Why the colour of KMnO4 at the end point of titration may disappear after some time?

The most likely reason for the colour of permanganate to vanish after a few seconds even though you thought you have the end point is: you have not reached the end point.

Acid/base reactions are typically extremely fast reactions; their rate is not even limited by the speed of diffusion since protons can be shuffled along via the Grotthuss mechanism much more rapidly. Thus, you can assume each acid/base reaction to instantly reach completion.

With practically all other reactions that is not the case. They are typically slower and can be much slower. The reacting species must meet each other in solution which can take time. At the beginning of the titration, there are still many iron(II) ions floating around ready to be oxidised by any incoming permanganate. Towards the end, iron(II) is getting more scarce. So a permanganate ion may linger around much longer before it meets an iron ion. You should give it a certain time in case it does find one. The value of $\pu{30s}$ is arbitrary. Ten seconds or two minutes would probably do equally well (except that ten seconds may or may not be slightly on the short side). The book/method just needs to present any ballpark value.

So the reaction occuring that causes the discolouring of the solution is indeed exactly the reaction you are observing: the oxidation of iron(II). It is merely slowed down so it can take a few seconds to proceed to completion when close to the equivalence point.


Biodegrability of straight chain and branched chain detergents

According to Scott and Jones1:

... problems arise with branched alkyl chains, a side chain methyl group or a gem-dimethyl-branched chain cannot undergo β-oxidation by microorganisms and must be degraded by loss of one carbon atom at a time (α-oxidation).

* text by the authors, highlighting and hyperlinking by me.
A study by Whyte et al2 on the degradation of alkanes by Rhodococcus yielded similar findings:

After 28 days of growth at 5°C in MSM containing 0.1% diesel fuel, Rhodococcus sp. strain Q15 degraded almost all the n-alkanes (C10 to C21) and, to a lesser degree, many of the branched alkanes found in diesel fuel.

P.s: both of these articles are open access, and explain the biodegradation process quite well. You can refer to them directly if you need more detail or specific examples.
References:

Scott, Matthew J., and Malcolm N. Jones. “The Biodegradation of Surfactants in the Environment.” Biochimica et Biophysica Acta (BBA) - Biomembranes, vol. 1508, no. 1–2, 2000, pp. 235–51. doi:10.1016/S0304-4157(00)00013-7.
Whyte, L. G., et al. “Biodegradation of Variable-Chain-Length Alkanes at Low Temperatures by a Psychrotrophic Rhodococcus Sp.” Applied and Environmental Microbiology, vol. 64, no. 7, 1998, pp. 2578–84. PMID: 9647833


Comparing the zero point vibrational energies of DCl and HCl

I don't know which equations you are using, but this is the one I was taught:
$$E(v)=hc\tilde{\nu}\left( v+\frac{1}{2}\right) -hc\tilde{\nu}\chi\left(v+\frac{1}{2}\right)^2.\tag{1}$$
Here, $\tilde{\nu}$ is the wavenumber of the vibration; $v$ is the quantum number that represents the vibrational levels $(v= 0, 1, 2,\ldots);$ $\chi$ is the anharmonicity constant:
$$\chi=\frac{\tilde{\nu}}{4\,D_\mathrm{e}},\tag{2}$$
where $D_\mathrm{e}$ is the depth of the lowest point of the potential well (not the depth of the $v = 0$ vibrational level!).
The zero-point energy is the energy when $v=0$ i.e. the molecule is in the lowest vibrational state. So you have to substitute that in the energy equation.
Now, we can assume that the well depth $(D_\mathrm{e})$ of the $\ce{H-Cl}$ and $\ce{D-Cl}$ bonds are same.
The only difference between $\ce{H-Cl}$ and $\ce{D-Cl}$ would be in the value of the vibrational wavenumber $(\tilde{\nu}),$ which you can get from the vibrational frequencies that you already have, as you wrote in the question. Once you substitute that in the equation, you will get your answer.


How can you simulate the effect of a solvent on a polymer in an MD simulation?

You have generally two choices a) implicit and b) explicit solvent. Presumably, you don't have good interaction parameters for explicit solvent, since you are likely doing a generic polymer melt.

There are a variety of solvation models; LAMMPS, for instance, implements a dielectric pair style for which you can attenuate the dielectric constant of your solvent (and a few other things). Other popular approximations are generalized Born models.

This paper uses GROMACS to simulate poly(styrene sulfonate) and poly(allylamine hydrochloride) in explicit and implicit solvent, as an example from the literature.


Why neglecting OH Ions from charge balance?

This is not totally unreasonable.

Calcium carbonate as solubility product of $K_{\mathrm{sp}} = 3.3\times 10^{-9}$ (source).

This means that if you're dissolving the solid, you'd expect a maximum calcium ion concentration of $\ce{[Ca^{2+}]} = 5.7\times 10^{-5}\ \mathrm{M}$. This is a 2.5 orders of magnitude larger than either hydronium or hydroxide concentration in neutral solution. You'd expect the carbonic acid to be mostly dissociated (at least 50%), so the bicarbonate concentration will be similar to the calcium ion concentration (maybe factor of 2 smaller).

The moment you step away from a neutral solution though, this analysis totally falls apart. You'd need more details on the assumptions of the model in that case.


Development of experiments to optimize HPLC-MS procedure using 2D central composite full factorial design

FCC is a variant of a CCD design. It's usual to run your 2 factorial design first and establish which factors are relevant.  In your case, it won't matter so much but if you have 5 or more factors, it does.  The 2 factorial establishes your range with an upper and lower limit (+1 and -1). When you move to CCD,  you now have options. Using alpha of root 2 gives your "response surface design" orthogonality and rotatability, meaning, loosely, it is easier to analyse for variance in your model. The reason for varying alpha usually comes down to operational reasons e.g if a negative time or negative mass would result.  This is when a FCC might be used instead. If you ignored your coded +1 and -1, in effect your root 2 values would become your new +1 and - 1. The rationale for this is when a formula is generated to model each response,  the one using coded factors (set at +1 or -1) will indicate the relative influence of each factor, which the actual factor values will not do. The formula using coded values will not base it on your alpha value.


General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $\mathrm{sp}^3$ hybridized.

No. One could say that, but $\mathrm{sp}^2$ is equally possible. In fact, it may seem more likely, we’ll get to that in a second.

Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2\mathrm{v}}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $\mathrm{a}_1, \mathrm{b}_1$, and $\mathrm{b}_2$.

Yes, and you did a nice job of drawing them.

There is only one non-bonding orbital ($1\mathrm{b}_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

And this is where it gets interesting.
I’ll once again point to Professor Klüfers’ web scriptum for the basic and inorganic chemistry course in Munich, section about localising molecular orbitals. If you don’t understand German, all you need to do is look at the pictures and understand that wenig means a little in this context.
You see, to do a discussion such as ‘this is the bonding electron pair’ we need to localise molecular orbitals. Specifically by linear combining them. It’s almost like in the linked pictures: $\Psi_2 (1\mathrm{b}_2) + \Psi_3 (3\mathrm{a}_1)- \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_1$. Invert the signs: $\Psi_2 (1\mathrm{b}_2) - \Psi_3 (3\mathrm{a}_1)  + \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_2$. These are your two bonding orbitals to hydrogen. We have some $\Psi_1$ remaining (and, theoretically, also some $\Psi_2$ and $\Psi_3$, since we didn’t use all of them). That (remaining linear combination) is our second non-bonding orbital.

Now why is $\mathrm{sp}^3$ a bad description? You see, we arrived at one lone pair that is definitely $\unicode[Times]{x3C0}$-shaped or p-shaped. The second is somewhat p-ish but also somewhat s-ish, so maybe some sp-hybrid. But there is no mistaking that $1\mathrm{b}_1$ is in no way influenced by all the others. It is antisymmetric to the $\ce{H-O-H}$-plain of symmetry while all the others are symmetric. $\mathrm{sp}^2$ describes this a lot better.


What is the range of values of n in [Cu(NH3)4(H2O)n]SO4?

Tetraamminecopper(II) Sulfate or cuprammonium sulfate can be synthesized as its monohydrate (Ref.1):

  A solution of $\pu{50 g}$ of finely divided $\ce{CuSO4 . 5 H2O}$ in $\pu{75 mL}$ of conc. ammonia and $\pu{50 mL}$ of water is filtered and precipitated by slow addition of $\pu{75 mL}$ of alcohol. After standing for several hours in the cold, the crystals are filtered on a Büchner funnel, washed with a mixture of alcohol and conc. ammonia (1:1) and then with alcohol and ether, and dried by suction.

$$\ce{CuSO4.5H2O + 4NH3 -> [Cu(NH3)4]SO4.H2O + 4H2O}$$

The properties of tetraamminecopper(II) Sulfate monohydrate is listed as follows (Ref.1):

  Deep blue crystals, $\mathrm{d}^{20}_4$: 1.81. Solubility ($\pu{21.5 ^{\circ}C}$): $\pu{18.5 g} / \pu{100 g}\: \ce{H2O}$. Decomposes in air. Loses $\ce{H2O}$ and $\ce{2NH3}$ on heating to $\pu{120 ^{\circ}C}$; the remaining ammonia is evolved at $\pu{160 ^{\circ}C}$.

The crystallographic study of tetraamminecopper(II) Sulfate monohydrate was done several times, and most recently by Mazzi (Ref.2). In the abstract of Ref.1, the author states that:

  $\ce{Cu(NH3)4SO4.H2O}$ is orthorhombie, $Pmcn-D^{16}_{2h}$, with four molecules in the unit cell. The determination of the structure was carried out by means of Patterson and Fourier projections on $(001)$ and $(100)$. The structure consists in layers parallel to $(010)$ formed by $\ce{Cu(NH3)4}$ squares and $\ce{SO4}$ tetrahedra held together by $\ce{H2O}$ molecules. Copper atoms, which lie on symmetry planes, have four nearest $\ce{NH3}$ neighbours ($\ce{Cu-N} = \pu{2.04-2.06 Å}$) and two more $\ce{H2O}$ neighbours ($\ce{Cu-O} = \pu{2.59-3.37 Å}$), which complete the distorted octahedral co-ordination group. Each $\ce{SO4}$  tetrahedron ($\ce{S-O} = \pu{1.40-1.43-1.46 Å}$) has two oxygen atoms linked with hydrogen bonds to adjacent $\ce{H2O}$ molecules ($\ce{OH...O} = \pu{2.67 Å}$).

All of these data directed to the comment of @andselisk who states that tetraamminecopper(II) sulfate is a monohydrate but, Probably why Wikipeda article suggesting it as  $\ce{[Cu(NH3)4(H2O)n]SO4}$ that this compound may be hygroscopic, and thus, its water content may deviate (or vary). The crystal structure has confirmed it is a monohydrate.

References:

Georg Brauer, Ed., Handbook of Preparative Inorganic Chemistry – Volume 2; 2nd Edn, P. G. Stecher, Translation Ed., Translated by Scripta Technica, Inc.; Academic Press: New York, NY, 1965, “Section 19: Copper, Silver, Gold, by O. Glemser & H. Sauer,” p. 1021.
F. Mazzi, “The crystal structure of cupric tetrammine sulfate monohydrate, $\ce{Cu(NH3)4SO4.H2O}$,” Acta Cryst. 1955, 8, 137–141 (https://doi.org/10.1107/S0365110X55000571).


How do I interpret a flux quoted in cm⁻² s⁻¹ in atmospheric chemistry?

The "strange" unit describes the photon flux, which is the number of photons (of unspecified wavelength) that hit an area (thus $\mathrm{cm^{-2}}$) in a time interval (thus $\mathrm{s^{-1}}$).

  It feels like it should be something like CCs of gas produced […]

Not really. Not every photon leads to an excited state, not every excited state triggers a chemical reaction.


If you live close to a research university, I would recommend you to look up if they have a surplus store. Often times when professors retire a lot of the lab supplies get sent to surplus, where they are sold very cheap. You can also build equipment out of basic stuff you can find at any thrift shop (pressure cookers, blenders, etc).

As for chemicals, it really depends on what you are interested in doing. Metals are all around you. You can take some pennies, dilute hydrogen peroxide, any carboxylic acid (vinegar, citric acid, amino acid supplements, etc.), and stick the mixture into a microwave for a few seconds. The copper will form complexes with the carboxylic acid you add. Different acids will give you different colors. Leave them standing for a bit and you will get crystals.

If you get to know the plants around you, you will have access to a huge array of naturally occurring small molecules and enzymes. Some palm trees have a high concentration peroxidases, onions and garlic undergo some pretty cool chemistry which results in the formation of pigments (blue, green, pink), there's pH indicators, capsaicin, etc.

Buying pure chemicals for organic synthesis is usually tricky, and for many of them you need to be affiliated with a research institute.

There are several YouTube channels with chemistry experiments which mentions where they obtained the chemicals. I recommend 'NurdRage'.

Good luck!


Enthalpy change for exothermic and endothermic reactions

According to the definition,
 Enthalpy = Enthalpy(products) - Enthalpy(reactants), this should be followed.

When you are saying that 40 J are used up it means that 40 J are used in transforming the reactants to an activated complex(reaction intermediate) and when you say that 50J are released then it means that 50J is liberated when this activated complex is transformed to the product.

Assume that the energy of the activated complex be 0, then the energy of the reactant must be -40J(because 40J are required to convert reactant to the activated complex) , and the energy of the products should be -50J( because energy is released in converting the complex to the product)

Enthalpy(products)=-50J

Enthalpy(reactants)=-40J

Therefore,
 Enthalpy = (-50)-(-40) J = -10J.


Comparison of basicity of o-phenanthroline and ammonia

A quote from here explains and give a good answer to your wondering:

Amines are the most basic of the common organic functional groups, but are still fairly weak bases. Protonation occurs on the non-bonded electron pair exclusively. The basicity of amines is directly dependent on the “electron density” at the nitrogen atom. Both inductive and resonance effects can alter the basicity of a nitrogen atom.
Hybridization on the $\ce{N}$ also affects basicity. An increase in $\mathrm{s}$ character on an atom increases the electronegativity of that atom which favors acidity and therefore disfavors basicity. Hence $\mathrm{sp^3}$-hybridized nitrogen is more basic than either $\mathrm{sp^2}$ or $\mathrm{sp}$ hybridized nitrogen.

The availability of this non-bonding lone pair is a factor of basisity. The 1,10-Phenanthroline is a pyridine derivative. Thus, both lone pairs of two $\ce{N}$ atoms in the ring system are contributed to the system's aromaticity. This make them not 100% available to incoming protons while the lone pair in ammonia is 100% available. Therefore,  in aqueous solutions, the basicity of $\ce{NH3} \ (\mathrm{p}K_\mathrm{a} = 9.3)\gt$ basicity of 1,10-Phenanthroline $(\mathrm{p}K_\mathrm{a} \approx 4.9)$. This is similar to $\mathrm{p}K_\mathrm{a}$ of pyridine, which is $5.2$ in water (comparison with piperidine is depicted in following scheme):

Keep in mind that there are some other factors are effecting the basicity of amine as well. For example, the ring sizes of cyclic amines (Ref.1):

The order is 5-membered $\ge$ 4-membered $\gt$ 6-membered $\gg$ 3-membered.
References:

Scott Searles, Milton Tamres, Frank Block, Lloyd A. Quarterman, "Hydrogen Bonding and Basicity of Cyclic Imines," J. Am. Chem. Soc. 1956, 78(19), 4917–4920 (https://doi.org/10.1021/ja01600a029).


Calculate the mole fraction of component B in this solution? (partial pressures)

Raoult's law is the correct approach to solving this problem. The definition of Raoult's law is explained well in this Wikipedia page as:

  a law of thermodynamics established by French chemist François-Marie Raoult in 1887. It states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.  Mathematically, Raoult's law for a single component in an ideal solution is stated as:
  $$P_i = P_i^*x_i$$
  where $P_i$ is the partial vapor pressure of the component $i$ in the gaseous mixture (above the solution), $P_i^*$ is the vapor pressure of the pure component $i$, and $x_i$ is the mole fraction of the component $i$ in the mixture (in the solution).

Because you will be solving for mole fraction, you can rearrange the equation to:

$$x_i = P_i/P_i^*$$
$$x_i=\pu{0.60atm}/\pu{0.80atm} = 0.75$$

Of course the value calculated above is for the volatile solvent A. Since the total mole fraction of both components combined is equal to unity, the mole fraction of the non-volatile solute B must be $1 - 0.75 = 0.25$.


What is pre-concentration (context: Breath analysis)?

Pre-concentration just refers to the process of concentrating a sample before analysis, so that trace components won't be overlooked. If you're looking for something that's only present at a few parts-per-million, you could easily miss it if you used the raw sample, so ideally you want a way to separate off either large, common sources of noise/error, or to just extract the exact components you're looking for.

A quick look suggests, for example, that a key first component to remove in breath analysis is moisture, as some volatile components will dissolve readily in the water and make it harder to detect them.

Anyway, this paper on a few pre-concentration and detection techniques might be informative: Mochalski P, Wzorek B, Sliwka I, Amann A; Chromatogr B Analyt Technol Biomed Life Sci. 2009 Jul 1;877(20-21):1856-66.


Sodium borohydride is a complex unit made from the let's call it molecule $\ce{BH3}$ and the salt $\ce{NaH}$. Group 3 compounds like boron only have what we call an electron sextett instead of an octett. That makes them good Lewis acids, hence they tend to react with Lewis bases to form adducts. A famous example is the $\ce{H3B-NH3}$ adduct. In your example the hydride anion $\ce{H^-}$ is the Lewis base that forms an adduct with $\ce{BH3}$ to give $\ce{Na[BH4]}$. This means that at least one $\ce{H}$ in $\ce{[BH4]^-}$ can be somewhat easily removed again. The others tend to react as well but let's just call the $\ce{BH3}$ unit a stable molecule and assume it's a combination of $\ce{BH3}$ and $\ce{NaH}$. It's just a good way to store this hydride-anion so that it can be easily removed again. So once it sees your carbonyl group that has a positive partial charge on the carbonyl carbon (as the more electronegative oxygen pulls the electrons closer to itself) it can attack there as nucleophile. As you drew it already the double bond will turn to a single bond giving a negative charge on the oxygen.

Next you have to assume that $\ce{B}$ and $\ce{O}$ form quite stable bonds, hence the formation of a $\ce{B-O}$ bond is somewhat favored. And of course once the hydride ion is at the carbon we end up with an electron sextett on the boron again. Therefore it's the same thing, the carbonyl oxygen with an extra negative charge behaves like a strong Lewis-base again forming an adduct with the $\ce{BH3}$ unit.

This is later treated with water or acidified water to get the alcohol. To the question on the final product. I once talked to a boron chemist about that. He doubts that all four hydrogens will act as reducing agent under normal conditions. It's more likely that they will get hydrolyzed during workup.

Edit: Thinking about it. If you aren't sure about what attacks first you could probably also consider a transition state. I drew in the arrows here but you shouldn't do that (although you see it too often). I guess this would just be an alternative view on the reaction.


Theoretical comparison of dipole moments of N,N-dimethyl-4-nitroaniline and N,N,2,3,5,6-hexamethyl-4-nitroaniline

The workbook's explanation could have been more specific. If you build a 3D model, you may see that the orbital interaction between the lone pair on the $\ce{NMe_2}$ and the $\pi$-system on the ring is somewhat hindered by bad overlap because the steric interaction of the methyl groups (on the ring and the $\ce{N}$) prevents perfect alignment. Same for the nitro substituent. A higher interaction would lead to more charge shift.


If a good choice of a unit cell should be the one of most symmetry ,then why keep body centered tetragonal if face centered cubic exists?

Body-centered tetragonal is face-centered cubic only if $c/a=\sqrt2$.  If you try your transformation with a $c/a$ value greater/less than $\sqrt2$, your "cube" will have lateral edges that are longer/shorter (resp.) than the basal edge and so really remains tetragonal.
You can also read this the other way.  If you are presented with a body-centered tetragonal lattice and $c/a$ seems to be close to $\sqrt2$, maybe it's really fcc.


Why is fluoride the most nucleophilic halide in aprotic solvents?

When talking about elements of a group, the size of the atom is more effective than the electronegativity factor.
As $\ce{F-}$ size is too short so its electron density will be much higher than other hallogens.

In aprotic solvent the order of nucleophillicity:
$\ce{F- > Cl- > Br- >I-}$(No hydration occurs) In protic solvant(like $\ce{H2O}$) its reversed because of hydration of ions having much electron density.


Why is atmospheric composition compared by volume?

The short answer is that (a) partial pressures have units of pressure, and (b) pressure changes with altitude and/or temperature.

  The composition of the Earth's atmosphere by its constituent gases (nitrogen, oxygen, argon...), is often given by percent volume. (See this Wikipedia page, for example). Why are these compositions given by volume? I understand that a volume composition shows the mole fractions, which is the real point of interest.

You are right that the mole fractions are the real points of interest.  You may additionally be interested to know that many techniques that measure the percent volume of a certain gas actually measure its mole fraction.  Thus, reporting the results in percent volume is correct only if the gas is ideal.

  Volume compositions compare the volume taken up by different amounts of gases under the same hypothetical pressure and temperature. However, in any "container," each gas occupies the same volume as another; what differs is the partial pressure of each gas. Thus, why not base compositions off of partial pressures?

Partial pressures have units of pressure.  So at 1 bar of pressure at 298 K, oxygen has a partial pressure of 0.21 bar and nitrogen has a partial pressure of 0.78 bar.  Let's say you had a closed rigid container of gas initially at this pressure and temperature.  Let's now say that it's a hot day so the temperature is 313 K.  Now the total pressure is $1\times\frac{313}{298}=1.05$  bar, and as a result the oxygen partial pressure is 0.22 bar and the nitrogen partial pressure is 0.82 bar.  But meanwhile the volume fraction (or equivalently the mole fraction) of each gas is unchanged.

  It is physically correct to say that oxygen accounts for 21% of the pressure of the atmosphere, rather than saying that oxygen accounts for 21% of the volume of the atmosphere (which is thankfully not true).

I don't fully understand this sentence but do not believe it is correct.  Oxygen does account for 21% of the volume of the atmosphere.

  I understand that comparing pressure under equal volume and comparing volume and equal pressure both show the same thing, but why not use pressure if it is more physically correct?

It is not any more physically correct than using volume fraction.  It additionally is complicated that partial pressures have units and will change with temperature or total pressure.


Conversion of carboxylic acid to alcohol

Directly using $\ce{LiAlH4}$ works, but borane is probably a better choice:

  Borane ($\ce{BH3}$) is particularly good for carboxyl groups and permits selective reduction of them in the presence of many other groups (although the reaction with double bonds takes place at about the same rate in ether solvents). For many years, borane was the reagent of choice for this reduction. Borane also reduces carboxylic acid salts. (March 7ed, p 1515)

Clayden 2ed, pp 531-2 also discusses the borane reduction. Here's a small portion of it:


What is the molecular formula of the nitrate of a metal Q that forms a carbonate of the form QCO3?

The carbonate anion is $\ce{CO3^{2-}}$. This tells us that the metal cation is doubly positively charged:
$$ \ce{Q^{2+} + CO3^{2-} -> QCO3} $$

The other two anions mentioned in your question are

the nitrite anion $\ce{NO2^-}$
the nitrate anion $\ce{NO3^-}$

Since the question you're concerned with asks for the formula of the nitrate of $\ce{Q}$, the correct answer would be $\ce{Q(NO3)2}$.

However, the compound $\ce{Q(NO2)2}$ is, at least from a charge perspective, also correct. But that is the nitrite of $\ce{Q}$.


How to measure the speed in a catalized reaction?

It depends on scenario.

If time is cheap, while measurement is expensive ( e.g.high fixed price for each value ), then constant time method is probably cheaper.

If time is expensive, while measurement cheap ( e.g. continuos measurement ), the constant products method may be cheaper.


First, let me say that I've enjoyed many times exploding soap bubbles of about one milliliter filled with hydrolysis gas. That is 1 cubic centimeter.
That will give you a sound that rings in your ears in a decent sized living room. You may wish to use ear protection for the experiment.

50 ml will have an effect in a lecture hall that not only wakes up everyone, but also may make people complain.

Now while the explosive limits of hydrogen in air range from about 18 -- 60 % the flammable limits are from 4 -- 75 %, in oxygen the limit of flammability goes all the way from 4% to 95% read: for practical purposes, hydrogen in oxygen is always at least a flammable mixture.
For comparison, gasoline in air is flammable roughly between 1.5 - 7%.

However, you start with a stochiometric mixture which is ideal for explosion (after all, that's why you do it, right?), and it is in oxygen, not in air. So even if you "only" have a flammable mixture reaching an ignition source (e.g. electric switch), chances are quite high that as the ignition proceeds through the mixture, it can reach a zone where the gas/air mixture is explosive. Also, because hydrogen is so much lighter than air, it tends to accumulate under the ceiling, so even if there is enough air in the room that an ideal mixture would not be explosive, chances are that there is an explosive layer of gas. (Same with gasoline or, practially more relevant: solvent from glue for parquet floor tiles)

The safety relevant points are:

the limits of explosion and flammability are, even in air, extremely wide.
limits of flammability are important even if there happens to be no explosive mixture: hydrogen is a small and light molecule that diffuses fast. If the air you breathe happens to have a hydrogen concentration within the flammability limits, any ignition will burn deep into your throat on its way to the lungs.


Monochlorination of 2-chloropentane and possible enantiomer products

2-chloropentane is optically active compound hence in some cases we will get only diastereomers. For example 1,2-dichloropentane is single product because it is formed by attack at carbon no. 1 hence no change in chirality of carbon no. 2.  So product formed by attack at carbon 1 (one chiral product) carbon3 (two chiral product i.e. diastereomers) carbon 4 (one chiral product, i.e. one chiral and one meso) carbon5 (one chiral product)


In what order should the compounds containing Ru and Os, or Mo and W be named?

When IUPAC rules refer to electronegativity (and electropositivity), they don't mean actual electronegativity measured by one or another scale. IUPAC decided, for nomenclature's sake, on a formal ordering of electronegativity, which goes like this:

This is Table VI, on page 260 of IUPAC's very own 2005 Red Book, aka Nomenclature of inorganic chemistry. The cases in which it is used are detailed in IR-2.15.3.1 (page 42).

The explanation of the changes from earlier IUPAC recommendations is in IR-1.6.3 of the Red Book:

  The element sequence of Table VI
  In Nomenclature of Inorganic Chemistry, IUPAC Recommendations 1990 (Ref. 11), the position of oxygen in certain element sequences was treated as an exception. Such exceptions have been removed and the element sequence of Table VI is now strictly adhered to. In particular, oxygen is treated as the electropositive component relative to any halogen for constructing compositional names (Section IR-5.2) and corresponding formulae (Section IR-4.4.3) for binary compounds. This results in, for example, the formula O2Cl and the name dioxygen chloride rather than the formula ClO2 and the name chlorine dioxide.

  In Ref. 11, the formulae for intermetallic compounds were also subject to an exceptional rule although no guidance was given for naming such compounds, and the term ‘intermetallic compound’ was not defined. The problem is to define the term ‘metal’. Therefore, no attempt is now made to make a separate prescription for either the formulae or the names of intermetallic compounds. It is stressed, however, that the present recommendations allow some flexibility regarding formulae and compositional names of ternary, quaternary, etc. compounds. Several ordering principles are often equally acceptable (see Sections IR-4.4.2 and IR-4.4.3).

  The element sequence of Table VI is also adhered to when ordering central atoms in polynuclear compounds for the purpose of constructing additive names (see Section IR-1.6.6).

For more on this topic, you've got to go read (and upvote) this excellent answer by Manishearth, which is where I first learnt that useful and intriguing nugget of IUPAC idiosyncrasy!


The question can be understood as:
Which one of the following oxidants has a greater n-factor as there's only 1 mole each of permanganate ion and dichromate ion.

The n-factor of permanganate ion is (7-2)=5

while

The n-factor of dichromate ion is (6*2-3*2)=6

So, the answer is the 2nd reaction.


Why is proton coupling through bonds and how does this affect identical protons?

How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?

Through space coupling is known as dipolar coupling and is actually much greater than through bond coupling (J-coupling).  However, in low viscosity liquids the chemical shift effects of dipolar coupling are averaged to essentially a single value by molecular rotation.  Relaxation effects of dipolar coupling are still observed.

  Why do identical protons not influence each other?

For two protons there will be four states $\alpha \alpha , \alpha \beta , \beta \alpha$, and $\beta \beta$.   In the general case, each state may have a different energy and there can be four possible one quantum transitions ($\alpha \alpha$ to $\alpha \beta$, $\alpha \alpha$ to $\beta \alpha$, $\alpha \beta$ to $\beta \beta$, and $\beta \alpha$ to $\beta \beta$), each transition having a unique energy and hence 4 peaks (two doublets).

However, when the two protons are equivalent, $\beta \alpha$ and $\alpha \beta$ will be the same energy, $E_2$ let's say.  So, letting the energy of  $\alpha \alpha$ be $E_1$ and $\beta \beta$ be $E_3$, and given a one quantum transition, the possible energies for transtions are only $E_3$ - $E_2$ and $E_2$- $E_1$.  But $E_3$ and $E_1$ are equally spaced above and below $E_2$, so there is only one energy of transition.

For a more-rigorous explanation see: http://www.chemie.uni-hamburg.de/nmr/insensitive/tutorial/en.lproj/energy_levels.html


Electron configuration and the 2p orbital

first of all electrons do not rotate but they have spins
next magnetic quantum number decides how many electrons an orbital can have
s can have 2(m=1){consider 1 box}
p can have 6(m=3){consider 3 boxes}
d can have 10(m=5){consider 5 boxes}
f can have 14(m=7){consider 7 boxes}
where each box can have at max 2 electrons and each box represents different orientation of the orbital
and all the orbitals(except s) have degenerate orbitals, i.e they have the same energy so all have to filled with atleast one electron before pairing up. this rule is known as Hund's Rule of Maximum Multiplicity
there are are some more reasons fro it. they are
to maximize exchange energy
to attain symmetry


How concentrated does the analyte solution need to be in order to perform a (chelatometric) titration

Sorry to tell you a bad news. Classical methods, like titrimetry or gravimetry do not work very well at low concentrations. Failure would be a strong word. The beauty of EDTA is that it forms 1:1 complex with most metals. Pb is not an exception. If your lead solution is 0.000001 M, your EDTA solution should be 0.000001 M. Such concentrations are never recommended in classical methods.

For ppm level, spectrophotometric methods can work quite well. Go to Google Scholar and search the methods in Google Scholar


Does there exist any relation between metallic property and oxidation potential?

The short answer is yes. The reason for this is that metals exhibit metallic properties because their electrons are relatively loosely bound to the atoms. A metallic bond is one in which the valence electrons are free to move throughout the lattice structure of the metal. Here's a schematic diagram from wiki commons:

The idea is that the positively charged metallic "cores" (which includes the nucleus and the more tightly bound inner core electrons) are sort of suspended in a crystalline structure within a "sea" of mobile electrons.

This model explains most of the  metallic properties we observe, from heat and electrical conductance, to malleability and ductility.

Now think about oxidation - in an oxidation-reduction (redox) reaction, one element gains electrons (it is reduced) and the other loses electrons (it is oxidized). Here's an animation, also from wikipedia:

So, for an element to be metallic, its valence electrons need to be relatively "loose" - it can't take too much energy to detach them. For an element to be oxidized, something else has to "want" the electrons more - the energy cost of removing the electron has to be lower then the gain from adding them to the reducing agent.

As a result there is a very strong correlation between the metallic nature of elements and their relative ease of oxidation. There is also a very strong correlation between both of these properties and the trends in ionization energy, electron affinity, and  electronegativity.

Of course, in chemistry, things are never that simple. There lots of details and it is tough to make general statements that are always true. For example, you can say that often times the more "metallic" an element is, the more easily it will be oxidized. However, when you compare elements using a standard electrode potential table (which gives us a way of measuring relative ease of oxidation) you find that it is easier to oxidize calcium ($-3.8 \space \rm{V}$) than it is to oxidize potassium ($-2.9 \space \rm{V}$), even though potassium is closer to the "metallic side" of the periodic table, it has a lower first ionization energy, and it is less electronegative.

This illustrates a limitation of thinking of elements in terms of metals and non-metals: it is really a shorthand way of describing all of the things that affect how tightly those valence electrons are bound to the atom, in comparison to how badly other elements "want" to remove them. With that amount of generalization, we lose too much detail to be able to make accurate predictions in every situation.


Is the E in Arrhenius equation the change of enthalpy?

The "E" is the height of the thermodynamic potential energy barrier. Which potential that is (enthalpy, Gibbs energy, Helmholtz, or internal energy) depends on the conditions under which the reaction takes place - which thermodynamic state variables are held constant.

For example, in a system where the number of moles, the pressure, and the temperature are held constant (called NPT), then the thermodynamic potential energy (and thus the activation energy) is measured using Gibbs free energy.


Why exactly is a fluoride anion a stronger base than iodine anion?

First off, I've learnt that stronger acids produce weaker conjugate bases (through Brønsted–Lowry acid–base theory).

That is correct.

  Then I looked at the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ and came to the conclusion that $\ce{HI}$ is the strongest among them and that explains why $\ce{I^-}$ is the weaker base.

I don’t agree with that explains; but the observation is correct, iodide ($\ce{I-}$ is the weakest base and $\ce{HI}$ is the strongest acid.

  But then I remembered that $\ce{F^-}$ is in fact the strongest oxdizing agent out there.

No. The fluoride anion is not an oxidising agent. If it were, it would take up electrons to become a fluoride dianion ($\ce{F^2-}$), inserting an electron into the 3s level—not happening.

What you probably meant is fluorine (the element; the diatomic molecule) being the strongest oxidising agent out there. That might be correct or it might not (I am not up to date with extremely strong oxidising agents) but it is not a statement I could dismiss at first glance. Certainly, fluorine is a stronger oxidising agent than iodine.

  So shouldn't that mean that $\ce{F^-}$ ion is more stable than $\ce{I^-}$?

Thermodynamic stability is sometimes a tricky beast, but in general a fluoride ion is more stable than an iodide ion which is prone to oxidation to give iodine or iodate. This is connected to the oxidising ability of fluorine (strong) versus iodine (weak), not to the strengths of the acids, though.

  And hence does not give away its electrons easily and therefore is a weaker base?

The conclusion is invalid. Not wanting to give away one’s electrons means not wanting to get oxidised. As mentioned, fluorine is a strong oxidising agent so to reduce fluorine to fluoride is a favourable process. But acid-base chemistry has nothing to do with gaining or losing electrons: it is an entirely electrostatic process and builds on different principles.

The high oxidising ability of fluorine can be thought of as a side effect of its high electronegativity: it has a tendency to strongly attract electrons since it has a highly positively charged nucleus that is shielded by only two core and seven valence electrons. Adding an eighth electron is comparatively easy. For iodine, there are four core shells and a valence shell between the nucleus and the incoming electron; this shielding effect can be thought of as reducing electronegativity and oxidising ability.

The high acidity of $\ce{HI}$ and low basicity of iodide is a direct consequence of its size: the negative charge is distributed across a much larger volume meaning it is more stabilised. In fluoride, the negative charge is confined to a much smaller volume meaning that positive charges such as a proton are attracted stronger. This is the underlying reason for the observed acidities/basicities.


What is the formula of bleaching powder?

The bleaching agent in all commonly available bleach powders is hypochlorite.  Calcium hypochlorite is a pure compound with the formula $\ce{Ca(OCl)2}$ and is the active component in most bleach powders.  (Note: active does not necessarily mean the most abundant component).

You presented three examples of formulations for bleaching powder that you have encountered:

  $\ce{Ca(OCl)2}$
  $\ce{Ca(OCl)Cl}$
  $\ce{CaCl2 + Ca(OH)2 + H2O -> CaCl2\cdot Ca(OH)2\cdot H2O}$

The third species, $\ce{CaCl2\cdot Ca(OH)2\cdot H2O}$, cannot be a bleaching agent of any type because it does not contain hypochlorite.  So I view that as plainly incorrect.

According to wikipedia,

  Bleaching powder is made with slightly moist slaked lime. It is not a simple mixture of calcium hypochlorite, calcium chloride, and calcium hydroxide. Instead, it is a mixture consisting principally of calcium hypochlorite $\ce{Ca(OCl)2}$, dibasic calcium hypochlorite, $\ce{Ca3(OCl)2(OH)4}$, and dibasic calcium chloride, $\ce{Ca3Cl2(OH)4}$ [7].

I was able to find the source cited by Wikipedia online; it seems quite authoritative.  It says:

  Calcium hypochlorite solutions are made by suspending lime or hydrated lime in water and adding
  chlorine. However, reacting lime with water generates lots of heat, and this heat must be allowed to
  dissipate before chlorination. The maximum concentration is ∼85 g/L
   of available chlorine, above
  which dibasic calcium hypochlorite begins to precipitate.
  Chlorination can be done in a batch or continuous reactor similar to those used to make sodium
  hypochlorite, except that agitation is needed to maintain the suspension. In a batch reactor, agitation
  can be achieved using a mechanical stirrer, or, more effi ciently, by circulation. A centrifugal
  pump is used to pull the solution from the bottom of a conical tank and then reinject it tangentially
  to create a strong vortex. Alternatively, the solution passes through a packed tower before it returns
  to the tank. The packing in the tower creates shear that breaks the particles apart before the chlorine
  is injected at the bottom of the tower.57 The reaction is as follows:
  $\ce{2Cl2 + 2Ca(OH)2 → Ca(OCl)2 + CaCl2 + 2H2O}$

  [...]

  Hemibasic calcium hypochlorite, $\ce{Ca(OCl)2\cdot 0.5Ca(OH)2}$, can be precipitated from a solution of calcium hypochlorite at 40–80°C.143 These solutions are commercially made by chlorinating a
  slurry of hydrated lime.5
   The precipitate is removed by filtration and dried. Although it may still be
  made in Asia, it has not been made in the United States or Germany since 1955.

  [...]

  Dibasic calcium hypochlorite, $\ce{Ca(OCl)2 · 2Ca(OH)2}$, is commonly used as an intermediate in making calcium hypochlorite, $\ce{Ca(OCl)2}$.   It has never been successfully marketed as a product because it dissolves much too slowly.

[...]

  Although calcium hypochlorite dihydrate, $\ce{Ca(OCl)2 · 2H2O}$, can be made by chlorinating a suspension
  of lime, it is difficult to filter, partly due to the presence of calcium chloride. Calcium chloride also reduces the stability of the dry product. [...] Sodium hypochlorite is commonly added to convert calcium chloride to calcium hypochlorite
  as follows: $\ce{CaCl2 + 2NaOCl → Ca(OCl)2 + 2NaCl}$

Pages and pages follow with detailed process descriptions for the manufacture of various formulations of bleaching powder.  Suffice it to say, an ideal powder would be something like pure $\ce{Ca(OCl)2 · 2H2O}$, but this is very difficult to prepare economically, because of contamination with various amounts of chloride salts, hydroxide salts, or (depending on the route of manufacture) sodium salts.


What is the role of a monochromator in AAS?

This question has lingered on the site for a long time, which is a shame because it is a good question to ask.  Atomic absorption spectroscopy (AAS) uses a hollow cathode lamp (HCL) as a source, and HCLs are line sources; therefore, if a line source is being used, why would an instrument need to be made more complex by adding a wavelength selector?

The first point to consider is what the source really looks like.  The emission spectrum from HCLs have multiple lines.  Here, for example, is the emission from a nickel HCL, taken from Spectrochemical Analysis by Atomic Absorption and Emission by L. Lajunen.

A wavelength selector is needed in order to ensure that only one of the emission lines is being used for the analysis.

Additionally, light can be scattered in the flame, which can introduce errors in the analysis if not eliminated.  If you compare the block diagram of an AAS vs a typical visible spectrometer, you'll note that the wavelength selector is typically placed after the sample compartment for AAS to address these concerns about scattered light; however the issue is not that significant in visible spectroscopy so the wavelength selector can be placed before the sample holder.

This representative block diagram for AAS is taken from here.

A representative UV/Vis spectrometer block diagram is from this source

Bottom Line AAS needs a wavelength selector to ensure that the analytical line of interest is the only radiation hitting the detector and it minimizes the negative effects of radiation scattered in the flame.  A good summary can be found here.


I learned that the absolute and relative error have only 1 significant figure

That really isn't  true. In your example you showed 5.34532g ± 0.001428g which could result from some kind of experiment. The point being that the digits 1428 could all be significant.

You then simplified to 5.345g ± 0.001g which is fine. My personal preference would be more like 5.345g ± 0.0014g since ± 0.00067g would round to ± 0.001g as well. So I'd prefer one more digit in the ± "precision" than the measurement itself.

  If there is an experiment with a lot of steps and error propagation wouldn't the rounding of all the errors in every single step change the result a lot?

Yes absolutely. Carry "extra" significant figures until the final step.

  Wouldn't be rounding the error just in the end make more sense?

That is the preferred method.

We typically use a calculator now to do the calculations. You can set the calculator to show 3 significant figures but it carries more internally.

There are various statistical techniques which allow all the measurements used in the calculations to have random errors injected and then the  calculation is simulated many times to see how the errors propagate through the calculations.


(a) is not true, because, in addition to being adiabatic, the process must be reversible.

(b) is not true, because the gas must be assumed to have constant heat capacity for this equation to apply


Will cyanide substitute bromine in bromobenzene?

Economically, your approach is probably a bad idea.

But in general, the conversion of aryl halides to aryl nitriles does work.

I have prepared large amounts of 2,3,5,6-tetramethylterephthalonitrile from 1,2,4,5-tetramethylbenzene in a two-step procedure

iodination
reaction of the diiodide with $\ce{CuCN}$

back in the days for my PhD thesis. For the cyanation, I followed the procedure described in Synthesis, 1974, 53-55, but used N, N-dimethylformamide (DMF) as a solvent.

The halogen-cyanide exchange is know as the Rosenmund-von Braun reaction and you might find further information in a review on Cyanation of aromatic halides (Chem. Rev., 1987, 87, 779-794).


Which formula is correct for calculating the heat of dissolution?

Your energy conservation equation is $$q_{rxn} + q_{soln} = 0$$ so $$q_{rxn} = -q_{soln} = -m_{soln}c_{soln}\Delta T_{soln}$$ Note the minus sign; if $\Delta T$ is negative, the energy will be positive, and vice versa. And yes, you need the mass of the solution here, and the specific heat capacity of the solution too.

If you only need a couple of significant figures and the solution isn't too concentrated, you can assume $c_{soln} \approx c_{\rm H_2O}$.

If you need more precision, though, you'll need to measure or look up $c_{soln}$. And you'll need to include the heat absorbed by the calorimeter in your energy conservation equation, too.


How do we know the size of atoms in body centered cubic system?

The easiest answer in my opinion; try it! If each atom has the same radius, it is simply not possible to fit one of them into the center of the cubic system (known as a cubic hole). It is, however, possible if the atom is sufficiently small. CsCl is an example of such a compound.

One can show this geometrically (as is done here), but you can also convince yourself that it must be the case using equally-sized tennis balls or something similar to try it out in practise.


Why can one product in an equilibrium system not be produced without the other?

In the first experiment, the dissolved $\ce{CuSO4}$ reacted with $\ce{NH3(aq)}$ to form the soluble, deep blue/purple complex via equation 1, $\ce{[Cu(NH3)4]^{2+}}$, just as you said.

1) $\ce{Cu^{2+} + 4NH3 <--> [Cu(NH3)4]^{2+}}$

Depending on the concentration of $\ce{NH3}$, you would also have precipitated some amount of insoluble $\ce{Cu(OH)2}$ from the hydroxide formed by the ammonia reacting with water as in equation 2:

2) $\ce{NH3 + H2O <--> NH4+ + OH-}$

The equilibrium of equation 2 lies far to the left, so when you add more ammonia, you aren't adding an equal amount of hydroxide.

The difference in the second experiment is that you explicitly added excess hydroxide.  In this case the high hydroxide concentration would have pushed the dominant product to $\ce{Cu(OH)2}$ via equation 3.

3) $\ce{Cu^{2+} + 2OH- <--> Cu(OH)2}$ (solid)

Regarding the statement "The goal of the experiment was to see if one of the products could be produced without the other":

The products of both experiments were the same, just to different degrees; in other words, under different pH conditions, different products predominated.  In the first experiment, you had far more ammonia than hydroxide, so equation 1 dominated; lots of $\ce{[Cu(NH3)4]^{2+}}$ and a small amount of $\ce{Cu(OH)2}$.  In the second experiment, you added excess hydroxide so lots of $\ce{Cu(OH)2}$ was formed an very little $\ce{[Cu(NH3)4]^{2+}}$.

So the upshot is that these reactions are in equilibrium, and because of equation 2 you always had both $\ce{NH3}$ and $\ce{OH-}$ in your reaction mixture so that both equations 1 and 3 always produced some small amount of their respective products, $\ce{[Cu(NH3)4]^{2+}}$ and $\ce{Cu(OH)2}$, respectively.


For isomers, what does the term "connectivity" mean?

Connectivity describes how the atoms are connected to each other. Consider a daisy chain (image courtesy of Wikipedia).

All the daisies are similar enough much like carbon atoms are similar enough between them; we cannot distinguish. Counting ten daisies in that chain, I could also represent it as (with the ends of the chain reconnecting):
$$\ce{-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$$
This compound would by cyclodecane.
Now imagine you replaced two daisies with violets. Of course, violets are immediately distinguishable from daisies like oxygen atoms are from carbon atoms. We now have a number of different options how to replace daisies with violets. They could be next to each other, separated by one, two three or four daisies. That would give rise to the following structures:

1-peroxecane
$\displaystyle\ce{-O-O-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,3-dioxecane
$\displaystyle\ce{-O-CH2-O-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,4-dioxecane
$\displaystyle\ce{-O-CH2-CH2-O-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,5-dioxecane
$\displaystyle\ce{-O-CH2-CH2-CH2-O-CH2-CH2-CH2-CH2-CH2-{}}$

1,6-dioxecane
$\displaystyle\ce{-O-CH2-CH2-CH2-CH2-O-CH2-CH2-CH2-CH2-{}}$

In connectivity terms, in the first case and the first case only the two violets are connected to each other. In the second case and the second case only, there is a single daisy connected to two violets. These should be the obvious cases of different connectivities, although each different violet placement gives rise to a new connectivity.
These compounds are different; especially the first (which is a peroxide) and the second (which is the formyl acetal of a dialcohol) have very different chemical (and physical) properties from the others (which are simply cyclic diethers).
Added difficulty arises for issues that I have purposefully ignored in my daisy chain analogy: it does not consider three-dimensionality. Of course, atoms in reality are connected in a three-dimensional manner. You want to be sure that the connectivity is the same in all three dimensions (but not necessarily in the same orientations). Taking butan-2-ol as an example, we speak about identical connectivity whether we are dealing with (R) or (S)-butan-2-ol; because the same carbons atom are still connected to the same other atoms (e.g. carbon 2: Connected to a methyl group, an ethyl group, a hydrogen and a hydroxy group). But a different connectivity would be butan-1-ol, where carbon 2 is suddenly connected to a hydroxymethyl group and a second hydrogen (in addition to the ethyl group and the hydrogen which remained unchanged).


What happens to the colour change of an equilibrium reaction when one of the reactants begins to form a complex? (specific problem included)

Going by the right hand picture, what was formed there is the very poorly soluble and lightly yellow $\ce{FePO4.2H2O}$. A precipitate and clear supernatant liquid can clearly be seen.

If so, the reaction is:

$$\ce{FeSCN^{2+}(aq) + HPO4^{-}(aq) + 3 H2O(l) \to FePO4.2H2O(s) + SCN-(aq) + H3O+(aq)}$$

This explains why on adding the $\ce{Na2HPO4}$ the red colour of the $\ce{FeSCN^{2+}}$ complex disappears and a yellowish, flocculant precipitate appears. This is not however, as the OP suspected, a complex. It's merely an almost insoluble salt.

Some info on $\ce{FePO4}$.

A true complex would have been formed by the addition of $\ce{CN-}$, which would have led to the formation of $\ce{Fe(CN)6^{3-}}$, which has a much higher formation constant ($K_f$) than the thiocyanate complex. It's also weakly coloured and would not form a precipiate. The reaction would have been:

$$\ce{FeSCN^{2+}(aq) + 6 CN-(aq) \to Fe(CN)6^{3-}(aq) + SCN-(aq)}$$

The solution would have evolved from bordeaux to weakly amber/red.


The idea behind this is that a weak acid like $\ce{CH_3CO_2H}$ will break apart, but won't break apart completely. If you have some $\ce{CH_3CO_2H}$ in water, most of it will remain as $\ce{CH_3CO_2H}$, so in the net ionic equation it is written whole, but some will separate into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$ ions. This relationship is quantified by the $K_a$ of the acid. Essentially, the dissociation reaction $\ce{CH_3CO_2H -> CH_3CO_2^{-}  + H^+}$goes both ways: $\ce{CH_3CO_2H}$ decomposes into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, and $\ce{CH_3CO_2^-}$ reacts with $\ce{H^+}$ for form $\ce{CH_3CO_2H}$. Since the forward reaction happens at a rate proportional to the concentration of $\ce{CH_3CO_2H}$, and the backwards reaction happens at a rate proportional to the product of the concentrations of $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, eventually the ratio $\frac{[\ce{CH_3CO_2^-}][\ce{H^+}]}{[\ce{CH_3CO_2H}]}$ becomes constant, and we call this constant the $K_a$ of the $\ce{CH_3CO_2H}$. For $\ce{CH_3CO_2H}, K_a=1.7\times 10^{-5}$.

Now its time to connect all this back to your question. When you add $\ce{Ba(OH)_2}$ to a solution of $\ce{CH_3CO_2H}$, the $\ce{OH^-}$ ions present from the dissociation of $\ce{Ba(OH)_2}$ react with the $\ce{H^+}$ ions from the (partial) dissociation of $\ce{CH_3CO_2H}$. This lowers the concentration of $\ce{H^+}$, and so more $\ce{CH_3CO_2H}$ dissociates to keep the ratio of concentrations constant at the $K_a$. Therefore, in the presence of enough $\ce{Ba(OH)_2}$, $\ce{CH_3CO_2H}$ dissociates almost completely, even though, on its own, $\ce{CH_3CO_2H}$ dissociates only to a very small extent.


What happens to acids/bases left out overnight?

First of all, blowing into the buret seems like a less-than-safe thing to do. $\ce{NaOH}$ could easily fly out into his/her mouth and nose.

I would expect that your acid and base solutions were already at the equilibrium concentration of $\ce{CO_2}$ when you got them, since they were exposed to air. I think that dissolving large amount of $\ce{CO2}$ into solution at atmospheric pressure is no easy task. I just don't think small amounts of $\ce{CO2}$ would account for the difference.

When you say "left them out" do you mean uncovered? If so, there's probably a fair amount of evaporation happening. The analyte solution could be affected depending on your experiment, and the concentration of your titrant would go up, which means you'd actually need less of it to reach the end-point.


Relation between thermodynamic reversible process and reversible reaction

In thermodynamics, a reversible process -- or reversible cycle if the process is cyclic -- is a process that can be "reversed" by means of infinitesimal changes in some property of the system without entropy production (i.e. dissipation of energy).[1] Due to these infinitesimal changes, **the system is in thermodynamic equilibrium throughout the entire process.**

Thermodynamic equilibrium is an axiomatic concept of classical thermodynamics. It is an internal state of a single thermodynamic system, or a relation between several thermodynamic systems connected by permeable walls. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems

dO YOU OBSERVE SUCH KIND OF equilibriums IN A REACTION. SO THEY ARE NOT RELATED AT ALL

ACCORDING TO WIKIPEDIA

A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as
 aA + bB ---> cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.


Does benzene vapor react with air in a closed barrel?

Combustion is occurring, but the kinetics are very very very slow at room temperature.  So, on a time scale of days, weeks, and months, the system will be closer to the first picture.  On a time scale of decades, centuries, and millenia, the system will be closer to the 2nd picture.


How to find the rate of effusion given rates?

HINT:

This question seems to be a simple application of Graham's Law (assuming you know Graham's Law), i.e the rate of effusion of a gas is inversely proportional to the square root of the mass of their particles. This empirical law can be formulated as follows:

$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$

Solution:

   I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.

 $R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in

 $M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$


This nomenclature comes from a long time ago. In the $19$th century, it was decided that the acidic compounds containing H, O and another atom X would be named from the name of this third atom X.
If only one molecule of such an acid was known, the name would be made by adding the ending -ic. Example: Only one acid is known with a Carbon atom, and it is $\ce{H2CO3}$, which is called carbonic acid.
If two acids were known differing in the number of oxygen atoms, the most oxidized would follow the preceding rule. In the least oxidized, the ending would be changed into -ous. Example : In the $19$th century, the two following acids were known containing a nitrogen atom. Well  !$~\ce{HNO3}$ was called nitric acid and $\ce{HNO2}$ nitrous acid.
If more than two acids were known differing in the number of oxygen atoms, the chemists would copy the case of the chlorine atom, where four acids were known : $\ce{HClO, HClO2, HClO3, HClO4}$. In this case, the two acids situated in the middle of the list are named according to the previous method : $\ce{HClO2}$ is chlorous acid, and $\ce{HClO3}$ is chloric acid. For the two extreme cases, it was decided to add prefixes, hypo- for the lowest number, and per- for the highest number. As a consequence, $\ce{HClO}$ became hypochlorous acid, and $\ce{HClO4}$ was called perchloric acid.
Today, the same regulations is applied, but the criterion of the number of oxygen atoms has been replaced by the oxidation number. The suffix -ous and -ic must be used if the corresponding acids have a difference of $+2$ in their oxidation numbers.
It should also be mentioned that the previous nomenclature is valid for acids having only one X atom. If an acid have two X atoms, the prefix di- should be added, and tri- for three X atoms. For example, $\ce{H2S2O5}$ is called disulfurous acid, and $\ce{H2Cr2O7}$ is dichromic acid. There are also other prefixes like ortho- and meta- to distinguish between acids having one X and the same oxidation number, but a difference in the number of H atoms.


Are electron orbitals in a helium atom degenerate?

The energy of orbitals for hydrogen atom is dependent only on the principal quantum number: $$E_{n}(\rm{eV})=-\frac{13.6}{n^2}$$
For atoms with more than one electron (all the atoms except hydrogen atom and hydrogenoid ions), the energy of orbitals  is dependent  on the principal quantum number and the azimuthal quantum number according to the equation: $$E_{n,l}(\rm{eV})=-\frac{13.6Z^{*2}}{n^2}$$
Where $Z^{*}$ is the  effective nuclear charge: $Z^{*}=Z-\sigma$.

where$Z$ is the atomic number and $\sigma (n,l)$ is the screen (shielding) constant and it depends on the principal quantum number and the azimuthal quantum number.

According to this, the atomic orbitals $2s$ and $2p$, that are at the same energy in the case of hydrogen atom, are no longer at the same energy for the atom of helium.


Carbon with two double bonds and oxygen name

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) the name of the given compound is the systematically constructed name for ketones.

Therefore, the preferred IUPAC name (PIN) is ethenone.

The name ketene is retained, but only for general nomenclature.


Just like phosgene or phosphorus oxychloride, p-TsCl is acting as a dehydrating agent. Its affinity for oxygen atoms is fairly apparent when you consider its common use as an alcohol protecting agent. The mechanism is as follows:

$\hspace{2.6cm}$


If you have a solution, so long as you don't have too much of your "good", more solubilising solvent,  you should be able to precipitate your terpene.  Maybe not crystallise, but at least bring it out of solution. Lower alcohols and water act as anti-solvents for long chain hydrocarbons like lycopene.
If this is for lab purification scale, another option is to simply evaporate your solvent and dissolve it in your crystallisation solvent of choice. For these longer chain polyenes,  it may require a two solvent system anyway to crystallise the carotene of choice.
Going back to one of your comments, the acetone will not interfere so long as it is the minimum amount. With the Soxhlet method you use, presumably you extract unwanted compounds also, so your experience of how easily these dissolve/precipitate will affect your choice of isolation method and specifically, the anti-solvent choice.


This question is impossible to answer without knowing which TPE you mean - TPE is a mixture of thermoplastic elastomers, which is a huge class of compounds.

PVC is polyvinylchloride, which is a specific compound. I found an MSDS here:

http://www.fishersci.com/ecomm/servlet/msdsproxy?productName=AC183320010&productDescription=POLY

My advice is to contact the supplier for each fabric you are considering and to ask them for the MSDS. In that you will see what the acute and long-term exposure effects are for people. "Environmental effects" is a very broad subject, and you are not very likely to find research for any proprietary blends, particulary if they don't come from the US or EU. However, sometimes MSDS list non-proprietary compounds, and you can then check the EPA website for those names to see if there are any reports on broad environmental effects.

If you are also interested in the environmental impact of the entire production chain (which you should be, since end-use is usually the least significant part of that), then you should also ask the suppliers about their manufacturing process, whether they have any "green" certifications, etc. You can look up the patents for each production process (most of the time they will be patented) to see what their raw materials are, and then check the MSDS and EPA reports for those. It is also a good idea to see if you can tour their production facilities, once you know what the potentially dangerous precursors are and how they should be handled. It is easy to temporarily hide bad working conditions, but it is harder to fake treatment and containment facilities. Of course, if there isn't good regulatory oversight, there is a very good chance that the producer will take shortcuts to reduce costs, so keep in mind that what they tell you isn't necessarily going to be the truth.


Finding time t in a first order reaction (JEE Adv 2019)

The basic law of chemical kinetics states for $\ce{aR -> bP}$, that
$$A = kR = -\frac{1}{a}\frac{\mathrm dR}{\mathrm dt} = \frac{1}{b}\frac{\mathrm dp}{\mathrm dt}$$
(where $A$ is activity and $k$ is called rate constant1)
Note that, in our case, the stoichiometric coefficient of reactant (i.e., $\ce{N2O5}$) is 2, therefore the differential rate equation becomes,
$$-\frac{1}{2}\frac{\mathrm dR}{\mathrm dt} = kR$$
or,
$$-\frac{\mathrm dR}{\mathrm dt} = (2k)R$$
On integrating and re-arranging, we get,
$$R = R_0 \mathrm e^{(2k)t}$$
which means, $$t = \frac{2.303}{2k}\log_{10}{\frac{R_0}{R}}$$
This factor of $\frac{1}{2}$ was what you missed, I think.

1: As per my first equation, it's clear how rate constant is specific to a certain reaction. The following is an extract from Concentration and Rates (Differential Rate Laws) (under the heading "Reaction Orders"):

The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent.


Is it possible for one of reactants to deplete before achieving equilibrium

No, that is not possible.

Consider the definition of equilibrium from the IUPAC gold book:

  chemical equilibrium
  Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy, $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:
  \begin{align}
\Delta G_\mathrm{r} &= \Delta G_\mathrm{r}^\circ + R T \ln K &&= 0\\
\Delta G_\mathrm{r}^\circ &= − R T \ln K\\
\end{align}
  The equilibrium constant, $K$, is given by the mass-law effect.

The important thing is that the reaction proceeds in both directions. It means, that as soon as some of the product is formed, it will react back towards the reactants. The equilibrium is reached when the rates of both of the directions are identical.
From the formula in the definition, you can also see, that the location of the equilibrium, i.e. the concentration of all species or $K$, is only determined by the difference in the Gibbs energy between products and reactants.
In other words, a system that forms an equilibrium will always form an equilibrium, independent of the concentration of the starting materials.


Calculation of molarity using dilution equation

You're mixing up the terms and formulae. Instead of rote learning the formulae, remember that the moles of the solute will remain the same in a solution - before and after dilution.

With that in mind, let's inspect your $\pu{7.5ml}$ stock solution. Before dilution, it had $\pu{7.5ml}\times\pu{0.1M}$ moles of the solute. After dilution, it has an unknown molarity. Let that molarity be $x ~\pu{M}$. Then, the new moles are $\pu{10ml}\times x~\pu{M}$. Equating and solving, you get $x=0.075$.

If you do that again for the second solution (I give it you as a homework), you'll get $x=0.05$. This easily verifies your own intuition that you should get "a higher molarity for the first dilution since it is higher in concentration".


I’m pretty sure that this depends on the atom attacked. Most attacks on double bond are stepwise to a certain extent. And even when they aren’t, the absolute configuration should be discussed on a per-atom basis.

But if you have a concerted reaction and want to label both: (2si,3re).


Water, when hit by radioactive particles, will most likely suffer ionization (or radiolysis). The ejected electron very quickly leaves the vicinity of the affected molecules (in the case of exposure to gamma or beta radiation) or a very chemically stable neutral helium atom is formed (in the case of exposure to alpha radiation), so you obtain short-lived species which will tend to react with most other substances nearby, even if neutral. The electron, after being slowed down considerably, becomes hydrated and can participate in reactions itself. This article has several examples of reactions triggered by radiolysis, and also separates them by their relevant timescales.

Some further examples of reactions involving radiolysis of water via beta particles and gamma rays are (from the very well-cited article "Critical review of rate constants for reactions of hydrated electrons, hydrogen atoms and hydroxyl radicals (⋅OH/⋅O−) in aqueous solution"):

Of course, not only are the decay particles damaging, but so are the entities produced by their interaction with matter in the vicinity. If the nuclear decay particle doesn't directly ionize organic matter, reactive species such as $\ce{OH^{.}}$ or $\ce{H^{.}}$ could then react with biomolecules, degrading cells and tissues.

This Wikipedia link has some details on the radiolysis of water. Interestingly, it seems that dissolved hydrogen in water can efficiently suppress the formation of unusually reactive species, presumably as the side reactions eventually cause the formation of hydroxyl anions (if electrons are present in excess) or hydronium cations (if positive charges are present in excess). This image (source) shows a collection of possible reactions in sequence. If you search for articles on water radiolysis, there are many other results, such as these two. It's interesting to note that the effect of radiation on water still doesn't seem to be entirely understood, in part due to our incomplete knowledge of the behaviour of hydrated electrons.

Here's an extra bit of background on radiation and its biological effects (at least for the most common types. There are many ways nuclei can decay).

Depending how far the radiation source is from you, gamma radiation is either the least or the most dangerous type of nuclear decay. If the radiation is coming from well outside your body, then the danger scale goes alpha < beta < gamma. If you happen to have a radioactive nuclide in your pocket, the scale may well become alpha < gamma < beta. However, if the radiation is coming from inside your body (avoid ingesting glowing substances), then the danger scale goes gamma < beta < alpha. Why does radiation behave like that? There are two main effects in play: penetration capacity and ionization capacity.

In collision theory, different types of projectile and target can be characterized by something called an collision cross-section (or absorption cross-section, in nuclear physics), which basically gives you a probability of how likely it is for the projectile to hit a certain target. If the combination of projectile and target have a high collision cross-section, they they are very likely to interact, and so almost any projectile will be stopped by even a thin target. If the collision cross-section is low, then the projectile has a low chance of hitting the target, and therefore will tend to pass through it entirely. To compensate, you need to make a very thick target so there are more chances for the projectile and target to interact.

The different types of radioactive decay have different penetrating capacities. All of matter as we know it is composed of atoms, which contain well-defined oppositely charged regions (the nuclei and the electron shells), and so we can expect them to interact strongly with other charged particles. Alpha and beta(-minus) decays emit particles with a charge of $+2e$ and $-e$, respectively ($+e$ for beta-plus decay). We then expect these particles to interact strongly with matter, being stopped easily. Indeed, most alpha particles are efficiently stopped even by air. Beta particles have half the charge of an alpha particle and are going quite a bit faster (having approximately 1/7000th the mass), so they manage to go further before interacting. A few centimetres of of metal shielding or a concrete wall are usually enough to stop them. However, gamma rays have no charge, so they interact comparatively weakly with matter. These highly energetic photons will go though dozens of centimetres of solid matter. To maximize the odds that they will hit something and be stopped, you want to use a shield made of a very dense substance (which would pack a lot of possible targets in a small space). This is why lead is the substance of choice for radioactive shielding; its high density combined with low cost allows the fabrication of relatively compact shields which are capable of stopping even gamma radiation, when around 10 cm thick.

Now, for the mechanism of how particles from radioactive decay are stopped. For uncharged particles, the absorption cross-section is low because interactions only happen when the uncharged particle comes very close to a target particle. For gamma rays (very high energy photons), when there is an interaction, it is usually due to Compton scattering, where an energetic photon interacts with an electron, giving it a large amount of kinetic energy, while the photon is re-emitted with a smaller amount of energy. The target electron (not necessarily a valence electron) often gains enough energy to be completely removed from the atom, causing ionization.

For collisions between charged particles (both same and opposite charges), interactions happen much more easily, as charges interact strongly even at comparatively much larger distances. Beta-minus particles (very fast electrons) can more efficiently repel electrons and shove away from atoms, creating cations. Alpha particles (bare helium-4 nuclei) are interesting because their positive charge means they can tear electrons from any atom or ion. In fact, it's defensible that alpha particles are the strongest Lewis acid one can commonly find; they are effectively free protons with twice the charge.

Notice that penetration capacity and ionization capacity are inversely correlated; if a radioactive decay particle is strongly ionizing, it interacts well with matter and is stopped in short order. If it is weakly ionizing, one of the main interaction mechanisms is suppressed, and the decay particle can go further before it is stopped. In order to not damage you, you want the nuclear decay products to not dump their energy in your body, i.e. you want them to pass right through you. Here, it is clear that alpha particles are the most dangerous, followed by beta and gamma radiation. However, alpha particles interact so strongly with matter that they literally need to be released closer than your skin to dump energy into living tissue, so you need to ingest, breathe or inject alpha emitters. When this happens, they are extremely deadly. It is a very rare occurrence, however, since radiation usually comes from far outside. In that case, gamma radiation has a larger chance of reaching you, so it is said to be the most dangerous.


Why doesn't cyclopropyl methyl carbocation stabilises itself by ring expansion?

Unfortunately, the premise of your question is wrong.

There is no cyclopropylcarbinyl or cyclobutyl cation. They're the same thing. Unlike cations which you might be more familiar with, for example, t-butyl cation, the cation you're looking at here, also called homoallyl, is a non-classical cation.

The open form and the two structures you drew in the second line are not distinct. They're just resonance forms of each other.

You ask does this ion undergo ring expansion? Well, yes and no. Yes, in that there is a contribution from the 4-member ring to the resonance structure. No, in that you can't expand something that doesn't exist.


Is the haloform reaction possible with fluorine and pseudohalogens also?

TL;Dr: Highly unlikely to be made using the general haloform reaction.
Long answer:
The mechanism of haloform reaction involves hypohalite ions resulting from the reaction between the base and the halogen.
$$\ce{X2 + NaOH -> X- + XO- + H2O}$$
If we use fluorine as halogen, the resulting reaction will produce hypofluorite ion but this ion is extremely unstable and as such fluoroform can't be made using this reaction.  Trifluouromethane is known and used industrially in the manufacture of polytetrafluoroethylene ("Teflon"), but it is commonly made by anion exchange with another haloform, typically chloroform plus $\ce{HF}$.

For many years, chemists have been unable to isolate cyanoform as a neat, free acid. However, in September 2015, it was successfully isolated by a team of scientists at Ludwig Maximilian University of Munich. It is stable at temperatures below – 40 °C. So, it requires drastic conditions to synthesize cyanoform and is only stable at low temperatures. So, it is highly unlikely to be made at ambient conditions using laboratory grade reagents.

Nitroform has been made in the laboratory by the hydrolysis of tetranitromethane under mild basic conditions. But can it be made using general haloform reaction? Yes, it can technically be made but there are face several problems. Let's start with the starting reaction i.e. reaction between base and nitrogen dioxide which produces nitrite and nitrate ions:
$$\ce{2OH- + 2NO2 -> NO2- + NO3- + H2O (+other nitrogen oxides)}$$

The nitrite that forms is not that great reactive oxidizer.
There might be an undesirable side reaction which is kind of a Meyer reaction to form undesirable nitrolic acid/psudonitrol which in basic conditions gives blood red/blue colored solution.
The reaction involves huge amount of nitrogen oxides in general which is not very economical to deal with.
Multiple SN1 reactions occurs that generates a lot of isomers and other nitro compounds of compounds along with a whole lot of nitrite esters.
Steric hindrance might play a role and the reaction would not proceed properly.
The nitroformate ion ($\ce{C(NO2)3^-}$) might ionize to form different ions: $\ce{CH(NO2)2^-}$, $\ce{CH2NO2^-}$.
Nitroform is itself a strong oxidizer.

There are discussions in forums regarding trinitromethane synthesis using general haloform reaction but this ended up being fruitless. It even discussed using copper salts as catalyst and even employed haloform itself to react with $\ce{Ag/NaNO2}$ in DMF/ether but these syntheses posed so much problems and the yield was generally very, very poor.


Is it more precise to measure a substance as a solid or as a liquid?

Weighing is always far more accurate than any volume measurement. The extreme example is that of atomic weight (mass) measurements, which required accuracy to 4-5 decimal places. All the work was always done on the basis of gravimetry or by weighing the substances on high quality balances.


In the pH system of acids and bases, are there 10^-7 loose ions per
liter (at pH of 7) or per mole of solvent? (Or per mole of total
substance?)

In a rigorous sense none of the options is correct, because pH being approximated by molarity is an approximation itself. The problem lies in the measurement itself and pH should be
$$pH= -log (a_{H^+})$$
The pH probe "senses" the activity of hydrogen ions not the molarity. This is why the pH of 5 M HCl is not equal to $$pH= -log (5)$$
There are no free or loose hydrogen ions in water, like a bare proton, rather this proton is attached to a network of water molecules.
Wiki is talking about an approximation. 1 ton is equal 1000,000 grams, so 10$^7$ mole/g x 18 g $\ce{H2O}$ is indeed 180 t. This part is correct so far. Note the author must be following metric ton. The whole world does not follow the US ton.
1 liter of water weighs ~ 1000 grams. Then 1000 g water contains 10$^{-7}$ moles of $\ce{H+}$, and 18x10$^{7}$ g water should have 1.8 x 10$^{-2}$ mol $\ce{H+}$ or 0.018 g $\ce{H+}$. I don't get 1 g.


why are halogens considered deactivating groups while methoxyl groups are activating?

Because fluorine has such a high electronegativity, its inductive withdrawing effect is much larger than any resonance electron-donating effect.

The other halogens have poor orbital overlap with the pi system of an aromatic ring, which precludes any appreciable resonance electron-donation (this is why you almost never see a C-X double bond). Therefore, their inductive capabilities lead to a net deactivation of the aromatic ring.

This is why an alkoxy group is more activating  than a halogen. Oxygen does have a higher electronegativity than does chlorine, bromine or iodine. However, oxygen's orbital overlap with carbon allows it to more readily donate into the ring's pi system. The larger halogens aren't able to do this (to an appreciable extent).


Why should duplicates compounds be removed from compound libraries before screening/activity prediction?

Let's consider a data set with three attributes for each compound: a chemical name, a CAS number (unique identifier for a compound) and a single empirically measured numerical interest, such as efficacy in lowering blood pressure in rats. If you had duplicate entries for a single experimental result, that could potentially bias the analysis. If you had two results for the same compound from two different experiments or measurements, you would surely want to include both in the analysis. If you had two results for the same compound (same CAS number) but named differently, you would want to keep them both in your data set, but use only the CAS number as the nominal variable for structure.


Reaction of sulfate ion in copper sulfate electrolysis

A typical solution for copper plating contains about $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{CuSO4}$ and $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{H2SO4}$:

$$\begin{align}
c(\ce{Cu^2+})&=0.5\ \mathrm{mol\ l^{-1}}\\
c(\ce{H+})&=1\ \mathrm{mol\ l^{-1}}\\
c(\ce{SO4^2-})&=1\ \mathrm{mol\ l^{-1}}\\
\end{align}$$

The positive ions $\ce{Cu^2+}$ and $\ce{H3O+}$ are attracted to the cathode. The redox potentials $E$ for $\mathrm{pH} = 0$ show that cathodic reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ is preferred over reduction of $\ce{H+}$ to $\ce{H2}$:

$$\begin{alignat}{2}
\ce{Cu^2+ + 2e- \;&<=> Cu}\quad\quad &&E^\circ = +0.340\ \mathrm{V}\quad\quad &&&E=E^\circ+\frac{0.059\ \mathrm{V}}{2}\log 0.5 = 0.331\ \mathrm{V}\\
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000
\end{alignat}$$

At the anode, the redox potentials $E$ imply that oxidation of $\ce{H2O}$ to $\ce{O2}$ is preferred over oxidation of $\ce{SO4^2-}$:

$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{S2O8^2- + 2e- \;&<=> 2SO4^2-}\quad\quad &&E^\circ = +2.01\ \mathrm{V}
\end{alignat}$$

Nevertheless, $\ce{SO4^2-}$ actually is oxidised at the anode:
$$\ce{SO4^2- -> SO4- + e-}$$
The reason for this reaction is that negative ions are attracted to the anode. Under neutral conditions, $\ce{OH-}$ ions would migrate to the anode where they would be oxidised to $\ce{O2}$. However, under the given acidic conditions, the concentration of $\ce{OH-}$ is very low. The predominant negative ions are $\ce{SO4^2-}$, which migrate to the anode.
Since the produced concentration of $\ce{SO4-}$ is low, the $\ce{SO4-}$ ions do not combine to $\ce{S2O8^2-}$. Instead $\ce{SO4-}$ reacts with water:
$$\ce{4SO4- + 2H2O -> 4SO4^2- + O2 + 4H+}$$

Therefore, the net reaction is the above-mentioned oxidation of $\ce{H2O}$ to $\ce{O2}$. The $\ce{SO4-}$ ions act as catalyst:

$$\begin{align}
\ce{4SO4^2- &-> 4SO4- + 4e-}\\
\ce{4SO4- + 2H2O &-> 4SO4^2- + O2 + 4H+}\\
\hline\ce{2H2O &-> O2 + 4H+ + 4e-}
\end{align}
$$


Thermochromic pigment that changes at around 50 °C / 120 °F?

The easiest way to obtain such a pigment is to mix a blue and red thermochromic pigment together, one that changes at 32 and another that changes at 50 dregrees. They aren't very easy to find, for the moment i have found the 32 degrees pigments on ebay, and a 45 degree pigment on aliexpress:


Can hydrocarbons undergo combustion without oxygen present?

They would certainly combust if meeting a worse adversary than oxygen. $\ce{ClF3}$ would oxidize them, with extreme certainty, rapidity and violence.

Here is a short excerpt from the Wikipedia article of $\ce{ClF3}$ which amused me:

  It is, of course, extremely toxic, but that's the least of the
  problem. It is hypergolic with every known fuel, and so rapidly
  hypergolic that no ignition delay has ever been measured. It is also
  hypergolic with such things as cloth, wood, and test engineers, not to
  mention asbestos, sand, and water — with which it reacts explosively.
  It can be kept in some of the ordinary structural metals — steel,
  copper, aluminum, etc. — because of the formation of a thin film of
  insoluble metal fluoride which protects the bulk of the metal, just as
  the invisible coat of oxide on aluminum keeps it from burning up in
  the atmosphere. If, however, this coat is melted or scrubbed off, and
  has no chance to reform, the operator is confronted with the problem
  of coping with a metal-fluorine fire. For dealing with this situation,
  I have always recommended a good pair of running shoes.


Where does the RT term come from in the derivation for the activation enthalpy from the Eyring equation?

Using vibrational partition functions to define the reaction rate constant produces an equation of the form $\displaystyle k=aT^be^{-\Delta U_0^\mathrm{O}/(RT)}$ where $a,b$ are constants independent of $T$ and $\Delta U_0^\mathrm{O}$ is the difference in zero point energies at the transition state compared to reactant, this is related to what is commonly called the activation energy. This approach allows molecules to have discrete vibrational levels and so a thermal population of these. The constants $a,b$ are found from the partition functions.

The (classical) Arrhenius equation is often written as $k_A=Ae^{-E_A/(RT)}$. Taking the log of each equation and differentiating wrt. $T$ gives

$$\frac{d\ln(k)}{dT} =\frac{bRT+\Delta U_0^\mathrm{O}}{RT^2}\equiv \frac{E_A}{RT^2}$$

thus the Arrhenius activation energy is $E_A\equiv bRT+\Delta U_0^\mathrm{O}$.

(Texts on statistical thermodynamics gives details of partition functions, some book on kinetics also give this detail.)


Why do the d orbitals have these notations?

The notation is shorthand for the hydrogen atom $\ell = 2$ wavefunctions in real form, in cartesian coordinates.  In other words, the Schrodinger equation is solved in spherical coordinates, in terms of spherical harmonics, which involve complex numbers; however, linear combinations of degenerate (equal energy) solutions are also solutions.  This permits the wavefunction solutions to be expressed using real numbers only. These real solutions can be transformed to cartesian coordinates.

$z^2$ represents that the wavefunction is proportional to $(3z^2-r^2)/r^2$.

$xy$ represents that the wavefunction is proportional to $xy/r^2$.

$yz$ represents that the wavefunction is proportional to $yz/r^2$.

$xz$ represents that the wavefunction is proportional to $xz/r^2$.

$x^2 - y^2$ represents that the wavefunction is proportional to $(x^2 - y^2)/r^2$.

See this Table of Real Spherical Harmonic Functions for the a list of functions in both spherical coordinates and cartesian coordinates, for s through f orbitals.


Why isn't auto ionisation of water considered when we find the pH during salt hydrolysis?

What happens in salt hydrolysis is enhancement of water ionization with salts. Try asking yourself where do $\ce{H+}$ and $\ce{OH-}$ come from in the solution of $\ce{AB}$, and you'll find that they come from $\ce{H2O}$. Or equivalently, hydrolysis of $\ce{A-}$ can be written as
$$
\ce{A- + H+ <=> HA,H2O <=> H+ + OH-}\stackrel{\text{add}}{\Longrightarrow}\ce{A- + H2O <=> HA + OH-}.
$$
In a word, ionization has been taken into consideration in your calculation. There are multiple equilibria in the system, so you cannot simply add $10^{-7}$.


How to construct a QST2 input file for Gaussian16 for multiple reactants and products?

TL;DR: QST2 will not work.

For QST2 calculations you need one reactant structure and one product structure, the program will then attempt a transition state guess. The essential point is, that the supplied structures have to have identical ordering (in Cartesian coordinates), or identical z-matrices, although I am not sure that this works.
For more on how to run these types of calculations, see Finding transition states in Gaussian – focus: Electrophilic Addition Reaction.

QST2 (and QST3 by extension) only make sense, if you know that you will have a single transition state - they are especially difficult to use for dissociation or association reactions because of the implied supermolecular approach. The minima $(\ce{CH3 + N})$, $(\ce{HCN + H2})$ will be very shallow and not well converged. From these structures it would be very difficult to guess the transition state, so you would at least have to supply a guess for it (QST3).

Additionally you are going from a "quintet" state (doublet + quartet) or "triplet" state (doublet + doublet, or doublet - quartet) to a singlet state (singlet + singlet). You could probably force your reactants into a broken singlet state (doublet - doublet), then it's a problem of how much sense this makes. In the other cases it is doubtful that the program will get a good guess for the electronic state of the TS.

DFT is maybe not the best methodology to investigate this kind of reaction in the first place.

In any case, here is an input file (tested with Gaussian16 rev. A.03) that will at least start running. SCF will not converge. Also reactants and products have not been optimised beforehand (which you must do). This is only for illustrative purposes.

%chk=bp86svp.chk
#P BP86/def2SVP
EmpiricalDispersion=GD3BJ
DenFit
opt(QST2,CalcFC,maxcycle=5)
gfinput gfoldprint iop(6/7=3)
nosym
pop=full

N + CH3

0 1
C        0.000000000      0.000000000      1.891780658
N        0.000000000      0.000000000     -2.108219342
H        1.029433990      0.000000000      2.255740536
H       -0.514716995     -0.891515987      2.255740536
H       -0.514716995      0.891515987      2.255740536

HCN + H2

0 1
C        0.649013613      0.189100642      0.000000000
N       -0.817143840      0.083131316      0.000000000
H        0.964811208     -1.929636661     -0.499461795
H        0.964811208     -1.929636661      0.499461795
H        1.696363276      0.452628980      0.000000000


How does Rosuvastatin 5 S-lactone form in human body?

Rosuvastatin is metabolized to a limited extent by CYP2C9 to form an N-desmethyl metabolite and the other metabolite is the rosuvastatin lactone.

The acid moeity (highlighted part) is likely to undergo glucuronidation first and subsequently lactonisation to form the rosuvastatin lactone

Unlike other statins, rosuvastatin undergoes relatively little metabolism by hepatic CYPs with no significant involvement of CYP3A4.

  The metabolism of statins is complex, involving the inter-conversion
  of acid/lactone by various pathways. The acid forms are converted to
  the corresponding lactones via a glucuronide intermediate.

Here is the proposed, generalised metabolic pathway of statins. The acid forms of statins are converted to the corresponding lactones via a glucuronide intermediate via an elimination reaction.

Sidenote for some statins: (After metabolism by CYP3A4 some of the resulting lactone metabolites are converted to the corresponding acid forms:

References

Focus on Statin research. B. A. Wong: Drug Metabolism and Transporter profiles of Statins.
Foye’s Principles of Medicinal Chemistry 7th ed


How does binding energy "manifest" itself?

If you want to break a bond you first need to excited the right degrees of freedom so the molecule can move along the reaction coordinate of interest. In other words, the correct vibrational modes need to be excited. At equilibrium distance this will be reflected in the kinetic energy of the atoms. As you move towards the dissociation limit this kinetic energy gets transformed to potential energy, much like you give a ball a push to move up a hill. Any excess energy (if you excite more than binding energy is required) will result in kinetic energy after the dissociation.

Forming bonds happens in reverse: Two fragment come closer and convert their potential energy into kinetic energy.

After such an reaction (bond forming or breaking) several options can happen:

the excited states go into lower states by emitting photons (e.g. fire, chemoluminescence)
the energy dissipates into other degrees of freedom (e.g. by momentum transfer in collisions with other molecules)
the reaction may happen backwards and you get your initial educts again
(there are probably further options ...)


Why boiling can't take place in closed containers

You can understand this well by thinking about why a liquid boils in the first place.  Imagine you have a pot of water, and it's just sitting there in your kitchen.  Intermolecular forces (specifically hydrogen bonding for H20 molecules) keep the water in liquid state.  The air molecules zipping around the pot and above the water's surface represent a certain pressure.  Let's say your kitchen has standard conditions, so 1 atm.  Then, you put the pot on a stove, and heat the water up.  This heat breaks the intermolecular bonds between the liquid molecules.  Now, the only reason the molecules rise up and become water vapor is because the kinetic energy of the water molecules matches that of the surrounding air.  Pressure moves from high to low, and as the water boils, certain molecules acquire the necessary speed (and therefore pressure) to enter the gaseous state.  Keep in mind that heating the pot negligibly changed the kitchen pressure.  The pressure in your kitchen should still be about 1 atm, since the vast air space should distribute the heat produced by the stove.

In a closed container, however, the story is different.  Let's say you put a lid over your pot.  Now, there is a relatively small amount of air trapped between the lid and the water's surface.  Heating the pot heats the air as well, and pressure does not remain at 1 atm.  That is because that small amount of air has to handle all that heat, whereas in the situation without the lid, a bigger volume (the kitchen) allowed for more heat dissipation.  So now the small amount of air molecules are moving really fast.  The heat is still breaking the intermolecular forces of the water, but this time, there is no (entropic) reason to move up into the gaseous state.  The air molecules are moving just as fast, and therefore have just as much pressure, as water molecules that have broken bonds.

The key to your question is rooted in why the water molecule (without a lid) moves up out of the surface of the water and joins the air molecules, effectively becoming water vapor.  Simply put, it is because the water molecule (because of the heat) has enough speed to zip out of the water.  Pressure is intrinsically based on speed of molecules, and so you can think about this situation as: the water molecule(s) has a higher pressure than the air molecules, and thus they move up to join the air molecules, in order to balance out the pressure.  A lid prevents this from happening because the water molecule(s) pressure will never be higher than the air molecules, since the air takes the same amount of heat as the water.  (The preceding statement is true only in usual cases.  There are caveats to this - see comments.)

Therefore, water does not (usually) boil in a closed container.  A couple of things: There are many more explanations/videos online about this topic (boiling water, pressure cookers, more altitude = lower pressure, higher boiling point etc..)in case you want to strengthen your understanding further and explore the core of this question from different angles.  Also, the intuition in this Khan academy video https://www.youtube.com/watch?v=hA5jddDYcyg is pretty good.


Yes, the melting point of ocean salt is going to be lower than pure NaCl at 801 °C.

In sea water you don't have compounds but rather dissolved ions. Other than the most abundant Cl- and Na+ (in this order), sea water contains Mg2+, K+, Ca2+ and SO42- (not in this order). Thus, we can assume that the solid residue would be NaCl, MgCl2, CaCl2 and KCl.

Let's have a look at several phase diagrams, taken from http://www.crct.polymtl.ca/fact/pdweb.php:

So yes, adding additional cations other than Na+ will reduce the melting point. Note however, that you will get a lower melting point (solidus) and a higher melting point (liquidus) that in between you will have an equilibrium mixture of solid crystals and melt.

Other than Cl- there's also SO42- and CO32- (as both carbonate and bicarbonate). This will also lower the melting point:

The melting point of the sea salt mixture will be lower than the decomposition of the sulfates and carbonates, so you will dissolve them into the liquid salt rather than degas them as SO2 and CO2 (at least initially).

It's also interesting to see that as you heat the mixture, you will end up with intermediate compounds, such as (Na,K)Cl and (Na,Mg)Cl. As to the exact melting point of the mixture, I couldn't find any experimental data in my quick search. It is possible that thermodynamic modeling would be able to give a prediction. But remember, as the mixture is eventually dominated by NaCl it will not be too far off from the melting point of the pure compound. My uneducated guess is that you will only start seeing any significant melting at around 700 °C and above.

Note of caution - if you do decide to actually melt sea salt, be sure to take the necessary precautions. Even though SO2 will mostly dissolve in the melt, some of it will degas. Molten halides are extremely corrosive and will react with atmospheric moisture, sometimes explosively. They will react with your crucible, possibly causing leakage of really nasty stuff. Unless you know what and how you're doing it, don't do it.