your menus will have different labels.)

In Avogadro, open a Python Terminal (In German, this is done via
Einstellungen -> Werkzeugleisten -> Python-Terminal, which should be something like Settings -> Tool bars -> Python terminal

When run in the Python terminal of Avogradro, the following snippet writes

atom index
atomic number
x, y, z coordinates

for all atoms of the currently opened molecule to a csv file into an existing tmp directory in the user's home.

import os.path
import csv

home = os.path.expanduser('~')
data_dir = 'tmp'
atom_table_name = 'atoms.csv'
atom_table_path = os.path.join(home, data_dir, atom_table_name)
# generate full path for file with atom data

with open(atom_table_path, 'wt') as fout:
    writer = csv.writer(fout)
    writer.writerow(('Index', 'Atomic number', 'X', 'Y', 'Z'))

    mol = Avogadro.molecule

    for atom in mol.atoms:
        index = atom.index + 1
        x, y, z = atom.pos
        writer.writerow((index, atom.atomicNumber, x, y, z))

EDIT

I mentioned the Python Terminal first because I didn't want to introduce too much different things at once.

Once you have played a bit with the build-in Python Terminal, you will want to go one step beyond and try a more comfortable IDE with logging, tab-completion, build-in help, etc. My suggestion is to try IPython!

With IPython (or any other IDE of your choice), you can make use of Avogadro through its API without actually running it.

Given that your data file propen.cml looks like this:

<molecule>
 <atomArray>
  <atom id="a1" elementType="C" x3="-6.736896" y3="1.198243" z3="-0.001854"/>
  <atom id="a2" elementType="C" x3="-5.625957" y3="2.233390" z3="-0.108371"/>
  <atom id="a3" elementType="H" x3="-6.748313" y3="0.730420" z3="0.988429"/>
  <atom id="a4" elementType="H" x3="-6.615420" y3="0.408802" z3="-0.751312"/>
  <atom id="a5" elementType="H" x3="-7.710837" y3="1.671184" z3="-0.163704"/>
  <atom id="a6" elementType="C" x3="-4.268024" y3="1.629800" z3="0.106393"/>
  <atom id="a7" elementType="H" x3="-5.801148" y3="3.017025" z3="0.637857"/>
  <atom id="a8" elementType="H" x3="-5.668681" y3="2.695849" z3="-1.101309"/>
  <atom id="a9" elementType="C" x3="-3.137190" y3="2.345874" z3="0.062502"/>
  <atom id="a10" elementType="H" x3="-4.203709" y3="0.562202" z3="0.306350"/>
  <atom id="a11" elementType="H" x3="-3.147760" y3="3.413685" z3="-0.133506"/>
  <atom id="a12" elementType="H" x3="-2.172936" y3="1.873521" z3="0.222912"/>
 </atomArray>
 <bondArray>
  <bond atomRefs2="a1 a2" order="1"/>
  <bond atomRefs2="a1 a3" order="1"/>
  <bond atomRefs2="a1 a4" order="1"/>
  <bond atomRefs2="a1 a5" order="1"/>
  <bond atomRefs2="a2 a6" order="1"/>
  <bond atomRefs2="a2 a7" order="1"/>
  <bond atomRefs2="a2 a8" order="1"/>
  <bond atomRefs2="a6 a9" order="2"/>
  <bond atomRefs2="a6 a10" order="1"/>
  <bond atomRefs2="a9 a11" order="1"/>
  <bond atomRefs2="a9 a12" order="1"/>
 </bondArray>
</molecule>

you can load the CML file and iterate over the bonds to display the index numbers of the atoms in the bond and the bond length using:

import Avogadro as av

mol = av.MoleculeFile.readMolecule('propen.cml')
# don't forget to adjust the path for the data file!

for bond in mol.bonds:
    first = bond.beginAtom
    last = bond.endAtom
    print first.index, last.index, bond.length


This can be explained in terms of orbital overlap and therefore bond strength.

Consider a single carbon in an ethene molecule with two sp3 hybridised carbon atoms. The orbitals on the carbon repel each other and arrange themselves into a tetrahedron. Two of the four sp3 hybrids are used to form sigma bonds with hydrogen atoms. The carbon is then left to share its remaining two sp3 hybridized orbitals with its neighbouring carbon. Draw this system and you can see that two sigma bonds can be formed, though they are "bent" as the sp3 orbitals on the carbon are at a 109.5 degree angle to one another.

Now consider a single sp2 hybridised carbon in an ethene molecules with two sp2 hybridised carbon atoms. Again two orbitals are used to bond with hydrogen. This time we have an sp2 orbital pointing directly at the neighbouring carbon with which to form a strong sigma bond. The remaining p orbital which is orthogonal to the C-C sp bond can also forms a C-C pi bonding interaction.

Draw these orbital systems and compare.

The sp2 hybridised system constructs a strong linear sigma bond.
The sp2 hybridised system better shields the nuclei of the bonding atoms.
The sp2 hybridised system provides more orbital overlap, leading to stronger bonding and lower free energy.


Is there a difference between anti-anxiety and anti-depression drugs?

[Basic]

Receptor: a molecule in a cell membrane that responds specifically to a neurotransmitter, hormone, antigen or other substance.
Agonist: Substance which initiates a physiological response when combined with a receptor.
Antagonist: Substance which inhibits the physiological response when combined with a receptor.
Allosteric: Substance which alters the activity of an enzyme by means of a conformational change induced by a different molecule.

[Response]

GABA receptors are related with anxiety [1]. Anxiety occurs because of low concentration of GABA in the brain. Therefore, Chordiazepoxide is an allosteric agonist.
Dopamine receptors are related with depression [2]. Depression occurs because of low concentration of Dopamine in the brain. Therefore, Phenelzine is a non-selective and irreversible monoamine oxidase inhibitor, which means that the brain does not recycle/degrade the monoamine neurotransmitters (Which dopamine is one of them).


How to prepare a Gaussian input file for a binary solvent?

I am unaware that this feature exists in Gaussian. It is certainly something that stretches the possibilities of a polarisable continuum model. I am quoting (liberally) from an email to the computational chemistry list (CCL, http://www.ccl.net/), because I found no evidence, that this has changed for Gaussian 16:

Actually, there is no standard approach for solvent mixtures to you use it directly in Gaussian03/09. A good suggestion is to define the solvent parameters as a linear combination of the solvents in the proportions of the mixture.  So for toluene:water in a 4:1 mix use 4/5 of the toluene values plus 1/5 of the water values.
It is recommended to use the SCRF=SMD approach here, but that is not recommended for properties other than solvation free energy. It may be give satisfied results in the case of geometry optimization or energetic properties.  It is especially not recommended for derivative properties, because the cavity for the SCRF=SMD model has a number of continuity issues.  So for the default SCRF=IEFPCM model with the smooth cavity you mainly need the static (or zero-frequency) dielectric constant and the dynamic (or optical) dielectric constant.  You would specify EPS and EPSInf in the additional SCRF input section after the blank line which terminates the structure input.

For example:

{%chk=}
#P {Method, Basis, additional keywords} SCRF=(Solvent=Generic,Read)

{title card}

{charge, multiplicity}
{molecule specification}

EPS=(Add the value of linear combination of both solvents)
EpsInf=(Add the value of linear combination of both solvents)

See the Gaussian Manual for more details.

Another approach is to first do an optimisation and related properties calculation in both pure solvents, and then compare these results. For example the difference in the structures, or spectra.
Since you are using nine parts DMSO and one part water, it might well be that the DMSO calculation is a well enough approximation to the mixture.
Solvent calculations are tricky and often enough unreliable. One should never ever expect a quantitative agreement. It is a qualitative description you should be after.
If you need more accuracy, you have to use an explicit solvent model with MD calculations, or you can look into COSMO-RS.


What you understand by "Order of reaction"?

Historically, the concept of reaction order came from Ostwald's famous Lehrbuch of 1890s. An abridged translation is available in Internet Archive as "Outlines of General Chemistry". In this book, surprisingly, when a single reactant undergoes a transformation he does not call it a first order reaction. He designated rate of reactions controlled by two species as reaction of the second order.

  Reaction of the Second Order. We come to a second kind of chemical process, regulated by a new law, when in its course two substances
  undergo change of quantity. We must again assume that the action is
  proportional to the concentration of each present.

  Reactions of Higher Order. We must now consider reactions in which three or more substances take part.

So roughly, you can think of reaction order as the number of species in a chemical reaction which control the rate of reaction.

Coming to modern times, you have fractional reaction orders, and zero order and so on. For understanding you will need to see the concept of elementary and overall reaction orders.
A reaction may be occurring in several steps. In such cases, you might end up in fractional exponents. However, once you are able to know the steps of that reaction, you will see that those elementary reactions don't have fractional exponents. You can then interpret, each elementary reaction in the original Ostwald's way.


You can't simply filter it and take the feed because all the vegetables fiber remain in the filter with your starch. If you want to make a starch with a decent level of purity you should start directly with the potatoes: grating the potatoes, putting them in hot water (preferably salted: this make more effective the separation from proteins) for a bit of time. Then take the filtered solution (not the residual) make the water evaporate now what you get is the starch with the potato protein, amino acids and minerals (what is called fruit water). There is this video if you want to see.

If you want to have a more pure starch also you can try to precipitate the proteins warming the solution and adding an acid. If you are at home acetic acid could be enough.
Eventually you can cool down the solution this make starch precipitate so you can remove the water with a filter and take the precipitate.

If you really want to isolate starch there are more sophisticated methods see here and here for references.


How to know the ionization power of an unknown metal? - difficult

I am guessing you know when a metal bar is becoming thinner, the metal is being oxidised, i.e $$ \ce{M_{(s)}} \rightarrow \ce{M^{n+}_{(aq)}}$$

$P^{p+}, Q^{q+}, s^{s+}, R^{r+}$
Let's start with beaker 3 and beaker 4

In beaker 3, you have $\ce{S_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and you observer the metal bar becomes thinner. Which means, $\ce{S_{(s)}}$ is being oxidised by $\ce{R^{r+}_{(aq)}}$, or in other words, the reduction potential of $\ce{R^{r+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{S^{s+}_{(aq)}}$

$$\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}}$$

In beaker 4, you have $\ce{Q_{(s)}}$ in $\ce{S^{s+}_{(aq)}}$, and again the same reasoning applies.

$\ce{Q_{(s)}}$ is being oxidised by $\ce{S^{s+}_{(aq)}}$, or in other words, the reduction potential of $\ce{S^{s+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{Q^{q+}_{(aq)}}$

$$\ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$

Thus, so far we have:

$$\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$

Let's look at beaker 1 and 2

In beaker 1 you have $\ce{P_{(s)}}$ in $\ce{Q^{q+}_{(aq)}}$, and oxidation doesn't take place. Which means the reduction potential of of $\ce{Q^{q+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$

$$\ce{P^{p+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$

And similarly, in beaker 2 you have $\ce{P_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and nothing happens. Which means that the Which means the reduction potential of of $\ce{R^{r+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$

$$\ce{P^{p+}_{(aq)}} > \ce{R^{r+}_{(aq)}}$$

So the final order becomes (in terms of reduction potential, or oxidising ability):

$$ \ce{P^{p+}_{(aq)}} >\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$


Solving two electron integral numerically

The canonical text on quantum chemistry is Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund. Every electronic structure theorist I have ever met keeps a copy of that book within arms reach of his/her desk.

As for software, I would recommend PySCF. It is an entirely free python library for quantum chemistry with the most critical routines written in C and is an excellent platform to develop on. It's also robust and fast enough to be used for large calculations should you want to scale up. (https://github.com/sunqm/pyscf)

PySCF uses libcint to perform the integrals, and contains a low-level interface to the integral evaluator. An example of how to use these methods to make a df-HF calculation is shown in the docs here: https://sunqm.github.io/pyscf/tutorial.html#molecular-integrals. A complete list of available integral routines from the libcint package is shown here (it's very comprehensive): https://sunqm.github.io/pyscf/gto.html#module-pyscf.gto.moleintor


Is the equilibrium constant calculation only applicable to homogeneous reactions?

Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types?

The latter. If you consider the distribution of a solute S in a mixture of water and octanol, this is a heterogenous system (in this case, two immiscible liquids). The equilibrium reaction (or process) is:

$$\ce{S(oct) <=> S(aq)}$$

There is an equilibrium constant for this process,

$$K = \frac{[\ce{S(aq)}]}{[\ce{S(oct)]}}.$$

If I dissolve some S in the aqueous phase, some of it will go into the octanol phase until equilibrium is established. If I dilute the aqueous phase, some S will go back to the aqueous phase to re-establish equilibrium.


What will "Efavirenz amino alcohol methyl carbamate" look like?

The "efavirenz amino alcohol methyl carbamate" is (S)-methyl 4-chloro-2-(4-cyclopropyl-1,1,1-trifluoro-2-hydroxybut-3-yn-2-yl)phenyl carbamate (CAS 211563-41-6),

according to this reference.


The figure shows the relationship between the solid and liquid solvent and a solution.

Instead of freezing at the freezing point of the solvent $T_1$ a solution freezes at $T_2$, where the vapour pressure of the solution equals that of the pure solid solvent.

Using the Clausius-Clapeyron equation we can show that the solution curve will always intersect the solid solvent curve. The vapour pressure of the solution follows the line $\displaystyle \frac{d\ln(p)}{dT}=\frac{\Delta H_{vap}}{RT}$ and the pure solid solvent $\displaystyle \frac{d\ln(p_{ss})}{dT}=\frac{\Delta H_{sub}}{RT}$ where $\Delta H_{sub}$ is the molar heat of sublimation. This in turn is given by $\Delta H_{sub}=\Delta H_{vap}+\Delta H_{fus}$ where 'fus' means heat of fusion. Thus $\Delta H_{sub} \gt \Delta H_{vap}$ which in turn means that the slope of the solid solvent curve is always greater than that of the solution $\displaystyle \frac{d\ln(p_{ss})}{dT} \gt \frac{d\ln(p)}{dT}$.  As the freezing point of the pure solvent is higher than that of the solution the vapour pressure curve of the solid solvent and solution must always intersect.


Fundamental physical forces behind osmosis

If you dribble a basketball, the force of your hand on the ball, that keeps it bouncing, is not (1) gravity; (2) the weak force; or (3) the strong force. It must be electromagnetic force--the electrons in the molecules of your skin repelling the electrons in the molecules on the surface of the ball. Also, your skin and the ball remain (largely) intact because of the electromagnetic force that hold them together, respectively. Same for osmotic "force". The motions of the solute and solvent molecules are not governed meaningfully by anything other than electromagnetic force.


To understand why non - ideal mixtures form azeotropes, we need to understand what is ideal mixture. If you have a liquid mixture of two components A and B, ideal mixture is a mixture in which heterogeneous interactions (A - B)  are the same as the average homogeneous (A - A and B - B). This means when you mix components A and B, they don't "feel" like their enviroment has changed at all and thus no component has higher nor lower tendency to escape the liquid phase and go into vapor phase. When interactions are largely different, in most cases heterogeneous interactions are less favourable than homogeneous (molecules of A in most cases like to be around molecules of A rather than B) than component will have higher tendency to escape from liquid phase to vapor since in vapor interactions are much less pronounced due to much bigger average distance of the molecules. If this scenario happens you got yourself a lower boiling azeotrope since due to higher tendency of molecules to escape liquid phase, vapor pressure is higher than in ideal case (Raoult's Law). In some cases heterogeneous interactions can be very favourable and this is usually due to some specific interactions which mostly originate from hydrogen bonding, if so molecules will tend to stay in liquid and you got yourself higher boiling azeotrope. Example of such azeotrope is HNO3 and water. If you have any questions, fire.


The primed version of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ can only occur if there is a double bond in the vicinity of the leaving group as in your example. I have drawn both possibilities for the $\mathrm{S_N2}$ case in the scheme below.

Scheme 1: Comparison of the reaction products in an $\mathrm{S_N2}$ (top) and $\mathrm{S_N2'}$ (bottom) pathway.

The $\mathrm{S_N2}$ reaction is as you expect it to be. The $\mathrm{S_N2'}$ reaction uses the double bond as an electron relay system. Instead of the nucleophile (here: $\ce{OMe-}$) directly interacting with the $\unicode{x3c3}^*(\ce{C-Br})$ orbital, the π system interacts with the σ* orbital. The nucleophile then interacts with the π system’s LUMO (corresponding to the middle orbital of the allyl π system) to perform the attack. Since the nucleophilic attack and the leaving group are on different carbon atoms, relayed by the π system, this is not a direct $\mathrm{S_N2}$ but a derivative of it ($\mathrm{S_N2'}$).

The same logic can be applied to $\mathrm{S_N1'}$: here, the intermediate carbocation is not a localised one but an allyl cation.

The primed versions here are observed because not only the bimolecular attack on the highly substituted tertiary carbon ($\mathrm{S_N2}$) is unlikely — this carbon is tertiary which usually already implies no $\mathrm{S_N2}$ — but also the capturing of the carbocation under $\mathrm{S_N1}$ conditions is more likely to happen at the sterically much less hindered carbon atom.

As to why methanol predominantly attacks according to an $\mathrm{S_N1}$-type mechanism while methanolate predominantly follows an $\mathrm{S_N2}$-type mechanism I will refer you back to your textbook; the reasons are stated pretty often.


Does the distance increase as the polarity increases?

I'm not sure if the guideline actually works out in empirical observations. There are at least two factors affecting bond length - atomic radii - and polarity of the bond. A more polar bond will have a less even distribution of electrons. The electron density will generally be closer to the more electronegative atom. This isolation of electrons may have the effect of "stretching" the bond.

Consider the following two molecules, both of which are linear and polar:

From polarity considerations alone, we might expect the $\ce{H-F}$ bond to be longer than the $\ce{H-Cl}$ bond because fluorine is more electronegative than chlorine. However, chlorine is a bigger atom than fluorine by nature of having additional sublevels of electrons.

Also, consider the below data: bond length actually increases with decreasing polarity (quantified through dipole moments).

Fluoromethane Bond length 136pm, Bond strength 108 kcal/mol, Dipole moment (D) 1.85

Bromomethane Bond length 178pm, Bond strength 84 kcal/mol, Dipole moment (D) 1.87

Chloromethane Bond length 193pm, Bond strength 70 kcal/mol, Dipole moment (D) 1.81

Iodomethane Bond length 214pm, Bond strength 56 kcal/mol, Dipole moment (D) 1.62

The unfortunate thing about your guideline is that I thought that it would work at least across periods - in which atomic size stays relatively constant. However, the facts do not fit the case. Consider $\ce{H-N}$ and $\ce{H-O}$. Oxygen is more electronegative than nitrogen; we'd expect the $\ce{H-O}$ bond to be more polar, and thus longer than the $\ce{H-N}$, at least according to your guideline. The opposite is actually true; the $\ce{H-O}$ bond is actually shorter than the $\ce{H-N}$ bond.


Why tautomers are considered to be the same chemical compound?

There are a number of types of rapid chemical equilibria that basically cannot be turned off. Tautomers are probably the best known examples, although fluxional behavior is observed in many inorganic and organometallic systems.

The origin for most tautomers is that protons are really quantum-mechanical. They can tunnel just like electrons can delocalize. Obviously different tautomers will have different stability, much like some electric resonance structures are clearly more stable.

The comments to your question above indicate that most chemists would just call it "acetone" but of course there's some prop-1-en-2-ol present too.

Yes, the reactivities might be different.

It's a partly a matter of the balance of the equilibrium (i.e., that for many tautomers, only one form predominates). There are exceptions, e.g.
phenols (from Wikipedia):

  In certain aromatic compounds such as phenol, the enol is important due to the aromatic character of the enol but not the keto form. Melting the naphthalene derivative naphthalene-1,4-diol 2 at 200 °C results in a 2:1 mixture with the keto form.

It's also a matter of the time-scale. For example, at room temperature if I take an NMR, do I see two sets of peaks? What about IR or other characterization techniques? If I can't tell the difference because there are one set of peaks, I think many chemists will consider there to be "one compound" even if multiple tautomers are present.

For an example of this (from the cheminformatics perspective), Marc Nicklaus of the NIH/National Cancer Institute gave a great talk at the 247th ACS national meeting "The NCI/CADD Group's InChI Usage and Analysis of Tautomerism for InChI V2".

The talk focuses on tautomers and what software is trying to do to derive simple rules (as opposed to lots of quantum calculations or experiments) to determine one standard tautomer. But they also measured several tautomers by NMR:

These would obviously be chemically quite different, but one form clearly predominates.


What is Inversion Temperature in terms of Joule Thomson Effect? Does Critical Temperature depend upon Inversion Temperature?

This is half-part answer to your question:

Joule Thompson Inversion Temperature is the temperature when Joule Thomson coefficient $\mu_\mathrm{JT}$ is null.

$$\mu_\mathrm{JT} =  \left(\frac{\partial T}{\partial p}\right)_h$$

It means that the gas does not chill ($\mu_\mathrm{JT}>0$) or warm ($\mu_\mathrm{JT}<0$) when it experiences expansion. Pressurized Hydrogen has a negative Joule-Thomson coefficient, which must be taken in account if you plan to stock it and expand it, because it will significantly heat when expended.

This factor is influenced by gas molecular interactions (repulsive and attractive). Therefore, for an Ideal Gas implies that $\mu_\mathrm{JT} = 0$.

You will find detailed explanation in Atkins, Physical Chemistry, Chapter 2: First Law - The Joule-Thomson Effect.


How would you measure the rate of this SN1 reaction?

If you want to measure it with an electronic device I would recommend some spectroscopic technique such as IR, Raman, 1H-NMR etc. or using a digital pH meter. For the spectroscopic techniques you should probably cool down the solution to minimize error during the measurement or alkalify it to avoid further reaction.
However if you want to avoid these methods and you have enough soultion-volume you could try titrating it with NaOH solution and a proper indicator.


Your second drawing would imply a $\ce{S=O}$ double bond, due to available orbitals it would be a d-p π bond. There is considerable debate as to whether these bonds even exist with most of the evidence presented here at chemistry.stackexchange.com pointing towards a ‘no’. That would rule out your second mechanism.

That said, your first mechanism suffers from comparing sulphur with carbon too much. You assume a rather short-lived transition state and direct displacement. I don’t know for sure, but I don’t think that is the case.

Rather, I would propose a two-step mechanism with a medium-lived tetravalent sulphur. In this mechanism, the attack of the nucleophile would turn pseudotetrahedral (one corner of the tetrahedron being a lone pair) into a pseudo-trigonal bipyramid or a see-saw with $\ce{Nu}$ and $\ce{Cl}$ forming a four-electron-three-centre bond. This would then collapse back into a pseudotetrahedral product liberating the better stabilised anion. Close to your first mechanism, but not identical:


Why is this named 3-phenyl-2-butanol instead of 2-phenyl-3-butanol?

You have already correctly identified the compound as a phenylbutanol. The only remaining problem is the numbering of locants. The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds (first to double bonds if there is a choice)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Clearly, the $\ce{-OH}$ group is the principal characteristic group in this case since you have already used it as principal characteristic group to find the correct suffix ‘ol’.

Therefore, a low locant is assigned first to the $\ce{-OH}$ group according to Rule (c), which yields phenylbutan-2-ol.

After that, a low locant is assigned to the phenyl group according to Rule (f); however, the only possible numbering left is 3-phenylbutan-2-ol.


Vapour pressure in an enclosed container with liquid at normal conditions

The kinetic theory of gases is based on the idea that particles are moving randomly. Even if all particles started motion with exactly the same speed, as they interact with each other and the walls of the container their velocities would become randomized. The kinetic energy of particles at a given temperature is an average; some particles are moving much faster, and can escape from a liquid. These faster-moving particles, removed from the liquid, leave the liquid a bit cooler. Thus, evaporation puts a liquid into a gas and chills the liquid.

For a more complete explanation, see this video.


What visual observations can confirm that a reactant is the limiting reagent?

The colour of the filtrate after filtering out the solid is still lightly blue in colour. This indicates that Copper (II) sulfate still is present in the solution, meaning that there was not enough iron to completely react with all the copper (II) sulfate, visually confirming it to be the limiting reagent.

A solution containing only Iron (II) sulfate is colourless, so if the solution is blue in colour, then it shows that Copper (II) sulfate still is present.

Credit to Ivan Neretin for helping me.


Why do biphasic systems of fluorous and organic solvents form?

Fluorous solvents have the odd property of being both hydrophobic and lipophobic (Ref) and thus are not miscible with either aqueous or many organic solvents as you've noted. Fluorine, as the most electronegative element, does odd things to a molecule (and is often used in pharmaceutical compounds just because of some of these odd properties).

In the perfluorinated hydrocarbons, the carbon-fluorine bonds are quite polar with the electron density higher toward the fluorine. The solvents themselves, however, are not polar due to free rotation about the C-C bonds, so they are not miscible with polar solvents and are hydrophobic.

Because the F is so electronegative, the electrons are quite tightly held and the molecules have a low polarizability and unusually weak London dispersion forces, so they aren't miscible with non-polar solvents either. These weak London forces are illustrated by the fact that the perfluorinated hydrocarbons have nearly the same boiling point as their corresponding normal hydrocarbon, despite having more mass. (Ref)

You mentioned heat as one way to render the organic and fluorous solvents miscible. Adding gaseous $\ce{CO2}$ to the biphasic system also makes the phases miscible in many cases due to the high solubility of $\ce{CO2}$ in both fluorous and organic solvents. (Ref)

This reference suggest that shape may also be a consideration when considering flourous solvent's miscibility with alkanes, also through poor/weak interaction of London dispersion forces in the two liquids.


Will a crystal of iodine lower the yield of an Grignard Reaction

No. Iodine is used to help magnesium react with the bromoderivate. If pure magnesium is added to a solution containing a bromoderivate, it should react quickly. But sometimes it does not react at all, because magnesium is covered by a thin and protective layer of magnesium oxide. Added iodine will etch this layer and the underlying metal. This reaction breaks the oxide layer, so that the metal can get in contact with the bromoderivate, and quickly react.


The selection rules are related to the transition dipole moment that determines the intensity of a given transition. If a given selection rule holds perfectly, all that it says is that a forbidden transition has zero intensity and an allowed transition has non-zero intensity.

If the selection rule does not hold exactly, just like the in the case of the spin selection rule for heavy atoms, a forbidden transition can be intense. Some examples: the $\text{B}^3\Pi_u-\text{X}^1\Sigma^+_g$ electronic transition in $\ce{Br2}$ and $\ce{I2}$, that are strong when compared to the same transition in $\ce{F2}$ and $\ce{Cl2}$; and the strong $^3P_1-^1S_0$ transition in $\ce{Hg}$, while the analogous one in $\ce{He}$ is very weak. As $L$ and $S$ are not good quantum numbers when there is strong spin-orbit coupling, the selection rules involving them can't be used.

The second answer you came up with makes sense, but not for spin selection rule. Consider the parity selection rule for molecules with an inversion center. As the molecule vibrates (in an asymmetric mode) it can be distorted into a geometry that does not have an inversion center, so the selection rule does not hold. In this case, this vibronic transition will be allowed.


Reverse washing soda buildup on stainless steel

If you use tap-water together with Sodiumcarbonate, chances are that the residue is Calciumcarbonate (unless your tap-water is very soft).

Calciumcarbonate is significantly less soluble in water, but can be dissolved in acid. I would try to put a bottle/arm in a hot 5% solution of acid (acetic acid is fine, personally I would prefer citric acid) to see if any (small) gas-bubbles are formed at the residues. If it does, it is carbonate and can be removed in an acid bath.

If not, the residues could be glue from the labels, and getting rid of this is probably more complicated.


Reduction of acetaldehyde from rectus and sinister faces

Just to keep it slightly more realistic from a chemistry point of view, let's consider the reduction of deuterated acetaldehyde (1-Deuterioethanal):

When we add from the si face, we get an (R) center, and if we add from the re face, it's the other way around (as it should be!) and we have (S) configuration.

Now, what if we reduce "normal" acetaldehyde? We then have to decide which of the proton substituents gets the higher priority when trying to assign pro-R or pro-S. If you decide that the added proton has higher priority (as you did in your question) then you get an equivalent of a pairwise exchange of the substituents (in comparison to the structures that I drew), and the stereocenter is exactly the opposite of what I drew in the figure above.

From a synthetic point of view it is much cheaper to use hydrogen gas for the reduction and have the isotope label already reside on the starting material.


You'll probably be better served asking an allergist about this.

While many things contain PG, that is not the same as a compound similar to PG, which could have PG (or part of its IUPAC name, propane-1,2-diol, not to be confused with propane-1,3-diol, which would be its possibly harmless-to-you although structurally similar isomer) as part of its name or could interact with your body in a similar way. Case in point, your confusion over sodium laureth sulfate, which is nothing like PG, although that's an irritant for everybody. However, it looks like PG allergy is sometimes co-sensitized with corticosteroids - which are themselves structurally nothing like PG!

Thus, the confounding variable here is the way your body decides what will or will not set off your immune system, which can have very little to do with the chemical behavior of the allergen and everything to do with what mechanisms your body is using to make that distinction.


Reaction-enthalpy for synthesis of 40 g hydrazine

You are on the right track. Basically, you are just missing the last step. Your answer for the enthalpy of reaction should be $ΔH_{rxn}=−393.8\frac{\mathrm{kJ}}{\mathrm{mol}}$. So all you have to do is find the moles of hydrazine
$n=\frac{40}{32.0452}=1.248\space\mathrm{moles}$ and multiply it by your $ΔH_{rxn}$.


It seems you were on right path - for example common feldspars are solid solutions and they are mixed because they share cations, while they easily substitute one another, otherwise they exsolve - become separate phases. Also you're right that van der Waals forces aren't important here as they are ionic compounds.


After looking further into indigo production, I came across a related structure called indirubin.

Indirubin:

Indigo:

Indirubin is a byproduct of indigo synthesis and has a reddish color. It may also be more soluble than indigo, and indirubin derivatives have some anticancer activity. If you trust this supplier, you can buy indirubin for pretty cheap to compare with your lab's indigo reactions.

I also found a study done on old native american dyes that studied indigo and indirubin in fabric fibers, among other dyes. They provide a UV-VIS spectrum of both compounds, with Ind = indigo and Inr = indirubin. We can see that indigo has an absorption peak in the red region, so blue is reflected. Indirubin has a peak at around 540 nm, which is yellow-green, allowing some blue and red to reflect, but probably more red than blue. This paper also references a 1910 paper** studying indirubin levels in commercially available indigo samples, but I can't find that article. They suggest that higher levels of indirubin are produced when more indoxyl is converted to isatin. If you can find this 1910 article, it may provide some insight into reducing isatin and indirubin levels.

** William Popplewell Bloxam and A. G. Perkin, J. Chem. Soc., Trans., 1910, 97, 1460-1475.


Nitrogen can have a variety of oxidation states ranging from -3 to +5. For example:

  -3

The nitrogen atom in ammonia $\ce{NH3}$ has an oxidation state of -3 (each hydrogen is +1).

  -2

The nitrogen atoms in hydrazine $\ce{N2H4}$ has an oxidation state of -2.

  0

The oxidation state of the nitrogen atoms in $\ce{N2}$ is 0.

  Nitrate and nitrite

The nitrogen atom in the nitrate ion $\ce{NO3-}$ has an oxidation state of +5. Each oxygen atom has an oxidation state of -2, and the overall charge on the ion is -1.

The nitrogen atom in nitrite ion $\ce{NO2-}$ has an oxidation state of +3.

Nitrogen can have such a range of oxidation states because of its electron configuration, which is $\ce{[He]}2s^2 2p^3$ (5 valence electrons). Nitrogen can "gain" up to three electrons or "lose" up to five electrons.


In the first reaction, the reaction doesn't take place at a stereocenter, so there's no way to observe the inversion.

In the second reaction, the stereochemistry in the product is R, R. The Gilman reagent approaches opposite the C-O bond. After bond rotation, we see the product in the indicated conformation.


Is 44.5 mg/L a measurement of weight or volume?

$50~\mathrm{mg~L^{-1}}$ is an expression of concentration by mass. It means that if you took one litre of the solution and extracted all the metal, the resulting metal would weigh 50 milligrams.

The other common option is to express concentration in terms of moles per litre ($\mathrm{mol~L^{-1}}$) which means that if you extracted all the solute from one litre of solution, you would have 50 moles of solute.


What polymer can be dissolved in ethanol and is insoluble in water?

As a preface, I'm not sure what the exact solubility levels are, so this answer is qualitative.  Hopefully someone else can improve upon it or suggest something better, but for now:

Poly(1,2butylene glycol) (PBG) of moderate molar mass (Mn ~1000) is soluble in ethanol but sparingly soluble in water. Most solids will wind up being soluble in water if they're soluble in ethanol, owing to the two solvents' similar intermollecular forces but water's smaller size and greater overall solvent strength. This said, materials with the ethyleneoxy- repeating unit like Polyethyleneglycol (PEG), Poly(1,2butylene glycol) (PBG), or crown ethers have enhanced solubility in ethanol due to ethanol's ability to line up along the backbone of the polymer. Meanwhile, the ethyl pendant group hanging off the backbone of Poly(1,2butylene glycol) strongly decreases water solubility. The end of the ethyl group is far enough away from the oxygen that it's completely non-polar and the shape winds up causing something called clathrate formation in the water.

This polymer is mostly used as a Solgel or a coating. It could also be used for chelating some metal salts in solution. However, it would make a rather terrible structural polymer.


Increasing temperature in system of dynamic equilibrium

Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction?

Yes, they made a mistake. For all the other scenarios, they paired a figure of the rate changes with a figure of the matching concentration changes. For this scenario (increase in temperature) they matched it with a correct figure of concentration changes when the temperature is decreased:

If you change the label in the figure posted by the OP, it would reflect what happens when increasing the temperature: Both rates increase, but the reverse rate increases by a larger factor, resulting in a net reverse reaction until equilibrium is reached again.

Here are sketches for the rate changes both for increase in temperature (low to high, upper panel) and decrease in temperature (back to low temperature, lower panel). As you can see, the reverse rate is more sensitive to temperature (higher activation energy because the product is lower in energy than the reactant for this exothermic reaction):


I am conducting a school-based research project into the antioxidative properties of green tea. Part of the procedure is to determine the concentration of antioxidant in green tea. This antioxidant is EGCG (a type of catechin).

First of all, you cannot determine the concentration of just EGCG (epigallocatechin gallate) alone, because any extract of green tea contains polyphenols other than EGCG, which include catechin, and/or other flavonoids such as flavanols, flavandiols, and phenolic acids. Most of the green tea polyphenols are flavonols, commonly known as catechins, which are flavan-3-ols (Ref.1):

As shown in the diagram, There are four kinds of catechins mainly find in green tea: epicatechin (EC), epigallocatechin (EGC), epicatechin-3-gallate (ECG), and epigallocatechin-3-gallate (EGCG). The amount of catechins varies in the original tea leaves due to differences in variety, origin, harvesting time, and growing conditions.  Also, the preparation of fresh green tea cannot totally extract catechins from the leaves, therefore, the amount determine by you may not reflect the absolute values determined through the complete extraction of leaves. Thus, your findings may differs from the accurate benefit of green tea.
Nonetheless, there are methods to determine polyphenol content in particular extract as the amount equivalent to a reference. Easily, your reference could be gallic acid. One easy method I found in literature is the estimation of total phenolic content and other oxidation substrates using Folin–Ciocalteu reagent* (Ref.2; PDF is attached for convenience):

Determination of Total Polyphenols by Folin-Ciocalteu Method: One milliliter of the tea extract was diluted with water to $\pu{100 mL}$. The diluted extract solution $(\pu{1.0 mL})$ was then mixed with $\pu{5.0 mL}$ of 50% Folin-Ciocalteu reagent. After $\pu{5 min}$, $\pu{4.0 mL}$ of 7.5% $\ce{Na2CO3}$ solution was added and mixed. Then the mixture was incubated at room temperature for $\pu{60 min}$. The absorbance of the solution was measured at $\pu{765 nm}$ with a Shimadzu UV-2550 UV-Vis spectrophotometer against a reagent blank. The determination was three times. The total polyphenol content was expressed as gallic acid equivalents (GAE, units: $\pu{g}/\pu{100 g}$ sample).

Preparation of Tea Extract: Each tea sample was ground to a fine powder and $\pu{1.0 g}$ of the powder was extracted with $\pu{25 mL}$ of methanol/water (7 : 3, v/v) in a water bath at $\pu{70 ^\circ C}$ for $\pu{10 min}$. The extraction was repeated twice and the extracts were combined. After filtration through a filter paper and a $\pu{0.45 \mu m}$ membrane filter (Millipore), the extract volume was adjusted to $\pu{100 mL}$.

According to the procedure, you need just a spectrophotometer in visible range (do not need an UV-Vis spectrophotometer, which is an expensive one). If you do not have a membrane filtration setup, you may extract leaves without powdering them, but need extra extracting time (may be pre-prepare extract overnight at $\pu{70 ^\circ C}$ using the same solvent system). If you want to check the green tea as we prepared them to drink, use that way but you may have to adjust reagent concentrations accordingly to test.
If you have a 96-well plate-reader and some micro-pipettors, there is a reliable high-input method is listed in Ref.3 for the same method (PDF of the paper has been attached to compare).

*Note: Folin–Ciocalteu reagent is a specific redox reagent containing phosphotungstic-phosphomolybdenum. The reaction forms a blue chromophore constituted by a phosphotungstic-phosphomolybdenum complex with phenols that can be quantified by visible-light spectrophotometry (Ref.4,5).
Gallic acid could be used as calibration standard for this determination and the total phenols could be reported as gallic acid equivalent (GAE). Other alternative is pyrogallol. Choose the concentration range enough to give absorbance 0.4-1.0 (at $\pu{760 nm}$) so that the assay is linear (Ref.5). Following is the published calibration curve for gallic acid using Folin-Ciocalteu Assay (Ref.6):

The procedure used: In brief, aliquots of a standard solution of gallic acid $(20, 40, 60, 80, \pu{100 mg L-1})$ were mixed with $\pu{500 \mu L}$ of Folin-Ciocalteu reagent (previously diluted 10-fold with distilled water) and $\pu{400 \mu L}$ of $\ce{Na2CO3}$ (7%). After $\pu{40 min}$ of incubation at room temperature $(\pu{23 \pm 2 ^\circ C})$, the absorbance was measured at $\pu{760 nm}$ using a Spectrophotometer against a blank sample.
Note that the temperature of $\pu{23 ^\circ C}$, time of $\pu{40-60 min}$, and  $\lambda = \pu{760 nm}$ are optimized condition for Folin-Ciocalteu assay (Ref.5). Also make sure that you add $\ce{Na2CO3}$ solution at least after $\pu{10 min}$ of the addition of Folin-Ciocalteu reagent (not before). Folin-Ciocalteu reagent is unstable under alkaline conditions.

References:

Sarah C. Forester and Joshua D. Lambert, "Antioxidant effects of green tea," Mol. Nutr. Food Res. 2011, 55(6), 844–854 (doi:10.1002/mnfr.201000641)(PDF).
Chunjian Zhao, Chunying Li, Shuaihua Liu, and Lei Yang, "The Galloyl Catechins Contributing to Main Antioxidant Capacity of Tea Made from Camellia sinensis in China," The Scientific World Journal 2014, Article ID 863984 (11 pages) (doi: https://doi.org/10.1155/2014/863984)(PDF).
Elizabeth A Ainsworth1 and Kelly M Gillespie, "Estimation of total phenolic content and other oxidation substrates in plant tissues using Folin–Ciocalteu reagent," Nature Protocols 2007, 2(4), 875-877 (doi:10.1038/nprot.2007.102)(PDF).
P. Schofield, D. M. Mbugua, and A. N. Pell, "Analysis of condensed tannins: a review," Animal Feed Science and Technology 2001, 91(1–2), 21-40 (doi: https://doi.org/10.1016/S0377-8401(01)00228-0).
Andressa Blainski, Gisely Cristiny Lopes, and João Carlos Palazzo de Mello, "Application and Analysis of the Folin Ciocalteu Method for the Determination of the Total Phenolic Content from Limonium Brasiliense L.," Molecules 2013, 18, 6852-6865 ( doi:10.3390/molecules18066852).
A. Bouyahya1, N. Dakka, A. Talbaoui, N. El Moussaoui, J. Abrini, and Y. Bakri, "Phenolic contents and antiradical capacity of vegetable oil from Pistacia lentiscus (L)," Journal of Materials and Environmental Sciences 2018, 9(5), 1518-1524 (doi: https://doi.org/10.26872/jmes.2018.9.5.167).


I think using acetone will be helpful. Acetone will evaporate fastly and it is polar and can form hydrogen bonds to the detergent powder like water. Though you need to use much acetone to wash all detergent powders and after that, wait for some minutes for acetone to vaporize from your machine.


What is meant by the numbers on reagents above and below arrows?

The numbers indicate steps. The only problem is that the definition of a "step" can be slightly different. Sometimes, separate steps mean that you do something, wait some time, and then do the next thing. Other times, you need to work-up or destroy the excess reagents that are left and isolate the desired product before you can proceed with the next step.

In your example, you need to keep the mercury and the borohydride separate. If you used any excess in the first step, you might need to destroy it before you add the borohydride.


Why is CO ligand higher than CN- in the spectrochemical series?

You're on the right track. It has to do with the energies of the frontier orbitals. As you rightly said, both species are isoelectronic, and the orbital energies in CO are lower than those in CN−.

The lower HOMO energy means that CO is a poorer σ donor orbital towards the metal than CN−. Likewise the lower LUMO makes it a better π acceptor.

These two factors are conflicting: stronger π acceptors are stronger-field ligands, but poorer σ donors are weaker-field ligands. Empirically, since CO is higher in the spectrochemical series, it stands to reason that the π acceptor effect is the most important factor here. This isn't really a surprise, as π-effects tend to outweigh σ-effects when it comes to the spectrochemical series.

I don't pretend that the above is a rigorous proof, but short of crunching the numbers, I don't think there's a much better explanation.


As Curt F. has correctly showed that, in general, if the rate of reaction is plotted against time for an autocatalytic reaction, it would be the shape of curve B (in your question). Since Curt F. has already worked with the mathematical part, only I can do is show you some literature evidence. One of research on an epoxy resin composite hardening (Ref.1) state that:

  Composite structures for aerospace applications are mainly made by the well-known prepreg technology. In order to achieve adequate prepreg processing schedules, and consequently maximum fiber strength utilization, one has to know in deep the cure kinetics of matrix, which held the fibers together. This work describes a procedure to study the cure kinetic and has as example how aromatic amine hardeners influence the cure kinetics of an epoxy resin used in advanced composites. The investigation was carried out by using the DSC technique and it was found that depending on the system used the cure kinetics of the formulation obeys order $n$ or autocatalytic order.

For Comprehensive details for autocatalyzation and review, see Ref.2 & 3.

References:

Michelle L. Costa, Luiz C. Pardini, Mirabel C. Rezende, “Influence of aromatic amine hardeners in the cure kinetics of an epoxy resin used in advanced composites,” Mat. Res. 2005, 8(1), 65–70 (http://dx.doi.org/10.1590/S1516-14392005000100012).
Donna G. Blackmond, “Asymmetric autocatalysis and its implications for the origin of homochirality,” Proc. Natl. Acad. Sci. USA 2004, 101(16), 5732–5736 (https://doi.org/10.1073/pnas.0308363101).
Andrew J. Bissette, Stephen P. Fletcher, “Mechanisms of Autocatalysis,” Angew. Chem. Int. Ed. 2013, 52(49), 12800–12826 (https://doi.org/10.1002/anie.201303822).


Wouldn't radiolabelled phosphorus in DNA break it apart as it disintegrates?

Of course it would break, just like you said; also, a high-energy $\beta$ particle would kill quite a lot of bystander molecules. Also, if not for other reason, the resulting molecule would no longer be DNA , since the decayed atom would no longer be $\rm P$. Also, the product would no longer be radioactive, so we wouldn't be able to detect it anyway.

The same applies to any other case of radiolabeling, in biochemistry or elsewhere, and you seem to be missing the very point of it: All this is not important.

Say, we have some $\ce{A + B -> C + D}$, and we label some atom in $\ce{A}$ so as to find out whether it will go to $\ce{C}$ or to $\ce{D}$. Then we run the reaction as usual. Now, radioactive atoms do not decay all at once; some decay earlier and some later. Those which decay earlier (that is, before and during the reaction) are lost to us, and we don't care about them one bit. Then we separate $\ce{C}$ from $\ce{D}$ and measure the radioactivity of each, and it is those atoms which decay now that matter.

Oh, and it is not like we label all atoms in $\ce{A}$. No, a very tiny fraction would suffice. Those viruses would probably get but a few radioactive atoms each. Some will be lucky enough to reproduce. Some will be killed by an early decay and remembered as martyrs of science.

So it goes.


IUPAC naming of 1,2-disubstituted cyclohexane derivative

Simple prefixes (simple substituent groups such as methyl and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘methyl’ is considered to begin with ‘m’; ‘isopropyl’ is considered to begin with ‘i’.

On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Thus, the correct alphanumerical order in your example corresponds to the name x-isopropyl-y-methylcyclohexane.

Furthermore,

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule f takes precedence over Rule g; however, Rule f does not permit a decision to be reached between ‘1-isopropyl-2-methylcyclohexane’ and ‘2-isopropyl-1-methylcyclohexane’ since both names correspond to the locant set ‘1,2’.

According to Rule g, your example is named as 1-isopropyl-2-methylcyclohexane rather than 2-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.

Note, however, that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.

Therefore, the preferred IUPAC name for your example is actually 1-methyl-2-(propan-2-yl)cyclohexane because ‘methyl’ is cited first in the name (since alphabetical order is used to establish the order of citation of substituent prefixes in the name).


Assuming ideal gas behavior, how can I tell when the change in enthalpy is equal to the change in the inner energy?

You got it right, and there is really nothing to add. Same number of molecules means no change in volume, hence no work is done, hence $\Delta H=\Delta U$.

On a side note, I find it hardly believable that burning of methane (b) will result in liquid water, but if the problem statement was given to you exactly like this, well, so be it.


0.125M of $\ce {Ba(OH)2}$ means 0.125 moles of $\ce {Ba(OH)2}$ per volume unit (so far I've only seen liters been used as volume unit) .
Assuming that all the $\ce {Ba(OH)2}$ you used was consume on the reaction with the acetic acid, and knowing the ratio of $\ce {Ba(OH)2}$ and acetic acid on the reaction is 1/2 you`d have this reaction:

$$\ce {Ba(OH)2 + 2CH3CO-OH -> Ba(CH3COO)2 + 2H2O}$$

All you need to do is calculate the amount of acetic acid consumed. Knowing the amount of acetic acid consumed based on the amount of moles of $\ce {Ba(OH)2}$ that are present on 37.5mL. Then you need to get the molar mass of acetic acid  and calculate the amount of mass of acetic acid you had. Then divide the total mass of vinegar by the mass of acetic acid  you found, multiply it by 100 and you'll have the mass percentage of acetic acid.


In addition to permeakra's comment, I want to add the following thoughts:

  1) How were these bond lengths determined?

Were these values determined computationally? If so, what is the standard error in the computational method uses? In other words, is the difference between the two values within the error of the method, and thus the difference is not significant. Additionally, a computational method operates on single molecules "in the gas phase", and there might be just enough interaction between the hydrogen atom and one of the other oxygen atoms. This interaction would also explain the disparity in bond angles:

Were they measured experimentally, for example from x-ray crystallography? If so, then dissymmetry arising from a preferential crystal packing arrangement can lead to differences. In particular, in the solid state, the position of the hydrogen atom can make a big difference in preferential hydrogen bonding, either intermolecularly or intramolecularly. For units like the following hyrogen-bond dimers, the hydrogen bonding could lead to dissymmetry between the N-O bonds.

  2) Bonds vibrate

They do, even in the solid state. Thus, an x-ray crystal structure has error its measurements as well. Again, is this difference in bond lengths within the experimental error?


A sample must not contain more than 2% contaminants to be pure. Are there instances where 2.12% is ok?

It depends mainly on the accuracy of your measuring instrument in practice and the accuracy required by your work.

For example:

If you want to make a solution with 2g solute in it, and to measure amount of solute you have an measuring instrument that have least count of 0.1g, then 2.05g is acceptable as the instrument will show 2.0g only but if the least count of intrument is 0.01g then 2.05g is not acceptable if you want 2g.

In real life the least count varies for different situations.

For example if you are manager of a water bottle manufacturing unit you won't need much accuracy to fortify the water with minerals, but if you are carrying out a titration in lab to find strength of some oxidising agent you need very high accuracy that is why burette are made such accurate.


FFC Cambridge Process and impure reactants

[FFC process][1]
[1]: https://www.youtube.com/watch?v=97dfa1GoF7w

next step is that the pellets are placed into a crucible or reduction vessel that's filled with very pure molten salt normally this will be calcium chloride.


Why does the carbocation in this scheme react under elimination conditions rather than being attacked by water?

The simple answer is that water is a very weak nucleophile and substitution does happen reversibly, but usually dissociates immediately. After which, the final reaction shown happens.

There is likely a (very) minor product that follows along the substitution pathway with an alcohol final product.


Fitting parameters of Lennard-Jones potential to properties of real materials

All molecular simulation force fields are essentially "fitted" mathematical functions. The parameters are adjusted until the simulation results match some physical properties we care about. As Geoff says, Lennard-Jones force fields are not particularly realistic, in the sense that you wouldn't expect a simulation of atoms interacting solely through Lennard-Jones potentials to behave exactly like a real system. This becomes very important for things like proteins, where conformation and charge-charge interactions are very important.

However, what might not be clear from the other answers and comments is that all force fields are not 100% realistic. The trick is to intelligently choose your level of realism. It could be that a Lennard-Jones force field that ignores  charge-charge interactions works perfectly well for your application. Without knowing the details, I can't tell you how suitable it would be.

I do know that for polymers all sorts of real phenomena can be modeled with very simple Lennard-Jones-like force fields. A famous example is this article by Kurt Kremer and Gary Grest:

Dynamics of entangled linear polymer melts:  A molecular‐dynamics simulation. J. Chem. Phys. 92, 5057 (1990)

Also, as Geoff mentioned, LJ potentials work pretty well for things that can be modeled by the van der Waals equation of state: noble gases, simple diatomics, and other small molecules at high temperature and low pressure.

Some coarse-grained simulations use LJ potentials for each "blob," and depending again on what you want to study, this can be very successful.

And while we are on the subject of unrealistic force fields, the hard-sphere model has been used to successfully model things like glass transitions (I don't have a good representative link for this, but you can search for "hard sphere glass transition" to find many studies).

So, in response to your actual question:

  Does it even make sense to differentiate between different materials when using a crude approximation like the Lennard-Jones potential?

The answer is:

Maybe - it depends on what the goal of your simulation is. What physical behavior are you trying to model? If you just want to see how a string of soft, slightly sticky beads sticks to another string, then you don't need to differentiate between atoms. If you want to see what happens when some of those beads are really sticky (sort of like hydrogen bonding sites on a protein), then it would make sense to use different parameters for very sticky vs. slightly sticky beads. It probably would never make sense to worry about which beads were carbon and which were oxygen, nitrogen, etc., or to try to select parameters for them that were based on published values (unless those studies were specifically looking at the same system and properties that you are). Most LJ parameters are derived from fitting simulated phase behavior to real phase behavior - which is pretty far removed from what you are doing.

If you want to get a better idea of what real proteins would do, then the problem becomes much more complicated. Real protein behavior has a lot to do with interactions with water, which means you would need to simulate that as well. Charge-charge interactions also become much more important (and the simulations much more computationally expensive)

My advice (and I think what the other responses are saying as well) is to choose your system and goals carefully so that you don't need this level of realism. Or, if you do need it, then do it right; don't try to make a simple model work for a system that it doesn't fit.


Why is the units of rate constant dependent on the order of the reaction?

The rate has units $\displaystyle \frac{concentration}{time}$ and is $\displaystyle \frac{d[C]}{dt}$ and the right hand side of the equation has always to have the same units as the left hand side, so for first order reaction consuming C;
$\displaystyle \frac{d[C]}{dt} = -k_1[C]$
and the rate constant $k_1$ therefore has units $\displaystyle \frac{1}{time} $ usually $\mathrm{s^{-1}}$ and for second order
$\displaystyle \frac{d[C]}{dt} = -k_2[C][C]$
the units of rate constant $ k_2$ are $\displaystyle\frac{1}{time \times\,concentration}$ which is $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$.
The units for other examples follows in the same way.
In the Arrhenius equation as the exponential is dimensionless the other terms must have the units of the rate constant, which means units of PZ in your notation. These units must always depend on the order of reaction as described above and if they don't then there is an error somewhere.


How can Hydrogen NMR be used to distinguish between these two isomers?

In the absence of any other configuration anchoring factors, I would not expect the conformation of the two structures to be reasonably stable at room temperature to be able to distinguish both by NMR. You may well need lower temperatures to freeze chair inversion and create a set of distinct, non-averaged peaks.

This issue aside, it is very easy to distinguish the two once you know you have a stable conformation. The key is the Karplus relationship which allows you to link a vicinal coupling constant to the dihedral angle between the two hydrogens. In general, a $^3J_\ce{HH}$ coupling constant will be very high ($^3J_\ce{HH} > \pu{9Hz}$) if the dihedral angle is either $0^\circ$ or $180^\circ$ — typically, if the hydrogens are anti-configured. On the other hand, the coupling constant becomes minimal for values around $90^\circ$ ($^3J_\ce{HH} < \pu{2Hz}$). For a $60^\circ$ gauche arrangement, a coupling constant of $^3J_\ce{HH} \approx \pu{3 Hz}$ is expected.

So the coupling between $\ce{H{1}}$ and $\ce{H{2}}$ would be around $\pu{9Hz}$ or $\pu{10Hz}$ in the first case but only around $\pu{3Hz}$ in the second — if the configuration is long-living on the NMR timescale.


How much heat is gained/released based off of equation?

I'm going to assume that the question is presenting the given change in enthalpy as a molar enthalpy change (so as to say that -325.0 kJ are evolved per mole of CaCl2).

Under that assumption, answering the question is then a matter of using the a molar fraction of CaCl2 for the amount you have present.

CaCl2 has a molar mass of 110.98 g/mol so with 15.0 g of CaCl2, you'd have roughly 0.1352 mol of CaCl2. 0.1352 mol * -325.0 kJ/mol = -43.94 kJ. The negative here means that the CaCl2 will release 43.94 kJ of heat.


Why does internal combustion engine smoke appear blue? Could there be "blue molecules" in it?

I can't really tell you the chemical difference but usually blue smoke indicates that oil is burned, too. Like some of the oil cycles get into contact with the engine and it is burned as well. And then probably heavier components of your gasoline once you start the cold engine. My Prof (lecture on Petrochemistry) once told us that there is indeed a difference between premium gasoline and the cheap one although the octane number could be similar. For the cheap ones they add a lot of stuff with a very low octane number and finally only a small amount compounds with higher octane numbers to get to 95. But it makes a difference in burning it. Maybe if the engine is still cold some of the "harder to combust" compounds don't burn that well and cause the blue color, too.

It's probably due to Rayleigh scattering as you mentioned. The sky is blue and clouds are white, the normal exhaust gases should be water and CO2 so similar to clouds and the blue particles are smaller rests which appear blue.


Using UV-C (254nm) light to cleave the amino acid off of vyvanse

Honestly, I wouldn't bet on a cleavage here.

In principle, excited state amides and imides can undergo electron transfer reactions. We did work a bit on that (DOI) and an early example on intramolecular electron transfer between an aromatic amide and a tertiary amine, resulting in amide cleavage, was given by J.D. Coyle (PDF).

In your case, however, the situation is a bit different. You don't have an aromatic amide but an arene and an aliphatic amide. The latter aren't supposed to exhibit a strong absoption at 254 nm, see an older article by Nielsen and Schellman (DOI).

Even if you'd manage to excite the amide, I'd exclude electron transfer from one of the primary amines in your starting material and hence the cleavage of the amide.