What was the lithium concentration in 1940's 7-Up?

Given that the formula was a trade secret and that lithium was removed from the product ~70 years ago, the likelihood of this question being answered by providing a detailed formula seems remote. Note in addition that the formula might have evolved to reduce cost or improve market reception. You might therefore want to settle for an estimate of the upper limit on the lithium citrate concentration, which is relatively easy to provide.

A basic estimate can be made based on (1) the historical limits of sodium concentration in sodas; (2) the current composition of 7-up; and (3) the relative saltiness (taste) of lithium citrate versus sodium.

First one can attempt to determine what range of sodium concentration makes for a flavorful soda. Here is a quote:

  The amount of sodium in soda varies depending on the type. A well-known brand of cola has 46 milligrams of sodium in a 12-ounce can. The same sized can of root beer has 71 milligrams; a can of lemon-lime soda has 64 milligrams; a 12-ounce can of all-natural soda contains 36 milligrams. These are just examples, and the amounts in particular brands of soda may differ. It's best to check the label of your favorite soda to make sure.

This leads to a range of ~150-200 mg/L sodium concentration.

Next, consider the current formula for 7-up (in the USA):

  in 12 fl oz (355 mL):
    40-50 mg  sodium,
    60 mg potassium

The current formula contains ~110-140 mg/L sodium and ~170 mg/L potassium.

Finally, consider the saltiness of lithium citrate versus sodium. Lithium citrate was the salt form of lithium in the 7-up recipe  according to a number of sources including the Wikipedia:

  The soft drink 7Up was originally named "Bib-Label Lithiated Lemon-Lime Soda" when it was formulated in 1929 because it contained lithium citrate. The beverage was a patent medicine marketed as a cure for hangover. Lithium citrate was removed from 7Up in 1948.[Ref 4]

and  this one (which cites Ref 1):

  As to why “7Up,” C.L. Grigg never explained how he came up with the cryptic name. Several theories exist about its origin:
     "7Up was the product of seven ingredients. (Which, in a way, was at least true with regard to the classes of ingredients in that original formulation: sugar, carbonated water, essences of lemon and lime oils, citric acid, sodium citrate, and lithium citrate.)"

Lithium and potassium are among few cations with a salty flavor similar to that of sodium, albeit weaker. Citrate reduces the saltiness further. According to Ref 2, the saltiness of lithium and potassium can be estimated to be ~0.4 and 0.6 that of sodium, repectively, while Ref 3 explains that citrate has an inhibitory effect.

Consider now the potassium concentration (~170 mg/L) in the current formula of 7-up. Potassium is added as the citrate salt, and could be replaced with the lithium salt with little change in taste perception, according to what is reported on the relative saltiness of the cations and citrate salts. This suggests that if the taste of the beverage has remained reasonably constant since inception, it is not unreasonable to estimate an upper bound of ~200 mg/L in the concentration of lithium citrate (replacing the potassium salt) in the orginal recipe.

Note this is based entirely on the taste perception and not on any medical implications of consuming lithium at such concentrations. Lithia water (a brand of spring water sold in the USA) contains only ~0.5 mg/L of lithium. For reference, an online medical advice site suggests that a dose for treatment of depression is ~60 mg of lithium daily. So a dose of ~60 mg lithium citrate per serving certainly seems high relative to lithium-rich spring water but is just about a typical clinical dose.

References

Rodengen, Jeffrey.   The Legend of Dr. Pepper/7Up.   Fort Lauderdale: Write Stuff Syndicate, 1995.   ISBN 0-945903-49-9.
McLaughlin, S., Margolskee, R.F.. The Sense of Taste. American Scientist, 82, 538-545 (1994).
C.M.D. Man. Technological functions of salt in food products. in Reducing Salt in Foods, Woodhead Publishing, 2007. ISBN 978-1-84569-018-2.
Gielen, Marcel; Edward R. T. Tiekink (2005). Metallotherapeutic drugs and metal-based diagnostic agents: The use of metals in medicine. John Wiley and Sons. p. 3. ISBN 0-470-86403-6.


Assigning the 13C NMR spectrum of bromobenzene and 4-bromobenzophenone

Bromine is an example of what is sometimes described as the 'heavy atom effect', where chemical shift follows an opposite trend than would be expected due to electronegativity. These types of atoms (Br to a lesser extent than for I and Te) have a large electron cloud distribution. The electron cloud has a significant effect on shielding of the ipso carbon, effectively immersing the carbon within a cloud of electrons. This contributes to the diamagnetic shielding around the carbon nucleus and results in an upfield shift. As an aside, because of their size, Br and I often have significant γ effects also, which are usually steric in origin.

For calculating 13C shifts at C1, the trend between halogen-substituted benzene series follows what is expected based on electronegativities. So, qualitatively, one might expect shifts to be in the order:

F  >  Cl  >  Br  >  I  >  H

However, you just need to be aware of the heavy atom effect for Br and I, and that these will result in an upfield shift for the ispo. Hence, one can expect, qualitatively, that the progression of shifts would look like:

F  >  Cl  >  H  >  Br  >  I

This heavy atom effect is also well demonstrated in the substituted methane series:

    CH4     CH3X    CH2X2   CHX3    CX4
Cl  -2.5    25.6    53.5    77.4    96.3
Br  -2.5    9.5     21.7    12.4    -29.4
I   -2.5    -24.0   -53.8   -139.7  -292.2

So, you can qualitatively predict shifts for C1, as long as you are aware of this contribution for Br. Predicting the order for C2/C3 is not so easy. Qualitatively, it is a reasonable assumption to say that the effect at C3 is negligible, that is, without any other knowledge, δ shift will be ~0. So you just need to look at the β carbon, or C2. Again, a reasonable guestimate might predict an upfield shift for those substituents with strong electronegativities. And, this holds for many cases. However, there are many exceptions to this, and the case becomes very complicated as soon as you start introducing multiple substituents. In this case, one should always assign peaks using 2D methods rather than just predictions.


When an electron returns to it's ground state does it release the same wavelength of light that promoted it in the first place?

Not necessarily... it can return to the ground state in multiple steps, releasing different wavelengths each time.

Also, the excitation could be caused by something other than light in the first place (heat, for example).

Also, relaxation isn't instantaneous, so the emitted photon wouldn't really be "cancelling out" the color of the absorbed one as some time has passed.


Is Km the concentration needed to keep the enzyme running at max speed?

$K_m$ is the lowest concentration of substrate needed to ensure that the system runs at $V_{max}$?

No. $V_{max}$ is never reached, it is the limit of the rate when you increase the substrate concentration indefinitely. Less dramatically said, when the substrate concentration is about ten times higher than $K_m$, the rate will be about 90% of $V_{max}$ (and so on, so hundred times higher - about 99%).
In practice it is often difficult to get to hundred times $K_m$ because the substrate might not be that soluble or there is substrate inhibition at non-physiologically high substrate concentrations.

Could someone explain, preferably intuitively, why this is true?

As I read in a textbook, imagine a bicycle repair shop with 5 people working in it. When the number of bikes brought in doubles from what it is usual, they will repair more bikes. It does not matter, however, if you park a million or a billion bikes in front of their shop, they will be working at capacity in either case.
If you like math, you can also answer your question by looking at the relationship between substrate concentration and rate described by the Michaelis-Menten equation set up by  Leonor Michaelis and Maud Menten:
$$\mathrm{rate} = V_\mathrm{max} \frac{[S]}{K_m + [S]} = V_\mathrm{max} \frac{1}{1 + \frac{K_m}{[S]}}$$
When $[S] \gg K_m$, the fraction approaches 1 and the rate approaches $V_\mathrm{max}$.

My lecturer has said

That is what you heard. If you think the lecturer misspoke or you misheard, it is a good idea to check a textbook or ask here.
The figure
The figure is a conceptual one with two conceptual errors. First, the rate in the figure does exceed $V_\mathrm{max}$.

Secondly, the figure shows a linear increase of the rate for substrate concentrations lower than $K_m$. In the graph showing the real shape of the curve, you can see appreciable curvature at lower substrate concentrations.

For the graph, $V_\mathrm{max}$ was set to 1 and $K_m$ to 0.2. The shape of the curve, however, is the same (if you choose the plotting range in an analogous manner) for any combination of the two parameters.
The problems with the figure are common, as the following gallery shows:


How does carbonic acid cause acid rain when Kb of bicarbonate is greater than Ka?

What is the ${K_a}$  of carbonic acid? Note that sources differ in their ${K_a}$ values, and especially for carbonic acid, since there are two kinds - a pseudo-carbonic acid/hydrated carbon dioxide and the real thing (which exists in equilibrium with hydrated carbon dioxide but in a small concentration - about 4% of what what appears to be carbonic acid is true carbonic acid, with the rest simply being $\ce{H2O*CO_2}$.

Either way, I find that the ${K_a}$ of the mixed carbonic acid is about $4.2 \times 10^{-7}$, which is greater than $1.0  \times 10^{-7}$, and this implies that a solution of carbonic acid alone should be acidic no matter what. This suggests to me that your numbers are wrong; would you mind sharing your numbers and their source if possible?

EDIT: I see that you have updated your numbers. Nonetheless, I believe that your ${K_a}$ for carbonic acid is wrong; that number looks suspiciously like the ${K_a}$ instead for hydrogen carbonate ion (or the bicarbonate ion). According to Wikipedia, the ${pKa}$ of carbonic acid, is 6.3 (and this is taking into account any aqueous carbon dioxide). Convert this to a ${K_a}$ value and we get about $5.0 \times 10^{-7}$. This is in-line with the value I obtained from a copy of Daniel C. Harris' Qualitative Chemical Analysis.

EDIT 2: I think you've realized your mistake; as you say, the values are for $\ce{HCO_3^-}$, which is the hydrogen carbonate ion.


Why would it not? C is a nonmetal, and it is common for nonmetals to form compounds with metals. Oxygen gives oxides, nitrogen gives nitrides (if you succeed in making him react, which may be not quite trivial, but that's another story), chlorine gives chlorides, etc. Now, C is less eager to take extra electrons than N or O, so it would form ionic carbides only with active metals. Well, Ca is pretty active.


Can different groups within a single molecule photoisomerize at different wavelengths?

In theory: yes

As soon as you have two distinct groups, you will have two different, discrete energy differences. If you manage to irradiate at exactly the correct wavelength, only one of the two will be excited.

In practice: possibly, but it depends

Two diazo groups are still rather similar. You will either need a beam that has a good wavelength focus or you will have to make their wavelengths sufficiently different.

By the way, note that both your average IR spectrum and your average UV/Vis spectrum show this property. At a specific wavelength, exactly one vibration (IR) or electron (UV/Vis) will be excited. With sufficient resolution, you should be able to excite the different nitro group vibrations separately. If this applies to one light wavelength band, it will work in others, too.


Conversion of benzenediazonium chloride to benzene using lithium aluminium hydride

I just wanted to add that it's up to my knowledge that $\ce{LiAlH4 /
H^{−}}$ can't reduce the double bond of Benzene, because when styrene
  reacts with $\ce{LiAlH4 / H^{−}}$ it gives ethyl benzene.

That answer is as good as any for the first part of your question.  It demonstrates that LAH will react preferentially with aliphatic pi bonds rather than the aromatic ring.  Similarly there are aliphatic pi bonds in $\ce{Ar - N^+\equiv N}$, and so the LAH will reduce the diazonium group rather than displacing it.

For the hypophosphorous acid reduction, the mechanism you found has a problem.  The principle of microscopic reversibility argues against three components coming off at once in what is shown as the first step (it would be equivalent to three independent bodies colliding simultaneously).  The first step may be correct but it likely has more than one component to it.


Which will be major and minor alkene formed if E1 reaction occurs on 2-chloro-2,4,4-trimethylpentane?

The major product in this case is favoured due to Steric hindrance in the product. Drawing the full structure of the molecule in the reaction makes this clearer

We can compare the degree of steric hindrance in the major product and the minor product by using A-values. The Wikipedia article gives the following A-values for common substituents:
$$\begin{array}{|c|c|}
\hline
\text{Substituent} & \text{A-value} \\
\hline
\ce{-CH3} & 1.7 \\
\ce{-CH2C(CH3)3} & 2 \\
\ce{-C(CH3)3} & >4 \\
\hline
\end{array}$$
Clearly, the major product has a lower set of A-values for it's substituents (2 & 1.7) compared to the minor product (4 & 1.7). Also, the minor product does not have any scope for forming stereoisomers, as both the groups on one end of the double bond are methyl. If one of them was a hydrogen group, then the E isomer would be the major product.


A vat of Hydrogen Peroxide ($\ce{H2O2}$) contains 455 grams of oxygen atoms.
  How many molecules of ($\ce{H2O2}$) are in the vat?

How many moles of oxygen atoms are in 455 gms of oxygen atoms? Let's call the answer $x$.
How many moles of $\ce{O2}$ are in $x$?  Let's call the answer $y$.
How many moles of $\ce{H2O2}$ could you make from $y$ moles of $\ce{O2}$?  Let's call the answer $z$.
1 mole of any substance contains Avogadro's number of molecules.  How many molecules are in $z$ moles?


Reliable automatic addition of hydrogens and charge assignment to small molecules at given pH?

If you're looking for comprehensive protonation and evaluation of $pK_a$, I suggest my answer to a previous question. There is specialized software for evaluating protonation and $pK_a$ based on large databases.

Avogadro has some simple rules, mostly for amino acid residues, but is not intended for automated evaluation of thousands of molecules.


How to dissolve bronze at room temperature

Supposing that the bronze is only 78% copper and 12% tin (and there are no other metals). Nitric acid will dissolve the copper... but this reaction is exothermic, which means it will heat up.

At boiling temperature. Nitric acid will also react with tin to make tin oxide which is soluble. However, if you kept it cool/room temperature with an ice bath, then you would have leftover tin which did not react.

Tin will react with aqua regia to form tin (IV) chloride, but copper is slower to react with aqua regia... so it would be better to use nitric acid first (to dissolve the copper) and then add hydrochloric acid to react with and dissolve the tin. Once again, the reaction between tin and hydrochloric acid (or aqua regia) will proceed much faster with heat, but it will happen at room temperature.

Aqua regia = 1 mole of Nitric acid and 3 moles of hydrochloric acid).


Is water stream deflecting in an electric field really based on water's polarity?

An electrostatic field will attract both conducting and nonconducting materials even though those materials have no initial charge through electrostatic induction. That is why you can pick up lint or aluminum foil with a charged comb or balloon. See this demo with a neutral aluminum can.


Reaction of aldoxime and a ketoxime with acetic anhydride

This might sound obvious, but since the reagent and the reaction conditions are identical, the reason for the different outcome of the reactions must be related to the starting materials.

In the case of the aldoxime, the carbon atom of the former $\ce{CHO}$ group still bears a hydrogen atom.

   Together with the $\ce{OH}$ group on the adjacent nitrogen atom, this allows (formally) for the elimination of water from the aldoxime under suitable conditions. High temperatures and the presence of a dehydrating agent facilitate exactly that.

The formation of a nitile from an aldoxime in boiling acetic anhydride is (formally) a dehydration reaction.

For obvious reacons, this pathway is not possible in the case of the ketoxime.

What is the role of acetic anhydride here?

I mentioned that the formation of nitriles from aldoximes is formally a dehydration. In reality, however, it is more likely that acetic anhydride actually reacts with both oximes.

Both the aldoxime and the ketoxime are

   acylated. For the ketoxime (R' = alkyl), the story ends here. In the case of the aldoxime (R' = H), a subsequent elimination (of acetic acid) eventually furnishes the nitrile.


How do the differences in the number of protons result in such great differences in elemental properties?

Unfortunately, I have to disagree with trb456's answer.  It is both the protons and the electrons that make up chemistry. (Neutrons are considered massive "spectators" though they are what keep protons (all + charged) together)

Bonds are formed due to the attraction of electrons and protons. (cf. Slater, Bader, et al.) (The precise shape and nature of the electron density is additionally determined by constraints such as Pauli Exclusion and electron-electron repulsion.)

So if you think about it, there is a very complex interplay upon mixing electrons up with protons. More positive charge in the nucleus tends to keep electrons drawn tight, but in some ways that is counter-balanced by electron-electron repulsion--and the fact that electrons are attacted to other nuclei as well.

So although we ascribe color ("electronic transitions") and reactivity to electrons, this is only part of the story. The protons are also there, and they provide some governance (via Coulombic interaction) over what electrons are permitted to do.


Method of allylic and benzylic oxidation

This is one of those heterogenous reactions occurring on a surface which are  mechanistically messy. As expected for a reaction that occurs on the surface of a reagent, factors such as surface area and reagent preparation (freshness, surface contamination) are significant variables.  A free radical mechanism is generally accepted and a reasonable mechanism is posted below.

image source

When an allylic alcohol with stereochemistry (cis, trans) around the double bond is oxidized, the stereochemistry around the double is preserved.

image source

This suggests that we are not really dealing with a free, long-lived radical intermediate, but rather a free radical held in close association with the manganese ion.


The strength of the bond is due to 2 things : Difference in electronegativity and atomic radius of the elements .Halogens except fluorine , have a very big atomic radius so less bond energy.And they have 0 difference in electronegativity so even less bond energy.They both "want" to get the bonding pair of electrons closer to them , so because they cant both atoms are "unhappy".


You haven't accounted for the stoichiometry of the reaction, and I suppose you wrongly converted minutes to seconds. Always start solving problems like this with writing down the chemical reaction:

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

By definition rate of consumption of hydrochloric acid over time $\Delta t$ is:

$$r=\frac{\Delta c(\ce{HCl})}{\Delta t}$$

Since all calcium carbonate reacted completely:

$$\Delta c(\ce{HCl}) = \frac{\Delta n(\ce{HCl})}{V} = \frac{2n(\ce{CaCO3})}{V} = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})}$$

where $m$ is mass, $M$ - molar mass, $V$ - volume.

And the average rate is:

$$r = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})\Delta t} = \frac{2\cdot\pu{3.45 g}}{\pu{1 L}\cdot\pu{100.09 g mol-1}\cdot\pu{4.50 min}\cdot\pu{60 s min-1}} = \pu{2.55e-4 mol L-1 s-1}$$

Also, be careful with notations. Use proper capitalization, and don't equate moles to grams! This is not tolerable in natural sciences.


Why so much color change in synthesis of triphenylmethanol?

It sounds like you might have generated a species with this structural moiety:

This can happen if your solution was concentrated enough or the boiling of your ether was vigorous, so the local overheating caused the formation of $\ce{MgO}$, the guy above, and $\ce{Br-}$, which gives a very distinct bright red-pink color, which is VERY bright and only requires a bit of the side product to form. Though this carbocation is kinda stable, it will find the $\ce{Br-}$ in the solution and form triphenylmethylbromide as a minute impurity.


In the lattice model, what's a nearest neighbor?

A nearest neighbour in general terms is literally that: Find the closest atom of any given element, that is your nearest neighbour distance for that element in the lattice. The number of nearest neighbours for that element is the number of atoms that are at this distance from your starting atom.

$N_{11}$ is the number of nearest neighbour pairs of atoms of element 1, and $N_{12}$ is the number of nearest neighbour pairs between any atom of 1 and any atom of 2 etc. For the $N_{11}$ or $N_{22}$ case, you have to consider that each atom of 1 or 2 in the equation has a contribution to $zN$, hence the term being multiplied by 2 when viewed from the perspective of element 1 or 2.

In the case of your thought experiment, your expected value neglects edges (you can't just neglect edges!) so will be too high by a good margin.

The dissertation of Alexander Zhou1 covers these equations (p.34) and explains them slightly differently (and more clearly I think) than de With's textbook.2

Alexander Zhou. Explicit Non-Random Contribution in Lattice Fluid Equation of State. Explizite nicht zufällige Wahrscheinlichkeitsverteilungen in gitterbasierten Zustandsgleichungen. Doktor Ingenieur, Technische Fakultät der Universität Erlangen-Nürnberg, Nürnberg, 2013. See UB Friedrich-Alexander-Universität Erlangen-Nürnberg (FAU)
Gijsbertus de With. Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications; John Wiley & Sons: 2013. See Google Books


Why is arginine's positive side chain classified as basic and not acidic?

Your textbook is imprecise. We consider whether a sidechain is acidic or basic based on the action of the sidechain when it is neutral.

Because arginine's sidechain is basic, it preferentially picks up a proton, and hence becomes positively charged. (In the same vein, carboxylic acid sidechains are acidic because they preferentially lose a proton, and exist as the negatively charged carboxylate anion.)


How should I name this branched alkyl substituent?

According to Section P-46 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), compound acyclic substituents consisting of a principal chain and one or more acyclic substituents can be named in two ways:

by using alkyl substituents according to method (1) of Subsection P-29.2

  (1) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups and for the mononuclear hydrides of silicon, germanium, tin, and lead. Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

by using alkanyl substituents according to method (2) of Subsection P-29.2

  (2) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ are added to the name of the parent hydride with elision of the terminal letter ‘e’, if present, when followed immediately by the letter ‘y’. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix ‘ylidyne’, the locant ‘1’ must be cited. This method is used to generate names of ‘alkanyl-type substituent groups’ that are simple substituent groups with free valences at positions other than ‘1’.

For both methods, the principal substituent chain has the greater number of skeletal atoms, i.e. the longest chain. Low locants are given first to the free valence; for method (1), the free valence always has the locant ‘1’.

The preferred names for simple substituent prefixes derived from saturated parent hydrides with a single free valence that is not at the end of the longest chain are formed according to method (2).

According to method (1), the alkyl-type name of compound substituent group given in the question is ‘1,2-dimethylpropyl’. The preferred alkanyl-type prefix according to method (2) is ‘3-methylbutan-2-yl’.

Therefore, the preferred name for the complete compound given in the question is ‘5-(3-methylbutan-2-yl)decane’.

Note that, according to current IUPAC nomenclature, locants are placed immediately before that part of the name to which they relate. Therefore, correct prefix is ‘3-methylbutan-2-yl’ (not ‘3-methyl-2-butanyl’).


Why is moist hydrogen made to react with chlorine to give hydrogen chloride?

H2O has h ions so it adds more + ions so that the cl- gets reacted faster towards the H+ ions so water is a catalysts or booster
I hope u understnd


In potentiometric analysis, there are two electrodes, an indicator electrode (platinum here) where reduction takes place and reference electrode (calomel electrode) where
oxidation takes place.

So here the reference is not $E_{STP}$ cell but the $E_{cathode}$ that is the calomel electrode, so there is drop to $-241mV$ after the equivalance point. Hence,
 $E_{cell} = E_{cathode}$


How do I determine the absolute configuration experimentally?

If your compound has been synthesized before, you can measure its optical activity and compare it with the data in the literature.

If your compound hasn't been synthesized before, it can be possible to differentiate between two enantiomers with NMR by measuring the spectra of your molecule in a chiral environment. Here's a review article on how to resolve enantiomers with NMR.

Otherwise, diastereomers can be differentiated easily. By looking at your molecule, you can (sort of) predict which spectrum it should give rise to by using the concepts of enantiotopic, homotopic, heterotopic and diastereotopic protons. We had to analyse and solve 2D-NMR spectra (1H and 13C) of diastereomers last year, and to a beginner, I have to say it was a challenge.

As always, it boils down to forming diastereomers.

EDIT

Take a look at these diasteromers:

(I've got finals coming up), it's easy to see that the hydroxyl will shift the ethyl's CH2 peaks towards lower field (higher ppm) in the second isomer. The interpretation of chiral media NMR spectra would have to be similar. My impression is that the acquisition of chiral media spectra is more involved, but the interpretation isn't. I might be totally off track, though.

Someone might want to share a more thorough answer.


What's the reason for discrepancy between DFT calculated image and NC-AFM-acquired one in this article?

1) Most definitely. Adsorbed molecules often are very distorted, up to chemical reaction with support, especially when reactive surface is used. Despite common misconception, even noble metal surfaces are quite reactive.

2) DFT, while being indeed inexact (as any calclation with limited size of basis set), catches major features of electronic density very well.

3) AFM stands for atomic force microscopy, which is not connected to electronic density directly. Now, naphthalene molecule features large $\pi$-system, which is very polarizable. Since nonvalent interatomic forces beyond electrostatic interactions usually increases with polarizability of interacting particles, I would expect $\pi$-systems  to be highlighted on the NC-AFM images.

4) I'm not an expert on atomic-resolved microscopy, but I never read about this possibility in the articles I worked with, though the works I usually works with used STM. and SEM mostly.


Complex structure

Modafinil is a pretty complex molecule and judging by its structure (see below) I think it will be extremely difficult to find a single reagent test that will identify it.

Post-consumption identification

Interestingly it is much easier to find after consumption, either by the strong sulfur smell of your urine or through chromatography techniques (also here).

'Sketchy' tests

There are apparently some simple tests you can try like dissolving it in hot water and vinegar (which should happen without formation of gas), but these tests are far from fool-proof and might even cause you to think something is a fake while it is not.

Combination of physical/chemical tests as a 'solution'

The best thing to do if you really want 'kitchen-level' tests is to test for a whole range of chemical/physical properties. For example this reference has some information on the pH, water solubility and melting point which are fairly easy to test. Passing the tests will certainly not guarantee anything, but at least it will take away a significant part of the doubts that the product is fake.

Something to consider when designing these tests is that there will always be fillers (e.g. lactose monohydrate, microcrystalline cellulose, pregelatinized starch, croscarmellose sodium, povidone, and magnesium stearate) in the sample so you have to find ways to correct for that.


Does heat loosen "corrosive bonds" in iron compounds... permanently?

If by corrosive bond they mean a physical bond not a chemical bond then yes, there is a very strong force of dry friction between the two parts.

Bolts and nuts are made so that the two fit together tightly so that they do not wiggle loose but not so tight that they are impossible to fasten. The problem is is that when iron corrodes the oxide has more volume than the iron. In fact when iron oxidizes, its size increases 70% (See: Pillings-Bedworth ration). This expansion puts a lot of force between the nut and the bolt. Since dry friction increases linearly with force, friction between the two increases also making the bolt harder to remove.

When you expand the nut and/or shrink the bolt some of this force is relieved and friction decreases making the nut easier to turn.


Difference between positional isomerism and metamerism (organic chemistry)

If functional group carbon is part of principal chain then metamers are also positional isomers. In case of polyvalent functional group like sulphoxide($\ce{>S=O}$), secondary amine ($\ce{-NH-}$) etc. structure formed by change in structure of alkyl groups will be metamers only.  For example, diethyl amine ($\ce{Et2NH}$) and methyl propyl amine ($\ce{MeNHPr}$) are metamers.


Mixing solvents and other compounds - safety issues

First, I would suggest you read the material safety data sheets (MSDS) for each of those products to be sure of any incompatibility that may arise. They are available throughout the web or on distributor web sites like Sigma or Fisher.

For your list of waste disposal (as it seems to be) most of your products are alcohols or water so they can be mixed without issues.
The ethyl acetate will react with water to create ethanol and acetic acid and acids may be an issue if you have great quantities. So I would suggest to react and neutralise with $\ce{NaOH}$ prior disposal.
Same thing for acetic acid: you should neutralise prior disposal.

You can also request an evaluation of price from your chemical waste service contractor, which will give you the price of each disposal category with the incompatibility management included.


What is the maximum molarity of aqueous HCl?

The HCl molecule is, at standard conditions, a gas. The reagent used in laboratories is HCl dissolved in water, which is why you'll find in the label that it is around 37% HCl in weight. The other 63% is water.

Considering 37% as the maximum solubility of HCl, you can calculate the molarity using the solution density (1.2 g/mL) and HCl's molar mass (36.46 g/mol).

For 1000 mL of solution, you will have 1200g of weight, of which 37% is HCl: 444g.

In 444g of HCl you have 12.18 moles (444/36.46), which means the concentration of P.A. HCl is around 12.18 moles per liter.

Note that because P.A. HCl is a solution saturated with a gas, vapors are likely to be released and escape the solution (which is why P.A. HCl shouldn't be handled outside fume hoods), which makes this concentration an approximation, and it should never be used to prepare a solution which you need to know the exact acid concentration.


Does mercury amalgamate with Brass or Steel?

Brass contains zinc and copper.  Both of those materials will readily amalgamate with mercury and thus it is not recommended to allow brass to contact mercury.

Steel will not form an amalgam and is perfectly fine for use with mercury. In fact is Approved as a container for mercury in transport.


There are a couple of reasons why Mercury is a liquid at STP, I will go through each of them with an explanation that won't be too complex for your understanding.

First of all understanding why most metals are metallic is a good place to start. This happens because of the sharing of valence electrons. The sharing of these electrons make metals malleable and ductile and thus not liquid. Generally the more the electrons shared means the higher the melting point and a harder metal.

OK, getting into the Mercury. Like others have probably mentioned above, Mercury doesn't like to share its electrons. It hangs onto them very tightly and thus Mercury-Mercury bonding is very weak and thus the compound melts and boils at very low temperatures. In other words, Mercury lacks the sea of electrons.

So then, why doesn't it like to share electrons? The $\ce{6s}$ electrons come very close to the nucleus and swing around it at near the speed of light. Due to this, relativistic effects occur. In other words, the $\ce{s}$ electrons behave as if they are more massive than other electrons. This lowers the energy of the electrons. To summarize, the electrons are buried deep in the atomic core and thus aren't able to participate in chemistry.

Overall, since you are a high school student, I would probably just say that Mercury doesn't readily bond well and that is due to the $\ce{6s}$ electrons being very attracted to the atomic core.


Ideally, the concentration of indicator is much, much lower than the concentration of the material to be titrated and of that used to titrate. Therefore, less than a drop of standard solution is needed to make the indicator entirely cross the equivalent point.

For most titrations, the equivalence point of the indicator is not the end point. It’s just somewhere near. And the only thing you can actually see and record is said indicator’s colour change. So it’s the only thing you can resort to.

The important choice in a titration is choosing the correct indicator. If you are titrating a strong acid with a strong base (so the titration’s equivalence point is approximately $7$), you want an indicator whose equivalence point is just behind that (at ca. $8$), so that the tiniest of drops added additionally will cause the colour change.


What are the microstates of the triplet oxygen molecule?

In a $(\pi2p)^2$ configuration the orbital $\lambda$ and spin s quantum numbers have to be tabulated and combinations removed which break the Pauli principle. The total angular momentum quantum number is $\Lambda$ and spin $\Sigma$, for example the combination

$\lambda_1 = 1, \lambda_2 =1 , s_1=1/2, s_2 =-1/2$

generates total angular momentum $\Lambda = 2, \Sigma =0$ and this is part of the $^1\Delta$ state.

There are only 6 valid combinations leading to $^1\Delta, ^3\Sigma^-,^1\Sigma^+ $ states of which $^3\Sigma^-$ is the ground state.

A vector spin diagram shows how a triplet can have spin multiplicity of zero. There is a picture of vector spin for singlets and triplets in my answer to this question; How does spin flipping of triplet carbenes occur?


How big is the influence of basis set and functional on ZPE and thermal corrections?

This is more of a comment, apologies.

Consider this paper (Ruth L. Jacobsen, Russell D. Johnson III, Karl K. Irikura, Raghu N. Kacker: Anharmonic Vibrational Frequency Calculations Are Not Worthwhile for Small Basis Sets, J. Chem. Theory Comput., 2013, 9 (2), pp 951-954). The magnitude of the computational-level-dependent scaling factors for the frequencies reported therein should give you a sense of the magnitude (I would also think that at least some other studies of the topic are cited).


The reaction rate when the rate-determining step has a back reaction

Thus, the rate is
  $$
\mathrm{rate} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}].
$$
  Though, I'm not sure if the reasoning is right. Is it correct?

This is correct but does not give you much insight into how [D] changes over time. This system will reach equilibrium eventually. If you call the equilibrium concentrations $[\ce{C}]_{eq}$ and $[\ce{D}]_{eq}$, and $\epsilon$ the difference between current and equilibrium concentration for C, you get the following:

$$[\ce{C}] = [\ce{C}]_{eq} + \epsilon\tag{1}$$
$$[\ce{D}] = [\ce{D}]_{eq} - \epsilon\tag{2}$$

Also, the rates are equal at equilibrium:

$$ k_2 [\ce{C}]_{eq} - k{_2}'[\ce{D}]_{eq} = 0\tag{3}$$

Plugging all of this into the rate equation gives:

$$\frac{d[\ce{D}]}{dt} = -\frac{d\epsilon}{dt} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}]\tag{4}$$

$$= k_{2}([\ce{C}]_{eq} + \epsilon) - k{_2}'([\ce{D}]_{eq} - \epsilon)$$

$$= k_{2} \epsilon - k{_2}'(- \epsilon)\tag{*}$$

$$ = (k_{2} + k{_2}') \cdot \epsilon$$

The simplification yielding the expression labelled '*' is because the net rate at equilibrium is zero (substituting (3) into the previous expression). As a result, you get a first order behavior of approching equilibrium with respect to the "excess concentration" $\epsilon$. When you integrate this, $\epsilon$ decays exponentially, with the sum of the forward and reverse rate constant in the exponent. This is somewhat surprising because we started out with these rate constants appearing with opposite signs in the (correct) rate expression given by the OP. However, this result is correct and well-known from temperature jump kinetics.

What about A and B?

The derivation above is for a two-species system $\ce{C<=>D}$. If A and B are in fast equilibrium with C, equation (1) or (2) has to be modified to account for A and B supplying more C as it reacts to D. One way is to define $[\mathcal{C}] = [\ce{A}] + [\ce{C}]$ and write (instead of (1)):

$$[\mathcal{C}] = [\mathcal{C}]_{eq} + \epsilon\tag{1'}$$

Because the first equilibrium is fast, you can calculate $\ce{A, B and C}$ from $[\mathcal{C}]$ at any point in time, and you would have to adjust equation (4) accordingly. The details depend on whether $\ce{A and B}$ are present at stoichiometric amounts or one of them is in excess. If one of them is in large excess, the result would just be different by a constant $\beta$, i.e.

$$ \mathrm{rate} = (\beta k_{2} + {k_2}') \cdot \epsilon$$

where $\beta$ depends on the equilibrium constant of the first (fast) reaction and the concentration of the reactant in excess.


What is the nucleophile in alkaline hydrolysis of a haloalkane?

Consider the reaction of a haloalkane with aqueous sodium hydroxide to produce the alcohol plus sodium chloride. The net reaction consumes one hydroxide ion for each alcohol molecule produced, so it is not really correct to refer to it as a catalyst.

(1) If this reaction occurs via the $\mathrm{S_N2}$ mechanism, a one-step displacement mechanism, you are correct to say that hydroxide (being a much better nucleophile than neutral water) is the nucleophile attacking carbon.

(2) If the reaction occurs via the $\mathrm{S_N1}$ mechanism (note that which mechanism is preferred depends on the structure of the haloalkane, and in some cases a haloalkane can react via both mechanisms), you have a two-step process in which the haloalkane initially dissociates (a slow step) to form a carbocation plus halide ion. The carbocation then reacts very rapidly with whatever nucleophile is handy to form the alcohol. Water is likely more abundant than hydroxide, and is a strong enough nucleophile to react instantly with a carbocation, so most of the product will be formed that way, with release of $\ce{H+}$, rather than by combination of the carbocation directly with hydroxide. The $\mathrm{S_N1}$ reaction of a haloalkane in this manner with excess water or other polar solvent is often referred to as a "solvolysis" reaction.


Is it possible for halogens to show oxidation states less than -1?

Yes, they can show oxidation states lower than -1. For example in hyperlithiated compounds, like $\ce{Li3Cl}$, the chlorine would formally have an oxidation state of -3.

However, a word of caution, oxidation states are a tricky thing. They are determined by adopting the hypothetical view that the compound you are looking at consists only of single-atomic ions, e.g. in the case of $\ce{Li3Cl}$ of 3 $\ce{Li+}$ and $\ce{Cl^{3-}}$. The hypothetical charges of these "ions" are then your oxidation states. Clearly, this view has nothing to do with the real bonding situation in the compound and oxidation numbers correlate very poorly with the real electron distribution. It is simply a useful little device for keeping track of redox reactions. So, the -3 oxidation state of chlorine in $\ce{Li3Cl}$ does not mean that chlorine has a 10-electron valence shell. The real bonding situation will be somewhat more complicated.


In a single cell battery, is voltage exactly equal to cell potential?

It is not uncommon for two identical phenomena to be described by different terminology in different contexts or scientific disciplines. In this case, voltage and electric potential difference refer to the same thing. According to this Wikipedia article:

  Voltage, electric potential difference, electric pressure or electric tension (formally denoted ∆V or ∆U, but more often simply as V or U, for instance in the context of Ohm's or Kirchhoff's circuit laws) is the difference in electric potential energy between two points per unit electric charge. The voltage between two points is equal to the work done per unit of charge against a static electric field to move the test charge between two points. This is measured in units of volts (a joule per coulomb).


Which of the following pair of molecules have identical bond dissociation energies (approx)?

The answer here has nothing to do with how the molecules relate to each other, e.g. similar bond order or length. This is actually a question that highlights an unexpected break in a trend.

The bond enthalpies for the halogens are:

$$
\begin{array}{|c|c|}\hline
\text{Halogen}&\text{Bond energy (kJ/mol)}\\\hline
\ce{F-F}&\pu{156}\\\hline
\ce{Cl-Cl}&\pu{243}\\\hline
\ce{Br-Br}&\pu{193}\\\hline
\ce{I-I}&\pu{151}\\\hline
\end{array}
$$
Plotting these data we would expect the bond enthalpy of $\ce{F2}$ to be the strongest of them all, something around $\pu{300 kJ/mol}$, but in fact it much weaker than for chlorine.  Indeed it is so weak, it as almost as weak as for iodine.

The reason for this is that fluorine is so small that it breaks the trend.  The atomic radius is so small that the electrostatic repulsion between the nuclei is significant, and the non-bonding electrons surrounding each atom also repel each other due to the small amount of space available.  These forces counteract the strong attraction between the nuclei and the bonding pair of electrons and weakens the bond compared to what might be expected.


Am I right that you are planning to simply calculate the $\ce{Fe}$ concentration using the Lambert-Beer equation?

The obvious proof would be to prepare $\ce{Fe}$ solutions with known concentration(s) to compare the absorbances. Using these solutions, you can also make sure that the assumption of linearity over the concentration range is actually fulfilled.


I need to report MoS2 values from Mo and Sulfur numbers

As the molecular mass of molybdenum is $95.95~\mathrm{g/mol} \approx 96~\mathrm{g/mol}$ and the molecular mass of sulfur is $32.06~\mathrm{g/mol} \approx 32~\mathrm{g/mol}$ one can construct a ratio between both molecular masses that in turn can be solved easily.
$$\frac{96~a}{32~b}=\frac{5.2}{5.3}=0.98\approx1$$
Which gives us after the multiplication with 32/96 on both sides
$$\frac{a}{b}=1\cdot\frac{32}{96}=\frac{1}{3}$$
So customer 1's determined "compound" is not $\ce{MoS2}$ but more $\ce{MoS3}$. Now I would guess that they don't analyze the copper ore for the compound $\ce{MoS2}$ but for both elements $\ce{Mo}$ and $\ce{S}$ separately which is why they came up with $5.2~\%~\ce{Mo}$ and $5.3~\%~\ce{S}$.

In my opinion, you should ask customer 2 to split up it's $12.34~\%~\ce{MoS2}$ into the percentages of each element and compare them with $5.2~\%$ and $5.3~\%$.

Using the ratio for $\ce{MoS2}$ which is $60:40$ one would estimate those $12.34~\%$ to split up into $7.40~\%~\ce{Mo}$ and $4.94~\%~\ce{S}$ or into $6.16~\%~\ce{Mo}$ and $6.18\%~\ce{S}$ if it is more like $\ce{MoS3}$.

Or to answer your question ... probably you can simply add both values to give you $10.5~\%$ which nonetheless will not be equal to $\ce{MoS2}$ based upon the data you are given!


Mechanism of hydrolysis of polycyclic oxetan

I have drawn a mechanism but I haven't checked for any references as I am at home now, so it's more like paper chemistry. The chlorine atoms should play some role and possibly disfavour formation of certain carbocations that you would expect in a standard Wagner - Meerwein. It is possible to draw a concerted mechanism (not shown here) but seems very unlikely

**EDIT: I have changed my initial mechanism so it explains better the methyl ester formation (see the comments below). The stereochemistry at C* has also been corrected.

As a general comment you cannot propose a non trivial mechanism without drawing one, it is handwaving. So I did.


Does sour milk have something else in it besides lactic acid that gives it such a horrific taste and odor?

Probably. Lactic acid bacteria produce other volatile compounds such as acetaldehyde and diacetyl that could be perceived as off-flavors. Both where fermentation is the objective and in the case of spoilage, the VOC profile can vary with the identity of the responsible bacteria but also of the milk and its treatment.
For instance, in Ref. 1 they discuss the effect of different lactic acid bacteria:

During fermentation, S. thermophilus and L. delbrueckii ssp. bulgaricus generate lactic acid and a variety of volatile organic aroma compounds. These compounds impart particular flavors to fermented milk (Beshkova et al., 1998). The overall flavor of any fermented milk product is formed by a large number of these volatile compounds. Among these compounds, carbonyl compounds and organic acids such as acetaldehyde, diacetyl, and acetic acid play a crucial role and can be used to evaluate the flavor quality of dairy products (Zha et al., 2015). Ott et al. (1997) found that acetaldehyde provided the typical flavor of dairy products such as yogurt and buttermilk. Erkus et al. (2013) also reported that diacetyl contributed the characteristic sour buttery flavor to some fermented dairy products.

The offending chemical could also be vinegar (acetic acid). Again from Ref. 1:

Acetic acid has been found previously as a volatile in yogurt by Panagiotidis et al. (2001). High levels of acetic acid contributes most to the tart flavor of yogurt and is particularly associated with a “vinegary, pungent, acidic” taste (Panagiotidis et al., 2001).

If you search for "chemical profile of spoiled milk gc" you'll find a wealth of references discussing the effect of different milk (due to species, handling or forage), milk treatment, and spoilage bacteria on the VOC profile.
In addition, what makes the milk taste bad may depend on the bacteria and on you. That makes narrowing the answer down to a particular compound impossible.
References

T. Dan, D. Wang, R.L. Jin, H.P. Zhang, T.T. Zhou, T.S. Sun, Characterization of volatile compounds in fermented milk using solid-phase microextraction methods coupled with gas chromatography-mass spectrometry, Journal of Dairy Science, Volume 100, Issue 4, 2017, Pages 2488-2500, ISSN 0022-0302, https://doi.org/10.3168/jds.2016-11528.


Conservation of angular momentum in electronic transition

I think you have misunderstood some concept. The angular momentum of spin is just $\sqrt{s(s+1)}\hbar$. The spin which we always say is spin projection quantum number in fact not the angular momentum. And I'm not familiar with chemistry, but I know that we can not add up the orbital and spin angular momentum directly because they are not in the same direction. Sometimes we can add up the quantum number. If the spin and orbital angular momentum in opposite direction, we should subtract.

I have some idea about your example. The electron transition is "electron jumps from one energy level to another". So the energy must change. Where does the energy come from? If it is photon, this example has no paradox. But can the energy come from other way? I have no idea.


Why are acetals stable to bases and nucleophiles?

Keep in mind that in these reversible reactions of carbonyl compounds, the mechanism for the forward and reverse reactions must go through the same intermediates. The result is that if you can't make a particular functional group under particular conditions, that functional group can't be destroyed under the same conditions.

Hydrates (as shown in the second reaction in ron's answer) and hemiacetals can be formed/decomposed under either acidic or basic conditions. In the hemiacetal formation under basic conditions, an alkoxide (methoxide below) attacks the carbonyl. Protonation gives the hemiacetal. In order to proceed to the acetal, the alkoxide would have to displace the hydroxide in an SN2 reaction. Hydroxide is not a good enough leaving group for SN2 reaction, so it is impossible to form the acetal under basic/nucleophilic conditions.

Hemiacetals can be decomposed to a carbonyl and alcohol under basic conditions. Deprotonation of the hydroxyl starts the mechanism, and the resulting anionic oxygen can kick out an alkoxide as a leaving group to give the carbonyl compound. Note that in this carbonyl chemistry hydroxide and alkoxides are competent leaving groups in an anionic elimination mechanism, such as the one here.

In an acetal, without the free hydroxyl group, alkoxide/hydroxide has nothing to deprotonate and there are no suitable leaving groups for SN2 reaction. There are no reasonable pathways for a nucleophile or base to react with the acetal.


How do I distinguish between internal energy and enthalpy?

Internal energy, $U$, is the total energy contained in a thermodynamic system. However, absolute internal energy is hard to determine, and even relative internal energy and changes to internal energy are hard to determine. Here's why:

Changes to internal energy usually occur through heat transfer $q$ or work done $w$. There are other ways to transfer energy also (like radiation), but these two are common. Thus,

$$dU = q + w$$

The heat transfer can be determined easily by calorimetry, where $q=cdT$, where $c$ is the heat capacity of the system and $T$ is the temperature of the system. Alternatively, it may be easier to measure the temperature change of surroundings with known heat capacity, since $q_{sys}=-q_{surr}$

Work is a little more challenging to determine, since work is defined as the change in volume at constant pressure. $w=-PdV$ (the minus sign is there to show that when the volume of the system increases, the system does work on the surroundings). However, unless you have a closed system, volume changes are difficult to measure. Additionally, pressure may not be constant.

Thus,

$$dU=q+w=cdT-PdV$$

Internal energy changes can be determined easily at constant volume $dV=0$ so $w=0$ so $dU=cdT$. Internal energy changes can also be determined in closed systems at constant pressure (so $PdV$ is evaluatable).

However, in open systems (like a beaker), $dV$ is hard to determine, making $w$ hard to determine. A new thermodynamic state function was devised to remedy this problem.

Enthalpy is a measure of the total energy of a system. More precisely, it is a thermodynamic potential. Enthalpy is defined as $H=U+PV$. $PV$ is a state function, even though $P$, $V$, $PdV$, and $VdP$ are not state functions. Thus the change in enthalpy is:

$$dH=dU+d(PV)=cdT-PdV+PdV+VdP=cdT+VdP$$

Enthalpy changes are easy to determine at constant pressure (most open systems are constant pressure systems, which is why enthalpy is so useful to most chemists), since $dP=0$ and $dH=cdT$. Enthalpy changes can also be determined in closed systems of constant volume where the pressure change can be determined.

To summarize, internal energy and enthalpy are used to estimate the thermodynamic potential of the system. There are other such estimates, like the Gibbs free energy $G$. Which one you choose is determined by the conditions and how easy it is to determine pressure and volume changes.

At constant volume: $dU=cdT$ and $dH=cdT+VdP$
At constant pressure: $dU = cdT-PdV$ and $dH=cdT$
At constant volume and pressure: $dU=dH=cdT$
At variable volume and pressure, $dU$ and $dH$ are both hard to
determine (but $dH$ may be easier since you may be able to numerically estimate $d(PV)$ from pressure-volume data).


Is there a replacement for glacial acetic acid in determining peroxide value

Your school is not allowing use of glacial acetic acid, but you insist of using glacial acetic acid in determining peroxide value, you may use a solvent system, which consists of glacial acetic acid. For example, you can use 60% ($v/v$) glacial acetic acid in chloroform. I give that example because it is commercially available online at Reagents. If you are determining peroxide value of fats, this is an ideal solvent to dissolve fat and resultant solution is soluble in water.

I'm suggesting this because there is a method recommended by International Fragrance Association (IFRA) in Belgium. Breifly, one of the IFRA analytical methods for determination of the peroxide value is described as follows:

  Transfer approx. $\pu{3.0 g}$ of the sample, accurately weighed, into a 250-mL Erlenmeyer flask with glass stopper. Add $\pu{50 mL}$ of the appropriate solvent mixture (glacial acetic acid : chloroform 3:2; glacial acetic acid : cyclohexane 3:2; glacial acetic acid : isooctane 3:2) and saturated potassium iodide solution, $\pu{1 mL}$, freshly prepared and allow to react for $\pu{60 ± 1 sec}$ and shaking thoroughly during this period. Then add water, $\pu{100 mL}$ and shake. Titrate with $\pu{0.01 mol/L}$ sodium thiosulfate solution, or $\pu{0.1 mol/L}$ sodium thiosulfate solution at higher usages, using $\pu{1 mL}$ starch solution or $\pu{0.1 g}$ of Thyodene indicator. The indicator should be added towards the end of the titration but while the pale straw color is still present. During titration shake until the blue color disappears. Carry out a blank titration under the same conditions. No more than $\pu{0.5 mL}$ of $\pu{0.01 mol/L}$ sodium thiosulfate solution should be consumed for this purpose.

Note: Use $\pu{0.1 mol/L}$ sodium thiosulfate solution when the expected peroxide values are $\gt \pu{20 mmol/L}$. Use $\pu{0.01 mol/L}$ sodium thiosulfate solution when the expected peroxide values are $\lt \pu{20 mmol/L}$.

Thyodene indicator is also commercially available in Fisher.


$$\Large\color{red}{\text{nucleophilicity} \ne \text{basicity}}$$

Please memorise firmly that nucleophilicity and basicity, albeit related in many ways, are different physico-chemical properties. Most importantly, basicity is a thermodynamic property while nucleophilicity is a kinetic property. Therefore, it is not uncommon for compound $\ce{A}$ to be more basic but less nucleophilic than compound $\ce{B}$.

This is the case here, too. It is unquestioned and supported by the corresponding $\mathrm{p}K_\mathrm{a}$ that tert-butanolate is the strongest base, followed by isopropanolate followed by ethanolate. This is due to the $+I$ effects you correctly outlined.

Nucleophilicity does not parallel this series since the it does not abide by the same rules. For a compound to be nucleophilic, the nucleophilic atom’s lone pairs must be able to actually attack, i.e. reach, the electrophilic site it wishes to attack. This means two molecules interacting and not just one molecule interacting with a proton which is so small it can even tunnel. In tert-butanolate, you have three α-methyl groups that take up space and make said interaction harder. Due to this steric interaction, tert-butanolate is an absolutely lousy nucleophile (while it is a great base).

It is similar for isopropanolate which still has too much steric hindrance to be an effective nucleophile although the conditions are better than for tert-butanolate.


Can flexible plastics have reflective coating (vacuum metalized)?

There are many techniques for metal-coated polymers, as linked by permeakra Generally, vacuum deposition is an issue because the metal is hot. Cold deposition techniques (e.g., sputtering, ink coating) often work better.

Common, commercially-available materials include:

Pyralux - copper-coated polyimide from DuPont.
Mylar - DuPont's name for biaxially-oriented polyethylene terephthalate (BoPET)

As mentioned in other answers, adhesion is an issue, and to my knowledge, the processes to bind Pyralux and metabolized Mylar are proprietary.

Chemistry.About.com does mention vapor deposition of metals into Mylar, though.


The number of geometrical isomers of complex of type [Ma₃b₂c]

First, let's establish that we're all talking about octahedral complexes to eliminate any possible confusion.
Fix $\ce{c}$. You can have either $\ce{b}$ trans or $\ce{a}$ trans.
If $\ce{b}$ is trans to $\ce{c}$, there's only one distinct way to arrange the remaining $\ce{b}$ and three $\ce{a}$'s. So that's one isomer.
If $\ce{a}$ is trans to $\ce{c}$, then the remaining two $\ce{a}$'s can be cis or trans to each other, i.e., all three $\ce{a}$'s can be facial or meridional, respectively. That's two more isomers. That's it.
I think your confusion comes about from symmetry because some of the combinations you calculated can be rotated into others.


Equalization of pressure in heat exchanger

Macroscopics

Imagine the tube is initially closed. Then one vessel is heated up. As $V$ and $m$ are constant:
$$p=k\,T$$

So if $T$ is increased, $p$ is increased. As $p$ is increased, there will be a net flow of mass, until they are balanced. That's what we observe. Now, we can think of it microscopically.

Microscopics

Initially, there are the same number of particles in each vessel. When heating up there is an increase in velocity. So, in the high-temperature vessel the same number of molecules are with higher velocity. Imagine a plane in the middle of the narrow tube, the number of collisions per time, because of velocity, will be higher in $400\ \mathrm K \to 300\ \mathrm K$ direction. It indicates a net flow of mass. This flow of mass will equal pressures on both sides.


Relating properties and structures of polymers

I think that the best thing for you to start with is to compare the radical formed low density polyethylene (LDPE) and the Ziegler-Natta high density polyethylene (HDPE).

The low density polyethene has plenty of side groups which are randomly arranged through the length of the chain, this can be regarded as a copolymer of a range of different alkenes which are randomly arranged.

The high density polyethene (HDPE) is an almost perfect homopolymer of CH2 units, this polymer has far better mechanical properties, it is a polymer where the chains pack closer to each other than the chains in LDPE.

There is a version of HDPE which has properties closer to that of LDPE, this is formed by using a mixture of alkenes using the organometallic Ziegler-Natta cataylsts.


When is the n+l rule followed when working with energy of different orbitals?

3s, 3p and 3d orbitals all have the same energy in hydrogen atoms ( or monoelectron atomic ions ) according to the nonrelativistic Schroedinger model.
Fine measurements and relativistic models of hydrogen atoms shows there is slight difference in their energies, related to the fine structure of spectra  and Lamb shift ( Wikipedia  Hyperphysics )
For multielectron atoms, s, p, d orbitals have different energies, as there come in the play 2 major effects, which affect electron energies:

Mutual electron repulsion
Screening/shielding of the nucleus charge by other electrons.


Superacids consisting of a mixture are going to have a certain range for the Hammett acidity function $\mathrm{H_0}$. For example, $\mathrm{H_0}$ for both Magic Acid $\ce{HSO3F-SbF5}$ and fluoroantimonic acid $\ce{HF-SbF5}$ relies on concentrations defined by the equilibrium of the acid's formation, which, in turn, is affected by how much $\ce{SbF5}$ has been added [1]:

  $$
\begin{align}
\ce{2 HF + 2 SbF5 &<=> H2F+ + Sb2F11-}\\
\ce{2 HSO3F + 2 SbF5 &<=> H2SO3F+ + Sb2F10(SO3F)-}
\end{align}
$$

  The acidity function of $\ce{HSO3F}$ increases from $-15.6$ to $-21.0$ on addition of $25$ mole percent $\ce{SbF5}$ ... as shown in Fig. 1. Extrapolation of the $\ce{HSO3F-SbF5}$ curve in Fig. 1 would lead to an $\mathrm{H_0}$ value of about $-25$ for Magic Acid. Fluoroantimonic acid is even stronger. As shown in Fig. 1, with $4$ mole percent $\ce{SbF5}$ the $\mathrm{H_0}$ value for $\ce{HF-SbF5}$ is already $-21.0$, a thousand times stronger than the value for fluorosulfuric acid with the same $\ce{SbF5}$ concentration. At present it is difficult to estimate the acidity of $1:1$ $\ce{HF-SbF5}$, but a value of $-28$ can be predicted ... on the basis of isomerization kinetics data.

  Fig. 1. Relative acidities of $\ce{HF}$ and $\ce{HSO3F}$ on addition of $\ce{SbF5}$.

That is why most textbooks (e.g. [2]) list $\mathrm{H_0}$ from $-21$ to $-25$ for Magic Acid and $\mathrm{H_0}$ from $-21$ to $-28$ for fluoroantimonic acid.

References

Olah, G. A.; Prakash, G. K. S.; Sommer, J. Superacids. Science 1979, 206 (4414), 13–20. https://doi.org/10.1126/science.206.4414.13.
Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, Fifth edition.; Pearson: Boston, 2014. ISBN 978-0-321-81105-9.


In rdkit, how to preserve stereochemistry in a SMARTS subgraph?

For anyone interested in this topic: I just learned that this is a bug that should already be filed on GitHub, but apparently the developers themselves do not find it anymore - probably that's why I didn't find it. Nonetheless, the bug is known and will be fixed in the next release.


How can we prove the inequality TdS > dU + pdV?

For a closed system with a single phase of a pure substance, when only expansion work is possible
$$\mathrm dU= -\mathrm p\,\mathrm dV + T\,\mathrm dS, \label{eqn:1}\tag{1}$$
where $p$ (the state variable) is the internal pressure and $T$ (the state variable) is the internal temperature of the system.
The first law, however, says more generally that
$$\mathrm dU = \mathrm dw + \mathrm dq. \label{eqn:2}\tag{2}$$
Equation \eqref{eqn:2} was derived by applying equation \eqref{eqn:1} to a reversible process.
For the reversible process, thermal and mechanical equilibrium between internal and external pressure means that $p_\mathrm{ext} = p$ and $T_\mathrm{ext} = T,$ and, combined with the second law, that
$$\mathrm dw = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV \tag{3}$$
and
$$\mathrm dq = -T_\mathrm{ext}\,\mathrm dS_\mathrm{ext} = T\,\mathrm dS. \tag{4}$$
More generally, the second law says that
$$\mathrm dS_\mathrm{tot} = \mathrm dS_\mathrm{sys} + \mathrm dS_{surr} \ge 0 \tag{5}$$
the equality holding for a reversible process. Dropping the label for the system
$$\mathrm dS \ge -\mathrm dS_\mathrm{surr} \quad\leftrightarrow\quad \mathrm dS \ge \frac{\mathrm dq}{T_\mathrm{surr}}. \tag{6}$$
Therefore for an isothermal process (thermal equilibrium condition applies)
$$\mathrm dS \ge \frac{\mathrm dq}{T}. \tag{7}$$
The inequality applies for instance to an isothermal free expansion.


Various mechanisms for interaction of gamma radiation with matter are possible. Depending on the energy and the composition of the absorbing material, the most important types for gamma radiation are photoelectric effect, Compton scattering, and pair production.

In absorption of gamma radiation of lower energy by the photoelectric effect, the photon is absorbed completely by the atom and an energetic electron is ejected. The energy of the emitted electron is the difference between the energy of the gamma-ray and the binding energy for that electron in the atom.

Gamma radiation of higher energy directly interacts with one electron by the Compton effect. Usually, this is the most probable interaction of gamma radiation. The incoming photon transfers a portion of its energy to the electron and is deflected through an angle with respects to its original direction. Since all angles of scattering are possible, the transferred energy varies from zero to a large fraction of the energy of the photon. The scattered photon may have sufficient energy to interact further by Compton effect or photoelectric effect.

If the energy of the gamma radiation is very large (i.e. it exceeds $1.02\ \mathrm{MeV}$, which is twice the rest-mass energy of an electron), the process of pair production is energetically possible. In the interaction, the photon disappears and is replaced by an electron and a positron. The process may be considered as the inverse of positron annihilation.

Your additional question about the equivalent dose is slightly misleading. The resulting equivalent dose from gamma radiation is not particularly low. By definition, the radiation weighting factor of gamma radiation is $w_\text{R}=1$. Thus, an absorbed dose of $D=1\ \mathrm{Gy}$ leads to an equivalent dose of $H=1\ \mathrm{Sv}$. However, you may ask why some other radiation types have a higher relative biological effectiveness.


Are the bonding orbitals in methane equivalent - photoelectron spectrum

This question goes along the line of what does it mean when it is said that an sp3 orbital has 25% s-character. It also intrigued me so I have tried to find answer, which would not break my hybridised orbital view.

What you clearly see from the spectrum is that there are two bands, corresponding to different energy levels in the molecule and their intensity is 3:1. You can say, it is $3\times 2p : 1 \times 2s$ and you are done. Not at all. The most important thing to deal with is the $T_{d}$ symmetry of the molecule. That means, all four $\ce{C-H}$ bonds are equivalent.

But it is this symmetry which has the answer. Leaving out the $1s$ orbital of carbon, the occupied orbitals must have the following symmetries:
$1\times A_1$ and $3\times T_2$. And this is what you see in the spectrum.

Now how it comes with the hybridisation? Upon simple HF/STO-3G calculation, the orbital populations are following:

$A_1$: 0.62 C 2s + 4x 0.18 H 1s  ... fully symmetric, great
$T_2$: 0.57 C pz + 0.3 1H 1s + 0.3 2H 1s - 0.3 3H 1s - 0.3 4H 1s
$T_2$: 0.57 C py + 0.3 1H 1s - 0.3 2H 1s - 0.3 3H 1s + 0.3 4H 1s
$T_2$: 0.57 C px + 0.3 1H 1s - 0.3 2H 1s + 0.3 3H 1s - 0.3 4H 1s

Well, as if you would copy the symmetry table ;) The downside of this picture is that all the bonds are completely delocalised (as you would expect from canonical MOs). Although observable through PES, they do not allow you to draw line between any C and H atom.

Where do we get the hybridised orbitals? If we want to localize the orbitals, whatever scheme you choose, you just form linear combinations of the existing MOs so that you have least number of atoms involved in any bond. One of such methods is NBO and as a result you obtain:

C - H1  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  1 s(100.00%)
C - H2  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  2 s(100.00%)
C - H3  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  3 s(100.00%)
C - H4  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  4 s(100.00%)

All with 2 electron occupation and energy -0.63479 Eh.

So where is the truth? ... $\Psi$ is the answer and you can massage it to your will, depending what you want to see. You want localized bonds? You will get them. You want orbitals with the proper symmetry required by spectroscopy? As you wish.

Does the hybridisation concept have sense? In localized bond picture yes. Is it observable? No, and was never meant to be.


What determines whether a double displacement reaction will occur?

The simple answer is whether the reaction leads to the formation of a solid (precipitate) or a gas otherwise there is really no reaction


How do I achieve same crystallinity when cooling polymers?

Is there a way of ensuring that the crystallinity of the polymer is
  the same for all of them?

First you need a method to assess the degree of crystallinity.  Samples of the same material that differ in degree of crystallinity should show different light transmission intensities due mainly to increased light scattering in the more crystalline samples.  If you have access to a spectrophotometer you could insert the films and measure the light transmission.  If you don't have access to a spectrophotometer, a "poor-man's" approach would be to use visual comparison of the transmitted light (use a black piece of paper and a pencil beam flashlight in a dark room).  It is important that the film thicknesses all be the same.  If there is variability in your film thickness, then cast thicker films so that the percentage variation will be smaller.

By cooling your sample more quickly, you will tend to preserve more of the amorphous nature found in the solution.  Conversely, cooling more slowly or annealing your sample will tend to increase the degree of crystallinity.

Adjust your rate of cooling on each of the films until they all produce the same transmission intensity, then the degree of crystallinity should be similar in your films.


Why don't we consider VΔP when we define Q?

Let's start by the fundamental ideas. Assume, you have a container with volume $V$ with a piston on it of  cross sectional area $A$ and container is filled with gas.
Suppose, you push the container by $\mathrm dx$ amount. So, the gas is compressed and work is done on the gas.
By the definition of work classically,
$$\mathrm dW=F\,\mathrm dx$$
Now if we use the fact that $P=\frac FA$, and $\mathrm dV=A\,\mathrm dx$, we will have $$\mathrm dW=\frac FAA\,\mathrm dx= P\,\mathrm dV$$ So, the classical definition of work only tells us the that work done on the gas should be $$\int_{V_1}^{V_2}P\,\mathrm dV=P\,\Delta V$$ if $P$ is constant.
So, the definition tells us there is no $V\,\Delta P$ term in the expresion for work. And, therfore, in the first law of thermodynamics, i.e. $\Delta Q = \Delta U - W_\text{on the system}$, also, that expression doesn't occur from work, and it is not also included in $\Delta U$ either.
But if you consider enthalpy change for a system, there $V\,\mathrm dP$ term will occur as $$\mathrm dH=\mathrm dU+\mathrm d(PV)=\mathrm dU+P\,\mathrm dV+V\,\mathrm dP$$


The most likely culprit is something acidic reacting with the baking soda (sodium bicarbonate).

Try mixing separately:

Baking soda and witch hazel
Baking soda and aloe vera gel

My guess is that one of these two has an acidic buffer in it that's causing the baking soda to turn into a science fair volcano on you.

You could try adding some calcium carbonate to the mix (crush up an antacid tablet, or buy powder from somewhere). $\ce{CaCO3}$ is more basic than $\ce{NaHCO3}$, so it might neutralize enough of the acid that it avoids the foaming.

Sodium carbonate, often sold as "washing soda", is essentially a more-basic version of baking soda and might also work.  You have to be careful not to add too much $\ce{Na2CO3}$, though, because it could make the product too alkaline and thus irritating to the skin. In particular, do not substitute washing soda for all of the baking soda—only add the amount of washing soda needed to prevent the frothing.

There are harsher chemicals that you could use to neutralize the acidity, but I wouldn't recommend trying them in a household project like this.


Is a highly oxidized element more electropositive than a lesser oxidized counterpart?

There are two definitions of electropositive:

  1:  charged with positive electricity

  2:  having a tendency to release electrons

The author is using the first definition, while you are thinking of definition 2, which is the one that usually appears in chemistry books.

Instead of "makes the atom more electropositive" it could say "makes the atom more electrically-positive".

For example according to the calculation data from Paul J. Schupf Computational Chemistry Lab

Starting from a -1 charge for Cl-

Adding one oxygen atom (ClO-) increases the charge on the Cl atom to -0.298.

Adding two oxygen atoms (OClO-) increases the charge on the Cl atom to +0.212.

Adding three oxygen atoms (chlorate) increases the charge on the Cl atom to +0.510

Adding four oxygen atoms (perchlorate) increases the charge on the Cl atom to +0.907.

Another example of a different author using the word "electropositivity" in the same way, but who makes the meaning explicitly clear is Interaction of indole derivatives with biologically important aromatic compounds. Part 22. Importance of simultaneous co-operation of hydrogen-bond pairing and stacking interactions for recognition of guanine base by a peptide: X-ray crystal analysis of 7-methylguanosine-5′-phosphate–tryptophanylglutamic acid complex

  MNDO calculations showed that the protonation of the
  guanine N(7) atom increases the electropositivity of the 8-H
  atom [ -0.014 e for guanine and 0.079 e for N(7)-protonated
  guanine],

(bracketed text in original document)

Specifically regarding chlorine, the term "electropositive chlorine" has been used to refer to Cl+, for example in the Canadian Journal of Research

Another example to look at is from the thesis Synthesis of functionalised silanes for use in the asymmetric allylation reaction:

  In silicon based C-C bond forming reactions the important factor key to their
  success is the ability of the coordination number of the silicon atom to be
  varied. As the number of bonds to silicon increases the electropositivity and
  consequently Lewis acidity on the atom increases.[reference 1]

  Upon expansion (Scheme 3; 11 to 13), and ensuing reduction in the s-character
  orbital composition at the silicon centre, the electron density decreases.

  Therefore the electropositivity and Lewis acidity at the silicon centre is
  increased. These tetravalent and pentavalent Lewis acidic silianes have been
  exploited in several Lewis acid-catalysed transformations. Once the shell has
  expanded to incorporate six subsituents it is unlikely for any further extension of
  the valence shell to occur.[reference 2]

  Silicon becomes more positively polarised with the addition of each ligand. As
  this occurs there is a shift in electron density. While the electron density is
  increasing at the ligand, it is decreasing at the silicon centre. The magnitude of
  this polarization is also dependent on the electronegativity of these ligands.

For a very old example, showing the original meaning of "electropositive" to be the first definition, see William Henry's  1823 The Elements of Experimental Chemistry, Volume 1 where on page 229 there is a section "Electro-positive Bodies" which defines electropositive bodies as bodies attracted to a negative electrode and whose inherent electrical state is therefore positive.

Also, there is an historically-interesting article "Electropositivity" and "electronegativity" J. Chem. Educ., 1938, 15 (6), p 291 discussing the inconsistent usage of the term electropositive and advocating for saying "electron donor" instead (if meaning #2 is intended) to avoid confusion.

There is also a 1949 textbook General Chemistry which uses both meanings in the same sentence, using the older meaning to define the newer meaning:

  The electropositive character (tendency to go to a more electropositive condition) of elements, in general, decreases across the periods

Overall, it is confusing to use the word "electropositive" contrary to the definition that usually appears in recent chemistry books, but not really wrong.


What properties of carbon dioxide make it a greenhouse gas?

According to the Intergovernmental Panel on Climate Change (IPCC):

  "Greenhouse gases are those that absorb and emit infrared radiation in the wavelength range emitted by Earth."

In order for a molecule to absorb and emit in the infrared (IR) region, its chemical bonds must rotate and vibrate in a manner that affects something called the molecule's dipole moment. It turns out that due to the symmetry of diatomic molecules like $\ce{O2}$ and $\ce{N2}$, this process cannot happen, and thus these types of molecules cannot absorb in the infrared, which is the wavelength range in which heat is radiated.

Because $\ce{CO2}$ and other greenhouse gases do not have this kind of symmetry, they can vibrate at specific frequencies within the IR in a manner that affects the molecule's dipole moment, and thus absorb this radiation resulting in the transfer of heat. Much of the heat radiated from the Earth's surface is of the proper wavelength (energy) to be absorbed by these gases, so the criteria given by the IPCC for a greenhouse gas are thus met.

As I'm not certain of your level of understanding of chemistry, I tried to direct my answer somewhere between purely technical and purely non-technical. If I've left anything unclear to you, please don't hesitate to leave a question in the comments.


Do electrons in covalent bonds always obey the octet rule?

The elements in borons row do not always follow the octet rule as in the case of boron fluoride as boron is in the center of the periodic table and will usually only bond to fill up its outer shell with 6 electrons. Additionally, elements with can form expanded octets which can share more than 8 electrons such as phosphorous pentachloride.  Of course hydrogen only has 2 electrons to bond with so it is also an exception to the octet rule.


Taking kinetic readings for the Briggs–Rauscher oscillating reaction

The comment by K. T. gives me a nice idea.

Set up your reaction in an environment with stable illumination and against a white background or far from any specific one. Then you can take ordered photographs that you print on transparent slides.

You can also set a cuvette in a white box and fill it with sample from the reaction vessel, to take better photographs.

After you cut them in stripes with the cuvette face size and by inserting them in the slits of the spectrophotometer you get the absorbance spectrum of a sample that mimics your reaction.

You should easily get stripes of the cuvette size, if not you can always arrange something to be measured against the appropriate background (baseline).

You can also make standards or at least checkpoints using a single component of the reaction, and in all cases measuring at a relevant wavelength.

The only hypothesis is that the A of the stripes scale linearly with that of the reaction solution, and this can be confirmed as in the above paragraph.

I never thought before of this especially because my work required some accuracy. But I could have tried it indeed when the kinetic study apparatus was already in use (just to identify a change in colour, for instance). For your essay, it might be more than enough.

If a student of mine would circumvent the given limitations in experimental set-ups using a trick as I have described, I'd be rather impressed.

edit I must add that once, for fun, I used the smartphone to photograph my finger at different levels of blood oxigenations, and after emphasising the differences looked for a trend, which I have found.  This made me consider possible colour picker applications or photography suites as a tool to quantify the color (absorbance) without even going to a spectrophotometer. Digging Internet with this in mind I have found that my idea isn't original anymore and found project similar to that in the Q, and indeed conducted using a smartphone. Examples can be easily found and downloaded searching for "smartphone as a colorimeter" and similar search keys.

https://pubs.acs.org/doi/abs/10.1021/ed300567p?src=recsys&journalCode=jceda8

is an introduction in J. Chem. Ed.


Identifying a Positive or Neutral Ligand Inside Coordination Sphere?

Books have been written on the topic of whether a certain nitrosyl ligand is the nitrosyl cation, the nitrosyl radical or — you forgot this one — the nitrosyl anion $\ce{NO-}$. This ligand is termed a non-innocent ligand, meaning we cannot just look at it and tell what it is.

You asked for an answer that does not require IR or Raman spectroscopy. Well sadly, that means one has to resort to something even more exotic such as Mössbauer spectroscopy and ESR techniques. If you get lucky, a crystal structure may also help. Because the only chances we have to somewhat determine what we are dealing with are answers to the questions:

what is the $\ce{NO}$ bond length?
what is the $\ce{M-N-O}$ angle?
what is the spectroscopic oxidation state of the central metal (not the formal one)?
what is the count of unpaired spins?

One of the most obvious data points is the bond angle. A nitrosyl cation will typically only coordinate in a practically linear matter with an angle of $\approx 180^\circ$ while the nitrosyl anion will typically coordinate in a ‘bent’ manner with an angle of $\approx 120^\circ$. However, we also have the nitrosyl radical which has been shown to adopt both bond angles so we just lost our key piece of evidence.

Even using ESR to determine the spin state (i.e. number of unpaired electrons) can be inconclusive. Taking the related example of the binding of oxygen to haemoglobin, the current consensus accepts a structure that does not immediately conform to the observed spin of $S = 0$. The answer here is antiferromagnetic coupling that keeps to spins antialigned even though they could also not couple and observe Hund’s rule.

Thus, without a plethora of experiments you cannot determine in which state non-innocent ligands such as nitrosyl, dioxygen (which can be $\ce{O2, O2^.-}$ or $\ce{O2^2-}$ etc. bond to the metal. You may want to look at the related question on the electronic structure of $\ce{[Fe(H2O)5(NO)]^2+}$ for which I went through the trouble of actually looking up literature.

Because the oxidation state of iron and nitrosyl are so unclear, a curly bracket notation has been suggested. In this case, we would be dealing with $\ce{\{FeNO\}^6}$ (if I did the maths correctly).

Thankfully, you can always name these compounds by IUPAC nomenclature. Because IUPAC nomenclature desires to allow naming even when the actual electronic structure is unknown, it has defined nitrosyl to always be a neutral three-electron donor for nomenclature purposes. Thus, we would be able to call the complex you are interested in disodium pentacyanidonitrosyliron(III).


Combining solutions with multiple concentrations - Mixing NPK fertilizer ratios

Let consider ratio of N/P.
Let fertilizer A has this ratio N/P=a.
Let fertilizer B has this ratio N/P=b.
Let you want the ratio c, where a < c < b.
Then you need to mix A : B in ratio $$\frac{ b - c}{c-a}$$
In the Central Europe with German scientific influence, it is called the "mixing cross rule".

Note that you can chose just ratio of 2 from 3 elements and only in the range between these 2 source fertilizers. The ratio of the 3rd element is determined by your prior choice for the other two elements.
You would need 3 fertilizers to mix to independently choose ratio of all 3 elements, again within the range of these sources.

A graphical chart method(*) may be more practical than a numerical method.
One can draw a equal-side triangular ternary chart. Each each corner of the triangular ternary plot represents formally a fertilizer with the solely N ( or P or K ) component.
If we take e.g. the N corner, than lines  parallel to the side opposite to N corner represent N relative content to the sum N + P + K. The same for other 2.
Within this triangle, one can draw 3 points, each representing 3 different particular fertilizers.
Then one can draw similar chart between these 3 points for these fertilizers.
Available ratios than fall into this smaller triangle.
Then one can on the big chart choose N:P:K ratios and than read from the small triangle chart the ratios of all 3 fertilizers.

There is small catch in all this. The amounts of fertilizers to be mixed should be normalized for their total NPK content.
So mixing fertilizers A, B, C would not affect the total NPK amount per volume/mass, only the component ratio.

(*) It is also called a ternary plot on Wikipedia. See also the listed referenced there:

an online tool to create ternary plots)
simple Excel template
Trip-Plot Excel package

Or you can used PDF chart template to be printed, provided by @MaxW.

Alternatively, I have prepared on Google docs the spreadsheet,

using linear algebra, which provides the mixing rations for given 3 fertilizer compositions, checking forbidden combinations.


Titration of barium(II) hydroxide by hydrochloric acid

Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid.
Unfortunately, this is wrong and not at all how stoichiometry works.
Let's focus on what's important.
In the nutshell, we are dealing with a typical neutralization reaction:

$$\ce{H3O+ + OH- <=> 2 H2O}$$

and note that $\ce{BaCl2}$ as a salt of a strong acid and a strong base won't undergo hydrolysis and, as a consequence, won't have a noticeable impact on $\mathrm{pH}$.

As you [semi-]correctly calculated, there is an excess of a base ($n_0$ values refer to the initial amounts):

$$
\begin{align}
n_0(\ce{H3O+}) &= n(\ce{HCl}) \\
  &= c(\ce{HCl})\cdot V(\ce{HCl}) \\
  &= \pu{2.4e-2 mol L-1}\cdot\pu{2.5e-2 L} \\
  &= \pu{6e-4 mol}
\end{align}
$$

$$
\begin{align}
n_0(\ce{OH-}) &= 2\cdot n(\ce{Ba(OH)2}) \\
  &= 2\cdot c(\ce{Ba(OH)2})\cdot V(\ce{Ba(OH)2}) \\
  &= 2\cdot\pu{3.4e-2 mol L-1}\cdot\pu{4.5e-2 L} \\
  &= \pu{3.06e-3 mol}
\end{align}
$$

One can see that there is a tremendous excess of hydroxide of the following amount:

$$
\begin{align}
n(\ce{OH-}) &= n_0(\ce{OH-}) - n_0(\ce{H3O+}) \\
  &= \pu{3.06e-3 mol} - \pu{6e-4 mol} \\
  &= \pu{2.46e-3 mol}
\end{align}
$$

and the leftover concentration is

$$
\begin{align}
c(\ce{OH-}) &= \frac{n(\ce{OH-})}{V(\ce{HCl}) + V(\ce{Ba(OH)2})} \\
  &= \frac{\pu{2.46e-3 mol}}{\pu{2.5e-2 L} + \pu{4.5e-2 L}} \\
  &= \pu{3.51e-2 mol L-1}
\end{align}
$$

Finally, as you rightfully suggested (use the correct value for $c(\ce{OH-})$, however):

$$
\begin{align}
\mathrm{pH} &= 14 - \mathrm{pOH} \\
  &= 14 + \log(\pu{3.51e-2}) \\
  &= 12.55
\end{align}
$$

Notes

Using the term "moles of X" is incorrect.
The proper terminology here is the amount of substance.
Never try to substitute a physical quantity with its unit.
Watch out for the proper capitalization, which is actually a big deal in natural sciences.
$\ce{HCl}$ is a formula for hydrogen chloride, whereas $\ce{HCL}$ is not and may denote "hydrogen-carbon-ligand" or be an abbreviation.
Same goes for math operators, including $\mathrm{pH}$ and $\mathrm{pOH}$: $\ce{Ph}$ is phenyl and $\ce{PH}$ can be read as "phosphor-hydrogen", or be yet another abbreviation.


There are a few different ways of making your own glue, here are four different versions you can try :

Simple Glue Ingredients:
1/2 cup flour
1/3 cup water

Paper Mache Glue Ingredients:
1 cup flour
1/3 cup sugar
1-1/2 cup water with 1 tsp white vinegar mixed in

Cornstarch Paste Ingredients:
1-1/2 cups cold water
2 tablespoons cornstarch
2 tablespoons corn syrup
1 teaspoon white vinegar

Simple No-Cook Flour Paste:
1 cup flour
1/2 cup water
1/8 teaspoon salt

Taken from http://www.wikihow.com/Make-Glue


Determine the molecular formula of the compound using percentage composition

You're actually on the right track. Looking at the percent composition, you've correctly identified that the ratio of $\ce{C}$ to $\ce{F}$ atoms is 1:1, however, you cannot assume that the formula is just $\ce{CF}$ (which isn't a known compound), it could be any compound with that ratio, $\ce{C2F2}$, $\ce{C3F3}$, $\ce{C4F4}$, etc.

The way to narrow it down is to use the gas data. You are right that you need to use the ideal gas law. If you look at the equation, you have everything except the number of moles, so you're interested in finding $n$:

$$n=\frac{PV}{RT}$$

As for the units, that depends on what units you have in $R$. If you have $R$ in $\mathrm{J/(mol\cdot K)}$, then $P$ needs to be in $\mathrm{Pa}$ and $V$ in $\mathrm{m^3}$, if it's in $\mathrm{L\cdot atm/(mol\cdot K)}$, then $P$ should be in $\mathrm{atm}$ and $V$ in $\mathrm{L}$. You can either convert what you have to match $R$ or convert $R$ to match your variables; it doesn't really matter.

Once you have the number of moles, you can calculate the molar mass of the compound from the sample mass:
$$m_\mathrm M=\frac{m_\mathrm s}{n}$$

Then, since we know the ratio of $\ce{C}$ to $\ce{F}$ is 1:1, the molar mass will be some multiple of the mass of $\ce{CF}$ which is $ 31.0091\ \mathrm{g/mol}$, so just divide the molar mass by that and you'll know $n$ in $\ce{C_nF_n}$.

Also, while not a sure way if fake compounds are possible answers, not all compounds with formula $\ce{C_nF_n}$ are known ($\ce{CF}$, for instance), so you can narrow down your choice of answer.


what is the cross aldol product of ethanal and acetone(major product)?

As pointed out by Waylander, the question essentially wants students to go for the most stable product, or thermodynamic product. This is sort of like a well-understood thing for aldol product determination questions. Why? Well... Simply because there is usually only 1 thermodynamic product while there are several possible kinetic products. It is therefore hard to argue that the kinetic product drawn is really the major product under kinetic control. Under kinetic control, you may just get a mixture of products with nothing formed in significantly higher quantity.

Based on this understanding, we can go on to determine the thermodynamic product. This reaction is a very straightforward aldol reaction due to there being only two possible reaction pathways since one reactant is a symmetrical ketone while the other is an aldehyde. So there really are only two possible products...

One possible product comes from aldehyde as a the electrophile and acetone behaving as the nucleophile. Upon dehydration, we obtain A, which is a ketone and has a di-substituted conjugated $\ce {C=C}$ double bond. The other possible product comes from aldehyde as the nucleophile and acetone as the electrophile. Upon dehydration, we obtain B, which is still an aldehyde but it has a tri-subustituted $\ce {C=C}$ bond.

Which one is the thermodynamic product? Usually, we will just look at the substitution of the conjugated double bond formed. Very clearly, the kinetic product must be A since the more nucleophilic enolate from acetone attacks the more electrophilic aldehyde. However, is B really more stable? No, it isn't. The explanation presented below is based on qualitative molecular orbital theory.

Well... The key differences between A and B are:

A, being a ketone, has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=O \pi^*}$ MO, compared to the B
B instead has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=C \pi^*}$ MO, compared to A

The former hyperconjugative interactions are stronger since the $\ce {C=O \pi^*}$ MO is lower in energy and thus more energetically-accessible to the $\ce {C-H \sigma}$ MO. This is due to the higher electronegativity of the oxygen atom. Thus, we would expect the thermodynamic stability of A to be higher than that of B.

To provide even more concrete substantiation for the higher stability of A over B, Karsten Theis has kindly provided a source of data which indicates the Gibbs free energy of formation of the two compounds. As stated on chemeo, here and here, A has $\Delta G_f = \ce {-57.48 kJ/mol}$ while B has $\Delta G_f = \ce {-36.63 kJ/mol}$. This very clearly indicates that A has higher stability than B.

Thus, A is the answer.


Counting basis sets in quantum chemistry

Yes, you have counted the number of basis functions correctly:

    #1s    #2s    #2px    #2py    #2pz    #Total
H1   1                                      1
H2   1                                      1
O    1      1       1       1       1       5

Which as you also counted is 7 basis functions.
As you also correctly state, each of the basis functions will be a linear combination of $n$ primitive Gaussians:

$$\phi(r)=\sum_{i=1}^nc_i\exp\left(-\alpha_i\left(r-r_0\right)^2\right)$$

As you have observed that is what the $n$ in STO-$n$G denotes.


How is the distance between m/z axis values in mass spectrums called?

I wanted to back up my comment and make sure it's justified, so I decided to flip through Gross'Mass spectrometry: a textbook [1] and it looks like there are indeed basically two possible variations, including distance between $m/z$ values, as you suggested:

{Distance | difference | mass difference} between {peaks | signals | peak tips}
{Difference in $m/z$ | $m/z$ difference | $Δ(m/z)$}

Both refer to describing isotopic distribution as well as to reading and interpreting the spectra.

Selected parts touching standardization (emphasis mine):

From [1, p. 11]:

  The distance between peaks on that axis has the meaning of a neutral loss from
  the ion at higher $m/z$ to produce the fragment ion at lower $m/z.$ Therefore, the
  amount of this neutral loss is given as “$x~\pu{u}$”, where the symbol $\pu{u}$ stands for unified atomic mass. It is important to notice that the mass of the neutral is only reflected by the difference between the corresponding $m/z$ values, i.e., $Δ(m/z).$ This is because the mass spectrometer detects only charged species, i.e., the charge-retaining group of a fragmenting ion.

Further summary from [1, p. 20]:

  1.7.1 Basic Terminology in Describing Mass Spectra

  […]

  Ranges in spectra or ranges set in operating a mass spectrometer are to be
  referred to in the form of “$m/z~10–100$” or “$m/z~10$ to $m/z~100$”.

  […]

  Neutral losses (rarely called dark matter […]) are exclusively recognized from
  the distance between peaks expressed in terms of the difference $Δ(m/z).$ The
  mass of the corresponding neutral is then given in units of $\mathrm{u}.$

References

Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.


What compound(s) contribute to bleach's smell and are they dangerous?

Bleach is alkaline sodium hypochlorite solution, that eventually soaked inside the wood and is gradually releasing gaseous chlorine.
It is merely just disturbing and eventually irritating mucous tissues in longer term, not really dangerous. Some chlorinated public pools have strong chlorine smell as well, being smelled from swimming suit for long time until washed in the wash machine.
A countermeasure could by applying suitable diluted reducing solution, like ascorbic acid, pyrosulfite, thiosulphate ( big crystals from photo fixers ) or similar, that would react with hypochlorite and chlorine.
Questionable can be applicability on wood surface for estetic reasons. But if bleach was applied, this should not make things worse.


Chart of Electron Affinities

The Electron Affinity can be thought of as the "electrical advantage" given by adding an electron to an atom.  So, if you have a halogen that gains an electron, it becomes more stable because now it has its octet, orbitals filled, etc.  This is why the 1st EA is large.

But if you want to add an electron, it needs to go to a whole new energy level, which will not be very desirable in terms of the electronic structure of the atom.  So the difference in electric stability or potential formed is nearly 0.  In other words, the halide ion has "0" want for another electron, since that would require adding another energy shell.

Another way to think about it is like this: the 2nd EA of a halide is basically the same as the 1st Ionization energy of the alkali metal to which it is isoelectronic (since adding an electron releases as much energy as removing that electron will take).  Since the 1st Ionization energy of a Group 1 metal is normally very small because the electron is so far away and shielded from the nucleus, so is the 1st electron affinity of a noble gas, and thus, the 2nd electron affinity of a halide.

Furthermore, now that the halogen has a negative charge, it experiences repulsion with the electron soon to be added. (Why does oxygen not like to be doubly anionized?)


Liquid with low specific heat and high thermal conductivity (relative to water)?

Most heat transfer through fluids is actually convection rather than conduction, so you need something that expands a lot with heat and has a low viscosity. Mineral oils or silicone are used as heat transfer fluids when temperatures above 100 °C are needed, but they are much more viscous than water.
However, for your system I guess that heat capacity is actually the most important number, as convection will be sufficient to ensure an equal temperature throughout the system (unless you actively impede it) even with a viscosity slightly higher than that of water.

Eyeballing some of the tables at http://www.engineeringtoolbox.com I’d suggest trying ethanol – its heat capacity is around half that of water, its viscosity is not much higher and it stays liquid over the desired range. Plus of course, unlike some of the more exotic solvents, it’s not a disaster if some of it does leak in to your food. If the food isn’t actually going to be eaten (just for demonstration purposes) then you could perhaps try toluene or petrol (gasoline).