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Why does paper crumble on getting wet?
Why does paper crumble on getting wet?
This is something I have noticed hundreds of times but cannot think of an explanation.
Does it have anything with water disturbing the intermolecular forces between cellulose micro fibrils in paper or something else?
Also, one peculiar observation is that it does not crumble immediately but only after the water evaporates? Why?

This is just a guess, but it may be due to water molecules occupying some of the intermolecular spaces, weakening the bonds that hold paper together.

Polymer properties in water often relate to the relative strength of polymer-polymer interactions versus polymer-water interactions. For example, the solubility of amylose > amylopectin > cellulose is attributed to how strongly molecules of each of these polymers interact with water molecules compared to other molecules of themselves.
In the paper making process, the cellulose fibers of paper dry out and interactions with water tend to get replaced by interactions with nearby polymer molecules. The resulting structure is strong because of polymer-polymer interactions as well as purely physical effects (e.g., "matting" of fibers, and surprisingly strong surface-surface interactions), all of which is enhanced by the compression and tension forces placed on the paper during its manufacture (google it to see photos).
When paper is later made wet, the entrance of water eventually disrupts these interactions, leading to local regions of what you might think of as micro solubility.  Then, when the paper dries out again as you describe, it probably does so without any compression/tension forces on the paper, thereby reducing the extent to which the matting of the fibers and chemical interactions between polymers takes place, and so it crumbles because of those micro regions of "solubility".
(Disclaimer: The above is just rationalizing off the top of my head. But since it is based on expertise as a PhD Chemist for nearly three decades, I'd give it a 70+% chance of being accurate ;)
The real science begins when you test the above ideas: For example, one test would be drying out paper under different degrees of compression and/or for different time periods and then testing the crumbling behavior.  Another might be to wet the paper while it's under different degrees of compression.  But the best test of the above ideas would be if there were some way to assay the extent of polymer-polymer versus polymer-water interactions -- no doubt such interactions result in very different vibrational modes. So there, I've outlined a Ph.D. thesis for you.

Welcome to Chemistry.se! These are quite thought inspiring arguments you present here. I am not sure how relevant the information that you have a PhD in Chemistry is for this answer, except for if your expertise is actually in paper processing for the last three decades, then you should rather mention this.

This is best understood as a macroscopic mechanical process: Most "everyday" paper is nothing more than cellulose pulp that is processed into an aqueous slurry, aligned and layered to a desired thickness, and then pressed to remove most of the water.  The pressing process creates mechanical links (or "entanglements") between cellulose fibers.
If you add water to this dry cellulose (which is hydrophilic) you are partially undoing the last step of manufacturing: As it absorbs water and expands, some of the mechanical links are torn.  (If you want to "recycle" the paper, you only have to mechanically agitate it with an excess of water to continue to break those links and return it to a slurry.)  Where those links are broken the paper is only held together by water-polymer bonds.  If you then let it dry without pressing or other action to restore the broken mechanical links, the paper crumbles along the lines of larger broken mechanical links.

All this is simply due to hydrogen bonds being created and destroyed. Cellulose is a rather long linear molecule with plenty of $\ce{-O-H}$ groups attached along the chain. When dry, hydrogen bonds are created randomly between $\ce{OH}$ groups from neighboring cellulose molecules. Not all $\ce{OH}$ groups are included in the final arrangement. The existence of a carbon framework products a more or less stiff network, hold together by hydrogen bonds.
When cellulose is wetted, all those cellulose-cellulose hydrogen bonds are replaced by hydrogen bonds between cellulose and water, plus multiple bonds water-water. The whole is not so rigid, as the carbon framework is not fixed any more to another carbon network. The cellulose molecules can easily move by breaking and creating all the time new hydrogen bonds with water. This is why dry paper is hard, and wet paper crumbles.


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Theoritical amount of Sodium Bicarbonate
I have to find the theoretical amount of sodium bicarbonate. The steps I followed in order to produce it are:

15g $\ce{NaCl}$ are added to a beaker. Then we add 87.5 mL of $\ce{NH3}(12M)$
We add 115g of dry ice(whose density is 1.5 g/mL)
Then the solid is taken with filtration and we rinse it first with 9mL of cold water and then with 6mL of acetone.

Then we put the solid in the oven. I know that the product after we take it off the oven will be sodium bicarbonate.
I believe that in order to find the theoretical amount of sodium bicarbonate I have to write down the reactions that are taking place.
$\ce{NaCl + NH_3 + CO_2 + H_2O → NH_4Cl + NaHCO_3}$
But I'm confused as to if there is another reaction where $\ce{NH_4Cl}$ and $\ce{NaHCO_3}$ react with water or acetone. Also, working with the stoichiometry of the reactions, from which reactant is it better to conclude about the moles of NaHCO3?

considering your second question, when looking at the reaction $\ce{2A + B -> AB + A}$, how much AB do you think you get? 1 or 2?

I get 1 AB. But we have to know the moles of A and B in order to see how much reacts from each and how much AB we will finally get.

yeah. So have you calculated how many moles you have of your reactants?

The moles of NaCl are 0.256, the moles of NH3  are 1.05. I'm confused about the moles of CO2(which is produced by the dry ice)

$\ce{CO_2}$ moles are not produced by dry ice. Dry ice is $\ce{CO_2}$.
You need 1 $\ce{NaCl}$ to get 1 $\ce{NaHCO_3}$, right? So if you have 0.256 mol of $\ce{Na}$, how much $\ce{NaHCO_3}$ could you theoretically synthesize (no losses, no by-products)?

Yes I didn't say it correctly, I meant that CO2 is dry ice. Theoretically we could synthesize o.256 mol of NaHCO3. But do we have to take into consideration the moles of NH3? Also, in order to fully conclude that the moles of NaHCO3 are 0.256, we should write the reactions between NH4Cl and NaHCO3 with water and acetone.

For the main reaction you need a 1:1 ratio of $\ce{Na}$ and $\ce{NH_3}$. How would you "take the moles of $\ce{NH_3}$ into consideration"??
Have you checked what reactions between your products and water & acetone can happen? What do you think could happen? Does it even matter? Have you done the reaction? How much bicarbonate have you got?

Since the moles of NH3 are 1.05 and the moles of NaCl are 0.256, it means that NH3 exceeds so my fault. I really don't know if they react with acetone, I just know that we used it in order to dry the solid.

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Rate of reaction with HBr

Rate of reaction is more when Carbon is more electrophile(less electron density)
So according to me the order should be D>B>C>A
As NO2 has -M -I effect , Cl has +M -I effect , while OCH3 has +M -I effect.
But the answer given as A>B>C>D

It's SN1 not 2.

@Mithoron then also the reason will be same.

Nooo... Simple check of carbocation stabilisation is what you need.

Stability of carbocation is directly proportional to +M effect .
So ,
 carbocation will more stable on A then on B and then C as NO2 has high -M. Effect it's last..

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How to compare the bond lengths of an peroxide complex before and after oxidation?

The complex $\ce{K6[(CN)5Co-O-O-Co(CN)5]}$ is oxidised by bromine into $\ce{K5[(CN)5Co-O-O-Co(CN)5]}$.

In this we have to comment on the change on length of $\ce{O–O}$.
But for that we have to find the oxidation number of $\ce{O2}$, so that we can know bond order.
But I am confused, how can I find the oxidation number?

We can't. It is a borderline case. With transition metals in unusually high oxidation states, you can never tell for sure by looking at the formula alone.

this is a case of the so-called "non-innocent" ligand: there is no way of knowing how $\ce{O2}$ as a ligand behaves without actually doing experiments on it (spectroscopy, diffraction, etc).

Exactly. So in fact, the reasoning goes the other way around: first we find the distances, then we use those to deduce the oxidation states.

@IvanNeretin I wouldn’t exactly call cobalt(III) ‘unusually high’ …

The short and unpleasant answer is that we cannot derive the (spectroscopic) oxidation states based on theory alone. We need experiments (preferably crystal structures and some way to determine the oxidation states of cobalt which are accessible by experiment).

Oxidation states are unfortunately only easy for inorganic molecular compounds or simple salts. In your case, you have a bidentate bridging dioxygen ligand. This could be any of the species $\ce{O2, O2^.-, O2^2-}$ and potentially even more (with an even weaker oxygen–oxygen bond). This is what is called a noninnocent ligand: it can undergo redox chemistry with the central atom. The difficulties of finding out what the actual oxidation state is, is outlined in this question concerning the nature of the $\ce{Fe-O2}$ bond in haemoglobin.
For these cases, IUPAC has defined a formal oxidation state which can be derived according to a set of rules so that one may at least name the complex in oxidation nomenclature. In these rules, both the cyanido and the peroxido ligand would be seen as anions, peroxido being $\ce{O2^2-}$ and cyanido obviously $\ce{CN-}$. By considering the total charge of the $\ce{[Co2(CN)10\mu{-}(O2)]^6-}$ we can allocate cobalt an oxidation state of $\mathrm{+III}$ in the reactant where the maths adds up nicely.
Upon oxidation two things can happen:

one of the two cobalt ions supplies the electron, resulting in a decacyanido-μ-peroxidodicobaltate(II,III)
the peroxido bridge supplies the electron resulting in a superoxido complex with again two cobaltate(III) nuclei.

The resulting crystal structures would be indistinguishable except if one is able to synthesise both products and compare the bond lengths. Therefore, one must resort to other methods of determining the oxidation states. The most practicable way in my humble opinion would be to examine the UV/Vis spectra of the cobalt centres. These should remain low-spin due to the ten cyanido ligands. If we happen to have two $\mathrm{d^6}$, then cobalt was not oxidised. If one of the cobalt centres turns out to be $\mathrm{d^7}$ after the reaction, this should result in a noticable change from the original reactant spectrum.
In the absence of experimental evidence we need to resort to hidden clues in the question. You were asked to consider the oxygen–oxygen bond length. If the metal centre was oxidised, this should not change significantly. However, if the peroxido ligand was supplying the electron for oxidation, the bond order would increase when the species transforms from peroxido to superoxido and this should result in a reduced bond length.
Only with this meta-knowledge we can a posteriori deduce that oxygen’s oxidation state must have gone from $\mathrm{-I}$ to $-\frac12$.

If you're interested in the electronic spectra, the oxidised form is analysed in Inorg. Chem. 1975, 14 (10), 2318–2321; it's consistent with two Co(III).

To further illustrate brilliant Jan's answer, here are the experimental crystallographic data for both structures; also notice that reduced form contains $\ce{[Co(CN)5O]}$ units tilted at about $45^\circ$ dihedral angle, whereas in oxidized form those are perfectly aligned:

Oxidized form $\ce{[(CN)5CoO2Co(CN)5]^6-}$, $d(\ce{O-O}) = \pu{1.447 Å}$, ICSD#6091 [1]:
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$.

Reduced form $\ce{[(CN)5CoO2Co(CN)5]^5-}$, $d(\ce{O-O}) = \pu{1.243 Å}$, ICSD#4093 [2]:
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$.

Bibliography

Fronczek, F. R.; Schaefer, W. P. Inorganica Chimica Acta 1974, 9 (Supplement C), 143–151 DOI: 10.1016/S0020-1693(00)89896-5.
Fronczek, F. R.; Schaefer, W. P.; Marsh, R. E. Inorg. Chem. 1975, 14 (3), 611–617 DOI: 10.1021/ic50145a035.

By the way - how did you search for the structures?

@orthocresol For inorganic structures I use (no $\ce{C-C}$ bonds) either ICSD or COD, for organic stuff (at least a single $\ce{C-C}$ bond) -- CCDC. Sometimes authors attach CIFs in ESI with the paper, but good luck finding that on a regular basis.


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How can I estimate how much humidity is present in a NaOH granulate?
I have a container with NaOH granulate. I want to use it for making soap again, as I did in the past.
When I checked it I saw big clumps because the seal was not good and humidity got in since last time. Yesterday I used it anyway, increasing the amount arbitrarily by 10%, and still was not enough.
I have the following tools:

scale (sensitivity 1 gram),
arbitrary glass or plastic volume meter calibrated with said scale,
temperature probe (up to 70C with 0.5C tolerance and 0.1C sensitivity),
white vinegar,
concentrated lemon juice,
concentrated HCl (I think 36%).

More common "kitchen" chemicals may be available, if you ask in the comments.
How can I estimate how much humidity is present in the NaOH granulate using household items/chemicals?
A measurement performed directly on the NaOH solution used for the soap (around 300 g NaOH and 750 g water) is acceptable.

could you not just dry it in the sun or in the oven at a moderate temperature?

Related: http://chemistry.stackexchange.com/q/32588/40029

@Fl.pf. well I was worried about unexpected reactions. I will do it.

I can't think of something bad to happen. 50 degree C should be enough to dry it in a short time and there shouldn't be any chemistry happening i think

For a rough estimate, you could do the following:

Weight a sample of it and heat it above 100°C for a while (say, one hour?).
Weight the sample afterwards.

The difference should be the amount of water given away and is an estimate (from below) of how much water there was in the sample.
But be aware that you may have some sodium carbonate depending on storage conditions.

Heating it in a steel pot in the oven is safe?

Stainless steel is okay, oven safe ceramic would be better. If you have parchment paper or silicone baking mats, you can use those for the surfaces that touch the NaOH.

For @Felipe Schneider's hint pointing to $\ce{Na2CO3}$, especially if the sample were openly exposed to air while being heated, I'm skeptic if simple heating will dry NaOH or rather worsen the problem. Not only worsen the problem by creating more $\ce{Na2CO3}$ (my oven is not purged by Nitrogen or Argon ...), but that NaOH, already hygroscopic at ambient temperature, will catch even more humidity.

As a speculation, perhaps following Armaregos' Purification of Laboratory Reagents may be adapted a little bit. By appropriate downscaling you might obtain a good sample of NaOH that then may used in new batch or a comparative analysis with the larger amount of NaOH kept aside.  To quote the entry in question:

"Sodium hydroxide (anhydrous) [1310-73-21] M 40.0. Common impurities are  water and sodium carbonate. Sodium hydroxide can  be purified by dissolving 100g in 1L of pure EtOH, filtering the solution under vacuum through a fine sintered-glass disc to remove insoluble carbonates and halides. (This and subsequent operations should be performed in a dry, CO2-free box.) The soln is concentrated under vacuum, using mild heating, to give a thick slurry of the mono-alcoholate which is transferred to a coarse sintered-glass disc and pumped free of mother liquor. After washing the crystals several times with purified  alcohol to remove traces of water, they are vacuum dried, with mild heating, for about 30h to decompose the alcoholate,  leaving a fine white crystalline powder [Kelly and Snyder JACS 73, 4114, 1951.]"

For the eventual comparison, you may engage a volumetric analysis.  Don't worry if you do not have a burette, a finely graduated cylinder will do so, too.  In absence of pH paper I would resort to red cabbage, too.
I mean, if the transfer of the filtrate into the vessel for evaporation of EtOH under reduced pressure is done rapidly; and not at least, if there is (borrowed?) equipment to perform all this, if a purchase of new batch NaOH isn't an option, and, and, and ...

The procedure is interesting, but do you call those "household items/chemicals"?

@OlafM It was called a speculation, as repeated soap making could be read like there were some special equipment available.  Hence all the "if" in the end, yet simultanously including the suggestion that proper disposal of the old and purchase of a new batch of NaOH were an alternative to consider -- both in investment of time, chemicals, and money.

Ok now I see what you meant.

Measure out a known volume of NaOH and weigh it, then compare this mass to the theoretical mass calculated using the density of dry NaOH granules and the known volume.

How can I measure the volume, since it contains voids between granules? if I put it in water to fill them, it dissolves.

You would have to measure out NaOH that hasn't clumped yet. The assumption that I made was that the extra water from the air was absorbed uniformly enough the mass difference will be noticeable from any sample taken from your jar.


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Diprotic Basic Buffer
So, as part of an experiment, I plan to make a diprotic basic buffer. For a monoprotic basic buffer, I used the standard combination of Ammonia and Ammonium Chloride. However, how can I make a diprotic basic buffer using commonly available reagents? If possible, could you also explain how the buffer behaves in the presence of an acid/base?

For a typical diprotic base buffer system you can use sodium bicarbonate. Adding hydrochloric acid to sodium carbonate solution
The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is:
$$\ce{Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + CO2(g) + H2O(l)}$$
If you had the two solutions of the same concentration, you would have to use twice the volume of hydrochloric acid to reach the equivalence point - because of the 1 : 2 ratio in the equation
Take a look at he sample titration curve:

The sense of diprotism of a base, is revealed through its ability to accept the two protons from hydrochloric acid to form water.
Referencing and diagrams from

pH (TITRATION) CURVES


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Why is the nitrile nitrogen sp hybridized?

This site, from which the above image was taken, indicates that the hybridization of the orbitals on the C and N atoms in hydrogen cyanide is sp. I can certainly see how that would be the case for carbon, but why need the terminal nitrogen atom have sp hybrid orbitals? Couldn't it form a σ bond with the carbon's sp orbital through its s or p orbitals and use two other p orbitals to form the two π bonds?

Where do the "lone pair" electrons end up in such scenario? Do they move into the s orbital and orbit closer to the N nucleus, or do they occupy the p orbital and spend some time between the carbon and the nitrogen?

@Max I'm not sure what you're getting at, but I'd say the former.


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Factors affecting yield of reaction

How will these changes affect the yield (increase, reduction, no change) of the products:

The product of a step doesn't get fully dried.

We weighed 14 g of a reactant instead of 15 g which are required for the experiment, but we noted down 15 g.

A reactant that has to be heated in order to produce another product, doesn't get enough heated.

If a solid gets rinsed with only cold water, but not with acetone.

For the first one, I believe that the yield will increase, because the product won't be fully dried and the water will add mass to the product.
For the fourth one, I believe that since we are not using acetone, the product won't be dry enough, so it will also increase the yield.
For the third one, I'm not sure. But I believe if the product is not heated enough, less product will be formed, so the yield will be reduced.
For the second one, I'm also not sure.

Come to think of it, real yield and apparent yield are different things, and I'm not sure which of them is referred to in the question.

I can't answer this, because it is not referred in the question.

1,2,4 seem to be asking about apparent yield. But yes, it is unclear.

Is it fair to say that for the second one, there will be no change?

1) You are correct for question one, since the product is still wet you will weigh both your product + the water, so the true mass will be less than that weighed.
2) Your yield will be lower. Since you you less reactant (14g instead of 15g), your yield will be less than it would be if you used 15g. Since you calculate your theoretical yield using what you wrote down (15g), you will expect more product, since you assume you started with 15g of reactant, not 14g.
3) You're right again. Since the reactant is not heated sufficiently, your reaction does not go to completion, and you end up with less product.
4) I think your reasoning is ok.This is a bit unclear though. Acetone is a pretty hard solvent (meaning is dissolves many materials) so depending on what you are washing you may want to avoid acetone. Since the question is rather unclear, I suggest you provide any answer that makes sense, and is supported by reasoning. For example: I believe the yield will be greater, because water is less volatile than acetone, so it will remain in my material, and cause the my product yield to increase, unlike acetone that would evaporate quickly/easily.
You could also say that assuming the material is dried afterwards, it does not matter, but removing water is more difficult than removing acetone, so suggest washing with acetone.

The acetone wash may be to remove an acetone-soluble impurity

Good point @Waylander. I don't think there is a right or wrong answer to question 4, anything that can be supported by some reasoning should be correct since the question provides too little detail to answer objectively.


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How can I generate 250 GPa to create metallic hydrogen?
In the preparation of metallic hydrogen, how does one conjure up such immense pressures?
I mean, I read that something like diamond anvils were used to contain the hydrogen, but now, in accordance with Newton's 3rd law, shouldn't the diamond also face the same extreme forces?
I mean a 10 on the Mohs scale is very high, but how does that suffice to withstand the immense force applied back by the compressed gas?

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

http://physicsworld.com/cws/article/news/2012/nov/02/improved-diamond-anvil-cell-allows-higher-pressures-than-ever-before

According to The Compressive Strength of a Perfect Diamond 410-480 GPa is possible.  So, yes, the diamond experiences the applied pressure, but it is strong enough to sustain it.
For more recent work, see Implementation of micro-ball nanodiamond anvils for high-pressure studies above 6 Mbar


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Why does permanganate oxidation of 2-nitronaphthalene not form phthalic acid?
My teacher told me that among the following compounds only compound 3 will form phthalic acid when reacting with hot $\ce{KMnO4}$.
I do understand why the other compound will not form phthalic acid.

Looking at the substituents; NH2- activates the ring, making it more reactive than dibenzene ring system (easily oxidised), and the NO2- substituent is very deactivating - (strongly electron withdrawing), making it less reactive (not easily oxidised).
Although  a powerful oxidising, optimised oxidising agent (KMnO4) is being used, it is only able to oxidise the 2-naphthylamine compound to produce phthalic acid.
and 2-nitronaphthalene is very unlikely or at least takes longer to completion with very little yield.

Could the nitro compound oxidize on the other ring, maybe?

@OscarLanzi even the second ring is already deprived of electrons (they are delocalised but spend most of the time towards the NO2- substitiuent due to its strong electron withdrawing nature, besides directly cleaving the arene bond and oxidising it is very difficult , which is the reason why reactions inlvoving arenes are determined by substiuents and **not ** on the arene ring itself. I hope it makes sense.


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Polymer synthesis with redox initiation
I made four different samples of poly acrylic nitrile (PAN) by going through a precipitation polymerization. The system was initiiated by the redox system $\ce{K2S2O8}$, $\ce{FeSO4}$ and $\ce{Na2S2O5}$.
The difference between the four samples was the used amount of $\ce{Na2S2O5}$. So my understanding was, that the $\ce{Na2S2O5}$ recovers the oxidized $\ce{Fe^{3+}}$ back to $\ce{Fe^{2+}}$. Therefore more initiations would take place. I noted the time each sample needed until they start to blur. To my surprise at a certain point, the more $\ce{Na2S2O5}$ was added the longer took the sample to precipitate. Now I wonder why this is the case. I'm thankful for every answer!


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Will Ethanol degrade Silicone or "RTV" specifically "RTV 734", or cause it to "leach" ingredients at temperatures below 250 degrees farenheit?
Will Ethanol degrade Silicone or "RTV" specifically RTV 734, a flowable sealant produced by Dow Corning, or cause it to "leach" ingredients at temperatures below 250 degrees Fahrenheit?
Such as in the distilling process.
Thanks!


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Why do the halides run in reverse orders in the spectrochemical series and the trans-effect series?
Spectrochemical series: $\ce{F^-}>\ce{Cl^-}>\ce{Br^-}>\ce{I^-}$
Trans-effect series: $\ce{I^-}>\ce{Br^-}>\ce{Cl^-}>\ce{F^-}$
This seems slightly odd to me because their relative ordering surely should be the same because both series depend on sigma donor ability. Does this have something to do with π donor ability?
In what order does σ donor and π donor ability go for halides in transition metal systems?

Yes I always found this rather irritating too. On top of the sigma donor ability, fluoride is also a better pi donor than iodide so you'd expect fluoride to be lower in spectrochemical series. Imo the only plausible explanation left is that I 5p orbitals are too diffuse and overlap poorly with TM 3d (despite the better energy match), compared to F 2p. (analogous to H-I bond being weaker than H-F)

The two features you cite — trans-effect and spectrochemical series, i.e. ligand-induced crystal field split — are based on two different phenomena, two different structures and two entirely different observations that warrant two different discussions. Thus, the probability for the two trends to be inverse is $0.5$.
Spectrochemical series
The spectrochemical series in itself is already a convolute of multiple different effects and it is defined for octahedral complexes. However, if we stick to small regions of it, we can make out trends based on general observations. Here, we are dealing with four very similar ligands which are all weak-field ligands, σ bases and π bases. To understand the effect of their π basicity and to relate them to each other, we need to take a look at the molecular orbital diagram of a typical $\ce{[ML6]}$ complex accounting for π effects. Figure 1 shows such an MO scheme.

Figure 1: Octahedral $\ce{[ML6]}$ complex including σ and π ligand-metal interactions.
We see that the $\mathrm p$ orbitals (or more precisely: that $\mathrm p$ group orbital with $\mathrm{t_{2g}}$ symmetry) interact with the metal’s $\mathrm{t_{2g}}$ orbitals in a bonding/antibonding manner of π symmetry — thus the interaction is weaker than the $\mathrm{e_g}$ interaction which features σ symmetry. In the case of all halido ligands, the ligand orbitals are fully populated meaning that the metal electrons end up in the $\mathrm{t_{2g}^*}$ orbitals which have been raised with respect to the neutral (or π-effect absent) state. This reduces the energy difference between $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ and thus the value of $\pu{10Dq}$ or the field split.
The different halides differ only in their electronegativity and thus in the relative positioning of their orbitals. The higher the electronegativity the lower the corresponding orbitals — i.e. fluoride will have the lowest-lying orbitals. Since these are all non-metals with relatively high ionisation energies (even for the anions that we are looking at here), they will all be below the metal orbitals. However, a larger interaction — and thereby a lower value of $\pu{10Dq}$ — is typically obtained if the interacting orbitals are close in energy.
Iodide is the one that is the least far away, meaning that the &#3c0; interaction is largest in an $\ce{[MI6]^n-}$ complex, meaning that $\pu{10Dq}$ is lowest meaning that it is on the very weak end of the spectrochemical series. The higher up we go through the group, the lower the halide’s $\mathrm p$ orbital energies will be, thus the weaker the interaction and the higher $\pu{10Dq}$. Naturally, the orbital energy decreases monotonously and therefore the ligand series of the halides will be as observed.
Trans-effect
The trans-effect is typically observed for complexes of platinum or related elements and is typically referred to for square planar geometries. It concerns ligand exchange which has been shown by a number of experiments to proceed via an associative mechanism:
$$\ce{[MX4]^n- + L -> [MX4L]^n- -> [MX3L]^${(n-1)-}$ + X-}\tag{1}$$
The initial and final complexes are square planar, the intermediate is trigonal bipyramidal. Going by the principle of least movement, the additional ligand $\ce L$ should approach via the distinguished axis (I’m calling it $x$) towards the inital complex and displace two ligands which will now simultaneously occupy the $xy$ plane (meaning that $z$ is the axis of the two non-moving ligands). Then, one of the three $xy$ ligands will get displaced; it has to be the one that is trans to the ligand which exercises a trans-effect and it has to be one of the two in the $xy$ plane — i.e. those that were originally on the $y$ axis.
Since the pentacoordinated transition state is obviously most important here, we should draw up its molecular orbital scheme. This is a scheme much less known than the octahedral scheme shown above so I will point out a few things. The trigonal bipyramidal complex’ point group is $D_\mathrm{3h}$. The metal’s $\mathrm d$ orbitals transform as $\mathrm{a_1' + e' + e''}$ with the orbitals within each irreducible representation being $\mathrm{d}_{z^2}, (\mathrm{d}_{xy}, \mathrm{d}_{x^2-y^2})$ and $(\mathrm{d}_{xz}, \mathrm{d}_{yz})$, respectively. The ligands’ σ-symmetric orbitals transform as $\mathrm{2a_1' + e' + a_2''}$. Broken down, the $z$ ligands are $\mathrm{a_1' + a_2''}$ and the $xy$ ones are $\mathrm{a_1' + e'}$. The π-symmetric $\mathrm p$ orbitals of the $z$-ligands should transform as $\mathrm{e' + e''}$ and the π-symmetric $\mathrm p$ orbitals of the ligands in the $xy$ plane are $\mathrm{a_2' + a_2'' + e' + e''}$. We can see a lot of $\mathrm e$ irreducibles which is where all the fun takes place. For completeness, the central metal’s $\mathrm s$ orbital transforms as $\mathrm{a_1'}$ and its $\mathrm p$ orbitals as $\mathrm{a_2'' + e'}$.

Figure 2: Qualitative molecular orbital scheme of a trigonal-bipyramidal $\ce{[ML5]}$ complex including both σ and π interactions.
The MO scheme is given in figure 2 above. The key to the analysis is to find out why certain ligands preferentially occupy the $xy$ plane but are not displaced themselves. The key lies within the orbitals that are labelled $\mathrm{e'}$: they bond in a σ fashion will all equatorial ligands in the intermediate being degenerate while in the reactant and product state they have different energies — one corresponding to a σ bond, one to a π bond. Thus, in moving from the reactant to the intermediate, we are generating a new interaction of σ symmetry which — again according to symmetry — affects all three equatorial ligands.
As I pointed out before, the lower a halide’s electronegativity the higher its orbitals energy levels and thus the greater the interaction. Therefore, by having the less electronegative halide participate in the new σ bond (i.e. by having it as part of the equatorial ligand sphere and thus by displacing the ligand trans with respect to it) will generate a more stable intermediate and will thus be a more favourable path.
Thus it is the same property of the halogens that create the two line-ups but the final effect is opposite in the two cases.

However, there is no way to know the relative ordering of ligand group orbitals and metal d orbitals a priori. The usual order is to place ligand orbitals below metal d orbitals, and if one assumes that this is the case for all four halides, then iodide would have the closest energy match with metal 3d.

I think I've read somewhere that the spectrochemical series is in order of the pi-abilities and they even give water a very small pi ability (not sure whether donor or acceptor) to add it to the list. If something is considered stronger in the spectrochemical series it's because it usually causes a bigger ligand field splitting but pi-donors decrease the gap between the split orbitals, so iodide, being weaker than fluoride (according to spectrochemical series) automatically means it has a higher pi-donor ability. And the trans-effect for 2 pure pi-donors like fluoride and iodide in comparison would then of course be stronger for the one that donates more so again iodide. At least this is how I would explain it to myself (might be wrong here).


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How do I find the moles of NaOH required to reach the equivalence point?

For each trial with the unknown acid, calculate the following:
  a) Moles of $\ce{NaOH}$ required to reach the equivalence point.

I need help with part A. I have already solved and figured out the concentration of $\ce{NaOH}$ ($.125$M). I've been reading my notes and it says moles of acid will equal moles of base. I've tried solving for both using the volumes and molarity and using the $M_1V_1 = M_2V_2$
I think that's wrong and I'm not sure how else to solve it.
Do I even have the correct equation?

Assuming you are titrating a monoprotic acid, the moles of acid and base should be the same at the equivalence point. To find the moles of base used to reach equivalence, you simply need to multiply the volume of base you added to reach equivalence by the molarity of the base (which you determined from your standardization runs). You also will then know the moles of acid, since they have to be the same at equivalence.


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According to the Molecular Orbital Theory, what makes a compound aromatic?
For example, benzene has 6 $\pi$ electrons. Which means the p orbitals will be occupied, and only the bonding molecular orbitals.
But 1,3-cyclopentadiene has 4 $\pi$ electrons, which means 2 of the 3 molecular bonding orbitals will be occupied. Why does this make it non-aromatic?

There's 4 criteria for aromaticity:
1. The compound must be cyclic. As in a ring structure of some kind.
2. The compound must be planar. All the atoms making up the ring must be on the same plane. The hybridization of bonds causes this which is why benzene is planar and cyclohexane is not.
3. The compound must be conjugated. Therefore it must alternate between single bonds, double bonds, lone pairs, or empty orbitals. For example, two double bonds in a row mean it is not conjugated. But a single bond connected to an atom with an empty orbital still satisfies the rules of conjugation. Another example, benzene alternates single bonds and double bonds, so it is conjugated.
4. Must satisfy Huckel's rule. So the number of pi electrons must fit in the equation 4n+2, with n being a whole number or 0. So 6=4n+2 works but 4=4n+2 would mean n=1/2 so that does not work.

This is correct, but I believe @terra2322 was asking how aromaticity is explained in the context of MO theory.

{4n+2} is such an obsolete way to describe the Huckel number. A compound satisfies Huckel's Rule if it contains an odd number of pi-electron pairs. Done. No more need to confuse the new students. (Not directed at Nick specifically, just trying to get the word out.)

@electronpusher I'm a little embarrassed that I never made that connection, but looking at the formula after reading your comment its fairly obvious now that is all the formula is saying.

It's ok, I felt that way too. It's still a little known simplification. An excellent example of how cumbersome formalities may persist for historical reasons, even in the face of more efficient and clear ways of presenting information.

Yeah, this Huckel number was very odd to me. I know it works, but i don't feel like it would be right to justify aromaticity using only this...

Many times i rather see if the carbons are all conjugated, if there is sp3 in the middle of them...

@electronpusher finally someone agrees with me! I also used to insist on using "odd numbers of electron pairs". More often than not, my peers would count individual electrons, try to use 4n+2; the really long way of finding.

@electronpusher Hückel's rules are derived from a very narrow set of compounds, it does not make sense to use that rule at all in the broader concept of aromaticity, as there is still no consensus of what actually constitutes aromaticity. Just drop the concept completely.

My statement was not about the validity or scope of Huckel's Rule, I just suggested improving the pedagogical presentation whenever it does happen to be invoked.


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What prevents everyday objects from reacting?
This may be a stupid question but I would like to know what prevents chemical reactions from happening among common objects in everyday life? The opposite would be, what requirements must be met for chemical reactions?
For example, as I am typing this what prevents no reaction between my fingers and the keyboard, or everyday objects such as chairs, tables, etc. In other words, why do we not constantly see chemical processes happening all around us constantly? Why don't my feet chemically react with the ground, my pencil reacts with the air, my fridge with the floor, etc...
There must be electron transfer such as conductivity, and charging but why no chemical reactions? Even as I am in AP Chemistry and understand Thermodynamics, Hess' Law, and so on, I am too embarrassed to ask my teacher with such a silly question.

Perfectly good question. Basically everything constantly reacts / interacts with it's surroundings, physically and chemically, trying to be in it's lowest energy state. Balls roll off the table to end up in a lower energy state on the ground. If you touch a pencil, there is no energy difference so nothing happens. If you soak your finger in bleach, the lower energy state is for some of the bleach to react with your skin, breaking open the cell walls, etc. until everything falls into it's lowest energy state, or chemical equilibrium. The picture is obviously more complex, but that's the basics.

I love this question!

The reason could be the spontaneity of the reaction, which can be determined by the change is Gibbs enery. For a reaction to occur freely in nature, the change in Gibbs energy must be negative.

definitly not a silly question. Prepare for a wall of text or even no answer though, because you're asking a pretty heavy question with lots of answers.

As I see it, two factors: thermodynamics and kinetics.
Thermodynamics: Many of the reactions that "don't happen" in everyday life are not favored thermodynamically, we call them non-spontaneous. This means that the products would be higher in energy than the reactants. Such reactions can certainly occur if you supply external energy (charging your cell phone battery, for instance, is a non-spontaneous process). But, without applied energy, non-spontaneous processes generally will not occur. Reactions that we observe in nature are generally spontaneous, favored by thermodynamics (with lower-energy products). Some thermodynamically spontaneous  processes are not observed to occur, even when we might naively expected them to. Example: Sugars decomposing to $CO_2$ and water is a very energetically favorable reaction, in fact the energy released from this process fuels nearly all known life. However, we do not observe our table sugar decomposing in its container because without outside help like enzymes and applied heat, the process is terribly slow kinetically.
Kinetics: For us to observe a reaction happening, it must proceed at rate that is fast relative to our human scale of time. Perhaps your fingers fusing to your keyboard is thermodynamically favored/spontaneous, but if it takes hundreds of years to occur then for our human purposes we may regard it as "no reaction". Indeed, the decomposition of most types of the biomolecules constituting our bodies is favored thermodynamically, but the kinetics are so slow (440 years for table sugar)°. There are a few ways to increase the rate at which a reaction occurs. To occur, a reaction requires a certain amount of activation energy. Think of this as hill you have to climb before you can sled down to the products (or up, potentially). The reaction does not occur if the starting material molecules do not possess enough energy to climb the activation energy barrier. You may help them overcome the barrier by supplying heat; thus, heating your table sugar will allow it to decompose into $CO_2$ and water before your eyes (the water will be released as steam of course). The other main way to increase the kinetics of the reaction is to lower the hill, instead of adding the energy to climb it. That is, use a catalyst like an enzyme to lower the activation energy barrier of the reaction. Without such external factors like applied heat and use of a catalyst, many reactions that are thermodynamically favored will be disfavored by kinetics, and not proceed at an observables rate.
So the reason the universe isn't a useless, constantly rearranging chemical soup boils down to thermodynamics and kinetics. Even if a process is thermodynamically spontaneous, it may be kinetically slow. I should point out that in a lot of places, such as the centers of stars, there is indeed a copious abundance of uncontrollable chemical and nuclear reactions. And planets like Jupiter pretty much are a useless chemical soup (no offense to Jupiterians). But in our favorite places on Earth, at room temperature and atmospheric pressure, the carbon in our bodies isn't going to undergo nuclear fission and become lead (thankfully).
By the way, this is in not a silly question. In fact, if a chemistry student doesn't wonder this, one may question if they are actually learning the subject. I applaud you for your curiosity.

This question is more than a bit ironic; without the millions and billions of chemical reactions that we each are composed of, it couldn't be asked. Two books come to mind here: The Watchmen (specifically Dr. Manhattan) and Godel, Escher Bach by Hofstadter, which you may (or may not) enjoy reading. Also this:https://en.wikipedia.org/wiki/The_Siphonaptera. There are only two thermodynamically stable minerals: quartz and darn it, I've forgotten the other one! So when you look around you, virtually nothing you see is chemically or physically stable. (Arguably, water is stable too but it is slowly being lost to space, although you and I won't have to worry about that.) So we need to be aware that most of what is going on around us we are unaware of. (Thank goodness, most of the creepy crawlies living on and in us are the stuff of nightmares..)
Anyway, while you are unaware of the sea of microbes that we are surrounded by, and of the countless chemical reactions going on around us (did you know that concrete generally has a lifetime of only 20-50 years?) that shouldn't imply that our environment is chemically inert.E-pusher mentioned activation energy, chemists generally think of it as the hill between two valleys, where the two valleys are "stable" chemical compounds and the hill is the "activated species".
For instance carbohydrate (like wood is mainly composed of) is one valley, and $\ce{CO2}$ is the other (lower energy) valley. Wood doesn't seem to do much until you light a match to it (providing enough energy to push some of it up the hill). A lot of our natural world is composed of polymers. Unlike small molecules like water or sugar, polymers (like carbohydrate, protein, plastic, paint, DNA, RNA, etc.) are typically very slow to react. It's also very difficult to get them to diffuse or dissolve (many of them, anyway).
At the other extreme, take a piece of aluminum - say an aluminum nail or screw. Cut it (or just scratch it). Within microseconds, the surface of the $\ce{Al}$ is oxidized to form a film of $\ce{Al2O3}$. Most of us don't ever realize that aluminum is a very reactive metal, much more reactive than iron or steel. One reason that we don't see many reactions occurring, is because if they did, then the object they were occurring on/in wouldn't be very useful. (Of course, cooking is just a series of (useful) chemical reactions, which allowed us to reduce the energy we needed to digest food and was probably one reason we developed intelligence).
Anyway, the basic reason why we can surround ourselves with all sorts of materials which don't seem to degrade, is due to the need for the reaction to get up that activation energy hill. Even a reaction as simple as $\ce{2H2 + O2 -> 2H2O}$ doesn't proceed very fast at room temperature (unless a catalyst, such as $\ce{Pt^0}$ (with a clean surface) is present and then whoosh!).
By the way, all of my kids are long grown up and gone, but I've still got some of their polyethylene toys around. The plastic looks almost like new, but squeeze it, or drop it, and it cracks and breaks, even though when it was new, the toy was very tough. Even polyethylene ages, and undergoes oxidation reactions. Makes you wonder about all of the plastic plumbing pipes in homes, huh?


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Why is 2s more stable than 2p, if 2p experiences greater effective nuclear charge?

I'm referring specifically to this graph, which appeared in a chemistry lecture but which the lecturer could not explain. We know that 2s is lower energy than 2p. But surely those electrons in 2p orbitals, which apparently experience greater effective nuclear charge for Z > 10, would have lower energy! Can someone reconcile my understanding of orbital energy levels and this diagram?

perhaps you could explain how $Z_{eff}$ was calculated?

Who says $2p$ has a greater effective nuclear charge than $2s$?  In elements B, C, N, O, F where it really counts in chemistry, it's the other way around!

One big issue here is that we generally don't talk about different ionization energies of anything but the outermost orbitals. Once electrons are in "core," we usually don't care since they generally do not participate in any kind of everyday chemistry.
We generally consider the $2s$ orbital to be lower in energy than the $2p$ because the $2s$ has some density inside the radial node that we think of as "nuclear penetration," i.e., the $2s$ electrons are able to get closer to the nucleus through the $1s$ core electrons. Now, once you add a $3s$ electron, I wonder if the innermost radial probability regions of the $3s$ actually repel the $2s$ electrons more than the $2p$ electrons which are generally farther away. The differential repulsion would have to happen in such a way so that it more than compensates for any nuclear penetration effects of the $2s$ electrons. This would certainly align with the observation that the $2p$ electrons are more stabilized only when we add a $3s$ electron.

Note that the x-axis is atomic number.  The graph does not allow you to compare the energy of Fluorine's 2s and 2p orbitals (Z=9) but rather shows the energy of Lithium's and Beryllium's 2s orbitals (Z=3 and 4) and the energy of Boron's through Fluorine's 2p orbitals (Z = 5 through 9).  The energies of the orbitals are not comparable on the graph because the atomic numbers are not the same.  To compare the energies of Fluorine's 2s orbitals to its 2p orbitals, you would need to calculate the effective nuclear charge felt by the 2s electrons in Fluorine and the effective nuclear charge felt by the 2p orbitals in Fluorine, both calculations using Z=9.  You can do this using Slater's rules for assigning shielding constants felt by the electrons in these orbitals.


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4-fold chemical compounds
I have been reading some papers and one mentioned "4-fold triethylamine based on PEG-diol end groups", "4-fold acryloyl chloride." What is the meaning of "4-fold" in those compounds? Also, can I find these chemicals for sale or must I synthesize them myself?

Can you link to some of these things you're referring to? There's not really enough context here to be able to help explain them.

Welcome to the Chemistry SE! You should take the tour and, if you have any further questions about the site, visit the help center. As written, your question may get closed for not providing sufficient context. You should edit your question to include links to the articles you are referring to and maybe what you hope to do with this compound. I have found a patent  that references what you are talking about. Best of luck in finding an answer!

Assuming the "4-fold" is not referring to "use four equivalents of X per molecular function Y to obtain the intended chemical reaction", it refers more often to a specical class of star shaped polymers, called dendrimers.  Commonly, around a nucleous, you have several times the same chemical group, as, for example, drawn here:

In this example, instead of a linear chain of polyethylene glycol (PEG), four branches are stitched together; the fourth example may relate the closest to your question.
Source of the picture: Zhu, Biomaterials 31 (2010) 4639-4656, doi: 10.1016/j.biomaterials.2010.02.044


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Properties of wood in furniture
What properties of wood make it useful for it to be in furniture?

Wood was always a prime building material for tools, structures, furniture and so on.
For the most part, probably because there is or was so much of it. Also, its easy to work with. You can shape wood quite easy and good.
Its also hard but not brittle, so it can withstand some pressure/weight.
Also, it grows literally on trees.
Negative side of course, it burns rather good.

Splinters are a negative too!!


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Why does internal combustion engine smoke appear blue? Could there be "blue molecules" in it?
Four-stroke and two-stroke gasoline engines as well as diesel engines can all emit dense clouds of blue-tinged smoke under various transient loads or incorrectly tuned conditions.
Raleigh scattering - in this case probably from particulates of a limited range in size - strongly favors short wavelengths, so there is a possible physical explanation.
But I would like to know if there are any chemistry effects that could lead to or contribute to the slightly blue color sometimes seen in engine smoke under these conditions.
Could there be large "blue molecules" in the smoke? I don't think so; usually gases are transparent, though there are a few notable exceptions like molecular iodine and chlorine, but I'm not sure.

above: Iodine gas, from here

above: Chlorine gas, from here

below: snapshot sitting on a bicycle, looking at adjacent scooter while holding breath and waiting for the light to change.

below: screenshot from the YouTube video Big Engines Starting Up.

Comment and not answer because I do not know if one can see it (anyway I am oriented on Rayleigh scattering as everyone here) but for sure there is a blue molecule which has been detected in exhaust and is azulene. It is a very fascinating molecules under several points of view. For sure is found in ppm in diesel engines exhaust. Take it as an answer about their presence in smoke whatever its colour is.

@Alchimista I've narrowed the language of the question to "blue chemicals" (horrible term, sorry) to cover that aspect, and to try to stay on-topic. I also assume it's Rayleigh scattering, but I don't know, nor do I even know how to ask if it's usually from unburned oil droplets, or particles or agglomerations of reaction intermediates. Nice molecule though, and first time I've seen a blue mushroom without having eaten one first!

Well azulene is a blue chemical and can be found in exhaust. My "whatever the colour is" refers to the smoke, to make clear that I am not sure if, when azulene is present in the exhaust, the latter looks bluish. I even do not think so and I think as well read that Rayleigh scattering  plays a role. It can be observed also with smoke rising gently rising from a cigarette tip, when sometimes a blu and a white streams are seen.

Very interesting, thank you for that image, I might even remember noticing that about cigarette smoke, though it could be the power of suggestion. I'll look into it.

@Alchimista it turns out I've been drinking chamazulene (also here) from time to time without realizing it! Blue anti-oxidant tea to counteract the blue cigarette smoke? ;)

:))  indeed. Another funny thing : camille tee  is well known to be a good topic remedy for eyes against inflammation. Azulene is in the composition of over the count medication for that. But if handle a small vial of azulene, it is better to open it with goggles else you cry! Paracelsus principle at work!

Interesting question. Real life question often have an open ended answer. Let us make some assumptions, smoke particles originate from incomplete combustion, however the temperature in combustion is high enough to burn most organics. This eliminates organic blue molecules, even if they exist, they cannot exist in the gas phase at those temperatures.
Now there are couple of things which give a blue hue to the smoke. One is scattering depending on the angle of view (Rayleigh scattering, Mie scattering). Secondly, another point which we our discussion might be missing is fluorescence from aromatic hydrocarbons in partially combusted diesel oil. These molecule glow really pure blue in UV. All you need is trace of a fluorescent molecule. I feel that there is some contribution of blue fluorescence in the exhaust. A quick search shows there are plenty of aromatic hydrocarbons in diesel exhaust which are fluorescent. BTW, petrochemicals can be naturally fluorescent. See this beautiful picture here
http://www.dakotatechnologies.com/learn-more/intro-to-lif/overview
If you ever wear a white shirt and expose it to UV it glows blue thanks to the aromatic dyes added to modern detergents.

hey I think we are on to something! Yes I'm familliar with bluing but I'd never thought about fluorescence in this kind of exhaust. It's theoretically testable but getting an engine blowing blue smoke and a bright UV light in a dark space together sounds like a bit of a challenge. I wonder if there's a better way to test this. Thanks for the insight! Any idea how far into the UV would be necessary? There are UV LEDs but they get more expensive the shorter you go in wavelength.

Not very far UV is needed. If it fluorescing in the sunlight how much UV do we need? Instead of doing the experiment, search the literature along these directions. Ordinary black light, is good enough for detecting fluorescence.

I've just stumbled upon Diesel exhaust; specific chemicals which lists Fluoranthene Pyrene at several thousand PPB. A four-cycle gasoline engine in poor condition as well as a two-cycle engine will be burning oil and so could also have such compounds. Fluorescence could indeed be all or at least part of the "blue" in blue smoke from such engines! I think you've almost nailed it.

It is a plausible hypothesis but I love the very phenomenon of fluorescence!

I can't really tell you the chemical difference but usually blue smoke indicates that oil is burned, too. Like some of the oil cycles get into contact with the engine and it is burned as well. And then probably heavier components of your gasoline once you start the cold engine. My Prof (lecture on Petrochemistry) once told us that there is indeed a difference between premium gasoline and the cheap one although the octane number could be similar. For the cheap ones they add a lot of stuff with a very low octane number and finally only a small amount compounds with higher octane numbers to get to 95. But it makes a difference in burning it. Maybe if the engine is still cold some of the "harder to combust" compounds don't burn that well and cause the blue color, too.
It's probably due to Rayleigh scattering as you mentioned. The sky is blue and clouds are white, the normal exhaust gases should be water and CO2 so similar to clouds and the blue particles are smaller rests which appear blue.

I have a hunch that you are right. If there is a bluish tinge it may just say something about the size distribution of particles, but if it has to do with with larger particles and their index of refraction, that could be chemical. Larger particles with a higher index of refraction at short wavelength might produce a color, and that might be related to chemistry in some way, but that might be too complicated for an answer here. Thanks for your insights on combustion as well!

The oil leaking into the engine and not being fully combusted is usually cited as the reason for the blue smoke. So you are exhausting oil droplets along with some steam; the oil droplets could be dispersed as a fine mist that facilitates Rayleigh scattering.

The blue colour is due to the particle size. In his wounderful book Heys (Chemistry Experiments for Boys and Girls) he wrote of an interesting experiment which requires you to stand near a person smoking.
If you look at the colour of the smoke from the end of a ciggertte which is emitting smoke such as one being held in the person's hand but not in their mouth it will emit smoke with a blue-white colour.
But if you look at the smoke which comes out of their mouths when they exhale it is white. The reason is that the average particle size increases inside their lungs. Please keep in mind that colour can be caused by either a coloured chemical or a physical (optical) effect caused by things like particle size.
By the way, smoking is a disgusting and carcinogenic habit.

Why does the average particle size increase? Are they growing, or is it just that the smallest particles get preferentially stuck inside the smoker?

Well I am not sure exactly why the average particle increases, the problem with Heys is that he did not explain much about why thing happen in his book. It is just a book telling you how to do chemistry experiments. I think that the smaller particles could be lost in the lungs or the particles could join together with each other or adsorb water from the water vapour in the lungs to increase in size.

I wish there was some data on this, I'd love to see it (and be disgusted as you point out). I have a hunch this is the right answer, but I can't quite click accept yet...

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Why can primary cells not be recharged effectively?

Primary cells cannot be recharged effectively because their reaction products are not in contact with the electrodes. True or false?

I got it correct just by rote-learning the theory so the answer is true. But when I tried to understand it properly I'm getting confused between a galvanic cell and a primary cell. They both work on the same principles (chemical energy is converted to electrical), but they are not the same thing? Since in galvanic cells are where there are 2 beakers with 2 electrodes, 1 in each, their products are always in contact with electrodes, opposed to primary cells where it is not modelled in the same way (i.e. just a battery where the products can move away from the electrodes)?

Galvanic/electrolyic refers to how the cell is being run. If you have an electrochemical cell, and you connect a load to it, and it produces a spontaneous current, it's a galvanic cell. If instead you connect a voltage source, and the current from the voltage source induces a reaction, it's an electrolytic cell. There's no requirement that a galvanic cell be made with two beakers and a salt bridge, that's just a very common setup for introductory chemistry experiments using the Daniell cell.
The defining characteristic of a secondary battery is that the net "electrolytic" reaction is the reverse of the net "galvanic" reaction. So a secondary battery is a galvanic cell when it's discharging and an electrolytic cell when it's charging. In a primary battery, whatever reaction happens during charge doesn't return the battery to its initial state. A common reason for this is that the galvanic reaction produces insoluble products, so the concentration is really low by the electrode. Then when you reverse the current and run it as an electrolytic cell, something else reacts instead, like water electrolyzing.


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What does the universal gas constant represent?
In the formula $$V=\frac{nRT}{P},$$ when we say $R$ is $\pu{0.082 L atm//mol K}$:

What do we mean? Specifically what does $\pu{K}$ mean here? Also $\pu{mol k}$? and $\pu{L atm}$?
Can you simplify it for a college level student who is taking a beginner course of chemistry.

Here K is the temperature in $Kelvin$ , mol K means moles Kelvin and L atm means Litre Atmospheres , they are respective units of the quantities in your formula. $n$= moles , $T$ in Kelvin , $V$ is Volume in litres and $P$ is pressure in atmospheres.

Thank you Piyush Raut . Why do they say mol kelvin ? what is the relationship between quantity and temperature?

Actually it's not like any relationship but it's the way to write the units of desired quantity like in this case $R$=$\frac{VP}{nT}$ so arranging your units for respective quantities we get $units-of-R$=$\frac{Litre-atm}{mol-Kelvin}$. I hope you got that.

@PiyushRaut, I think what the OP is asking is what is the physical representation of R, i.e. where or what is it in real life? We know what a volume is, we know what pressure is, we know what temperature is and we know what quantity is, but what does this NUMBER that ties them all together really mean to us? Most people's immediate response would be: this is a "conversion factor" or a "constant of proportionality", but these are only mathematical definitions. In other words, can R be found anywhere in this world? At least, that's my interpretation, and I think it's a great question.

@Don_S I agree that would be a great question, and that was what I was hoping for when I clicked on the question. However, it seems that OP's real issue is over what the unit symbols of $R$ actually mean. (Note that the title is not OP's.)

@Don_S I agree but the way the question was framed before being edited was like, related to units of the quantities mentioned . I agree it would have been really a good question if it would be something related to physical  interpretation of Universal Gas Constant.

R is work that a mole of ideal gas can do if heated by one degree Kelvin. It also pops up in expressions of molar heat capacity of ideal gases. However, if you are puzzled by how absurd units of R look, fear not. There are more obscure things. You should not be afraid until things like ln (m) or sin (sec) start appearing, then you may be sure something is wrong.

The universal gas constant arises because the units we use for the quantities in the ideal gas law (pV=nRT) are completely arbitrary. A harmonious set of units could be specified that allow R to be unity.
K is kelvin, the SI unit of temperature. It's based on the phase changes of water.
L is litre, or cubic decimeter
atm is atmospheric pressure
mol is the mole, defined as the number of atoms in 12g of carbon-12

note that the mole is the amount of substance that contains the same number of atoms/molecules/ions as 12g of carbon -12.

The $R$ is basically a empiric factor so that the formula gives you the right answer. The units are chosen so that the units in the formula are balanced. The $\mathrm K$ stands for kelvin (temperature).
I don't know where you got your definition of $R$ from, but in SI units it is $8.314\ldots\ \mathrm{\frac{J}{mol\ K}}$.
I would propose to always use SI units when possible.

Interesting question, I always assumed it was just a constant to adjust it to real values but there is also a formula for R I never head before. According to Wikipedia (if you trust these sources) R is also the difference of the heat capacity at constant pressure and constant volume and as we know R is the same for all ideal gases. So perhaps if you really want to describe and compare theoretical gases where you say they should behave ideal you don't want too much difference in your results once you change parameters like constant pressure or constant volume which is why you say this value, which might arise there is always the same for ideal gases but has to be included because we see that cp-cv is not equal to zero and therefore there has to be some, I just call it "intrinsic" property we have to add?
This would at least be my interpretation of this specific formula and the way how I would explain the necessity of having it to myself, like we assume an ideal gas has this perfect relationship where can bring volume and pressure together and easily calculate things but then suddenly we see that in fact keeping one of them constant causes differences in some thermodynamical calculations and we have to include something that corrects it.
Might be awefully wrong here though.


text:
Distillation of aniline/water mixture
I read from a source that aniline and water can be separated by steam distillation. But isn't this senseless as steam distillation will simply add more water and make separation more difficult?

Can't we just distill the aniline off?
The boiling point of aniline is around 180 °C. Remember that water boils at 100 °C.
Do water and aniline mix?
Yes, but the solubility of aniline in water is rather low (around $\pu{30 g/l}$).
What's the advantage of the steam distillation?
Aniline and water form an azeotrope with a boiling point around 98 °C.

Can't we just distll the water off?

@AaronJohnSabu No, because the boiling point of the azeotrope is lower than that of water.


text:
Silver Nitrate Electrolysis
i have a solution of silver nitrate at 1M. the aim of my experiment is to determine in 40 mins, how much Ag will be deposited. Should i use copper or graphite electrodes.

How much will deposit? Nothing. You forgot to turn on the current.

I'd suggest you do it with graphite electrodes.
Because:

Its  more inert than copper
If copper is used, it readily reacts and more copper will be deposited at the cathode compared to silver
the yield of silver will be greatly reduced

Also, Ag might migrate into the copper electrode... Hg certainly does.

Also, If i have a solution of AgNO3 (is diluted with water), what will oxidise and what will reduce. What are the half reactions?

@DrMoishePippik


text:
Why alkanes containing even number of carbons can be packed more efficiently than alkanes with odd number of carbon atoms?
Variation of boiling and melting points of alkanes with increase in the number of carbons

The odd-numbered alkanes have a lower trend in melting points than
  even numbered alkanes. This is because even numbered alkanes pack well
  in the solid phase, forming a well-organized structure, which requires
  more energy to break apart. The odd-numbered alkanes pack less well
  and so the "looser" organized solid packing structure requires less
  energy to break apart.
Source: https://en.wikipedia.org/wiki/Alkane#Melting_points

Why is that the alkanes containing an even number of carbons can be packed more easily compared to the alkanes containing an odd number of carbons?
Both types of alkanes can be laid next to each other; hence, why should the number of carbons even matter?

I started drawing and noticed something.

The diagram makes it pretty obvious that the alkanes with odd numbers cannot be packed efficiently.
I am not sure if this is the correct answer.

Related: Chaotic melting points of n-alkyl carboxylic acids

On that related question, DHMO shared a great article (http://pubs.rsc.org/en/Content/ArticleLanding/2004/NJ/B307208H#!divAbstract). "Several of the key structural features may be rationalised using a simple two-dimensional model that describes the packing of modified parallelograms and trapezoids, representing the even and odd acids, respectively."

@electronpusher Right on, I noticed the same thing while I started drawing the 2D structures a few minutes ago. I'll try making 3D models and see how it affects packing efficiency. I am not sure if my reasoning is correct but I guess that research paper should answer it.

A recent article in PubMed (Angew Chem Int Ed Engl. 2016 Nov 2;55(45):14090-14095.) suggests that this odd-even effect also affects translation diffusion in the liquid state.
In the introduction of the PubMed article, it also addresses your particular question and confirms your intuition that it relates to the symmetry of the molecule. Interestingly, this was only experimentally confirmed in the last decade. To quote the article:

Although this striking odd–even effect was known hundreds of years ago, its molecular origin was only revealed in the last decade. With the help of single-crystal diffraction and controlled crystal growth, the crystalline lattice of n-alkanes was determined.[5] It was confirmed that the intermolecular distances between the end groups, namely, the CH3 groups, are responsible for the alternation in the densities and melting points. In short, n-alkanes with an even number of carbon atoms pack more efficiently into ordered periodic crystalline structures and thus exhibit higher densities and melt at higher temperatures, whereas n-alkanes with an odd number of carbon atoms do not pack as effectively, and thus their densities are lower, and lower temperatures are needed for melting or freezing. A “packing effect” has been established to explain the odd–even effect as well as the related physical properties in the solid state.

Edit: For completeness, this is the article they link to that originally proved the effect experimentally: http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-3773(19990401)38:7%3C988::AID-ANIE988%3E3.0.CO%3b2-0/full


text:
The safety of homemade slime
Disclaimer: I am not a chemist, but I figured this may be the place to ask.
So, recently, there has been a lot of news going around about the homemade slime burning a child's hands and I got curious as to whether the ingredients were harmful and if the parents just didn't know any better:

Borax (Sodium Borate)
Elmer's Glue-All
Warm Water

And realized that none of these are really harmful to handle on your own.
Would mixing these create a harmful compound?
Here's one of the news articles, they're mostly all about the same incident: 11-year-old girl's hands covered in third-degree burns after making "slime"

I'm going to leave this open, because I don't think this is a personal medical question. As for 'too broad', hmm, might be a good close reason, but meh

What can I do to improve my question? I intended it to be targeted as asking "Does the mixture cause a harmful substance? If not what are some possibilities that could have caused burns in the child?" - I have since removed the speculation portion of my question.

I think this question is OK and will reopen it. It is really not a personal medical question, and if anybody wants to discuss this further, please take it to meta. [Please note the difference between, for example, "Is cyanide toxic?" and "Should I drink small doses of cyanide to cure my cancer?"] OP, I saw yesterday that you had a link to the news article - it would also be good if you could put that back in to provide some context.

I think this question is okay.

According to this article, the girl's doctors concluded that it was "an extreme" reaction to the borax.

Short Answer(as it applies to humans):
Not in short exposures.
Longer Answer:
Polyvinyl acetate (PVA or PVAc) is the substance that makes up the active portion of Elmer's / school / wood glue -- the rest being water. Sodium borate (borax) can be used to cross-link the polymers present in the emulsion to form a slimy substance.
Some polymer cross-linking processes can be exothermic, but I'm going to set that aside for now because... extended exposure to borax alone can cause the symptoms described. Some of the articles relating to the incident linked to in the question state that the child had been making and handling this material for weeks.
There are a few studies in which volunteers were given lotion/shampoo containing >1% sodium borate. After about a week, some experienced the described symptoms.
DOI:10.3109/10915818309142004

A cumulative irritancy test was used to study the effect of repeated exposures
  of a hair preparation containing 3.2% Sodium Borate and a cleansing cream containing
  1.7% Sodium Borate on the skin of 12 and 14 subjects, respectively (Table
  2). The test material was applied under an occlusive patch to the backs of subjects
  daily for 21 consecutive days. Sites were scored one hour after patch
  removal. Applications 4 to 21 of the hair preparation produced erythema and
  papules in most subjects; the total cumulative irritancy score was 571 (maximum
  score = 630). The cleansing cream caused slight erythema in two subjects only,
  resulting in a total irritancy score of 6.4. The investigators concluded that, under
  the conditions of the study, the hair preparation was a "mild to moderate"
  cumulative irritant, whereas the cleansing cream was practically non irritating.
In a similar study, the contact-sensitizing potential of a cleansing cream containing
  1.7% Sodium Borate was tested in 22 subjects (Table 2). Preliminary irritancy
  testing revealed no irritation to a 48 h patch containing this product;
  therefore, skin sites were pretreated with 5% SLS for 24 h prior to application of
  the initial induction patch. The product was applied under occlusion for 48 h,
  every other day for 10 days (five applications). Following a 10- to 14-day rest,
  48-hour occlusive patches containing the test material were applied to fresh sites
  (with and without SLS pretreatment). SLS controls were also applied. Sites were
  scored at 48 and 72 h. More than half of the subjects reacted to SLS treatment
  (with or without application of test material). No significant irritation was observed
  at sites tested with the cleansing cream alone. The product containing 1.7%
  Sodium Borate was nonirritating and nonsensitizing.

In summary, this material should be pretty safe for short term handling with unbroken skin. Continuous use increases the risk of mishandling (broken/compromised skin contact) leading to boron absorption which may result in skin irritation (redness, bumps).
Note: Abstract does not mention the specific study referenced.

Can you quote the relevant parts of the study? From what I saw, it reports it as "mildly or moderately irritating", which doesn't seem to match the symptoms in the OP.


text:
Regioselectivity in opening of cationic three-membered rings
The mechanism of electrophilic addition on alkenes often involves an electrophile attacking the double bond and causing an electromeric shift of electrons. The electrophile then forms a bridged cation if possible.

The carbon on the left side has an alkyl group which shows inductive effect (and hyperconjugation effect) which increases the electron density on the left carbon. The right carbon is not bonded by any such alkyl group. Therefore, the electron density on the left carbon must be higher than that on the right carbon.
Why does the nucleophile $\ce{Nu}$ attack the carbon which has higher electron density?

There are three factors that you should consider in nucleophilic attack.

Steric effects
Electronic effects
Stereoelectronic effects

Two of these are related to lowering the overall energy of the transition state. The third relates to how productive a collision can be.
The one that matters less here (or is already accounted for) is stereoelectronic effects. Notice that you proposed a backside attack for a $\mathrm{S_N2}$ reaction. Therefore, you have considered the proper alignment of orbitals, which is what is meant by stereoelectronic effects.
Steric effects government repulsion of large groups. In your example, attacking the more substituted side would result in greater steric repulsion since the $\ce{R}$ group is larger than $\ce{H}$. Steric arguments therefore argue for attack at the other carbon.
Frequently, the three factors do not say the same thing, and you need to use some intuition and/or experience to figure out which one dominates (in reality, the relative energies will determine the "winner").
In this case, it turns out with a carbocation and a negative nucleophile, perhaps not surprisingly, electronic effects dominate. There are actually two related electronic effects. One is that the partial positive is greater on the more substituted carbon (cf. Markovnikov rule). This carbon is hyperconjugatively stabilized (relative to the other carbon) by greater substitution. In addition, the greater positive charge also implies a weaker $\ce{C-E}$ bond, which further implies a lower energy $\ce{C-E}$ $\sigma^{*}$ orbital. We can then argue that the transition state energy will be lower given that the two interacting orbitals are closer in energy. And indeed, the electronic effects dominate this reaction giving the reactivity shown in your diagram.


text:
What is the bond energy of Carbon-Deuterium bond?
The difference in bond strength of a CH single bond and CD single bond can be explained by Kinetic Isotope Effect. I was wondering if it is possible by bond energy. I could not find the value of bond strength of CD bond. For CH bond it is 410. What is the energy for CD bond? Experimental data if available is appreciated.

related: Would C bond with H or D?

i guess it comes out to be 3.52 eV (3.51673, was the cited value), btw, heres the link:-http://www.millsian.com/summarytables/SummaryTables022709S.pdf

Bond strength is the same as bond energy. To explain one by the other is equivalent to saying "this bond is stronger, because, well, it has more strength".

@Supernova the table gave the information required.

@YAHB, glad it helped...!!!


text:
Why does one of Dalton's compound elements with oxygen and phosphor consists of four atoms instead of two?
I stumbled across a peculiar detail while reading the chapter about the  Chemical Revolution in Bowler and Morus' book Making Modern Science.
According to this book, Dalton argued that compound elements are created in the simplest possible way. For instance a compound consisting of equal number of X- and Y-atoms would then consist of one X-atom and one Y-atom. Still, in this illustration from Dalton's New System of Chemical Philosophy, oxygen with phosphor (compound 51) consists of two oxygen and two phosphor atoms. What is the reason for this?

The plate is from Dalton's book A New System of Chemical Philosophy.
The explanation of the plate is on pages 546-547 of the book.
The explanation says that "51" is "phosphorous acid" with weight 32.
(keep in mind that Dalton is basing this on O=7 and P=9)
"52" is "phosphoric acid" with weight 23.
The discussion of phosphorous acid is mostly on pages 408-410.
Initially, he contradicts the plate saying "I am inclined to believe" that phosphorous acid is one O and one P with a weight of 7 + 9 = 16.
But then he says:

If this be the case, it may appear singular that none of the other elements exhibit acid properties when combined with 1 atom of oxygen ; but it should be observed, that the phosphoric oxide is in a liquid form, and disposed to separate into phosphorus and phosphoric acid, circumstances that do not combine in regard to the other oxides.  In fact, phosphorous acid may be considered as phosphoric acid holding phosphorous in solution, rather than a distinct acid.

So, if phosphoric acid (one phosphorous atom and two oxygen atoms accord to Dalton) "holds" another phosphorous atom, the result is what is show on the plate.


text:
How to calculate molar mass of (6)UO2+2D and (6)UO2OH+D?
For a hydrogeochemistry project, I want to calculate the molar masses of some compounds. I came across a problem when encountering species like (6)UO2+2D (aq) and (6)UO2OH+D(aq). I don't know what the (6) and the +2D mean.
I started off with IPC-MS results for several trace elements and fed them to VMinteq for the speciation modelling. I tried to google the species, but that query produced no useful results. I am no chemist and I don't know "how" or "where" to search efficiently for an answer.
EDIT: another (6) appeared in FA1-UO2(6)(aq)

Welcome to the Chemistry SE! You should take the tour and, if you have any further questions about the site, visit the help center. I'm unfamiliar with the topic, are you saying you got these species from the modeling program or you saw them in a paper?  Best of luck in getting an answer!

I'm completely unfamiliar with both those (6) and +D notations. Must be field-specific.

Those species were produced by vMinteq, the modeling program. Thanks for the links, I'll make sure to study the help sites to cozy into the mores and fashion of chemSE.

Did the software provide the $m/z$ values to accompany those formulas? That might help to know. (Also, did you mean ICP-MS?)

What are m/z values, @hBy2Py? I have to go, will be back in about 3h.

$m/z$ is the ratio between the mass and the charge of a particle hitting the MS detector. That would allow to guess whether the (6) means "six of these fragments", or "a fragment that has six of these things in it"

The MS might have produced that data, but the project coordinator never talked about anything like that. I send him an e-mail to ask him what I should do of the numbers and the D's. It seems those symbols are rather unconventional. Thanks anyway. I'll report my findings on the matter when I have more info.

The (6) for uranium in this Minteq output refers to the valence number for uranium here (+6), which means that $\ce{UO2}$ is really $\ce{UO2^{2+}}$. The D is very likely deuterium, but I cannot confirm that from the user's manual.

These notations signify someone performed calculations in Visual MINTEQ accounting for dissolved organic matter (DOM) using a NICA-Donnan modeling method.
From the manual nicadonnan.pdf bundled with the installer:

In Visual MINTEQ, a “D” suffix is used to identify a diffuse-layer species such as a Donnan species.
[…]
The numbers 6, 7, 8 and 9 have no conceptual significance; they are only used by Visual MINTEQ to distinguish different humic components, and they reflect the order in which the components appear on the NICA-Donnan menu (with 6 as the starting number).

Considering the default numbering layout

and that FA1 and FA2 refer to fulvic acids containing carboxylic (FA1) and phenolic groups (FA2), OP's species can be interpreted as such:
(6)UO2+2D(aq): Weakly (electrostatically) bound uranyl $\ce{UO2^2+}$ to fulvic acid in the aqueous phase.
(6)UO2OH+D(aq): Weakly (electrostatically) bound uranyl monohydroxide $\ce{UO2(OH)+}$ to fulvic acid in the aqueous phase.
FA1-UO2(6)(aq): Organically complexed uranyl $\ce{UO2^2+}$ to carboxylic fulvate in the aqueous phase.
Several examples from the manual:

FA1-Zn(aq), FA2-Zn(aq): Organically complexed Zn to dissolved fulvic acid. Sites 1 and 2  refer to carboxylic and phenolic functional groups, respectively.
FA1-Zn(s), FA2-Zn(s): Organically complexed Zn to fulvic acid in the solid phase.
HA1-Zn(s), HA2-Zn(s): Organically complexed Zn to humic acid in the solid phase.
(8)Zn+2D: Weakly (electrostatically) bound Zn to dissolved fulvic acid.
Zn+2D(s)(6): Weakly (electrostatically) bound Zn to fulvic acid in the solid phase.
Zn+2D(s)(7): Weakly (electrostatically) bound Zn to humic acid in the solid phase.


text:
The reason for Hydrogen bonding
Why does hydrogen bonding take place? Is there any kind of electron donation? Or does something like an ionic bond take place?

http://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-does-it-really-need-fluorine-oxygen-or-n

http://chemistry.stackexchange.com/questions/35488/is-hydrogen-bonding-a-type-of-dipole-dipole-interaction


text:
Oxidation of ethanol to acetaldehyde
In my book, it is written the best method to oxidise ethanol into acetaldehyde is a three step process:
$\ce{(i) TsCl (ii) DMSO (iii) NaHCO3}$
Why is this better than other methods, such as the following:
$\ce{(A) H+/KMnO4, heat}$
$\ce{(B) H+/K2Cr2O7, heat}$
$\ce{(C) MnO2, heat}$

Well, actually getting ethanal looks like major plus here.

@Mithoron then in others what are other products

Why it has got too many down votes

Because you didn't say anything about your knowledge, simply putting random oxidants of alcohols is, welll...

Because none of the other methods quoted will work. Permanganate and Dichromate will overoxidise and give acetic acid. Manganese dioxide won't touch it.

why MnO2 will not touch it ?

MnO2 only oxidises benzylic and sometimes allylic alcohols.

I stand corrected. Manganese dioxide WILL oxidise alcohols to aldehydes under solvent free conditions. Haven't actually used it myself for thus. Tet. Letts 43 2002 6149

I heard that dichromate could work too in controlled situation.

Dichromate plus pyridine so that you have in-situ PDC should be OK but that's not what the OP specified


text:
Hydrohalogenation vs Hydration of Alkenes
So, I am doing an investigation involving the hydration and the hydrohalogenation of ethylene. Essentially, to produce the ethylene, I heat ethanol, in the presence of concentrated sulfuric acid, to obtain ethylene gas, which I plan to bubble into a solution of hydrochloric acid at varying concentrations.
Now, when the ethylene is bubbled in, I imagine that it will react in the solution and there are two possible products it could potentially form; Ethyl Chloride (from the hydrohalogenation mechanism) and ethanol (from the hydration mechanism). The problem is that I want to make a prediction of which product would be found in greater proportion, though I'm not very sure how I might go about doing this.
I'm thinking that, because the HCl dissociates to form hydronium ions and chloride ions, there are more hydronium ions than HCl molecules and therefore, hydration is likely to take precedence.
However, I've also thought that, in both mechanisms, the first step is protonation. In that case, it would also be possible that the chloride might add to the positively charged ethyl carbocation to a greater extent than the water molecules. Again, this is because both the carbocation and the chloride are charged and attract more easily. The oxygen in water is only partially charged and therefore, does not act as a good nucleophile in this case. However, I'm not sure how valid my arguments are.
Could someone help me out with pointing out flaws in this scheme of work?

"I'm thinking that, because the HCl dissociates to form hydronium ions and chloride ions, there are more hydronium ions than HCl molecules and therefore, hydration is likely to take precedence."
Definitely don't go by the that logic. The first step of hydrohalogenation requires the dissociation of $HCl$ and produce $H^+$ (/hydronium), so that the $H^+$ can activate the alkene forming the carbocation intermediate. This is also the first step of the competing hydration mechanism, as you ended up realizing.
The key here is the second step, where a nucleophile attacks the carbocation to form the neutral product. The nucleophilicity of water and chloride aren't extremely different. One might argue that "in a vacuum", chloride is a better nucleophile due to the charge (though that charge is quite stabilized over a large volume). But, practically speaking, water tends to attack instead of chloride in most situations. This is because when both are present, water tends to be in much higher concentration which increases the rate of the water attack pathway.
So consider your conditions, and if water is in much higher concentration than chloride, I would assume the hydration pathway will dominate.


text:
How can I calculate the pH of the solution created by reacting sodium with water?

$\pu{23mg}$ of sodium was added to $\pu{100cm3}$ of water.

What is the pH of the solution produced?

$$\ce{2Na + 2H2O -> H2 + 2NaOH}$$
So firstly to find the moles, I've divided $\pu{0.023g}$ by $23$ to reach $\pu{0.001 mol}$.
Am I correct that the concentration of $\ce{NaOH}$ would be $\pu{0.01mol\,dm^{-3}}$?
$[\ce{OH-}] = 0.01$ which should be multiplied by $2$ due to the balancing numbers.
So $K_\mathrm{w}$ divided by $0.02$ gives $5\times10^{-3}$
$$
\mathrm{pH} = -\log[\ce{H+}]\\
\mathrm{pH} = 12
$$

Is this correct?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

You answer is right, but your reasoning is not.
Amount of $\ce{Na} = \frac{0.023}{23} = 0.001$ mole
$[\ce{NaOH}]= \frac{0.001}{0.100} (\mathrm{moleL^-1}) = 0.01$ M
$\downarrow$
$p\ce{OH} = 2 \to p\ce{H} = 14 - 2 = 12$


text:
Methyl Orange & Sodium bicarbonate
I found a old bottle of methyl orange in my microscope kit, in solid form. I added water to it to form methyl orange solution. I tried putting 2 drops of methyl orange on a spot plate, then adding sodium bicarbonate. I got pink foam, which is not what i expected, as methyl orange is suppose to turn yellow on sight of a base. Does anyone know what this "methyl orange" might be instead of methyl orange? This did come from the manufacturer of the microscope kit, which has a label saying Methyl Orange, so I am confused on why it does not act as an indicator. It was originally in the kit to act as a dye for biological samples.

mightve been oxidised due to exposure , since you mentioned its old.

@Supernova In some dye collections, the samples of methyl orange, kept in glass jars, and hence maybe exposed to a little of air, are decades old and still work fairly good as pH indicator.

"Formation of foam" upon addition of NaHCO3, a weak base, may be indicative for the liberation of CO2.  Methyl orange is a sulfonate dye, i.e. an acid, and could ‒ if not delivered as a sodium salt ‒ in principle trigger the decomposition of NaHCO3.  But its solubility in water is rather low and I assume your freshly prepared solution was a diluted, one, too. (Which colour spot?) That it is sometimes intentionally mixed with xylene cyanol (https://en.wikipedia.or/wiki/Methyl_orange)  doesn't match with your likely addition of a lot of  NaHCO3  (compared to the number of molecules of dye), too.

so is this because it is old?


text:
Why does chromium(III) have the highest tendency to form complexes, comparatively, among Mn(III), Cr(III), V(III) and Ti(III)?
I was looking at some quick revision notes for d&f block elements and it had this random fact: Out of $\ce{Mn^3+}$, $\ce{Cr^3+}$,  $\ce{V^3+}$, and $\ce{Ti^3+}$, chromium readily forms complexes. I did not understand which properties of chromium make it advantageous for it to form complexes, was it due to its size? Its electronic configuration? Were there any other properties at play?

The premise of the statement is skewed. All four of those have the same (rather strong) tendency to form complexes since naked metal cations in aquaeous solution simply do not exist. However, a point may be made that the complexes of chromium(III) are the least labile (or most inert) ones of the series.
Th reason for this lies in the typical octahedral arrangement observed in transition metal aqua complexes. This arrangement causes splitting of the $\mathrm d$ orbitals (which are degenerate in ground state) into two sets of $\mathrm d$ orbitals: $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In very basic crystal field theory, this can be understood as the $\mathrm{e_g}$ orbitals being those pointing towards the ligands approaching along the coordinate axes and bearing a point-shaped negative charge; thus, these orbitals are destabilised (raised in energy). This affects the orbitals $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2 - y^2}$. On the contrary, the other three $\mathrm{d}$ orbitals — $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ — are considered stabilised since the entire transformation should preserve the total energy. (This concept is not physico-chemically accurate in the way it is presented here but close enough.)
The energy difference between the two energy levels is defined as (the relative value of) $\pu{10Dq}$. Again, using the ‘conservation of total energy’ principle, the higher-up orbitals are defined as having an energy of $\pu{+6Dq}$ each while the stabilised orbitals are defined as having $\pu{-4Dq}$ each. A ligand field stabilisation enthalpy or LFSE can be calculated by taking the electrons in each orbital and adding up the corresponding $\pu{Dq}$ value to give a total energy; the higher this is, the more stable the configuration is relatively.
All complexes in question are high-spin complexes, meaning that even though the $\mathrm d$ orbitals are no longer degenerate they are still filled according to Hund’s rule. Since the order of the elements in the periodic table is $\ce{Ti, V, Cr, Mn}$ and since each is in the same oxidation state ($\mathrm{+III}$), we can instantly see that they differ by the number of $\mathrm d$ electrons only: they have the configuration $\mathrm{d^1, d^2, d^3}$ and $\mathrm{d^4}$, respectively.
If we add between one and four electrons into the five $\mathrm d$ orbitals according to Hund’s rule and then calculate the LFSE, we will get the following result:
\begin{array}{cr}\hline
\text{conf.} & \text{LFSE}\\\hline
\mathrm{d^1} & \pu{-4Dq}\\
\mathrm{d^2} & \pu{-8Dq}\\
\color{red}{\mathrm{d^3}} & \color{red}{\pu{-12Dq}}\\
\mathrm{d^4} & \pu{-6Dq}\\
\hline\end{array}
The highlighted value is the configuration with the lowest energy, i.e. the one which is most stabilised during complex formation. Therefore, chromium(III) complexes are the most inert complexes of the series.


text:
How do nylon and polypropylene blend together in the production of a toothbrush?
Why is nylon (nylon 6) and polypropylene chosen together to make a toothbrush? Specifically, how are they blended together during the production process, and what determines their price?

A toothbrush need not be made of the two materials you mention.  The two materials are choosen because of their low price of production*), ease of manufacture of the intended tooth brush (heat molding of the frame), moderate / tuneable flexibility of the material for the bristles (nylon).  In comparsion to bristles of natural origin, nylon equally is easier to cleanse by rinsing with water, and dries rapidly ‒ offering less ground for bacteria to grow on them.
*) Waste disposal is not accounted here. While the brush, consisting of polymers, may be incarcerated, why do we throw away the whole toothbrush away if apparently only the heads / bristles wear out?


text:
HCl + NaOH: Why H prefers to move to O instead of staying with Cl?
The question is - fundamentally why does H "decides" to replace Na? Why is it more preferable than staying with Cl?

This "replacement" teaching should be banned, it seems. This is acid-base reaction!

I know that this is an acid-base reaction. The question is why (fundamentally, at low level) it happens.

Because hydroxide is a stronger base than chloride. Chloride is a much larger ion, and thus its negative charge is spread over more volume, stabilizing it. The negative charge on the oxygen in hydroxide is spread out over less volume, leaving it more reactive. Thus, when a $\ce{H^+}$ ion is available, it will preferentially be retained by the hydroxide rather than the chloride.

We can also turn to the final arbiter of spontaneity: thermodynamics. This reaction at the standard state (1 M concentration, 1 bar pressure) is spontaneous to the tune of $\Delta_r G^\circ = -80.4\ \mathrm{kJ/mol}$ at room temperature.

The quantitative information is nice, thanks for sharing. I think the spontaneity can be accounted for (qualitatively) by simply saying which base is stronger, because basicity is defined in terms of thermodynamics in the first place.

Thanks! One piece that I still don't get though: why does NaOH give away Na in exchange for H?

Honestly, the most practical way to think of it is the Na+ wasn't there in the first place. Na+ acts here as what we call a spectator ion, playing essentially no role other than to balance the charge of its negative partner. When in solid form as NaOH(s), this role is important (see Ionic Bonding and Crystal Lattice). However, when NaOH hits water, it instantly splits up into Na+ and OH- (we say the sodium hydroxide compound dissociates into these ions). We rationalize this by supposing the ions would rather interact with water molecules than each other (see Solvation and Lattice Enthalphy).

However, OH- would rather interact with an H+ ion than with water molecules, because it can form a chemical bond and thus both ions (hydroxide OH- and the proton H+) become a more stable product: water. The forming of a covalent bond is not observed with Na+, not by hydroxide or any other compound that I've heard of. So you could think of when you put hydroxide in a solution of water and H+ (and Cl-, also a nonreactive spectator ion here), the hydroxide ditches the sodium ion it was settling for (dissociates), and also happens to then find a partner "for life" (reacts chemically with H+).

Adding to electronpusher's answer, one must keep in mind that when HCl is dissolved in water, it doesn't exist as HCl molecules anymore. Rather, HCl ionises to form $\ce{H+}$ and $\ce{Cl-}$ ions. So in the aqueous state, the H isn't actually bonded to Cl, rather you can think of it existing as $\ce{H+}$ ions floating freely in water (its not really like that). The same goes for NaOH, in the aqueous state it really exists as $\ce{Na+}$ and $\ce{OH-}$ ions. However, when $\ce{H+}$ and $\ce{OH-}$ ions meet each other, they react to form $\ce{H2O}$, which actually exists as a molecule rather than as separate ions. And like mentioned in electronpusher's answer, the reason why they react in the first place is because $\ce{OH-}$ is a stronger base than $\ce{Cl-}$.