Why is solubility often listed as a chemical property? Isn't the action of dissolving just a physical change?

The classification of physical and chemical changes has done more harm to students than good. This topic was taught in the 19th century but it still lingered in modern general chemistry textbooks which serve to repel good students from taking up chemistry as their field. For example, if we call the charge of electron red, and the charge of proton blue, will it change any fundamental thing?

You can try to understand physical vs. chemical change as follows: If you are changing the physical form (liquid, solid, gas) of one or more component, then you can safely label it as a physical change provided you can bring the original form back.

You can melt NaCl crystals into a liquid, this is a physical change because we can cool it and bring back the original material.

In the same way, solubility is just a property of matter. It is a very complex phenomenon. Whether you call it physical or chemical it does not matter in real life (most scientists will not worry about it and nobody will ever ask you these questions once you pass this course-so much so for the utility of this concept). For example, HCl is very soluble in water, but also reacts with water in such a way that it protonates water. Dry HCl gas is "neutral" but its solution in water is a powerful acid. There is no simple way to extract the gas back out of water. This solubility involved a chemical change.

On the other hand, salt NaCl when dissolved in water, it will not chemical react with water per se and you can recover the salt by evaporation. Should I call this a chemical change or a physical change? It does not matter.


Yes, congo red bind the cellulose because is a substantive (or direct) dye these dyes don't need a mordant so it binds suddenly to the cellulose.
 There are others dyes for example safflower, cochineal with this property. In biology Congo Red is frequently substitute with Sirus Red F3B and Tioflavone S. At this link you can find some alternatives to Congo Red use for demonstrating amyloid, however I don't know if you can apply these dyes to your case, because I don't find any reference about their use as pH indicator. Although is reported that many substantive dyes are used as pH indicator I can't find any list.

Take in account that acid pigment are not substantive to cellulosic fiber and most of the pH indicator are acids.

I think that maybe you can try to use a mordant to obtain the same result with a pH indicator normally used, however I'm afraid that using a mordant you risk to change the behavior of your pH indicator.


Molecular orbital explanation of why the protonation of the oxygen atom makes a carbonyl group more electrophilic

Ignoring the presence of the positive formal charge on the oxygen

I think this is exactly what is responsible, even from the MO viewpoint. From the MO perspective, the electrophilicity of the carbonyl group can be viewed as how energetically-accessible is the $\ce {C=O \pi ^*}$ MO to the nucleophile's HOMO. The protonation of the oxygen atom effectively decreases the energies of all MOs of the carbonyl, including the $\ce {C=O \pi ^*}$ MO. Why? The proton is positively-charged. By attaching to the oxygen atom, there is a net decrease in the potential energy of the molecule due to there being more electron-nuclei attractive forces in the resultant entity. Hence, protonation makes the $\pi ^*$ LUMO more accessible to the attacking nucleophile, increasing electrophilicity.


In milk, casein exists as a colloidal aggregate (casein micelle) with the properties you state, but purified casein is different. It is rich in proline (shown below), has no disulfide bridges, and hence little tertiary structure to hide the hydrophobic residues in the core of the protein with the hydrophilic residues on the surface to bind to water. In addition to having a hydrophobic side chain, proline stiffens the aminoacid backbone, hindering tertiary structure. The net effect is a molecule with too much exposed hydrophobicity to be water soluble. Casein micelles are complex structures and quite different from lipid micelles.

PROLINE


Is this product description wrong and a simple battery and not a fuel cell?

Your analysis seems correct, i.e. it is only a fuel cell if the magnesium is being oxidized by the air. You can prove if that is the case, and do so as part of your classroom demonstration, by excluding any air, and therefore oxygen, from the fuel cell.

Use distilled water and fairly pure $\ce{NaCl}$ (rather than table salt) to make the salt solution.
Boil the salt water for a minute or so to remove air bubbles (some oxygen from air dissolves into water).
Put the cell with electrolyte in a plastic bag sealed around the wires and a narrow tube such as a plastic coffee stirrer or soda straw that you'll use later to admit air, carefully squeezing out air bubbles. Seal both bag and tube with plasticine modeling clay or soft wax.
Observe the open circuit voltage of the cell.
Then run it for a while with a load (e.g. a 75 mA 1.5 V lamp) or a 100 ohm resistor, which will draw 1.3 mA), depending on the current output of the cell. You would expect some voltage drop over time, either due to cell polarization or due to any residual oxygen being used up.
To prove that this is a true fuel cell oxidizing the magnesium using air, bubble some air into the plastic bag through the (unplugged) tube. *If the voltage goes back up, then drops down until yet more air is added, you've shown that oxygen is being used and it's a true fuel cell.

Please accept my compliments for your thoughtful question; do post your findings here!


The principle of solubilization — in the ideal case — is closely related to that of melting. Think of solubilization as a two step process:

melt the solute;
mix the solute and the solvent.

Of course, step 1 involves raising the temperature to the melting point, or at least adding energy (in the form of heat) at the melting temperature in order to break attractive bonds between solute molecules. We will consider only the case where we are already at the melting point (an isothermal process), so you only need to add heat in order to turn the solid into liquid.

Step 2, on the other hand, is spontaneous if the solute and solvent mix ideally. You don't have to add heat (energy) for this step to proceed, assuming both substances are liquids and mix ideally.

Now what happens when you solubilize something is that you need to add less heat than you would have needed in step 1 to melt all of the solute, since some of the required (free) energy comes from the process of mixing.

By a similar line of argument, based on the properties of ideal solutions, you can relate (approximately) the solubility to the melting point and the heat of fusion as follows:

$$\log(x_2) = \frac{\Delta H^\circ_\mathrm{fus}}{R}\left(\frac{1}{T_\mathrm{fus}}-\frac{1}{T}\right)$$

Since $\Delta H^\circ_\mathrm{fus}$ is generally positive (heat is required to melt something) then increasing $T$ tends to increase the solubility of solids.

Crystallization is just the reverse of solubilization! And yes, there is always an equilibrium between soluble/insoluble solute, it's just that sometimes the soluble or insoluble part is negligible, depending on the values of $\Delta H^\circ _\mathrm{fus}$, $T_\mathrm{fus}$, and $T$.


The "classical" way

Direct addition of $\ce{H-X}$ to alkenes proceeds via the addition of a proton to the $\ce{C=C}$ double bond in the initial step, i.e via a carbocation. This works nicely for strong mineral acids, but not for the weak acid $\ce{HCN}$ with the rather weak nucleophile $\ce{CN-}$.
Synchronous addition?
Q: Can't $\ce{H-CN}$ add synchronously add to $\ce{C=C}$?
A: NO! Such a reaction would involve the interaction of the HOMO (highest occupied molecular orbital) of one reaction partner with the LUMO (lowest unoccupied molecular orbital) of the other. The LUMO of an alkene has a nodal plane orthogonal to the $\ce{C-C}$ bond, while there is - obviously - no nodal plane but a $\sigma$ bond between hydrogen and carbon in $\ce{HCN}$.
The synchronous addition of $\ce{H-CN}$ to $\ce{C=C}$ is symmetry-forbidden!
Catalysis?
Yes! Since the first catalytic hydrocyanation using $\ce{Co2(CO)8}$ (see Philipp's excellent comment above), other catalysts have been developed, namely $\ce{Ni(0)}$ based species using a trypticene-derived chelating phosphine ligand (see image).


MgO as food supplement - solubility in water

MgO will react with stomach acid, not dissolve in it. I suppose that's what the manufacturers go for.

This is not factually right at all. Stating the metal content as an oxide is an century old tradition of quoting the metal content as oxides. Magnesium percentage quoted as percent MgO does not mean that the tablet actually contains magnesium oxide. The links OP provided in the comments shows that the tablet actually contains magnesium stearate. This is done to avoid upsetting the stomach because a very water soluble magnesium salt can induce diarrhea in a few hours.
Magnesium steareate is a salt of a fatty acid, which is rather water insoluble as expected but it has some fat solubility. So when you take it with food which has some fats along with this fatty magnesium salt, there must a be slow absorption of the fatty salt by the body hence the tummy is happy.
Basically, you are eating a magnesium soap!


Calculating heat of combustion in a bomb calorimeter

In a constant volume bomb-calorimeter, we can make chemical reactions take place. If we let chemical reactions freely take place in it, then the total internal energy change is zero. This is due to the fact there is no expansion work since the system is a constant volume by definition and the heat transferred is zero since the bomb calorimeter is well insulated. The heat released in a reaction is automatically absorbed by the bomb calorimeter device. Hence the total internal energy change is zero.
Now, consider another path of the reaction. We start with reactants and turn them into products under constant volume and constant temperature conditions (*) and then these products we raise the temperature as to reach the final temperature we got in the previous path.
The energy associated with moving from reactants to products isothermally is $\Delta U_{rcxtn}$ and the energy change for raising up the products to the temperature of first path is $ \Delta U_{heat}$
Now, we can write:
$$ \Delta U_{heat} = C_{v}^{cal} \Delta T$$
And,
$$ \Delta U_{rcxtn} + \Delta  U_{heat} = 0$$
Or,
$$ \Delta U_{rcxtn} = - C_{v}^{cal} \Delta T$$
Now, consider the enthalpy change of reaction:
$$ \Delta H_{rcxtn} = \Delta U_{rcxtn} + \Delta (PV)$$
Now, if we were to assume our reactants as ideal gases a, then $PV=nRT$ and also considering constant temperature, we get:
$$ \Delta (PV) = \Delta (nRT) = RT \Delta n$$
Using the above result and expression for internal energy, we finally arrive at the following equation:
$$ \Delta H_{rcxtn} =  - C_{v}^{cal} \Delta T + RT \Delta n$$

Refernces:
Around 42:49 of this video
Page-2 of these notes


How many atoms are represented in this equation?

The referenced atoms on both sides of the arrow are the exact same atoms? I like your thinking!

But, chemical equations do not refer to individual atoms.

The formula $\ce{Pb(NO3)2}$ can be read as "A molecule of this substance consists of six atoms of O, two atoms of N, ...". That is correct.

But the equation $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$, strictly speaking, does not talk about single molecules. The numbers (including the 1's that are nor written) are called "stoichiometric coefficients". The equation means: "Any number of $\ce{Pb(NO3)2}$ molecules react with the double amount of $\ce{KI}$ molecules to yield ...".

Wikipedia supports my view:

  In lay terms, the stoichiometric coefficient (or stoichiometric number in the IUPAC nomenclature) of any given component is the number of molecules that participate in the reaction as written.

  ...

  In more technically precise terms, ...

... the stoichiometric coefficient is the number of molecules divided by the progress variable or extent of reaction.

  ...

  The (dimensionless) "units" may be taken to be molecules or moles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules.


Balancing Redox Equations - Half Reactions

You are right for the half reaction for oxidation is:
$$\ce{Zn(s) \rightarrow Zn^{2+} + 2e^-}$$
Now, you have to write properly the  half reaction for reduction:
 $$\ce{2H^+(\mathrm{aq}) + 2e^- \rightarrow H_2(g)}$$

This means, that the electrons given by zinc are taken by ion hydrogen.
So, if we add the two half reactions:
$$\ce{Zn(s) + 2H^+(\mathrm{aq}) \rightarrow Zn^{2+}(\mathrm{aq}) +  H_2 (g)}$$
Notice, that ion chloride doesn't change its oxidation state. Its role is to keep the medium neutral electrically. So, we'll add two ions chloride to the two members of the equation:
$$\ce{Zn(s) + 2(H^+_{\mathrm{aq}} + Cl^{-}_{\mathrm{aq}}) \rightarrow (Zn^{2+}_{\mathrm{aq}} +2 Cl^{-}_{\mathrm{aq}}) + H_2(g)}$$
Remember, redox reaction are reactions of electrons exchange between an oxidant and an reducer.


Both papers you refer to used methods which were basically state of the art at their time (Gaussian 98/03 compared to Gaussian 16 today). But that is almost twenty (or fifteen) years ago. Nowadays we thankfully have more developed methods available, and one should check whether the results found back then are still consistent. An (admittedly also a bit outdated) overview can be found here: DFT Functional Selection Criteria.

Long story short: You cannot reliably perform calculations on complexes where non-covalent interactions are dominant. You need to to use a method that treats dispersion at least at an empirically derived level. A popular example is Grimmes DFT-D3 program, which is also available in many ESS codes. (Starting point is The Journal of Chemical Physics 2010, 132 (15), 154104.) In the semi-empirical method xtb a newer version is already implemented.

For that reason I guess (since I only read the methods sections) they used MP2 for molecular structure optimisations, as it recovers dispersion interactions to some degree.

Whether or not DFT is suitable for your investigations, I cannot say. A thorough research of the recent literature will certainly offer far better insight.

The Pople basis sets, while an excellent example to understand the structure of them, are generally a terrible choice for almost everything practical. There are multiple basis sets available, which produce more consistent results, and are better optimised to perform in modern ESS codes. If implementations are missing from you ESS program of choice, you have a high chance to find it in the Basis Set Exchange Library.

For ORCA there are most (if not all) Ahlrichs' def2-variations are available, see the ORCA input library.
I would expect augmented basis sets (ma-def2-XVP or def2-XVPD) are necessary to investigate such complexes.


Which element has the highest binding energy per nucleon

Can't be expertly comment on the plot without knowing exactly how the plot was constructed. But I can make the following observations...

User Mithoron makes a good point that it depends on how the Y-axis of the plot is taken. You can plot average mass per nucleon in which case iron-56 is the heaviest atom per nucleon. However if you plot binding energy per nucleon, then nickel-62 has the highest binding energy per nucleon.
For elements with stable isotopes I'd guess only the stable ones are shown. For elements with unstable isotopes than some selection process is necessary so as not to clutter up the plot too much.
I'll point out that the "peak" is marked as "Iron Group." It isn't entirely clear what isotopes that means. Wikipedia notes that in astrophysics The iron group in astrophysics is the group of elements from chromium to nickel which are substantially more abundant in the universe than those that come after them – or immediately before them – in order of atomic number.
In terms of stellar production much more iron than nickel is produced. The iron  produced in stars is about about 92 % iron-56. Nickel is only about 3.6% Nickel-62.
I'd guess that the plot is a "best fit line." Here is a different plot showing more detail. You can see that not all isotopes fall on "the line."

Here is another plot at link pointed out by user IanBush in a comment. The "best fit" curve doesn't pass through all the data points.


Enthalpy of a system is a state function, meaning, it does not depend upon the path you take to reach the end state. What that means in this case is that the change in the enthalpy of combustion will be the heat required to raise the temperature of the combustion material to the new temperature and to increase the pressure. Essentially, think of it as if the combustion is occurring at STP, and you must construct a path from the beginning state to the end state which will pass that combustion at STP. In this case, that would be:

Initial state of reactants -> STP reactants
Heat of combustion at STP
STP products -> final state of products

Your final state will be the same as your initial state, 650.9 K/1779 kPa. The important point is that the heat in the change of the reactants to STP will be different from the change of the products back to the initial state. This difference would be reflected if you found a heat of combustion at some condition other than STP.

Nitrogen is a spectator to the reaction, so you can completely ignore it for the sake of reaction. However, you do need to account for raising it to the final pressure/temperature if either is changing. I'll assume you've got that under control.

Now, the issue comes in at the fact these are somewhat extreme conditions, at least in pressure. To calculate this properly would take a great deal of time and effort. That said, there may be corners you are capable of cutting in the process. The general process you need to follow is:

Take octane and $\ce{O2}$ from 1779 kPa to 1 atm, 650.9 K to $T_{1,\mathrm i}$.
Take octane and $\ce{O2}$ from $T_\mathrm i$ to 25 °C at constant pressure (using $C_p$).
Use the heat of combustion.
Take $\ce{H2O}$ and $\ce{CO2}$ from 1 atm to 1779 kPa, 25 °C to $T_{2,\mathrm i}$.
Take $\ce{H2O}$ and $\ce{CO2}$ from $T_{2,\mathrm i}$ to 650.9 K.

The overall enthalpy change of this process will be the heat of combustion at your conditions. Now, you'll have to decide for yourself about steps 1 and 4, as you need to be considering intermolecular interactions, and thus need to pick an equation of state (not ideal gas). I feel like I am not necessarily knowledgeable enough in this specific field to offer suggestions in that regard. I personally would just pick something like VDW, but there might be a specific standard for combustion processes. Further, only one book that I have access to right now that has the proper temperature-dependent-equation form for $C_p$, and it doesn't have one for octane. These are the ones for the other gases involved:

$\ce{O2}$: $C_p = 25.46 + 1.519 10^{-2} T - 0.715 10^{-5} T^2 + 1.311 10^{-9} T^3$

$\ce{CO2}$: $C_p = 22.243 + 5.977 10^{-2} T - 3.499 10^{-5} T^2 + 7.464 10^{-9} T^3$

$\ce{H2O}$: $C_p = 32.218 + 0.192 10^{-2} T + 1.055 10^{-5} T^2 - 3.953 10^{-9} T^3$

From Chemical and Engineering Thermodynamics (Sandler).

I wish I could help more. Hopefully this explanation at least points you in the right direction.


Shouldn't the pH at the equivalence point always be 7?

Two kinds of reactions have to be considered:

Neutralisation of the weak acid with the strong base
Secondary reactions from the products

The neutralisation is fairly straightforward:
$$\ce{HA + NaOH -> NaA + H2O} $$

However, and here comes number two from the list: It doesn't just stop there because we still have unreacted acid in the solution and lots of water. So what will happen? The conjugated base of the acid can also react (as a base):
$$\ce{A- + H2O <=> OH- + HA}\; ,$$

thus raising the pH of the solution. This depends on the equilibrium constant $\mathrm{K_b}$ of the conjugated base.

So although you have reacted exactly equal amounts of a (strong) base with the (weak) acid, you also have to consider the (weak) conjugate base of the acid, which also swims around in your titration soup.


Volume of nitrogen gas produced during taking off

Your next step is to identify the limiting reagent.

Observe that six moles of ammonium perchlorate react with ten moles of aluminum.  So by taking ratios you find that:

1)  The $10^5$ moles of ammonium perchlorate you started with react with approximately $1.7×10^5$ moles of aluminum.

2)  The $10^5$ moles of aluminum you started with react with $6×10^4$ moles of ammonium perchlorate.

Now there is a problem with (1) -- you have only $1.0×10^5$ moles of aluminum available so not all of the ammonium perchlorate can react as indicated in (1).  So we now look at (2) -- it indicates that all of the aluminum can react and you still have $4×10^4$ moles of ammonium perchlorate remaining.

Thus we can realize condition (2) rather than (1) because the amount of reaction is limited by the aluminum you have available.  Thus aluminum is the limiting reagent.

Now that you know the limiting reagent you can infer how much product is formed.  With $10^5$ moles of aluminum all being consumed, you should be able to see that $3×10^4$ moles of nitrogen molecules are formed as gas.  The volume under normal conditions should then follow.


Well as santiago has pointed out, they are all in fact metals.

The way you can determine if it is a metal or not is that when any metal reacts with acid, hydrogen gas is released. This is shown in this general equation (it is not balanced): $$\ce{M^{m+} + HCl -> MCl + H2}$$

Meanwhile, generally non-metals don't react with acid. This because when a substance reacts with acid, it electrons to the H+ ions created by the acid. Non-metals are generally electron-accepters, rather than electron donors so there is no way for them to donate electrons to the hydrogen ions in the acid.

In your experiment, the bubbles that formed when reacted are indication that hydrogen gas is being formed, therefore the substance is a metal. Therefore, you should be able to tell that Fe, Al and Zn are metals.

Therefore, you might assume that Sn and Cu have to be non-metals. However they are actually metals. The reason why no bubbling is observed is because these metals are unreactive and don't get oxidised by the acid.


DNA dye: Are there any natural dye for DNA (instead of artificial ones)?

Though not a naturally occurring dye, methylene blue has been used as a DNA stain. It is inexpensive and readily available. It is also comparatively nontoxic, and is used as a medication in pisciculture, so it can be found in pet stores.
If you choose to use methylene blue, save some for the blue bottle demonstration, a sensitive test for oxygen. Vitamin C or other reducing agents can be used instead of glucose.


One of the explanation for answering this question can be provided as, moving from lower to higher Fullerenes the number of atoms in the structure increase, increasing the number of atoms in the system results in reduction of the discreteness of energy states, which results in decrease of the gap between HOMO and LUMO, thus the band gap is reduced.
You can also check for the particle in box model of the Quantum Physics for better understanding.


Mechanism for oxidation of primary alcohols to carboxylic acids

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

  Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

The actual mechanism for the oxidation step is as follows (Clayden, p. 195):

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


Comparing the stability of geminal diols

Your reasoning is correct, but I think you'd mistakenly compared the ring strain of the rings, while I would have compared the ring strain in the corresponding ketones.
Note that, you are asked about the stability of gem-diols, which could convert into the corresponding ketone. The point is, the ring isn't going to change, so the stability of rings don't matter.
Now, as you know the corresponding cyclic ketones would be $\ce{sp^2}$ hybridized. So, the ring angle should be $120^\circ$ in order to minimize the ring strain in ketone. The most closest is in C and least is in A.
Hence, the order.


As a starting point, THC testing is advertised online for the following prices:

--Green Leaf Lab: $75

--Pharm Labs : $75

These companies both use HPLC. Without looking anything up, I'd guess that: The machine is around $100,000. It measures a sample every 30 minutes. (There might be modern marvels of engineering that run several samples in parallel and run way faster - idk.) Solvents and consumables are cheap. Maintaining the HPLC is expensive. You need a trained technician to prepare samples, run the analysis, and troubleshoot. That's expensive. You also need someone to deal with customers, which can be time-consuming as well.

You are not the only person thinking about THC testing. Google finds a handful of articles on what a business opportunity it is, and equipment manufacturers are ready to sell you stuff.

Good luck!


What are the molecular orbitals shaped like in diberyllium (mono)cation?

Short answer: the molecular orbitals will be $\sigma$ and $\sigma^*$ and thus cylindrically symmetric.

First off, your electron configuration is wrong. $\ce{Be2+}$ will have 3 electrons, so the $\ce{2p}$ orbitals need not be considered.

Here's a quick MO diagram:

So there aren't an $\pi$ bonds at all. There's a $\sigma$ and a $\sigma^*$ orbital, for a bond order of 0.5.

If I generate orbitals (using Avogadro) then I see something like this for the $\sigma$

And this is the $\sigma^*$


MO-Scheme of SCN- and its bonding properties when used as a ligand

One issue with SCN-, is that it can bind either at the nitrogen or at the sulfur. The way you presented the formula, $\ce{[Co(en)2(NCS)2]SCN}$, suggested that the SCN- ligand is bound via the nitrogen. If it were bound by the sulfur, the formula might be written  $\ce{[Co(en)2(SCN)2]SCN}$.

SCN- should have a similar MO diagram to OCN-, which is isoelectronic with CO2, so your idea to compare to CO2 is not  a bad one. However, since SCN- has a different symmetry group, you cannot trust that comparison too much. Your comparison to CO is also a good one, since SCN- and CO have the same point group.

However, what you really need to do is determine the symmetry group for $\ce{[Co(en)2(NCS)2]SCN}$. Then, assign symmetry labels to the SCN- ligands. The MO that is most important for SCN- is, qualitatively, the "lone pair" on the atom pointing at the metal center, which can form a "$\sigma$-bond" with cobalt. I would only consider using those $\pi$ orbitals on SCN- if the complex needs more electrons to get to 18. When you assign symmetry labels to the SCN- ligands, consider them as a pair: i.e. assign a symmetry label to the in-phase pair and then assign a symmetry label to the out-of-phase pair. The in-phase pair should probably have the same symmetry label as the cobalt $d_{z^2}$ orbital.


Why does Xe have a higher boiling point than Ne?

The main interaction between noble gases is Dispersion. (inert gas that's where the name comes from)
Dispersion is larger in heavier atoms. (Considerable larger, 18+18+8 more electrons difference is massive)

an example:
When Interaction energy passes zero point and thermal energy you can have the solid phase. This is why you will not have solid helium.


The charge on CH2- doesn't say it's electronegative. The direction of inductive effect is from a negatively charged atom to the rest of the uncharged molecule (if there's only one negatively charged atom in a given (canonical) structure) and the effect decreases as we go away from the negative charge. The electronegativity of carbon atoms in organic molecules is more closely related to their hybridization; sp-hybridized carbons are  the most electronegative, followed by sp2- and sp3- hybridized carbons. Thus, your textbook is right.


The mechanism is the same with the generation of the alkynyl anion as the abstract of this paper  ref 1 makes clear:

  Tetrabutylammonium triphenyldifluorosilicate (TBAT) can be employed as a fluoride source to cleave silicon−carbon bonds thus generating in situ carbanions that coupled with a variety of electrophiles, including aldehydes and ketones, in moderate to high yields. Among the examples reported is the first instance of fluoride-induced intermolecular coupling between allyltrimethylsilane and imine derivatives. Also, of particular note is the TBAT-initiated coupling of primary alkyl halides with allyltrimethylsilane. TBAT is an easily handled crystalline solid that has several advantages over tetrabutylammonium fluoride (TBAF) as a fluoride source; it is anhydrous, nonhygroscopic, soluble in most commonly used organic solvents, and less basic than TBAF.

Bear in mind that the $\mathrm{p}K_\mathrm{a}$ of an alkynyl proton is approximately 25 (ref 2) so this is a far more stable anion than any possible alternative. Clearly with $\ce{MeOH}$ as the solvent it will be quenched immediately.


Aluminium oxide can be isolated from feldspar by first acid digestion with a mixture of hydrofluoric acid and sulfuric acid (get rid of silicon by formation of silicon tetrafluoride), then digesting the residue with sodium-potassium carbonate, dissolving the obtained mass in dilute hydrochloric acid and then precipitating $\ce{Al(OH)3}$ by adding ammonia. The $\ce{Al(OH)3}$ can be calcined to aluminium oxide by heating to over 1100°C.

Note that the described procedure requires a suitably equipped laboratory .


What exactly is causing analyte protonation in electrospray ionization MS?

[ This is a short answer, as a complete answer would require going through multiple references and books on the topic. ]
First of all, the ESI process does not require the pH to be acidic : positive ESI is routinely done in ammonium acetate or ammonium bicarbonate media, which have a pH of 7.4 or 8.1. But indeed, to produce protonated analytes it indeed requires a source of protons. So what is the source of protons?
One should go back to the initial stage of ESI, which is the production of positively (or negatively) charged droplets under the influence of a strong electric field. Let's assume that we focus on positive ESI and that the droplets carry an overall positive charge. These charges are often simply represented as + signs, but their molecular nature is important. In the case were an electrolyte is present in high concentration and in its ionic form, the major charge carrier will be the positive charge of this electrolyte. So when considering an organic acid at low pH, the positive charges will be mostly solvated protons. If the electrolyte was ammonium acetate at neutral pH, it would be ammonium ions. Even at higher pH, if the electrolyte is ammonium hydroxide, the droplets would still bear a net positive charge borne by an excess of ammonium ions.
Then come the evaporation / Coulombic fission stages whereby smaller droplets are formed from the larger one, with an overall concentration step. The concentration effect of this stage will affect the overall pH of the droplet if initially acidic, but it is not required to lead to protonated species. The important point being that within the droplets, there is still an excess of positive charges, mostly in the form of solvated proton. At some stage, as concentrations of both the analyte and the solvated proton increase, the proton will be placed in close proximity with the analyte (both not being volatile do not evaporate). At this stage, the system is no more in the bulk solution thermodynamic conditions, but not really either in the gas phase thermodynamic conditions.
But we can still at this stage approximate the situation as "in the middle" by considering the thermodynamic parameters for both ends: on the bulk side, we have pKa of the analyte, with decreasing pH, and on the gas phase side, we have the proton affinity (PA) of the analyte that should be compared to the PA of the other species competing for the proton. The two scales are related but not strictly: for instance water has a quite low proton affinity (692 kJ/mole) compared to most polar molecules (ketones are in the 800 kJ/mole range). Thus is the proton located on the analyte is more stable than on molecules from the mobile phase, it will end up protonating the molecule, without considderation of the pKa.


Number of geometrical isomers when a cyclic ring is involved along with axis of symmetry

The answer in your textbook is correct. Here are all possible isomers:

I added a symmetry axis to the right, so you can see that these compounds are mirror images and are not superimposable. The formula you used for calculations is too complicated. The way to calculate it is simple: there are four stereogenic centers so: $$2^4=16$$There is one axis of symmetry so:$$16/2=8$$Have more faith in your books:)

EDIT:
Because my answer has raised an interesting discussion, I will try to present my point of view. First of all, the cis/trans isomerism is a descriptive way to represent a stereochemistry of a molecule, but its scope is limited. It is useful for teaching organic chemistry, but not for a systematic nomenclature (especially for a ring chirality). I think that the main source of confusion here is understanding the difference between a local and global (absolute) stereochemistry. In the above picture the controversy is whether the four enantiomers in the two rightmost columns can be treated as geometrical isomers. Let's simplify the problem and consider 1,2-dimethylcyclopentane:

The situation is clear for the first two isomers (a and b) - these are geometrical isomers, no doubts about it. What about b and c? We see that in both isomers the methyl groups are in the trans orientation, so one may conclude that they are not geometrical isomers. Period. However, cis/trans notation describes only the local geometry (E/Z notation obviously cannot be used in this example). Isomers b and c have the same relative orientation of methyl groups, but in a global scope these are two different molecules, and thus two different geometrical isomers. Therefore, 1,2-dimethylcyclopentane has one cis and two trans isomers. This may be counterintuitive at first, because there are no such situations in case of double bonds, where the cis/trans notation is more frequently used, but in rings we have an additional layer of complexity. I am aware that some people may disagree with this explanation, but as far as I know there are no official (i.e. IUPAC) directives to resolve this controversy.


When did the atom Theory Become indisputable?

Short Answer: Henry Moseley provided the first experimental evidence for the nuclear model of atoms through X-ray observations in 1913.

Long Answer:

Proving something is a rather nebulous task when there is no judge or jury and consensus is arrived at gradually over a century rather than instantaneously. Many scientists would consider the atom "proved" decades apart.

Avogadro was the first to distinguish atoms from molecules, which alleviated much confusion regarding atoms and allowed quantitative description/experimental evidence as relating to Advogadro's number (a mole). That was during the early decades of the 1800s (19th century)

Various scientists became convinced at various times over the next century.

Take Max Plank for an example. Around 1898 Plank became persuaded that the molecular approach was the only way forward for physics. He had formerly ascribed to the "pea soup" model where there were not discreet atoms. It was the molecular approach that allowed Plank to discover energy level quantization that is the foundation for quantum mechanics. E=hv

Lewis dot structures were some of the earliest electronic approaches to molecular geometry (circa 1910), but long before that there was an understanding of basic molecular geometries as demonstrated in the correct postulation of Benzene's hexagonal ring structure in 1865. Carbon was known to form 4 bonds with other carbons in the late 1800s. Linus Pauling was the first to calculate bond angles based on quantum mechanics in 1931.


Tl;DR - You do not have enough information.

I cannot help but think that this question is not answerable. You have four samples: Huy Fong Sriracha and three "knock-offs". There are six lanes on the plate. All six samples contain potassium sorbate.

If the question was something more like: You have five mysterious hot sauces and one of them is Sriracha... Then the TLC plate makes sense. One of those poorly labeled lanes is sriracha and the others are the mystery sauces. You just need to pick the one that gives the same pattern of spots with the same Rf values.

You don't have that. You have a mismatch between number of lanes and number of samples. You don't know which one is the authentic standard. All of the sauces contain the potassium sorbate (and they should - its a common preservative). You've determined the Rf values and that is all you can do. Well, maybe not all.
I can tell that lane 1 and lane 4 probably contain the same sauce and likewise lanes 2 and 5. Other than that, Jan's comment is correct. The potassium sorbate should always be at the same Rf (and since it is ionic, that likely means at the bottom of the TLC plate).


How to remove wash anthracene oil/creostae oil from stainless steel tanker?

Anthracene oil (CAS registry number: [90640-80-5]): Both European Chemical agency (ECHA) and US Toxicology Date Network (TOXNET) classify anthracene oil as:

A complex combination of polycyclic aromatic hydrocarbons obtained from coal tar having an approximate distillation range of $\pu{300 ^{\circ}C}$ to $\pu{400 ^{\circ}C}$ ($\pu{572 ^{\circ}F}$ to $\pu{752 ^{\circ}F}$). Composed primarily of phenanthrene, anthracene, and carbazole.

Anthracene oil  is also classified as a carcinogen the risks to human health and to the environment as well (though the relevant symbol is not issued).

Creosote oil (CAS registry number: [8001-58-9]): Creosote oil is a synonym for coal tar creosotes. It is a distillation product of coal tar. Creosote oil is an oily liquid consistency and range of color from yellowish-dark green to brown.  It is a mixed compound, composed primarily of aromatic hydrocarbons, anthracene, naphthalene, and phenanthrene derivatives. At least 75% of the coal tar creosote mixture is polyaromatic hydrocarbons.
Creosote oil is heavier than water (specific gravity: 1.07-1.08) and has a continuous boiling range beginning at about $\pu{200 ^{\circ}C}$  (range: $\pu{194\!-\! 400 ^{\circ}C}$).  They are slightly soluble in water, but miscible with alcohol and ether (Ref.1).
Koppers' Safety Data Sheet has following warnings for creosote oil:

Hazard Statements:

Causes eye irritation.
May cause cancer.
May cause respiratory irritation (No wonder your servicing person had severe respiratory problem when he went inside the tank).
Very toxic to aquatic life.
Toxic to aquatic life with long lasting effects.

It also listed following precautionary statements for prevention:

Avoid breathing dust/fume/gas/mist/vapors/spray.
Wash thoroughly after handling.
Wear protective gloves/protective clothing/eye protection/face protection.
Obtain special instructions before use.
Do not handle until all safety precautions have been read and understood.
Use only outdoors or in a well-ventilated area.
Avoid release to the environment.

I put all this effort to show that how dangerous your situation is and how ignorant you have been. Your first mistake is not receiving all special instructions for handling these kind of chemicals from responsible people (see above bold-face precaution). In your case, I assumed the manufacturer of the chemicals should provide these details. What you should have done now is go back to the manufacturer, and request the instruction for clean-up procedure safely. And be aware that whoever the person following that instructions, should wear with all personal protective equipment (PPE) required, specially suitable breathing equipment.
Alternatively, since the both oils are viscous, you may slant your tanks backward to make them drain out to safety container (for long time) before starting to wash them (assuming drainage is in the back). Once drained, a person wearing PPE can go inside and wipe out inner walls with cloths soaked with alcohol several times, since both oils are miscible in alcohol. Also, kerosene or gasoline or even coconut oil may work as well (please dispose those cloths accordingly). Finally, you should use appropriate detergent (or soap) with water to remove trace of remaining hazard chemicals (if there any) and washing organic material used.
References:

B. F Hibbs, J. Georges (Agency for Toxic Substances & Disease Registry), Toxicological Profile for Creosote (updates):  Wood Creosote, Coal Tar Cresote, Coal Tar, Coal Tar Pitch, and Coal Tar Pitch Volatiles; Forwarded by Dr. D. Satcher; Prepared for US Department of Health & Human Services, 1996.


Highest oxide of an element is the oxide compound of the element which has highest possible value of oxidation state of the element to which the oxygen is bonded. Particularly, the highest oxide of period 3 is chlorine heptoxide, $\ce{Cl2O7}$ in which chlorine's oxidation state is +VII.

In general, the highest oxide of transition metal which has oxidation value of + VIII is ruthenium tetroxide, $\ce{RuO4}$ and osmium tetroxide, $\ce{OsO4}$.


Structural determination of carbohydrates(sucrose)

First of all, lets be familiar with what per-iodic acid does to organic compounds.

It selects the carbon-carbon bonds that have either an -OH group, a =O(carbonyl) group or if it is the functional group carbon of a carboxylic acid. Both the carbons of the C-C bond must satisfy this condition. Here are some examples of the selection points of per-iodic acid:

Now that per-iodic acid has chosen the center, it cleaves the C-C bond and adds an -OH at each carbon. For every C-C bond, one per-iodic acid molecule is consumed. I am avoiding going throughout the actual mechanism to keep things simple. If any of the carbons has two -OH groups, they will spontaneously dehydrated to give a =O bond instead. For the examples shown above, the final products of the cleavage-oxidation would be:

Now, coming to your main question, the structure of sucrose would be:

By what we have seen, lets mark the C-C bonds that per-iodic acid can cleave:

Since we have 3 centers, so thats why 3 molecules of per-iodic acid are consumed per molecule of sucrose. Now you can even proceed and find out the oxidation products.

I hope this answer helps. Happy carbohydrate breaking!


How do you determine the hybridisation state of a coordinate complex?

Again, I feel a bit like a broken record. You should not use hybridisation to describe transition metal complexes.

Instead, you should discuss transition metal complexes using molecular orbital theory; a simplified version of it such as the crystal field theory may also suffice for simple complexes. Below, I have attached the orbital scheme of an octahedral coordination complex of a random transition metal not including π effects.

Figure 1: molecular orbital scheme of an octahedral $\ce{[ML6]}$ complex. Originally taken from Professor Klüfers’ coordination chemistry course’s web scriptum (in German).

Forcing hybridisation onto this system is very bad and very wrong; look at the huge energy differences between the 3d, 4s and 4p orbitals. Hybridisation would also lead us to assume six equal bonds, which is not the case: the six $\ce{M\bond{->}L}$ bonds transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ which leads to three different energies. And finally, hybridisation is not able to explain crystal field split, and why $\mathrm{e_g}$ — the orbitals that should take part in hybridisation and thus form bonds — are raised in energy (they are $\mathrm{e_g^*}$ in MO terms).

In most cases, coordination complexes would default to octahedral or derived. The main catch case is low-spin $\mathrm{d^8}$ in which case square-planar should be assumed. This is relatively independent of the type and number of ligands and most exceptions are due to sterics. Thus, octahedral complex is a safe first guess. Note that the only difference between two complexes of the same metal and ligands but with different metal oxidation states is typically whether a Jahn-Teller distortion is relevant and how strong it is.


Doubts about IUPAC nomenclature of 4-ethyl-1-fluoro-2-nitrobenzene

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

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Furthermore,

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; (…)

Therefore, the example is named as 4-ethyl-1-fluoro-2-nitrobenzene rather than 1-ethyl-4-fluoro-3-nitrobenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.


How can water be electrolysed if it's not ionic compound?

Pure water contains only small amounts of $\ce{H3O+}$ and $\ce{OH-}$ which are generated by autoprotolysis (the reaction given by Klaus Warzecha in his comment to your question). Reduction of $\ce{H3O+}$ at the cathode produces molecular hydrogen. At the anode, water is in turn oxidized, yielding molecular oxygen and $\ce{H3O+}$ (reference).

$$\mathrm{Cathode:}\ \ce{2H3O+ + 2e- \rightarrow H2 +2H2O}$$

$$\mathrm{Anode:}\ \ce{6H2O \rightarrow O2 + 4H3O+ + 4e-}$$

$$\mathrm{Overall\ reaction:}\ \ce{2H2O \rightarrow 2H2 + O2}$$


Photochemistry of beta, gamma-unsaturated ketones

Beta, gamma-unsaturated ketones can undergo Norrish photoreactions from the excited singlet state.  For example, in the case (1) below $\ce{R'}$ would be an allyl radical if we had started with a beta, gamma-unsaturated ketone.

From this solvent caged biradical several things can happen.  The acyl radical can decarbonylate (lose carbon monoxide) and then recombine with either end of the allyl radical to produce hydrocarbons.  Also the radical pair could simply recombine.  This would regenerate the starting material, or if the allyl radical has different substituents on each end, then it could recombine at the other end of the allyl radical to produce an isomeric beta, gamma unsaturated ketone.

However, the really interesting photochemistry observed with beta, gamma-unsaturated ketones occurs from the triplet state and is known as the oxa-di-pi-methane rearrangement.  The following figure (2) provides an example showing the "characteristic" 3-membered ring in the final product.

It is just the "oxo" analogue of the all-carbon photo di-pi-methane rearrangement.


What is the state of aggregation of iodine formed during electrolysis of molten nickel(II) iodide?

Basically, during the electrolysis process of molten nickel(II) iodide ($\ce{NiI2}$), $\ce{I– (l)}$ ions would be oxidized to $\ce{I2 (g)}$ at the anode, and $\ce{Ni^2+ (l)}$ ions would be reduced to $\ce{Ni (s)}$ at the cathode. My state assignments were based on following facts assuming the temperature of electrolytic cell is kept below $\pu{900 ^\circ C}$:

Melting point of $\ce{NiI2}$ is about $\pu{797 ^\circ C}$. Therefore, electrolytic cell must be kept above $\pu{800 ^\circ C}$. Thus following equlibrium will be maintained during electrolysis: $$\ce{NiI2 (l) <=> Ni^2+ (l) + 2I- (l)}$$
Boiling point of $\ce{I2}$ is about $\pu{134 ^\circ C}$. Hence, $\ce{I2}$ would be released as a gas at anode.
Melting point of $\ce{Ni^\circ}$ metal is about $\pu{1455 ^\circ C}$. Hence, $\ce{Ni}$ would be deposited as a solid at cathode. If you want to keep nickel metal in liquid form, your electrolytic cell must be kept above $\pu{1500 ^\circ C}$.

I think I have given enough information you to figure out the rest.


Electrolysis with baking soda solution, a penny as the cathode and graphite for the anode

Yes, I agree you are getting hydrogen an oxygen as you expected.
However, the copper anode is undergoing anodic corrosion and forming CuO (which is characteristically black) and possibly also some Cu2O (yellow or red, depending on the size of the particles, where the color combination could be described as orange).
Now, in the presence of HCO3- and O2, a naturally occurring mixed salt, copper patina (which forms on the copper roofs of houses) could also be forming. Wikipedia on patina, to quote:

patina is a coating of various chemical compounds such as oxides, carbonates,... formed on the surface during exposure to atmospheric elements (oxygen, rain, acid rain, carbon dioxide,...

Patina has been described as a blueish-greenish naturally forming on a copper surface (here, your copper penny electrode).
A closely related version of your electrolysis experiment has been previously discussed elsewhere.


Why add acid to rid of soluble carbonate or hydroxide impurities? Will it not also react with halide I'm testing for

Hydrogen halides are strong acids, meaning that their conjugate bases $\ce{X-}$ are extremely weak. Thus, adding $\ce{H+}$ will neutralize stronger bases such as $\ce{OH-}$ first before $\ce{X-}$.


You say that you have

10 mL 0.9% w/v NaCl solution

This is an unusual specification and as user M. Farooq pointed out it is ambiguous. So let's assume that this is a reagent that I am to add for some reaction. Is this:
$\quad$(1) 10 ml water (10 grams) and 0.09 grams of NaCl
$\quad$(2) 9.91 ml water (9.91 grams) and 0.09 grams NaCl
$\quad$(3) 10 ml aliquot of 99.1 ml water and 0.9 grams NaCl
Each of these would be slightly different. The gist is that (1) and (2) are different because the densities of the solutions will be slightly different as well as the volumes. (2) and (3) are different because the volumes will vary even though the densities are the same.
Pragmatically such reagent solutions are generally used in a large excess for whatever their use is, so the slight variations don't make any difference in the reaction. However it is just sloppy to give an ambiguous specification when a specific specification could and should be given.
Significant Figures
The other inconsistency with the question posed is with significant figures. The OP has specified 0.9% NaCl, but calculated "mole fractions of 76.43% H2O and 23.56% NaCl". You can't get four significant figures for the mole fraction when the NaCl concentration only has one significant figure.


Light is an electromagnetic wave. A metal has a large cloud of relatively free electrons (electrons that are loosely bound to the metal surface). When a beam of light is incident on a metal surface, it polarizes the electron cloud, i.e. some regions on the metal become relative more "positive" while some regions relatively become more "negative". Thus, this induces a field which makes the electrons to start oscillating. This oscillation generates another electromagnetic wave which opposes the incident radiation, (An ideal metal will completely oppose the incident light radiations), and hence our incident light rays get reflected.


Solvents for tartrates / tannins that won't attack glass

The CRC Handbook indicates that potassium d-tartrate and dl-tartrate are soluble to the extent of 100-150 grams per 100 mL H2O vs about 0.5 grams for the bitartrate. Rather than using acid to clean the decanter, alkali would neutralize the bitartrate to the much more soluble tartrate . Washing soda (Na2CO3) should not be too alkaline, but even sodium bicarbonate (NaHCO3) might work. Neither is expected to be harmful at the level that might exist after rinsing.

The tannins or other organic might slow things down. Bleach (NaClO) is alkaline, and although bleach is toxic if ingested, it should present no problem if the decanter is rinsed. Water could be flushed around inside the decanter, then tested with litmus paper. Or the decanter could be rinsed with vinegar and a little honey to remove the bleach: the vinegar makes the liquid acidic (HClO) and the honey gets oxidized to gluconic acid, consuming all the bleach - leaving a sweet, tart taste.


How does chromium(III) react to form chromium(VI) with hydrogen peroxide?

The hydrogen peroxide is a strong oxidiser. In basic conditions, this is partly due to $$\ce{H2O2->H2O + O\ \ }[*].$$ The monooxygen is very eager to gain electrons. Therefore, $$\ce{Cr^{III}->[\ce{[O]}]Cr^{VI} +3e-}.$$Chromium(III) hydroxide$^{[**]}$ is amphoteric, it reacts with bases, too.

$$\ce{Cr(OH)3 + 3NaOH <=>Na3[Cr(OH)6]}$$

This complex is then oxidised. For common chromium(VI) compounds, one would expect to see the equilibrium $$\ce{2CrO4^{2}^- <=>[\ \ \ce{H3O+}][\ \ \ce{OH-}]Cr2O7^{2}^{-}} \ \ \text{[***].}$$

Thus, in (strongly) basic conditions, the chromate ion $\ce{CrO4^{2}^-}$ would be favoured.

Extra

One thing that might be confusing are the equations themselves. In other words, they simply show

the most likely process (not always the case),
the end result (there are exceptions),
the formulae are not true to the actual structure (cannot be sure).

To make inorganic chemistry more tolerable, it is best to accept that there are multiple reactions occurring concurrently. Then it is easier to arrive at the "correct" end result by taking into account reaction conditions and stability.

[*] This is a simplified explanation taken only as a model for basic understanding. The actual decomposition mechanism is more complicated, depending on pH, other ions in solution, etc. For more information on specific situations, look at Kinetic Studies and Mechanism of Hydrogen Peroxide Catalytic Decomposition [---]. There you can find many other great references as well.

[**] Chromium(III) hydroxide might actually be $\ce{CrOOH*H2O}$, or $\ce{Cr2O3*}$ $n$$\ce{H2O}$. Even if we were to be dealing with the hydroxide, it actually comprises of $\ce{Cr(OH)6}$ octahedrons, and is claimed to be inactive (active form is the oxide hydrate). [N. S. Ahmetov's Inorganic chemistry]

I would make the claim that the hydroxide does not actually exist at all, and is, like for iron, mistaken for the oxidehydroxide hydrate. Furthermore, the reaction might not be direct, it could be a case of Le Châtelier – Braun principle.

[***] The overall transition is $$\ce{2CrO4^{2}^{-} +2H3O+ <=>Cr2O7^{2}^{-} +3H2O}$$

from where it is easy to see that adding hydroxide ions tilts the equilibrium to the left. Also, you might wish to look up different aqueous complexes of chromium and the structures of the ions above. That also helps to visualise the process.

Suggestion: Chromium complexes tend to polymerise ;).

Even if $\ce{CrO3}$ would form, as you suggested, it reacts rather well with water via

$$\ce{CrO3 + H2O \rightarrow H2CrO4, \\ 2CrO3 + H2O \rightarrow H2Cr2O7,}$$

and hydroxides:

$$\ce{CrO3 + 2NaOH \rightarrow Na2CrO4 + H2O}.$$


Electrolysis of salt water into sodium hydroxide with YSP

If you want sodium hydroxide for soap making you should just buy some: it is a cheap and common chemical. This is because there are effective industrial processes to make it in vast quantities.

Trying to make it yourself is not easy and will cost you far more effort than just buying it. The least of your problems will be what happens to contaminants like the hexacyanoferrate (which is present in such low concentrations it isn't really significant). Have a look at the way industry makes bulk hydroxide and chlorine (e.g. here); it isn't as simple as just electrolysing brine. All the processes involve sophisticated ways to separate the products produced at the different electrodes. You don't get decent quantities of either product if you don't maintain this separation (one method uses flowing mercury; another expensive ion-exchange membranes). In addition to this, uncontrolled electrolysis is dangerous as it produces toxic and explosive gases. You really don't want to try it at home.

Since the most basic components for an effective reaction will cost you far more than the couple of dollars a large bottle of pure sodium hydroxide will cost, buy the chemical and ignore the electrolysis as a route to what you want.


TL;DR. Are you asking a question about energy (thermodynamics) or a question about rate? What possible relation is there between boiling HCl(aq) and HF at room temperature as far as rate? I see no value in comparing those two reactions, but I'm probably missing something. So, if I understand you correctly, you are claiming that once you see some "purple" color, that therefore the Ti° is "bare", right? That, of course, doesn't (necessarily) follow. I'm not familiar with the reaction of Ti° + H+ → Ti(IV) + H2. (I'm assuming the +4 oxidation state) but almost certainly (I'd guess) it is not a single step reaction. Where the intermediate steps occur and how the presence of F- or Cl- ions impact one or more of those steps is the question. They could be absorbed on the metal surface, form complexes with intermediates, compete with oxide or hydroxide or oxyhalide species. It seems to me that the slow reaction rate of HCl is telling you that there continues to be something (some barrier) on the surface of the Ti° (by "continues" I don't mean to imply that it was there prior to the oxide films dissolution). About the only thing the energy can tell you about a reaction is that if its exothermic then you'd expect hotter and if hotter then faster (but there are (fairly rare) exceptions to this where hotter means slower). Since the value of comparing reaction A at 25°C to reaction B at 100°C isn't particularly informative (except if B is slower at 100° than A at RT, then we'd predict A will be even faster that B at 100° and B be even slower at RT) - especially if the reaction conditions are so different that we can't assume the mixing is the same (viscosity, acid dissociation, diffusion (turbulence?, H2 behavior?) all will be substantially different). Again, I'm not familiar with this reaction, but since it is a redox reaction, you can't necessarily even assume that the reduction is occurring near (on the molecular scale) to the oxidation, electrochemical cell formation is another possible complication... Could be complicated, IDK.


How to make a CP2K input file with Avogadro?

If you look at the CP2K website, you'll see that the coordinates are only a small fraction of the input file. Most of it is text describing how a calculation should be run.

The key parts from a coordinates perspective look like this:

&CELL
  ABC 12.4138 12.4138 12.4138
&END CELL
&COORD
  O      12.235322       1.376642      10.869880
  H      12.415139       2.233125      11.257611
  H      11.922476       1.573799       9.986994
&END COORD

As it looks, these are the Cartesian coordinates of the atoms (in regular space) and the unit cell vectors. You can either save the file in .XYZ format or use an input generator like GAMESS or Gaussian to get these coordinates.

If you want a periodic boundary calculation, you just need to estimate the bounding box. You can do this manually, or you can go to Crystallography | Add Unit Cell and then tweak the volume. In Avogadro v1.1.x, you'll see the ABC lattice vectors needed for that CP2K block.

As it might seem, Avogadro2 will be v2.0 of Avogadro. It's still in development.


Why the bond angle of PH3 is lesser that that of PF3?

VSEPR works by accident in the cases of $\ce{NH3}$ and $\ce{NF3}$. In reality, there are much more things to consider as shown in this answer. All four compounds should have bond angles of $90^\circ$ if there were no other effects present.

For both nitrogen compounds, the effects are the short $\ce{N-X}$ bonds which lead to steric clash of the three substituents. Thus, the bond angle is broadened from the original, hypothetical $90^\circ$ to whereever it may seem comfortable to the molecule. Since fluorines are larger atoms than hydrogens are, the $\ce{N-F}$ bond is longer than the $\ce{N-H}$ bond ($137~\mathrm{pm}$ versus $102~\mathrm{pm}$) and therefore the atoms can agree on a smaller bond angle (less crowding).

For phosphorous, we can initially assume that the bond angle of $\ce{PH3}$ is close enough to $90^\circ$ for no hybridisation to be necessary (semi-proven by this answer of Martin) and likewise for $\ce{PF3}$. The bond lengths are $142~\mathrm{pm}$ and $156~\mathrm{pm}$, respectively. The difference in bond lengths is only half of that of the nitrogen compounds ($14$ versus $35~\mathrm{pm}$). Therefore, we can assume the bond length to be mainly dictated by the central phosphorous atom, while the bond angles must be dictated more by the outer atoms. Since again fluorine is larger than hydrogen, this time the better way to stabilise the molecule is not to lengthen the bonds but to expand the angle.


What are the fundamental parameter by which lubricants are graded?

Lubricants are usually graded for a host of different properties, but ultimately what you are looking for is it's thermal stability, and a viscosity index related to the temperature your mills parts run at. These properties are optimized for different parts of the machine because they run at different speeds and as a consequence temperatures. I could delve into more, but I would need more specifics about the machine itself. If you wish to pursue the problem further on your own I would suggest This website as a reference. Also take a look at this cautionary article regarding accurate measurements on viscosity index as it can be a bit tricky.

The consequence of using the wrong lubricants varies based on how wrong you are... But ultimately the lubricant will thin out to the point that is does not decrease friction and or cool the parts it is lubricating if it degrades at the temperature the part runs at.


The product is optically active compound. However reaction is not stereoselective and therefore both stereoisomers are formed without one being in excess to the other. Product is a racemic mixture of both isomers and therefore should not expose optical activity because effects of both isomers just cancel out.

Reagents or substrate must be optically active in first place in order to get optically active mixture of products.


Does a diatomic gas have one or two vibrational degrees of freedom?

I think this confusion arises from quite different definitions of degrees of freedom. Here are three different definitions that could reasonably arise:

Positional degrees of freedom. This leads to the common $3N$ degrees of freedom based on movement in three directions. Refer to Jan's answer. (Question: Bonds constrain atoms, so why don't they reduce the number of degrees of freedom? Answer: Bonds are not rigid constraints; the atoms don't necessarily remain a fixed distance apart.)
Phase space degrees of freedom. Here we have $6N$ degrees of freedom: $3N$ from the positions of all particles, $3N$ from the velocities of all particles. Positions and velocities are independent, so to specify a system of $N$ particles exactly requires $6$ degrees of freedom per particle. (We also presume no internal structure; as example, for rigid bodies, there are $12$ degrees of freedom: the $6$ mentioned above, $3$ more for specifiying the orientation of the body, $3$ more for describing the speed of rotation of the body.)
Quadratic degrees of freedom. This may seem like a strange distinction to make, but is a key requirement of the equipartition theorem, which is one of the areas where the concept of degrees of freedom is first introduced. These degrees of freedom are the subset of the phase space degrees of freedom whose elements all contribute quadratically to the system energy.

We will clarify these degrees of freedom with the example of a diatomic gas. In this case, there are $6$ positional degrees of freedom, $12$ phase space degrees of freedom, and $7$ quadratic degrees of freedom. Jan's answer describes the $6$ positional degrees of freedom well; specifying the $6$ analogous velocity degrees of freedom gives the $12$ phase space degrees of freedom. Let us enumerate the $7$ quadratic degrees of freedom: $$\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}, q, \dot{q} \quad \Longleftrightarrow \quad \frac{1}{2}m\dot{x}^2, \frac{1}{2}m\dot{y}^2, \frac{1}{2}m\dot{z}^2, \frac{1}{2}I\dot{\theta}^2, \frac{1}{2}I\dot{\phi}^2, \frac{1}{2}kq^2, \frac{1}{2}\mu\dot{q}^2$$ Positional degrees of freedom do not play into this, because there is no quadratic energy contribution associated with them. Instead, we have the six velocity degrees of freedom from the two atoms, $\{\dot{x}_1, \dot{y}_1, \dot{z}_1, \dot{x}_2, \dot{y}_2, \dot{z}_2\}$, which we replace by the conventional center-of-mass coordinates $x, y, z$ and the two angles $\theta, \phi$ that describe the orientation of the molecule, $\{\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}\}$. One degree of freedom has "gone missing", because by symmetry there is no (quadratic) energy associated with rotation about the bond axis. The bond also produces two more quadratic degrees of freedom, represented by the vibrational coordinate along the bond, $q$, corresponding to the potential and kinetic energies of vibration.

Hence in one sense there exists only one vibrational degree of freedom; in another sense, two.

Remark. One might argue that the two vibrational degrees of freedom $q, \dot{q}$ are not independent, as they should be related by the equation of motion for a harmonic oscillator; this is important for the statement of the equipartition theorem. This is true for an isolated system, but presumably collisions decorrelate $q$ and $\dot{q}$---I don't have a better answer for this.

At any rate, experimental evidence (heat capacities) does corroborate seven quadratic degrees of freedom for diatomic molecules at high enough temperatures, when vibrational modes are active.


How can a dilute acid dissociate more than a concentrated one? (Ostwald dilution law)

A dilute acid has more solvent molecules to interact with than a concentrated acid, and thereby has more frequent collisions with the solvent. You can think of the dilute acid as "drowning" in the solvent.

A concentrated acid, by contrast, interacts more with other acid molecules, and therefore less with the solvent which dissolves it.

A concentrated acid will typically have more total dissolved molecules than a diluted acid simply because there are more of them, but it will have less as a fraction of total number of acid molecules; i.e. a weaker affinity for dissociation.

In short, $\alpha$ describes affinity for dissociation, not total dissociation. "Better", not "more".

Note that a higher $K_a$ corresponds to a stronger acid, so it makes sense that stronger acids will have a higher affinity for dissociation.


Why do certain groups withdraw/ donate electron to double bond or conjugated system in resonance effect?

Generally, it is observed that if there are lone pair(s) present on the atom of function group directly attached to to the conjugated system, then $+R$ effect is observed.

If the same atom has double bonds or dative bonds with another atom, then $-R$ effect is observed.

Example: –$\ce{OH}$ shows +R due to 2 lone pairs on Oxygen.
— $\ce{NO2}$ shows -R due to one double bonds between Nitrogen and Oxygen, and one dative bond from Nitrogen to another Oxygen.


Changes in freezing point of sodium acetate

My assumption  is that the addition of water (other than water of hydration) lowers the temperature - consider that the heat of crystallization is shared among the sodium acetate, the added water and even the packaging. Perhaps the manufacturer adds water to lessen the chance of burns.

You'd need to experiment with lab-grade chemicals to be sure, such as this ~99% pure sodium acetate trihydrate.


What mixing ratio of ethanol and acetone has the lowest freezing point?

its me again, I  want to post the answer I found because I think it might be useful for somebody in the future; it appears a 2:3 molar acetone:methanol proportion gives the lowest freezing temperature ~158 K (see attached picture below).

It appears the difficulty finding these curves was because they are usually called "solid-liquid equilibrium" curves instead of "freezing point curves", as its the same physical situation (Explanation Here)

There is a large database of these in the following link.

EDIT: As commented by "Ivan Neretin", ethanol-acetone mixtures are also good for low freezing temperatures:


Is it possible to have electric field in water without having electrolysis?

Well, if the voltage is lower than needed for any possible electrolysis reaction, there would be no reaction, just the field. So the answer may seem to be yes.

Now, in fact there is a catch. Once you turn on the voltage, even if the reaction is not possible, the ions will rush to the respective electrodes and form the so-called electric double layer. Its thickness, known as Debye length, is usually quite small. As for the rest of the solution, it will be effectively shielded from the electric field.

Whether or not this counts as a positive answer is up to you.


How is Cu(OH)2 a base if it can't dissolve in water?

Copper (II) hydroxide has some small solubility in water, determined by its solubility product constant.  More importantly, it will dissolve freely and act as a base if a suitable acid, such as acetic acid, is is available to be neutralized.  See also the comment from @f".


Does hydrazine crystallize when freezing?

I would take a different position on answering the question, and look at the water-hydrazine binary phase diagram. Looking at J. A. McMillan and S.C. Los, J. Chemical Physics 42(1) 160-161 (1965) one finds the experimentally determined phase diagram. With enough water in the mix, one proceeds smoothly from the pure hydrazine melting point down to a slight eutectic near the monohydride, but only 3K lower in temperature.

Further, the same authors have another paper, J. Chem Phys 42(3) 829-834 (1965)  that discusses the nonequilibrium phase transformations. The first sentence in the abstract says

  Although neither hydrazine nor water supercool to a large extent, their mixtures do. Aqueous solutions of
  hydrazine may be strongly supercooled and stabilized as rigid glasses.

In the introductory paragraph, one also finds

  Since the phase diagram of this system
  is known, it was tempting to look for a correlation
  between the irreversible processes of crystallization and
  the equilibrium phase transformations. The hydrazine-water
  system is particularly useful for this purpose,
  since crystallization can indeed be induced in the entire
  range of concentration. Samples of different hydrazine
  content were then supercooled until glass stabilization
  occurred and then warmed up at different rates. The
  processes of glass transformation and crystallization
  were studied, and values of the free-energy barriers
  could be obtained for the latter process.

The rest of this second paper is devoted to measuring and describing thermodynamically the crystallization curves at different water-hydrazine compostions.

So, yes, you can definitely get a glass, if cooled quickly enough, but it just might decide to phase separate on you if it warms up a bit.

For the case of (only) hydrazine-1,1-dimethylhydrazine, the phase diagram is covered in J. A. McMillan, Journal of Chemical and Engineering Data 12(1) 39-40 (1967). This shows a very slow decrease in the liquidus temperature as 1,1-dimethylhydrazine is added, with the eutectic occurring at 94% mole percent 1,1-dimethylhydrazine at 214K, only 61K below the pure hydrazine melting point. There are no compounds. Further, the article states

  Unlike the systems of the components with
  water, the mixture hydrazine-1,1-dimethylhydrazine
  does not supercool

which indicates that phase separation and crystallization of the two components will readily occur. On the other hand, the samples were sealed in glass under nitrogen, and if kept in the dark were stable over a 6 month period.


Calculating the ionization energy for hydrogen

I assume you refer to Bohr's atomic model.

The energy of an electron in Bohr’s orbit of Hydrogen atom is given by the expression:

$$
\begin{align}
E_{n} &= \frac{2\pi^{2}me^{4}Z^{2}}{n^{2}h^{2}\left(4\pi\epsilon_{0}\right)^{2}} \\
&= \color{\navy}{-13.6\frac{Z^2}{n^2}\pu{ eV}}
\end{align}
$$

Since $Z = 1$ for hydrogen, the above equation can be further simplified. Now just plug in the required values.


Defintions

Occur on Earth differentiates between all isotopes in the universe and isotopes which exist or have existed on Earth. The passage identifies that there are a total of 308 isotopes occurring on Earth, but it is worth noting that there are more than 308 total isotopes in the universe. This discrepancy is because some isotopes have extremely short half-lives, making their total "lifetimes" extremely short. As an example, an isotope generated in a supernova which has a half-life of 1 second will have totally degraded in at most 32 seconds (32 half-lives degrades the sample completely no matter the original amount). That is not nearly enough time for the isotope to travel from the exploded star to Earth, so that isotope does not occur on Earth (and it never can). This hypothetical supernova isotope would exist but it would not be part of the 308 Earth occurring isotopes.

Extinct means that the isotope is no longer naturally occurring within a given environment. Isotopes have differing half-lives, and a currently extinct isotope would have degraded so completely as to no longer be present at all. The isotopes would have originally been seeded into the environment by processes like supernovas, which produce ejecta containing the elements produced by nuclear fission and nuclear fusion. When that ejecta reaches the Earth the isotopes within it begin to occur on Earth, but eventually those isotopes will disappear due to half-life degradation. There are processes which can replace degraded isotopes in a self-regenerating manner and thus those isotopes never go extinct due to self-regeneration. An example of this type of self-regenerating isotope is Carbon-14; in fact, Carbon-14 will likely never become extinct. Self-regeneration is not the only process to replenish isotopes; the aforementioned supernovas are another. Unlike self-regeneration, which is a continuous process in which degraded isotopes continuously get replaced, supernovas deposit isotopes intermittently and comparatively infrequently. In fact, if a supernova deposit is delayed for an extended amount of time, isotopes which come only from supernovas can actually disappear from Earth entirely- they would become extinct. These extinct isotopes would stay extinct on Earth until a supernova redeposited the isotope.

Clarifying the Passage

As far as your question about an extinct isotope occurring on Earth- for the isotope to be extinct it cannot be currently occurring on Earth, but it was present in the past (logically, if an isotope never appeared on Earth it couldn't be part of the 308 isotope total, so for some of that 308 total to be extinct means they were present on Earth in the past). That question is sort of unrelated to the passage however, but the passage was written poorly enough that I think you misunderstood that they were trying to say that 121 isotopes should hypothetically be extinct, but that some of them are not actually extinct. To reduce the confusion the passage is causing, I have rewritten it to be far more clear: "Among all the elements that have have ever occurred or currently occur on Earth there are 308 total isotopes. Those isotopes are split into 150 stable isotopes and 158 unstable isotopes. Of the 158 unstable isotopes, 121 should be or currently are extinct, however, there are isotopes not actually extinct because they are constantly renewed, like carbon-14."


Why is the solid product washed with ice-cold solvent during in the last stages of recrystallisation?

Possible sources of impurities in the recrystallization process can be described in terms of phenomena occuring during coprecipitation and which are the basis of gravimetry:

inclusion (impurity occupies positions in the crystal lattice);
occlusion (impurity is physically captured within the crystal as it has been growing);
adsorption (impurity is adsorbed on the surface layer of the crystal);
postprecipitation (impurity forms secondary solid phase once recrystallization of the target stopped/slowed down).

Washing crystals with a fluid only addresses the last two phenomena, adsorbance and postprecipitation. Postprecipitation is rarely a problem as it occurs on a large time scale or when a really poor solvent has been chosen.

Properties of the washing liquid vastly depend on the properties of the compound being recrystallized. In the simplest case it's the same solvent (as in your example), but it can also be an electrolyte, hydrolysis suppressor or another solvent (e.g. distilled water) altogether.

As for the temperature of the washing liquid, it's actually better to use many small portions of the hot solvent as filtration of the hot solutions happens much faster, surface adsorption decreases with the temperature and removal of mother solution from the surface is faster. However, if solubility of the recrystallized product is significantly increasing with the temperature (and this is typically the case), then one has to sacrifice the efficiency of the washing process for the better yield and use cold solvent instead.

Impurities, insoluble in the cold solvent, are practically not an issue. First, recrystallization starts from a homogeneous solution. Second, before crystals are washed, they are separated from the mother solution so that only minor amount of it is left on the crystals' surface, and if there were any impurities, they don't get a chance to nucleate as their concentration in washing solvent is going to be negligible.

As for your perception of the recrystallization as of a "weird" process, I'd say it's probably more complicated than one would expect from looking at the beaker with some crystalline mess. In fact, underlying processes are described with more or less appropriate models, and once a person
 gets experienced enough, it all starts to make much more sense.


The "soap molecules" are so called amphiphiles: They've got one end that likes water ("polar" or "lipophobic") and the other ("lipophilic" or "fat-loving") doesn't.

These amphiphile molecules like to form a single-molecule thick film on water, all standing up in an orderly fashion (foot in the water), so if you put a drop of it on a water surface, it spreads very wide, until the film is monomolecular.

Or, until something pushes back. In your case the pepper, when its particles are pushed together so they also form a tightly packed surface. Waiting longer, the soap will likely act as a detergent, wrap the pepper particles in a bilayer, and drag them under the surface.

Search for, e.g., surfactants, detergents for some pictures.

As to your last question: Some water gets dragged along with the surfactant, but no, there is no large lateral movement of water on the surface. The spreading front you see is just the soap.


How to determine methane percentage at home?

A couple of ideas:

If you think the diluent gas is non-flammable, you should be able to use the adiabatic flame temperature of the gas when you burn it in air to estimate the composition.  You will need to carefully adjust your stove to give the optimal fuel/air blend (that maximizes the flame temperature).  Then you will need to accurately measure the flame temperature.  If you assume a heat capacity for the unknown diluent, you can estimate its concentration in the gas by comparing the measured temperature to the theoretical adiabatic flame temperature for pure propane burning in air.
Use a vacuum pump to empty a container of known volume, and then allow the cooking gas into the container at ~1 atm pressure.  Measure the density of the gas and compare to the density of pure propane.
If you think the diluent gas does not contain carbon: hook up flow meters to the gas bottle (and an air or oxygen stream) to form a torch flame with known gas flow rate.  Collect all of the the exhaust produced by the flame, pass it through a cold trap to freeze out water vapor, and then pass the cold flame exhaust over a bed of anhydrous sodium hydroxide of known initial mass.  It will absorb the carbon dioxide from the exhaust.  Measure the mass increase of the sodium hydroxide.  By comparing the moles of gas that flow into the flame to the moles of CO2 absorbed (as calculated from the mass increase) you can estimate gas purity.
Find a spectrometer and record the IR or Raman spectrum of the gas!  This will probably be more accurate than any of the other methods.


proving an unknown organic compound using reactions

The task you had being assigned is called identification by derivatization. It's one of the strongest proof for the confirmation of the structure of a molecule.
If you have an hard-to-identify compound X, you can make it react with Y and then analyze the product of the reaction, X-Y, instead.

I once synthesized a polysubstituted aromatic compound and wasn't sure whether or not a iodine atom was present on the ring. So I performed a Heck reaction on the molecule obtaining the corresponding substituted olefin, which was a known compound, present in the literature and could be identified by its NMR spectra.

Chlorobenzene is quite an anonymous compound and its derivatization does not lead to significative compounds. Just an example to answer your question, you can perform a nitration to get o-chloronitrobenzene which is a solid and can be identified by its melting point.


Why doesn't frozen sugar solution taste sweet?

Where is the sugar?

When you freeze a dilute aqueous sugar solution pure water freezes first, leaving a more concentrated solution until you reach a high concentration of sugar called the eutectic concentration.  Now you have the pure water that's frozen out, called proeutectic water, and the concentrated eutectic sugar solution from which the sugar is finally ready to freeze along with the water.  Upon freezing this eutectic composition forms a two-phase eutectic mixture, in which the sugar may appear as veins or lamellae (like veins of some ores among Earth's rocks, though these typically form form a different process).  If that structure is in the interior of the ice cube, likely since you cooled the solution from the outside, then licking the outside you got only the pure water proeutectic component.

See https://en.wikipedia.org/wiki/Eutectic_system for more about this process.

Addendum:  I tried this with store-bought fruit juice which was red in color.  Poured it into an ice tray and froze it overnight in my household freezer.  It appeared to be a homogeneous red mass and tasted sweet, but was also mushy implying some liquid was still present (after overnight freezing for an ice cube sized sample).  The juice was roughly 10% sugar by weight.


What is the refractive index of poly(methacrylamide)?

As reported in  Gooch JW (Jan W. Encyclopedic Dictionary of Polymers. Springer; 2011. doi:10.1007/978-1-4419-6247-8. Poly(N-methyl methacrylamide) has a refractive index of 1.5398


How may copper acetate ligands be manipulated to change colors?

I may have accidentally created copper cloride in addition to the copper acetate. I decided to just let the solution (made form Copper, vinegar, and salt) evaporate with a small fan speeding the evaporation for the last few weeks. The tallest copper acetate crystals started breaking the surface and then the surface of the liquid started to form a skin (pic 1) so I poured off the now green liquid into another jar. Copper acetate is blue, which is why the green solution seems to suggest that there is copper cloride.

Evaporation! Note the blue-green layer of copper carbonate below the dark copper acetate crystals.

Top View

The green liquid on the left is what was left from evaporation (and I think is copper chloride) - next to another jar of copper acetate.

The copper acetate crystals were covered in a thin layer of copper carbonate because last week I disturbed the solution and stirred up a cloud of copper carbonate accidantally. I rinsed them with vinegar and they cleaned nicely.

The largest crystal was just over 1 cm wide. These crystals were grown from a solution that started evaporating 5 years ago.

The copper carbonate goo in the bottom needed a use so I added more vinegar to it and it formed the copper acetate solution three pictures up (next to the green copper chloride).


Does intramolecular hydrogen bonding occur with carbonic acid and geminal diols?

Is this possible?

Probably not possible. An elementary explanation would be "too much ring strain."

https://en.wikipedia.org/wiki/Ring_strain

  I'm a bit confused: What are the requisites to occur a intramolecular hydrogen bonding?

5 and 6-membered rings are more favorable. As a matter of fact, beta-ketoacids, when warmed, readily decarboxylate. Why? The intermediate to decarboxylation is stabilized through hydrogen bonding.

Note the 6-membered ring in the picture below.


Can a product be alcohol free while containing isopropyl myristate or isopropyl palmitate?

Isopropyl myristate and palmitate are esters formed from isopropyl alcohol and myristic and palmitic acids. These products will have been purified before being used in the cosmetics to remove any residual isopropyl alcohol. If the labelling does not mention isopropyl alcohol as a component then it will not be present.

Note. I don't know your reason for wanting to know about the presence of isopropyl alcohol but if you are concerned for ethical reasons about the presence of alcohol then you should be aware that what is generally referred to as 'alcohol' is ethanol. Isopropyl alcohol is a different substance.


Amphiphiles spontaneously form a variety of self-assemblies in water, such as
micelles, vesicles, and bilayers. The simplest form of self-assembly for single chain amphiphiles are micelles.

The concept of packing parameter proposed by J. Israelachvili allows a simple
and intuitive insight into the self-assembled aggregates of amphiphiles in water

The packing parameter $P$ is defined as:
$$P = \frac v{ l \times a_0}$$

where $v$ and $l$ are the volume and length of the surfactant hydrophobic tail, respectively, and $a_0$ is the optimal surface area per amphiphile.

Let me briefly explain the concept of optimal surface area per amphiphile ($a_0$). It is based on the principle of opposing forces; the hydrophobic interactions act to pack the soap molecules closer together to avoid exposing their hydrophobic tails to water, while the repulsive energy between the charged headgroups of the soap molecules increases as molecules are pushed closer together. The surface energy of the self-assembly is minimized when the amphiphilic molecules occupy the optimal area.

Single-tailed amphiphiles (like your soap molecules) have a large head-group area and have a packing parameter $P < \frac13$. Thus, they form spherical micelles. Ionic surfactants such as sodium dodecyl sulphate (SDS/NaDS) generally forms spherical micelles in solution at low salt concentrations. Single-tailed amphiphiles with a small head-group area ($\frac13 < P < \frac12$) will form elongated or cylindrical micelles. Non-ionic surfactants and ionic surfactants such as
SDS and CTAB in high salt are known to form elongated micelles. Double-tailed
amphiphiles with a large head-group area (phosphatidylcholine, phosphatidylserine, phosphatidylinositol etc) have a packing parameter approaching unity and tend to self-assemble into planar bilayers and vesicles/liposomes.

The figure here will explain this schematically:


Why are compounds of V5+ and Cr6+ colored?

Firstly, a note about terminology. As Mithoron noted, the naked ions $\ce{V^5+}$ and $\ce{Cr^6+}$ do not exist because they require a prohibitively large amount of energy to make (just take the sum of the first 5 or 6 ionisation energies of V/Cr to see why). As such your question should really be asking about the vanadium(V) and chromium(VI) oxidation states, which can exist in acidic solution as the oxoanions $\ce{VO2+}$ and $\ce{Cr2O7^2-}$ respectively. The same idea can be extended to the manganese(VII) oxidation state, which exists as $\ce{MnO4^-}$.

Now, on to the question itself. As you rightly note, the central atom in these oxidation states have the same electronic configuration as argon; we usually call it a $\mathrm{d^0}$ configuration to indicate that there are no d electrons. The colours of these species therefore cannot arise from d-d transitions, which are the transfer of electrons from one metal orbital to another metal orbital. They are actually associated with electron transfer from the ligand (in this case, $\ce{O^2-}$) to the central metal atom. The name for this is ligand-metal charge transfer (LMCT).


Molality of solution having more than one solute?

The boiling point elevation fomula assumes that all particles are equal, only the number of particles in the solution matters.

So I would use:

m = (0.2 + (3)(0.8))(molality of $\ce{MgCl_2}$) + (0.2 + (2)(0.8))(molality of $\ce{NaCl}$)

To find molality of $\ce{MgCl_2}$ and molality of $\ce{NaCl}$, consider a 1kg sample of the solution.  There are 95g  $\ce{MgCl_2}$, 117g $\ce{NaCl}$ and 788g water.

Calculate the number of moles $\ce{MgCl_2}$ in 95g $\ce{MgCl_2}$, and divide this number by 0.788kg.  This is the molality of $\ce{MgCl_2}$ in the solution.

Repeat for $\ce{NaCl}$


How can the equilibrium shift, while Kc remains constant?

$K_c$ is related to the ratio of reactants to products at equilibrium.

If the reaction is currently at equilibrium, and you add more products then the reaction is now out of equilibrium and the reverse reaction will happen until it is back in equilibrium.

I don't like the wording of "equilibrium shifts to the left" myself, I would say that reverse reactions occurs to restore equilibrium. But $K_c$ doesn't change based on reactants/products concentrations since its the ratio at equilibrium. $K_c$ will normally depend on temperature though.

Based on the options you have, $K_c$ not changing and the reverse reaction occurring, the closest answer would be D. (If you take equilibrium shifting to the left to mean the reaction goes in that direction.)


The geometry optimization routine does not appear to have a method to output frames to an .xtd file for CASTEP in Materials Studio. Usually a movie makes more sense for a molecular dynamics simulation.

However, if you construct a script that performs a sequence if geometry optimizations with MaxIterations = 3 (must be >2), you can output each result to .std (study table) or .xtd. You would do this by outputting a geometry optimization after three iterations, then repeat the calculation on the result, until convergence. Then you take the frames and make a video

The help file in Materials Studio is rather a good resource, be sure to dig around in there, especially for the scripting API


This reaction proceeds through a reverse Friedel-Crafts alkylation followed by a forward Friedel-Crafts alkylation. The last step of Friedel-Crafts alkylation is deprotonation, which means that the first step of the reverse process will be protonation. Next, the carbon-carbon bond breaks, with 1,2,4,5,-tetrabromobenzene acting as a leaving group. The resulting t-butyl cation can undergo the typical electrophilic aromatic substitution steps, attack by an aromatic ring (benzene here) and deprotonation to regain aromaticity.

One might wonder why aluminum chloride is present in the reaction mixture, since we can write a reasonable mechanism that excludes it. Likely it is present in case chloride (from HCl) traps it to form t-butyl chloride. Aluminum chloride will catalyze the reformation of t-butyl cation.


How to convert benzene to N,N-dimethylbenzamide?

The scheme you have written will satisfy your teacher in my opinion but it can be shortened. The suggestion of @Eli Jones of the V-H formylation to give benzaldehyde is a good one. However, it can be done in 2 steps using reasonably well-known chemistry:

Step 1 - Brominate with $\ce{FeBr3/Br2}$ to give bromobenzene

Step 2 - Either lithiate by Li-Halogen exchange using $\ce{n-BuLi}$, or form the Grignard (directly with Mg or by Knochel exchange with $\ce{i-PrMgBr}$) then react the resulting benzene organometallic with dimethylcarbamoyl chloride ($\ce{(Me)2NCOCl}$) to give the benzamide.

There are a couple of ways of doing this in one step from benzene by the use of biscarbamoyl diselenides/Lewis acids$\ce{^{[1]}}$ and by $\ce{(Me)3SiOTf}$ activated carbamoyl chloride$\ce{^{[2]}}$.

References

Biscarbamoyl diselenides as new carbamoylating reagents. Lewis acid promoted carbamoylation of aromatic compounds by Shin-Ichi Fujiwara, Akiya Ogawa, Nobuaki Kambe, Ilhyong Ryu, Noboru Sonoda, Volume 29, Issue 47, 1988, Pages 6121-6124, DOI: https://doi.org/10.1016/S0040-4039(00)82282-0
Jin-Wei Yuan, Qian Chen, Chuang Li, Jun-Liang Zhu, Liang-Ru Yang, Shou-Ren Zhang, Pu Mao, Yong-Mei Xiao and Ling-Bo Qu, Silver-catalyzed direct C–H oxidative carbamoylation of quinolines with oxamic acids, Organic & Biomolecular Chemistry, 10.1039/D0OB00358A, (2020).


Calculating yield percentage of magnetite synthesis

Add up the reactions (double the second one so there is sufficient hydroxide) to get a net reaction with your starting materials as reactants and your desired product as one of the products:

$$\ce{3FeSO4.7H2O(aq) + 6KOH(aq) + KNO3(aq) -> Fe3O4(s) + NO2-(aq) + ...}$$

Then, find the limiting reactant, calculate the theoretical yield from it, and compare it to the amount of product to figure out the percent yield.


Which conductivity law should I use to calculate the conductivity of a solution with high/low concentration?

You can use the Debye-Hückel-Onsager limiting law. For a strong electrolyte without the formation of ionic pairs, with $z_+ = |z_-|$, just like $\ce{NaCl}$:
$$ \Lambda_m = \Lambda_m^\infty - (az_+^3 + bz_+^3\Lambda_m^\infty)(c/c^\circ)^{1/2}$$
For aqueous solutions at $25^\circ$C and $1$ atm: $a = 60.6 \, \Omega^{-1}\mathrm{cm}^2\,\mathrm{mol}^{-1}$, $b = 0.230$. For $\ce{NaCl}$, $\Lambda_m^\infty = 126.4 \, \Omega^{-1}\mathrm{cm}^2\,\mathrm{mol}^{-1}$.

References

Ira N. Levine, Physical Chemistry, 6th Ed.


How do self-relighting (trick) birthday candles work?

In normal candles, as you blow them out, you will see burning embers in the candles, which vaporises the wax and thus causes a ribbon of paraffin wax to rise up (this looks like a wisp of smoke). These embers are often not hot enough to light the ribbon of wax, so the candle goes out.

However, in trick candles there is magnesium put in the wick. When normally burning, the liquid wax shields it from oxygen so it cannot ignite. However, when the candle is blown out, this magnesium is able to react with oxygen (ignite) and light up the paraffin wax ribbon rising, reigniting the candle (or rather, the volatile materials above the candle). The magnesium is often powdered or in thin strips for low ignition temperature and easy ignition.

Sometimes regular candles do reignite, in these cases, the embers are hot enough to light up the paraffin wax vapour