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Solubility of NOx gases
Are the oxides of nitrogen (such as $\ce{NO}$ and $\ce{NO2}$) emitted from the exhaust of a diesel engine soluble? If so, in what solvents is it soluble?
I want to analyse the nitrogen oxides in the exhaust gas of a diesel engine. Is it possible to collect the gas by solvent extraction method?

$\ce{NO_x}$ can be quantitatively removed from such combustion streams by absorption/dissolution in basic aqueous solutions containing an appropriate oxidizing agent. While $\ce{NO2}$ is readily soluble in aqueous solution, $\ce{NO}$ is not. However, $\ce{NO}$ is easily converted to $\ce{NO2}$ when an oxidant is added to the solution.
The PhD thesis by Chen-Lu Yang at NJIT, "Aqueous Absorption of $\ce{NO_x}$ Induced by Sodium Chlorite Oxidation", concluded the following:

$\ce{NO2}$ is essentially quantitatively absorbed in $\ce{NaOH}$ aqueous scrubbing solutions.
$\ce{NO}$ is not absorbed by $\ce{NaOH}$ aqueous scrubbing solutions.
$\ce{NaClO2}$ containing aqueous scrubbing medium quantitatively oxidizes $\ce{NO}$ to nitrate [which is highly water soluble].

The net effect of these solubility/reactivity properties is that $\ce{NO}$ and $\ce{NO2}$ can be quantitatively removed from a gas stream using an aqueous solution of $\ce{NaOH}$ and $\ce{NaClO2}$
Additionally, according to this PubChem entry, $\ce{NO}$ is soluble in sulfuric acid, ethanol and carbon disulfide, while according to this Wikipedia article, $\ce{NO2}$ is soluble in carbon tetrachloride, nitric acid and chloroform.


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Experiment using hydroxides of alkaline earth metals
In three different test tubes we add a very small amount of:
1)In the first test tube, we add oxide of magnesium
2)In the second test tube, we add hydroxide of calcium
3)In the third test tube, we add hydroxide of barium
Then we add deionized water in the three test tubes. We shake them and after we filtrate them, we measure the pH of each filtrate.
I'm curious why we use hydroxides of calcium and barium but we use oxide of magnesium instead of hydroxide of magnesium?

I would attribute this two 2 reasons:
Reason 1: Insolubility of Magnesium hydroxide.
Magnesium hydroxide is almost insoluble in water and will not easily dissolve in water.
Reason 2: Safety is one of the reason of the choice of compounds.
The formation of hydroxides of alkaline earth metals are generally exothermic and this intensity increases down this group.

Reaction of magnesium oxide with water will produce the
corresponding hydroxide:
$$\ce{MgO + H2O → Mg(OH)2  (very slight reaction)}$$
(The reaction quickly stops because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction).
The reaction of calcium oxide to produce its hydroxide is
exothermic:

$$\ce{CaO (s) + H2O (l) ⇌ Ca(OH)2 (aq) (ΔH_r = −63.7 kJ/mol of CaO)}$$

Barium oxide is  harmful to human skin and may cause irritation on exposure for prolonged periods or in large quantity.

Its reaction with water is also know to be violent.

So in my opinion it is much better to user hydroxides of Ba and Ca.

Probably a matter of availability/cost.  Magnesium oxide is easily obtained by calcining the hydroxide or carbonate and can be kept in its oxide state, whereas with calcium and heavier alkaline earth metals the anhydrous oxide is much harder to obtain; the hydroxides or carbonatesrequire higher temperatures to calcine in the first place.  Thus for the heavier alkaline-earth metals the hydroxide is more readily available instead of the pure oxide.
But do not fuss too much over this.  Both oxide and hydroxide should react the same way.  Magnesia should give a slightly alkaline solution, lime will give a more alkaline solution and the barium compound will act as a full-fledged strong base.

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Have the shapes of synthesized anthropomorphic molecules (nanoputians) been verified experimentally?
Have the shapes of synthesized anthropomorphic molecules (nanoputians) been verified experimentally?
X-ray diffraction, atomic force microscopy imaging of single molecules, anything?
I was tempted to use the title Has anyone ever seen a nanoputian?
Edit: The IBM video World's Smallest Movie is not of a molecule, but it does contain an anthropomorphic shape built from individual atoms, visualized at atomic resolution using scanning atomic force microscopy.
below x2: Simple illustration of AFM imaging of organic molecule shape. From here. For more recent experiments and videos, see 30 years of atomic force microscopy: IBM scientists trigger and observe Reactions in an individual molecule and also IBM & Warwick Image Highly Reactive Triangular Molecule for the First Time.

below: Lewis structure of a nanoputian; "Chemical structure of 2-(2,5-bis(3,3-dimethylbut-1-ynyl)-4-(2-(3,5-di(pent-1-ynyl)phenyl)ethynyl)phenyl)-1,3-dioxolane." From here.

below: Lewis structure of a nanoputian; "Ball-and-stick model of the nanoputian molecule, also known as nanokid, an synthetic compound whose molecules are human-shaped." From here.

In analogy to Jonathan Swift's Lilliputians, shouldn't the correct spelling be Nanoputians?

@aventurin it seems you are right! Perhaps subconsciously I was thinking nanoputan (sic) sounds more like a chemistry term, e.g. furan.

While I doubt the represented conformation is the only one to consider (free rotations mainly around C($sp^3$)-C($sp^3$) as well around C($sp$)-C($sp$) bonds), maybe user https://chemistry.stackexchange.com/users/18002/nanoputian does follow up their properties closely?

@Buttonwood you've made a helpful point. I've asked this follow up question about the molecular structure, rather than the shape. Still, I'm surprised no-one, not even someone at IBM Research–Zurich  has let a drop evaporate on a flat surface and made some AFM images (see edit).

@uhoh Regarding the molecular structure (in terms of how the atoms are connected with each other), the SI of the article  are accessible freely (just answered to you more recent question addressed).

@Buttonwood I'm usually reading papers in other disciplines where there is rarely SI. If I'm going to try to learn anything at all about organic synthesis, I'd better spend more time checking the SI carefully. Thanks for the advice.

I think to solve the question "seeing one of the nanoscholars, nanotexan, etc." in this particular conformation shown requests (at least) two steps:

The deposit of a single molecule on a clean, flat, and cold surface.  From the outsider's perception of work done @IBM Zurich and other groups, preparing such a pad accommodating an organic molecule is something understood well; perhaps some adjustments in how the molecules are then deposited are needed.  Are nanoputains as substance well sublimable, would be OMBE feasible?
The cold surface tightens the contact between molecule and substrate, lowers thermal vibrations of the molecule deposited; contributes that the then collected AFM is crisp.  Provided the earlier mentioned conformational flexibility of the material (not necessary flat molecule, either), how do you -- if necessary -- decoil the members of a nanoputain?  To move small molecules entirely, optical tweezers come to my mind, but here it were moving just a portion of a molecule representing a leg, redressing bow tie and hat that I assume as even more delicate.

So my speculation is the second part is the more challenging one.  Maybe it was already addressed as the original work was published in 2003, a citation analysis may shed some light on this aspect.

I think this is the closest an answer can come in a situation where the answer is probably "no". Non-existence proofs are tough :)

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Anitaromatic compounds: Why do they have lower stability?
What is the cause for some compounds like cyclobutadiene to be antiaromatic?
I know they don't satisfy Huckle's rule but I don't understand why that leads to the decrease in stability of the compound. Whenever we analyse stability we look at steric hindrance, delocalization, etc. Delocalization of electrons does take place, then why does the stability decrease?


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How to calculate Gibbs free energy from pKa?
Is it possible to calculate Gibbs free energy ($ΔG = -RT\ln K_\mathrm{eq}$) of an acid dissociation reaction using the $\mathrm{p}K_\mathrm{a}$? If so, how?

Are you trying to find Gibbs free energy of the reactants and products or just the change in Gibbs free energy from the reaction?

Just the change in Gibbs free energy from the reaction.

To find the change in Gibbs free energy $\Delta G$ for the reaction of an acid, you can simply use $K_\mathrm{a}$ in the formula you gave in place of $K_\mathrm{eq}$ ($K_\mathrm{a}$ is an equilibrium constant).

If you want to find the Gibbs Energy of the acid disassociation, then sure you can just use Ka. But if you want to find the Gibbs Energy for some other reaction involving the acid, you can’t.

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Can alkenes have triple bonds?
Or is it characteristic of alkynes only ? also, can they (the alkenes) have double and triple bonds in the same molecule ?

If it has a carbon-carbon double bond and a carbon-carbon triple bond the molecule is both an alkene and an alkyne.

Alkenes in the IUPAC Gold Book

A compound containing both a C-C double and triple bond is both an alkene and an alkyne. This is readily apparent in such a compound's systematic name, which will be something like alk-#-en-#-yne. Such a compound can be referred to as an enyne, or (equivalently) alkenyne, especially if the bonds are conjugated.

Alkenes can have triple bonds but, they should be between carbon and any other hetero elements and not between carbon and carbon. Otherwise, the compound ceases to be an alkene.
Feel free to correct me if I am fallible.


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How to know the kinetic rate of one reaction from 3 tests with different concentrations?
Recently, I've run an experiment about the effect of reducing concentration on rate of a reaction.
Here's the chemical reaction:
$\ce{10 NaHSO3 + 4 KIO3 ⇌ 5 Na2SO4 + 2 I2 + 3 H2SO4 + 2 K2SO4 + 2 H2O}$
I ran three tests:

$10$ cc $\ce{KIO3}$ Solution + $10$ cc $\ce{NaHSO3}$ solution _ Time: $163$ sec

$10$ cc $\ce{KIO3}$ Solution + A $6$ cc $\ce{NaHSO3}$ and $4$ cc water solution _ Time: $320$ sec

$10$ cc $\ce{KIO3}$ Solution + A $4$ cc $\ce{NaHSO3}$ and $6$ cc water solution _ Time: $713$ sec

I calculated the molarity of $\ce{NaHSO3}$ in each test and also calculated the rate of each test using the formula : $\frac{[\ce{NaHSO3}]}{t}$
Now the question is how can I know the kinetic order of this reaction?
I mean with only 3 numbers of concentration, I cannot draw a curve. Is there any other way to know the kinetic order?
Edit: I hope you already know that the curve of concentration according to time shows the kinetic order of reaction. It's just with 3 data you cannot draw an accurate curve.

Is cc an abbreviation for cubic centimetre, i.e. mL?

Yes cc is cubic centimitre, maybe I should've  written mL (mililitre).

What do  the times you give represent ? How did you obtain them?

The time to see the I2 blue colour. It means by reducing the concentration of NaHSO3, the rate of reaction also reduces. In other words, it takes more time to see the result. I've already said I ran the tests myself.

Are you allowing for the possibility of a complex reaction mechanism and thus a fractional or negative reaction order? If not and you are just trying to choose between zero, first, and second order, it would seem to be closest to linear.

zero order is linear, but first and second are not. In first order log[A] and in second 1/[A] is linear. The point is my data is only 3 concentration, few in my opinion, and on the other side the numbers are experimental, I mean resulted from a simple kinetic test. Based on 3 numbers, I cannot draw an accurate curve to make sure of the order of reaction above.

@Crimson-Ruby I realize I was misreading your data as rates instead of times. This would change my conclusion to closer to second. You can compute reaction order without a graph, its a commonly assigned problem in general chemistry courses (see example 4on chemlibre). The main problems you have see seem to be that experimental error can skew the results away from a perfect linear fit and you don't consider the possibility that the reaction has a complex.

(continued) dependence on the concentration. Your reaction could just as easily have a fractional, negative, or mixed order dependence on concentration (see the wikipedia page for reaction order).

I see. So actually my data may be nearly accurate and still the graph may not be linear. There was a formula by which I could calculate the order of reaction in proportion to [NaHSO3] : log(R1/R2)/log(A]1/[A]2 But that doesn't give me the order of whole reaction, unless and as I think it is, order in proportion to [KIO3] is zero.


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Why is the ortho isomer a major product in the nitration of toluene?
Usually when an ortho-para directing substituent is present on the benzene ring for an electrophilic aromatic substitution reaction, the para product is the major product (exceptions can be there when hydrogen bonding or ortho effect of COOH group makes the ortho product a major one.) But I don't understand why the ortho product in nitration of toluene is the major product (and the para product isn't). The ortho position is supposed to have more steric crowding, right?

Now I know nothing about this particular nitration but one thing that tends to make ortho products favorable is the inductive effect, due to which ortho position of toulene is slightly more activated than para.

i read the answers and now I'm confused . If this is the reason then what if there is an ethyl group instead of methyl . Ortho position should be major because of +I effect but this is not the case as para is major .why is that

@utkarsh Then steric hindrance increases and the nitro group will go to the para position.

Related: https://chemistry.stackexchange.com/questions/88752/why-is-the-rate-of-reaction-of-polar-eas-favored-by-inductive-effects-and-the-ra

A methyl-group is not t-butyl, so it isn't all that crowded.
There are two ortho-positions.
There is also the possibility of an ipso-attack with the nitronium donor attacking the position of the methyl group, then shifting to the ortho-position.

Given all that without any orientation effect, the ratio between ortho and para should be around 3/1 - 2/1. It is commonly observed around 3/2, so the crowding does have an effect.

There are three effects: sterics, electronics and statistics.
Yes, there is more steric crowding in the ortho position, but a methyl group isn't that big. In addition, electronic effects are stronger in ortho position than in para, since it is much closer to the +I substituent. And last but not least: there are two ortho positions which can react, but only one para-position, which can have a rather big influence if reactivity towards ortho and para position is similar.

One other factor favoring the ortho isomer: hydrogen bonding. Yes, hydrogen bonding to the methyl group.
Following the molecular orbital description of hydrogen bonding, an electron pair from the nitro group oxygen atoms can overlap the antibonding orbitals in the methyl groups, delocalizing what are formally the carbon-hydrogen bonds in the methyl group. We can represent this interaction with an additional contributing structure that no octet-deficient atoms:

A para, or for that matter meta attack would not allow this delocalization to take place due to the remoteness of the nitro and methyl functions.

Ortho nitrotoluene is a major product while para nitrotoluene is a minor product .
Reasons:
In case of reactivity:
Ortho product is more reactive comparing with para product because electron density is more enriched at ortho position compared to the para position.
In case of stability:
Para product is more stable because is away from substituent group so it does not affected by steric hinderence as the ortho position I'd more steric hindered.


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How to find pH in the following titration/buffer problem given concentration of base and acid

A $100. \text{ mL}$ sample of $0.10 \text{M } \ce{HCl}$ is mixed with $50. \text{ mL of } 0.11 \text{M } \ce{NH3}$. What is the resulting pH? (Kb for $\ce{NH3} = 1.8 × 10–5$)

I was thinking more along the lines of trying to find out the moles of acid and the moles of bases that we have and subtracting the moles and finding the concentrations so I did the following:

I first make the chemical equation:
$\ce{NH3 + H2O <-->OH- + NH4+}$
then I find that there are 0.0055 moles of NH3 and I set up the Kb problem:
$\mathrm{Kb}=\frac{X^2}{0.0055-x}$
then I find $x$ to be $3.15*10^{-4}$
then I subtract that with the moles of $\ce{H+}$ ions from the $\ce{HCl}$ and divide that with $.15$ liters to get $.065$
when I take the negative log of that I get $1.189$ but that is none of the answers.
Why is this and what source of error have I made and is there a concept that I do not understand? Thus I request for assistance.
*I do understand that we can use the Henderson Hasselbach but I was wondering how to solve this without the equation?

This is a titration problem - not necessarily a buffer problem. While you have $0.0055$ moles of $\ce{NH3}$, you have $0.01$ moles of $\ce{H}$ from $\ce{HCl}$ dissociating. $0.01 - 0.0055 = 0.0045$ moles of $\ce{H}$ left. Take H/total L to find M. Take the $-\log$ of $\ce{[H]}$.


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Calculating the pH of soft drink
Me and my classmates are doing an experiment in order to determine the pH of Sprite. We are going to conduct a titration with NaOH.
We are taking under consideration that the citric acid in Sprite is the only thing that is causing the drink to be acidic.
Our question is how do we calculate the pH of the drink.
We determined that it took 26,9 ml of NaOH to make the solution of sprite (50ml) and water (50ml) neutral (pH=7).

Recall what the definition of pH is, and apply that definition to the system (Sprite before titration).

Do you know the concentration of the NaOH solution?

the concentration of NaOH is 0,1 M

Do a test when cold and bubbly and then boil to drive the bubbles out, allow to cool, and re-test.  You will be able to determine the effect of the carbonic acid.


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Why is this molecule considered a cis isomer of 2-bromobut-2-ene?

I know that we have two CH3 groups on one side and restricted rotation around the double bond but aren't we supposed to have two similar groups on the other side as well ? Does this mean that we only need one side of the molecule to have identical groups/atoms in order to call it a cis isomer ?

Cis/trans is not applicable here, you have to go with E,Z.

You may say that the two methyl groups are cis to each other, but saying that the alkene is cis is wrong.

With the same reasoning as provided here, I recommend that you name this compound (2 E)-2-bromobut-2-ene.
Using (cis/trans) may be useful around an isolated double bond with only two substitutents. While in the example presented by you the two methyl groups are on the same side of the double bond, they are not the only ones to be considered here. Triple and quadruple substituted dienes should be named in accordance to the CIP rules only.
Edit by 02-Jan-2019:
Agreeing with a comment by @Loong, the previously suggested name, "(E)-bromo-2-butene", was replaced by "(2 E)-2-bromobut-2-ene".  This edit removes simultaneously the ambiguity about the position of the bromine, as well as the double bond.  The interested is referred to the illustrated examples of rule R-7.1.2 of the corresponding Blue Book here.

The suggested name “(E)-bromo-2-butene” is ambiguous and not in accordance with IUPAC nomenclature. The systematic name is (2E)-2-bromobut-2-ene.

@Loong I agree with your comment completely. +1.


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Chemistry Calculation Dilutions of Acid
I was presented with the following problem in my lecture and I am confused as to what to do.

You are required to prepare $\pu{250 cm3}$ of $\pu{0.100 mol dm-3}$ HCl by diluting $\pu{0.600 mol dm-3}$ HCl with water.
Calculate the volume of $\pu{0.600 mol dm-3}$ HCl that must be diluted with water.

This was my attempt.
Moles of the $\pu{250 cm3}$ of $\pu{0.100 mol dm-3}$ HCl (target acid)
= $\frac{250}{1000} \times 0.100 = 0.0250$ moles.
Hence
$\frac{\pu{mol}}{\mathrm{conc}} = \mathrm{vol}_{\pu{cm3}} = \frac{0.0250}{0.600} \times 1000 = 41.66... \approx \pu{42.00 cm3}$
I'm not sure if the second step is right.
If I am right (hopefully) can someone please explain why.
Many thanks.

The clue is the moles in the dilute solution is the same as the moles of the concentrated stock solution - it has to be constant.
The method I have used is (therefore) correct.


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Polarity and stability
Why is  cis-1,2-cyclohexadiol less polar than trans-1,2-cyclohexadiol?  I know it has something to do with stability and chair conformations, but I'm not sure how it relates to polarity.  I know cis-1,2-cyclohexadiol has a more stable chair conformation than trans.

Does the cis compound have an internal hydrogen bond which decreases the polarity of the molecule?

Could you indicate what the difference in dipole is, and perhaps add pics of the structures?

Did someone say VDW? In some cis, trans (geometric) isomerism you may experience lack of polarity. But that doesn't mean there's none as much of polarity rise from intermolecular bonding composed of dipole-dipole and van der Waals dispersion forces.

If I assume you refer to 1,2-cylohexanediols, the two isomers to consider were (cis), and (trans)

Assuming the cyclohexane ring as a plane, an (axial, equatorial) orientation of the two O-substitutents is puts them both on the same side of this reference plane; the relative orientation of the two substituents is (cis):

This contrasts to the case of the (trans)-configuration, where the two substituents may be either both in axial, or (likely preferred by thermodynamics) both in equatorial orientation:

From the later picture, taking into account i) the relative orientation of the two hydroxyl groups towards each other as well in respect to the cyclohexane moiety and ii) the (assumed) conformational preference for this form over the conformer with two axial oriented hydroxyl groups, I would assume the two individual vectorial contributions along $\ce{C -> O}$ are better lined up than in the instance of the (cis)-configuration to yield a larger (global) dipolar moment.

The dihedral angle of the cis- structure is -28.2° while that of the trans- is 77.6° when drawn in Avogadro and geometry optimised using the molecular mechanics software installed.  Wouldn't vector algebra suggest that the cis- structure has a greater dipole moment?  Hence my question about internal hydrogen bonds mitigating the polarity of C-O bond.

I found some values on the web which are 2.33 D and 2.39 D for cis and trans, so probably identical within error (not quoted with data). I calculated (v. simply using bond dipoles from textbook and cyclohexane geometry) and got 1.99D cis and 2.1D trans (so also similar and surprisingly accurate considering model used) but only 0.16D for trans a-a isomer in your bottom figure. So it would seem trans o-e is most stable as expt values are similar.

@porphyrin I refrained from bringing all three structures into Avogadro. Beside it is currently not accessible for me, I am not aware if applying MMFF94 (as maybe more suitable than UFF) and its implementation in Avogadro are tools allowing to discern the situtation between the three instances (especially trans-configuration as in axial, axial vs. equatorial, equatorial orientation) and hence addresses the question "correctly" within its allowances by theory. The small differences (experimentally, and by calculation) you fond suggest me, here "no" is more likely than anticipated by mine.

You switched the pictures. Cis isomer should have both subsituents on the same side, NOT on opposite. For instance, the 2-D-molecule you have drawn for the trans isomer has two directional bonds in front of the molecule, but you have drawn the correct conformer for the name. Note that on the top of the left hydroxy group there is a hydrogen which truely is in front!

@冰淇淋 Thank you very much for identifying the problem.  The figures are redrawn from scratch, the text edited slightly to correct this entry.

No problem, I also wasn't meant to write an exclamation mark, I didn't want to come off as rude xD


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Can olivine be used to make glass?
This concerns ongoing work to find a plausible approach to large-scale production of clear glass on the Moon. It's for an ultra-hard science fiction project and I really want to get it right.
Olivine of good purity can be gotten from deposits of dunite in the lunar highlands. You need to get to the bedrock, but it's there. If you melt the dunite you can use fractional crystallization to improve its purity - I have asked about that on Earth Sciences but it seems to be true. If that works it would also suggest you can do the same thing again to separate the olivine into forsterite ($\ce{Mg2SiO4}$) and fayalite ($\ce{Fe2SiO4}$).
If the forsterite is pure enough, could that be used to make decent clear glass?
Could the fayalite be processed into iron, water, and silicon dioxide with something like the reaction below?
$\ce{Fe2SiO4 + 2H2 -> 2H2O + 2Fe + SiO2}$

The previous answer is not really helpful because you would need huge amounts of CO2 which are not available on the moon. You need 2 moles of CO2 to generate 1 moles of SiO2. You simply do not have that amounts of CO2.
Furthermore, assuming you do someone get that CO2, you need to physically separate the Mg and Fe carbonates from the silica. One way would be dissolving the carbonates in acid, which is also not available on the moon. Another way would be to melt the thing and separate the two immiscible carbonate and silicate liquids by centrifuging. This will require very high pressure to keep the carbonate in, otherwise, this thing will decarbonate and just make olivine again.

If the forsterite is pure enough, could that be used to make decent
  clear glass?

No. Forsterite has a melting point above 1800 °C. It also has the problems I mentioned in my previous answer to you:
Can glass be made with anorthite?
The problem of quench crystals forming in your liquid once you cool the forsterite is much worse than what I described for anorthite in the answer given in the link. You will have to find a way to cool it down even faster, and I doubt you can do it. It's also probably going to crack and break when cooling down.
Then there's the question of if you can even make it pure enough, but the answer for that will be in your other question here:
https://earthscience.stackexchange.com/q/10110/725

Could the fayalite be processed into iron, water, and silicon dioxide...

Yes. It still doesn't solve the problem of forsterite being a terrible glass material.
EDIT
There's a recently published paper:
Schleppi, J., Gibbons, J., Groetsch, A. et al. J Mater Sci (2019) 54: 3726. https://doi.org/10.1007/s10853-018-3101-y
Manufacture of glass and mirrors from lunar regolith simulant
It is open access. You might find it informative.

The process temperatures on the Moon would have to be considerably higher than on Earth, but that is much easier to do on the Moon thanks to the sun and the vacuum. If forsterite isn't useful as a glass, it may still be useful as a crystal, under the circumstances that could be practical. Or, fluxes can be added to lower its freezing point, as you mention. Your answer as it stands to the question on Earth Sciences doesn't clarify for me if that process would work, because it doesn't seem like it deals with a molten process.

I am marking this as the accepted answer because it addresses the matter much better than the other answer, but is not likely now to be voted up enough to rise above it. However, i wish i could discuss the issues i've raised in my comment above.

You could proceed from either end member of the olivine solid series and yield $\ce{SiO2}$ as you suggest.
However, I'd consider mechanisms that have been researched in the course of studying so-called mineral sequestration in addition to what you've written, especially considering the energy requirements you propose: the mineral sequestration reactions are (slightly) exothermic and would thus also yield heat as a by-product which I suspect would be desirable in a cold place like the Moon. That said, the reactions are slow at standard conditions and can be accelerated at the cost of supplying energy. How that balance works in your scenario is ultimately up to you, of course.
Additionally, carbon dioxide is a reactant (but needs to be in a supercritical state, which costs energy), which you'd get from human respiration and other organic sources "for free," which relieves you of having to have a lot of hydrogen gas on-hand.
The idea with mineral sequestration is to allow supercritical carbon dioxide to react with certain minerals to yield carbonates that are stable over some long time period:

Mineral carbonation reactions are known to geologists and occur spontaneously on geological time scales. For example, the reaction of $\ce{CO2}$ with common mineral silicates to form carbonates like magnesite or calcite is exothermic and thermodynamically favored.

An example is:
$$\ce{Mg2SiO4 + 2CO2 -> 2MgCO3 + SiO2}$$

(which) illustrates the transformation of forsterite, which is the end member of the common silicate mineral olivine. One ton of olivine can dispose of approximately two-thirds of a ton of $\ce{CO2}$. Again, the reaction is
  exothermic and releases 90 kJ/mole of $\ce{CO2}$.

In summary: There's nothing wrong with your chemistry or geology and there is more than one way to yield silica from mafic/ultramafic minerals.
The reference I have quoted above and cited below also details the process in general terms for minerals such as olivine and serpentine, and illustrates schematics for implementing such schemes on industrial scales, which might also be of interest to you.
Reference from the National Energy Technology Laboratory within the DOE:

Goldberg, P. ,Chen, Z. Y. ,O’Connor, W. ,Walters, R., and Ziock, H. $\ce{CO2}$ Mineral Sequestration Studies in US. Technology. 1 (1): 1–10 (2000).

After some thought, i'm not sure that CO2 is the better fit. The minerals have to be melted anyhow to purify the forsterite and fayalite,  processing with hydrogen would yield iron as well (which is highly useful), and the hydrogen can be recycled. Storing hydrogen is a pain, but the process yields water, so the hydrogen could be split from that when needed. And it should be possible to do most of the heating with sunlight, which is constant and strong during the 2 weeks of daylight each month. Also worth considering - in a place with no carbon, you avoid taking it out of the system.

I was suggesting an alternative process in addition to answering your question (see the first sentence of my reply) in the affirmative. Thank you for expanding on the other consideration that need to be accounted for in your scenario.

Fair enough. It's just that i'd accepted the answer, but decided that goes too far and undid it. The additional information is valuable, but since at this point i think it would be better to stick with the hydrogen method, i don't want to discourage people from posting more answers.


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Is there a free online resource where I can get the solubility of a solute in a solvent?
I'm looking for a online resource that lets me find the solubility at of any solute (e.g Choline Chloride, Sodium Hydroxide, Dextrose) in any solvent (e.g. Ethylene Glycol, Oleic Acid, Isopropanol, Water). Preferably, I'd like a webpage that lets me enter solute and solvent, then gives the answer, but even just lookup tables would be helpful.

You need to narrow down your question.  Currently, the potential solutes include NaOH (inorganic salt), dextrose (organic neutral compound), and choline chloride (organic quaternary ammonium salt) on one hand, and on the other hand a wide range of potential solvents.  It is unlikely a single database will (ever) cover such a  broad range of possibilities.
Nevertheless, there are some resources that may be useful to consult.  For example solubility tables like this or the properties tables for some of the compounds in wikipedia, like here.  If you know both solvent, and solute, NIST-IUPAC's solubility database here may be worth to visit.
If your institution / library provides access to Reaxys/Beilstein (a commercial database by Elsevier), you may find solubility data in the category of physical properties. This database covers both organic and inorganic (including metal organic) compounds and will provide you both the numerical value, as well the primary reference.

Reaxys/Beilstein sounds really interesting, but I don't have an institution / library that provides access. is there any kind of public portal to it?

@Cassie http://chat.stackexchange.com/rooms/57110/solubility-data

@Buttonwood: This is a great link / idea but it seems to be dead (when I click on it at least).

@Hand The first three links point to Wikipedia, all functional.  The fourth (nist.gov) requires e.g., NoScript for the Firefox web browser to enable JavaScript.  The link to the chatroom still is accessible when NoScript is temporarily switched-off, too, with pointers to materials.springer.com, CRC Handbook of Chemistry and Physics, Perry's Chemical Engineers' Handbook, etc.


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Identify a molecule by structure
I have a drawing of a molecule, I want to know its name. Is there a software, or website that can help me identify the chemical? Im not a chemist, its been a decade since I had chemistry in college, and I cannot simply guess the name of the chemical.
The model goes like this:
C4H9-Benzine-N=CH-Benzine-CH=N-Benzine-C4H9
(also if there is a simpler form of this molecule, Im not sure how to simply it) I'm hoping It is a known chemical and that I can get details about it.

Is it Terephthalylidene Bis(p-butylaniline)?

@Joh If you have a drawing of the structure in question, may you add this drawing?  This could facilitate to provide you a rapid and unambiguous answer.  Should you do not have access to a sketcher to draw the structure again, you may consider uploading a photo (for example by a mobile), reasonably in dimension and resolution, too.

Joh,
Please let me know if you referred to this compound:
Also, if this is your compound, please find the name.

Hey Che Mistry. In the compound shown above, a prefix 'styryl' is used. Is that valid according to IUPAC? Im sure that some terms are allowed be IUPAC like benzene and pyridine, but not sure about stryrene.

Thanks Pritt for pointing out
Oh, I missed that
I too am not very much sure
I actually generated the name with some software...
In that case, it could have been 'phenylethenyl' instead of styryl

ChemExper has a free-to-use structure search, which would usually give you a name and access to safety information from the supplier. Caveat: this will necessarily only work for commercially available and indexed chemicals.
If you have access to university resources or similar, you may be able to use SciFinder or Beilstein.

There are two altogether different aspects to this question.
First, you may want to look for your structure in the databases which contain known molecules. This aspect is covered (partially) by TAR86.
Second, you may want to know the name of a molecule of which you have a drawing, and which may or may not be known, or even exist in nature. In this case you may resort to some automatic naming engine, which are built into many popular molecular drawing programs, and also available online. What's the point of knowing a name (but not the properties) of some strange molecule is another question. If you intend to use it to search for the molecule in some database, then look again: maybe the said database comes with an interface for search by structure, and you don't need the name in the first place.

I would suggest visiting www.organic-chemistry.org/chemicals/structuresearch.htm

It's a pretty good website, I must say. Simply stretch and draw carbon bonds, replace atoms of carbon with other heteroatoms, or even add double and triple bonds. It usually gives you the name of the compound as well as sources to buy it upon clicking the Exact Search button.
Happy structure searching!


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Why does the electrolytic solution need be electrically neutral in order for galvanic cells to work?
According to my book (and various other sources), the salt bridge in a galvanic cell serves to keep the solution electrically neutral. Yet I was not able to find any information on why the solution needs to be electrically neutral. As long as the oxidation reaction takes place in the anode (thereby dumping electrons and rendering the anode negatively charged) and the reduction reaction continues in the cathode, the charge disparity should create a potential difference right?
What happens when the solution isn't electrically neutral that prevents the oxidation and reduction reactions from taking place, because if they were taking place the electrodes should have a non-zero potential difference right?
So my question basically is, why does the electrolytic solution need to be electrically neutral in order for a potential difference to develop between the electrodes of a galvanic cell?

Then your book does a poor job at explaining this phenomenon. Remove the salt bridge, and the potential difference will develop all right. The problem is, there will be no current.

Then if you have a solution of soluble copper salt with solid copper rod dipped into it and solution of a soluble silver salt with solid silver rod dipped into it, and you try to measure the potential difference between the two metal cathodes with a volt meter, you would read a non-zero potential difference?  There are no salt bridges, only cathode and anode

Take a cell, for example, Zn/ZnSO4 & Cu/CuSO4
Initially, Zn is at more negative potential (high electron density) than at Cu. So electrons move through external circuit (the conducting wire, if the 2 electrodes are connected) from Zn to Cu.
At the same time Zn2+ ions pass into solution, making it successively more +ve; while at Cu electrode - Cu2+ ions from solution combine with these electrons (came from Zn) leaving behind excess SO4 2- ions in the solution which will make the originally neutral CUSO4 successively -ve
Zn losses electron = anode (oxidation)
Cu2+ gains electron = cathode (reduction)
There will be a situation (if no other provisions like salt bridge are made) that +vely charged ZnSO4 solution will not be ready to accept further Zn2+ ion from Zn rod & at the same time -vely charged CuSO4 solution to loose Cu2+ ions
This is when we say, the potential difference will soon be zero, as the electron density at the two electrodes will be zero.

The oxidation and reduction will take place. But remember we want the electrons to pass through the wire only, not through solution. The salt bridge won't let the electrons to reach the other side by going through the solution itself. Thus, we will have electric current through the wire and can see the potential difference.


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Definition of "soluble"
How to define "soluble" in chemistry?
For example, in 1:1 v/v, benzene does not mix with water, i.e. insoluble. Common sense.
However, in an analytical procedure, it called for making a "saturated" aqueous solution of benzene, which is about 1.5 g/L. So, benzene actually  has some solubility in water.
So, my question is, is there an official definition of "soluble"?

This is a qualitative description, much like "heavy" or "hot". When used in the solubility tables, it may have precise definition, like "more than 1g per 100g of water", but it doesn't count as official outside of that particular scope.

The general notion is that polar substances are soluble in polar solvents, while non polar substances are soluble in non polar solvents. According to this statement benzene is not soluble in water (as you've mentioned) while methanol is. However, when benzenze is added into a volume of water, no matter the amount, a small percentage of benzene molecules will eventually dissolve in it. I think that solubilty is a broad way of describing the amount of substances dissolved in a solvent. As Ivan nicely mentioned, it is a qualitive description.

Hot:soluble::temperature:solubility

see https://chemistry.stackexchange.com/questions/70247/what-means-fairly-in-terms-of-solubility

We consider if a solute is completely spread out through the solvent in the maximum possible extent, we consider the solute to be dissolved in the solvent. We can identify a solution by a few characteristics:

Solutions are uniform and have no precipitation or coagulated mass in it.
Solutions are clear, no granules or anything.
Solutions exhibit the phase of the solvent.

Solutions come under a more general category called mixtures. Mixtures can be either solutions, colloids, or suspensions. Lets see each one at the molecular level.
Solutions:
These have the solute finely dispersed as individual molecules throughout the solution.
Colloids:
These have the molecules in small aggregates and dispersed in a medium.
Suspension:
These have the molecules in very large granules. They are usually unstable and result in precipitation.
Now, coming to your question.
If a solute can disperse itself completely in a given solvent, its consider as soluble. In fact, every substance has some solubility in every solvent. Even benzene dissolves in water to a very very small extent. You may have heard of phrases like "soluble", "partially soluble" or "insoluble". Soluble salutes completely dissolve in the solvent, as well as amount of solute that can exist within the solvent is high. Acetone and Water are such an example. Some substances are what we would call as "partially soluble", they dissolve if a small but still perceptible amount of solute is added, but they do not dissolve if too much is added. Aniline would be described as "partially soluble" in water. Now, coming to insoluble ones, as I mentioned even they have a very little solubility, typically of the order of nanograms per liter. If you were trying to dissolve, say, benzene in water, you would probably take a few grams of benzene and mix. You won't be able to perceive the few nanograms that did dissolve in water, but the remaining large majority would be easily visible as undissolved. From the experimenter's point of view, he would consider it as insoluble since he can't really know that a few nanograms did dissolve. It's mostly based like this. However with more sophisticated equipment,you could measure the few nanograms, and then the they would say that benzene has a "solubility".
I hope I have clarified your doubts.

IUPAC defines "Solubility", but not "soluble". There is no single definition because there is no single meaning. The reason there isn't a single meaning is because it is useful to have different meanings in different contexts. (This is true of almost all words.) According to Wikipedia, 0.1g per deciliter is a common ceiling for insolubility. I've seen various values for the lower value for labeling a compound "soluble". I've seen 1g, 2g, 3g, 5g and possibly more claimed to be the (arbitrary) floor. I don't see any reason to call for standardization, because the use for such a property is varied and diverse.

If you refer to specific resources (e.g. IUPAC definitions or Wikipedia pages), it's helpful to provide a link (or reference) to those resources. (e.g. Which Wikipedia page did you get the 0.1 g per deciliter figure from?)

authors: Αντώνιος Κελεσίδης


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Why isn't mercury electrode used to produce commercial element sodium?
I know that while electrolysing a solution of brine, if we use mercury cathode, we get sodium amalgam instead of having hydrogen gas liberated. Even the Castner-Kellner process uses this to produce sodium hydroxide.
However, wouldn't the mercury electrodes be used to produce commercial elemental sodium? I feel it's much easier to electrolyse aqueous NaCl than to melt NaCl. Then why don't they use aqueous NaCl decomposition with mercury cathode to get Na metal?
Edit: Its seems even more topsy-turvy now. Regular electrolysis of Brine itself produces NaOH, then why bother with the mercury?

Perhaps because the production of Na is a small subset of the production of Cl, so the process is optimized for the latter, not the former.

Two thoughts: 1. Are you trying to make elemental sodium in the presence of water? 2. Hg(Na) amalgam is not the same thing as elemental sodium.

@Zhe Yes Sodium amalgam isn't the same as sodium, but I'm sure that it can be separated from mercury, since mercury boils easily, unlike sodium. I think sodium amalgam is more stable in water, and it can be isolated

Are you going to distill it? Once the mixture is highly concentrated in sodium, the distillation of the remaining mercury is probably non-trivial.

Why is distillation not feasible? We could do it in an air-tight container, so that sodium doesn't explode. After all, mercury is a liquid, and can be easily separated right?

Elemental mercury is a liquid, but most other forms, including amalgams, are not. Also, although elemental Hg is a liquid, it has a relatively low vapor pressure / high boiling point (357C) (that doesn't mean it can't be distilled off, it's just not as easy as you might think).

Among other things, sodium amalgam is not really sodium dissolved in mercury in the manner that an aqueous sugar solution is sugar dissolved in water. Evaporate the water from the sugar solution and you get back the sugar, but sodium amalgam forms intermetallic compounds instead.
Deiseroth et al.[1] give a review; even in the 1990s several such compounds were known:

A summary of recent results concerning the binary system NaHg is given. Characterized by six thermodynamically stable solids: $\ce{NaHg4, NaHg2, NaHg (α,β,γ), Na3Hg2, Na8Hg3 (α, β, γ)}$ and $\ce{Na3Hg(α,β)}$, their crysta[l] structures could be solved by single crystal X-ray studies, except those of $\ce{NaHg4}$ and $\ce{β-NaHg}$.

So, attempting to purify the amalgam by the usual physical means will give intermetallic compounds instead of elemental sodium. Sodium is but one of many metals whose amalgams with mercury form various intermetallic compounds, and so to get the pure metal you need to use a product other than the amalgam.
Reference

Hans-Jörg Deiseroth, Erik Biehl, Matthias Rochnia (1997).
"Sodium amalgams: phase diagram, structural chemistry and thermodynamic data, a summary of recent developments",
Journal of Alloys and Compounds,
246, 80-90,
ISSN 0925-8388,
https://doi.org/10.1016/S0925-8388(96)02460-7.

Also, of course, a process that avoids the use of mercury would minimise the emissions of mercury. There is also a reason why non mercury-using alternatives to the Castner Kellner process were developed.


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How do gelatin and aluminium sulfate strengthen paper?
When wet, the hydrogen bonds between the cellulose chains in paper are disrupted due to the formation of hydrogen bonds with water. This leads to the paper becoming extremely weak. This is the case with most paper, however, additives such as gelatin and aluminium sulfate are added to the paper to strengthen it.
I can't see how the addition of either of these strengthens the paper. I would guess that it aids the bonding between cellulose chains, perhaps by creating a 'bridge' between them, but I'm not sure if this is correct or how it would happen. Another idea is that the aluminium could form a complex ion, and then perhaps form a chelate to strengthen the paper, though I have little confidence in this.
So, how do the gelatin and aluminium strengthen the paper - particularly filter paper, which is almost pure cellulose.

The process of adding gelatin and such to paper (cellulose) is called sizing. The idea behind using gelatin is that it has both a hydrophilic end and a hydrophobic end to the molecule. The aluminum sulfate (alum) is added as a hydrogen bonding linker between the cellulose and the hydrophilic end of gelatin (or some other similar molecule). Mostly this is just to prevent the paper from getting wet in the first place.
Your second paragraph is spot on. Typically other reagents are used to cross link ( the cellulose in order to retain strength upon getting wet (wet strength). Those reagents are a bit more aggressive. Like alkyl ketene dimer

or urea-formaldehyde polymer.


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Degrading Bakelite
I'm aware of the fact that Bakelite is a thermosetting polymer, and is infusible. But what happens if you keep on heating a Bakelite piece to ridiculously high temperatures. I'm pretty sure it would break apart at very high temperatures.
Aside from high temperatures, is it possible to degrade Bakelite chemically? Perhaps using ozone could cleave off the benzene rings and produce possibly a soup of various carboxylic acids.
Is it possible to degrade Bakelite by irradiating light of high energy like ultraviolet or x-rays to break the bonds and degrade it.
In short, is there anything we can do to destroy Bakelite?

Although Bakelite has extraordinarily high resistance to electricity, heat and chemical action$^1$, it is ultimately just an organic hydrocarbon polymer, strengthened by organic binders and some surfactants. Heating under oxygen at temperatures greater than $\pu{400^oC}$ should effectively convert it to $\ce{CO2}$, water and possibly traces of ash and gases of sulfur or nitrogen based on the surfactant composition.
Regarding non-thermal breakdown of Bakelite, one could easily achieve a similar result by digestion with an oxidizing acid such as $\ce{HNO3}$ or $\ce{HClO4}$.
1) Meikle, Jeffrey L. (1995). American Plastic: A Cultural History. New Brunswick,NJ: Rutgers University Press. ISBN 0-8135-2235-8.

Thank you for your response, it has really helped me. Do you know any reason as to why recyclers don't try to degrade the Bakelite into carbon dioxide? I feel it's much easier than to keep it lying there, scraps of broken Bakelite.

No problem, glad my answer was helpful. It's kind of energy intensive (e.g. expensive) to get rid of it the way I described and cheap to pile it up. I don't know much about possibilities for recycling it though.


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Cotton and synthetic clothes
Why do cotton clothes take a long time to dry as compared to synthetic clothes?
I read in a book that cotton fibres form hydrogen bonds with water molecules and thus, the rate of evaporation of water molecules is reduced in the case of cotton clothes. Is that so? How are the synthetic and cotton fibres different chemically? What is the structure of these two fibres like?

Do you have any particular synthetic fiber you are interested in? There are many synthetic fibers, so it is not a simple comparison of cotton v.s. all synthetic clothing.

Note that there are  many kinds of synthetic materials used in clothing: nylon, orlon, polyester, etc. Even different fabrications of these materials can have very different properties, and may contain additional synthetic materials like water repellents, i.e. in polyester fleeces and such.

Well, I don't think it is that much broad. I am just asking for the structure of the two fibres and why cotton takes more time to dry.


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Acid-Base Titration, why does for a weak acid indicator, [HA] = [A-]
I am reading about indicators and their behaviour as weak acids.
I know that the end-point is when exactly all of the H+ ions from an acid have neutralised the OH- ions of the base.
The book later says

At the end point of a titration, the indicator contains equal concentrations of HA and A- and the colour will be inbetween the two extreme colours e.g. for methyl orange, the colour at its end point is orange.

It continues to say

For an indicator HA ⇌ $H^+$ + $A^-$
  At the end-point, [HA] = [$A^-$] and $$K_a = \frac{[H^+][A^-]}{[HA]} = [H^+]$$

I am confused why this is true. why is the expression [HA] = [$A^-$] valid?

In an Acid titration, we want the end-point to be as close to the equivalence point as possible.
Different indicators have different end-points (they will change colour at different pH values).
When an end-point is reached, the colour of the indicator will change with respect to the [$H^+$] concentration of the solution it is measuring.
As [$H^+$] increases, there will be a point when [HA] = [$A^-$] and depending on the indicator used by the person doing the titration, this would be close to the calculated equivalence-point.
Note, the difference in volume between the end-point and equivalence-point is very very small, hence we can approximate the end-point as the same as the equivalence-point in any experiment.


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Are alkynes more electrophilic than alkenes?
I think that since alkynes have a greater electron density, they would be less electrophilic than alkenes.
Which functional group is more prone to nucleophilic attacks and why? An example of a known reaction would be greatly appreciated.

Alkynes are less reactive than alkenes in general, however, alkynes are more reactive towards hydrogenation.

Here are what my notes from Bruice's Organic Chemistry textbook says:
"The clouds of electrons surrounding the sigma bond makes an alkyne an electron-rich molecule. They are therefore nucleophiles that react with electrophiles.
Thus alkynes, like alkenes, undergo electrophilic addition reactions because of their weak pi bonds. The same electrophilic reagents that add to alkenes also add to alkynes.
Despite being less stable than an alkene, an alkyne is less reactive than an alkene, since reactivity depends on ΔG ‡ , which depends not only on the stability of the reactant but also on the stability of the transition state.
When a proton adds to an alkene, an alkyl cation is formed. When a proton adds to an alkyne, a vinylic cation is formed. A vinylic cation has a positive charge on a vinylic carbon. It is less stable than a similarly substituted alkyl cation because the positive charge is on an sp carbon, which is more electronegative than an sp 2 carbon of an alkyl cation. This makes it less able to bear a positive charge."

You say that an alkyne is less stable than an alkene. Why is that?


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Gromacs: What can be reason of getting final potential energy greater than the starting one?
I wanted to optimize potential energy of a molecule using Gromacs software (v. 5.0.4, single precision). I was really surprised when I discovered that after running Conjugate Gradients integrator, the final potential energy was greater than the starting one. Does anyone know what can be a reason of that?
Perhaps I don't understand what is optimized because potential energy went down and up during simulation.
Fragment of logs:
Step 0, Epot=4.840216e+06, Fnorm=2.899e+05, Fmax=4.849e+07 (atom 672)
....
Step 56, Epot=8.949906e+03, Fnorm=2.368e+03, Fmax=4.351e+05 (atom 783)
...
Step 244, Epot=8.550481e+07, Fnorm=2.406e+05, Fmax=4.658e+07 (atom 54660)

Energy minimization has stopped, but the forces have not converged to the
requested precision Fmax < 10 (which may not be possible for your system). It
stopped because the algorithm tried to make a new step whose size was too
small, or there was no change in the energy since last step. Either way, we
regard the minimization as converged to within the available machine
precision, given your starting configuration and EM parameters.

Polak-Ribiere Conjugate Gradients converged to machine precision in 244 steps,
but did not reach the requested Fmax < 10.
Potential Energy  =  8.5504808e+07
Maximum force     =  4.6581140e+07 on atom 54660
Norm of force     =  2.4062566e+05

One thing to note, though I'm not certain this applies to your software, many geometry optimization techniques are designed to find stationary points rather than minimum. This means the procedure could reasonably converge to a saddle point depending on where it starts on PES.

You are right, in a normal simulation using conjugate gradient (i.e. an energy minimization) the total energy should gradually decrease towards a plateau.
Based on the numbers in the log I guess that there are some atom clashes. The potential energy of the system is too high and the forces also. I would check atom 54660 (as indicated in the log) to see if you have it too close to other atoms or in a strange configuration. If you don't see anything wrong, double check the dimension of the simulation box, maybe it is too small and it causes problem with the periodic boundary conditions.
Anyway, energy minimization using conjugate gradients (CG) should be able to solve these kind of problems. You say you run gromacs in single precision. Try to use gromacs compiled in double precision, at least at the beginning of the energy minimization. If the initial conformation of the system is very bad, single precision can be unstable and unable to minimize the energy.


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Na(s) vs NaCl(s) Entropy
I would think the crystalline, ionic solid would have the lower entropy, however, the tablature indicates otherwise. Why does $\ce{NaCl(s)}$ have a greater entropy than $\ce{Na(s)}$?

there are two types of ions vs one type.

Somewhat related: https://chemistry.stackexchange.com/questions/14660/entropy-wiggle?rq=1

If you don't mind me asking, why would you expect the $\ce{Na}$ to have a higher entropy than the $\ce{NaCl}$? The entropy of a state is defined as,$$S=k_b\ln W$$where $W$ is the number of possible microstates which can make up that macrostate.
In this case, we are interested in the number of possible ways that we can arrange $\ce{Na}$ into a lattice or the number of ways we can arrange $\ce{Na}$ and $\ce{Cl}$ ions into a lattice. The latter has many, many more microstates for a given macrostate, and hence a larger entropy.
The standard entropies can be found in this table.
It is worth noting that what I have written above is not exactly the same as the standard entropy $S^\circ$ which is defined as,
$$S^\circ=\sum_{i=1}^N\int\frac{dq_i}{T}dT$$which just says that the standard entropy going from zero kelvin to 298 kelvin depends on the heat put into all $N$ atoms of a system. Thus, there is a dependence on the energy of any state the system passes through.
For the solids you're talking about, if the energetics of the systems were exactly identical, then the ratio of standard molar entropies would be the same as the ratio $\frac{\ln W_1}{\ln W_2}$. But, because sodium has much weaker bonds than sodium chloride, it has a higher entropy change per unit change in temperature than sodium chloride. This is because for weaker bonds, the atoms have greater mobility in space which increases entropy, and it is more likely for there to be lattice defects when the bonds are weak which also increases entropy.

I like jheindel's answer, but I think it's worth having a more conceptual (admittedly a bit hand-wavy) answer  for the more math-averse.
Intuitively, I would expect NaCl to have more entropy than Na because it is composed of more unique components. As shown in jheindel's answer, entropy deals with the number of unique ways a system can be arranged, so it makes sense that having more unique pieces would lead to more arrangements.
To look at a very simplified model, consider arrangements of colored particles between two boxes. Let's say we are arranging two blue particles. In this case, there are 3 arrangements: both in one box, both in the other box, or one in each box. Now, what if we instead arrange a red particle and a blue particle. In this case, there are 4 arrangements shown below: both in one box, both in the other, blue/red, and red/blue.

By analogy, Na has lower entropy than NaCl because, if you compare the same amount by moles of each, NaCl has  additional arrangements due to incorporating Cl. This is a simplification because it doesn't account for things like crystal structure, but just going off intuition rather than actual calculated/measured values, this is probably the best explanation.
So just as general guideline, entropy increases with the number of particles, the number of unique particles, and the amount of space they can occupy.


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How to measure absorbance at 882 nm?
I am trying to measure the absorbance of a solution at 882 nm, but this seems to be beyond the range of many UV-Vis spectrometers. Is this a job for the IR spectroscopy, or something else? I do not have much experience with IR spectroscopy.

No this is not a job for an IR machine. It all depends on the machine you are using, but provided the monochromators will extend that far all that may be necessary is to change the detector for one that is more red sensitive. If your machine will not do the job, and if its a one-off type of measurement  try to get time on another machine somewhere else. If this is not possible or you need to measure many samples over a long period you may need to experiment by using a lamp , interference filters and a detector etc and make your own machine to measure transmittance.

To second @porphyrin, this is still requesting UV-Vis spetroscopy. Perhaps a group specialized in analytical chemistry, or physicists active in material science either i) may provide you access to UV-Vis spectrometers like the Lambda900 by Perkin Elmer (example) or a Cary / Agilent (example), or ii) may record the data for you.
Depending on the solvent deployed and the question addressed by the analysis, it is not this uncommon to characterize newly prepared materials by UV-Vis spectroscopy in a range of 200 or 250 nm to 1000 nm; because frequently you do not want to restrain your analysis in the determination of the maximal $\varepsilon$, but to record the absorption band  completely. In addition, such an extended data set may be helpful to prepare / compare further characterisations, like ellipsometry, for example. If you are starting to record UV-Vis data on your own, I suggest consultation of a primer like this as a helpful reference.
Routine IR spectra are recorded in a different range of wavelengths, typically expressed in wavenumbers, of 4000 1/cm (about 2500 nm) to 550 1/cm (about 18200 nm).  There may be some overlap if, for example, a polymer coating is characterised firstly with a research UV-Vis spectrometer mentioned earlier in principle capable to collect data up to 3000 nm, and subsequently with an IR spectrometer.  This however requests some additional preparation of the data collection.


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Are lithium bicarbonate and magnesium bicarbonate both soluble in water?

Are lithium bicarbonate and magnesium bicarbonate both soluble in water?

Source: JEE Mains 2017 (National Level Examination of India)
Please do cite a reputed source along with your answer. Currently I have found a line from JD Lee (Concise Inorganic Chemistry 5th ed):

https://srdata.nist.gov/solubility/IUPAC/iupac.aspx this is the source behind the Wikipedia article airhuff posted.

Unfortunately, those data are not freely traceable.  When I attempt to follow the link to the source data I hit a paywall :-(.

According to this Wikipedia solubility table, lithium bicarbonate is soluble in water to $\pu{5.7 g/100mL}$ at $\pu{20^oC}$, while this Wikipedia page gives the solubility of magnesium bicarbonate as $\pu{0.077 g/100mL}$.
However, another source (found by Oscar Lanzi in his good answer), "The Solubility of Magnesium Carbonate in Aqueous Solutions of Certain electrolytes, The Journal of Physical Chemistry, Volume 7, pub. 1903, reports making two measurements of the solubility of magnesium bicarbonate and finding $\pu{3.1 g/100mL}$ and $\pu{2.0 g/100 mL}$.  Note that this source is from 1903, but also that the the Wikipedia link above does not give a source for their value at all. At least one reason for the dearth of information regarding magnesium bicarbonate is that it easily decomposes and it cannot readily be isolated outside of aqueous solutions. My feeling is that the Wikipedia value seems low, and that the values reported in the Journal of Physical Chemistry seem more reasonable.

Lithium bicarbonate is also not available was a solid.  See the OP' s own citation.

I beg to differ on magnesium bicarbonate.  In https://books.google.com/books?id=bCfzAAAAMAAJ&pg=PA583&lpg=PA583&dq=magnesium+bicarbonate+solubility&source=bl&ots=3c8f8q4ocI&sig=XAY4Z50JRHq3FBRe8VPZjDf10sA&hl=en&sa=X&ved=0ahUKEwj91byQ3qPTAhVCOSYKHdNuDAw4ChDoAQgcMAE#v=onepage&q=magnesium%20bicarbonate%20solubility&f=false the solubility of magnesium bicarbonate is $3.06$ g/($100$mL).

+1 for the good find. I think the values reported in this paper make more sense than the values given by Wikipedia. Note that I modified my answer to include this source, and gave credit to you and your answer for finding it, for whatever that's worth. Again, good find.


text:
Molar ratio of two oxides of the same metal

A certain "metal oxide" is made of the oxides $A$ and $B$ of the same metal. In the oxide $A$, the metal is divalent, and in B, the metal is tetravalent. By reduction with hydrogen of a quantity $a$ of the "metal oxide", $B$ is transformed into metal, and $A$ remains unchanged, and $a$ drops by $0.15 g$. By treating the same mass $a$ of the "metal oxide" with dilute sulfuric acid, only $A$ reacts and becomes an insoluble sulfate and the mass $a$ increases by $0.75 g$.
  What is the molar ratio between $A$ and $B$?

So far, I have come up with this solution:

Let $Me$ be the metal.
The reduction reaction of $B$ goes like this: $$\ce{MeO2 + 2H2 -> Me + 2H2O}$$
The mass drop of $0.15 g$ must be given by the loss of oxygen that goes into water.
When reacted with dilute sulfuric acid (for $A$): $$\ce{MeO + H2SO4 -> MeSO4 + H2O}$$
The mass increase must be given by the addition of $0.75 g$ $\ce{SO4^2-}$.

Am I right up to this point? What should I do next? How do I use this information?


text:
Singlet or triplet carbene
On reaction of diazomethane with ethene a carbene is formed as a intermediate . This carbene then attacks on the double bond and form cyclopropane . However I read that both singlet and triplet carbene can react in this reaction but according to me since the carbon of carbene is overlapping with two different orbitals hence each electron of carbene must be in different orbital hence must be triplet. I can't find a reference to this on the internet, perhaps you all could help me .

part of the answer may be given here https://chemistry.stackexchange.com/questions/50277/how-does-spin-flipping-of-triplet-carbenes-occur/54259#54259

I think a triplet carbene reacts in a stepwise fashion rather than a pericyclic concerted cyloaddition as in the singlet case. The triplet adds to one side then a spin flip must occur before radical combination. This leaves the possibility for stereochemical change

This earlier answer may be helpful.

Ron I understood that part and thanks for help but my question is still unanswered that how come if carbon of carbene  has to overlap with two different orbitals of two carbon atoms

Then how can singlet carbene be possible .( Please see the previous three comments including this one as one )

@user3650050  Look at the diagram in my earlier answer.  Even in the singlet process the 2 bonds are formed sequentially, not at the same time.  It's just that the second bond forms much more rapidly with the singlet than the triplet so that stereochemistry is preserved with the singlet.

Diazomethylene is singlet therefore carbene is formed in the singlet state.
singlet carbene is highly reactive therefore it goes through reaction probably before intersystem crossing (depends on competitive speeds). this is why singlet state of carbene is usually invisible. Triplet carbenes react like radicals unlike singlet state (besides some aryl carbenes) therefore the products are different. you can look up Skell-Woodworth for reaction of different spin states of carbenes.
based on the products you can find which spin state of the carbene has been involved in the reaction.


text:
A titration mixture having pottasium oxalate and pottasium hydrogen oxalate!
I came across this question, this is how it goes-
A solution is prepared by dissolving a solid mixture of $\ce{K2C2O4}$ and $\ce{KHC2O4}$. A $\pu{10 ml}$ portion of this solution required $\pu{10 ml, 0.05 M \ce{KOH}}$ solution for titration reaction. In a seperate analysis $\pu{10 ml}$ of the same stock solution required $\pu{10 ml, 0.06 M}$ acidified $\ce{KMnO4}$ solution for titration.
I  have  to  find the ratio of $\ce{K2C2O4}$ to $\ce{KHC2O4}$ in the initial reaction mixture.
I know that(I think), 1 mole of $\ce{KHC2O4}$ would require 1 mole of of $\ce{KOH}$ for neutralization. While $\ce{K2C2O4}$ would not react with $\ce{KOH}$ (please correct me if I am wrong).
Also, $\pu{31.8 g}$ of $\ce{KMnO4}$ would oxidise $\pu{63.5 g}$ of $\ce{K2C2O4}$. But when I don't know how (differently) it shall oxidise $\ce{KHC2O4}$.

You are right, $1$ mole of $\ce{KHC2O4}$ reacts with $1$ mole of $\ce{KOH}$. From that first piece of information in your exercise, you can find out that, in $\pu{10mL}$ of solution, you have $\pu{0.05 * 0.01 = 5 * 10^{-4}}$ moles of $\ce{KHC2O4}$. The next piece of information is meant to let you how much substance there is in the solution. $10$ moles of both substances react with $4$ moles of acidified $\ce{KMnO4}$. Which means the sum of the quantities of both substances in that solution is $\pu{(0.06 * 0.01 * 10) / 4 = 15 * 10^-4}$. So you got $\pu{a = 5 * 10^{-4}}$ and $\pu{a + b = 15 * 10^{-4}}$. Do the math.
Please tell me if I was clear enough.

Bang on! So oxalate and hydrogen oxalate is oxidised to CO2. Yes you were very clear. Thanks.


text:
Why can't HCl be used in chlorination of alkane?
Is it because the  HCl bond is strong or it is because we have to form HCl at the product side during the propagation step?

HCl can't be used because it just won't chlorinate alkanes, no matter how hard you persuade it. Why would it, really?

$\ce{HCl}$ is not used in chlorination of any alkane as the bond between them is highly polar which dissociates to give $\ce{H+}$ and $\ce{Cl-}$ that will not be able to initiate the formation of free radicals in alkane. In this case $\ce{H+}$ will be released from the alkane as carbon is more electronegative than hydrogen. Chloride ion formed will again react with $\ce{H+}$ ion to form $\ce{HCl}$ as it will feel a replusion from carbon due to the lone pair. Hence no chlorination takes place so $\ce{HCl}$ is generally avoided.


text:
Calculating conductivity of a solution given molar conductivities and concentration
From my understanding the conductivity of a solution ($K$) is given by the sum of the conductivities of ions in solution:
$$K = c(K_{m,1}+K_{m,2})$$
Where $c$ is the concentration, and $k_{m1}$ and $k_{m2}$ are the respective molar conductivities of the two ions. For example, with a solution of $\pu{0.1 mol L-1}$ $\ce{NaOH}$:
\begin{align}
K &= 0.1(K_{m,\ce{Na+}} + K_{m,\ce{OH-}})\\
K_{m,\ce{Na+}} &= \pu{50 S cm2 mol-1}\\
K_{m,\ce{OH-}} &= \pu{197.9 S cm2 mol-1}\\
K &= 0.1(247.5)\\
K &= \pu{24.75 S m^-1} (?)
\end{align}
However I think the units are throwing me off here as my calculations aren't making much sense - I was led to believe the conductivity was much lower.
Could you please offer some help with these calculations?

One question,  if $K_m$ is already molar conductivity, Then Why  are you multiplying the concentration with it?

Conductivity = molar conductivity * concentration
at least that was my understanding from wikipedia

Just be explicit with all the units, including for the molar concentration, and be sure they cancel out properly.  You have $\pu{0.1 mol L-1}$ for the solution concentration, but your $K$ values are in $\pu{S cm2 mol-1}$.
So convert your molar concentration to cubic centimeters (there are $\pu{1e3 cm3}$ in one liter) and you get $\pu{1e-4 mol cm-3}$.
Then when you multiply both values and your units cancel nicely to give you
$$\pu{1e-4 mol cm-3} \times \pu{247.5 S cm2 mol-1} = \pu{2.475e-2 S cm-1}$$


text:
Regioselectivity in addition of hydrogen iodide to vinyl chloride
I encountered a question which was stated as:

The reaction of vinyl chloride with hydrogen iodide to give 1-chloro-1-iodoethane is an example of anti-Markovnikov addition. Is this true or false?

The way I could see it, vinyl chloride has a dipole moment as shown below, since chlorine is more electronegative than the carbons in the double bond.

Therefore the carbon on the left would have a greater partial positive charge than the carbon on the right. I therefore concluded that in a Markovnikov addition, the iodine atom would add to the CH2 carbon.
Since the question states that the product is 1-chloro-1-iodoethane, with the iodine atom adding to the CHCl carbon, that would thus be an example of anti-Markovnikov addition, according to the argument above.
I was told, however, that the reaction shown above was indeed Markovnikov addition. How is this so?

"Therefore the carbon on the left would have a greater partial positive charge than the carbon on the right" That's not correct. Dipoles don't work that way.

But, due to chlorine pulling the electrons towards itself, wouldn't they move towards the right, hence causing a partial depletion of electrons on the carbon? Please explain.

Why do you think that effect extends through the entire molecule?

What does Markovnikov's rule state? Does it mention hyper conjugation or does it mention dipoles?

It mentions neither of those things. The product created is via the more stabilized intermediate. But this is just a generalization of the Hammond Postulate for a transition state that is endergonic.

As far as I can see, that reaction follows Markovnikov's rule. The hydrogen goes where there is more hydrogen and the halide goes where there is less hydrogen.
If you are confused as to why the product of that reaction is 1-chloro-1-iodoethane, it's because the electromeric effect the chloride atom exhibits trumps the inductive effect. We need to take the electromeric effects into consideration because the chloride atom is attached to a $\mathrm{sp^2}$ carbon. The inductive effect makes it attract electrons to itself, but the electromeric effect makes it push them away, like a $\ce{CH3-}$ would.

It's the reason why $\ce{CH3-}$ and $\ce{Cl-}$, when attached to an aromatic nucleus, orient new substituents in the same positions, ortho and para (though $\ce{CH3-}$ has a stronger effect than $\ce{Cl-}$).


text:
What pressure is created by the electrolysis of water?
I am interested to know what is a reasonable maximum working pressure I could expect against the evolution of oxygen and hydrogen when electrolysing a typical water electrolyte.
My aim is to make small quantities of isolated $\ce{H2}$ and $\ce{O2}$ for a hobby hydrogen torch and I am curious how tall I can make the middle branch of the Hofmann Voltameter before the back pressure stops the production of new gas.  Having a predictable gas pressure would allow me to use the collection spaces for gas storage. I do not like the idea of mixing the gases before the torch nozzle like some brave people do so plan to use two collection volumes.
What parameters would be most likely to effect this pressure, electrolyte makeup, temperature, voltage?
Bonus question, is the pressure of $\ce{H2}$ and $\ce{O2}$ expected to be the same in theory and/or practice?
EDIT:
It looks like there is a high pressure electrolysis process out there that operates at many atmospheres of pressure and has been used to pressurise the gas products.  I missed it in my first readings on Wikipedia and further research has confirmed it though information is limited, it seems to be a specialised process usually associated with proton exchange membrane systems I think but I am just hoping to use two metal electrodes.

Not sure how to link this as a Related question, sadly without a conclusive answer.  -  https://chemistry.stackexchange.com/questions/15118/whats-the-effect-of-pressure-on-water-electrolysis-of-sea-water-with-increase-in

The maximum gas pressure you can expect in a Hofmann Voltameter is the atmospheric pressure plus the pressure from the water column in the middle cylinder ($1~\pu{atm} / 10~\pu{m}$).
For a reasonably sized apparatus, you will not reach conditions where you had to consider "back pressure" effects that might stop production of hydrogen or oxygen.
Given that the cross section of all three cylinders is the same, the pressure of hydrogen is slightly higher than oxygen pressure since the water column that adds pressure is higher (as can be seen in the Wikipedia article you've linked.)

I am trying to find out where the threshold for the reasonably sized apparatus you allude to is reached.  What is the maximum pressure that can be achieved in a closed electrolysis apparatus?

Let's assume the middle cylinder is 100 m higher and filled with water. Then the excess pressure would be 10 atm. 2. In a closed electrolysis apparatus pressure could grow very high (you can calculate the theoretical limit using the ideal gas law and the density of water). Such an apparatus will most probably be destroyed.

aventurin, you suggest that all the water will turn to gas.  Admittedly I have not found any reference to a limit to the pressure that can be reached by electrolysis but that that seems hard to grasp.  Is there no reverse reaction that will reform (liguid) water from the evolved gas at the electrode with the free complementary gas in solution due to the high pressure?

A point of concern producing hydrogen under high preasure.
Under increasing preasure, both hydrogen and oxygen will increasingly diffuse into the water, then into the opposing columbs,  and excape mixing with the gases in the other columb. This will cause it to get harder to keep the respective gas columns purity high, and more importantly, safe.
One would not want to make a potential bomb with explosive amounts of hydrogen in the oxygen columb, nor explosive amounts of oxygen in the hydrogen columb, under normal preasures let alone at high preasures.
Mater of fact, even at low pressures, given enough time, diffusion will cause a mixing of the gases. The relative purity is related to both the rate of production and the rate of diffusion.


text:
Why does soda explode if artificial sweetener or other powders are added?
I opened a can of soda and spilled some generic artificial sweetener into the soda and it exploded. Why did this happen? I am not talking about mentos, I mean artificial sweeteners like Splenda.

Possible duplicate of How does the Mentos-Coke explosion work?

This answer is applicable to your question also.

I would suggest tbat the added fine powder provided a lot of nucleation sites for the dissolved CO2 in the soda very quickly hence the foaming.

I don't think a more general question should be considered dulicate.

Artificial sweetener powder causes nucleation of the carbonation in the soda/pop/soda-pop. The carbon dioxide concentration in the soda is higher than the equilibrium concentration for the partial pressure of $\ce{CO2}$ in the atmosphere thus over time the soda will off-gas $\ce{CO2}$ and if allowed to sit long enough will become "flat".
The off-gassing is spontaneous but not instantaneous, thus it is subject to kinetic rates. To form a bubble, enough $\ce{CO2}$ molecules need to coalesce at one time to form a large enough bubble to be stable and not redissolved in the soda (nucleating).
When a rough surface is introduced, the number of $\ce{CO2}$ molecules that must come together to form a stable bubble is far less that the pure drink, so the off-gassing occurs much faster causing the observed "explosion".
You can observe this when certain glasses (ussually beer glasses) etch a small ring in the glass, creating a rough surface and the formation of bubbles at the ring.  you can also observe this if you put a larger rough object in the soda. This could be a rock, a piece of wood (like a tooth pick), or even a Mentos.
To solve this you might be able to dissolve it in a small amount of water first. This may not completely solve issue though, as most sweeteners use silicon oxide or titanium oxide powder to prevent clumping, neither of which are soluble and will still nucleate bubbles.


text:
Bonding of Lithium and Argon
I saw a meme that was joking around about Lithium and Argon bonding (see pic below). It got me wondering: Can Lithium and Argon bond in any circumstance?

https://www.nature.com/articles/srep16675

@orthocresol The world LiAr is all throughout that paper. :P

It's no joke; a gas-phase lithium-argon diatomic molecule does exist. However, it (and about two dozen other di-atoms having various elements bonded to the argon) is considered to be bound only by a van der Waals interaction. See the list of diatomic van der Waals molecules here.

Can you justify your statement ? Do you have a reference ?

Did I not link to the Wikipedia article? They cite many references for the claimed molecules.

Argon is a noble gas. This means that it does not simply "bond" with anything. These gases have a very strong resistance to joining with other atoms. So, to answer your question, no, you can not bond argon and lithium under any circumstances. Anyone who says they have done so is a LiAr.
Also, since lithium is highly reactive and flammable and argon is used in neon, I would not suggest trying to combine these two without a proper location and gear.

I suggest taking a look at orthocresol's link in the comments to the OP, and doing a search for noble gas chemistry and compounds. There are too many compounds to name, including LiAr.

Yes, as I said, it is a Noble gas so it will not simply bond, please point me out in the areas that I was misinformed on so that I can correct myself. Thanks.

@Nathanael the comment is meant to be helpful. The first lines in the abstract of the paper: "High pressure can fundamentally alter the bonding patterns of chemical elements. Its effects include stimulating elements thought to be 'inactive' to form unexpectedly stable compounds with unusual chemical and physical properties." If you take the stackexchange tour you'll see that SE is a little different than other sites. Welcome!

"High pressure can fundamentally alter the bonding patterns of chemical elements. Its effects include stimulating elements thought to be “inactive” to form unexpectedly stable compounds with unusual chemical and physical properties."- From the abstract, ortho sent.

@NilayGhosh and uhoh Thank you for your input, I forgot about pressure. I appreciate your help! :D It's nice to know when I messed up on something so I can go and relearn it! Thanks again!

@airhuff, if I sounded offenseful I apologize, I could have worded that differently. Since I overlooked this, how much force (estimated) would be needed to change the bonding patterns? Thanks again and sorry if I came out rude!

@Nathanael you can avoid loss of the reputation points by deleting this answer. It's the way stackexchange works; the idea is to leave a set of questions and helpful answers for the benefit of future readers as well as for the OP (original poster) of the question. Down votes aren't towards the poster of a question or answer, they are votes on the content only, so never take them personally even when they feel like it. Theoretically we should all be working together to learn from each other and leave a nice set of Q&A's behind for future readers, and eventually for AI when it takes over ;)

I do not understand "and Argon is used in neon".

@Nilay I accidentally approved your flag - I shouldn't have. Flags should not be used to indicate wrong content, that's what voting is for.


text:
Nomenclature of Halogen substituted Alcohol and longest chain

For the following compound, the correct IUPAC Name is "4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol" but according to what I know, Chlorine is a substitutent and alcohol is a functional group. So including alcohol is necessary but not $\ce{Cl}$. And if we see the longest chain including alcohol then it would be a pentanol rather than a butanol. Why are we not naming it like that? Is it because the substitute $\ce{Cl}$ is necessary to include? Or what?

Related: How to name the following alcohol?

Your rationale is correct.
The proposed name “4-chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol” is unambiguous and describes the correct compound; however, it is not in accordance with IUPAC nomenclature.
The important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes (here: “-ol”)
longest chain
lower locants for suffixes
greater number of prefixes (e.g. “chloro-”)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

Therefore, the preferred IUPAC name for the compound that is given in the question is 3-(chloromethyl)-2-(propan-2-yl)pentan-1-ol.

The position of the chloro group or other substituent groups that are cited as prefixes is not relevant for the choice of a principal chain in this case. Only if there is still a choice, such substituent groups are taken into account according to P-45.2.


text:
Comparing first, second, and third ionization energies between atoms
Among the elements boron, carbon, nitrogen, and magnesium, what are the first ionization energies from least to greatest? What about the second and third ionization energies?
I understand the first ionization energy is given by the following formula: $\ce{X -> X+ + e-}$, the second ionization energy is given by: $\ce{X+ -> X^2+ + e-}$, and the third ionization energy is given by: $\ce{X^2+ -> X^3+ + e-}$.
The electron configurations of the elements listed are:
$B: 1s^2 2s^2 2p_x^1$
$C: 1s^2 2s^2 2p_x^1 2p_y^1$
$N: 1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$
$Mg: 1s^2 2s^2 2p^6 3s^2$
I know that magnesium has the lowest first ionization energy because it undergoes the most electron shielding; it is shielded from the positive nuclear charge by the ($1s^2 2s^2 2p^6$ electrons).
I can deduce that nitrogen has a higher I.E. than carbon because they are both shielded by the same electrons, the $1s^2 2s^2$ electrons, but nitrogen has a higher nuclear charge. The same reasoning can be used to conclude that carbon has a higher first ionization energy than boron.
I did the above reasoning with respect to the first ionization energies. How do I apply a similar logic with the second and third ionization energies? Do the second and third ionization energies follow periodicity with respect to periods and groups? What is that trend, if any?

This table from Wikipedia should help a lot. You can typically follow periodic trends, but there are other factors (like half and fully filled shells) that might cause exceptions.

One thing I would note in addition is that it is difficult to compare across columns and rows at the same time. For example, phosphorus should have a lower first ionization energy than nitrogen because it is in the same column and below it. But comparing nitrogen and say chlorine is more difficult because there are competing trends (decreasing down a column, increasing along a row).

If you know the trends for $1^\text{st}$ ionization energies, a good first approximation for the trends of $n^\text{th}$ ionization energies at any given point is the $1^\text{st}$ ionization trend $n$ elements earlier.
As an example, look at the first and second ionization energies for the second row elements (Table).
From $\ce{Li}$ to $\ce{Be}$, the $1^\text{st}$ ionization energy increases due to the greater effective charge on the outermost electron. From $\ce{Be}$ to $\ce{B}$, the $1^\text{st}$ ionization energy slightly decreases, as the electron we removing is in the p-subshell and is shielded by the inner s-subshell electrons. From $\ce{B}$ to $\ce{C}$, it increases again.
What about the $2^\text{nd}$ ionizations? Well from $\ce{Be}$ to $\ce{B}$, it increase, as the $2^\text{nd}$ electron from each is pulled from the s-subshell, but $\ce{B^+}$ has a greater effective charge than $\ce{Be^+}$. From $\ce{B}$ to $\ce{C}$, it decreases, as the $2^\text{nd}$ electron pulled from $\ce{C}$ is still in the less shielded p-subshell.
The $\ce{Be}$ to $\ce{B}$ trend in $2^\text{nd}$ ionization matches up with the $\ce{Li}$ to $\ce{Be}$ trend in $1^\text{st}$ ionization. The same is true for $\ce{B}$ to $\ce{C}$ (2nd) and $\ce{Be}$ to $\ce{B}$ (1st).


text:
Calculations for making a buffer from a weak base and strong acid

How many moles of $\ce{HCl}$ must be added to $\pu{100 ml}$ of a $\pu{0.100 M}$ solution of methylamine ($\mathrm pK_\mathrm b = 3.36$) to give a buffer having a $\mathrm{pH}$ of $10.00$?

(The answer is supposed to be $\pu{8.1 mmol}$.)

My thought process:
First, I found the $\mathrm pK_\mathrm a$ of methylammonium ion:
$$\begin{align}
\mathrm pK_\mathrm a(\ce{CH3NH3+}) &= 14 - \mathrm pK_\mathrm b(\ce{CH3NH2}) \\
&= 10.64
\end{align}$$
Substituting this into the Henderson–Hasselbalch equation
$$\begin{align}
\mathrm{pH} &= \mathrm pK_\mathrm a + \log\left(\frac{[\ce{CH3NH2}]}{[\ce{CH3NH3+}]}\right) \\
10.00 &= 10.64 + \log\left(\frac{\pu{10 mmol}}{x}\right)
\end{align}$$
where $x$ is the amount of $\ce{HCl}$ that must be added. However, when I solve for $x$ I find $x = \pu{43.7 mmol}$, nowhere near the correct answer. What am I missing?

Since $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$, you get $\mathrm{p}K_\mathrm{a} = 10.64$ for the methylammonium cation.
$\ce{HCl}$ protonates methylamine. The amount of methylammonium increases by the same amount methylamine decreases. So
$$\mathrm{p}K_\mathrm{a} = \mathrm{pH} + \log\left(\frac{[\ce{HA}]}{[\ce{A-}]}\right) = \mathrm{pH} + \log\left(\frac{x}{10~\pu{mmol} - x}\right)$$
lets you calculate the value $x$ of $\ce{HCl}$ needed to obtain the desired $\mathrm{pH}$.


text:
First order, parallel reaction integration
I have parallel equations:
\begin{align}\ce{A &->[k_b] B},&\ce{A &->[k_c] C}.\end{align}
I understand how to determine formation of $\ce{A}$ (with the integration) but I really keep going in circles with integrating $\ce{B}$. If I can have some assistance in how to integrate $\ce{B}$, then I will be fine for integrating $\ce{C}$.
This is what I have:
\begin{align}
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}
&= k_b[\ce{A_0}] \cdot \mathrm{e}^{(-k_b+k_c)t}\\
\int_{[\ce{A_0}]}^{[\ce{A}]}\frac{\mathrm{d}[B]}{\mathrm{d}t}
&= \int k_b\cdot[A] \mathrm{d}t\\
\ln\frac{[\ce{B}]}{[\ce{A_0}]}
&= \int k_b\cdot[\ce{A_0}] \cdot \mathrm{e}^{-(k_b+k_c)\cdot t}\mathrm{d}t\\
\end{align}
To save the embarrassment, once I raise to the "$\mathrm{e}$" of both sides, I get
\begin{align}
\frac{[\ce{B}]}{[\ce{A_0}]}
&= \mathrm{e}^{\frac{k_b\cdot[\ce{A_0}]}{(k_b+k_c)}t} \cdot \mathrm{e}^{\mathrm{e}^ {-(k_b+k_c)t}}\\
[\ce{B}] &=[\ce{A_0}] \mathrm{e}^{\frac{k_b\cdot[\ce{A_0}]}{k_b+k_c}t} \cdot \mathrm{e}^{\mathrm{e}^{-(k_b+k_c)t}}\\
\end{align}
Can someone show me the steps of integrating the $\int k_b\cdot[\ce{A}] \mathrm{d}t$?
Very basic steps are available in solutions, but it still does not fully show the integration/algebra and I am just very lost.

I submitted an edit to improve the formatting of your post. Let me know if everything was  rewritten as you intended.

I have only corrected some of the bad MathJax,  there might still be errors in the constants introduced by @Tyberius, please check again.

Start by solving the rate equation for $\ce{[A]}$:
\begin{align}
\frac{\mathrm{d}\ce{[A]}}{\mathrm{d}t}
&= -k_\mathrm{B}\ce{[A]} - k_\mathrm{C}\ce{[A]}&
\Longleftrightarrow&&
\ce{[A]}
&= \ce{[A]_0}\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t).
\end{align}
We can then solve the rate equation for $\ce{[B]}$:
\begin{align}
\frac{\mathrm{d}\ce{[B]}}{\mathrm{d}t} &= k_\mathrm{B}\ce{[A]} \\
&= k_\mathrm{B}\ce{[A]_0}\exp\left(-(k_\mathrm{B} + k_\mathrm{C}\right)t)\\
\ce{[B]}-\ce{[B]_0}
&= \int_{\ce{[B]_0}}^{\ce{[B]}}\mathrm{d}\ce{[B]}' \\
&= k_\mathrm{B}\ce{[A]_0}\int_0^t\mathrm{d}t'\,\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t')\\
 &= \frac{k_\mathrm{B}\ce{[A]_0}}{k_\mathrm{B} + k_\mathrm{C}}[1-\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t)].
\end{align}
I use primes to distinguish the variable of integration from the limits of integration.

The expression given is not correct, one limit has been missed, I have edited please check

Just to add, the term in front of the exponential  ($kB/(k_B+k_C)$is the fraction going towards B.

Thank you @a-cyclohexane-molecule. Can you just lastly explain the [B]}-[B]_o and where that comes from? I know the ln[B]/[A]0 = ln[b] - ln[A]o but if you "e" both sides, I don't see how you get your equation, and then I am back to the original problem so there must be a step I do not know. And in the integration you ended up with a one, but I dont understand how that comes about.

Sorry about the mistake; thanks for fixing it. @juliodesa, $\ce{[B]}$ and $\ce{[B]_0}$ are the bounds of our integral -- we start with $\ce{[B]_0}$ concentration of species B, and have $\ce{[B]}$ concentration of species B at time $t$. The one comes from evaluating the integral at the lower bound of $t=0$.

and also note that the rate constant at which B appears is $k_b+k_c$ and similarly for appearance of C and disappearance of A, this is true even though the rate constant to B is $k_b$ and that to C, $k_c$.

Why is ka=kb+kc? Can you explain this thought to me

And why did you use primes....I don't understand this logic

@Abhinav, 1. What do you mean by ka? 2. Primes denote dummy variables of integration.

By ka ,I mean the rate of dissapearence of reactant A

I have never heard about dummy variables of integration....I am new to calculus so maybe that's why

@Abhinav, A yields B with a rate kb and A yields C with a rate kc---in other words A is consumed at a rate kb in forming B and at a rate kc in forming C. Then the overall rate of consumption of A is kb + kc.


text:
What is the mechanism that dihydrolipoic acid undergoes to be oxidized by FAD?
The mechanism for dihydrolipoic acid oxidation in the pyruvate hydrogenase complex is not clearly demonstrated in any Lehninger textbook, other than the fact that FAD becomes FADH2. I am wondering what intermediates the FAD/FADH2 undergoes, my textbook just does skips over the details.
In the diagram below, the mechanisms are mostly all clear, except for the arrow I've highlighted with orange.

related https://chemistry.stackexchange.com/questions/47375/why-cant-electrons-be-transferred-using-hydrogen-ions

Interesting question. Let's look at the overall reaction first:

This reaction is catalysed by the dihydrolipoamide dehydrogenase in 6 steps and as you see from the schematic below (ref.1) the mechanism involves also four aminoacids near the active site (Cys48, Cys43, Hys451, and Glu456):

References:

1.https://www.sciencedirect.com/science/article/pii/B9780080453828001374

First image attribuition: By Jkallini - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=24958884

The redox steps are hydride transfers, i.e. hydrogen with two electrons. A description of the structural aspects is here.