Thermodynamics: ΔU = Q + W or U = Q? Heat is the total kinetic energy of all atoms of the system. When work is done on the system it means that a part of system kinetic energy is used to do the work, and this work makes the surrounding warmer.
So "$\Delta U$" of the system is equal to "$Q$".
And now, why we use the work of the system in: $\Delta U = Q + W$?

The questions is a good one, and answering it winds up being central to a lot of key concepts in thermodynamics. A brief correction before starting, though I don't think it's central to the question,  "When work is done on by the system it means that a part of system kinetic energy is used to do the work..."

The disconnect between your understanding and that used in thermodynamics is that in thermodynamics, the system and its surroundings are separate. So, when our system does work on the surroundings that energy won't come back to heat our system. If we do as you say and use our system to do work on the surroundings, turn that work into heat, and then apply that heat back into our system, the result is just as you say! The kinetic energy in our system isn't lost. Or put in another way when we're allowing heat to flow in and out of our system, all of that heat ultimately winds up as kinetic energy and $ΔU = Q.$

Let's look at a couple examples of putting $\pu{1 J}$ of heat into a system with different surroundings to see how this works out. Our system will be an insulated piston filled with an ideal gas (say helium), which can expand and push on one of 3 things in its surroundings: a mass, a vacuum, or a generator.

Mass: This follows $ΔU = Q - W$.  Of the $\pu{1 J}$ of heat we apply, the portion that goes into warming up our system and increasing internal kinetic energy is $C_v$ (the molar heat capacity is $1.5R),$ whereas the portion of that instead goes into doing work on the surroundings is $R$ (the universal gas constant). So our system gains

$$[\pu{1 J}]\cdot \frac{C_v}{C_v + R} = \pu{\frac{3}{5} J}$$

of thermal energy, and does

$$[\pu{1 J}]\cdot \frac{R}{C_v + R} = \pu{\frac{2}{5} J}$$

of work on the surroundings by expanding and pushing up the mass. This $\pu{\frac{2}{5} J}$ of work energy won't return. These portions $C_v/(C_v + R)$ and $R/(C_v + R)$ show up a lot later on in thermodynamics as their inverses, $k$ and $k/(k - 1)$, respectively.
Vacuum: This follows $ΔU = Q$. There is no surroundings to do work on, so all of the expansion energy goes into slamming the piston against its backstop, and after everything settles this ultimately results in the full $\pu{1 J}$ of heat becoming thermal energy inside our piston.
Generator: This also follows $ΔU = Q$. It's the same situation as the mass, with $\pu{\frac{3}{5} J}$ of thermal energy and $\pu{\frac{2}{5} J}$ of work done on the surroundings, but that $\pu{\frac{2}{5} J}$ of work can be turned into electricity and then be used to heat our piston.

Just like in the vacuum case, all $\pu{1 J}$ of heating ultimately winds up as thermal energy in the system, though it spent some time as work on the surroundings during the process. So, $ΔU = Q - W + W.$ I think this final case is the closest to matching your question.


How do I know which atoms react during hydrolysis? In the hydrolysis of propanamide ($\ce{CH3CH2CONH2}$) how do I know which molecules react with the water?

$$
\ce{CH3CH2CONH2 + H2O -> ?}
$$

I don't know how to split the molecule. In the videos I've watched (example) it seems as they just assume you know this.

I think you mean which atoms react with the water, as clearly the molecule that reacts is propanamide (it's the only non-water molecule in your question).

The product of the reaction is propanoic acid and ammonia.

$$\ce{CH3CH2CONH2 + H2O -> CH3CH2COOH + NH3}$$

The carbonyl carbon in amides (and esters and acyl halides) has a partial positive charge because of the electron withdrawing effect of the carbonyl oxygen.  (Oxygen is very electronegative.)  This makes the carbon atom is activated toward nucleophilic attack.

This Wikipedia page has a good explanation and different mechanisms for acid vs. base conditions.


Animate molecular vibrations as multiple xyz coordinate files I have a frequency calculation by Gaussian09. Now I would like to extract vibrational modes as a set of xyz coordinate files. I would like to manually recompute numerically the points along a transitional vector, since IRC calculations all fail. (I suspect it is no real saddle point, flat PES.)

Consider the formation of the Lewis-Acid-Base complex $\ce{H3N\bond{-}BF3}$ with a transition state given in $$\ce{NH3 + BF3 -> [H3N\cdots BF3]^{TS} -> H3N\bond{-}BF3}$$ where this procedure would equal a (rigid) scan of the $\ce{N\cdots B}$ bond. This would be easy to accomplish.

Things get more difficult considering the pseudo vibration of ammonia, pyramidal to planar to pyramidal. Here it is not only one variable that has to be changed. (With dummy atoms it is still manageable.)

If molecules get more complicated, the manual specification of the bond parameters is too complicated and a source of many errors.

In the days when some of the gui programs were not able to calculate the vibrations and showing them in that fashion it was common to produce a set of multiple xyz files, each of them a snapshot of the coordinates along a specified vibrational vector. (This is now done internally and you can look at fancy pictures.) One of those programs capable of doing so is XVibs. For various quantum chemistry packages they calculated the distorted geometries. Unfortunately that is now ancient. (It was possible with a Gaussian09 output.)

For Turbomole there is a tool distributed along called screwer.

Unfortunately I have no access to that program package and I also do not want to change all the computations so far done to this software.

Lordstryker already posted in the comments, that it might be done by hand with chemcraft. I tried that and it is not very satisfying. One produces too many errors.

Are there any other programs/scripts that might be able to produce these files?

If I remember correctly, frequency data are in the stdout clusterfuck usually named log file, as well as in the (formatted) checkpoint files. There shouldn't be any need to prepare a cube file for this.

If GaussView is not available, GM's comment has the answer: Avogadro can visualize vibrations and I'm sure I've done this. Unfortunately, I don't have an output file at hand here.


Why does this compound only have one pair of enantiomers? I've been doing some reading about enantiomers, which are apparently chiral molecules that are non-superimposable mirror images of each other. An example I found of this in some problems that I've been attempting seems to have stumped me somewhat:

Take the compound $\ce{CH2=CH-CHD2}$, where $\ce{D}$ is deuterium, or hydrogen-2, and react it with deuterium gas. A pair of enantiomers is formed, both with the condensed formula $\ce{DCH2-CHD-CHD2}$. The answers provided say that these enantiomers are as follows:

...where $\ce{A}$ is standing in for deuterium.

Why don't you get enantiomers where the orientation of the $\ce{C-D}$ bonds at both ends of the molecule varies? (ie. where those $\ce{C-D}$ bonds are also in varying planes, represented by different combinations of hashed and solid wedges on paper.) Here's an example of what I mean:

Why aren't these two structures considered to be another pair of enantiomers also?

Is this - the orientation of bonds at the end of carbon chains counting effectively for nothing - the case with all stereoisomerism in alkanes?

Usually it doesn't as rotation around single bonds is generally very quick in room temperature. However, sometimes it matters because rotation can be blocked by bulky substituents, or other parts of the molecule. Even double bonds usually connected with E/Z isomerism, can in certain situations cause optical activity.

These are all examples of so called axial chirality. In the first case (rotation blocked by substituents) it's called atropoisomerism. Second (blocking by other parts of molecule) is present for example in helicenes. Third case (cumulated double bonds) occurs in allens.


How to name the tautomers of hexane-2,4-dione correctly according to IUPAC rules? Today I was considering to answer the question "Which tautomer of hexane-2,4-dione is more stable?" I wanted to run a few calculations and while preparing a directory structure I wanted to make sure that I used correct and unique names for the compounds. I am citing the picture in the named question:

I was going to derive the names of the structures from the parent compound hexane-2,4-dione. Hence I would have named compound 1 (Z)-2-hydroxyhex-2-en-4-one and 2 (Z)-4-hydroxyhex-3en-2-one.

When I recreated the above picture in ChemBioDraw and used it to autogenerate the name, the result for 1 was different: (Z)-5-hydroxyhex-4-en-3-one. See below.

I am no expert on nomenclature and I do not have the blue book handy. Usually I rely on the names generated through software. Most of the times they are accurate enough to name a compound unambiguously and that is often what I am going for.
I understood from this answer, P-14.3.5 Lowest set of locants, that I should use the 4,2,2 set over the 5,3,2.

Is my thinking correct and the auto-generated names are wrong, or is the software correct and my thinking is wrong?

You numbered all locants together in a series and you found the lowest set of locants according to Subsection P-14.3.5 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) since the locant set ‘2,2,4’ is lower than ‘3,4,5’.

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

This approach would be correct if all locants refer to detachable prefixes (like ‘2-hydroxy’). However, in the given compounds, the locants refer to prefixes (‘hydroxy’), endings (‘ene’), and suffixes (‘one’). The assignment of low locants to such different features is stipulated in Subsection P-14.4 as follows:

  P-14.4 NUMBERING

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

  (…)

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

  (…)

  (e) saturation/unsaturation:

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

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

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

  (…)

Therefore, the resulting name is (4Z)-5-hydroxyhex-4-en-3-one (not (2Z)-2-hydroxyhex-2-en-4-one) since the locant ‘3-one’ for the principal characteristic groups is lower than ‘4-one’.


Gases and physical chemistry PV=nRT. Hence, T=PV/nR.Let's consider a gas containing in a
bulb in a way that  its pressure,volume and number of moles of gas can be changed.So if I want to attain a certain temperature for n moles of a gas in the bulb,I can adjust the PV value in the equation so as to fit to the value of T.Theoretically this can be done with above equation.But is this experimentally possible?

I will try to summarize what Karl is saying about the ideal gases.
Suppose that you measure the temperature of a gas contained in a canister. Suppose that you measure the temperature $T_i$

Using your equation that you derived from $PV=nRT$, we should have $$T_i=\frac{P_iV_i}{nR}$$
Suppose that we then apply work to the canister. Since it is an ideal gas the only way that we can change its energy is through a change in volume ,pressure, or number of moles. So that is what we are going to do. Suppose that we compress the canister so that the pressure of the gas changes. Thus, our new canister will have a new pressure $P_f$ and a new volume $V_f$.
This volume $V_f$, however, is related to $V_i$ through the ideal gas equation $$P_iV_i=P_fV_f$$
Hence we have that $$V_f=\frac{P_iV_i}{P_f}$$
Lets substitute this equation to find the new temperature of our compressed canister:

$$T_{compressed}=\frac{P_f\frac{P_iV_i}{P_f}}{nR}$$
Which gives us the initial equation. Hence the temperature of the ideal doesn't change when it is compressed.

I will post on what happens when you increase the number of moles.


Hydrogenation at atmospheric pressure in amateur environment I was planning on doing some small scale hydrogenation practicals at atmospheric pressure (probably from Vogel) and ran into a couple of questions. I will be using the "balloon method" (described in Procter's Advanced Practical Organic Chemistry book) as I consider it an easier technique for small scale reactions than constructing a Brown hydrogenation apparatus. Note that I'm working in an amateur environment (i.e. my home lab which does not have a fumehood). Now for the questions:

What rule of thumb can be used to determine the amount of catalyst (in these cases $\ce{Pd/C}$) needed per mole of hydrogen used (I suppose these two variables are depending on each other)?
There are different types of $\ce{Pd/C}$ out there: 3%, 5% or 10% $\ce{Pd}$ by weight, with or without a variable amount of added $\ce{Pb}$, different solvents (mostly I see is water) and also there are different bases available, not just palladium on carbon. I understand that $\ce{Pb}$ is added to poison the catalyst such as with Lindlar catalyst, but if total reduction is required then why would this be necessary? Also how would I decide between 3%, 5% or 10% $\ce{Pd}$ by weight?
What is an effective method for generating the hydrogen to fill the balloon with? I suppose that normally a pressurized hydrogen tank would be used for this, but I find this overkill for the reaction I'm planning. Furthermore, the use of  pressurized hydrogen tanks is not quite without danger and I'm not trained to handle them safely.
Why would it be required to flush the reaction mixture with an inert gas after the hydrogenation has completed? I understand that this is necessary because the dissolved hydrogen gas can ignite and possibly explode when, while filtering the catalyst, the catalyst spontaneously ignites (e.g. when the catalyst is not suspended enough and direct contact with the atmosphere is possible). However, why is an inert gas advised? What is wrong with ordinary air?
What is the function of Celite when filtering the catalyst?
How many times can the $\ce{Pd/C}$ catalyst be reused? How would you spot whether or not your catalyst has become too poisoned to be successfully used in hydrogenation?

I've adddressed a part of your question here.

As far as the catalyst is concerned, 3-5% Pd/C is fine.

  What is an effective method for generating the hydrogen to fill the balloon with? I suppose that normally a pressurized hydrogen tank would be used for this, but I find this overkill for the reaction I'm planning. Furthermore, the use of pressurized hydrogen tanks is not quite without danger and I'm not trained to handle them safely.

In the lab, I only used pressurized tanks. While commercial hydrogen generators are available - they are mostly used to generate hydrogen gas as a carrier gas for GC/MS - these things cost a fortune and really would be overkill.

Actually, pressurized tanks are pretty safe if you follow some rules:

Obviously, you can't attach the pressure regulator and operate the whole thing unless the steel cap over the main valve is removed. Don't start anything unless the tank is secured to the lab bench or the wall with a clamp, a chain or another sturdy contraption that prevents the tank from falling over! You want hydrogenation, not a rocket ;)
Get familiar with the pressure regulator.
When you're done, remove the regulator again and put the cap over the main valve. Keep the regulator at a dry and clean place.
No grease on the threads!

When operated with a bit of common sense, a small pressurized tank is the most convenient option.

  What is the function of Celite when filtering the catalyst?

When you filter off the catalyst, you'll typically use a suction filter with a base of sintered glass (fritted glass). Palladium on charcoal is a very fine powder that will clog the fritted glass. A pad of Celite in the suction will prevent this. Celite is cheap, suction filters are not!


Meaning of charges in a group? In my class, my professor said that since hydrogen is in group one, it has a charge of 1+. Does this refer to its ionic charge or to its oxidation state?

To go from being confused to less confused, here are some facts.

Hydrogen has one valence electron which can participate in reactions so it is a group 1 element.
Hydrogen is not a group 1 alkali metal, nor is it a group 1 compound, so simply stating hydrogen is in group 1  is ambiguous and confusing.  It is a group 1 element.  The word element needs to be included for the statement to make any sense.
Elemental hydrogen has no charge at standard conditions because it exists as dihydrogen gas, $\ce{H2}$.  If hydrogen gas were to be ionized it could hypothetically break into two ions, one with a +1 charge and one with a -1 charge.  If no ions are involved then the concept of ionic charge is irrelevant.
Elemental hydrogen, again as dihydrogen gas, $\ce{H2}$, has a zero oxidation state.  The oxidation state refers to a hypothetical charge of an element within a compound. If no compound is involved then oxidation states are irrelevant.
Elemental hydrogen has an oxidation state of +1 in most compounds and -1 in metal hydrides because of an agreement within the International Union of Pure and Applied Chemistry (IUPAC). -- http://goldbook.iupac.org/O04365.html --  Oxidation states, as many of the concepts in basic chemistry, should be viewed as a tool to gain a rudimentary understanding of how compounds form.


Why is thiourea a monodentate ligand? I was going through a list of monodentate ligands, and I came across examples like $\ce{H2O}$ (which has 2 lone pairs), $\ce{Br-}$, $\ce{I-}$ (both of them have more than one lone pair). Similarly, thiourea has 4 lone pairs in all (2 from nitrogen and sulphur each).

When I searched about it, I realized that the oxygen in water can't simultaneously form bonds using both the lone pairs (due to some obvious geometrical reasons, it said).

However, I couldn't figure out any such (obvious) reasons in thiourea.

Can anyone please help?

This is not really a standalone answer, rather a pointer or an exeption. There are few complexes where deprotonated thiourea $\ce{S=C(NH2)NH}^-$ acts as a bidentate ligand. One example is (aminomethanethioamido)-hydrido-tetrakis(trimethylphosphine)-technetium(III) $\ce{[Tc(P(CH3)3)4(SCN2H3)]+}$ complex[1]:

The fact that thiourea ligand is deprotonated and is coordinated asymmetrically within equatorial plane in $\eta^2-\ce{N,S}$ bonding mode has been proven by NMR and SCXRD studies. From the article [1, p. 4642] (emphasis mine; $\ce{tu}$ = thiourea $\ce{S=C(NH2)2}$):

The $\ce{Tc-N}$ distance, 2.190(13) Å, is long when compared to
known $\ce{Tc(III)}$ complexes. The range for $\ce{Tc-N}$ bonds was found
to be 2.041 Å ... to 2.124 Å ... . This reflects delocalization of negative charge within the bidentate deprotonated thiourea ligand. This elongation is also seen in the $\ce{Tc-S}$ bond, 2.543(4) Å, which is significantly longer than $\ce{Tc-S}$ bonds in the parent compound $\ce{[Tc(\textit{S}-tu)6][Cl]3}$ ... .

As Martin pointed out, the reasons for the observed denticity are not always transparent. Simply put, it seems like in this case in order to be a bidentate ligand, thiourea molecule needs to increase its nucleophilicity and remove steric hindrance caused by $\ce{H}$-atoms lying within the $\sigma_\mathrm{h}$ plane.   Both can be achieved by eliminating a single proton from the thiourea molecule.
Reference

Watson, P. L.; Albanese, J. A.; Calabrese, J. C.; Ovenall, D. W.; Smith, R. G. Inorg. Chem. 1991, 30 (24), 4638–4643. DOI: 10.1021/ic00024a035.


Is isobutyl chloride or ethyl chloride more reactive in an SN1 reaction? Compare the rate of $\mathrm{S_N1}$ reaction in the following compounds:

We see that both compounds form primary carbocations. However, there is more hyperconjugation in the second compound as compared to the first one.
Also as the step in which the leaving group leaves is rate determining step (slowest step), which is before the rearrangement of the carbocation, I concluded that the second compound must have a greater rate of $\mathrm{S_N1}$ as compared to the first one.
But the answer says that the first has a greater reactivity than the second. Is the answer correct? If yes, then why?

The inductive effect of the methyl groups should help stabilize the carbocation formed from 1.
The answer may be a bit clearer if you add equilibrium to your thinking.

The initial dissociation of the 1 to form carbocation A is the rate determining step.  However the rearrangement of A to B should be relatively quick, and will drive the initial dissociation forward by Le Châtelier's Principle.

In contrast, 2 has less stabilization by the inductive effect, and there is no possibility of a rearrangement.


Change of potential difference during electrolysis while adding salt I was conducting a simple experiment at home on electrolysis. Obviously, when you build a circuit with a 9V battery, a lamp, and put two electrodes in water you measure a potential difference of ~9V on the two electrodes while the potential difference on the lamp is 0 (really close to 0). Then as I kept adding salt in water, I first measured that the potential difference across the lamp got higher and higher (reaching 6V), then suddenly, as I got from 40g /250ml of salt in water to 50g /250ml, the potential difference across the lamp fell to 4V.

Why is that? Maybe is it because of the ion association occurring in water as the concentration of salt gets higher?

4.5 V lamp. 8.6 e.m.f battery.

Tungsten lamps change resistance as they warm up. At room temperature, the resistance of a small flashlight lamp (e.g. #222) might be 10 ohms when brightly lit, but only 1.5 ohms at room temperature.

At first, the tungsten filament acts like a short circuit, so most of the voltage of the battery is dropped across the series resistance of the electrolytic cell. Once enough current goes through the lamp, though, its resistance abruptly rises and more of the voltage is dropped across the filament.

If you use an LED instead of a tungsten-filament bulb, the change in voltage would be more linear as you add salt. Note that an LED will be effectively an open circuit until a minimum voltage is reached (~1.6 V for a red LED, ~3 V for blue or white LED), but then current increases linearly until ~20-50 mA, at which point some may burn out.


How do you track Iodine on a black surface? While spraying with starch solution seems to be standard procedure, blue-black colouring seems not to stand out from a black (charcoal) surface.

I would spray some lead nitrate solution to the surface if I didn't care about it at all. It would form nice yellow crystals of lead iodide that could be seen on a black surface. Never tried it though.


Why dont we use potassium hypochlorite for bleach I have been reading wikipedia and according to the article on potassium hypochlorite it can be used as a disinfectant. But it is not listed anywhere in the bleach article. I know we typically use sodium or calcium hypochlorite and some others but is there a specific reason we dont use potassium?

Potassium hypochlorite can be used as a bleach as an alternative to sodium or calcium hypochlorite. It is not that it is not used but its uses as bleach is not common as compared to sodium or calcium hypochlorite. Enviro Klor™ is a potassium based bleach (12.5% Potassium Hypochlorite Bleach alternative). As written here:

  Enviro Klor™ potassium hypochlorite is a very versatile EPA registered
  (#63838-10, #63838-CA-01) biocide with all the power of a hypochlorite
  bleach and is approved for use wherever typical sodium hypochlorite
  bleach is, including:

  Recirculating cooling water systems
  Sewage and wastewater treatment
  Disinfection of drinking water
  Public water systems
  Pulp and paper mill process water systems
  Swimming pool water disinfection
  Sanitization of food contact services
  Laundry sanitizers
  Agricultural applications

  ADVANTAGES

  Bleaching and Laundry

  Allows claims for a more environmentally friendly household bleach
  Provides a sodium-free solution to laundry discharges

  Industrial & Institutional (I&I)

  Chemical Compounding – allows creation of sodium free formulas
  Replacement chemical in CIP cleaners
  More stable in certain formulations

  Agriculture

  Useful as a disinfectant for wastewater applied to land for    irrigation, conserving water
  Since potassium is one of the three required soil nutrients, there is    little objection to the introduction of potassium ions onto
  land.

Here's a table of comparison:

You can see that available $\ce{Cl2}$ (wt %) is more in case of sodium hypochlorite for which sodium hypochlorite is preferred as bleaching agent. Also, byproduct of potassium hypochlorite is potassium chloride for which potassium hypochlorite is used in agricultural purposes where soil needs potassium. Another thing is that potassium hypochlorite may disproportionates to potassium chlorate which is a strong oxidant and may react vigourously with water and ignite or explode.


How to convert -OH to terminal -SH group in nucleolipids: I am recently working on some synthesis of Nucleolipids and I want to transform $\ce{-OH}$ to terminal $\ce{S}$ group.

Can anyone help?

An alcohol can be converted to a corresponding thiol using Lawesson's reagent (LR; 2,4-bis(4-methoxyphenyl)-1,3,2,4-dithiadiphosphetane 2,4-disulfide). However, the reaction is messy and could give you a mixture of products in addition to intended thiol, including eliminated product, alkene (Ref.1). For example, treatment of acyclic 1,4-diols with LR gives the 1,3-dienes (Ref.1).

However, the substitution versus elimination outcome with LR is depends on the structural features of the substrate (Ref.2), and hence, worth considering. The computational mechanism is depicted below:

Alternatively, you can use highly efficient Mitsunabu type reaction with diisopropyl azodicarboxylate (Ref.3). The abstract summerize the reaction as:

  Various alcohols were converted to their corresponding thiolacetates by treatment with triphenylphosphine and diisopropyl azodicarboxylate in the presence of thiolacetic acid. The overall conversion was both highly efficient (89–99% yields) and stereoselective (99.5% inversion).

References:

T. Nishio, “Direct conversion of alcohols into thiols,” J. Chem. Soc., Perkin Trans. 1 1993, (10), 1113–1117 (DOI: 10.1039/P19930001113).
M. A. Chiacchio,  L. Legnani,  P. Caramella,  T. Tejero,  P. Merino, “Pivotal Neighboring‐Group Participation in Substitution versus Elimination Reactions – Computational Evidence for Ion Pairs in the Thionation of Alcohols with Lawesson's Reagent,” Eur. J. Org. Chem. 2017, (14), 1952–1960 (https://doi.org/10.1002/ejoc.201700127).
R. P. Volante, “A new, highly efficient method for the conversion of alcohols to thiolesters and thiols,” Tetrahedron Letters 1981, 22(33), 3119–3122 (https://doi.org/10.1016/S0040-4039(01)81842-6).


Does propylene glycol dissolve plastic? According to Médecins Sans Frontières propylene glycol can dissolve plastic:

  As propylene glycol can dissolve plastic... (Source).

Is this claim true? This seems rather spurious to me. If it matters, for a specific "plastic" I am particularly interested in evaluating the claim in regards to polypropylene and polyethylene as both are materials common for making syringes.

As propylene glycol can dissolve plastic...

This claim is absurd. Propylene glycol cannot dissolve poly(olefin) polymers especially polypropylene or polyethylene. It is possible for glycols to cause paneling of containers made to such, but not dissolve; and certainly wouldn't happen in the amount of time it takes to perform an injection.


How do I measure out these salts using the proper method? I'm planning on preparing a water solution using the following salts of reagent grade with approximately these concentrations:
\begin{array}{lr}
  \ce{CaCl2}  & \pu{10 mg/L}\\
  \ce{Na2SO4} & \pu{1 mg/L}\\
  \ce{NaHCO3} & \pu{20 mg/L}\\
  \ce{KHCO3}  & \pu{5 mg/L}\\
  \ce{NaF}    & \pu{0.25 mg/L}\\
  \ce{NaNO3}  & \pu{1 mg/L}\\
\end{array}
I suspect some of these salts are hygroscopic. Do any of them need to be oven dried prior to weighing? What is an appropriate procedure for this (i.e. what temperature, for how long, etc.). Are there any other requirements in order to correctly produce a solution of the above composition?
It's for an experiment where I'm needing to replicate a natural water source so I'm needing to be as accurate as possible.

You are dealing with very low concentrations and for those you will need very accurate equipment and a very clean lab. I am not sure that reagent purity is enough. In some parts of the world ordinary tap water might already have higher concentrations.
Unfortunately you have not given enough context for what you are doing, so I will outline some procedures which should guarantee quite good accuracy.

First thing you will need is bidistilled water probably lots of it.

For very accurate (and reproducible) results I would recommend purchasing stock solutions and the doing serial dilutions.

If the above is no option you will have to do some more research by yourself, because it depends on the quality and variation of the chemicals you buy. (And preparing a list of all that for your list of chemicals is beyond the scope of this summary.)
For example calcium chloride comes on merck in at least four variations: $\ce{CaCl2}$, $\ce{CaCl2.xH2O}$, $\ce{CaCl2.2H2O}$, $\ce{CaCl2.6H2O}$, and solutions. Depending on your starting material you will need to look up the decomposition temperatures. For calcium chloride you may find them on Wikipedia.

You first need to dry your container with which you will measure the mass of your reagent. Make sure you use gloves as fats and skin can change the result at these low concentrations. Measure it multiple times until you have a constant value (within the limitations of your scale). You need to dry the container at the same temperature which is appropriate for drying your chemical.
Then you follow the same procedure with the chemical. Obviously you will not hit the concentration exactly, but in science that hardly matters, as long as you know the correct value. Once you started the drying process you should keep the container closed until you are ready to transfer it into the solution.
You shouldn't store your prepared stock solutions for a long time and at below room temperature in appropriate flasks.

The other way would be checking the concentration against a primary standard. For calcium chloride you can probably use silver nitrate, which you have to dry like explained above. (It doesn't have the nasty crystal water, which is a plus as $\pu{50 - 90 ^\circ C}$ should be sufficient.)

For setting up the solutions you will need a good assortment of volumetric flasks, pipettes, micro pipettes, thermometers, and probably a bit more I just can't think of right now.

You might want to check with this open access publisher what is common in the field and adjust appropriately. It very much depends on your goal; approximately can cover a pretty large range of desired accuracy.


Regioselectivity in Carbometallation I am a student of organic chemistry and I frequently watch online open course lectures by various professors around the globe. Of these, in https://www.youtube.com/watch?v=OYAMls5x4EI, at about 27:10, I had a question, as mentioned in the image below.

Mechanism

What factors determine where the palladium will attach?

My initial thoughts were to find out the electron-rich regions of the $\pi$ bond, and the palladium atom in the $\mathrm{+II}$ oxidation state will act as an electrophile. But I wasn't able to figure it out exactly.

I understand that the ratio of products shown in the figure can be controlled by using various ligands, especially based on their steric effects, in this case.

Your mechanism is missing an important step: between the oxidative addition of palladium into the $\ce{C-Br}$ bond and the alkene insertion there should be an alkene annealing forming a palladium-alkene complex. This is displayed in the mechanism scheme of a generic Heck reaction below, which I sheepishly borrowed from Wikipedia.

Scheme 1: Catalytic cycle of a Heck reaction. Image taken from Wikipedia, where a full list of authors is available. Note that the β hydrode elimination in step C should be syn and not anti or gauche as displayed.

The annealing step and the π complex are important because they influence the Heck reaction’s selectivity. Due to the avoidance of steric stress, both residues will adopt an anti-type conformation as depicted in structure 3. On the (syn-selective) insertion of the alkene into the $\ce{C-Pd}$ bond, this geometry is kept resulting in the palladium catalyst ending up on the higher-substituted alkene carbon. For the inverse insertion, both organic residues — in your case two phenyl groups — would have to occupy the limited space on one side of the catalyst.

The subsequent β hydride elimination can theoreticly occur in both directions in your molecule. However, eliminating towards the hydroxy group allows keto-enol tautomerism with further stabilisation. Since β hydride elimination is a reversible process, the reaction will select the ketone over the allylic alcohol that would be formed upon opposite β hydride elimination.

The alkene insertion into the $\ce{Pd-C}$ bond, is not reversible. Two $\ce{C-C}$ single bonds are formed at the expense of a palladium-π interaction resulting in a notably downhill process.


Why are negatively charged ligands behind the neutral ligands in spectrochemical series? I had doubt that shouldn't negatively charged ligands be better electron pair donors than neutral ligands like $\ce{CO}$ or $\ce{PPh3}$. But its not so. Why?

I think you are confused by what the spectrochemical series actually shows. It is not a rating of stronger or less strong ligands to mean they can bind more or less strongly to the metal. It is only a rating of which ligands induce a high crystal field split in the resulting complex.

The splitting derives from the metal $\mathrm{e_g}$ orbitals — $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ — actually being antibonding to the metal-ligand interaction and thus higher in energy than the (unmodified in a complex with only σ type interactions) $\mathrm{t_{2g}}$ orbitals.

For systems more complex, i.e. including π interactions, the $\mathrm{t_{2g}}$ orbitals (the other three d-orbitals) interact in a π manner with ligand p-orbitals. If the ligand has populated p-orbitals such as halides have, the $\mathrm{t_{2g}}$ energy is raised as it is turned into a slightly antibonding interaction (the bonding one being on the ligands’ side). Since this brings $\mathrm{t_{2g}}$ closer to $\mathrm{e_g}$, the field split is reduced and the ligands are termed weak-field ligands. In simplified Lewis schemes, one could represent this by a double forwards bond from the ligands to the metal.

In the case of empty p-type orbitals such as in $\ce{CO}$, metal to ligand back-bonding can occur by populated metal orbitals interacting with unfilled ligand orbitals; now it is the metal’s side of things that is stabilised. Since this π backbonding stabilises the $\mathrm{t_{2g}}$ orbitals, their energy is lowered and the field split becomes greater. Hence they are termed strong-field ligands. In simplified Lewis structures one could represent that by drawing a donor-acceptor arrow from the ligand to the metal (σ forwards bond) and one the other way (π back bond).

This all has nothing to do with the overall strength of metal-ligand interaction, as I mentioned earlier. In fact, strong complexes are typically obtained for certain metal-ligand combinations and the same ligand can be very bad in interacting with other metals. Classic example: $\ce{CO}$ ligands are extremely good at stabilising low metal oxidation states such as in $\ce{[Fe(CO)4]^2-}$, while they are rather lousy in stabilising high ones: $\ce{[Fe(CO)6]^2+}$ can only be synthesised in the extreme absence of all other ligands, i.e. in $\ce{SbF5}$ from $\ce{FeF2}$ where the fluorides will be drawn towards antimony to generate the even more stable $\ce{[SbF6]-}$ complex.


Is Bond Formation "Strictly" Exothermic? Is bond formation "strictly" exothermic? The IUPAC definition of exothermic doesn't make any reference to bond formation. However, I have seen the aforementioned statement before - that bond formation is "strictly" exothermic.

I suspect that the answer is "it depends" and that "it depends" at least on how accurately bonds are characterized. For example, the first reaction below is endothermic, while the second one is exothermic:

$\ce{2NO -> ONNO}$.

$\ce{2NO_2 -> O_2NNO_2}$.

Lewis structure analysis for the first reaction at best misleads because one might suspect the nitric oxide molecule as having a bond order of 2.

However, the lone electron is actually delocalized. So the actual $\ce{N-O}$ bond order should be 2.5.

In the $\ce{ONNO}$ molecule however there is no delocalization of electrons and the bond order of the $\ce{N-O}$ bond is 2. So the $\ce{N-O}$ bond is weakened, and even the formation of the $\ce{N-N}$ bond cannot compensate for this weakening.

Also, the above example reminds me of certain unstable homonuclear species. Such as $\ce{Ne_2}$, which according to MO theory has a bond order of 0 - i.e. there is no bond. So I suppose that bond formation is not always exothermic. On the other hand, if the bond order is 0, is a bond really "formed"?

Bond formation is alway strictly exothermic in the sense of the change of enthalpy.

  exothermic reaction A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.

A bond can only exist, if it needs energy to break it, i.e. the bond dissociation energy is always positive.

  bond-dissociation energy, $D$ The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

This has absolutely nothing to do with a reaction being exothermic/endothermic or exergonic/endergonic, because this is defined by the rearrangements of bonds.

Regarding noble gas diatomics, it is quite clear from MO-Theory, that there is no bond. However, even these non bonded elements have a non-zero dissociation energy. Please refer to "Why are noble gases stable" and to answers and comments within.


Polyatomic Reaction Charges When a polyatomic ion reacts with another atom to create an ionic compound, often, we know the charge of the polyatomic ion. But how can we infer the charge of the atom it is reacting with?

For example, the question states, 'What is the correct formula for Calcium Phosphate?'

I know that the charge of the polyatomic ion named Phosphate is 3-. But does that mean the charge of Calcium, in this case, must be 3+ in order for it to react with Phosphate? If not, how can we figure out the charge of the Calcium ion in this ionic bond?

When a Calcium atom loses two electrons, it attains the stable electronic configuration of the noble gas Argon. Thus, Calcium cannot have a $\ce{+3}$ charge in any ionic compound. So, the ions forming your salt are $\ce{Ca^2+}$ and $\ce{PO4^3-}$. We need to balance the charges to obtain a neutral salt. We can see that the least common multiple of the charges (2 and 3) is 6. A total charge of $\ce{+6}$ can be obtained by taking three Calcium cations, and this can be balanced by taking 2 Phosphate anions.

Calcium Phosphate is $\ce{Ca3(PO4)2}$


How to oxidise iron in spinach to iron(III) I'm planning an experiment to test the iron content of spinach. I plan to ash spinach, grind it, then dilute it in a solution which would hopefully oxidise the iron ions in it to $\ce{Fe^3+}.$ This is in order to get a reaction with ammonium thiocyanate solution in the end, which will allow me to determine concentration by colorimetry.
What should I dilute my ashed spinach into to oxidise the iron in it? Preferably something safe with the thiocyanate.

Use ceric ammonium sulfate.[1,2]  Quoting from reference 2:

2 Reagents
(i) 0.25 per cent ceric ammonium sulphate in 8.8 N sulphuric acid.
(ii) 40 per cent potassium thiocyanate solution.
(iii) (a) Stock standard ferrous iron solution - 100 μg iron per ml.
(b) Working standard ferrous iron solution - 0.3 to 10 μg iron per ml
2.2 Procedure
To five ml of clear water sample in a test tube 0.5 ml of ceric ammonium sulphate solution is added and mixed thoroughly by shaking. Finally, 0.5 ml of potassium thiocyanate solution is added and again mixed properly. Exactly 5 minutes after addition of thiocyanate solution, the absorbance of ferric thiocyanate complex is measured at 480 nm in a spectrophotometer (Spekol, Carl-Zeiss, Jena).
A series of freshly prepared working standard ferrous iron solutions (upto 10 ppm iron) were treated similarly and the absorbances measured at 480 nm. From the calibration curve, plotted with concentrations of different standard iron solutions against absorbances, the iron content of the unknown sample is evaluated.

References

Willard, H. H.; Young, P. Ceric Sulfate as a Volumetric Oxidizing Agent. Ii. Determination of Iron. J. Am. Chem. Soc. 1928, 50 (5), 1334–1338. DOI: 10.1021/ja01392a013.

Goswami, D. C.; Kalita, H. Rapid Determination of Iron in Water by Modified Thiocyanate Method. DSJ 1988, 38 (2), 177–182. DOI: 10.14429/dsj.38.4835.
Also available via CiteSeerx. (The pdf icon has a cached version of the article.)


How to effectively wash the precipitate or crystals on the filter and why? My lab manual says that the precipitation on filter should be washed with a few small portions of the washing liquid. Why can't I just completely fill the filter funnel with the solvent instead of steadily add new tiny portions in it?

Also, how effective this washing process is? How many times should I wash the precipitation before it's completely free from impurities?

What filter funnel of what diameter of the paper filter should I use? Isn't the biggest one the is also the most effective?

Finally, why do I have to study limits when I want to become a chemist, not a mathematician?

Every time the precipitation on filter is washed, there is a certain constant volume $V_\mathrm{ret}$ of washing fluid retained on the filter. If the initial concentration of impurity is $c_0$, then initial amount of impurity on filter is
$$n_0 = c_0V_\mathrm{ret}$$
and after washing with a new portion of the fluid with the volume $V_\mathrm{add}$ concentration of impurity is reduced to
$$c_1 = \frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}c_0.$$
Analogously, after $n$-th washing act
$$
c_n = \left(\frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}\right)^n c_0
\qquad\text{or}\qquad
c_n = \left(\frac{1}{1 + \frac{V_\mathrm{add}}{V_\mathrm{ret}}}\right)^n c_0
$$
and, assuming we have a total volume of washing liquid of $V_\mathrm{tot} = nV_\mathrm{add}$, the former formula becomes
$$c_n = \left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0$$
Now, lets assume we have a moderate amount of impurities (e.g. $c_0 =\pu{0.1 mol L-1}$) that we are trying to get rid of by washing with different amounts of the same volume of the same solvent (say, $V_\mathrm{tot} = \pu{100 mL}$). However, we use different filters, for instance:

a large one (retains $1/10$ of the washing liquid, or $V_\mathrm{ret} = \pu{10 mL}$);
a middle one (retains $1/20$ of the washing liquid, or $V_\mathrm{ret} = \pu{5 mL}$);
a small one (retains $1/100$ of the washing liquid, or $V_\mathrm{ret} = \pu{1 mL}$),

and, of course, different portions of the liquid (various $n$). Lets plot the concentration of the impurity for each case as a function of washing frequency:

First and foremost, it's obvious that it is always better to wash with small portions many times, rather than two or three times with large quantities of washing liquid. Second, the smaller the filter, the less impurity will be left after the same number $n$ of washing acts.
Now, can we get rid of the impurity once and for all by washing with an extremely large number of small portions? In fact, lets use an infinite number and take the following limit
$$\lim_{n\to\infty}c_n = \lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0},$$
which is, in fact, one of the notable special limits, sometimes called the second remarkable limit:
$$\lim_{x\to\infty} \left(\frac{1}{1+\frac{a}{x}}\right)^x = \frac{1}{e^a}$$
so that in our case we can conclude it's impossible to completely get rid of the impurity only by washing:
$$\lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0} = c_0\exp{\left(-\frac{V_\mathrm{tot}}{V_\mathrm{ret}}\right)} \neq 0$$
Conclusions

It is impossible to completely remove all impurities by washing the crystals/precipitate on the filter even by washing it infinite number of times.
It is more efficient to wash with small amounts of liquid many times than only a couple with large ones.
Use filter of the right size: not too big to retain the traces of impurities, but big enough to maintain efficient liquid transport through its surface. When possible, use pleated filter paper.

Appendix
The figure have been plotted with pgflots. Feel free to play around with it or reuse for your lab manual. Source code:
\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
    \begin{axis}[
        title={Effectiveness of washing methods ($c_0 = \SI{0.1}{\mole\per\liter}$)},
        xmin=0, xmax=20,
        ymin=0, ymax=0.001,
        xlabel = $n$,
        ylabel = {$c_n$/\si{\mole\per\liter}},
        minor tick num=4,
        grid,
        legend cell align={left},
    ]

    \addplot [
        domain=1:20,
        samples=200,
        solid,
        thick,
    ]
    {0.1*(1/(1 + 10/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.10$}

    \addplot [
        domain=1:20,
        samples=200,
        dashed,
        thick,
    ]
    {0.1*(1/(1 + 20/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.05$}

    \addplot [
        domain=1:20,
        samples=200,
        dotted,
        thick,
    ]
    {0.1*(1/(1 + 100/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.01$}

    \end{axis}
\end{tikzpicture}

\end{document}


Why is change in volume of the solution zero in case of an ideal solution? In my textbook it is written that $ΔV_{mix} = 0$ for ideal solutions i.e. $V_{solution} = V_{A} + V_{B}$. But why is this true? There is literally no explanation in my textbook as to why this is true. I am so confused. Please can someone explain it rigorously. Please help.

If you have two component mixture of A and B, most important reason for volume of mixing being zero is that when solution is ideal heterogeneous interactions (A - B) are the same as homogeneous (A - A and B - B). If interactions are the same than volume doesn't change when you mix components. Other reasons which contribute to non - ideality of mixing is size and shape of the molecules. If you mix water with some liquid polymer, you can expect that  due to big difference in size and shape of water and polymer molecules, volumes will not be additive because water molecules being much smaller than polymer molecules can easily fit between them in free volume of polymer liquid. Such packing/fitting of water molecule is quite different than if you just look at water molecules in pure water or if you look at pure polymer since bulky polymer molecules can't just fit easily in free volume of other polymer molecules like water can.


How can I dispose of chlorine and sodium hydroxide as table salt and water? If I used chloralkali electrolysis of brine to produce hydrogen, and can't use the other products (or not all of them), what route(s) can I take to revert them back to brine? I figured, since I have to put in energy to take the brine apart, it should have some desire to reform.

With the $\ce{H2}$ taken away, the remaining products are:

$$\ce{2 NaOH (aq) + Cl2}$$

If I just mix them, I get sodium chlorate or hypochlorite. Can I process those further?

Both $\ce{H2O}$ and $\ce{NaCl}$ have a negative $\Delta H_\mathrm{f}^\circ$ ($\pu{-285.8 kJ/mol}$ and $\pu{-407.0 kJ/mol}$ respectively), and $\ce{Cl2}$ is already in its standard state, so we would need to overcome the $\Delta H_\mathrm{f}^\circ$ of the $\ce{NaOH}$, which is negative as well ($\pu{-470.1 kJ/mol}$). Can I use ion filters to some benefit?

Routes that allow retrieval of some of the energy spent for the electrolysis are favored (i.e. the more exothermic, the better).

It looks as if the simplest way would be to mix the products, and let the mixture warm up, so we get sodium chlorate in water. We can then have that crystalize and heat it until it disintegrates into $\ce{NaCl}$ and $\ce{O2}$. That would need no additional substances and have a clean end result. The only remaining question would be the energy budget, but the forming of the chlorate is exothermic and additional heat might be available.

Alternatively, we might find something useful to oxidize with the $\ce{NaClO3}$.


What are stereospecific and stereoselective reaction? I have read the defination but couldn't grasp the concept, what is the key factor that says this is stereospecific or this is stereoselective and what are the main intuitions i can get if i am given that a reaction is stereospecific or stereoselective.

If you have a question, ask it. You say you've read the defination[sic]. What it is that you fail to understand?

IUPAC Gold Book

stereospecificity (stereospecific)

A reaction is termed stereospecific if starting materials differing only in their configuration are converted into stereoisomeric products. According to this definition, a stereospecific process is necessarily stereoselective but not all stereoselective processes are stereospecific. Stereospecificity may be total (100%) or partial. The term is also applied to situations where reaction can be performed with only one stereoisomer. For example, the exclusive formation of trans-1,2-dibromocyclohexane upon bromination of cyclohexene is a stereospecific process, although the analogous reaction with (E)-cyclohexene has not been performed.
The term has also been applied to describe a reaction of very high stereoselectivity, but this usage is unnecessary and is discouraged.

stereoselectivity
(Also contains definition of: diastereoselectivity)

The preferential formation in a chemical reaction of one stereoisomer over another. When the stereoisomers are enantiomers, the phenomenon is called enantioselectivity and is quantitatively expressed by the enantiomer excess; when they are diastereoisomers, it is called diastereoselectivity and is quantitatively expressed by the diastereoisomer excess.


Do reverse reactions occur in combustion? I am currently reading a textbook on combustion. They assume that all reactions go to completion, such that post combustion only unburned fuel/air and actual products remain.

Assuming that combustion is lean or perfect (no fuel left unburned) and not neglecting dissociation effects, can reverse reaction actually occur to reform fuel at equilibrium? In other words, is the equilibrium constant infinity in real life?

An equilibrium constant is never $\infty$ in theory. In practice, large values of $K_{eq}$ (and slow reverse reaction rates) lead to the equivalent of complete combustion.


Which are the Geometrical isomers of the complex [Cr(NH3)2Cl2(en)]+?Is it one or two? I was recently studying coordination compounds and i came across optical isomerism shown by complexes.I think that the complex i gave above have 4 geometrical isomers(I think this is where I am wrong, that it only has three geometrical isomers),but among those four geometrical isomers,how many of them are optically active?.According to what i studied only two of the three geometrical isomers should be optically active,other two having plane of symmetry(I am not too clear about the plane of symmetry concept),but all  the solutions i looked up for the question said the complex only have 3 Geometrical isomers.Where am i wrong?.Is it that among the four geometrical isomers, two geometrical isomers are same?,if yes then how?

You have an octahedral complex where two of the adjacent positions are occupied by the ethylenediamine ligand. Let's make these equatorial for a clearer picture.

There are only two positions in the remaining four positions that are trans (C and D). You can choose to put two chlorides there, two ammines there, or one of each. Those are the 3 geometric isomers.
If the trans ligands are the same, then there is a plane of symmetry that goes through C-M-D and bisects the ethylene diamine ligand, so the complex will not be chiral. But if the two ligands are different in positions C and D, then there will not be a plane of symmetry and the complex will be chiral. You can draw them out to see what they are.
To summarize, there are three geometric isomers, one of which is chiral. If you have trouble with them types of problems, sometimes, it's helpful to just consider the cis/trans or mer/fac relationships between groups of ligands.


Intermolecular forces in Carbon Nanotubes? What are the intermolecular forces that can be found in Carbon Nanotubes?

There is known, but poorly understood $\pi-\pi$ stacking. While weak compared to covalent, dipole or ionic interactions, it still beats usual van-der-vaals interactions.

Carbon nanotubes are, essentially, polycondensed aromatical systems, so they can and would interact with each other as aromatic systems via $\pi-\pi$ stacking


Negative and Zero Enthalpy? I've been investigating chemical kinetics and stumbled across the CHEMKIN program manuals and the JANAF tables. In the original CHEMKIN database they list 'standard enthalpy'. I've checked and this is not in fact enthalpy of formation as this is a separately listed value. What I'm confused about is that I cannot find any reference to molar enthalpy any where else except in the the introduction to the JANAF tables and the CHEMKIN database. What's more it lists the enthalpy of $\ce{H2O}$ and $\ce{H2O2}$ as zero and other substances have negative enthalpy. Questions:

Is specific molar enthalpy in comparison to water or some other substance?

If it is still a relative measurement of enthalpy what advantage does it provide over heat of formation in practical use?

How can the two substances both have the same enthalpy even if it is relative if one has an extra atom and an extra bond?

Am I completely wrong and the whole thing is something else entirely?

At a glance, the values listed in the table in the CHEMKIN database are standard enthalpies of formation at 298 K, $\Delta _fH^○(T)$, and standard entropies at 298 K. Why? If only because CHEMKIN's $H^○(298)$ for the elements in their stable states are presented as equal to zero.

You may have just mixed up different enthalpies. Along with the heats of formation, thermodynamic tables (including NIST JANAF) often present the enthalpy increments (sometimes called simply enthalpies), $H^○(T)-H^○(0)$ or $H^○(T)-H^○(298)$. These are not to be confused with the enthalpies of formation $\Delta _fH^○(T)$.

By definition, the standard enthalpy of formation (at a certain temperature) of a complex substance AB is equal to the enthalpy of a chemical process A + B = AB, where A and B are elements in their most stable modifications (at a certain temperature). By convention, $\Delta _fH^○(T)$ for the elements in their most stable modifications is zero.

The enthalpy increment $H^○(T)-H^○(0)$ is an integral of the heat capacity at constant pressure, $\int\limits_0^T C_p(T)dT$. It is no less important than the heat of formation. For example, knowing $\Delta _fH^○(298)$ for our hypothetical substance AB and all the enthalpies $H^○(T)-H^○(298)$ for A, B and AB, we can calculate $\Delta _fH^○(T)$, an enthalpy of formation at different temperature:

$\Delta _fH^○(T) = \Delta _fH^○(298) + [H^○(T)-H^○(298)]_{AB} - [H^○(T)-H^○(298)]_{A} - [H^○(T)-H^○(298)]_{B}$

Note that similar formulae can also be found in the introduction to NIST JANAF (at least to the 4th ed. of it).


Behavior beyond the critical pressure I have seen some phase diagrams that look like this:
https://www.learner.org/courses/chemistry/images/text_img/phase_diagram_water.jpg

And also some like this:
https://en.wikipedia.org/wiki/Phase_diagram#/media/File:Phase-diag2.svg

My confusion is related to the Critical Point. The first picture doesn't really explain where the gas/supercritical fluid boundary is, and the right angle indicating the supercritical fluid boundary at the critical point seems unusual to me. I have some general questions about behavior beyond the critical temperatures and pressures.

I think my specific question is best summarized by this picture.

More specifically, suppose you have an arbitrary compound X with a Liquid/Gas critical temperature of 320 K and critical pressure of 40 Bar which follows the general shape of the above phase diagrams. I have the following questions

1) At 330 K  and 10 bar (basically beyond the critical temperature but below the critical pressure) would X be a gas? or would it be a supercritical fluid? Would we need more information to be able to decide?

2) As a more general part of 1), is it possible that a gas exists beyond its critical temperature?

3) At pressures above its critical pressure, a liquid can still exist so long as it's temperature is lower than the critical temperature correct?

In order to be in the "critical state", both temperature (x-axis) and pressure (y-axis) must be above the critical point. So answers to your questions become quite obvious:

Very hot but below the critical pressure, it is a gas. You don't need any more information than the temperature and the pressure.
Yes if its pressure is less than its critical pressure
Correct.


Is there a substance that would react with seawater to produce a dense, insoluble, nonporous precipitate? Is there a substance that would react with seawater to produce a dense, insoluble, nonporous precipitate? Alternatively, is there a hygroscopic substance that will absorb water and expand into a dense/tough non-permeable mass? This is a material to seal off a small inner diameter pipe once water intrudes into it

A superabsorbent polymer sounds appropriate for this application, and the wikipedia article suggests they are used for "blocking water penetration in underground power or communications cable".

Will the water be removed from the pipe again after it has intruded, and if so do you need the seal to remain impermeable? I would expect this kind of polymer to dry out and shrink with time in the absence of water and it would not necessarily then be able to swell again with the same performance.


Minerals in water solidify when heated? Is it possible for minerals in drinking water to solidify when high temperatures are applied? The minerals in mind are: iron, copper, aluminum, magnesium.

Background: I have small, brittle, darkish flakes appearing in my hot water system. The manufacturer of my tankless hot water heater said it's possible for these elements to solidify in this manner over time.

Looking at their website, I found this brief explanation:
http://support.noritz.com/article/faq-what-is-water-hardness-46.html

  "Water hardness is the measure of dissolved calcium, magnesium, and other minerals in your drinking water. It is typically measured in grains of hardness which equates to 17.1 mg/L of dissolved calcium carbonate. If not treated hard water can cause mineral build-up in your heater. Mineral build-up reduces flow and produces flakes at your fixtures."

We got out water hardness measured, and it came back as 68 mg/L.

The temperature of the water heater is set to 130F. I believe that the heater has a higher internal temperature.

I'm not sure why they are dark/black in color, all the pictures I'm seeing for calcium carbonate are white.

Thank you

Limescale (calcium and magnesium carbonate) precipitates in the heater, the brown colour is most likely some iron compound that co-precipitates. A fresh surface of limescale likes to adsorb other ions, which then get trapped in it as the scale grows. If you had a water kettle, you'd see the same stuff in there after using it a few times. (Not exactly, because carbonate precipitation depends on the pressure, but iron (hydroxide?) not. I wouldn't want to predict.)

The limescale first appears on the surface of the heater cell, and then breaks off during temperature changes, i.e. when you open the faucet. As you have no tank, the flow rate of the water is high in the heater and carries the scale out of there.

Very convenient, that. In a tank, the scale builds up and can fill the whole thing after a few years, depending on your water source.


What does Pb(NO₃)₂ + KI form? What does

  $\ce{Pb(NO3)2 + KI}$

form?

This is just a metathesis (double replacement) reaction.

We know that nitrates are soluble, and alkali metals are soluble (at least in these general cases).

Let's ignore balancing for now, and do that last.

$\ce{Pb(NO3)2 + KI -> Pb^{+2} + NO3^- + K+ + I-}$

$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + KNO3}$

But the $\ce{KNO3}$ is also soluble, and would dissociate.

$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + K+ + NO3-}$

Evidently, we can cancel out the spectator ions $\ce{K+}$ and $\ce{NO3-}$ and are left with a net ionic equation (which is easily balanced).

$\ce{Pb^{+2} + 2I- -> PbI2}$


Carbon 11 radiolabeled inhaled Beta-2 agonist or corticosteroids, is it possible? Background
Positron emission tomography (PET) is a highly sensitive imaging modality that can provide in vivo quantitative information of biological processes at a biochemical level. PET relies upon the administration of a chemical probe, often called a radiopharmaceutical, that is labeled with a short-lived positron-emitting radionuclide, examples of which are $\ce{^{11}C} \ (t_{1/2} = \pu{20.4 min})$ and $\ce{^{18}F} \ (t_{1/2} = \pu{109.7 min})$. Carbon-11 is one of the most useful radionuclides for PET chemistry, since its introduction into a biologically active molecule has minimal effects on the (bio)chemical properties of the compound (Ref.: Dahl, et al., Clin. Transl. Imaging 2017, 5, 275–289)
Biologicals of interest
Beta-2-Agonists are effective bronchodilators due primarily to their ability to relax airway smooth muscle (ASM). They exert their effects via their binding to the active site of b2-ARs, which are densely located on ASM. The presumed cellular mechanism of action involves the canonical signaling pathway via activation of Adenylyl cyclase (AC) and generation of intracellular cAMP,
which in turn can activate the effector molecules cAMP-dependent protein kinase A (PKA) and Epac, a Rap1 guanine nucleotide exchange factor. However, there is still little known regarding these pathways in airway cells (Ref.: Am J Respir. Crit. Care Med. 2013, 187(7), 690–696).
Question
Currently radiolabeled lung studies use labeled drugs that are produced for inhalation are "association” products, that is, the label is associated with the drug but not firmly bound to the drug. Without a firm bond between the active pharmaceutical ingredient (API) and the radiotracer, images must be acquired immediately post-inhalation before the dissociation of the label from the drug leads to a circulating level of absorbed radioactivity that could introduce an unacceptable error in the measurement of radioactivity in the lung.
Carbon-11 radiolabeled ICS or Beta-2 agonists would enhance our understanding of the cellular mechanism of action. Anyone aware of work in this area?

References:

Kenneth Dahl, Christer Halldin, Magnus Schou, "New methodologies for the preparation of carbon-11 labeled radiopharmaceuticals," Clinical and Translational Imaging 2017, 5, 275–289 (DOI: https://doi.org/10.1007/s40336-017-0223-1).
Mario Cazzola, Clive P. Page, Paola Rogliani, M. Gabriella Matera, "$\beta_2$-Agonist Therapy in Lung Disease," Am. J. Respir. Crit. Care Med. 2013, 187(7), 690–696 (DOI: https://doi.org/10.1164/rccm.201209-1739PP).

Positron emission tomography (PET) has been taken using $\ce{^{11}C}$- and $\ce{^{18}F}$-labelled reagents, although they are not $\beta_2$-agonists (Ref.1). The abstract states that:

Positron emission tomography (PET) provides three-dimensional images of the distributions of radionuclides that have been inhaled or injected into the lungs. By using radionuclides with short half-lives, the radiation exposure of the subject can be kept small. By following the evolution of the distributions of radionuclides in gases or compounds that participate in lung function, information about such diverse lung functions as regional ventilation, perfusion, shunt, gas fraction, capillary permeability, inflammation, and gene expression can be inferred. Thus PET has the potential to provide information about the links between cellular function and whole lung function in vivo. In this paper, recent advancements in PET methodology and techniques and information about lung function that have been obtained with these techniques are reviewed.

It is clear that unto 2007, there are no evidence for the use of radiolabeled $\beta_2$-agonist or corticosteroids for PET in literature. However, there are other radiolabeled reagents have been use for the imaging. For instance, the mixed-venous content is obtained by imaging the right heart during the infusion, and the alveolar concentration is obtained by imaging the lung, after the blood volume is subtracted from the image using a separate $\ce{^{11}C}$-labeled carbon monoxide ($\ce{^{11}CO}$) inhalation.
References:

R. Scott Harris, Daniel P. Schuster, "Visualizing lung function with positron emission tomography," J. Appl. Physiol. 2007, 102(1), 448-458 (DOI: https://doi.org/10.1152/japplphysiol.00763.2006)(PDF).
C. G. Rhodes, S. O. Valind, L. H. Brudin, P. E. Wollmer, T. Jones, P. D. Buckingham, J. M. Hughes, "Quantification of regional V/Q ratios in humans by use of PET. II. Procedure and normal values," J. Appl. Physiol. 1989, 66(4), 1905–1913 (DOI: https://doi.org/10.1152/jappl.1989.66.4.1905).


Prime in substituted benzene I was sent data for a molecule referred to as 1,2'-bis(4-aminophenoxy)benzene. I haven't been able to figure out if that is a real molecule.

If I drop the prime, I know 1,2-bis(4-aminophenoxy)benzene should look like that:

and Google agrees. But with the prime, I have no idea what the molecule could be. Is there such a molecule, or is it a typo?

This is a typo. The prime 2' would refer to position 2 in a separate ring, but you are only numbering the central ring.

Note that you have drawn the structure of 1,4-bis(4-aminophenoxy)benzene.

You can test the names of chemicals and their structures using a tool such as OPSIN.


Is the carbon atom in the carbon dioxide molecule partially positive? I had a question about nonpolar molecules that have symmetrical dipole vectors.

Let's take $\ce{CO2}$ as an example. Each of the $\ce{C=O}$ bonds are pulling in the opposite way. My teacher says that this causes all atoms in $\ce{CO2}$ to be equally charged as there mustn't be any net 'force'.

However, I disagree. Intuitively, it seems that the oxygen atoms would pull electron density away from the central carbon atom and make the carbon atom slightly positive and the oxygen atoms slightly negative, like so:

$$\large\ce{\overset{\small\delta-}{O}=\overset{\small\delta+}{C}=\overset{\small\delta-}{O}}$$

This process should make the carbon atom slightly positive and the oxygen atoms slightly negative. However, if I was right, then why don't we say $\ce{CO2}$ has a dipole (there is a separation of charge)? Perhaps I may have the wrong definition of a dipole.

You're correct in assuming that the carbon atom in $\ce{CO2}$ has a partial positive charge. This is because the oxygen atoms are much more electronegative, so they pull the electrons away from the carbon atom. However, this molecule is still nonpolar. This is because, when you draw a dipole moment, you have to take all bonds into account. Take water for example:

In this molecule, there are two bonds, each with their own dipole. But these cancel out like any other vectors, leaving you with a vertical net dipole. The dipoles in carbon dioxide cancel out in a similar way; however, they cancel each other completely, because the bond is linear, not bent like in water:

This produces a zero net dipole, rendering the molecule nonpolar.


Is purple paint more susceptible to UV damage than other colors? I was recently sorting some Mardi Gras beads. These are cheap plastic beads on strings, coated in cheap metallic paint. While sorting the beads into different colors, I noticed that the purple beads seem to have suffered more fading than the other colors. This is an effect I have noticed in the past: if the beads sit in the sun for a day or two, the colors fade rapidly. I'm guessing that the cause of that is that the UV radiation from the sun is breaking the molecules used for the pigments in the paints.

My first thought was that I had a pile of beads exposed to sunlight, and the purple beads in that pile had been at the top of the pile. But I noticed this happening in a few different piles that I had stored in different places and for different lengths of time. I have also noticed that the red beads and the purple/pink beads show some fading, though not as much as the purple beads. The yellow, silver, and green beads show almost no fading.

My current hypothesis is that the manufacturers use a pigment in the red and purple beads that is more susceptible to UV damage than the other pigments. So my question is: what is the chemical in these purple and red paints that is breaking down? (I guess there's also an implicit question here: is my thinking here chemically plausible?)

I would also agree with you and the comment from ssavec. The only point that I would make is that it shouldn't be because of the colour of the pigment only the stability of the particular dye it contains (almost certainly organic). In fact the colour of the paint should make it less susceptible to fading as it is reflecting colours at the violet end of the spectrum (subject to your definition of purple) rather than absorbing them.


Why is an S-S bond stronger than an O-O bond? I'm wondering why exactly the single bond between two sulfur atoms is stronger than that of two oxygen atoms. According to this page, an $\ce{O-O}$ bond has an enthalpy of $142~\mathrm{kJ~mol^{-1}}$, and a $\ce{S-S}$ bond in $\ce{S8}$ an enthalpy of $226~\mathrm{kJ~mol^{-1}}$. This one reports the $\ce{S-S}$ bond enthalpy to be $268~\mathrm{kJ~mol^{-1}}$, but I'm not sure which molecule they mean, or how they measured it. Anyway, it's still higher than that of $\ce{O-O}$.

Searching the Net, the only justification I could find was something similar to concepts they apply in VSEPR, like in this Yahoo Answers thread with such remarkable grammar. Quoting the answer, which might have borrowed some stuff from a high school textbook,

  due to small size the lone pair of electrons on oxygen atoms repel the bond pair of O-O bond to a greater extent than the lone pair of electrons on the sulfur atoms in S-S bond....as a result S-S bond (bond energy=213 kj/mole)is much more stronger than O-O(bond energy = 138 kj/mole) bond $\ldots$

Other variations of the same argument can be seen here, but it doesn't make sense, since one couldn't apply the same argument to $\ce{O=O}$ and $\ce{S=S}$. The first reference documents the $\ce{S=S}$ and $\ce{O=O}$ bond enthalpies to be $425$ and $494~\mathrm{kJ~mol^{-1}}$, respectively.

It's a bit shaky, and I'm looking for a solid explanation using MO or VB, or anything else that actually works. So, why is an $\bf\ce{S-S}$ single bond stronger than $\bf\ce{O-O}$, despite $\bf\ce{O=O}$ being obviously stronger than $\bf\ce{S=S}$?

TL;DR: The $\ce{O-O}$ and $\ce{S-S}$ bonds, such as those in $\ce{O2^2-}$ and $\ce{S2^2-}$, are derived from $\sigma$-type overlap. However, because the $\pi$ and $\pi^*$ MOs are also filled, the $\pi$-type overlap also affects the strength of the bond, although the bond order is unaffected. Bond strengths normally decrease down the group due to poorer $\sigma$ overlap. The first member of each group is an anomaly because for these elements, the $\pi^*$ orbital is strongly antibonding and population of this orbital weakens the bond.

Setting the stage

The simplest species with an $\ce{O-O}$ bond would be the peroxide anion, $\ce{O2^2-}$, for which we can easily construct an MO diagram. The $\mathrm{1s}$ and $\mathrm{2s}$ orbitals do not contribute to the discussion so they have been neglected. For $\ce{S2^2-}$, the diagram is qualitatively the same, except that $\mathrm{2p}$ needs to be changed to a $\mathrm{3p}$.

The main bonding contribution comes from, of course, the $\sigma$ MO. The greater the $\sigma$ MO is lowered in energy from the constituent $\mathrm{2p}$ AOs, the more the electrons are stabilised, and hence the stronger the bond.

However, even though the $\pi$ bond order is zero, the population of both $\pi$ and $\pi^*$ orbitals does also affect the bond strength. This is because the $\pi^*$ orbital is more antibonding than the $\pi$ orbital is bonding. (See these questions for more details: 1, 2.) So, when both $\pi$ and $\pi^*$ orbitals are fully occupied, there is a net antibonding effect. This doesn't reduce the bond order; the bond order is still 1. The only effect is to just weaken the bond a little.

Comparing the $\sigma$-type overlap

The two AOs that overlap to form the $\sigma$ bond are the two $\mathrm{p}_z$ orbitals. The extent to which the $\sigma$ MO is stabilised depends on an integral, called the overlap, between the two $n\mathrm{p}_z$ orbitals ($n = 2,3$). Formally, this is defined as

$$S^{(\sigma)}_{n\mathrm{p}n\mathrm{p}} = \left\langle n\mathrm{p}_{z,\ce{A}}\middle| n\mathrm{p}_{z,\ce{B}}\right\rangle = \int (\phi_{n\mathrm{p}_{z,\ce{A}}})^*(\phi_{n\mathrm{p}_{z,\ce{B}}})\,\mathrm{d}\tau$$

It turns out that, going down the group, this quantity decreases. This has to do with the $n\mathrm{p}$ orbitals becoming more diffuse down the group, which reduces their overlap.

Therefore, going down the group, the stabilisation of the $\sigma$ MO decreases, and one would expect the $\ce{X-X}$ bond to become weaker. That is indeed observed for the Group 14 elements. However, it certainly doesn't seem to work here. That's because we ignored the other two important orbitals.

Comparing the $\pi$-type overlap

The answer for our question lies in these two orbitals. The larger the splitting of the $\pi$ and $\pi^*$ MOs, the larger the net antibonding effect will be. Conversely, if there is zero splitting, then there will be no net antibonding effect.

The magnitude of splitting of the $\pi$ and $\pi^*$ MOs again depends on the overlap integral between the two $n\mathrm{p}$ AOs, but this time they are $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. And as we found out earlier, this quantity decreases down the group; meaning that the net $\pi$-type antibonding effect also weakens going down the group.

Putting it all together

Actually, to look solely at oxygen and sulfur would be doing ourselves a disservice. So let's look at how the trend continues.

$$\begin{array}{|c|c|c|c|}
\hline
\mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} & \mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} \\
\hline
\ce{O} & 144 & \ce{F} & 158 \\
\ce{S} & 266 & \ce{Cl} & 243 \\
\ce{Se} & 192 & \ce{Br} & 193 \\
\ce{Te} & 126 & \ce{I} & 151 \\
\hline
\end{array}$$
(Source: Prof. Dermot O'Hare's web page.)

You can see that the trend goes this way: there is an overall decrease going from the second member of each group downwards. However, the first member has an exceptionally weak single bond.

The rationalisation, based on the two factors discussed earlier, is straightforward. The general decrease in bond strength arises due to weakening $\sigma$-type overlap. However, in the first member of each group, the strong $\pi$-type overlap serves to weaken the bond.

I also added the Group 17 elements in the table above. That's because the trend is exactly the same, and it's not a fluke! The MO diagram of $\ce{F2}$ is practically the same as that of $\ce{O2^2-}$, so all of the arguments above also apply to the halogens.

How about the double bonds?

In order to look at the double bond, we want to find a species that has an $\ce{O-O}$ bond order of $2$. That's not hard at all. It's called dioxygen, $\ce{O2}$, and its MO scheme is exactly the same as above except that there are two fewer electrons in the $\pi^*$ orbitals.

Since there are only two electrons in the $\pi^*$ MOs as compared to four in the $\pi$ MOs, overall the $\pi$ and $\pi^*$ orbitals generate a net bonding effect. (After all, this is where the second "bond" comes from.) Since the $\pi$-$\pi^*$ splitting is much larger in $\ce{O2}$ than in $\ce{S2}$, the $\pi$ bond in $\ce{O2}$ is much stronger than the $\pi$ bond in $\ce{S2}$.

So, in this case, both the $\sigma$ and the $\pi$ bonds in $\ce{O2}$ are stronger than in $\ce{S2}$. There should be absolutely no question now as to which of the $\ce{O=O}$ or the $\ce{S=S}$ bonds is stronger!


Sugar: A molecule or crystal Is sugar a molecule or a large crystalline structure composed of many molecules? What do we mean when we say sugar?

Terminology can be confusing especially when some terms are only vaguely defined.

"Sugar" is one of those vague terms. There is a whole family of sugars recognised by chemists. In casual use what people usually mean is sucrose, a specific sugar molecule made from two other sugars, fructose and glucose, joined together. this is produced from sugarcane or sugar beet and is the dominant form of sugar seen by non-chemists.

But is sugar the molecule or the crystal? Both. Many pure chemicals form crystals, indeed crystallisation is a very common way of producing purer chemicals. Those crystals consist of a regular 3D array of a single molecule (for a pure chemical). So "sugar" describes both the molecule and the crystals of that molecule which consist of a large number of them arranged in a regular structure.


Why is the molozonide adduct thermally unstable? In the mechanism of alkene-ozonolysis, the intermediate molozonide adduct, which is a cyclical but symmetric molecule (shouldn't symmetry about the C-C single bond bring stability?), decomposes thermally into a ketone and a peroxy-zwitter ion via cleavage of the O−O and C−C bonds.
However, it seems unlikely, and is against the symmetry argument. Is there some better explanation? Or we don't know the correct mechanism? Please add some references.
Knowing that there are two symmetrically placed O−O bond in the adduct, the cleavage of the only one bond is weird.

Why is the molozonide adduct thermally unstable?

The molozonide adduct (1,2,3-trioxalane; the initial adduct by 1,3-dipolar cycloaddition of alkene and ozone) is thermally unstable because it contains two weak $\ce{O-O}$ bonds (Ref.1). Mechanistic studies reveal that the ozonide (1,2,4-trioxalane) is formed by a sequence of three pericyclic steps involving 1,3-dipolar cycloaddition, retro-1,3-dipolar cycloaddition, and again 1,3-dipolar cycloaddition as shown in following scheme:

Also, I'd like to point out that the initial molozonide adduct is not always symmetric. It is symmetric only if the alkene is symmetric. And, the rearrangement of 1,2,3-trioxalane to 1,2,4-trioxalane is not about the stability of $\ce{C-C}$ in 1,2,3-trioxalane as OP argued. It's about thermal stability of consecutive $\ce{O-O}$ bonds in 1,2,3-trioxalane (recall that one $\ce{O-O}$ bond in peroxy compounds is not thermally stable, e.g., $\ce{HO-OH}$).

Knowing that there are two symmetrically placed $\ce{O-O}$ bond in the adduct, the cleavage of the only one bond is weird.

The accepted mechanism (Criegee Mechanism) says one of either $\ce{O-O}$ bond in the initial adduct can be broken to give a corresponding carbonyl oxide (a dipole) and carbonyl compound (dipolarophile), which undergo second 1,3-dipolar cycloaddition to give stable ozonide (see above scheme). None of the mechanism says one $\ce{O-O}$ bond cleaves selectively. it could be either one:

However, the second 1,3-dipolar cycloaddition of either set of carbonyl oxide (a dipole) and carbonyl compound (dipolarophile) gives identical ozonide. This second dipolsar cycloaddition $([_\pi2_\mathrm{s} + _\pi\!4_\mathrm{s}])$ has also been confirmed by $\ce{^{17}O}$-NMR studies (Ref.2):

The mechanism of ozonolysis was revisited with the use of $\ce{^{17}O}$-NMR spectroscopy. In a crossover experiment with $\ce{^{17}O}$‐labelled benzaldehyde and the ozonides of styrene and ethylidenecyclohexane it was shown that only the ether bridge of the secondary ozonides is carrying the $\ce{^{17}O}$ label. This is contrary to results reported earlier and confirms the Criegee mechanism.

References:

Dipak K. Mandal, “Chapter 4: Cycloadditions 1: Perturbation Theory of Reactivity, Regioselectivity and Periselectivity,” In Pericyclic Chemistry: Orbital Mechanisms and Stereochemistry; First Edition, Elsevier Inc.: Amsterdam, Netherlands, 2018, pp. 107-190 (ISBN: 978-0-12-814958-4).
Christian Geletneky, Stefan Berger, “The Mechanism of Ozonolysis Revisited by $\ce{^{17}O}$-NMR Spectroscopy,” Eur. J. Chem. 1998, (8), 1625–1627 (DOI: https://doi.org/10.1002/(SICI)1099-0690(199808)1998:8<1625::AID-EJOC1625>3.0.CO;2-L)).


Why acyl halides are not named as 1-substituted aldehydes? For acyl halides, IUPAC uses preferably the "acyl halide" type naming (e.g. acetyl chloride). But they can be theoretically named as aldehydes, where a hydrogen atom in the $\mathrm{CHO}$ group (carbon 1) is substituted by a halogen atom (e.g. acetyl chloride $\mathrm{CH_3COCl}$ has the same structure as acetaldehyde (or ethanal) $\mathrm{CH_3COH}$, so it can be named 1-chloroethanal - the name is probably valid, some sources list it as a possible one).

Of course, functionally acyl chlorides and aldehydes are very different (since acyl chlorides are much closer to corresponding carbonic acids), but structurally they aren't (and the systematic name should probably express the structure of a compound at first). Consider for example the naming of alkyl halides (halogen-substituted alkanes), where IUPAC switched to the substitutive nomenclature (for example, from "ethyl chloride" to "chloroethane", despite that chemical properties of chloroethane and ethane are very different).

The question may be incorrectly stated, since I'm not a (professional) chemist. Thanks in advance!

Well, the rest of the substituted ethane in the chlorethane is still like an alkane, but the rest of to be 1-chlorethanal is not an alkanal in the first place. An alkanal must have the aldehydic group.

If there would be 1-chlorethanal, why not 1-hydroxyethanal ( aka known as the acetic acid  ), or 1-methylethanal ( aka acetone )  ?  :-)


Chemical equation to represent the hydrogen peroxide scavenging experiment, used to determine antioxidant ability of tea I am currently investigating the antioxidant ability of tea using hydrogen peroxide scavenging. Basically, I will look at absorbance of the tea samples with and without the addition of hydrogen peroxide at 230nm using a UV-Vis spectrophotometer and the difference between both is the amount of antioxidants which were able to reduce the hydrogen peroxide. However, although I've been looking theough many papers, none show me the chemical equation representing this reaction. Could someone help me with this?

I start by quoting a source (which also goes into great detail on explaining the underlying mechanism of chromophores excitation):

  Photosynthesis is initiated by the absorption of light by the chromophores, which excites the molecules from the ground state to an electronic excited state.

So, simply:

$\ce{Chromophores + hv -> *[Chromophores] -> [Chromophores](e-,h+)}$

with the creation of an electron and an electron hole. Now, absence the H2O2, the action of the electron-hole on OH- on water creates an hydroxyl radical:

$\ce{OH- + h+ -> .OH }$

Also, in the further presence of H2O2, a path to the so-called one-electron reduction of hydrogen peroxide from the action of a formed electron (e-) on H2O2, resulting in an additional hydroxyl radical:

$\ce{H2O2 + e- (aq) -> .OH (aq) + OH- (aq) }$

Now, as the tea acts acts as a radical scavenger, it reacts with the hydroxyl radical usually via an H atom extraction reaction forming water. For example:

$\ce{RH + .OH -> .R + H2O}$

The hydroxyl radical continues to attack the chromophores until there is just simple colorless products like CO2 and H2O. In time, the tea solution becomes colorless.

Now, in the combined chromophores and H2O2 experiment, the UV absorbance by tea solution should measurably display more decline with time due to the increased hydroxyl radical concentration.


Is it appropriate to call this material a hybrid composite? The material I've synthesised is made up of two primary parts at a 1:1 ratio:

Part 1: agar, glycerol, water.
Part 2: polyurethane resin.

The agar mixture on its own was too soft, so I added the resin to harden the material. Would this material be classified as a hybrid composite? If not, why?

There is a slight deviation in the literature as to what the hybrid composites are.
Originally, materials were considered hybrid composites if both inorganic (e.g. clay/silicates as the hydrophilic component) and organic (e.g. organic polymer as the hydrophobic component) constituents were involved in building up the matrix (see e.g. a review [1]).

Another review [2] and the more recent definition from Encyclopedic Dictionary of Polymers [3, p. 373]:

  Hybrid Composite n Advanced composite with a combination of different high-strength continuous filaments in the matrix.
  Also, composite in which continuous and staple fibers are used in the same matrix.

provide a more permissive definition, according to which hybrid composite materials are the materials in which a combination of reinforcing layers from two or more types of fibers is utilized.

With this in mind, in order to provide a definitive answer for yourself, you probably want to do surface imaging (e.g. with AFM or SEM) to prove that really are more than one type of fibers present.
If you are planning to publish this data, be prepared that you will be asked to provide rheological data as well.

References

Wahab, M. A.; Kim, I.; Ha, C.-S. Silica- and Silsesquioxane-Containing Polymer Nanohybrids. In Macromolecules Containing Metal and Metal-Like Elements; Abd-El-Aziz, A. S., Carraher, C. E., Pittman, C. U., Zeldin, M., Eds.; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 2005; pp 133–160. DOI: 10.1002/0471712566.ch6.
Perov, B. V.; Khoroshilova, I. P. Hybrid Composite Materials. In Polymer Matrix Composites; Shalin, R. E., Ed.; Soviet Advanced Composites Technology Series; Springer Netherlands: Dordrecht, 1995; pp 269–304. DOI: 10.1007/978-94-011-0515-6_6.
Encyclopedic Dictionary of Polymers; Gooch, J. W., Ed.; Springer New York: New York, NY, 2011. DOI: 10.1007/978-1-4419-6247-8.


The eigenvector matrix C I was wondering what the difference between $c_n$ and $c_{jn}$ is? In the context of molecular orbital construction, for example, if $c_{jn}$ is the coefficient of the basis functions $\phi_j$ (contribution of the constituting atomic orbitals), what would $c_n$ be?

$\mathbf{c}_n$ is a vector with $N$ entries (I switched to boldface to indicate - hopefully slightly more clearly - that it is a vector):

$$\mathbf{c}_n = \begin{pmatrix}c_{1n} \\ c_{2n} \\ \vdots \\ c_{Nn}\end{pmatrix}$$

This effectively means the same thing as writing

$$\psi_n = c_{1n}\phi_1 + c_{2n}\phi_2 + \cdots + c_{Nn}\phi_N = \sum_j c_{jn}\phi_j$$

I'll try to contextualise it. The columns of the eigenvector matrix are simply the eigenvectors themselves, so if (for example) the eigenvectors of an operator $\mathbf{A}$ are

$$\mathbf{c}_1 = \begin{pmatrix}p \\ q\end{pmatrix}; \qquad \mathbf{c}_2 = \begin{pmatrix}r \\ s\end{pmatrix}$$

then the corresponding eigenvector matrix $\mathbf{C}$ is

$$\mathbf{C} = \begin{pmatrix} \mathbf{c}_1 & \mathbf{c}_2 \end{pmatrix} = \begin{pmatrix} p & r \\ q & s \end{pmatrix}$$

[As an addendum: If the eigenvectors $\mathbf{c}_i$ are normalised, then you will find that

$$\mathbf{C}^{-1}\mathbf{A}\mathbf{C} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$

where $\lambda_i$ is the eigenvalue of $\mathbf{c}_i$, i.e. $\mathbf{Ac}_i = \lambda_i\mathbf{c}_i$.]


Lassaigne's test for nitrogen Why is it that benzene diazonium salts don't give the Lassaigne's test for nitrogen. After all they've got both carbon and nitrogen.

Benzene diazonium chloride is stable upto 5 degree Celcius, while in Lassaigne's test we need to heat the organic compound with sodium till red hot, before breaking it by plunging into cold water. All the nitrogen will be liberated before it finds a chance to combine with Na and form any NaCN, which is responsible for the test ultimately.


What would be the intermolecular forces between the molecules of methanetetrol? Methanetetrol has the following chemical formula: $\ce{CO4H4}$. It is also referred to as orthocarbonic acid.

Since the molecule has a molecular geometry of a tetrahedron, similar to that of a molecule of methane, I would assume that the dipoles at the bonds cancel each other out. Hence, making the molecule non-polar with no dipole or charge separation overall. Thereby, the intermolecular force between the molecules of methanetetrol should be London dispersion forces.

However, I've been told that the intermolecular force between the methanetetrol molecules is hydrogen bonding since the four $\ce{OH}$; hydroxyl groups in the molecule are capable of making 4 hydrogen bonds with 4 subsequent methanetetrol molecules.

Hence, my question is how can methanetetrol have hydrogen bondings as the intermolecular force between its molecules if the molecules do not have a dipole; or so I believe, i.e. what do you think is the intermolecular force between methanetetrol molecules?

Let's try to guess a reasonable structure for $\ce{C(OH)_4}$. The oxygen atoms and the central carbon would probably form a tetrahedron, therefore, those bond dipoles would basically cancel (as in e.g. $\ce{CCl_4}$). However, the carbon-oxygen-hydrogen angle would probably not be 180°. Whether the hydroxyl group would rotate freely about the carbon-oxygen bond or there would be some intramolecular hydrogen bonding preventing that, I would assume that intermolecular hydrogen bonding could still occur. (Which would be the first step towards eliminating water and forming $\ce{CO_2}$.)


Sodium thiosulfate formation I have a problem solivng this chemical problem. Any help would be highly appreciative.

Sodium Thiosulfate can be made by the reaction

$\ce{Na_2CO_3 + 2Na_2S + 4SO_2 ->  3Na_2S_2O_3 + CO_2}$

Question: How many grams of $\ce{Na_2S}$ are required to react with 25.0 g of $\ce{Na_2CO_3}$ if $\ce{SO_2}$ is present in excess?

My attempt:

$25.0 \text{ g } \ce{Na_2CO_3}\frac{1 \text{ mol } \ce{Na_2CO_3}}{105.99}\frac{2}{1}\frac{78.05}{1}$

The answer from the book states: 25.9g$\ce{Na_2S}$

You are completely correct in your dimensional analysis and stoichiometry.

  36.8g is the correct answer.

Molar mass $\ce{Na2CO3 = 106g/mol }$

${25g \rightarrow 25/106 = 0.236}\text{ mol } \ce{Na2CO3}$

This will react with ${0.236*2 = 0.472} \text{ mol } \ce{Na2S}$

Molar mass $\ce{Na2S} = 78.0 \text{ g/mol}$

$0.472 \text{ mol } = 0.472*78.0 = 36.8 \text{ g } \ce{Na2S}$ will be needed.

I will now prove your book wrong

25.9g $\ce{Na2S}$ = 0.332mol $\ce{Na2S}$

0.332mol $\ce{Na2S}$ reacts with 0.166mol $\ce{Na2CO3}$

0.166mol $\ce{Na2CO3}$ = 17.5g of $\ce{Na2CO3}$, which is different from our starting quantity of 25.0g.


What is the ionic equation for reaction of potassium permanganate with potassium iodide? The half equation for the reduction of the permanganate would be

$$\ce{5e- + 8H+ + MnO4- -> Mn^2+ + 4H2O}$$

But with regards to the iodide, I am very confused. I have seen some sources say that $\ce{I-}$ is oxidised to $\ce{IO3-}$ in this reaction, and others saying that it is oxidised to $\ce{I2}$.

How stable is $\ce{IO3-}$ in the presence of $\ce{I-}$ under acidic conditions?

It would end up being in an equilibrium with very few $\ce{IO3-}$ anions and lots of iodine, due to the following equilibrium:

$$\ce{2 IO3- + 12 H3O+ + 10 I- <=> 6 I2 + 18 H2O}$$


Why does delocalization lead to stability? If the electrons are relatively more stable when they are delocalized than when they are localized between two atoms, then why do we have bonds formed in the first place? Forming of bonds between two atoms means that electrons localized between the two atoms, if electrons tend to delocalize then why not all electrons (including the sigma bond electrons) are just delocalized? What is the driving force for delocalisation ? How does delocalisation the energy of the system?

Electrons "want" two thing: be close to a nucleus and have the freedom to move. Also, being fermions, they are "forbidden" by the Pauli exclusion principle to have exactly the same state (e.g. same set of quantum numbers in an atom) as another one.

The electron distribution in a molecule is a compromise of the two "wants" while not trying anything "forbidden". In the $\ce{H2}$ molecule, electrons get to enjoy the attraction of two nuclei while having more space available than when bound to an isolated hydrogen atom. Because of the Pauli exclusion principle, only two electrons get to enjoy that low energy.

Electrons associated with conjugated double bonds also have "more room" without being further away from the nuclei than electrons associated with isolated double bonds.

Sorry for the anthropomorphizing... for a more rigorous treatment, look up particle in a box in a quantum chemistry book.


Energy required to remove an electron from He The energy required to remove both electrons from the helium atom in its ground state is $79.0\;\mathrm{eV}$. The energy required to ionize helium (i.e. to remove one electron) is -

  (A)   $24.6\;\mathrm{eV}$

  (B)   $39.5\;\mathrm{eV}$

  (C)   $51.8\;\mathrm{eV}$

  (D)   $54.4\;\mathrm{eV}$

My thoughts: What I actually did was to take half of the given energy value since energy required to remove electrons depends only on the principal quantum number(and also on atomic number). But, I found my answer to be wrong.

On the other hand just calculating the energy required to remove one electron from $n=1$ state gives the correct answer. Why is it so?

We can use Bohr's atomic model to calculate the ionization energy of $\ce{He+}$.
The ionization energy will be the amount of energy we must give to the electron of singly ionized helium to remove it apart from the nucleus to infinity. That gives $IE_2=E_{\infty}-E_1$ where $E_1$ is the energy in the first bohr-orbit (principal quantum number=$1$). Conventionally taking $E_{\infty}$ as $0$ we use the Bohr's equation to find the energy $E_1$:-
$$E_n=-\frac{Z^2me^4}{8n^2h^2\epsilon_0^2}$$
 where $Z$= atomic number, $e$= charge on electron, $n$= principal quantum number, $h$= Planck's constant, $m$= mass of electron, $\epsilon_0$= permittivity of free space.

Using $n=1$, $Z=2$ and the appropriate values for other constants we get $IE_2=-E_n=\pu{54.4eV}$.
Now the energy for removing both electron $IE_{net}=IE_1+IE_2$. Using $IE_{net}=\pu{79eV}$and $IE_2=\pu{54.4eV}$ we get, $$IE_1=\pu{24.6eV}$$
which corresponds to option (A), which I assume is typed wrong by you in the question.


What is the reason behind a good smell? I know that volatile substance creates smell but how a good smell produce? Is there any common characteristics of the compounds having good smell?

There is a very wide variety of chemicals that have a good smell - and the flavor and fragrance industry has discovered and profited from thousands of them. To have a good smell, a compound first of all has to be low molecular weight and able to easily evaporate into air so it can be carried to your nose--i.e. a "volatile organic compound". Esters are a good class to look at for starters, and are responsible for many of the fruity odors. For example, amyl acetate is a simple compound, easily made in the lab, that smells like bananas. Another important class is terpenes: hydrocarbons made from 5-carbon units that include pine and cedar odors. Some odor components like those of chocolate are more complex, have many components, and are hard to reproduce exactly.

One way to think of good (and bad) smells is how over time animals and humans have evolved to appreciate the smells of some things (like fruit or cooking onions) as good and other things like rotting flesh as bad, since the former is good for you and the latter definitely bad. If you got it backwards, you would not survive for long. But flowers, for example, smell good but do not materially benefit us. Maybe we have some residual bee genes in us.