I think there might be a mistake in your question...
Around $\pu{2680 cm-1}$, there is the $S_1$ band (attributed to overtone or combination scattering). I think the $2D$ band you are referring to is the band shouldering the $G$ band (also called $D'$) which is around $\pu{1680 cm-1}$.
About the ratio, there are two ratios that are often used to characterise $\mathrm{sp^2}$ carbon based materials such as the CNT.
$$\text{Intensity ratio} =  R_1 = \frac{I(G)}{I(D)}$$
$$\text{Area ratio} = R_2 = \frac{A(G)}{A(G)+A(D)+A(D')}$$
Both these ratios help characterise the degree of organization of your sample.
For more information or more details, I would recommend checking out Spectrochim. Acta, Part A 2003, 59 (10), 2267–2276


Are CaOCl2 and Ca(OCl)2 same or different?

$\ce{CaOCl2}$ is a calcium chloride hypochlorite. There are two different types of anions: chloride $\ce{Cl-}$ and hypochlorite $\ce{ClO-}$. So its structure is:

$\ce{Ca(ClO)2}$ is a calcium hypochlorite (also called bleaching powder or bleach). There is only one type of anions: hypochlorites $\ce{ClO-}$. So its structure is:


The retained names “tert-pentyl”, “isopentyl”, and “neopentyl” were still included in the 1993 IUPAC recommendations.
However, according to Subsection P-57.1.4 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), these retained names are no longer recommended. The preferred prefix names are as follows:

“tert-pentyl”: 2-methylbutan-2-yl
“isopentyl”: 3-methylbutyl
“neopentyl”: 2,2-dimethylpropyl

P-57.1.4 Retained prefixes no longer recommended
The retained names phenethyl (2-phenylethyl) for $\ce{C6H5-CH2-CH2-{}}$; benzhydryl (diphenylmethyl), for $\ce{(C6H5)2CH-{}}$; isobutyl (2-methylbutyl) for $\ce{(CH3)2CH-CH2-{}}$; sec-butyl (butan-2-yl, 1-methylpropyl) for $\ce{CH3-CH2-CH(CH3)-{}}$; isopentyl (3-methylbutyl) for $\ce{(CH3)2CH-CH2-CH2-{}}$; tert-pentyl (2-methylbutan-2-yl, 1,1-dimethylpropyl) for $\ce{CH3-CH2-C(CH3)2-{}}$; and neopentyl (2,2-dimethylpropyl) for $\ce{(CH3)3C-CH2-{}}$; are no longer recommended; the first name in parentheses is the preferred prefix name.


Why is the smell of car exhaust much stronger in winter?

While car engines might look simple there is a vast amount of technology installed to ensure as complete a combustion as possible. Complete combustion according to the equation

$$\ce{C_{$m$}H_{$n$} + \frac{2$m$ + $n$}{2} O2 -> $m$~CO2 + \frac{$n$}{2} H2O}$$

(not accounting for oxygen, nitrogen or other elements present in the fuel) requires a precise amount of oxygen added. Too little oxygen will result in uncombusted hydrocarbons while too much oxygen will result in $\ce{NO_{$x$}}$. A small oxygen sensor is thus installed to improve the oxygen regulation. (It is a very interesing compound because it functions as an oxide ion conductor.†)

The thing about this oxygen sensor is that its function is temperature dependent: It needs to be heated to $\approx 500~\mathrm{^\circ C}$ to function properly. This is achieved quickly due to the heat of the exhaust, but does take a few minutes from starting the car. So throughout the first few minutes, the oxygen sensor does not function and incomplete combustion results in $\ce{NO_{$x$}}$ and uncombusted hydrocarbons that cause the odour. Once working temperature has been reached, the smell disappears mostly.

In cold weather it will take longer for the oxygen sensor to function properly and thus more incomplete combustion is observed. This becomes more pronunced the colder it gets.

†: Its German name is Lambdasonde from the Greek letter $\lambda$ which is used as a variable to determine if the correct amount is present or not. At the correct oxygen concentration, $\lambda = 1$.


Why is dimethylglyoximate a bidentate ligand and not a tetradentate ligand?

You are right. I took this two pictures from Wikipedia to compare EDTA (a tetradentate) to dmg (a bidentate).

For EDTA you got this:

For dmg, on the other hand, you got this structure:

You can notice that for both of the cases, the metal complex is connected to the ligands through five-membered rings. This observation is general in chemistry: to have stable rings, you need five (sometimes six) members for those rings.

Now it is clear to see a general rule of thumb for multidentate ligands: the two donating atoms must be connected through 2 (sometimes 3) atoms. For dmg, you can see that the two nitrogens satisfy this rule, while the oxygen-nitrogen pairs do not.

For EDTA, you can see that going through the molecule, you can find an oxygen, a nitrogen, a second nitrogen and a second oxygen, all bridged by 2 atoms to each other (while the two oxygens of the carboxyl group are not connected by 2 bridge atoms, otherwise you might even have a hexadentate ligand).


Term symbols for nitrogen: determining J

Firstly, regarding the extra question: Atkins' Molecular Quantum Mechanics (5th ed.) uses "term" for $^2\!S$, and "level" for $^2\!S_{1/2}$.

Back to the main question. It's been a long time since I did term symbols, so I am happy to be corrected, but if I am not wrong, your $^4\!P$, $^4\!D$, and $^2\!S$ terms are not allowed because of the Pauli exclusion principle. The argument is pretty similar to the one in the comments on my answer, which you're already aware of.

A term with $S = 3/2$ must have a set of states with $M_S = +3/2, +1/2, -1/2, -3/2$. Likewise, a $P$ term with $L = 1$ must also have a set of states with $M_L = +1, 0, -1$. Consequently, the total number of states associated with a single term $^{2S+1}\!L$ is $(2S+1)(2L+1)$. (The total number of states in a level, $^{2S+1}\!L_J$, is $2J+1$, and you can also show that the sum of this over the allowed values of $J$ is equal to $(2S+1)(2L+1)$.)

One of those twelve states in the purported $^4\!P$ term must have $(M_S, M_L) = (+3/2, +1)$. However, $M_S = +3/2$ can only be achieved if all three electrons have the same spin (spin up in this case), which in turn necessitates that the three electrons are all in different p-orbitals, such that $M_L = \sum m_l = +1 + 0 + -1 = 0$.

Ergo, a state with $(M_S, M_L) = (+3/2, +1)$ cannot exist, and the entire term $^4\!P$ cannot be possible. The same argument also applies to the $^4\!D$ term.

I suspect that showing that the $^2\!S$ term is forbidden is somewhat trickier. I had a similar issue before with carbon, which is an easier system to understand (only two electrons to consider instead of three), so maybe you will find this useful: Pauli-forbidden term symbols for atomic carbon.

Out of the three surviving terms ($^2\!P$, $^2\!D$, $^4\!S$) you already see from NIST that all the possible $J$ values (from $|L-S|$ to $L+S$) are allowed, so there is no issue with your calculation of $J$.


Confusion in the line written in my book

Ok let me explain you expecting you are from a cbse affiliated school.

Group oxidation state = (group number - 10)

This will give you the group oxidation state of that group. Group oxidation state is the maximum oxidation state shown by the elements of the group. For example in 15th group, the group oxidation state = (15-10) =5
That means the maximum oxidation state which the group 15 elements would show is 5.
Generally moving down any p-bock group, the tendency of occurence of the maximum oxidation state (i.e. +5 for group 15 elements, +3 for group 13 elements) decreases. Rather the most occuring oxidation state for heavier elements is the oxidation state 2 units less that the maximum/group oxidation state. The reason behind this is intervening of d and f orbitals for heavier elements i.e. in one word we say the inert pair effect.

Now what is inert pair effect? As we move down the group, gradually for heavier elements d and f orbitals start filling up. The shielding by electrons in d and f orbitals is very poor due to which moving down the group the nuclear charge increases, but there is not much increase in shielding due to which effective nuclear charge becomes effective and hence its not possible for the heavier elements to attain the maximum oxidation state because to attain the maximum oxidation state, they have to loose the more number of electrons. But that is not possible for heavier elements because the nuclear charge is now effective. It wont allow loosing much of the electrons. Hence the maximum oxidation state/ group oxidation state is not achievable for the heavier elements.

What they can achieve is two units less than that maximum oxidation state and that also makes sense. For example in case of group 15 elements, the maximum oxidation/ group oxidation state i.e +5 becomes unstable for heavier elements but +3 state (i.e. two units less than the maximum oxidation state/ group oxidation state) becomes increasingly stable for the heavier elements. Now you might think why only two units less than the maximum oxidation state stable for the heavier elements? Why not 3 units less? Because this is a fact i.e. because we have found the more observing/occuring oxidation state for p block elements is two units less than their corresponding group oxidation state. Now the statement was written because it was observed.

Negative oxidation state for heavier elements is also not achievable because down the group the shell size is large. Though I said effective nuclear charge is bit effective for heavier elements, but that doesn't mean it is stronger enough to gain an electron. It is just strong to get hold of some valence electrons i.e. not allowing loosing much of the electrons. But that strength is not that strong to gain an electron. Hence negative oxidation is not shown for heavier elements.

So basically stable oxidation state for heavier elements is 2 units less that the maximum oxidation state shown by that corresponding group.


The textbook answer must be wrong, since $\ce{IF5}$ is a sufficiently stable compound and fluorine is more electronegative than chlorine.

Rather, I think that the reason for $\ce{ICl5}$’s lesser stability is the lower electronegativity of chlorine. Creating this type of hypercoordinate compounds requires two sets of four-electron three-centre bonds of the $\ce{X- \bond{...}{I+}-X <-> X-{I+}\bond{...}X-}$ type. This type of bond requires a sufficient electronegativity on the outer atoms to stabilise the partial negative charge better — something fluorine is more capable of than chlorine. $\ce{ICl5}$ should only be possible because of iodine’s low electronegativity and $\ce{BrCl5}$ should be orders of magnitude less stable again.

A second aspect could be the larger size of chlorine atoms compared to fluorine atoms causing greater steric stress, longer bonds and therefore weaker interactions.


Using Mathematica, I obtained $T=546 K$, as follows.  One of the nice things about using Mathematica for physical calculations is that it has the ability to understand/keep track of/cancel out/convert units.

The volume is non-physically small -- about 750,000 x smaller than what we'd expect for a mole of, say, ideal gas at $P=100 Pa, T = 546 K$).  One does expect a non-physically small volume, since this is at a non-physically low point on the isotherm, but I would not have predicted it would be this much smaller.

If you were doing it by hand, you'd probably want to use a different approach:


Why does magnesium deposit during electrolysis of a molten mixture of MgI2 and AlBr3?

Yes, it is true that you cannot use the standard electrode potential to determine which cation would be reduced and vise versa during an electrolysis of mixed molten salts, in your case, $\ce{AlBr3}$ and $\ce{MgI2}$. That's because most tabulated values of standard electrode potentials are determined for aqueous solutions. Therefore, they can be used only as approximate guidance. Instead, electronegativity values (EN) can be used to estimate the stronger oxidizing and reducing agents. For example, EN values of $\ce{Br}$ and $\ce{I}$ are 2.8 and 2.5, respectively. Therefore, we can conclude that $\ce{I-}$ is the stronger reducing agent between two elements (because $\ce{I}$ is less EN than $\ce{Br}$). Similarly, between $\ce{Al}$ and $\ce{Mg}$, $\ce{Al^3+}$ is the stronger oxidizing agent because $\ce{Al}$ is more EN than $\ce{Mg}$ (1.5 vs 1,2, respectively), so $\ce{Al^3+}$ gains electrons easier.

Therefore, possible cathode half-reaction (reduction): $\ce{Al^3+(l) + 3e- -> Al(l)}$
And Possible anode half-reactions (oxidation): $\ce{2I-(l) -> I2(l) + 2e-}$


How to name binary (inorganic) compounds given their chemical formula, and vice-versa?

Prerequisites

These are the necessary prerequisites for this topic. If you're uncomfortable with any, please first head over to the corresponding links before continuing.

Chemical symbols are a shorthand method of representing an element. Instead of using the full name of an element, we simply refer to it with one or two letters. $\ce{N}$ for nitrogen and $\ce{Na}$ for sodium. More details on the Wikipedia page.
Polyatomic anions/Radicals: anions with more than one element, like nitrate ($\ce{NO3-}$) or sulfate ($\ce{SO4^2-}$). More details on the Wikipedia page.
Oxidation state: is an integer or decimal number assigned to an element in a chemical species. It is a tool that helps us do nomenclature easily. Read a detailed introduction here.
Ionic and covalent compounds: You must understand what ionic and covalent compounds are. You must also know the few elementary examples of each. For example, you should know that $\ce{N2O4}$ would be a covalent compound, while $\ce{NaCl}$ would be ionic. Here's an introduction by LibreTexts if you need a refresher.

Introduction

There are two separate cases here for ionic and covalent compounds.[1]. Some of the logic overlaps, but some does not, and care must be taken to avoid using one's rules for another.

Naming an ionic compound from its formula

Name the cation and then the anion. ("sodium sulfate", but not "sulfate sodium"). (How to identify the cations and anions in an ionic compound? The cation is usually a metal and the more electropositive element, while the anion is a non-metal and more electronegative element)
Keep everything lowercase (details) ("sodium sulfate", but not "Sodium Sulfate" or "sodium Sulfate" or "Sodium sulfate")
Do not apply numeric prefixes to either the cation or the anion. ("sodium sulfate", but not "disodium sulfate" or "sodium monosulfate")
Remember the suffixes for anions. The single atom anions of group 15, 16, and 17 have the suffix ending "-ide". Examples: fluoride, oxide and phosphide. However, polyatomic anions generally do not end in an -ide. See the Appendix for details.
Transition metal cations (and some p-block metals) also need their oxidation-state to be specified in parentheses, as they exhibit variable oxidation states.  Hence, "mercury(II) chloride", but not "mercury chloride". Notice that the oxidation state is denoted in capital Roman numerals.[2]

Let's take an example: $\ce{MgCl2}$. We can identify it's composed of magnesium and chlorine. Thus, the name would be magnesium chloride – in all lowercase and with the cation (metal) preceding the anion (non-metal). There is no need to denote the oxidation state as magnesium isn't a transition metal.

Examples:

$\ce{CsF}$, $\ce{NaNO3}$, $\ce{Al2(SO4)3}$, $\ce{FeCl2}$, $\ce{Pt3(PO4)4}$

Answers: (hover to peek!)

   caesium fluoride, sodium nitrate, aluminium sulfate, iron(II) chloride or ferrous chloride, platinum(IV) phosphate

Writing the formula for an ionic compound from its name

First write the cation and then the anion ($\ce{KBr}$ but not $\ce{BrK}$)
Use parentheses to denote multiple polyatomic anions ($\ce{Mg3(PO4)2}$ but not $\ce{Mg3PO42}$). Avoid them for a single polyatomic anion ($\ce{Na2SO4}$ but not $\ce{Na2(SO4)}$)
Use the trick of cross-transferring oxidation states between the ions to determine the count of each cation and anion involved. This trick works because the formula should denote an electrically neutral species. (you can verify this mathematically) So, to write the formula for aluminium sulfate:

Note that:

In case both oxidation states have a common factor, you should cancel them. Example: magnesium oxide:
.
drop the number 1. Hence, potassium sulfide is written as $\ce{K2S}$ but not $\ce{K2S1}$.

Examples:

potassium phosphate, calcium chloride, magnesium cyanide

Answers: (hover to peek!)

   $\ce{K3PO4, CaCl2, Mg(CN)2}$

Naming a covalent compound from its formula

Name the less electronegative element and then the more electronegative element ("carbon monoxide" but not "oxygen monocarbide")
Apply suffixes to anions in exactly the same way as we did for ionic compounds. ("sulfur hexafluoride" but not "sulfur fluoride") Also, keep the name all lowercase as before.
Use numerical prefixes.[3] Hence "dinitrogen tetroxide" for $\ce{N2O4}$ but not "nitrogen oxide". Here's the first ten suffixes: $$\begin{array}{|c|c|} \hline
\mathrm{prefix} & \mathrm{number}\\ \hline
\mathrm{mon(o)} & 1 \\
\mathrm{di} & 2 \\
\mathrm{tri} & 3 \\
\mathrm{tetr(a)} & 4 \\
\mathrm{pent(a)} & 5 \\
\mathrm{hex(a)} & 6 \\
\mathrm{hept(a)} & 7 \\
\mathrm{oct(a)} & 8 \\
\mathrm{non(a)} & 9 \\
\mathrm{dec(a)} & 10 \\
\hline
\end{array}$$ Note that:

the letter in parentheses is dropped when the anion begins with a vowel ("tetroxide", but not "tetraoxide")
we do not specify the mono- prefix for the first element ("carbon tetrachloride" but not "monocarbon tetrachloride")
we can construct similar numerical prefixes for larger counts by using the Greek system. Example: "undeca" for 11. Look at a larger list on Wikipedia.

Let's take an example: $\ce{P4O6}$. In one molecule of $\ce{P4O6}$, we can see that it has four atoms of phosphorus and six atoms of oxygen. So, its name should be of the form "Xphosphorus Yoxide". The table of numerical prefixes tells us that we should use the prefix "tetra" for nitrogen and "hexa" for oxygen. Since the name "oxygen" begins with a vowel, we shall write "hexoxide" (and not "hexaoxide"). Thus, the correct name for this chemical formula is "tetraphosphorus hexoxide".

Examples:

$\ce{CO2}$, $\ce{P4O10}$, $\ce{N2O}$, $\ce{N2S5}$

Answers: (hover to peek!)

   carbon dioxide, tetraphosphorus decoxide, dinitrogen monoxide, dinitrogen pentasulfide

Writing the formula for a covalent compound from its name

This is much simpler from the previous case, rather it follows an exactly reverse process.

We'll just take a look at two examples:

boron triiodide: we see that its formula would be: $\ce{B_$x$I_$y$}$. Now, $x=1$ since boron lacks any numerical prefix, and $y=3$ due to the "tri" prefix. Hence, the formula is $\ce{BI3}$.
dinitrogen pentoxide: we see that its formula would be: $\ce{N_$x$O_$y$}$. Now, $x=2$ and $y=5$ due to the "di" and "pent(a)" prefixes. Hence, the formula is $\ce{N2O5}$.

Examples:

hydrogen bromide, iodine heptafluoride, diboron hexahydride

Answers: (hover to peek!)

   $\ce{HBr}$, $\ce{IF7}$, $\ce{B2H6}$

Naming of hydrogen compounds

Several hydrogen compounds have non-systematic names that have persisted through the years in standard nomenclature and are now well recognized. Note that for alkali metals and alkali earth metal hydrides, the name is simply "metal + hydride", like sodium hydride, on the basis of rules described earlier.

However, covalent hydrides have varying names. There are not much rules around them, but rather "patterns" that the following list would show you:

Group 13: $\ce{BH3}$: borane, $\ce{AlH3}$: alumane, $\ce{GaH3}$: gallane, etc. Pattern: half of the element name suffixed by "-ane".
Group 14: $\ce{CH4}$: methane, $\ce{SiH4}$: silane, $\ce{GeH4}$: germane. Pattern: half of the element name suffixed by "-ane". Methane is a remnant of organic chemistry, and is a part of nomenclature of hydrocarbons.
Group 15: $\ce{NH3}$: ammonia, $\ce{PH3}$: phosphine, $\ce{AsH3}$: arsine, $\ce{SbH3}$: stibine, $\ce{BiH3}$: bismuthine. Pattern: half of the element name suffixed by "-ine". "Ammonia"'s origins, though, are a bit more interesting (read the Wikipedia page).
Group 16: $\ce{H2O}$: water, $\ce{H2S}$: hydrogen sulfide, $\ce{H2Se}$: hydrogen selenide, $\ce{H2Te}$: hydrogen telluride. Note that while "oxidane", "sulfane", "selane", etc. are acceptable, in the case of oxidane, that name is only used when denoting water derivatives. Read Do chemists refer to water as "dihydrogen monoxide"? and Does water have a chemical name?.
Group 17: they are simply named as "hydrogen halide". Example: "hydrogen bromide" $\ce{(HBr)}$. It is not named "hydrogen monobromide", as you may believe from earlier examples, and that it because hydrogen can only form a single halide with the halogens (see Note 3 below).

The curious students would also notice that – when writing chemical formulae – for groups 13, 14, 15, the hydrogen atom is written after the group atom $\ce{(NH3)}$, while in groups 16 and 17, it is written before the group atom $\ce{(H2O)}$. See this answer to understand why it is so.

Notes:

It is true that no compound is 100% ionic or 100% covalent. The above rules for "ionic compounds", thus, apply to compounds with dominant ionic character. Similarly, rules for "covalent compounds" apply to compounds with dominant covalent character.
Another way to write such a name is to use an "-ic" suffix for a higher oxidation state and "-ous" suffix for a lower oxidation state. As in "mercuric chloride" for "mercury (II) chloride". But, according to the IUPAC Recommendations 2005, it is no longer acceptable.
Why are numerical prefixes necessary? Note that in covalent compounds, a given anion can be present in more than one count. For example, the $\ce{F}$ in $\ce{BrF_x}$ can be present as $\ce{BrF}$, $\ce{BrF3}$, or even $\ce{BrF5}$. Hence, it is necessary to indicate if it is a monofluoride, a trifluoride, or a pentafluoride.
Several covalent compounds and ionic compounds have their own unique, common/trivial names. These names do not follow any systematic rules and are based on the source (location/discoverer) or properties of that compound. Learning these common names and their formulae comes by experience over time. Examples include: diborane $\ce{(B2H6)}$, Epsom salt $\ce{(MgSO4)}$, oil of vitriol (sulfuric acid), etc. Read more on the Wikipedia page.

Appendix: polyatomic anion names

Apart from the regular -ide suffix for monoatomic anions, there are some more suffixes for the polyatomic anions.

most pseudohalides retain the "-ide" suffix. (except cyanate derivatives)
the other two suffixes are "-ate" and "-ite". "-ate" is used when the central atom is in a higher oxidation state, whereas "-ite" is used for the lower oxidation state.
the "per" prefix is used when the central  halogen atom is in its highest oxidation state, while "hypo" is used when it is in its lowest oxidation state.

A few examples would make these rules clearer:

Sulfate $(\ce{SO4^2-})$ and sulfite $(\ce{SO3^2-})$: both are sulfur and oxygen anions, but in sulfate, sulfur has higher oxidation state as compared to that in sulfite. Similar case applies to nitrate, nitrite and phosphate, phosphite.
Hypochlorite $\ce{(ClO^-)}$, chlorite $\ce{(ClO2^-)}$, chlorate $\ce{(ClO3^-)}$, perchlorate $\ce{(ClO4^-)}$: have chlorine in increasing oxidation numbers. Similar series exist for other halogens (except fluorine).

Of course, there are many more polyatomic anions in chemistry, but these ions are relevant to our discussion at an elementary level.


Which sodium citrate is used in silver nanoparticles production?

You can use hydrated tri-sodium citrate, it's readily soluble in water. As for its chemical reactivity, it's identical to tri-sodium citrate.
You should take into consideration the increase in the molecular weight in the hydrated form due to the presence of water molecules.


Fragment #1 Preparation

1) An exact preparation of an aryl imidazolidinone (54% Yield)1
2) A modified preparation of a tetrazole from an aniline derivative (90% yield with pure aniline)2
3) A modified acetal formation (protocol modifications based on urea) (no reported yield)3

Fragment #2 Preparation

1) An exact preparation of an alpha-bromophenone (Hell-Volhard-Zelinski Reaction) (97% yield --likely prior to purification)4
2) Modified coupling of 1,2,4-triazole and alpha-bromophenone (100% yield with chloroacetone -- likely prior to purification)5
3) Wittig reaction modification (73% yield with phenyl methyl ketone)6
4) A modified triazolium formation (82% yield with methyl iodide)7
5) Stereoselective Jacobsen epoxidation (92% yield with 1-phenylcyclohexene)8

Final Fragment Coupling

1) Deprotonation of the amine with LDA
2) Epoxide ring opening followed by the protonation of the alkoxide and hydrolysis of the acetal to yield the final product

Stereochemical and General Considerations

The use of the Jacobsen epoxide formation allows for the use of the (L, L) Jacobsen catalyst in the final steps of the synthesis to yield the enantiomer of TAK-457. Synthesis of the other stereoisomer would require modification of earlier steps in the reaction.

Longest Linear Sequence: 5 steps
Total Number of Steps: 9 steps (I consider the final two steps as one since they are performed without subsequent purification)

Comment on 1,2,4-Triazole Ring: Yes, I did not prepare the triazole ring directly; however, this seems to be the more efficient route.

Comment on Wittig Reaction: This is the step I am probably the most worried about. The steric bulk from the phenyl group may be enough to slightly favor the desired enantiomer; however, this cannot be determined until further experimentation is performed.

References

Stabile, P.; Lamonica, A.; Ribecai, A.; Castoldi, D.; Guercio, G.; Curcuruto, O. Mild, Convenient and Versatile Cu-mediated synthesis of N-Aryl-2-imidazolidinones. Tetrahedron Lett. (2010) 51 (24). 3232–3235. DOI:10.1016/j.tetlet.2010.04.064
Kundu, D.; Majee, A.; Hajra, A. Indium Triflate-Catalyzed One-Pot Synthesis of 1-Substituted-1H-1,2,3,4-Tetrazoles Under Solvent-Free Conditions. Tetrahedron Lett. (2009) 50 (22). 2668–2670. DOI:10.1016/j.tetlet.2009.03.131
Shuai, F.; Wang, X.; Zhang, J. Preparation of Dipyridamole Active Pharmaceutical Ingredient. CNIPA Patent 108069972, Nov 16, 2016
Alonso-Alija, Cristina; et al. Preparation of Amino(monocyclic aroyl)pyridinones that Inhibit p38 Map Kinase for use as Antiinflammatory Agents. WIPO Patent 2003076405, Sep. 18, 2003
Sun, Y.; Huang, N.-Y.; Ding, M.-W. Efficient Synthesis of 6-(1H-1,2,4-Triazol-1-yl)-thieno[2,3-d]pyrimidin-4(3H)-ones via an Iminophosphorane. Synth. Commun. (2010) 40 (13). 1985–1991. DOI: 10.1080/00397910903219351
Pratsch, G.; Overman, L. E. Synthesis of 2,5-Diaryl-1,5-dienes from Allylic Bromides Using Visible-Light Photoredox Catalysis. J. Org. Chem. (2015) 80 (22). 11388–11397. DOI:10.1021/acs.joc.5b01962
Myles, L.; Gathergood, N.; Connon, S. J. The Catalytic Versatility of Low Toxicity Dialkyltriazolium Salts: in Situ Modification Facilitates Diametrically Opposed Catalysis Modes in One Pot. Chem. Commun. (2013) 49 (46). 5316. DOI:10.1039/c3cc41588k
Brandes, B. D.; Jacobsen, E. N. Highly Enantioselective, Catalytic Epoxidation of Trisubstituted Olefins. J. Org. Chem. (1994) 59 (16). 4378–4380. DOI:10.1021/jo00095a009


Heat is the transfer of energy to or from the body in forms other than matter flow or work (organized energy transfer, such as pushing).

Temperature is only a well-defined property for a collective body (you wouldn't be able to tell me the temperature of a single atom, for example). Like you said, it's the property of matter describing the amount of kinetic energy of the particles in the body. As to why this is, I'd ask: what happens at absolute 0?

   At absolute 0, heat has transferred out of the system so much so that you cannot lower the energy of the system any more.

(Note for the knowledgeable reader: there is indeed still a quantum mechanical phenomena -- zero point energy-- that prevents some energy from leaving the molecule, but that's a conversation for another time).

From a thermodynamic definition, temperature is the description of how the internal energy changes with entropy for a closed (no matter flows in or out) system of constant volume:

$T=\left(\frac{\partial{U}}{\partial{S}}\right)_{N,V}$

By increasing the entropy of the system by a fixed amount, the temperature of the system tells me by how much the internal energy will increase. Now, this is not a very useful form, as you cannot directly increase the entropy of a body (you must add energy and then let the entropy indirectly increase). It's much more useful to consider the inverse temperature:

$\frac{1}{T}=\left(\frac{\partial{S}}{\partial{U}}\right)_{N,V}$

At absolute zero, everything is in the lowest energy state. Any small transfer of energy to the system will result in a large increase in entropy. But this was only a small change, so the internal energy doesn't increase by much. Compare to a system at room temperature, where I must transfer much more energy to the system to achieve the same magnitude of increase in entropy.


How are electronic configurations including hybrid orbitals notated?

The electronic configuration concept does not really extend itself well to hybrid orbitals, because these are best considered mathematical concepts only. However, the problem is easily solveable:

Always use brackets when dealing with hybrid orbitals.

Even your $\ce{CO2}$ case is potentially misleading: I misread it once as $\mathrm{2s^1\ 2p^2}$. It is best to show what belongs together with: $\mathrm{[He]\ 2(sp)^2\ 2p^2}$ or, for methane: $\mathrm{[He]\ 2(sp^3)^4}$.


Sign of the work done during a reversible compression

PV work is defined as
$$w = -p\Delta V,$$
where $p$ is pressure and $\Delta V$ is $V_\mathrm{final} - V_\mathrm{initial}$.

It has this form because in chemistry, work is defined from the perspective of the system - if it is compressed, we say that we did work on the system and so the sign of work is positive.

In this example, $V_\mathrm{f}$ is less than $V_\mathrm{i}$, so $\Delta V < 0$. If $\Delta V$ is negative, then $w$ must be positive.

This means your answer should have a positive sign. However, I get a different number from $1.8$ - you should double-check that.


When you increase the temperature such that change in Gibb's free energy becomes zero, does Keq = 1 at that temperature?

Yes and no.

$K = 1$ is true, and it follows directly from the equation $\Delta G^\circ = -RT \ln K$.

However, there is no reason for it to be "an equilibrium in which reactants and products are in standard conditions". $K = 1$ does not necessarily mean that the concentrations (or activities if you prefer) of all reactants and products are also equal to $1~\mathrm{mol~dm^{-3}}$ (or $1$, in the case of activities). Remember that $K$ is a fraction. For the reaction:

$$\nu_{r1}\ce{R1} + \nu_{r2}\ce{R2}\cdots \longrightarrow \nu_{p1}\ce{P1} + \nu_{p2}\ce{P2} \cdots$$

we have:

$$K = \frac{[\ce{P1}]^{\nu_{p1}}[\ce{P2}]^{\nu_{p2}}\cdots}{[\ce{R_1}]^{\nu_{r1}}[\ce{R_2}]^{\nu_{r2}}\cdots}$$

Just like how the fractions $\displaystyle \frac{2}{2}$, $\displaystyle \frac{4}{4}$, or $\displaystyle \frac{4000\hbar\pi e}{4000\hbar\pi e}$ are all equal to $1$, the individual concentrations $[\ce{P1}], [\ce{P2}], [\ce{R1}], [\ce{R2}]\cdots$ might not be equal to $1~\mathrm{mol~dm^{-3}}$, but the result will still be $1$.


Can we process waste just by decomposing it to homogenous mass and extracting the elements from it?

Sure, this absolutely can be done.  In particular, you cannot destroy elements using conventional means like burning, so all the initial elements are still in the mass.  The question is whether or not this is a good way to extract what you want.  There may be better chemical techniques to do the extraction that are short of burning.  The branch of chemistry which deals with the optimal conversion of elements and chemicals to other useful substances is chemical engineering, and figuring out which process is "best" (e.g. fastest, cheapest, highest yield) is crucial for industry.

There is a very funny story related to this idea on the Periodic Table of Videos.  A scientist was trusted with the entire UK supply of plutonium (10mg, a very small amount) during World War II.  He accidentally spilled it on to his lab bench.  He then took a saw, sawed a large circle out of bench, burned that wood, and recovered nearly all the plutonium!  So yes, this works.  Here's the video, and this story begins around 15:30: Plutonium Video

EDIT: As to your update, there are lots of methods available, but it depends greatly on the starting conditions and materials.  For example, burning is a great way to extract carbon: burning hydrocarbons or carbohydrates burns off the hydrogen or water (respectively) and leaves mostly pure carbon.

But extracting metals might take more work.  Often, metals exist as ionic compounds.  For example, silver nitrate, $\ce{AgNO3}$, contains siver $\ce{(Ag)}$ in the +1 state.  While it might be possible to dissociate the silver with high enough temperature, it is far easier to extract metallic silver by reducing silver in the presence of a more reactive metal.  "More reactive" simply means a metal that will more easily give up an electron to the less reactive sliver, changing the silver ion $\ce{Ag+}$ to metallic $\ce{Ag}$.  Silver nitrate is highly soluble in water, and if you simply drop in a piece of metallic copper $\ce{(Cu)}$ metallic silver will slowly come out of solution and plate the copper with silver, while copper dissolves and creates copper nitrate.  The equation is:

$\ce{2AgNO3 + Cu->2Ag + Cu(NO3)2}$

but the crucial part is the ionic equation:

$\ce{2Ag+ + Cu->2Ag + Cu^2+}$

which shows that the silver ion is reduced to silver, and the copper metal is oxidized to copper ion.

Again, Periodic Videos has the footage of this exact reaction: Silver Video

Now I picked this example precisely because of this footage, but it's not the best example because silver is so low in reactivity that it exists metallically in nature (i.e. you can find lumps of silver if you are lucky!)  But many metals are so reactive that it takes a huge amount of energy to extract them.  Examples are sodium or potassium, which are so reactive they explode in the presence of water (go search YouTube and have a field day!).  To extract them as metals commercially usually requires electrolysis, which uses electricity to force electron transfer which would not otherwise occur spontaneously.

And on and on with other methods.  There is no universal extraction method, at least not yet, and this is why we have chemcial engineers to figure out better and better procedures to use.


Conversion of Z-matrix to Cartesian coordinates

I will try to answer the question in bold from a mathematical perspective.

Basically, the first two lines in get_local_axes are to build the vectors $\vec r_{12}$ and $\vec r_{23}$. The next line builds the cross product $$\frac{\vec r_{23}\times\vec r_{12}}{|\vec r_{23}\times\vec r_{12}|},$$

The next vector is
$$\frac{\vec r_{12}\times(\vec r_{23}\times\vec r_{12})}{|\vec r_{12}\times(\vec r_{23}\times\vec r_{12})|},$$

This is a vector perpendicular to $\vec r_{12}$, and lies in the plane where $\vec r_{12}$ and $\vec r_{23}$ lie.

The next lines are defining the three axes of the system. $\vec e_z$ is just $\vec r_{12}$ normalized, $\vec e_y$ is chosen so that $yz$ is the plane where the three points (atoms) locate. and $\vec e_x$ is chosen so that $(\vec e_x, \vec e_y, \vec e_z)$ forms a orthonormal set and are the versors of a right-handed coordinate system. This system is the local axis system.

So this is the meaning of "getting local axis system from 3 coordinates".


Can calcium hydroxide + sulphuric acid be a way of preparing insoluble calcium sulphate?

No.

The outside layer of the calcium hydroxide reacts with the sulfuric acid to form the calcium sulfate that you want. All good, until you realise that the newly-formed calcium sulfate is now coating the rest of the calcium hydroxide that didn't get to react. In the end you just get a big mess.

From a practical standpoint, even if your method worked, it's just much easier to just mix two solutions together and filter. No need to stir stuff and wait for insoluble powders to dissolve. On top of that, you're dissolving a white powder in order to make another white powder - how would you know when your reaction has finished?

Conclusion: Use calcium chloride and sodium sulfate.


Is the name of this chemical 3-oxopropanoic acid?

The name ‘3-oxopropanoic acid’ for the compound $\ce{OHC-CH2-COOH}$ given in the question is correct.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  15. aldehydes
  (…)

Thus, the carboxylic acid group has priority to be cited as a suffix.

Systematic names of carboxylic acids are formed as follows:

  P-65.1.2 Systematic names

  Carboxylic acids are named substitutively using the suffix ‘oic acid’ or ‘carboxylic acid’ and the prefix ‘carboxy’ to describe carboxy groups that cannot be included as the principal characteristic group or to express the carboxy group in the presence of a higher principal characteristic group.

  P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; (…). Except for (…), systematically formed names are preferred IUPAC names; (…)

Therefore, the preferred IUPAC name (PIN) of the unmodified carboxylic acid is ‘propanoic acid’. (Note that the retained name ‘propionic acid’ may be used in general nomenclature.)

Aldehydes can be systematically named in various ways. Monoaldehydes derived from alkanes are preferably named substitutively using the suffix ‘al’. In this case, however, this method cannot be used since the carboxylic acid group has priority to be cited as a suffix. Therefore, the aldehyde group has to be expressed by a prefix as follows:

  P-66.6.1.3 In the presence of a characteristic group having priority to be cited as a suffix or when present on a side chain, a $\ce{-CHO}$ group is expressed by the preferred prefix ‘oxo’ if located at an end of a carbon chain, or, otherwise, by the preferred prefix ‘formyl’.

Thus, the PIN for the complete structure is ‘3-oxopropanoic acid’.

The corresponding name that is formed by using the prefix ‘formyl’ is ‘formylacetic acid’. Such names are not acceptable as PINs but may be used in general nomenclature.

Furthermore, a special rule for aldehydic acids applies to the given compound:

  P-65.1.6.3 Aldehydic acids

  When a dicarboxylic acid has a retained name (…) and when one of its carboxy groups is replaced by a formyl group, $\ce{-CHO}$ (…), the resulting structure is called an aldehydic acid and, in general nomenclature may be named by replacing the ending ‘ic acid’ of the name of the dicarboxylic acid by the ending ‘aldehydic acid’. Preferred IUPAC names for aldehydic acids derived from all dicarboxylic acids are constructed systematically. The prefix ‘formyl’ is used in preferred IUPAC names, except for a $\ce{-CHO}$ group at the end of an acyclic chain, which is designated by the prefix ‘oxo’.

For the compound given in the question, the PIN of the corresponding dicarboxylic acid is ‘propanedioic acid’.

This dicarboxylic acid also has the retained name ‘malonic acid’, which may be used in general nomenclature. Hence, according to P-65.1.6.3, the corresponding aldehydic acid may be named ‘malonaldehydic acid’ in general nomenclature.

By way of comparison, for the analogous example $\ce{OHC-CH2-CH2-COOH}$, all three corresponding names are explicitly mentioned in the Blue Book:

4-oxobutanoic acid (PIN)
3-formylpropanoic acid
succinaldehydic acid


You need to reckon with how strong a base the nucleophile is compared with what you are trying to displace.  Only if the nucleophile is the stronger base can you expect the reaction to go based on thermodynamics.

Thus, both of the following reactions can go:

$\ce{RCl + OCH3^- -> ROCH3  +Cl^-}$

$\ce{RCl + CN^- -> RCN  +Cl^-}$

because both nucleophiles are stronger bases towards carbon than the chloride ion that would be displaced.  You can then say that the second reaction will go faster with a better (kinetic) nucleophile.  But if you have

$\ce{ROCH3 + CN^- -> RCN  +OCH3^-}$

where you are trying to displace the strongly basic methoxide ion rather than adding it as a nucleophile like the first reaction above, then cyanide ion is too weakly basic to push through.  Cyanide ion can be a good nucleophile only when displacing a weaker base or adding to an appropriate unsaturated compound (such as a carbonyl compound) without displacement.

Try it now.


Basic thermodynamics question about work

Well, the answer is tricky. But before the really tricky part, you may want to take a step back and consider a general statement of the 1st law,

$$dU = \sum_i \left(\frac{\partial U}{\partial q_i} \right)_j dq_i$$

where $q_i$ is an extensive coordinate of the system. Note this is consistent with our standard description of the relation between generalized forces and energy (the derivatives wrt coordinates are forces).

In the case of pV work, $q_i = V$ and

$$ \left(\frac{\partial U}{\partial V} \right)_j = -p $$

The partial derivative in question is evaluated holding all other extensive variables (including number of particles) constant. If there is a change in chemical composition of the system we can define a chemical force (although referred to as the chemical potential of substance i) as

$$ \left(\frac{\partial U}{\partial n_i} \right)_j   = \mu_i  $$

and an infinitesimally small amount of chemical work as

$$ \left(\frac{\partial U}{\partial n_i} \right)_j dn_i  = \mu_i dn_i $$

As noted in the comments, the formulation combining the first and second laws of thermodynamics presented in the OP is a particular representation of the first law for the case of reversible pV work, a more general formulation of the 1st law being

$$\begin{align} dU &= dq + dw\end{align}$$

Note that while $dU$ is exact, the differentials on the right-hand-side of the previous equation may be inexact. Distinguishing between pV and other types of work for a reversible process (for which the differentials $dw$ and $dq$ become exact) we can write

$$\begin{align} dU &= dq_{rev} + dw_{pV,rev} + dw_{nonpV,rev}   \\ &= dq_{rev} + dw_{pV,rev} + \sum_i \mu_i dn_i   \\ &= TdS + dw_{pV,rev} + \sum_i \mu_i dn_i \end{align}$$

The second equality follows from the first using the 2nd law. Then since $dw=-p_{ext}dV$ in the case of pV work  and since $p_{ext} = p_{sys}=p$ when the work is carried out reversibly, you arrive at the expression in the OP.

  I can imagine a situation where the volume is fixed, but number of particles increases, thus increasing the pressure. I would think that this increase in pressure would also increase the internal energy.

You have to be careful how you define the process that alters the number of particles. Do you have an open system that allows the particles in from the surroundings?

Assume you transfer an ideal gas at constant T from the surroundings into the system, which consists of a vessel at lower pressure with a small orifice which we seal at some point. Since the gas is ideal there is no change in energy with pressure at constant T. There is no mechanical work done, since there is no volume change, but the change in the number of particles clearly amounts to a transfer of energy. This is accounted for in the term $\sum_i \mu_i dn_i $ of $dU$, provided the entropy is constant.  Changing the number of particles amounts to "chemical" work.

  Secondly, in the same manner I also don't understand why dQ is defined as TdS and not as TdS+SdT. I would think that a change in temperature would surely change the internal energy.

A change in temperature can change the internal energy, but that does not require a change in the definition of heat or entropy. The equation $dQ=TdS$ holds only for a reversible transfer of heat, it is not general.

Now to the tricky part.

You can write

$$dU = -pdV-Vdp +SdT + TdS + \sum_i \mu_i dn_i + \sum_i n_i d\mu_i  $$

since

$$U = -PV + TS + \sum_i n_i \mu_i$$

The thing is you can show that

$$ 0 = -Vdp +SdT + \sum_i n_i d\mu_i  $$

This is known as the Gibbs-Duhem relation.


When dissolving, the ionic crystal’s lattice enthalpy must be broken while at the same time the hydration shell is generated around the cations and anions. The latter process is typically exergonic while the former is strongly endergonic.

How well a salt dissolves depends not only on the strength of solvation but also on the strength of the crystal lattice. If the lattice is strong, the solubility will be lower because there is an increased tendency to reform the crystal lattice.


How to explain the periodic trends in boiling points in groups?

I think it is wrong to say there should be a trend as we study the trends of individual atoms, not molecule and compound itself. Boiling points depend on the intermolecular force which depends on the polarity of the whole molecule, which in turn depends on the constituting species, so there should not be specific trend.
It is only the matter of comparison between the species in a molecule. Moreover, there is not a single type of intermolecular forces which bonds molecules such as dispersion, metallic or hydrogen: all have different strength.


Why does zinc plate copper in silver penny lab

Electricity is involved: the copper (well, copper plated zinc in the US) penny is in direct contact with the zinc, forming a short-circuited connection. Forcing the plating action "backwards" (copper being more electronegative than zinc) requires current.

Try the same experiment but with a separator, such as a piece of filter paper, between zinc and copper; there should be no plating.


Dieckmann condensation's final protonation step

why does the enol tautomerizes to the keto form?

In 1,3-dicarbonyl systems, like the one pictured above, we would generally expect to have a measurable amount of the enol form present.  The keto-enol interconversion is an equilibrium and ratio of keto and enol will be guided by thermodynamic factors.  Let's take a look.

If we compare the structural elements in the keto and enol forms, we see that the differences between the two tautomers are as follows:

The keto form has an extra carbonyl (178), a $\ce{C-C}$ single bond (83) and a $\ce{C-H}$ bond (99)
The enol form has a $\ce{C-O}$ bond (86), a $\ce{O-H}$ bond (111) and a $\ce{C=C}$ double bond (146)

The numbers in parenthesis are standard bond energies in kcal/mol.  If we add up the keto values we get 360 kcal/mol; while for the enol it totals to 343 kcal/mol.  The numbers strongly favor the keto form (by 17 kcal/mol).  That is why is most simple systems the keto form strongly predominates.

However as you point out, both resonance and hydrogen bonding can further stabilize the enol form in the current example.

The hydrogen bond strength in $\ce{O..H-O}$ systems is typically 5-6 kcal/mol. In this system, with an intramolecular bond and a 6-membered transition state the hydrogen bond strength is likely closer to 9-10 kcal/mol. Resonance stabilization in 1,3-pentadiene is approximately 6.2 kcal/mol. Due to increased polarization in the enol from the carbonyl replacing a carbon-carbon double bond, the resonance stabilization probably increases to something around 8 kcal/mol.

A 9 kcal/mol effect due to hydrogen bonding and a 8 kcal/mol push due to resonance, shifts the 17 kcal/mol stabilization of the keto form to something very close to where the keto and enol forms have comparable energy. Therefore, in  1,3-dicarbonyl compounds we would expect a delicate balance between keto and enol forms, where both forms may be detectable.


How does the Hammett acidity function work and how to calculate it for [H2SO4] = 1,830? Any references?

I have a solution which has $[\ce{H2SO4}]$ = 0.915M (I assume $[\ce{H+}]$ = 1,830 M).

While $\ce{H2SO4}$ is a strong acid, $\ce{HSO4-}$ is not ($\mathrm{p}K_\mathrm{a}$ around 2). So your pH should be about zero, and you might not need to use the Hammett acidity at all.


Sodium is normally stored as cubes under petrol or similar solvent. As a metal it is very soft, you can easily cut it with knife. For the reaction you need fresh reactive surface of the metal, so the wire is one of best choices. For lab-scale preparation see e.g. https://www.youtube.com/watch?v=btvQdW4fXJE


Let's take a (2-3) spinel ($\ce{A^{II}B^{III}_{2}O4}$) as an example. If it is normal or inverse can be predicted from the total crystal field stabilization energies. For a normal spinel, the total CFSE wold be: $$\Delta_{T}(A)+2\Delta_{O}(B)=\Delta_{T}(A)+\Delta_{O}(B)+\Delta_{O}(B)$$ and for an inverse spinel $$\Delta_{O}(A)+\Delta_{T}(B)+\Delta_{O}(B)$$ with $\Delta_{T}$ being the tetrahedral and $\Delta_{O}$ being the octahedral splitting energy of $\ce{A^2+}$ and $\ce{B^3+}$, respectively.

If $\Delta_{T}(A)+\Delta_{O}(B)>\Delta_{O}(A)+\Delta_{T}(B)$, the normal spinel structure should be favored, as the resulting CFSE is higher. If it is the other way around, an inverse spinel is to be expected.

To sum it up, if the CFSE gained by placing $\ce{A^2+}$ in the octahedral holes outweighs a potential destabilization caused by putting a part of $\ce{B^3+}$ in the tetrahedral holes, an inverse spinel can be formed. However, a normal spinel is favorable when $\ce{B^3+}$ clearly "prefers" the octahedral and/or $\ce{A^2+}$ the tetrahedral holes.


Role of THF in oxymercuration of alkenes

$\ce{THF}$ is a polar aprotic solvent, that is, it can stabilize positive charges but not negative ones since it won't form hydrogen bonds with the help of any of its own hydrogen atoms.

Thus the carbocationic intermediate, that is the mercurinium ion is still stabilized, allowing the reaction to proceed.

And in Hydroboration, $\ce{THF}$ acts as the solvent as well but plays the additional role of shifting the equilibrium between diborane and borane to the side of the latter. As you might know, the Boron atom of $\ce{BH3}$ is electron deficient, which plays a key role in the formation of the trialkylborane intermediate of the Hydroboration reaction. Formation of diborane quenches this electron deficiency of borane by the formation of two 3-centred, 2-electron bonds in the dimer.


Why is hexafluoridocobaltate(III) the only common high-spin cobalt(III) complex?

This is a question that the crystal field theory cannot answer because of its inherent shortcomings.

The crystal field theory assumes point-type charges approaching the central metal from the coordinate axes leading to the destabilisation of those $\mathrm d$ orbitals that do not feature a nodal plane in the coordinate axes: $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$. The others have zero contribution in all $x,y$ and $z$ direction and are thus not affected by point-sized ligands approaching from there. As simple as this model is, it only gives a crude approximation of why field splitting occurs and which orbitals are likely to end up higher in energy. For everything that involves quantitative discussions, the full molecular orbital approach is needed.

Below you will find the fullest molecular orbital scheme of an $\ce{[ML6]}$ complex including σ and π interactions. A simplified version without the π interactions can be found here; you should attempt to understand it before adding the π interactions to the picture. In this question, however, we need to understand the π effects and thus I cannot leave them out.

Scheme 1: Molecular orbital scheme of an $\ce{[ML6]}$ complex featuring both σ and π interactions.

The system we are looking at — $\ce{[CoF6]^3-}$ contains the π-basic fluoride ions (meaning that their $\mathrm p$ orbitals that can interact with the metal in a π symmetric manner, are fully populated). The interaction of the $\mathrm{t_{2g}}$ fluorine group orbital with the metal’s $\mathrm{t_{2g}}$ orbitals will create a bonding-antibonding interaction. With fluoride most likely having lower-lying orbitals, this interaction should raise the energy levels of cobalt’s $\mathrm{t_{2g}}$ orbitals which thereby turn into $\mathrm{t_{2g}{}^*}$ orbitals.

In this case, fluorido is the only strongly π-basic ligand you are considering in your comparison. And chloride does not form a hexachlorido complex like fluoride would but a tetrahedral tetrachloridocobaltate(III) for steric reasons. Thus, we don’t know whether the fact that fluoride is π-basic enough to raise the $\mathrm{t_{2g}}$ levels high enough so spin pairing does not occur is because of its π basicity alone or whether further elements come into play. Oxalato, ammin and aqua ligands have only weak π-basic effects while cyanido, if it displays π effects, is a π acid.


How to arrive at the preferred IUPAC name for 1,2-diphenylethane?

The essential problem in this case is finding the parent structure. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature.

1. Substitutive nomenclature

In many related simple cases, the compound consists of just a ring and a chain so that the parent structure could be the ring or the chain.

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

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

These rules apply to compounds that consist of one ring and one chain. The compound that is given in the question, however, consists of a symmetrical chain with two rings. Extending the use of the above-mentioned rules to this case, we could pick one of the rings as the parent structure in accordance with Rule P-52.2.8. This approach would result in the correct name (2-phenylethyl)benzene. Since this method usually yields the preferred IUPAC in many simple cases, it should not be a surprise if some computer-generated names follow this pattern also in this case. Note, however, that the name “2-phenylethylbenzene” suggested in Pubchem is not correct and ambiguous (“2‑phenylethylbenzene” could be interpreted as “(2‑phenylethyl)benzene” or “2‑phenyl‑1‑ethylbenzene”).

In accordance with P-44.1.2.2, the context may favour the chain as parent structure. For the compound that is given in the question, the context may favour the chain in order to highlight the symmetrical shape of the structure, in order to treat the equivalent rings alike, or because a following reaction happens at the chain. This approach would result in the correct name 1,2-diphenylethane.

2. Multiplicative nomenclature

The IUPAC recommendations explain when multiplicative nomenclature is used to generate the preferred IUPAC name (PIN) as follows.

  P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES

  P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
  (1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
  (2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
  (3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.

  When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The compound that is given in the question fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

  P-15.3.1.3 Multiplicative name formation
  Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
  When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:

  (a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 1,1'

  (b) the name of the linking multiplicative substituent atom or group;

Here: ethane-1,2-diyl

  (c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

  (d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: benzene

  The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Therefore, the preferred IUPAC name is 1,1'-(ethane-1,2-diyl)dibenzene.


Which chemical is used in drunken driver test?

The test is administered by breathing through a tube, and the results become apparent immediately. Indeed, this has something to do with oxidation of alcohol. As soon as the alcohol in the exhale reaches the contents of the test tube, the latter changes its color, and the police guys put you in the handcuffs. Or it doesn't, and they don't. That's a wonderful example of chemistry improving our life.

Now to the point of comparing different reagents.

Permanganate itself is way too dark. When looking through its solution, you won't be able to detect any trace of any other color, let alone the products of reduction. Besides, the solution is rather sensitive and won't last long.
Chromyl chloride is too dangerous. It is liquid, and the suspect is going to breath in some tiny droplets (and/or possibly vapors) of it, and that's not something you would want to happen to a human being.
Hydrogen peroxide... have you ever seen it, really? It has the almost same color as water (that is, no color at all), and the reaction products have the same color as well. You won't be able to see a thing. Besides, this one is also too sensitive. Besides, it won't oxidize ethanol right away.

So it goes.


How is temperature constant in enthalpy change formula?

The internal energy of a system is in general not only a function of temperature. This means that holding the temperature constant does not guarantee that the energy will be constant. The statement that the energy can be expressed as a function of only the temperature applies to the particular case of ideal gases.
Assuming that gases are ideal simplifies expressions for the change in internal energy and enthalpy. For ideal gases at constant temperature and composition ($\Delta T=0$ and $\Delta n=0$):
$$\Delta U = 0\\\Delta H = 0 \tag{1}$$
These expressions are not general. They apply to ideal gases at constant temperature and composition.
In general,
$$\Delta H = \Delta U + \Delta (pV) \tag{2}$$
For an ideal gas at constant temperature,  $ \Delta (pV)= \Delta (nRT)= RT\Delta n$
so that
$$\Delta H = \Delta U + RT\Delta n \tag{3}$$
It is also worth remembering that at constant pressure $w=-p \Delta V$ so that  equation (2) becomes $\Delta H = q_p$ (in general for any substance, provided p is constant and only pV work is done). But what is $\Delta U$?
We can show from the ideal gas law that it is only a function of T for a constant amount of a pure ideal gas, and, from the kinetic theory of gases, one can also show that for one mole,
$$U_{m,i} = f_iRT$$
where $f_i$ is a constant that depends only on the type of gas (not on thermodynamic variables). Then
$$\Delta U = \sum_i U_{m,i}\Delta n_i = RT \sum_i f_i \Delta n_i $$
for a mixture of ideal gases.


Why do we need three equations to find the pH of NaCN, given Ka(HCN)?

So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:

  Equation 1: $\ce{HCN + H2O <=> H3O+ + CN-}$

As you correctly note, this is to remind you of the definition for $K_a$

  Equation 2: $\ce{NaCN -> Na+ + CN-}$

Note that I've changed $\ce{<=>}$ to $\ce{->}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.

Value of writing this out:
In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.

  Where does Equation 3 come from?

Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:

$$\ce{CN- + H2O <=> HCN + OH-}$$


Up to which dilution can one assume that 1 mg/l = 1 ppm?

"Parts per …" are dimensionless entities: to derive $\pu{ppm}$ from mass concentration $γ$ (e.g. $\pu{1 mg L-1})$ one uses density of water $\rho(\ce{H2O}) = \pu{1E6 mg L-1}$ for normalization:

$$\frac{γ}{ρ(\ce{H2O})} = \frac{\pu{1 mg L-1}}{\pu{1E6 mg L-1}} = \pu{1E-6} = \pu{1 ppm}$$

Once the solution is too concentrated, you obviously are not allowed to use $ρ(\ce{H2O})$ and have to use the real concentration of the solution.
However, this occurs at the concentration levels way higher than those to be expressed with $\pu{ppm}$ anyway: that's why "parts per …" are so ubiquitously used in biochemistry. Higher molar masses, lower molar concentrations, primarily aqueous media: no need to account for the error stemming from the difference in the densities of pure water and the solution.

Note that "parts per something" units are deprecated and are not recommended for use, see e.g. Is 1 ppb equal to 1 μg/kg?.


Production of acetone from egg shells and vinegar

Calcium acetate decomposes starting at around 160 degrees Celsius, according to Wikipedia. So yes, it is definitely possible to do so in a gas flame.


Can magnetic fields affect a chemical reaction?

Here is a nice review paper on the subject: Rodgers, C. T. Magnetic field effects in chemical systems. Pure Appl. Chem. 2009, 81 (1), 19–43. DOI: 10.1351/PAC-CON-08-10-18.

From the abstract:

  Chemical reactions that involve radical intermediates can be influenced by magnetic
  fields, which act to alter their rate, yield, or product distribution. These effects have
  been studied extensively in liquids, solids, and constrained media such as micelles. They  may be interpreted using the radical pair mechanism (RPM). Such effects are central to the field
  of spin chemistry of which there have been several detailed and extensive reviews.


How do you experimentally calculate the ionization constant of the carbonate ion?

So far, so good. You'll have to graph the $\mathrm{pH}$ of your system with the volume of titrant you've used to construct what's called a titration curve.

When you do the titration curve for this experiment, depending on the relative strengths of the bases involved, you will see one or two points of inflection points that will signal that the equilibrium conditions have changed. This will tell you if you need to take into account the second ionization. If there is a second inflection point, that's the second ionization.

Without actually going ahead and looking at the $\mathrm{p}K_\mathrm{a}$ of $\ce{H2CO3}$, I think for most computations, it could be omitted, since it is a second ionization which has a much lower constant than the first. Not unless you're looking for zwitterionic states or something like that.


Words like "homogeneous" and "heterogeneous" do not have a well-defined meaning.  Different authors use them differently.  The fact is, all matter is made of atoms, and atoms are not homogeneous.  Thus, from a certain point of view, nothing is homogeneous.  Matter is lumpy!

But suppose we don't care about things on the scale of atoms, i.e. angstroms.  Then the question is "what scale do you care about?"  Nanometers?  Hundreds of nanometers?  Chemists are making materials that have lumpiness on more more less all of these scales.

Back before folks invented centrifuges, filtration, and microscopes, bacterial cultures, with cells as small as 1 μm in diameter, probably seemed homogeneous.  But van Leeuwenhoek peered through a microscope and dispelled that notion.


Formal charges can mislead when writing arrow-pushing mechanisms

Arrow pushing is a tool that helps you think about the possible reactions available for a given species or set of reactants: where electrons are likely to move in order to form new bonds and break old ones. But you still have to look at the species that you get after doing the arrow pushing and consider whether they will be favorable or not. Sometimes you can get a reaction between a (formally) negatively charged atom and a positively charged atom - for example between methoxide and tert-butyl cation. Do the arrow pushing and see how that works out.

The reactions you have labeled as incorrect can be seen (via arrow pushing) to produce species that have too many bonds or electrons around a single atom (that was what was meant by the term "hypercoordination" - you could also say hypervalent - for example, having 10 valence electrons around oxygen), whereas the ones labeled as correct lead to stable or at least normally bonded species.


A different type of photosensitive material

There are many types of photoresist that might suit your purpose. Many are commercially available from electronics suppliers; even Amazon carries some. A thin film of photoresist could be used to separate two liquids, and dissolve when exposed to light, if at least one of the liquids is transparent to the wavelength needed.

You might also consider a photo-sensitive valve using a plastic that changes shape in light.


Well, let's see. As you know, the equilibrium constant is determined by the reaction's $\Delta_rG:\;K=e^{-\frac{\Delta G}{RT}}$, so a discrete jump in the former means a discrete jump in the latter. What is that $\Delta_rG$? That's the sum of $\Delta_fG$ of products minus that of starting compounds. Now what is the $\Delta G$ of formation of a compound? A continuous function, unless there is a phase transition, which by your problem definition is not the case. Well, a linear combination of a few continuous functions is still a continuous function.


Why is dp/dv zero for a real gas in critical conditions?

This is merely a shard of a fact which does not make much sense in and by itself. After all, in systems with gas/liquid equilibrium there is nothing really special about $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. On the contrary, this is pretty typical. See all those points where the blue lines (isotherms) are horizontal? They make up a whole huge area on the diagram.

(source)

Indeed, if you have a liquid in equilibrium with its saturated vapor, then the pressure does not depend on volume (in an isothermal processes, that is). As you compress it a little more, some gas turns to liquid; as you release it, some liquid evaporates. The pressure still remains the same - namely, the saturated vapor pressure at given temperature. Things start changing only when you pull the piston all the way out or push it all the way in.

Now, the two-phase area, huge as it is, does not comprise all diagram. If you heat the gas past the critical temperature, you'll get an isotherm of a different sort (it is not shown on the picture, so just imagine one, positioned above the rest of them). It will be all smooth and monotonic, without those horizontal parts.

Well, then there must be a temperature where this behavior changes, and that's what we call the critical temperature. Sure enough, it produces a very special isotherm: there is only one point on it where $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. (That's what we call the critical point, BTW.) Moreover, since it is an inflection point, it also has $\left(\dfrac{\partial^2\mathfrak p}{\partial V^2}\right)_T=0$. Now that's the fact one should remember about the critical point.

The peculiar behavior of pretty much all physical properties around the critical point gives rise to various weird phenomena like critical opalescence, but that's another story.

So it goes.


How to calculate molality, molarity and mole fraction given density (g/mole)?

First let me correct you, m/m means mass/mass % I think you assumed it to be mole/mole.

Assume you have 100g of solution in which Cu is 17g and solvent is 83 gm. 17 gm copper means 0.268 moles of Cu.

Now, density = Total weight/Total moles = 100/(0.268 + x) = 1.367 where x is moles of solvent, from here you will get x = 72.8849

Thus, mole fraction of Cu is 0.268/(0.268+72.8849) = 0.00366

Molality = Moles of solute/Weight of Solvent(in Kg) = 0.268/0.083 = 3.229 m

Since, density in terms of g/l is not given I don't think it's possible to calculate molarity.


What does an endpoint saturated signal mean. How does it relate to equilibrium constant?

$EC_{50}$ is concentration of a bioactive molecule which gives half the maximum response. You should note that the response is bound by a maximum which means this dose-response curve follows a saturation kinetics.

Saturation kinetics can be modelled in many ways and Hill and Michaelis-Menten kinetics are the most commonly used models in biochemistry. Again, a point to note is that many mathematical functions/models do not explain the underlying mechanism. They just mimic the behaviour of the system.

Lets assume that the antibody-response follows a two-step kinetics like Michaelis-Menten.

$$\ce{Ab + Ag <=>[k_a][k_d] AbAg ->[k_{res}] Response }$$

Lets call $R_{max}$ as the maximum response rate. Like in the case of MM kinetics, the response rate $R$ would be (see the wiki page for the derivation):

$$R=R_{max}\frac{[Ag]}{K_m+[Ag]}$$

Here $K_m$ is the concentration of antibody which would give half the maximum response rate (half response rate times times a sampling time would give you the $EC_{50}$).

Now, what $K_m$ really is depends on the approximation of the model:

Equilibrium model: this assumes that the binding and unbinding reactions are much faster than the response reaction and are hence at equilibrium (wrt response dynamics). In this case $K_m$ would be same as the dissociation constant.
Quasi Steady State model: this assumes that enzyme-substrate complex is at a quasi steady state (QSS) i.e. it doesn't change with time. In the case of enzyme substrate reactions, the substrate gets consumed and so the QSS would result from both the binding-unbinding reactions and conversion reaction. In this case $K_m=\dfrac{k_d+k_{res}}{k_a}$. However in this case the antibody is not consumed to give the response. So QSS would be equivalent to equilibrium approximation.

This would also apply for the Hill kinetics in which response can be described as:

$$R=R_{max}\frac{[Ag]^n}{{K_m}^n+[Ag]^n}$$

where $n$ is an integer which represents co-operativity in binding i.e. one bound antibody enhances the binding of the next (total number of binding sites is the upper bound of $n$).

Note that equilibrium model may not be applicable for drugs that can get consumed in a reaction. Also, not all drug induced reaction would follow this simplified model. There are other models for co-operative binding too.

Therefore, before concluding anything about kinetics from an empirical measure like $EC_{50}$, we need to understand (or at least assume) the underlying model.


Spin interaction term for a two-particle system of fermions with a potential that contains spin-spin interaction

The answer to ' why is the spin-spin interaction more intense for a singlet than triplet'  is due to the fact that any internal interaction cannot change the overall energy  of the system. Put informally its a sort of 'centre of gravity' thing and applies in all cases. The total energy change is $-E_S + gE_T$, where the $g=3$ is the multiplicity of the triplet. The energy change is zero as $E_S = 3E_T$.

The spin-spin interaction has a standard calculation for example, for the hyperfine structure splitting of hydrogen atom ground $1$s $^2\mathrm S_{1/2}$ ground state, which produces the 21 cm line of radio astronomy and the hydrogen maser.  The potential in which the spins exist (a proton and electron in this case) only comes into the calculation in determining the overall magnitude of the splitting and then as $|\psi(0)|^2$ sometimes called the Fermi contact interaction.

(By standard methods I mean the interaction is of the form $H=A\vec I\cdot \vec J$ where the magnetic moment is generated by nuclear spin angular momentum $\vec I$ and $\vec J$ that due to electron spin. The total angular momentum of the atom is $\vec F= \vec I + \vec J$ is constant and the expectation value is the energy $E=A<\vec I\cdot \vec J>= \frac{A}{2}(F(F+1)-J(J+1)-I(I+1))$. In the case of two spin $1/2$ particles $J=1/2$ and $I=1/2$ and the energies are $A/4$ for $F=1$ and $-3A/4$ for $F=0$. The scaling term A contains constants and  $|\psi(0)|^2$ )


Experience with PTFE tubing & glass connectors?

In short: Do not attempt a direct connection of the PFTE hose with the glass.
Instead (assuming the tubing shall transfer gases with little pressure only):

Because to the (compared to silicon) lesser flexibility of the tubings made from PFTE, you should consider to use glassware with threaded joints with detachable connectors. This a theme already seen with condensers, e.g.

(source)
but may be equally applied to impingers

(from here)
or overhead exhausts:

(from here)
These connectors may be replaced with ones made mostly out of PFTE, too, e.g.

(source)
Thus you may connect the tubing to the connector at the bench (gentle warming of the tubing with a heat gun may help), secure this end with the clip, and then mount the connected hose to the reactor.  It is faster and more secure (should you need a bit of force, damages are to the tubing and the connector, but not to the reactor) than the direct mount of the tubing to the glass.

The other side of the PFTE tubing will connect with the metal ferrules to your gas supply, e.g.

(screen photo at 1:04 min from this video)
Depending on the dimensions, the fittings by Swagelok are sold in sets of 10; a typical mechanical workshop of your university may have them already in the shelf. The less pricey Bowden connections used for 3D printers (e.g., here) are designed for compressed air.

As for any equipment working with gas: monitor the gas pressure in your reactor and use a safety release / pressure valve.


I ran into the same problem myself when I bought (foolishly, it turned out) $\ce{H4EDTA}$ instead of $\ce{Na2H2EDTA}$ or $\ce{Na4EDTA}$ for some of the experiments for my doctorate.  It really confused me, because carboxylic acids are generally quite soluble and, with four of the things, you'd think it'd dissolve right away.  But, no.  There are even amines in there, too, but they apparently don't help much either.

Based on the abstract of a 1973 article by Ladd and Povey1, my guess is the apparent poor solubility of $\ce{H4EDTA}$ is actually a kinetic effect, due to a network of intra- and inter-molecular hydrogen bonds present in the solid (emphasis added):

  The [EDTA] molecule exists in a cis conformation and possesses two-fold symmetry, the diad axis passing through the central $\ce{C-C}$ bond. The nitrogen atom is protonated, and takes part in a bifurcated intramolecular hydrogen bond. A short $\mathbf{\left(2.46~Å\right)}$ hydrogen bond links the oxygen atoms of adjacent molecules.

I don't have the necessary thermodynamic data at hand to argue with certainty, but my guess is that the activation energy for solvent penetration into this hydrogen-bond network is rather high, and thus the rate of solvation is slow even though the overall process is (presumably) exergonic $\left(\Delta G < 0\right)$.

Reference: Ladd, M. F. C.; Povey, D. C. Crystallographic and spectroscopic studies on ethylenediaminetetraacetic acid (edta). J. Crystal Molec. Struct. 1973, 3 (1), 15–23. DOI: 10.1007/BF01270899.


Minimum heating temperature for dehydration of Cobalt Chloride

The minimum I could find is 110℃(Chemical forums) although not verified

Next is 150℃ from this link. Although, two temperature are written(110℃ and 150℃).

  mp 86 °C; loses $\ce{6 H2O}$ at 110 °C

  Anhydrous cobalt chloride is blue in color. It can be prepared by
  dehydration of the hexahydrate either at 150–160 °C in vacuo.

Next is 160-170℃. (Chemiday)

  $$\ce{CoCl2•6H2O ->[\Delta] CoCl2 + 6H2O}$$
  The thermal decomposition of hexahydrate cobalt(II) chloride to
  produce cobalt(II) chloride and water. This reaction takes place at a
  temperature of 160-170°C in a flow dry $\ce{HCl}$.

Next is 200℃. From this paper:

  Thermal deaquation of cobalt chloride hexahydrate has been subjected
  to investigation by numerous authors because it is one of the salts
  that shows distinct steps in the loss of water of crystallization in
  the ambient temperature range  of  200℃.


What chemical reaction can cause this blue-ish deposit?

Let's speculate a bit ;)

Colestyramine is a twodimensional polymer network, build from styrene and and 1,4-divinylbenzene.
The polymer has a lot of benzyltrimethylammonium groups. That means it can act as an anion exchanger.
It is used to strongly bind bile acids
Neither the polymer itself, nor the adduct with bile acid are supposed to degrade in the bowel. The loaded polymer is supposed to be excreted.

The dishwasher tablets, on the other hand, contain the sodium salt of a maleic acid/acrylic acid copolymer.

That means we have two polymers, one with a lot of $\ce{Ph-CH2N^+(CH3)3}$ groups and another one with lots of $\ce{-COO-}$ groups.

Sounds like a mass wedding of two chain gangs :D

Edit

It is conceivable that this deposits a thin film on the spoons and the glassware.

In order to explain the bluish colour, we have two options:

The polymer aggregate absorbs in the 580 nm range - we observe the complementary colour. Due to lack of conjugation, I don't see how this could happen here.
The colour results from some dispersion effect related to opalescence, as suggested in Michiel's comment.


While homologous series are a very nice thing for teachers to quiz students about, they practically don’t matter to modern lab chemists. Sure, we all know that methanol, ethanol and propan-1-ol form a series and that a $\ce{CH2}$ unit is added when going from one to the next but in terms of actual lab properties the three are different enough to be remembered individually. As for most synthetic applications, linear molecules (the only ones for which a homologous series makes some sense) are not really any target of relevance so they won’t be thought of.

That said, I see nothing against applying a term that works well for many functional groups (e.g. $\ce{OH}$ or $\ce{NH2}$) to another functional group ($\ce{Cl}$) so the only argument against calling them one is ‘my teacher said so and it’s important that I know that in the upcoming exam’.


Iron nomenclature differences between Iron(III) and Ferric

There is no difference, in both the examples you cite.  The forms Iron(II) and Iron(III) are preferred today, versus the older forms ferric and ferrous, respectively, because they more explicitly give the oxidation states of Fe, +2 and +3 in this example.


The galvanic cells from the same material are possible. They are called concentration cells and they use two electrolytes of the different concentration of the same electrolyte.

But their voltage is low. In the case of $\ce{CuSO4}$

$$U = 0.0295 \log\frac{c_1}{c_2}\ \mathrm V$$

Note that the electrode naming is based on the direction of the electron flow, and their mutual polarity are switched for electrolysis versus cells (primary cells or by convention discharging mode of secondary cells ).

Anode = electrons flow upward (anabasis), an anode is the place of oxidition. (Xenophon, Anabasis, "The way upwards")

Cathode = electrons flow downwards, a cathode is the place of reduction (katabasis - the way downwards)

During electrolysis, the anode is the positive electrode and the cathode is the negative electrode.

During the cell discharging, the anode is the negative electrode and the cathode is the positive electrode.


Predicting the product of bromination of acetanilide

This is a classic $\alpha$-bromination of a carbonyl. You can achieve this with an electrophilic source of bromine like in the reaction conditions provided or something like N-bromosuccinimide.

The reason you're not getting the EAS product is because that's a much higher energy pathway since you need to temporarily break aromaticity. Given more time, heat, and bromine, you can probably get the ortho- and para- products, but you will still $\alpha$-brominate on the amide first.


Why is it necessary to consider infinitesimal changes in p,V,T for H,U and G given that they're state functions?

It is important to make clear any requirement that a variable be constant for each equation you write.

To go from

$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T$$

to

$$\mathrm{d}G = V\,\mathrm{d}p,$$

you would need to assume temperature is constant.

And to go from

$$\mathrm{d}G = V\,\mathrm{d}p$$

to

$$\Delta G = V(p_2 - p_1)$$

you would need to assume volume is constant.

If temperature and volume are both constant, the pressure is constant also.

So your final equation is only saying: when temperature, pressure and volume are constant, there is no change in Gibbs free energy.


What is the difference between galvanization and sacrificial protection?

They are used in different media. Galvanization is used to protect iron exposed to non-corroding medium (air), while sacrificial electrodes are used to protect iron in electroconductive and corrosive medium (sea water or moist soil).

Protective zinc oxide layer on galvanized surface is desired to stay undisolved,  while the purpose of sacrificial electrode is to keep being oxidized to assure electron flow to other metal surface. In latter case medium electrconductivity is a must for ion transfer process, while corrosivness is needed for oxidized zinc surface layer disolution.


Is not it just the case?

Looks like it is indeed not the case. I think the phrase OP quoted from the manual, which says that the optimization stops when

  The Maximum Force and RMS Force are two orders of magnitude smaller
  than the thresholds shown, regardless of the values of the
  displacements.

can be interpreted a bit differently, but I guess that algoritmically it is implemented in a pretty simple way: the above mentioned quantities should be at least 100 times smaller that the corresponding thresholds. And this is not the case for the calculation under consideration: Maximum Force value (0.000005) is not 100 times smaller (but only 90) than the threshold (0.000450).


Acid strengths for HBrO4 vs HIO4, and HBr vs HI

"We know that HI is a stronger acid that HBr because I is a much larger atom"

That is only half the story. It is about the strength of the H-Br versus H-I bond. Since I is larger than Br, the H-I bond is longer, weaker, and therefore more prone to dissociation. See here for more in-depth discussion.

  $\ce{HBrO4}$ and $\ce{HIO4}$

both have the acidic proton bound to oxygen:

The H-O bond is approximately of equal strength in both acids and therefore you turn to electronegativity arguments.


Distinguishing propan-2-ol from ethanol using IR spectroscopy

Here is the spectrum for ethanol:

Here is the spectrum for propan-2-ol:

Here are the tabulated peak positions ($\pu{cm^{-1}}$) and percent transmittance (%T) for the two. I've tried to match the appropriate peaks.

$$
\begin{array}{cccc}%
\hline%
\text{ethanol position} & \text{ethanol %T} & \text{propan-2-ol position} & \text{propan-2-ol %T} \\
\hline%
3358 & 12 & 3346 & 9  \\
     &    & 3334 & 9  \\
2974 & 13 & 2972 & 5  \\
2927 & 38 & 2933 & 19 \\
     &    & 2907 & 23 \\
2887 & 35 & 2884 & 20 \\
     &    & 2722 & 60 \\
     &    & 2669 & 60 \\
     &    & 2521 & 74 \\
     &    & 2408 & 79 \\
     &    & 2387 & 81 \\
     &    & 2198 & 84 \\
     &    & 1903 & 85 \\
     &    & 1766 & 84 \\
1455 & 59 & 1467 & 24 \\
     &    & 1409 & 29 \\
1381 & 54 & 1379 & 16 \\
     &    & 1368 & 17 \\
1330 & 68 & 1341 & 28 \\
1274 & 74 & 1309 & 23 \\
     &    & 1162 & 10 \\
1090 & 22 & 1130 & 9  \\
1050 & 7  & 1110 & 15 \\
     &    & \color{red}{954}  & 4  \\
881  & 37 & 818  & 13 \\
669  & 67 & 660  & 37 \\
     &    & 654  & 37 \\
     &    & 490  & 66 \\
     &    & 436  & 68 \\
\hline%
\end{array}%
$$

One way of differentiating between the two is that I don't believe the peak at 954 wavenumbers in propan-2-ol appears in ethanol. Another better way that doesn't require looking at the fingerprint region is to see that the hydrogen-bonding peaks (the broadest ones to the left) are red-shifted in propan-2-ol compared to ethanol.


Does burning graphite give off carbon monoxide?

It could do - but I doubt you had enough carbon in the rod for any significant risk . Hence no detector signal.
( carbon monoxide is normally formed where oxygen is restricted in the combustion process )


How there are 7 enolizable hydrogens in 2‐methylcyclohex‐2‐en‐1‐one?

Actually there are exactly 7 alpha hydrogen in the given structure.  Just look that there is also a double bond and it is also considered as functional group.  So, you should count alpha hydrogen w.r.t  double bond and they are 5 .

Finally 2 + 5 you get the required answer = 7


How do we get rid of dangerous explosive acetone peroxide and nitroglycerine?

I don't know if it's practical to dissolve or otherwise stabilize acetone peroxide, but I could see absorbing nitroglycerine with sawdust or trapping it in gelatine, rendering it more stable, which is how dynamite is made. Certainly if it can be done safely, detonating in place is be the preferred solution for dealing with unstable explosives. It's done frequently with unexploded munitions if they're deemed too dangerous to move, e.g. WWII bomb in Munich. I also visited a potash mine once. Old dynamite tends to "sweat" nitroglycerine, but the explosives lockers are designed such that the contents can be detonated in place without causing damage, if this happens as it's safer than attempting to remove them.


According to this Wikipedia entry:

"[electronegativity] is a chemical property that describes the
tendency of an atom to attract electrons (or electron density) towards
itself.

then according to an electronegativity scale called the Allen electronegativity:

"...neon has the highest electronegativity of all elements, followed
by fluorine, helium, and oxygen."

By any scale, oxygen is among the most electronegative elements. More to the point, electronegativity decreases as you go down a period, so  $\ce{Te}$ is much less electronegative, much more able to give up its electrons, than oxygen.  Thus, $\ce{H2Te}$ makes a better reducing agent than $\ce{H2O}$.


How to identify an unknown compound with spectroscopic data?

Have a look at typical IR bands (other source) and identify them.

   - 3424 phenolic $\ce{OH}$ stretch
 - 1640 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch
 - 1264 aromatic $\ce{C-O}$ stretch
 -  747 aromatic $\ce{C-H}$ bending

Have a look at you spectrum, identify your residual solvent signal.

    7.15 is chloroform $\ce{CDCl3}$

Have a look at the typical $\ce{{}^{1}H-NMR}$ shifts table and identify the protons.

   These are aromatic protons

What is the summary of your first conclusions?

   It's an aromatic compound with an hydroxyl group.

How many protons?

   multiple of 5

Calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, hence $\nu_0=500~\mathrm{MHz}$.)
Example: Doublet at 7.55 ppm:
$$J=(7.5615-7.5461)\cdot500~\mathrm{MHz}=7.7~\mathrm{Hz}$$
Solutions:

   7.55 doublet, $2\ce{H}$, $J=7.7~\mathrm{Hz}$
 7.40 doublet of doublet (fake triplet), $2\ce{H}$, $J=7.6~\mathrm{Hz}$, $J=7.5~\mathrm{Hz}$
 The last one is tricky because there is one peak not picked
 It should be at 7.30 a doublet of doublet, $1\ce{H}$,  ${}^3J=7.7~\mathrm{Hz}$, $J=7.6~\mathrm{Hz}$

Disclaimer:
Please read long's answer and follow his thoughts. They make a lot of sense and hint that the compound we are looking at is not the one I am proposing. But due to my lack of current knowledge, I cannot make another suggestion.

Combine all you see and you will get a clue which substance it is.

   7.55 ortho
 7.40 meta
 7.30 para
 Phenol.


Nernst equation: half cell vs overall reaction

It is coincidence, since
$$\ce{AgCl -> Ag+ + Cl-}$$
is not a half cell reaction.

The silver chloride electrode functions as a redox electrode and the reaction is between the silver metal (Ag) and its salt — silver chloride (AgCl). There are two ways to represent the processes (half reactions) happening on the electrode:
$$\ce{AgCl + e- -> Ag + Cl-}$$
or
$$\ce{Ag+ + e- -> Ag}$$
$$\ce{AgCl -> Ag+ + Cl-}$$

And you should be applying Nernst equation to one of the two half cell reactions shown above.

The reason why you got the same result/answer is because the Nernst equation for silver chloride electrode has the dependence on the activity or effective concentration of only chloride-ions.

$$E = E^0 - \frac{\mathrm RT}{\mathrm F}\ln(a(\ce{Cl^-}))$$


What is the mole fraction in mixture before combustion?

It's always helpful to show as much work as you can so that we can see where you went off track,  In any case it appears you made an effort, so let's start at the beginning.

First we need to write out the balanced equations
\begin{aligned}
\ce{&C2H6(g) + $3.5$~ O2(g) -> 2 CO2(g) + 3 H2O(g)} \\
\ce{&C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)}
\end{aligned}

Second we're given that

mass ($\ce{CO2}$) = 1.819 mass ($\ce{H2O}$)

we can converting this to a molar ratio

$$\ce{\frac{[mass ~(CO2)]}{[$44.009$]}~=~ $1.819$ * \frac{[mass~ (H2O)]}{[$18.015$]}}$$

and we find that the molar ratio of
$$\ce{\frac{[moles~(CO2)]}{[moles~(H2O)]}~=~ $0.744604$}$$

Now, just to get a feel for things, let's look at the ethane case and we see that we have a mole ratio of $\ce{CO2~/~ H2O}$ of 2/3, whereas in the propane case the ratio is 3/4.  The observed molar ratio of 0.74 indicates that the propane reaction is the major contributor. We can write a general equation
$$\ce{\frac{[2/3 + (r * 3/4)]}{[1+r]}~=~$0.744604$}$$
where "r" is the molar ratio of propane to ethane being burned. Solving for "r" yields
$$\ce{r~=~$13.59$}$$

In other words $$\ce{moles~\frac{[C2H6]}{[C3H8]}~=~ \frac{[1]}{[$13.59$]}}$$


When I place a straw in carbontated drink and blow, more gas bubbles rise to the surface?

When you blow into a solution, your warm breath (~98.6°F) warms the solution up. As a solution warms up, less $\ce{CO2}$ is soluble.  Therefore, blow into a solution, warm it up, less $\ce{CO2}$ is soluble, $\ce{CO2}$ escapes (bubbles out) from the warmer solution.

(image source)


Whatever the nature of the acids, weak or strong, dilution of an acidic solution produces an increase of pH. The dissociation coefficient may increase, but the pH will always increase. The expression "like a Y letter" is pure fantasy.
Your calculations about acetic acid are correct. Just a small remark : in the second part, you say :"if  you double the volume". It should be stated that this operation is made by adding pure water. In this case, the total concentration of acid is divided by 2. Surprisingly, the concentration of $\ce{H+}$ is not divided by $2$ : $\ce{[H^+]}$ = $2.99 ·10^{-3} mol L^{-1}$. So it has been divided by $√2$ (and not by $2$).
Now, let's speak about the number of moles. During the dilution, the total number of moles of acetic acid has not changed, but the total number of moles of $\ce{H^+}$ has slightly increased during the dilution, which means that the acid is more dissociated in the diluted solution.


Is it possible to "dope" glass to change it's refractive index?

Glass doesn't crystallize at all, that's the point. Consequently, unlike crystalline materials, it doesn't have fixed composition. In a way, it can be said that glass consists of impurities. And yes, of course, you may change its composition this way or that, so as to tune the refractive index to your purposes.

Further reading: crown glass, flint glass.


Why is the melting point of magnesium nitride less than that of beryllium nitride?

Melting point in a compound with a lattice structure correlates more closely with lattice energy than with ionic character, and lattice energy tends to be greater when you use smaller atoms.  Among alkaline earth nitrides it is not just beryllium versus magnesium that shows this effect; it keeps going.  Melting point data below are from the Wikipedia articles as of April 30, 2021:
$\ce{Be3N2} = 2200°C$
$\ce{Mg3N2} = 1500°C$
$\ce{Ca3N2} = 1195°C$
$\ce{Sr3N2} = 1200°C$
The alkaline earth oxides show a similar trend, for the same reason.  Here the lattice energy of beryllium oxide is slightly depressed because each atom has a coordination number of only four whereas magnesia and its heavier congeners have coordination number six; thus the peak melting point is at magnesia:
$\ce{BeO} = 2507°C$ -- four-coordinate
$\ce{MgO} = 2852°C$ -- six-coordinate, same for the following oxides
$\ce{CaO} = 2613°C$
$\ce{SrO} = 2531°C$
$\ce{BaO} = 1923°C$


You are on the right track.

The Brockhaus Konversationslexikon (Brockhaus Encyclopedia, 14th edition, 1894-1894)

explains the German term Kupferwasser (copper water) as

  veraltete Bezeichnung für unreinen (kupferhaltigen) Eisenvitriol

which means outdated name for impure (copper-containing) iron vitriol.

Note that iron vitriol is an old term for iron(II) sulfate,  such as vitriolic oil is an outdated name for sulfuric acid.

If you take a look at English wikipedia page on iron(II)sulfate, you'll see that one of the common minerals is a heptahydrate, melanterite. If you follow the link, you will recognize that one of the associated minerals is chalcanthite, which is $\ce{CuSO4·5H2O}$.

To cut the long story short, the term copperas stems from the fact that natural iron(II)sulfate was often found to be associated and maybe contaminated with copper(II)sulfate.


Is it absolutely necessary to use centrifuge to separate ethylene glycol from precipitate?

What they do in the paper is what i was going to suggest even before reading it: Addition of ethanol in the ethylene glycol suspension lowers the viscosity of the solution making it more filterable. Since their nanoparticles are insoluble in ethanol they can add as much is necessary. After filtering the solid out they wash with plenty of ethanol to remove the leftover ethylene glycol (which since it has a high boiling point cant be dried easily). Hence they end up with their product contaminated with ethanol left over from the washings. However, ethanol is very easy to get rid off by applying vacuum to the powder due to being reasonably volatile. So the ethanol contaminated solid can be dried in a schlenk line or even a rotary evaporator overnight to remove any ethanol left. Note that you can do this not only with ethanol but any solvet that is volatile (b.p up to 110 deg C, you can do more but is a pain), miscible with the solvent you are trying to get rid of, and wont react with the compound that is being isolated. Therefore the answer to your question is: no, in general it is not necessary. However, if your precipitate is too fine for the filter you have available, and you cant get it to crash out as larger particles or if it is a very small amount and you cant scale it up you might actually need to use one.