text:
What is this piece of glassware called?
Here's an image:

The left side is designed to be fitted into a beaker, and it captures gas evolved during a reaction.  The right side can be filled with a liquid to then react with that gas.  We use it to capture ammonia from a decomposition reaction, with an indicator in the right half as part of an identification test.  I'm sure it must have a name, it's a commercial piece of glassware, but the box we keep it in in the lab doesn't identify it.

It might be part of a Kjeldahl kit???

One form of a bubble trap, with valve.

@blacksmith37 It is better if the blubberer (with all its oil inside) is a one-piece equipment only, and they don't come with a mechanical valve. The metal fork-shaped object is just a clip to keep the two pieces across the spherical joints together.  As in the accepted answer, though, the clip should be rotated by 90 degrees such that the hinge of the clip is parallel to «the seam» between the two glass pieces (thus still providing a little bit of mechanical flexibility), and not (as shown in the photo) across.

The picture shows parts of an arsine generator.

The apparatus can be used to determine trace amounts of arsenic by the silver diethyldithiocarbamate photometric method.
A typical method description can be found here (this is also the source of the drawing):
https://hmc.usp.org/sites/default/files/documents/HMC/GCs-Pdfs/c211.pdf
or
http://hmc.usp.org/sites/default/files/documents/HMC/GCs-Pdfs/GC_pdf_USP38/c211.pdf

That's not what I used it for, but if that's what it's called then thank you

I thought arsenic was detected by the Marsh test (Marshflugkörper) ;)

@Jan The methods are related; both generate arsine gas. However, the reaction with silver diethyldithiocarbamate certainly has a higher degree of selectivity than the silvery-black deposit of the Marsh test, and it also facilitates the quantitative determination.


text:
Difference in calculated pH and the real pH of a phosphate buffer
I want to prepare a buffer system with $\mathrm{pH} = 7.00$ and a total concentration $c_\mathrm{tot} = \pu{0.10 mol l^-1}$ using sodium hydrogen phosphate $(\ce{Na2HPO4})$ and sodium dihydrogen phosphate $(\ce{NaH2PO4})$ salts. I am calculating the amounts to prepare a phosphate buffer as follows:
$$\ce{H2PO4- <=> HPO4^2- + H+} \qquad \mathrm{p}K_\mathrm{a} = 7.20 \tag{R1}$$
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]} \tag{1}$$
$$\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}} = 10^{7.00-7.20} \approx 0.632 \tag{2}$$
$$
\begin{align}
c_\mathrm{tot} &= [\ce{H2PO4-}] + [\ce{HPO4^2-}] \\
  &= [\ce{H2PO4-}] + [\ce{H2PO4-}]\times 0.632 \\
  &= [\ce{H2PO4-}]\times 1.632 \tag{3}
\end{align}
$$
$$[\ce{H2PO4-}] = \frac{c_\mathrm{tot}}{1.632} = \frac{\pu{0.10 mol l^-1}}{1.632} \approx \pu{0.061 mol l^-1}\tag{4}$$
$$[\ce{HPO4^2-}] = c_\mathrm{tot} - [\ce{H2PO4-}] = \pu{0.10 mol l^-1} - \pu{0.061 mol l^-1} = \pu{0.039 mol l^-1} \tag{5}$$
For the preparation of $V = \pu{500 ml}$ buffer solution I used the following masses of corresponding salts:
$$
\begin{align}
m(\ce{NaH2PO4}) &= [\ce{H2PO4-}]\cdot V\cdot M(\ce{NaH2PO4}) \\
  &= (\pu{0.061 mol l^-1})(\pu{0.500 l})(\pu{119.97 g mol^-1}) \\
  &\approx\pu{3.66 g} \tag{6}
\end{align}
$$
$$
\begin{align}
m(\ce{Na2HPO4}) &= [\ce{HPO4^2-}]\cdot V\cdot M(\ce{Na2HPO4}) \\
  &= (\pu{0.039 mol l^-1})(\pu{0.500 l})(\pu{141.97 g mol^-1}) \\
  &\approx\pu{2.77 g} \tag{7}
\end{align}
$$
When I mixed the calculated amounts, the obtained $\mathrm{pH} = 6.5$ deviated quite a lot from the desired value. I thought it was my mistake in calculations or an error when I weighed the amounts, but I repeated everything several times and still not get the $\mathrm{pH} = 7.00,$ and the value always was $6.4$ or $6.5.$
I would like to know the reason. I think if the concentration of the monobasic salt is greater than the one of the dibasic salt, the $\mathrm{pH}$ tends to decrease, but in the calculations the relation [dibasic]/[monobasic] is always less than 1 and this again causes [monobasic] > [dibasic].

The flaw in your calculations is the assumption that the $\mathrm{p}K_\mathrm{a}$ is a constant independent of concentration.  In reality, $\mathrm{p}K_\mathrm{a}$ is a function of the ionic strength $I$.
At infinite dilution $I = 0$ and $\mathrm{p}K_\mathrm{a} = 7.20$ as you assumed. However at the concentration $\pu{0.10 mol l^-1}$ of phosphate buffer $\mathrm{p}K_\mathrm{a} = 6.81.$ See Florida State University — Phosphate Buffers and the paper by Green [1].
Reference

Green, A. A. The Preparation of Acetate and Phosphate Buffer Solutions of Known PH and Ionic Strength. J. Am. Chem. Soc. 1933, 55 (6), 2331–2336. DOI: 10.1021/ja01333a018.

It depends, if activity coefficients are managed explicitly or implicitly.

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text:
Why does the energy difference of a reversible process not equal that of an irreversible process in an adiabatic expansion?
Suppose that one mole of an ideal gas at $P_1$ and $V_1$ adiabatically expands to $P_2$ and $V_2$, reversibly and irreversibly (two separate processes). Since $\Delta U$ is a state function, why is it wrong to say that $\Delta U_{rev} = \Delta U_{irrev}$? (Since in an adiabatic process, there is no heat transfer, we know that $\Delta U = w$)?

Related: Reversible and Irreversible adiabatic expansion

I can't answer this question but I still want to nitpick (how typical of me). $U$ is the state function, not $\Delta U$.

Both $U$ and $\Delta U$ are state functions.

A state function is a property of a system. $U$ is a property of a system, and $\Delta U$ is, loosely speaking, a "property" of a process that said system undergoes.

Please do not use MathJax for question titles due to searching issues. Also welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to find out more about it.

But $\Delta U$ is defined in terms of state functions so it itself is also a state function....at least from what I recall

The correct application of that phrase is: $H$ is a state function because $H = U + pV$ and $U$, $p$ and $V$ are all state functions. $\Delta U$ with the $\Delta$ is not a state function. A state function is something that only depends on the current state of a system. $\Delta U$ depends on a process, defined by two states of a system, the initial and final.

@orthocresol Here state functions are defined as "a property whose value does not depend on the path taken to reach that specific value" which is the definition I am familiar with. This would imply that as notorious said $\Delta U$ is indeed a state function.

@Philipp I learnt, and I've only seen, the specific phrasing: "$U$ is a state function" $\Leftrightarrow$ "$\Delta U$ is independent of the path taken". The way I interpret the phrase "state function" is that it is a function of a state: $f = f(p,V,T)$. But $\Delta U$ is a function of two states: $\Delta U = \Delta U(p_1,V_1,T_1,p_2,V_2,T_2)$ (generally speaking). Analogously a path function is a function of a path (i.e. a line integral). But that's just my own interpretation, which is just my attempt to be logically consistent with what I was taught, so I might very well be wrong.

@orthocresol I understand your point of view and am not perfectly sure about the actual definition. So I am not able to say who is right or wrong. All I remember from my thermodynamics courses is that we treated the "$\Delta$-versions" of the thermodynamic potentials the same way we did the potentials themselves, i.e. treated them like state functions (e.g. used Maxwell relations and such).

I think the premise of the question itself is false: if you start with a given state (P1, V1) and undergo a reversible, adiabatic (hence isentropic) expansion, you can get to state P2, V2. But if you don't have the requirement of reversibility, I don't think it is possible to gain the same second state - there will be either internal heating/cooling, causing the expected final volume (or pressure, depending on the mechanism of expansion) to not be P2 or V2.

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text:
How can the gauche-effect be explained?
Generally, anti-conformations are more stable. But in the case of 2-fluoroethanol, the gauche confirmation is more stable.
It's given in March's advanced organic chemistry book that intramolecular hydrogen bonding is not the correct reason for the stability. Wikipedia explains it using hyperconjugation, but I am not able to understand it. What is the actual reason for the gauche effect?

Overlap of the sigma C-H orbitals with the sigma antiboding orbitals of C-O and C-F causes the stabilization, as their energy difference isn't so big (therefore the overlap is much bigger than the overlap of the sigma C-O with the antibonding sigma C-F orbital  or vice versa in the anti conformation).

So the overlap is much effective in gauche confirmation? Do you have any pictures on this?

Here it is https://en.wikipedia.org/wiki/Gauche_effect#/media/File:Gauche_effect_hyperconjugation.png.

How can a bonding orbital overlap with an anti bonding orbital? Won't they cancel out?

The overlap will certainly lengthen the C-F bond but they can't cancel out. As electrons are delocalized over a larger region in space, the overall effect is stabilizing.Bonding orbitals overlapping with antibonding ones is a common phenomenon.

The gauche effect is commonly explained with LCAO-based bond orbitals. LCAO is short for linear combination of atomic orbitals and implies that we can take two atomic orbitals $\phi_1, \phi_2$ and create two molecular orbitals $\psi_1, \psi_2$ from them by linear combination: $\psi_1 = a_1 \phi_1 + b_1 \phi_2$ and $\psi_2 = a_2 \phi_1 + b_2 \phi_2$.
Considering two neighbouring $\ce{sp^3}$ hybridised carbon atoms, we can create a $\unicode[Times]{x3c3}$-orbital between them by overlapping a hybrid orbital of each; and if $a_1 = a_2$ and $b_1 = - b_2$ (or vice-versa) we can create an antibonding $\unicode[Times]{x3c3}^*$-orbital; both are shown in figure 1 (the bonding in the bottom row, of course). Note that the lobes on each side are the same size.

Figure 1: $\unicode[Times]{x3c3}$ and $\unicode[Times]{x3c3}^*$ orbitals of $\ce{C-C}$, $\ce{C-H}$ and $\ce{C-F}$ bonds (schematically).
If we consider a $\ce{C-H}$ bond rather than a $\ce{C-C}$ bond, we need to take electronegativity into consideration. Compare with the middle column of figure 1: The more electronegative element, carbon, contributes more to the $\unicode[Times]{x3c3}$ orbital, while the less electronegative element hydrogen contributes more to the $\unicode[Times]{x3c3}^*$ orbital. (Note that hydrogen uses a 1s-orbital for binding.)
Similarly for the $\ce{C-F}$ bond, although now fluorine is more electronegative than carbon. And note how fluorine uses a p-orbital for bonding: No energy or anything else would be gained by hybridising fluorine’s orbitals. Neither the size difference between hydrogen’s and carbon’s nor that between fluorine’s and carbon’s orbitals is to any scale and both are grossly exaggerated.
The gauche effect, as a special form of hyperconjugation, relies us on looking at how these molecular orbitals can (or cannot) interact with each other. If we look at the relevant part of the molecule (compare figure 2), we can see that if hydrogen and fluorine are anti to each other, we get a $\unicode[Times]{x3c3}_{\ce{C-H}}$ bond and a $\unicode[Times]{x3c3}^*_{\ce{C-F}}$ bond that have favourable geometries and favourable lobe sizes for overlap.

Figure 2: Potentially overlapping orbitals in the gauche and anti cases of 2-fluoroethanol.
On the other hand, as also depicted in figure 2, the overlap would be bad if oxygen and fluorine be anti due to the small contribution of the less electronegative carbon to the $\unicode[Times]{x3c3}_{\ce{C-O}}$ orbital. (For simplicity, I have rendered oxygen using a plain p-orbital. Oxygen usually has orbitals somewhere between p and $\ce{sp^2}$.)
All this so far tells us why the interaction is geometrically favourable, but we yet did not deduce why it is energetically favourable. And here is where we are, more or less, linear combining the two molecular orbitals we created by linear combination earlier. The combining we are doing is a very general thing, observed not only in the gauche effect here or in hydrogen peroxide, but also is the basis for the anomeric effect in carbohydrates and hyperconjugation. The principle is: If we combine an occupied orbital with an unoccupied one, the occupied orbital gets lowered in energy while the unoccupied gets raised. Because only the occupied orbital contains electrons and their energy is lowered, the entire system gains stability. (You can read the two bolded half-sentences as one.) Compare figure 3.

Figure 3: Stabilisation by combining occupied and unoccupied orbitals
What we have done, is lower the energy level of an already lowered (because bonding is favourable) occupied orbital; the entire system gains. We often state this by saying things such as:

Electron density is transferred into the antibonding orbital.

But it really boils down to stabilising occupied orbitals. However, that transfer of electron density helps us understand part of the consequences associated with this favourable mixing. If both the bonding and the antibonding orbitals of a certain bond are occupied, you can think of that bond as non-existant; as if both sides had an electron pair. If, on the other hand, neither orbital is occupied, the bond does also not exist, but due to lack of electrons. We can use that analogy to create a mesomeric description of the gauche effect with the two resonance structures depicted in scheme 1.

Scheme 1: A possible depiction of resonance stabilisation corresponding to the gauche effect.
Many chemists out there will vigorously refute this depiction as invalid. Indeed, the resonance structure on the right-hand side of the mesomery arrow shows charge separation and a non-optimal number of bonds. We can therefore assume that it only contributes ever so slightly to the complete picture — which is in line with the electron density only minutely being transferred and also the new bonding orbital (figure 3) being a lot closer in shape and size to the $\unicode[Times]{x3c3}_{\ce{C-H}}$ orbital rather than $\unicode[Times]{x3c3}^*_{\ce{C-F}}$ orbital. But the depiction certainly helps us understand why oxygen cannot do the job: Imagine $\ce{HO+}$ instead of $\ce{H+}$. Looks ugly and unstable from a chemical point of view.

I have a doubt. Before the favorable overlap, the C-F bonding orbital becomes antibonding orbital. But isn't this unfavorable?

@AdityaDev I cannot follow.

@AdityaDev I think you are thinking that the C-F bonding molecular orbital (bonding MO) somehow converts into an antibonding orbital. That is not the case. Both, the bonding and the antibonding orbitals exist simultaneously, though not in the exact same region. They exist, if I must say so, "side by side". Only the bonding MO is occupied.

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text:
Does benzene's resonance structure allow it to enter DNA?
According to this link, benzene is able to insert itself into the human DNA. It isn't an authoritative source and appears to be quite biased, so I'm wondering if there's any truth to this.

The resonance hybrid structure of the benzene molecule causes it to have electrostatic potential, be very stable, and insert itself into human DNA. Taken together, these molecular and electrical characteristics of benzene make it carcinogenic to humans.

I will start my answer with a preface that the website linked to in the question is a pseudoscience website (and I am glad that it has vanished from the face of this earth, only accessible via the Wayback Machine). These people use scientific-sounding jargon that sounds impressive to the lay reader, but any actual scientist will know that it is simply rubbish. The objective is usually to fear monger, or to promote some product of theirs that has no special properties. Unfortunately, these seem to be becoming more and more prevalent recently.
However, the premise of the question is sound: benzene itself is known to cause cancer in humans. It is listed as a Class 1 carcinogen, meaning that there is evidence for carcinogenicity in humans.
As with many things in molecular biology, we do not actually know the full mechanism by which benzene causes cancer. However, there have been a number of papers written on the topic: some of them are linked at the bottom of this post.
Exactly which metabolite of benzene causes cancer is not known. It is likely to be a combination of them. The first step is oxidation of benzene by the enzyme cytochrome P450 2E1 (CYP2E1) to benzene oxide, which exists in equilibrium with oxepin (via a 6π electrocyclic reaction). The cytochrome P450 enzymes are found in the liver, and their role is generally to insert oxygen atoms into molecules, making them more polar and water-soluble, so that they can be excreted. For more information about this process, see Drug metabolism on Wikipedia.

These undergo further reactions, catalysed by various other enzymes, to give a large range of metabolites:

In particular,

The benzoquinones have been shown to inhibit DNA topoisomerase II, an enzyme that makes temporary cuts in double-stranded DNA in order to "unwind" DNA that has been entangled. It plays a key role in many cellular processes such as DNA replication, DNA repair, and chromosome segregation (during cell division); therefore, inhibition may lead to chromosome breakage or failure to segregate.
The same quinones can undergo a process called redox cycling, where they undergo an enzymatic reaction in which a single electron is added to them to form a radical anion. These species are then released, and react with molecular oxygen $\ce{O2}$ to give the superoxide anion, $\ce{O2^-}$... and then the process repeats itself. The buildup of $\ce{O2^-}$ (and other reactive oxygen species) leads to oxidative stress and DNA damage.
(E,E)-muconaldehyde has been recently shown to form an adduct with two molecules of deoxyguanosine, i.e. the guanine bases in DNA. Here, R represents the rest of the deoxyribose sugar. Intra- or inter-chain links can be formed via this method, which then lead to inaccurate replication or chromosomal aberrations.
A mechanism was proposed in reference 3. It is not reproduced here but it is not difficult to imagine how such a reaction might happen: nitrogen atoms in guanine are nucleophilic, and literally every carbon in muconaldehyde is electrophilic.

The literature on the topic contains much more information than I can write in here. Reference 4 is a relatively recent review on the topic, which would be a decent starting point to find further information.
Regardless of what mechanism it is, one thing is certain: benzene itself and its molecular properties are not likely to be the cause of its carcinogenicity. It is almost certain that multiple metabolites of benzene are the culprits.

References

Chen, H.; Eastmond, D. A. Topoisomerase inhibition by phenolic metabolites: a potential mechanism for benzene's clastogenic effects. Carcinogenesis (Oxford) 1995, 16 (10), 2301–2307. DOI:10.1093/carcin/16.10.2301
Rappaport, S. M.; Kim, S.; Lan, Q.; Li, G.; Vermeulen, R.; Waidyanatha, S.; Zhang, L.; Yin, S.; Smith, M. T.; Rothman, N. Human benzene metabolism following occupational and environmental exposures. Chem.-Biol. Interact. 2010, 184 (1–2), 189–195. DOI:10.1016/j.cbi.2009.12.017.
Harris, C. M.; Stec, D. F.; Christov, P. P.; Kozekov, I. D.; Rizzo, C. J.; Harris, T. M. Deoxyguanosine Forms a Bis-Adduct with E,E-Muconaldehyde, an Oxidative Metabolite of Benzene: Implications for the Carcinogenicity of Benzene. Chem. Res. Toxicol. 2011, 24 (11), 1944–1956. DOI:10.1021/tx2002838.
Hartwig, A. The role of DNA repair in benzene-induced carcinogenesis. Chem.-Biol. Interact. 2010, 184 (1–2), 269–272. DOI:10.1016/j.cbi.2009.12.029.

I am not sure about benzene but larger, multi-ringed aromatic compounds like phenanthrenes can stack themselves between DNA base pairs and cause mutations (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC137070/).

Um, what?
Toluene — $\ce{C6H5CH3}$, a relatively more reactive derivative of benzene — is highly water-insoluble, so can't be easily disposed of body. A group of six (so is reported) cytochrome P450 enzymes in liver are responsible for oxidizing it to benzoic acid, hence making it more disposable.
Benzene — $\ce{C6H6}$ — doesn't have any methyl groups and the enzyme changes it into some really reactive chemicals species, which can react with and alter your DNA, causing mutation. That's why benzene is carcinogenic.
Unfortunately, the paper discussing its metabolism is behind a paywall, but its title clearly states induction of cytochrome P450s.
The site seems like rubbish to me; something like a 10-year-old grabbing an introduction to organic chemistry and doodling what they understood from the first two lines of the introduction to aromatic compounds; he seems to imply that "aspirin", "naproxen", "celexib" and many other famous drugs are "sources" of benzene and carcinogenic. So$\,\ldots$!

First off, the linked site is a complete load of rubbish, claiming that any substance with a benzene ring is toxic. Sodium causes burns and chlorine is a toxic gas, so by the same logic we should be completely avoiding sodium chloride.
M.A.Ramenazi correctly points out that relative solubility in water and fat are important. Both toluene and benzene, once dissolved in fat, are difficult to excrete, but toluene is more easily converted into an excretable water soluble compound (benzoic acid) due to its methyl group, which may account for its being less harmful than benzene. The detergents mentioned in the are article are about equally soluble in water and fat, so they are excreted fairly easily (ironically, the hydrophilic group on these molecules is based on benzenesulfonic acid!)
Solubility obviously isn't everything, the different metabolic pathways of different substances are also massively important.
However, this idea that aromatic ring systems can intercalate with DNA didn't come from nowhere. See https://en.wikipedia.org/wiki/Intercalation_(biochemistry) http://www.jonathanpmiller.com/intercalation/ If we substitute the odd term "resonance hybrid structure" for the more accurate "flat aromatic structure" we see that these links indicate that some compounds with multiple benzene rings are believed to owe their toxicity to their ability to intercalate with DNA.
Flat aromatic molecules are able to intercalate between the bases of DNA, much as graphite is formed of flat aromatic layers with weak secondary bonding interactions between the layers. But note that all the toxins mentioned in the above link also have polar groups, which are probably just as essential to their mode of action as the flat aromatic structure.
For benzene itself, the mechanism of toxicity explained in Orthocresol's answer seems much more plausible.

Yeah, I always thought it had to do with that actually, but whatever brief research I did the other day told me that it was more of the metabolites. I'm sure there is some paper out there that shows that benzene itself can intercalate with DNA. Benzopyrene is a well-known example of an aromatic compound that does so.


text:
Comparing C=O bond length in CO2 and CH2O
Why is the C=O bond in $\ce{CO2}$ shorter than that in $\ce{CH2O}$? How can I use hybridisation theory to explain this?

First of all, what are the hybridisation schemes?

Drawing the Lewis structure for CO$_2$ and CH$_2$O shows that the C atom in CO$_2$ is sp hybridized and therefore forms a double bond (which is composed of a $\sigma$ and $\pi$ bond) with either O by overlapping one of its sp orbital with one of the sp$^2$ orbitals of O (since O is sp$^2$ hybridized) to form the $\sigma$ bond and one of its p orbitals with the p orbital of O to form the $\pi$ bond.
In CH$_2$O, the C atom is sp$^2$ hybridized, meaning the C$=$O bond is formed by an overlapping sp$^2$ orbital from C and an sp$^2$ orbital from O (since O is sp$^2$ hybridized) for the $\sigma$ bond and an overlapping p orbital from C and a p orbital from O for the $\pi$ bond.
So, the difference in the C$=$O bonds in the two compounds is in the constituent orbitals of their respective $\sigma$ bonds. The $\sigma$ bond of the C$=$O bond is shorter in CO$_2$ because it involves the overlap of an sp orbital with an sp$^2$ orbital. An sp orbital has more "s character" than an sp$^2$ orbital (you can think of an sp orbital as 50% s character, 50% p character and an sp$^2$ orbital as 33% s character, 67% p character), therefore the electron density in an sp orbital is on average closer to the nucleus of the C atom than in an sp$^2$ orbital, leading to a shorter $\sigma$ bond length. Because the $\pi$ bonds are formed from identical (up to orientation) p orbitals in each compound, the shorter $\sigma$ bond in CO$_2$ gives it a shorter C$=$O bond length than CH$_2$O.

Welcome to chem.SE. Your answer is perfectly correct, but we would really prefer if homework questions are not answered in full - it's the proverbial "giving a man a fish". This is especially so if no discussion or thoughts are provided by the OP in the question. You can check out the site's homework policy under the help centre.


text:
What causes the yellowing of the linear alkylbenzene (LAB) cocktail?
In our lab, we use the liquid linear alkylbenzene (LAB) cocktail mixed with 2g/L 2,5 diphe-nyloxazole (PPO) as the organic scintillator for radiation detection. The new LAB cocktails are stored in the transparent glass bottles provided by the company. Although they are exposed to the daylight lamp in the lab, they are always very clear(see picture 1, the sample was put in a small glass vial). However, when we put some liquids into a transfer system, they turned to be very yellow after about two weeks (see picture 2). In the system, the LAB liquid was first stored in a large Acrylic vessel (acrylic is proved to be compatible with the LAB) and then went through the stainless steel tubes and valves (from Swagelok company), finally went into a small acrylic container or a glass jar for preventing spilling. There is no heating during the experiment, just some pressure changes created by a pump to make the LAB transfer from the large vessel to the small container. The LAB liquid may also contact with the rubber from O-ring for vacuum sealing, the Teflon tape, the plastic(PTFE) gas tube, copper lid, etc. So in our case, what may cause the yellowing for the LAB?

The role of LAB in your cocktail is partly just a solvent. Usually, the cocktail also contains at least one primary scintillator (e.g. PPO = 2,5-diphenyloxazole) and maybe a secondary scintillator. The yellowing may be caused by degradation of these scintillators. Could you add the name or composition of your cocktail? In addition to heat and UV, cocktails can be sensitive to various chemicals, including atmospheric oxygen. In my experience, all cocktails turn yellow sooner or later, except when stored in flame-sealed glass vials in the dark.

Thanks Loong ! We have 2g/L PPO in the LAB. I may try to let one sample unsealed and expose to the air.

After sample testing, we know that it is the Buna O-ring causing the yellowing. The mechanism may be the triple bond Nitrile affects the cocktail.

Try to add one drop Bromine water to your LAB. If the color remains, it means that the LAB is pure enough. If not, it means that the LAB contains a trace of double bonds C=C. And these bonds are known to polymerize with the light to produce colored polymers.


text:
What temperature is needed for carbon monoxide to reduce iron oxide?
What temperature is needed for carbon monoxide to reduce iron oxide?
That is at what temperature the following reaction will take place:
Fe2O3 + CO -> Fe + CO2

Based on an experiment, carbon monoxide dose reduce iron oxide at 1200°C (2192°F).
But probably, this is not the minimal necessary temperature.
A better answer is still welcome.


text:
What is this Lab equipment?
I want to know the name of this equipment with a sealed graduated tube and the purpose it's being used.

I went through all the results of Google, but I was unable to find this exact equipment. I believe someone here should know this very well.

What's the piece you're holding?

@Beerhunter What do you mean? I'm also asking what's that :)

The glass piece, just out of shot, being directly held

Oh, it's the base stand of the equipment. Just a circular glass base which helps you to keep it on a table, you know. Nothing is hidden in the pic. Suppose my hand is the table, it's like now it's kept on the table :)

This lab equipment is called 'Fermentation Tube'.

If you want to see how an experiment is carried out with this equipment, please watch the video below.
Mark Garcia: Lab Protocol - Fermentation in Yeast (Unit 9 Respiration); YouTube

We use these in one of our Nursing Chemistry labs on carbohydrates.  They show that not all carbohydrates will ferment.

The piece is a type of eudiometer also called a gas measuring tube.  The graduations are for measuring gas collected and the bulb is used to hold liquid to assist in filling the tube with liquid.

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text:
Mathematical proof of ΔQ=ΔH at constant external pressure
Consider a system with an ideal gas. A process is being carried out in which the gas is being compressed at a constant external pressure $p_\mathrm{ext}$. For this process $\Delta Q$ should be equal to $\Delta H$, but this is where I am stuck.
Work done by the system will be: $$\Delta W=-p_\mathrm{ext}\,\Delta V$$ and
$$\Delta H=\Delta U+\Delta(p_\mathrm{sys}V)$$Now we can write: $$\Delta Q=\Delta U-\Delta W=\Delta H-\Delta(p_\mathrm{sys}V)-(-p_\mathrm{ext}\,\Delta V)$$ In this expression $\Delta (p_\mathrm{sys}V)$ and $(p_\mathrm{ext}\,\Delta V)$ are not equal as the process is irreversible and the so the external pressure will remain constant throughout the process but the internal pressure varies. I know that at the initial and final stage the external and internal pressure will be equal as the system will be then at equilibrium with the surrounding but the magnitude of the internal pressure will be equal to the constant external pressure $p_\mathrm{ext}$ only at the final stage. So how will $\Delta Q$ be equal to $\Delta H$ when the process is carried out at constant external pressure?

In an irreversible process, the total force per unit area exerted by the gas on the piston face is comprised of the local thermodynamic gas pressure at the piston face (as determined by the local gas density and temperature at the piston face) plus a viscous stress related to the rate of deformation of the gas in close proximity to the piston. So the thermodynamic pressure is not uniform throughout the gas and there are viscous stresses present.  By comparison, in a reversible process, the pressure is uniform throughout the gas and the viscous stresses are negligible.  In both cases, if the piston is massless, the force per unit area exerted by the gas on its side of the piston is exactly equal to $p_{ext}$, the force per unit area on the outside of the piston, and the work will be the same, as calculated from the force on either side of the piston.
If a frictionless piston has mass and the cylinder is horizontal, the piston will oscillate about its final position until the viscous stresses damp out its motion (i.e., its kinetic energy dissipates).  In the end, the final gas pressure will then be equal to $p_{ext}$, and, from a transient force and energy balance on the piston, one can show that the total work done by the gas will still integrate to $p_{ext}\Delta V$.
If the compression takes place irreveribly by, at time zero, suddenly increasing the external force per unit area from the initial gas pressure $p_0$ to the constant external force per unit area $p_{ext}$, the change in enthalpy of the gas from its initial state to its final state will not be equal to Q.  We will have $$W=p_{ext}(V_f-V_0)$$ and $$\Delta U = Q-W=Q-p_{ext}(V_f-V_o)$$so$$Q=\Delta U+p_{ext}(V_f-V_o)$$while$$\Delta H=\Delta U+(p_{ext}V_f-p_0V_0)$$

So in which type of the process will ΔQ be equal to ΔH? When we say that ΔH is equal to ΔQ at constant pressure,which pressure is being taken constant during the process(internal pressure of the gas or the external surrounding pressure)?

Q will be equal to $\Delta H$ if the gas compression is carried out  by deliberately removing heat from the system while maintaining the external pressure equal to the original gas pressure (i.e., $p_{ext}=p_0$).  But, if you try to control the compression by applying any other external pressure than the original gas pressure, Q will not equal $\Delta H$.

In response to the second part of your comment, as I said in my original answer, for a massless piston, the internal force per unit area exerted by the gas on the inner piston face (which is comprised of thermodynamic pressure plus viscous stresses) is always equal to the external force per unit area on the outer piston face.  From Newton's 2nd law, what else is possible?

So can we say that for an irreversible process ΔQ will never be equal to ΔH?Also please tell me in one line that constant pressure refers to which pressure as I cannot find answer to this question in any literature.

No.  If the heat is removed irreversibly by, say, plunging part of the cylinder into a cold bath, but $p_{ext}$ is held constant at the original gas pressure $p_0$, then Q will be equal to $\Delta H$.  Plunging part of the cylinder into a cold bath is irreversible because it creates temperature non-uniformities and gradients within the gas, and the average gas temperature is different from the bath temperature at the wall of the cylinder during the re-equilibration.  Regarding your question about which pressure is constant, there is ambiguity in the literature.

I ran out of words on my previous comment, so I want to continue.  Regarding the ambiguity, sometimes they mean that $p_{ext}=p_0$, while at other times they just mean that $p_{ext}$ is constant, but not necessarily equal to $p_0$.  In the former case, $Q=\Delta H$, while, in the latter case, it is not.  You just need to consider the context of the discussion.

One last doubt: if an ideal gas is compressed steeply(one step process) then will Q=ΔH? If yes then why?

What do you mean by "steeply."  Are you forcing it to compress by applying a higher external pressure than the original pressure of the gas?

Yes and that pressure is constant during the process.

Let us continue this discussion in chat.


text:
Partial hydrolysis of aluminium chloride
What is the meaning of partial hydrolysis of anhydrous aluminium chloride? What does the partial term signify?

Partial means that it is not fully hydrolyzed.
A partially hydrolyzed solution of aluminium chloride will be slightly acidic.
The chemistry behind the hydrolysis of AlCl3 is quite interesting due to the fact that many different complexes can be formed.


text:
Why are tertiary carbon atoms of polymers more prone to photo-oxidation and -degradation?
In Polymer Chemistry by Seymour-Carraher:

Saturated linear polymers such as HDPE are resistant to degradation, but slow degradation will occur in presence of $\ce{O2}$, UV, heat. Since tertiary carbon atoms are more readily attacked, PP is less resistant to degradation than HDPE.

It means a tertiary carbon atom makes polymer less resistant to photo-oxidation and photo-degradation. What makes tertiary carbon atom more likely to be attacked?

In Polymer Chemistry by Seymour-Carraher: "Saturated linear polymers such as HDPE are resistant to degradation, but slow deg will occur in presence of O2, UV, heat. Since tertiary carbon atoms are more readily attacked, PP is less resistant to deg than HDPE." It means tertiary carbon atom makes polymer less resistant to photo-oxidation & photo-degradation. What makes tertiary carbon atom more likely to be attacked?

You can always [edit] your question, to clarify it or add more information. I have hence inserted your comment.

This question is not specific to just polymers and could be answered by considering in general,  why are tertiary carbon atoms more likely to be oxidised.

This is admittedly a very brief and somewhat tempting answer and it might not be complete.
Photo-oxidation and -degradation occurs mainly via a radical mechanism. This is outlined on Wikipedia in more detail. In principle the initial step produces bond cleavage via UV radiation and with it radicals. Let $\ce{[P]_{$n$}}$ be a polymer, then $\ce{.[P]_{$n'$}}$ and $\ce{.[P]_{$n''$}}$ are decomposition products of this polymer, both of radical nature.
$$\ce{P ->[h\nu] .[P]_{$n'$} + .[P]_{$n''$}}$$
In a saturated non chained polymer a homolytical bond cleavage would lead two primary radicals, i.e. $\ce{R-CH2.}$, which are very unstable. In a chained polymer, like PP, if the bond break occurs at the tertiary carbon, there would be a secondary radical $\ce{R-\overset{\bullet}{C}H-CH3}$ and another primary radical. Homolytical bond cleavage of a hydrogen carbon bond can even lead to a tertiary radical, $\ce{R-\overset{\bullet}{C}(CH3)2}$.  All in all this can lead to a higher concentration of radicals, and therefore more access points for oxygen to attack, hence propagating the chain. One of the driving forces is the gain in entropy of this process.
$$\begin{align}
\ce{.[P]_{$n''$} + O2 &-> [P]_{$n''$}OO.}\\
\ce{[P]_{$n$} + [P]_{$n''$}OO. &-> [P]_{$n''$}OOH + .[P]_{$n$}}\\
\ce{[P]_{$n''$}OOH &->[h\nu] [P]_{$n''$}O. + .OH}\\
\ce{[P]_{$n$} + [P]_{$n''$}O. &-> [P]_{$n''$}OH + .[P]_{$n$}}\\
\ce{[P]_{$n$} + .OH &-> H2O + .[P]_{$n$}}\\
\text{etc.}&\phantom{\ce{->}~~}\text{pp.}
\end{align}$$

Tertiary carbons require a low amount of energy to scission. The energy contained in UV rad up to 370 nm could cause chain scission in tertiary carbons. Other polymers have lower activation maxima ( like 340 nm for HDPE) or 315 nm for Nylon. PP at 370 is weaker. Use Planks formula to see energy contained in 370 nm UV and compare to the bond formation ( or scissioning ) energy.


text:
How to assess the strength of the acid?

Which is the stronger acid in the given pairs?

$\ce{HClO3}$ and $\ce{HBrO3}$
$\ce{HClO2}$ and $\ce{HClO}$
$\ce{H2Se}$ and $\ce{H2S}$

given that the number of oxygen atoms is the same, the most electronegative atom should result in a stronger acid, thus $\ce{HClO3}$ is stronger.
$\ce{HClO2}$ is stronger since it has more oxygen atoms.
Here's where my logic fails, since the structure is the same (both have $\ce{2H+}$) then it boils down to 2 things to account for: electronegativity and the size of the atom. Going with electronegativity, sulfur is more electronegative thus stronger acid  which was my original answer. But, the correct answer is $\ce{H2Se}$ is the stronger acid which implies in this case atomic size ($\ce{Se}$ is larger than $\ce{S}$) outweighs electronegativity. I do not understand why in this case $\ce{H2Se}$ is stronger than $\ce{H2S}$ given sulfur is more electronegative, and most importantly why the same logic used in 1. can't be applied here and how do I discern where to account for atomic size specifically or when to account for electronegativity?

Any detailed-dumbed down explanations  on the logic are highly appreciated.

Let's start with $\ce{HXO_3}$.
In $\ce{HXO_3}$, after removal of proton, an $\ce{XO_3-}$ ion is formed which is resonance stabilised. For example, if X was $\ce{Cl}$,

Now, this shows that the $X$ needs to have a tendency to pull the electrons towards itself which decreases in the order :$\ce{Cl>Br>I}$. So more resonance is present in $\ce{ClO_3^-}$ than in $\ce{BrO_3^-}$ making the conjugate base of $\ce{HClO_3}$ more stable. Hence it is more acidic.
So acidic strength decreases as: $\ce{HClO_3>HBrO_3>HIO_3}$
Now let's compare $\ce{HXO_3}$ and $\ce{HXO_2}$. You should now know that the acidic nature depends on the stability of the resulting conjugate base. Here, the number of resonance structure for $\ce{HXO_3}$ is more. Hence, the acid strength decreases in the order:   $\ce{HClO_3>HClO_2>HClO}$. So its actually the number of resonance structures that affects the acidic strength.
Finally, let's come to hydrides of 16th group. Let me represent any 16th group element as $Y$. Now, as size of $Y$ increases, the $\ce{H-Y}$ bond length increases. This means, the $\ce{H-Y}$ bond becomes weaker and removal of $\ce{H+}$ ion becomes easier. The size of atom decreases in the order :$\ce{O>S>SE>Te}$.
Hence, the acidic strength decreases as : $\ce{H_2Te>H_2Se>H_2S>H_2O}$
This shows that electronegative is not the only reason. For example, if you considered $\ce{HF}$ and $\ce{HCl}$, then you may think that $\ce{HF}$ is more acidic because $\ce{F}$ is more electronegative and hence will be happy to get a negative charge. But, charge on a small atom is unfavourable. As atomic size increases, the charge can spread over a larger area. Hence, $\ce{Cl-}$ is more stable.
The same trend is seen in case of hydrides of 16th group. The charge in $\ce{Se-}$ is spread over a larger area. Hence, its more stable.

in the last part you mean the charge in Se- is more stable right? Btw, your explanation made so much sense!!! thank you so much  Aditya my professor could not  explain it at all!

Yes. I will edit it.

$-1$ The resonance concept much like the concept of double bonds for these structures (which would imply d-orbital contribution) is challenged and likely wrong.

I had the same doubt. But this is how my teacher explained it to me. It does give the correct order of acidic strength.

Just because a concept gives the correct predictions sometimes, it does not mean it is a good one.

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

Wikipedia mentions resonance.

The acidity of an acid is directly proportional to the stability of its conjugate base: The more stable a conjugate base is, the more stable the acid is. Thus for the first pairing, we need to consider the following more accurate Lewis structures (remember that main group elements observe the octet rule and the central halogen has a free electron pair resulting in an overall trigonal-pyramidal structure of the anions):

Comparing these two, we arrive at the conclusion that chlorine is more electronegative than bromine, thus allowing the oxygens to draw less charge towards them resulting in an overall lower charge density at each oxygen and therefore a greater tendency to displace a proton to gain more electron density.
For the second pairing, our picture is as follows:

In the chlorate anion, each oxygen effictively sees a net $+ \frac{2}{3}$ charge from the central chlorine atom. In chlorite’s case, they only see a $+\frac{1}{2}$ charge. Since $\frac{2}{3} > \frac{1}{2}$, the former is stabilised because of a greater partial charge at the central atom.
Finally, let’s consider hydrogen selenide versus hydrogen sulphide.
$$\ce{H-S-}~~~~~~\ce{H-Se-}$$
It is tougher here to decide which acid is stronger. Electronegativities do not help us, since sulphur and selenium only differ by $\Delta \mathrm{EN} \approx 0.03$ (values from Wikipedia) — almost nothing when compared even to the difference between chlorine and bromine which Wikipedia gives as $\Delta \mathrm{EN} \approx 0.2$. Both anions also have the same charge for the same number of atoms. However, selenium has an additional shell and is thus much larger. In a charge per volume description that means that the charge is more delocalised in selenium than it would be in sulphur and thus the hydrogen selenide anion is more stable and $\ce{H2Se}$ is the stronger acid.
Note how little electronegativity has to do with acidity: $\ce{HF}$ is the weakest hydrohalogenic acid and $\ce{H2O}$ is the weakest hydrochalcogenic acid with $\mathrm{p}K_\mathrm{a}$ values of $3.2$ and $15.7$, respectively. For comparison, the $\mathrm{p}K_\mathrm{a}$ value of $\ce{HI}$ is estimated to be $-10$.

Is this correct that less is the difference between size of atoms better is the resonance?

Like in case of ClO3- and BrO3- , resonance is better in former one due to less difference in size of O and Cl, so that is more stabilised.

@user161158 It is correct that orbital overlap and thus resonance is improved if the atom size is similar — which is usually read as ‘all elements are 2nd period’. So the resonance participation of sulphur in thioacetate is less than that of oxygen in acetate. However, you have managed to choose two examples (bromate and chlorate) in which there is no resonance to be observed. All three — as drawn in my answer — display only single bonds.

is this because of the same reason that resonance is not observed in these molecules?

But the resonance should be there to stabilise these anions.


text:
Energy of eclipsed and staggered conformations of ethane and propane

Which points on the given energy diagram represents eclipsed confirmation of ethane and staggered confirmation of propane respectively?

Attempt: Eclipsed configuration is unstable because of repulsions between filled $\sigma$ or bonding orbital and the anti or staggered configuration is more stable due to interactions between filled $\sigma$ orbital and empty $\sigma^*$ orbital.
From the given diagram, either R or S could represent eclipsed confirmation of ethane and P or Q could represent staggered confirmation of propane.
Since they are different compounds, they should lie on different energy diagrams. So the answer could be (R,Q) or (S,P). Which one is correct?

That diagram is horribly drawn.  The way it is drawn, the 13.6 kJ/mol number is meaningless - it measures from the maximum of one curve to the minimum of the other curve.
I think that you are supposed to assume that for one of the molecules, the one for the lower curve, the torsional strain is 12.6 kJ/mol and for the other molecule the torsional strain is 13.6 kJ/mol.
Do you have some reason to think that the barrier to free rotation for propane would be higher or lower than for ethane?  Perhaps your book gives the value for the torsional strain of ethane/propane? (through a quick google search I was able to find the torsional strain for both of these)

Its not given. So the question has insufficient data.


text:
Why the boiling point decreases with increase in branching in alkyl groups?
I found out the answer is that

with increase in branching, the molecules attain a spherical shape with less surface area. As a result, interparticle forces become weaker resulting in lower boiling point.

My problem is that if it attains a spherical shape then how is its surface area less because sphere has the largest surface area?
So please anyone can help me by answering in simple language?

It's not just how much surface area, but, crucially, the surface area in contact with other molecules. Bear in mind that we are looking especially at van der Waals' forces here (or London dispersion forces, or instantaneous dipole forces, or whatever other name you have for them).
If you have a lot of long chain molecule together, almost all of their surface area is very near the next molecule. Examine these sausages. See how almost their entire length on the left and right side of each sausage is in contact with the next sausage. Going back to molecules, this represents a large surface over which the intermolecular forces can act.

On the other hand, branched molecules do not fit as tightly together. See this image (no pun intended)

Extra branches on the molecules mean that they do not stack as well, so for a given molecule, there is less surface area of that molecule in close contact with another molecule (therefore weaker van der Waals).


text:
What is an example of a biodegradable polyethylene?
I would like an example of a Biodegradable Polyethylene.
I am using polylactic acid (PLA) as an example of a biopolymer/ bioplastic but I can not find a biodegradable polyethylene, that is a substance that contains polyethylene but has additives added so it biodegrades quicker

One word: "Impossible". Biodegradation has its special use cases, and is a fun subject, scientifically, but what you really need is thorough waste management system.

In a short research I found little to no evidence, that a there is  polypropylene/polyethylene, that can actually be called biodegradeable. I have found a couple of articles that have this as a topic. I don't have the time to actually read all of them and since this is going to be to long for a comment, I'll share it as an answer.

S. Bonhomme, A. Cuer, A-M. Delort, J. Lemaire, M. Sancelme, G. Scott, Polymer Degradation and Stability 2003, 81, 441–452.
This is a general overview of the mechanisms of biodegradation of such materials.
Ignacy Jakubowicz, Polymer Degradation and Stability 2003, 80, 39–43.
This article investigates the degradation of polymers, which contain pro-oxidant additives. It mainly focusses on the kinetics of the thermo-oxidative degradation. I am certain you can learn from this article which kind of additives are used to make braking long polymers into oligomers easier and hence complete degradation faster.
Prasun K. Roy, Minna Hakkarainen, Indra K. Varma, Ann-Christine Albertsson, Environ. Sci. Technol. 2011, 45 (10), 4217–4227.
This article is very promising as it also deals with some long-term effects of polymers with pro-oxidative additives. The abstract already lays out the main problem, when talking about biodegradation: entering the eco-cycle harmlessly.

However, the “degradable polyethylene” which is presently being promoted as an environmentally friendly alternative to the nondegradable counterpart, does not seem to meet this criterion. [...] It appears from the existing literature that our search for biodegradable polyethylene has not yet been realized.
(Emphasis by me.)

All in all I think it is very important to carefully outline, what biodegradation is or should be. I hope these few articles can be a starting point for your further research.

The only type of PE I know that may degrade faster than the regular material is linear low density polyethylene (LLDPE).  The branching in the chains would make it more susceptible to oxidation and subsequent scission of the chain like polypropylene. If you're a purist you might argue this is a copolymer and not true PE though.

it's ok if it is a copolymer, however i dont think linear low density polyethylene would be considered as 'biodegradable'
I think it still takes a long time


text:
Curious about the chemistry of Carbon-14 after it's produced in the atmosphere
It's often explained how $\ce{^14C}$ is formed in the upper atmosphere, a neutron hits a nitrogen atom and ejects a proton.  Source.  Since this is likely to happen to an $\ce{N2}$, I'm curious what the chemistry that follows is.
Does the energy of impact split the molecule, leaving a lone $\ce{^14C}$ and a lone $\ce{^14N}$ (and a lone hydrogen), (each highly eager to bond with something) or does it leave a $\ce{C-N}$ which might also readily bond with $\ce{O2}$ or maybe something else.    I'm mostly curious what the immediate chemical reactions are after the $\ce{^14C}$ forms cause it seems to me that it should be initially pretty reactive.
Edit & maybe a partial answer:
It occurs to me after giving it some thought that if an $\ce{N2}$ is split into a free $\ce{C}$ and a free $\ce{N}$, each would be most likely to bond to an $\ce{O2}$, or a single $\ce{O}$ from $\ce{O3}$ if the split happens near the ozone layer (which is possible given the listed height where most $\ce{^14C}$ forms, "altitudes of 9 to 15 km" - same link as above).
Initially I was thinking it might be more exotic than that, but having given this some thought I think that's what happens.   The $\ce{C}$ and the $\ce{N}$ split and mostly each binds with $\ce{O2}$, 2nd most common they bind with $\ce{O}$.  $\ce{CN}$ is probably pretty uncommon, now that I think about it.

My initial guess would be that the nitrogen molecule would form cyanogen radical.

wikipedia article cites: "After production in the upper atmosphere, the carbon-14 atoms react rapidly to form mostly (about 93%) 14CO (carbon monoxide), which subsequently oxidizes at a slower rate to form 14CO2, radioactive carbon dioxide. The gas mixes rapidly and becomes evenly distributed throughout the atmosphere (the mixing timescale in the order of weeks)."

The main reaction forming C-14 in the air is the reaction of N-14 with neutrons to form C-14.
I know that if fast neutrons from fission events are allowed to strike a nitrogen nucleous that a very high energy gamma photon (higher than 1.6 MeV) are emitted. Now if we consider for a moment the emission of a gamma photon. We must conserve momentum.
The momentum of a gamma photon is given by planks's constant divided by the wavelength. According to J. M. Hendrie (The Journal of Chemical Physics 22, 1503 (1954); https://doi.org/10.1063/1.1740449) to break a N2 molecule into two nitrogen atoms requires about 8.8 eV.
I have calculated the energy of the recoiling atom of C14 which would be for a moment inside a CN radical when the reaction occurs. Thus we will break the molecule up to form carbon atoms.

I imagine that a carbon atom will be able to react with oxygen with ease to form a carbon oxide.


text:
Polycarbonate resistant to chemical
Why polycarbonate couldn't stand some chemicals i.e. acid and base? What reaction will occur if polycarbonate meets acid or base?

Strong acids or bases will hydrolyze polycarbonate resin--specifically poly(bisphenol A carbonate)--at the ester linkages, first giving shorter polycarbonate chains, and eventually down to bisphenol A and carbonate ion or carbon dioxide.


text:
Calculate the equilibrium constant
So I have this question that I was assigned, and I am having trouble with step 2:

Hydrogen sulfide can be removed from natural gas via the following process:
  $$\ce{2H2S(g) + SO2(g) <=> 3S(s) + 2H2O(g)}\ \ \Delta H = \pu{-145 kJ mol^-1}$$

Write the equilibrium constant expression for this reaction.
Calculate the equilibrium constant, when $\pu{1.00 mol}$ of $\ce{H2S}$ and $\pu{1.00 mol}$ of $\ce{SO2}$ in a $\pu{1.00 L}$ vessel at $\pu{373 K}$ to give $\pu{0.50 mol}$ of water vapour under equilibrium conditions.
Identify FOUR factors that would maximise the removal of $\ce{H2S(g)}$ in this reaction.

I have the answers for this, and this is the step I got to:
$$\frac{[0.5]^2}{[0.5]^2[0.75]}$$
So, I know in the top part of the fraction, it is given that at equilibrium $\pu{0.5 mol}$ of water is present. I get that.
But it also says that $\pu{1 mol}$ of $\ce{H2S}$ and $\pu{1 mol}$ of $\ce{SO2}$ reacted. I'm not sure how these lead to values of $0.5$ and $0.75$?
It says there are $\pu{0.5 mol}$ of water at equilibrium. Does that mean that the $\pu{1 mol}$ of $\ce{H2S}$ and $\ce{SO2}$ are not all used? Thus since the ratio of water to $\ce{H2S}$ is $1:1$ it is $0.5$ in the fraction, but can nothing explain the $0.75$?

Firstly, let's write the equilibrium expression for this reaction: $$\ce{ K =\frac{[H2O]^2}{[H2S]^2[SO2]}}$$ So we know that initially $\ce{[H2O]}$ = $\ce{[SO2]}$ = $\mathrm{1~M}$ and $\ce{[H2O]}$ = $\mathrm{0~M}$. We also know that at the end, $\ce{[H2O]}$ = $\mathrm{0.5~M}$.
Now the stoichiometric ratios of $\ce{~H2O:SO2:H2O} = 2:1:2$. Therefore the decrease in the concentration of $\ce{H2S}$ should equal the increase in the concentration of $\ce{H2O}$. Also the decrease in the concentration of $~\ce{SO2}$ should be half of the decrease in the concentration of $\ce{H2S}$. Therefore $\ce{[H2S]}$ = $\mathrm{0.5~M}$ and $\ce{[SO2]}$ = $\mathrm{0.75~M}$
Therefore, plugging in these values into the equilibrium constant, you should get the required answer.


text:
Which part of the reaction of potassium with water releases more energy?
When potassium is reacted with water, hydrogen is given off, which is normally ignited by the heat of the reaction and combusts. How much energy is given off by the burning hydrogen compared to the potassium reaction?

I think this question contains every information necessary to calculate it yourself. If there are any open questions, you can [edit] your post and give the context here.

$$\begin{array}{c}
\ce{2K(s) +2H2O (l)->2KOH(aq) +H2(g)}\\
\ce{2H2(g) + O2(g) -> 2H2O(g)}\\
\end{array}$$
We have the two equations above, and we want to know which releases more energy. The answer is relatively simple to find: it's the standard enthalpy of formation of the products minus the standard enthalpy of formation of the reactants. This works because enthalpy is a state function and a systems enthalpy change independent of the path taken from reactants to products. So the way we do this is by imagining that we take our products, break them into their constituent elements (which have a $\Delta H_\mathrm{f}=0$ by definition), and convert those into the products. So:
$$(-482.37\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}-(-285.83\ \mathrm{kJ \ mol^{-1}})\cdot 2\ \mathrm{\ mol}=\frac{-393.08\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{K}}=-196.54\ \mathrm{kJ \ mol^{-1}}$$
$$(-241.82\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}=\frac{−483.64\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{H2}}=-241.82\ \mathrm{kJ \ mol^{-1}}$$
$$\begin{array}{c}
\ce{2K(s) +2H2O (l)->2KOH(aq) +H2(g)} & \Delta H=-196.54\ \mathrm{kJ \ mol^{-1}}\\
\ce{2H2(g) + O2(g) -> 2H2O(g)} & \Delta H=-241.82\ \mathrm{kJ \ mol^{-1}}^{[1]}\\
\end{array}$$
So the combustion of hydrogen gas produces more energy per mole than the reaction of potassium with water.

$^{[1]}$ Formation Enthalpies


text:
Custom-defined functionals in Gaussian 09
As stated before I want to try out custom-defined functionals. Besides the question about B3LYP itself, there is another related question.
Short excerpt:

$$\text{XC-Functional}=a E_x^\text{local}+(1-a)E_x^\text{HF}+b E_x^\text{non-local}+c E_c^\text{non-local}+(1-c) E_c^\text{local}$$
The Gaussian documentation for DFT inputs specifies how one could change the amounts of each "sub-functional" by using the IOp-Statements through:
IOp(3/76=mmmmmnnnnn) IOp(3/77=mmmmmnnnnn) IOp(3/78=mmmmmnnnnn)
Where mmmmm and nnnnn will get divided by 10000 and 3/76=$P_1P_2$, 3/77=$P_3P_4$ and 3/88=$P_5P_6$. Those $P_{1,..,6}$ values are part of the following equation, which is a modified version of the equation above.
$$\text{XC-Functional}=P_2 E_X^\text{HF} + P_1 \left(P_4 E_X^\text{Slater} + P_3\Delta E_x^\text{non-local}\right) + P_6 E_C^\text{local} + P_5 \Delta E_C^\text{non-local}$$

Now Gaussian's description site about DFT tells that one simply has to choose one exchange and one correlation functional, i.e. B and LYP. But then B would be used for $E_x^\text{non-local}$ and LYP probably for $E_c^\text{non-local}$.
Does that mean, that the $E_c^\text{local}$-part

cannot be defined,
is always the VWN functional as in B3LYP, or
is also the chosen correlation functional?

Take a look at the IOp manual. Under IOp(3/74),

IOp(3/74) Type of exchange and correlation potentials.

-5 Becke3 using VWN/LYP for correlation.
02 Lee-Yang-Parr correlation.
05 VWN 80 (LSD) correlation.
200 Hartree-Fock-Slater exchange (Alpha = 2/3).
400 Becke 1988 exchange.

So 100 is Hartree-Fock, 200 is Hartree-Fock-Slater, 205 is Local Spin Density, and 402 is BLYP.

They're unclear about it, but this means you can take the "sum" of a 1- or 2-digit number (which specifies the correlation) and a 3- or 4-digit number (which specifies the exchange) to build an exchange-correlation functional. So, IOp(3/74=-5) == IOp(3/74=402), and an identical way to define B3LYP is
IOp(3/74=402) IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)

Does that mean, that the $E_{c}^{\text{local}}$-part

cannot be defined,
is always the VWN functional as in B3LYP, or
is also the chosen correlation functional?

Looking at the available correlation functionals (incomplete, these are the lowest few),

01 VWN5
02 LYP
03 P81 (PL in keyword list)
04 P81 + P86
05 VWN3
06 VWN3 + P86
07 OS1
08 PW91
09 PBE
10 VSXC
11 Bc96
18 VWN5 + P86
19 LYP + VWN5 for scaling

your choices for local correlation functionals are rather limited. Either you choose a correlation functional that's purely local (VWN3, VWN5, or P81), or a non-local/gradient-corrected functional which in Gaussian all seem to have the local component as part of their definition (as in LYP), or they're forcibly combined (as in P86).


text:
How exactly does the reaction between alkyl halides and silver nitrite occur?
$$\ce{R-X + AgNO_2 -> RNO_2}$$
Now depending on the $\ce{R}$ and solvent the reaction can occur via $\mathrm{S_N1}$ or $\mathrm{S_N2}$.
When the reaction proceeds via $\mathrm{S_N1}$ the product formed is $\ce{R-O-N=O}$ whereas for $\mathrm{S_N2}$ it is $\ce{R-NO_2}$.
Why is this so? According to me as the $\ce{Ag-O}$ bond is covalent the lone pairs on nitrogen must attack in both the cases forming the product $\ce{R-NO_2}$.

The concept of hard and soft acids and bases (HSAB) proved to be
  useful for rationalizing stability constants of metal complexes. Its application
  to organic reactions, particularly ambident reactivity, has led
  to exotic blossoms. By attempting to rationalize all the observed
  regioselectivities by favorable soft–soft and hard–hard as well as
  unfavorable hard–soft interactions, older treatments of ambident
  reactivity, which correctly differentiated between thermodynamic and
  kinetic control as well as between different coordination states of ionic
  substrates, have been replaced. By ignoring conflicting experimental
  results and even referring to untraceable experimental data, the HSAB
  treatment of ambident reactivity has gained undeserved popularity. In
  this Review we demonstrate that the HSAB as well as the related
  Klopman–Salem model do not even correctly predict the behavior of
  the prototypes of ambident nucleophiles and, therefore, are rather
  misleading instead of useful guides. An alternative treatment of
  ambident reactivity based on Marcus theory will be presented.
— H. Mayr’s, M. Breugst’s and A. Ofial’s introductory remark in their review pulished in Angew. Chem. Int. Ed. 2011, 50, 6470. DOI: 10.1002/anie.201007100.

Within the review, in the section on the nitrite ion, the authors give two competing effects that control product distribution.
For electrophiles with an electrophilicity parameter $E < -3$ (the bis-(4-methoxyphenyl)carbenium ion has $E = 0$, lower values are more stabilised carbocations such as bis-(4-dimethylaminophenyl)carbenium) the reactions proceed at equilibrium under thermodynamic control and the product yields are equivalent to their relative stabilities. Since nitro compounds are more stable than nitrite esters, they predominate.
$$\ce{R-ONO <=> R+ + NO2- <=>> R-NO2}$$
For electrophiles with a parameter $E > 0$, the reactions are essentially diffusion controlled, i.e. the free electrophile will capture the first nucleophile it can find. This includes tertiary alkyl carbocations ($E \approx 7{-}8$). The reactions do not go through classical transition states, and any means of describing them via frontier orbitals must fail. That explicitly includes describing them according to $\mathrm{S_N1}$ or $\mathrm{S_N2}$. In these reactions, kinetic control determines that the $\ce{O}$-attack is slightly more likely with $k_\ce{O} / k_\ce{N} \approx 3.4$. (This corresponds to computational models that show a slightly smaller activation barrier for the oxygen attack.)
Reality is more complex than these simple models imply especially for the diffusion controlled reaction. While the attack of oxygen is favoured kinetically, the attack of nitrogen results in less required reordering of the molecule (the two $\ce{N-O}$ bonds would still be identical while they become $\ce{O-N=O}$ in the nitrite ester) which reduces the expected favourism for oxygen. Effects such as solvent and counterion slightly affect the product distribution, and the highest ratio of nitromethane : methyl nitrite can be achieved using $\ce{Bu4N+ NO2-}$ and $\ce{MeI}$ in $\ce{CDCl3}$ (a product ratio of $70\,:\,30$ in favour of nitromethane). Note that $\ce{AgNO2}$ and $\ce{MeI}$ in DMSO yield more methyl nitrate ($54\,:\,46$) although the same electrophile is used. The authors conclude the nitrite section with the statement:

It should be noted, however, that the selectivities
  shown in [the table presented in the review] are also not related to the hardness of the electrophiles.

Note
Throughout this answer, electrophilicity parameters according to Mayr’s equation are used:
$$\lg k_{20~\mathrm{^\circ C}} = s(E + N)$$
$E$ is a nucleophile-independent electrophilicity parameter while $s$ and $N$ are electrophile-independent nucleophilicity parameters.

Bibliography
A paper related solely to studying the nitrite anion: A. A. Tishkov, U. Schmidhammer, S. Roth, E. Riedle, H. Mayr, Angew. Chem. Int. Ed. 2005, 44, 4623. DOI: 10.1002/anie.200501274.
The review mentioned in the opening, concerning all sorts of ambient nucleophiles often explained via the HSAB concept: H. Mayr, M. Breugst, A. Ofial, Angew. Chem. Int. Ed. 2011, 50, 6470. DOI: 10.1002/anie.201007100.

You say that the attack with N would require less re-ordering of the molecule. But how would this be a hindrance to the attack?

@TanYongBoon If I can just put two things together, forming a bond is quickly achieved. If upon putting two things together I also need to distort one internally, there’s a slightly higher barrier to be overcome.

Pearson's HSAB theory provides a simple explanation. When the reaction proceeds via $S_N1$, a hard acid (electrophile) is firstly generated (the carbocation) which interacts with the harder part of the nitrite anion an that's the oxygen-hard nucleophile. The $S_N2$ proceeds smoothly for the pair soft nucleophile–soft electophile.
In the given example the soft electrophilic carbon reacts with the soft center of the nitrite anion, and that's the nitrogen atom. In the former case, the silver coordinated to the oxygen definitely weakens its nucleophilicity. The formed carbocation has to be pretty hard and not so much stabilized by the solvent for the O-attack to be operative.


text:
Calculate the partial pressure of NO2
Calculate the partial pressure of $\ce{NO2}$.
$$p_{\ce{N2}}/p_{\ce{O2}}=5$$
$$\ce{N2(g) + O2(g) <=> 2NO(g)}$$
Let’s assume $K=0.1$
$$K= \frac{p_{\ce{NO}}^2}{p_{\ce{O2}}\cdot p_{\ce{N2}}}$$
$$K= \frac{p_{\ce{NO}}^2}{p_{\ce{O2}}\cdot p_{\ce{O2}}\cdot5}$$
But how do I proceed?

Ok.
Now you have 3 variables and 2 equations.
To solve the problem you need 3 equations.
Third equation you'll get from the partial pressure law:

The total pressure of an ideal gas mixture is the sum of the partial pressures of each individual gas in the mixture.

Tip: suppose total pressure is 1.

so? p(o2)+p(N2) =1
Aha so I can choose whatever sum I want?

Don't forget about NO.

When applying partial pressures please do not forget the partial pressure that the product itself exerts (NO)
So we have: p(O2)+ p(N2) + p(NO) = total pressure of the system.


text:
Confusion on enantiomers and meso compounds
Question: Are these compounds enantiomers?

Attempt: The compounds given are mirror images. But if I rotated the three bonds in the third carbon (configuration will not change), I will get a meso compound (plane of symmetry) :

So are these compounds enantiomers or meso compounds?

Aren't there some restrictions for flipping/rotation of Fischer projections .

Answer given is meso. At room temperature, rotation about any one single bond can take place. Interchanging 3 groups will not change configuration.

Realistically for compounds with more than 1 chiral centre, fisher projection is require to determine if compound is mesoisomer or enantiomers. If someone draws like that then dont rotate carbons to avoid confusion and just assume if it is meso or enantiomers based on diagram alone.

An unfortunate drawing of the original projections Place H at top and bottom easy to see it is an R,S configuration hence meso.

You are right, the 2 structures are equivalent and the substance has a plane of symmetry and is a meso compound. When in doubt, make a 3D model if possible.


text:
Mass Spec - What could a loss of 28amu as the first fragmentation be?
I guess 28 could be CO, C2H4 or N2. However I can't work out a mechanism by which CO could be the first fragmentation of any common organic compound. It seems like a diazonium salt could lose N2 as the first fragmentation but I've reason to believe this is not the unknown compound I'm trying to work out.

You should remember that the species that undergo fragmentation aren't common compounds but excited radical cations.

Hi Mithoron thanks for your reply. I understand that the molecular ion is a radical cation, but I don't understand how the molecular ion could lose just CO. COH I understand but not just CO.

If you had an aldehyde which formed a radical cation and had loss of a hydrogen radical, a subsequent fragmentation could easily be loss of CO.  You to need to look at homolytically splitting your bonds. The other option is a Mclafferty rearrangement for loss of ethene

Hmm, it's possible IMO for some of cyclic ketons to lose CO as first, with contraction of ring.


text:
Concentration of dissolved lead(II) bromide
Calculate the concentration of $\ce{Pb^2+}$ and $\ce{Br-}$ when $\ce{PbBr2}$ dissolves in water. $K = 7.9 \times 10^{-5}$
$$\ce{PbBr2 <=> Pb^2+ + 2Br-}$$
In most cases we have the concentration of the solute, but here we don’t. How should I be thinking now?
if it’s $1~\mathrm{M}$ for example
$$K = \frac{[x][x]^2}{[1-x]}$$

The given value $K=7.9\times10^{-5}$ actually is the value of the solubility product constant $K_\mathrm{sp}$ of $\ce{PbBr2}$
$$K_\mathrm{sp}=a_{\ce{Pb^2+}}\cdot a_{\ce{Br-}}^2=7.9\times10^{-5}$$
which approximately is (when incorrectly neglecting activity coefficients $\gamma$, the difference of molality $b$ and concentration $c$, and the correct units)
$$K_\mathrm{sp}\approx \left[\ce{Pb^2+}\right] \cdot \left[\ce{Br-}\right]^2$$
When $\ce{PbBr2}$ dissolves in water according to the given chemical equation, the resulting concentrations of $\ce{Pb^2+}$ and $\ce{Br-}$ correspond to
$$\left[\ce{Br-}\right]=2\cdot\left[\ce{Pb^2+}\right]$$
These equations can be used for calculating a value for $\left[\ce{Pb^2+}\right]$.


text:
Finding equilibrium constant from standard free energy
This question is asking to calculate equilibrium constant for oxidation of glucose from thermodynamics quantities. So the first expression came to my mind was $-\Delta G^\circ=RT\ln K$. I found the value of $\Delta G^\circ=-2\,278.78$ but as I proceed I've got $\ln K=1\,161.94$ which I couldn't reach to the $K$ value. Is there any other ways to solve this question?
$$\ce{C6H12O6(s) + 6O2(g) <=> 6CO2(g) + 6H2O(l)}$$
$\Delta G^\circ_\mathrm r=-2878.78\ \mathrm{kJ}$
At equilibrium, $\Delta G=0$; $Q=K$,
$\Delta G=\Delta G^\circ+RT \ln K$
$2878.78\times10^3\ \mathrm J=\left(8.314\ \mathrm{J/(mol\ K)}\right) (298\ \mathrm K) \ln K$
$\ln K=1161.93596$
Answer from the book is $K=5\times10^{504}$

Plugging the values into $K = e^{\tfrac{-\Delta G^0}{RT}}$ gives me an equilibrium constant of 2.51. I am unsure what your question is. Do you have problems solving the mathematical equation?

@Anders MB I've posted the question together with my workouts.

Your answer is pretty close to the book answer.  When you found the change in free energy, did you use the free energy of formation for liquid water or for water vapor?

@chester miller, I plugged in formation of liquid water to find the free energy change. I'm not sure if my answer is close to the book, but I've no idea on how they can reach to such a large K value because we wouldn't get an answer from natural log of 5e+504. Same goes to my own answer, there's no way to natural antilog 1161.94. So I'm wondering if there's any other way to relate free energy with equilibrium constant.

e^1162 is about 10^504 so what's the problem really? It's terribly simple calculation

Your problem seems to be one of calculation using large numbers, so look at it this way:
$$
\ln K = 1161.93596
$$
So
$$
\log_{10}K = \ln K \log_{10}(e) = 504.65\\
\Rightarrow K = 10^{504.65} = 10^{0.65}\times 10^{504} = 4.5 \times 10^{504}
$$
without your calculator exploding.


text:
What are the steps to integrate the common rate law to find the integrated rate law for any order
I cannot find any sources that actually list the steps for integrating a rate law, and it’s been driving me crazy. My Ap chem 2 teacher doesn’t know how to do it and just wants me to memorize the common rate laws, however I don’t memorize equations unless I know how to write them out and where they come from. I have the equation
$$ \text{rate} = \dfrac{\mathrm d[A]}{\mathrm dt} = k \cdot [A_0]^{n} $$
where $n$ is the order of the equation
$[A_0]$ is the concentration at time $0$
$[A]$ is the concentration with respect to time
$k$ is the rate constant
I know that for a first order ($n=1$) reaction the equation becomes
$$ [A] = [A_0] \cdot \mathrm e^{kt} $$
and that a second order ($n=2$) reaction the equation is
$$ [A] = \dfrac{[A_0]}{1 + kt \cdot [A_0]} $$
But I don't know what the steps are to find these equations. If I find that the order of my equation needs to be 1.5 ($n=1.5$), I will have no Idea what the equation will be.
From these equations I will know most of the time $[A_0]$, $[A]$ and sometimes Rate.
Edit:
After following the links I learned a portion of the process to solve the equation for any $n$. However after testing it out I am either getting a incorrect answer of a incomplete answer
$$ [A] =\int_{[A_0]}^{[A]}{\frac{1}{[A_0]^2}\,\mathrm d[A]} = \frac{[A_0]^{1}}{1} = \int_{0}^{t}{k\,\mathrm dt}$$
the answer should be
$$ [A] = \dfrac{[A_0]}{1 + kt \cdot [A_0]} $$
what parts am I missing?

Discussion about this post in chat.

For starters your rate equation is not right. $[A_0]$ is a constant and the solution would be trivial as written. $[A_0]$ is the value of $A(t)$ for $t=0$. Unless it is zero rate, your reaction rate should depend on the concentration at any time $t$ and not only $t=0$. Check the wikipedia article.

The integrated rate laws can be derived by equating two expressions for the rate of the reaction. This gives us a first order differential equation which can be manipulated to get the integrated rate law. The steps that you need to follow are actually quite simple and structured. Here is an example for deriving the first order integrated rate law:
Step 1: Equate the two expressions for the rate of the reaction
For first order reactions, the two expressions for the rate of the reaction are: $$ r = -\frac{\mathrm{d}[A]}{\mathrm{d}t}$$ and $$r = k[A]$$ Equating them we get: $$-\frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]$$
Step 2: Rearrange so that all the same variables are on the same side to form an first order differential equation
So we rearrange this expression to get all of the same variables on the same side (note that k is a constant, not a variable, so it doesn't matter which side it is on) to get: $$\frac{\mathrm{d}[A]}{[A]} = -k\mathrm{d}t$$
Step 3: Integrate both sides
Now we integrate both sides: $$\int_{[A_0]}^{[A]}{\frac{\mathrm{d}[A]}{[A]}} = -\int_{t_i}^{t_f}k\mathrm{d}t$$
$$\mathrm{ln}[A] - \mathrm{ln}[A_0] = -kt$$
Step 4: Using some algebraic manipulation, rearrange expression to get an equation for [A]
$$\mathrm{ln}\frac{[A]}{[A_0]} = -kt$$
$$\frac{[A]}{[A_0]} = \mathrm{e}^{-kt}$$ Therefore we get as required:
$$[A] = [A_0]\cdot e^{-kt}$$
Using this guide, you should be able to derive the other integrated rate laws. If you have any trouble, feel free to ask.
Edit:
Since you have also mentioned that you are struggling with deriving the second order integrated rate law, I will show you how to derive it, using the steps that I have outlined above.
Step 1:
$$r = - \frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]^2$$
Step 2:
$$\frac{\mathrm{d}[A]}{[A]^2} = -k\mathrm{d}t$$
Step 3:
$$\int_{[A]_0}^{[A]}\frac{\mathrm{d}[A]}{[A]^2} = \int_{t_i}^{t_f}k\mathrm{d}t$$
$$\frac{-1}{[A]} + \frac{1}{[A]_0} = -kt$$
Step 4:
Okay now everything up to here has been exactly the same thing we did for deriving the first order integrated rate law. We have gotten an equation relating $[A], [A]_0$ and $t$. To make $[A]$ the subject, we multiply by $[A]$ and put all the terms containing $[A]$ on one side:
$$\frac{[A]}{[A]_0} + kt[A] = 1$$ Now we factorise $[A]$ out on the left hand side and divide both sides so that the left hand side equals to $[A]$:
$$[A]\left(\frac{1}{[A]_0} + kt\right) = 1$$
$$[A] = \frac{1}{\frac{1}{[A]_0} + kt}$$
Now we multiply the right hand side by $[A]_0$/$[A]_0$ and we get the required equation:
$$[A] = \frac{[A]_0}{1 + kt\dot~ [A]_0}$$


text:
Why and how does carbon hybridize itself according to its need?
Sometimes carbon forms sp3 orbital and sometimes sp2 and sp. Also, carbon chains are more stable with sp3 orbitals. So, why and how does carbon hybridize itself to other orbitals?
My question is more about stability. It's not about how it is significant to organic chemistry. For example, Methane which has 'sp3' orbitals is stable because of its sigma bonds. But, ethene, which has a pi bond is not stable and is more reactive than methane. So, why does carbon form this 'pi bond' while on the other hand has a better option of a sigma bond.

Hybridization is just a mathematical construct. It has no physical meaning.

Re your edit: No, ethene is as stable as methane (or ethane) is. In fact, they are both thermodynamically unstable towards combustion into $\ce{CO2}$ and $\ce{H2O}$. Their ‘stability’ is actually only kinetical inertness.

On the other hand, ethene cannot form a set of two sigma bonds unless it resorted to diborane-like banana bonds. But a set of sigma and pi is so much more favourable than two banana bonds (whatever Chuck Boldwyn might say).

I hesitate to reopen this question. It is not a strict duplicate anymore, but to me it is unclear what you are actually asking. Hybridisation has absolutely nothing to do with stability, like bon already said. It is also not very good manners to remove the formatting that other caring members of this community have introduced.

@Jan The formulation of two banana-bonds is equal to sigma and pi bond. The big but here is, that a banana bond is no sigma bond, so there are not two sigma bonds versus one sigma and one pi bond. True is: ethene cannot form two sigma bonds.

Hybridization is part of the language chemists use to speak to one another.  When someone says that a carbon atom is $\ce{sp^2}$ hybridized, it's a shorthand way of telling others

that there is a p-orbital on that carbon
that the angles around the carbon are roughly 120°
that the bonding orbitals (the $\ce{s^1p^2}$ orbitals) have 1 part in 3, or 33%, s-character and 2 parts in 3, or 67%, p-character
that the bonds around that carbon are likely to be a bit shorter than the bonds around an $\ce{sp^3}$ hybridized carbon (because the $\ce{sp^2}$ bonds have more s-character and are therefore closer to the nucleus, e.g. shorter).

So when someone says that a carbon atom is $\ce{sp^2}$ hybridized, they're really saying a lot more.
Using hybridization to describe an atom is similar to using VSEPR or MO theory to describe a molecule.  Hybridization is just another way to try and describe reality; just like the other methods, often it works well, but sometimes it doesn't.
Also keep in mind that there is a continuum of hybridizations available to a molecule all the way from pure s to pure p with everything else in between.
$\ce{s^1p^0...s^1p^1...s^1p^2...s^1p^{2.2}...s^1p^3...s^1p^4...s^0p^1}$
Of course, we normally drop the "1" and the "0" superscripts and write those hybridizations as
$\ce{s...sp...sp^2...sp^{2.2}...sp^3...sp^4...p}$
As to how carbon (or any other element) "decides" which hybrization to use, it adopts the hybridization that produces the overall lowest energy for the molecule. Let's take the case of a carbocation.  Suppose we start with an $\ce{sp^3}$ hybridized carbon in a molecule like $\ce{R_3C-I}$ and subject it to conditions where it ionizes to an $\ce{sp^2}$ hybridized carbocation and $\ce{I-}$.

(image source)
Your question is basically, why did the central carbon rehybridize from $\ce{sp^3}$ to $\ce{sp^2}$? The answer is, the ionized molecule (the carbocation) will reoptimize its geometry (rehybridize) in order to attain the lowest energy possible. In this example note that the carbocation p-orbital is empty - it contains no electrons. So why keep that orbital $\ce{sp^3}$ hybridized, why waste a part of a valuable, low-energy s-orbital in a bond with no electrons in it to stabilize? Instead, why not remove the s-character from the $\ce{sp^3}$ bond making it a pure p-orbital and mix that s-character into the other 3 orbitals (or bonds) to convert them from $\ce{sp^3}$ to $\ce{sp^2}$ thereby lowering the energies of the electrons that are in those 3 bonds. In fact, this rehybridization, or geometry reoptimization, is exactly what does occur in this reaction in order to stabilize the electrons as much as possible and produce a molecule (or ion) with the lowest possible energy.


text:
Discerning Sodium bicarbonate
I have 'sodium bicarbonate' powder , which is supposed to be (NaHCO3) , is there a way for me to do a test on it at home , without some special equipment and determine whatever its 100% pure or it has some additives?

Simple tests--for example, dissolving some of the material in water and looking for undissolved matter--can only tell you so much about its purity. If it is food grade and in its original brand package, it is probably OK to use, but if you really need to be sure, you are better off buying a new box. If you purchase analytical or reagent grade material, it comes with a certificate of analysis listing levels of impurities--still pretty cheap relative to having it analyzed.

Didn't bought the box in the first place , bought it in bulk as 500gr for like 30cents, that why there is no label to make it certain.
However disregarding this , even on those small packages of 20gr there is no label such as NaHCO3 only E500.

No. Without having some real idea of what impurity your looking for, or some idea of how much on an impurity is tolerable you can't make any judgement. For instance radioactive polonium would poison you at trace levels but it would be readily detected by a Geiger counter. As another example a mixture of $\ce{NaOH}$ and $\ce{NaCO3}$ would give a yellow flame test and a ppt much like pure $\ce{NaCO3}$. So never ever eat something unless you KNOW it is "food grade."

Without additives its a bit hard.
You can test easely if it must be it or not, if you put a bit of it on a flame you must see a yellow flame which means there is sodium $\ce{Na}$ in the compound. Like this.

And to be more sure if you pour some acid (like ethanoic acid or vinegar) it must makes gas. Now if you want to verify that this gas is carbon dioxyde you can try to do the reaction on a glass or whatever you want, light a candle and pour the gas (which is inside the glass) on the top of the flame, if the flame disappear it must be carbon dioxyde. Exactly like in this video obviously you don't use backing powder but your product..
You can also use a solution of limewhater but you musn't have this at your home...but if you have you must obtain this: (but you need to pour the gas only in your solution!) so its not really easy at home.

So if you do all this experiments you can say its $\ce{NaHCO3}$ but its impossible to be sure at 100% whitout doing an Infra-Red exams for example. It can be mass spectroscopy or RMN.. but obviously not at home, and you can detect all additives if they are.
Hope it will helps you a bit,
Friendly.

NB : If you are not sure don't use it for your food.. :)

Thanks for your time answering , well i am sure its NaHCO3 , what worries me is it only NaHCO3. As i suspected there is no simple test to make sure its 100% clean. I will take it tomorrow to lab.

@Sutekh this compound is use to absord smells so you may find a lot of things inside.


text:
Stoichiometry in Gas-Solid Flow System
I am a bit confused on how to account for the stoichiometry of a reaction as follows:
$$\ce{A}(s) + \delta \ce{B}(g) \ce{->} \ce{C}(s) + \delta \ce{D}(g)$$
The solid $\ce{A}$ is stationary in the reactor, and gas $\ce{B}$ flows through it with known inlet molar flow rate. Since 1 mol of $\ce{A}$ produces $\delta$ mol of $\ce{D}$, but the solids are static, how can I determine the amount of $\ce{D}$ formed (molar flow rate)? The other known piece of information is the normalized extent of reaction, $\alpha$, which can be defined as the amount of $\ce{D}$ formed at time $t$ divided by its total amount.
Please let me know if it's not clear enough. It's been difficult for me to find clear explanations about the modeling of such type of reaction in the papers and theses I've read. The closest I've found to a definition of $\alpha$ is as follows:
$$\alpha = \int \frac{\ce{D}(t)}{\ce{D}_\text{total}} \mathrm{d}t$$
(yes, written exactly like that)
My Approach So Far
I believe I have to somehow consider both the flow rate of $\ce{B}$ and the amount of $\ce{A}$ in the reactor. If the molar flow rate of $\ce{B}$ is "too large", then the stoichiometric coefficient $\delta$ will limit the flow rate of $\ce{D}$. I've thought of doing something like this:
$$\dot{n}_{\ce{D}} = \min\{\dot{n}_{\ce{B}},\delta n_{\ce{A}}\}$$
where $\dot{n}_i$ is the molar flow rate of species $i$ and $n_\ce{A}$ is the number of moles of solid $\ce{A}$, but as you can see one of the arguments of the $\min$ operator is a molar flow rate and the other is just moles, which is problematic. Any suggestions?
Edit
Clearly, I have to use $\alpha$ somewhere. But I'm still unsure what to do here. Generically, I'd write something like,
$$\dot{n}_i = \dot{n}_{i,0} + \dot{n}_0 \nu_i \alpha$$
but I'm not sure if this is correct in this case (again, how do I use the amount of solid in the reactor?).

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text:
Intensive properties clarification - Chemical thermodynamics
I just started learning chemical thermodynamics and have come upon the definitions for extensive and intensive properties. I had a great deal of confusion over the exact meaning of intensive properties, but I believe I have come to a proper understanding of it after doing some extra reading. I will present my understanding below, and I just need to clarify that this is exactly what intensive properties are.

Intensive properties are defined differently for homogeneous and heterogeneous systems. For homogeneous systems, intensive properties are those properties that possess the same value at different points within the system and for the system as a whole. For heterogeneous systems, it no longer has anything to do with the value of the property for the system as a whole, but instead transforms into a local property, concerning only small volume elements within the system.
Is this correct, or is there some brushing up needed? As of now I am only concerned with the most general understanding of intensive properties. Hope you could shed some light on this. Thank you.

I think of intensive properties as not depending on the amount of material that is present, and consistent with your characterization for a heterogeneous system, except applying to both homogeneous and heterogeneous systems.  Examples are temperature, pressure, specific volume, specific internal energy, specific entropy, etc.

I would not say that "intensive properties" are defined differently for single-phase and multi-phase materials; rather, I would say that an intensive property is, per definition, a physical property of a single phase that does not depend on the amount of that phase. If your material is multi-phase, then at the phase boundaries the intensive properties suddenly change.

@Anders MB You said what I was trying to say so much better than I did.  But I will add that temperature and pressure don't typically change at phase boundaries, except for a curved boundary at which surface tension is important.  There, the pressure does experience a discontinuity.


text:
Why does latex tubing become white and brittle over time, especially where it's bent?
Here's a picture of some latex tubing we have in the lab, bought many years ago and stored in a drawer.

What's responsible for the white corners near places where the tubing is bent? Is this dehydration, oxidation, an aging effect accelerated by stress, or something else?

Natural rubber instantly crystallises (reversibly!) under mechanical deformation. That is actually very desireable, because it gets tougher under load! High quality tyres (for trucks, not regular car tyres) today still contain quite a large amount of the world's natural rubber production for that reason.
You can try to pull at a lenght of the rubber tube, it should become pale, and go back to brownish-opaque when you release the pull.
The drawback is this also makes the material a bit brittle, if deformed beyond a certain limit. I believe this is what you see with your rubber tubing. If you unbend the kinks, the white colour should fade, but the tube is probably damaged at this point.