text:
Interpreting straight lines in a graph of isotherms of carbon dioxide
I am having difficulty in understanding pressure-volume graphs of $\ce{CO2}$ at different temperatures which describes liquefaction of $\ce{CO2}$.

I know that at the critical temperature of $\pu{31.1^\circ C}$ it is possible to liquefy the gas at the critical pressure of $\pu{73atm}$.
But I am having difficulty in understanding some statements about the graph.

(i)At temperatures below $\pu{30.98^\circ C}$ the behaviour of $\ce{CO2}$ towards compression is quite different.For example at $\pu{21.5^\circ C}$,  $\ce{CO2}$ remains a gas up to point $\mathrm B$. At point $\mathrm B$ a liquid of certain volume appears.Further compression do not change pressure but volume decreases till point $\mathrm C$ is reached. In fact, the gas and liquid $\ce{CO2}$ coexist along the horizontal line BC.The decreases of volume represents condensation of more and more $\ce{CO2}$ gas till the point C is reached.

I know that if we are compressing a gas  its pressure should increase but why its pressure is constant along the line $\mathrm{BC}$?

(ii)At point C all the gases has been condensed and further application of pressure simply compresses the liquid .The line $\mathrm{CD}$ represents the compressibility of liquid Carbon Dioxide.

I have studied in my earlier classes that liquids cannot be compressed but how can the liquid $\ce{CO2}$ has a tendency of compression here?

(1) As we compress the gas along the line BC, some of it turns to liquid, and we're left with a smaller amount of gas which exerts the same pressure. (2) True, liquids cannot be compressed... unless you press really hard.

Why should it exert same pressure as smaller amount of gas molecules means less collisions with the walls of container so it should have lower  pressure.

You said if we are compressing a gas its pressure should increase. Why?

The blue area is a mixture of gas and liquid,  and Ivan Neretin has explained what happens here. The area roughly V3 to H to E to V3 is liquid, the rest gas. The graph is poorly drawn and the line C-D, and similar, ones should be almost exactly vertical because as you mention liquids are effectively incompressible at the sort of pressures used here.

@Ivan Neretin: From Boyle's law we know that pressure is inversely proportional to volume.

@ porphyrin :This is the exact graph I have seen in my text book(it maybe incorrect as you are saying).

The graph is good enough for our purposes. (Also, it is repeated in pretty much every book on the subject.) Now to the point. The pressure decreases because there are fewer molecules, and at the same time increases because old man Boyle said so. As a result, it stays the same.

Yes the graph is not that accurate, this is often the case in textbooks. Both Atkins and Engel & Reid give more accurate graphs with the line CD shown effectively as a vertical line.

I will expand on the key points made in the comments. There are two issues that have been brought up:

Pressure is constant along line segment BC.

This is most easily intuited from a direct mechanical example. Consider a cylinder in which one face is a piston at vapor-liquid coexistence. Line segment BC then represents isothermal compression or expansion of the piston. As we push on the piston, the system volume decreases, and vapor condenses into liquid. As we pull on the piston, the system volume increases, and liquid evaporates into vapor.
That these processes occur just enough to maintain line segment BC as an isobar can be understood more quantitatively from thermodynamics. Phase coexistence actually places significant constraints on the system, as is quantified by Gibbs' phase rule---a system of a single chemical species at two-phase coexistence is specified by a single parameter, here the temperature $T$, because we are considering an isotherm. This implies that the pressure is known given $T$; that is, we have a relationship $P = P(T)$, and of course this is just the liquid-vapor coexistence line on a phase diagram. This pressure is, in particular, independent of the number of moles of substance in each phase, and so we must have the same pressure at phase coexistence; i.e., over the line segment BC.

Liquids cannot be compressed. (False!)

Liquids cannot be compressed easily. They are effectively incompressible for small changes in pressure (hence the steepness of the curve WX), but can certainly be compressed when subject to large pressures.


text:
Anti-adhesive agent for soot particles in the exhuast gas
I would like to know if there is/are any anti-adhesive agent for the soot particles (soot particles can be considered as the group of carbon particles) i.e., the agent which will prevent two carbon particles sticking togther and leading to agglomeration.
Thank you.

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text:
The molecular orbitals of lithium hydride
So my understanding of molecular orbitals is as follows: the molecular orbitals of $\ce{LiH}$ can be thought of as being formed from the interaction/overlap of two atomic orbitals, one from lithium and one from hydrogen in this case. I also understand that the atomic orbitals have to be quite close in energy for any significant interaction to take place.
The occupied atomic orbitals of lithium are 1s and 2s, with energies of around -66 eV and $\pu{-5.3eV}$ respectively. The occupied atomic orbital of hydrogen is 1s, with an energy of -13.6 eV. I believe that it's safe to assume that the 1s orbital of lithium is far too low in energy to interact with the 1s orbital from hydrogen, meaning that only the interactions are between the 2s orbital of lithium and the 1s orbital of hydrogen, producing a bonding MO and an anti-bonding MO.
Wikipedia states that "Bonding MOs are lower in energy than the atomic orbitals that combine to produce them." In other words, the 2$\sigma$ bonding MO should have an energy less than -13.6 eV, which is the energy of the 1s hydrogen orbital. However, computer  calculations of the MOs of LiH show that this bonding MO has an energy of -8eV. which would seem to contradict the statement from Wikipedia.
So assuming there is no fault in my reasoning so far, we can safely say that this Wikipedia statement is only generally true, and false in this particular case. So why is this?
I was trying to rationalise this as follows: generally, the reason why bonding MOs are lower in energy than the atomic orbitals used to create them is because bonding MOs concentrate electron density between the two nuclei, resulting in the electrons being attracted by both nuclei (as opposed to just one), leading to the MO being lower in energy than the original AOs. However, in LiH, due to the energy of the 1s AO on hydrogen being significantly lower than that of the 2s AO on lithium, the 1s AO will be the major contributor to the bonding MO, meaning that the electron density in the bonding MO will be significantly closer to hydrogen's nucleus than lithium's nucleus.
Now, originally, there is one electron each in the 1s hydrogen AO and 2s lithium AO, and when they come together, two things change: Firstly, there is electron-electron repulsion between these two electrons. Secondly, the electrons are now bound by two nuclei rather than 1. However, due to the high polarity of the bond, this second change is minimal, such that the first change (introduction of repulsion) actually outweighs the second change, meaning that the resulting bonding MO is actually higher in energy than the hydrogen 1s AO.
Is my rationalization of this accurate/correct? If not, please suggest alternative explanations/ improvements to my thought process because I'm frankly not fully convinced by myself.

That is not how orbitals energies should be used. I wish I had the link to the paper, but @orthocresol might still have it since he originally linked it to me.

@HunnyBunch Please note edits and consider revising. I do not have time to edit entire post.

@Zhe, are you talking about this one? https://pubs.acs.org/doi/abs/10.1021/ed200673w

@orthocresol Exactly!

We may want to consider also the energy change in the lithium 1s orbital.  Formation of the molecular orbitals also changes the energy level of the core orbitals even though these do not participate appreciably in the bonding.
This is because what I call "reverse shielding" occurs.  In a lithium atom in the ground state, there are of course two 1s electrons and one 2s electron.  We commonly suppose that the 1s electrons, packed close to the nucleus, partially shield the 2s electron from the nuclear charge and raise the energy level of the latter.  But, because the probability distributions of the 1s and 2s clouds overlap, there is a small probability that the 2s electron is instead shielding one or both of the 1s electrons.  The effect on the 1s wavefunctions is small, due to the small probability, but the effect on lithium 1s energy can still register in electron volts because the 1s wavefunction is crowded into a region of strong Coulonb potential.
Then, when the hydrogen atom is added and the bonding orbital forms, what had been the 2s electrons are drawn away from the lithium atom and thus the reverse shielding of the lithium 1s electrons is diminished.  Therefore, the lithium 1s orbital is lowered in energy, possibly enabling a stable molecule even if the bonding molecular orbital is higher than the isolated hydrogen 1s orbital as the OP suggests.

generally, the reason why bonding MOs are lower in energy than the atomic orbitals used to create them is because bonding MOs concentrate electron density between the two nuclei, resulting in the electrons being attracted by both nuclei (as opposed to just one), leading to the MO being lower in energy than the original AOs

That's not how bonding works. Actually having the two electrons "crunched together" in the middle is not that great at first because although they are now being attracted by both nuclei, they also repel one another and because the average distance between the electrons is smaller than their distance to the nuclei that repulsion is very noticeable. Therefore the overall potential energy of the electrons actually increases.
The gain in energy is due to the fact that the kinetic energy is being lowered due to the electrons now having "more space to move in" (if you refer to the particle in a box you'll see where that is coming from). A more concrete explanation is that the overall wave-function describing the two electrons has less curvature and as the curvature of a wave-function is directly connected to its kinetic energy (see Schrödinger-equation) this results in an overall lower kinetic energy.
Now that this is out of the way let me comment on your original question:

Bonding MOs are lower in energy than the atomic orbitals that combine to produce them.

That is true as long as you are viewing orbitals with roughly the same energy level (covalent edge case). If you are viewing two orbitals with a high energy difference (ionic edge case) you'll get a binding MO that is roughly at the level of the lower AOs. Therefore you could say that the electrons are getting transferred from the higher AO into the lower AO creating two ions that now bind together in a crystal-structure.
However that does not fit to the values you have provided either, so I would ask you to reveal where you got them from and unless someone has a better explanation I'd guess that one of those values is wrong.
I'm sorry for writing an answer that doesn't really answer your question but it was too much for a comment but (in my opinion) still something worth pointing out...

I think it qualifies as an answer. I don't think it can really be "answered" completely as it stands, the numbers seem off and this is quite possibly because they come from different sources, so it's fair to ask for where they come from.

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text:
Whether for all zero order reactions rate is independent of concentration of reactants
For example,for the reaction $\ce{A + B \longrightarrow C}$,If rate law expression is:
$$\mathrm{Rate=k \times [A]^{(0.5)}\times [B]^{(-0.5)}}$$
It is a zero order reaction but whether it is independent of concentrations of reactants?

It is not zero order; moreover, reaction order in general is not a thing at all. It is order 0.5 in A, and order -0.5 in B, period.

But overall order is sum of exponents of concentrations and here it is 0

@kanishkansenthil Yes, and the product of exponents is -0.25, but that doesn't mean it's a number that means anything useful.

There is no such thing as overall order.

Rate=[A]^x[B]^y.     [A] and [B] express the concentration of the species A and B (usually in moles per liter (molarity, M)). The exponents x and y are the partial orders of reaction for A and B and the overall reaction order is the sum of the exponents. {From Wikipedia}

That's why one shouldn't blindly rely on Wikipedia.

Even in my school book it is written that overall order is sum of exponents.but please clarify me with this confusion and the question that I have asked initially .please..

I think the issue is not that you can't define overall order, but that it isn't a meaningful quantity. This gets at the point that Zhe was making that while we can define the total order of a reaction as the sum of the exponents, it doesn't tell us anything about how the reaction occurs.
Your question gives a very good example of why it isn't a useful quantity: you might expect a zeroth order reaction to have no dependence on the concentration of reactants, but your reaction clearly depends on the concentration of $\ce{A} \text{ and } \ce{B}$, so the total reaction order isn't actually telling us anything about the kinetics of the reaction.
If the order of a particular reactant is zero however, you can say that its concentration won't affect the rate of reaction (within certain limits, as no reaction can be entirely independent of the concentration of an involved reactant).

The overall order of reaction helps to determine to some extent how to proceed with the mechanism of a particular reaction. For example, in the question you asked, the overall order zero for sure implies that the reaction is complex, it can never be in a single step. Also suppose there is a particular rate with overall order 3, it signifies that the reaction can be both single as well as multi step.


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How does a molecule find the lowest energy state?
I understand that in a crystal lattice the molecule will tend towards the structure that is the lowest in energy (fcc, hcp, ccp, etc.). How does the molecule find these lowest energy states? Is it possible for a molecule to be caught in a local minima?

Yes, local minima are totally possible. It happens all the time. Ever seen a diamond? That's an example.

@IvanNeretin Could you explain why a diamond is an example of a local minima? Wouldn't all of the carbons want to be as far away as possible, forming a cubic lattice where each carbon has only major interations with its 6 nearest neighbors?

(It is 4, not 6 nearest neighbors, but that's irrelevant.) Because the global minimum is graphite, we know that for sure.

Isn't it 6 nearest neighbors in diamond? Because you have the four around the sides and then the two above and below? So you're saying that a pure carbon molecule is metastable at diamond and stable at graphite?

No, it's 4, and they are not around the sides. A complete explanation would deserve a separate question on your part; then again, it has been asked and answered multiple times before. Carbon doesn't form molecules (except fullerenes), it's a covalent crystal. Yes, diamond is metastable at normal conditions and graphite is stable.

The answer is it doesn't. The molecule transfers from state to state, whenever it can get over the excitation energy barrier between them.
Once it lands in the lowest energy state (or a relatively low one), it just has a hard time getting out again. And if the lowest energy state is on the surface of a crystal, the hole gets deeper and the walls get higher the more other molecules (or atoms) arrange themselves around it.
If the arrangement (the crystal lattice) only represents a local minimum, that's likely still quite (=meta)stable. Actually those metastable states tend to be created with a high probablility, see https://en.wikipedia.org/wiki/Polymorphism_(materials_science)#Ostwald's_rule


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How does Boudouard reaction influence the temperature?
I've a question that is giving me a lot of troubles:
I'm depositing and gasifying Carbon by varying a flow rate of $\ce{CO}$ and $\ce{CO2}$ and by using the Boudouard reaction ($\ce{2CO -> C + CO}$). When I'm depositing carbon, I notice from my Temperature sensor at the outlet of the test-house, an increase of temperature (and I associate this both to the exothermic nature of the reaction but also to the fact that part of $\ce{CO}$ that didn't react is then burned into $\ce{CO2}$ in a furnace that surrounds my test-house. The problem is that I have a Temperature sensor also at the inlet and this shows a decrease of T when I deposit carbon and an increase when I am gasifying it (which is the contrary of what is expected). Do you know any possible reason for this strange behavior?

drawing of this setup? I dont't get it.


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Why are moles used in the ideal gas law and not simply mass?
The crux of my question is, why is just the amount of particles included in $PV=nRT$ and there is nothing to account for the mass of the individual molecules? Or is mass accounted for in temperature, the average kinetic energy which includes mass in $\tfrac{1}{2}mv^2$.
Thanks in advance!

In reference to the title, If you do it, R would depend on the identity of the gas...; it doesnt happen in the ordinary equation (R is universal).

Thanks, that makes sense. any idea on the answer to my second part however

R would depend on the identity of the gas, and n would be the mass PV=mR'T

Why in the world should the formula account for the color of my eyes? It shouldn't. Ditto for the mass of the molecule.

Change the formula to $pV=(m/M)RT$ and you have your mass in it. $m/M=n$ ! Pressure depends on the average impulse of every gas particle.

The overall notion is that gases exhibit "ideal" behavior so that the identity of the particular gas molecule doesn't matter. Thus regardless of what compound the gas molecule is, or if the gas is a mixture of different compounds, a mole of gas molecules at standard temperature and pressure (STP, 0 °C and $1.000\times10^5$ Pa) occupies a volume of 22.7 liters. So the equation for an ideal gas is
$$PV = nRT$$
In addition, there is the Maxwell–Boltzmann distribution which is for the speed of molecules. The gist is that regardless of the compound, gas molecules have the same kinetic energy spread. Since kinetic energy is given by $mv^2/2$, that means that lighter molecules have faster speeds than heavier molecules.
If you try to define the relationship by the molecular weight of the gas, then the value of the gas constant, $R$ would be different for different compounds and/or mixtures. Now in reality the ideal gas law works reasonably well, but there are empirical equations which fit the relationship between P, V, and T better. For those empirical equations the extra coefficients are fitted to the experimental data for particular gas compound under consideration.


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Why is gunpowder urinated on in Blood Meridian?
An interesting section of Blood Meridian (1985) by Cormac McCarthy has the gang of "Indian fighters" surrounded and out of powder, so one of them manages to make it from scratch, making charcoal from wood, saltpeter from bat guano and finding sulfur in a volcanic area, but the final step has them urinating on the mixture and then drying it in the sun. I understand that urine can in fact be a source of nitrates but given that they already had saltpeter, I am puzzled. At the same time, I understand that gunpowder is made wet to form larger grains but this can be done with water (I think), which they had access to.
The gunpowder was first dry so urinating on it was not for safety; it may have been an old-fashioned idea about how to make gunpowder and an unnecessary step or at least water would work as well for "corning" – making granules from it.

Yes, it can be done with water... but in the book, is water scarce?

Forget the grains. Gunpowder is made wet so as not to explode during production.

firstly, what Ivan says, and secondly, salpeter dissolves in water and after drying coats the other particles, which makes for a much higher contact area and greatly increases explosivity. If you just mix the dry stuff, your bullets will limp out of the barrel like dead fish.

and yes, they urinate on it to make it sound more tough. ;-)

@DrMoishePippik: yes, water was available and used in an earlier stage of the process -- there was a stream. They urinated on the final mixture, spread it on rocks to dry and were in a desperate race for it to dry in time which it did. So the urine stage if unnecessary put them all in jeopardy -- clearly The Judge thought urine was needed to make usable gunpowder.

Classic black gunpowder was often produced by a wet mixing process, after which powder was dried and then ground to fine powder or grains.
The reason for it was better homogenisation of the final powder and increased powder efficiency due faster deflagration.
The urine might not be used then for being urine, but for being water based, as they could be short of water supplies, not to waste water for it.

In Literature stackexchange we have this answer which indicates firstly that urine is something that can be used and secondly that saltpeter probably can't be extracted as fast as portrayed in the story (although that may or may not be true). https://literature.stackexchange.com/questions/24314/does-blood-meridian-accurately-describe-gunpowder-manufacture


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Is sodium "also known to turn purple and combust (like lithium)," when exposed to water?
I understand that indicator solutions can change color depending on pH, but the following statements in the Phys.org article Super cheap earth element to advance new battery tech to the industry has me baffled.

Most of today's batteries are made up of rare lithium mined from the mountains of South America. If the world depletes this source, then battery production could stagnate.
Sodium is a very cheap and earth-abundant alternative to using lithium-ion batteries that is also known to turn purple and combust if exposed to water—even just water in the air.

Later it says:

They minimized sodium's exposure to the moisture that would make it combust by making the sodium powder in a glovebox filled with the gas argon. To make the powder, they used an ultrasound—the same tool used for monitoring the development a fetus — to melt sodium chunks into a milky purple liquid. The liquid then cooled into a powder, and was suspended in a hexane solution to evenly disperse the powder particles.

Phys.org news items are generally of very high quality, so I'm wondering if this is just a mistake (perhaps a reference to the use of a pH indicator) or if there is some reaction between sodium, water and perhaps air that involves a purple color.
The "purple" statement is echoed on Purdue University's web page Super cheap earth element to advance new battery tech to the industry as well.
So far I've only found the video Sodium gas... omg it's purple!.
The paper in question is Ultrasound-assisted synthesis of sodium powder as electrode additive to improve cycling performance of sodium-ion batteries. Tang,  Kye and Pol, Journal of Power Sources, 396, pp 476-482, https://doi.org/10.1016/j.jpowsour.2018.06.067

I found the paper in Researchgate.
I'm going to leave this tentative answer in place but hopefully I or someone else will be able to investigate further and post a better answer.
These are particles of molten sodium (>98C) dispersed in mineral oil. The linked references suggest that the ultrasonic dispersal technique can be used to make quantum dots, so if these particles are similar size, then the color could be some quantum mechanical effect for the conduction electrons in the liquid sodium nanoparticles.
Since that is a size-dependent QM effect, one would not see it in experiments on bulk sodium.
This doesn't completely rule out Rayleigh scattering effects, but since the color changes as the particles cool (and presumably solidify) then that suggests it's the properties of the particles themselves, rather than there dispersal in oil.

Sodium Powder Preparation: Sodium powder was prepared via ul- trasonic heating, melting, and subsequent fragmentation of solid sodium chunks in an organic solvent. A similar technique was utilized in previous studies involving Sn nanoparticles 32 and dispersed sodium 33. Ultrasonication was generated from a Sonics VCX500 probe equipped with a stepped microtip. In a typical synthesis, 15 mL of mineral oil was first degassed by ultrasonication for 15min inside a 150 mL cone-shape sonochemical reactions vessel; then, about 100 mg of fresh metallic sodium chunks were added into the oil. A continuous argon flow was fed to the vessel to maintain an air/moisture-free environment. The Na/oil mixture was then exposed to ultrasonic irradiation at 40% amplitude. Pulsed ultrasonic irradiation was applied in 4 cycles of 59s-sonication and 30s-rest intervals. The formation of sodium powder took place in two stages. At the first stage, solid Na chunks were melted into liquid sodium when the mixture temperature rose above the melting temperature of sodium (mp = 98 °C). The temperature rise is a result of ultrasound induced heating to the oil. At the second stage, the molten Na particles were dispersed into mineral oil via ultrasound to form a homogenous mixture with purple coloration. Once cooled, the solution color changed to grey. A simplified synthesis method is illustrated in Fig. 1a (A video of the sodium powder synthesis can be found in Movie S1 in the supporting information). The mixture was then washed with anhydrous hexane and centrifuged three times to obtain a dispersion of Na powder in clean hexane. All washing procedure except the centrifugation were completed inside an Ar-filled glovebox to minimize air exposure.


text:
How can I define a torsion force field for $n$ points in space?
I would like to define a model for torsion force field between $n$ point in space.
I thought to generalize the OPLS model:
$$
\sum_\mathrm{dihedrals} \left( \frac {V_1} {2} \left [ 1 + \cos (\phi-\phi_1) \right ]
                + \frac {V_2} {2} \left [ 1 - \cos (2\phi-\phi_2) \right ]
                + \frac {V_3} {2} \left [ 1 + \cos (3\phi-\phi_3) \right ]
                + \frac {V_4} {2} \left [ 1 - \cos (4\phi-\phi_4) \right ] \right)
$$
How can I make them to lie in the same plane allowing fluctuations of $\pm10^\mathbf{o}$?

It is just a Fourier series; it could have as few or as many terms as you want.

@IvanNeretin, so the idea is to construct the potential I want and then fit it to a Fourier series?

Looks like so.$,$


text:
Mismatch of coefficients in rate law expression
For example, for the reaction $\ce{A + 2B -> C}$, we have
$$\text{rate} = -\frac{1}{2}\frac{\mathrm d[\ce{B}]}{\mathrm dt} \quad \text{and} \quad \text{rate} = k[\ce{A}]^x[\ce{B}]^y.$$ Together, this means that
$$-\frac{1}{2}\frac{\mathrm d[\ce{B}]}{\mathrm dt} = k[\ce{A}]^x[\ce{B}]^y$$
However, in many books it is also written simply that
$$-\frac{\mathrm d[\ce{B}]}{\mathrm dt} = k[\ce{A}]^x[\ce{B}]^y.$$
Which is correct?

Both are.$,!$

@IvanNeretin can you please elaborate? Are we taking $2k$ As $k$ ?

Yeah, sort of. What is $k$? Some number. What is $2k$? Some number. You may use either form, as long as you know which one are you using.

The rate may be written with both ways. But it is much more frequent to use the rate with the coefficient 1/2 in front of $\pu{\frac{d[B]}{dt}}$


text:
Titration of Na2CO3 against HCl

$\pu{25 mL}$ of $\pu{0.125 M}$ $\ce{Na2CO3}$ is titrated with $\pu{0.100 M}$ $\ce{HCl}$. Given that $K_\mathrm{a1} = 4.3\times 10^{-7}$ and $K_\mathrm{a2} = 4.8\times10^{-11}$ for the diprotic acid $\ce{H2CO3}$, calculate the $\mathrm{pH}$ values of the two equivalence points in the titration.

My method was to assume that all $\ce{CO3^2-}$ initially would be neutralized to form $\ce{HCO3-}$, and from there use the $K_\mathrm{a2}$ value to calculate the concentration of $\ce{H3O+}$ via equilibrium, but this gives me the wrong answer (by about $3$ $\mathrm{pH}$ units).
Could someone help me understand what I'm doing wrong, and how I should do it correctly? (The correct answer apparently should be $8.34$ for the 1st equivalence point.)

Use $K_\mathrm{b1}=\frac{K_\mathrm{w}}{K_\mathrm{a1}}$ in addition to $K_\mathrm{a2}$ in order to calculate $[\ce{H3O+}]$

I think the main confusion you're having here is that you misinterpretted the composition of the system at the first end point.
While you are correct that an equivalent non-equilibrium system is just all $\ce{HCO3-}$, it is not correct to think of this as an acid. This is bicarbonate. If you take sodium bicarbonate and dissolve it water, it is basic, not acidic. Therefore, you should have converted your $K_{\mathrm{a1}}$ value into a $K_{\mathrm{b}}$ value and treated the system as a base instead.
The second equivalence point is when you have the equivalent concentration of $\ce{H2CO3}$. Treat this as a normal acid dissociation problem.

At the first equivalence point:
$$\ce{Na_2CO_3 +HCl ->NaHCO_3 + NaCl}$$
$$\text{moles of}~ \ce{Na_2CO_3} =25\times{0.125}\times{10^{-3}}=3.125\times{10^{-3}}\pu{mol}=\text{moles of}~ \ce{HCl}=\text{moles of}~\ce{NaHCO3} $$
$$\text{moles of}~ \ce{HCl} =3.125\times{10^{-3}}=\pu{V_\ce{HCl}\times{0.1}} $$
$$\pu{V_\ce{HCl}}=31.25\times{10^{-3}}$$
$$\text{Total volume}=25\times{10^{-3}}+31.25\times{10^{-3}}=56.25\times{10^{-3}}~\pu{L}$$
$$[\ce{NaHCO3}]=\frac{3.125\times{10^{-3}}\pu{mol}}{56.25\times{10^{-3}}~\pu{L}}=5.6\times{10^{-2}}\pu{M}=\ce[{HCO_3^-}]$$
$$[\ce{H_3O^+}]=\sqrt\frac{K_\mathrm{a1}K_\mathrm{a2}[\ce{NaHCO_3}]+K_\mathrm{a1}K_\mathrm{w}}{K_\mathrm{a1}+[\ce{NaHCO_3}]}$$
$$[\ce{H_3O^+}]=\sqrt\frac{(4\cdot{3}\times{10^{-7}}\times{4\cdot{8}\times{10^{-11}}}\times{0.056})+(4\cdot{3}\times{10^{-7}}\times{10^{-14}})}{(4\cdot{3}\times{10^{-7}}+0.056)}\approx{5.35\times{10^{-9}}}\pu~{M}$$
$$\pu{pH}={8.27}\approx{8.3}$$

At the second equivalence point:
$$\ce{Na_2CO_3 +2HCl ->H_2CO_3 + 2NaCl}$$
$$\text{moles of}~ \ce{Na_2CO_3} =25\times{0.125}\times{10^{-3}}=3.125\times{10^{-3}}\pu{mol}=\text{moles of}~\ce{H_2CO3} $$
$$\text{moles of}~ \ce{HCl} =2\times{3.125\times{10^{-3}}}=\pu{V_\ce{HCl}\times{0.1}} $$
$$\pu{V_\ce{HCl}}=62.25\times{10^{-3}}$$
$$\text{Total volume}=25\times{10^{-3}}+62.25\times{10^{-3}}=87.25\times{10^{-3}}~\pu{L}$$
$$[\ce{H_2CO3}]=\frac{3.125\times{10^{-3}}\pu{mol}}{87.25\times{10^{-3}}~\pu{L}}=3.58\times{10^{-2}}\pu{M}$$
Treat carbonic acid solutions as if $\ce{H2CO3}$ were monoprotic, because the first
acid dissociation constant is much greater than either$ K_\mathrm{a2}$ or $ K_\mathrm{w}$  , so this becomes a standard monoprotic weak acid problem:
$$ K_\mathrm{a1} = 4.3\times{10^{–7}} =\frac{[\ce{H^+}][\ce{HCO_3^− }]}{[\ce{H_2CO_3}]}$$
Assume : $ [\ce{H+}] = [\ce{HCO_3^-} ]$ , the equilibrium expression becomes:
$$K_\mathrm{a1} = 4.3\times{10^{–7}} =\frac{[\ce{H^+}]^2}{0.0358-[\ce{H+}] }$$
As: $0.0358 >> K_\mathrm{a1}$,justifies the further approximation of dropping the $ [\ce{H+}]$ term in the denominator.
$$ K_\mathrm{a1} =4.3\times{10^{-7}} =\frac{[\ce{H^+}]^2}{0.0358}$$
$$[\ce{H^+}]=1.24\times{10^{-4}}\pu{M}$$
$$\pu{pH}=3.92$$

$pK_{a1} = -\log{(4.3\times10^{-7})} = 6.3665$
$pK_{a2} = -\log{(4.8\times10^{-11})} = 10.3188$
eq. pt. $ = \dfrac{6.3665+10.3188}{2} = 8.34265 \ce{->[Rounding]} 8.34$

but why does this work? also this cant be extended to calculate the second equivalence point

At eq. pt $\ce{[H2CO3] = [CO3^{2-}]}$. // What would you average with 6.3665 to get second eq. pt?!?

This relationship is also reasonably easy to derive using the corresponding Henderson-Hasselbalch equations, if you've learned though already.

At second eq. pt. (for $K_{a1}$) $\ce{[H+] = [HCO3-]}$. I'll leave the math to you.

$$\pu{pH}=\frac{1}{2}\times{(pK_\mathrm{a1}+pK_\mathrm{a2} )}$$
THis foumula suitable when $K_\mathrm{a1}<<[\ce{HCO_3^-}]\text{with concentrated solution,but neglegt water autoionization and assume} [\ce{H2CO3}] = [\ce{CO3-} ]$


text:
What is the mechanism of alkane combustion?
I understand that combustion of hydrocarbons (such as alkanes) produces energy by breaking bonds:
$$\ce{C_nH_{2n} + $\frac{3n}{2}$ O2 -> nCO2 + n H2O}$$
And I can find dozens of explanations of how combustion works in an engine.
What does the chemical reaction (not equation) of hydrocarbon polymer combustion actually look like?
In other words, how is a hydrocarbon polymer actually broken down to form energy?

What process actually breaks up the polymer into its respective monomers (methane?) for combustion to take place?

Please explain. I would also love a visual (like this or this ) if someone could find one!

Related: Is there a reaction type/name for the splitting of octane into its parts prior to combustion, or does this process all occur at once in an engine?

@theforestecologist combustion reaction mechanisms can be quite complex. Here is a paper about a simplified model of propane combustion: https://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/37_2_SAN%20FRANCISCO_04-92_0960.pdf

All common combustions at not too high temperatures if I remember correctly are radical chain reactions. Those should be known from your high school chemistry class, or you should have caught up in college. But, there is wikipedia https://en.wikipedia.org/wiki/Combustion#Reaction_mechanism to help you out. ;-)

@Karl radical chain reactions may very well have been taught in my HS/college chemistry classes, but it's been a few too many years to remember. FYI I'm not a chemist, I'm a biology professor -- so excuse my unspecific language as I'm barking up a tree I [clearly] remember less than optimal about. [btw, I would suspect you would not be treated so rudely as a chemist asking Q's over at Bio.SE]

I'm sure I wouldn't, because I still know my high school biology. Mostly. ;-) No, sorry. Its just that the answer here would require a solid chemical base knowledge to understand. But, your question makes it clear you don't have that available. Please forgive if that leads to some exasperation. Look it up, there's a good chance you have learned about this at university and will remember.

@Karl, I actually went back to my college chem textbook, and after a reasonably thorough search found no evidence of radical chain reactions or the discussion of combustion of hydrocarbons in any great detail. So it's not that I don't remmeber, it's that It was not included in my course. The only mention of chain reactions (1x) was in reference to atmospheric ozone, which I know well. So now I have an idea of what you mean by chain reactions. Lack of knowledge in this specific topic is not indicative of my base chemical knowledge anyay. I simplified the question because of a previous comment.

It is not a polymer, because there are no reasonable constituent "monomers" which you could put together to form these long alkanes, unlike e.g. a protein where you can join amino acids via condensation reactions. They are just standalone alkanes. I edited to reflect this and to put a more descriptive equation. Anyway, there is a reason why the mechanism isn't typically covered (even at undergraduate level): it's very messy. There are radical chain reactions which are easy to understand, but this is not one of them. [Now, that doesn't mean you can't get an answer - it is just a warning.]

@orthocresol thanks for the edit, clarification ,and heads up. However, I do want to point out that some sources (e.g., encyclopedia Britannica) do refer to alkenes as monomers and their chains as polymers. Given the repeatable nature of these chains, you could see where my confusion arose... Still, I'm curious how these chains are broken (and what we call that reaction) prior to combustion of the constituent carbons.

Higher alkanes, neglecting their end groups,  are not per se distinguishable from polyethylene. @orthocresol The messiness is what makes these reactions so fascinating. At least for people who are into unorderly (=polymer) chemistry. ;-) Btw. thanks for fixing up my answer.

The following, very general radical reactions apply to any common combustion reaction of organic material. Details vary largely.

Step 1

An energy spark of whatever kind turns triplet oxygen into its high energy singlet state$^*$, which typically strips a hydrogen atom off of some organic molecule
$\ce{R-CH3 + O2^* -> R-CH2. + HOO.}$
Now you have two radicals.

Step 2

The perhydroxy radical strips off another hydrogen somewhere,
$\ce{R-CH3 + .OOH -> R-CH2. + HOOH}$
while the alkyl radicals react with more oxygen (self-catalysed)
$\ce{R-CH2. + O2 -> R-CH2-OO.}$
$\ce{R-CH3 + .OO-CH2-R -> R-CH2. + HOO-CH2-R}$
, this incorporates $\ce{-OOH}$ hydroperoxide motives into more and more molecules. This is in itself a linear process, with a constant number of radicals. Similar to free-radical polymerisation or organic radical-substitution reactions, and very much the same as formation of the infamous ether hydroperoxides.

Step 3

These peroxides are not stable at elevated temperature, and their O-O bond tends to split homolytically
$\ce{R-OO-H -> R-O. + .OH}$
Now the number of radicals grows rapidly, the combustion is gaining speed.

Step 4

As temperatures rise (see step 5!), esp. larger organic molecules start to break up also by themselves
$\ce{R-CH2-CH2-R -> R-CH2. + .H2C-R}$
(That's also a possible initiation step.) You get more radicals, which can react with more oxygen to form peroxides. Now the chain reaction snowballs!

Step 5

$\ce{HOOH -> HO. + .OH}$ gives more hydroxy radicals (also from step 3), which turn into water
$\ce{HO. + H3C-R -> H2O + .H2C-R}$
leaving another radical behind. Producing water is highly exothermic, this fuels the reactions creating radicals (in steps 3+4+5). Making radicals costs a lot of energy.
Steps 2-5 (it's really not a sequence) run in parallel and repeat as they produce more radicals and peroxides all the time. The reaction is usually limited by the inflow of fresh oxygen.

Step 6

Termination: Two carbon radicals can recombine, but at high temperature this does not really happen, because the energy from the recombination has to go somewhere. When temperatures are too low, this makes coke and smoke.
This picture is by no means complete. $\ce{CO2}$ would likely occur via decarboxylisation of $\ce{R-COO.}$ radicals, etc.  Organic radicals at high temperatures have more internal reactions where they split off smaller molecules, e.g. $\ce{R-CH2-CH2. -> R. + H2C=CH2}$. (The latter can be seen as depolymerisation of polyethylene.) A radical in the middle of a chain splits it in half ($\ce{R-CH2-CH2.-R -> R. + H2C=CHR}$, increasing the volatility.
And remember that this picture is terribly simplified. I'm by no means an expert on this subject, and the experts also have not agreed on all details yet, especially since it varies a lot.

My selection of "steps" is not very convincing. Feel free to arrange the reactions in a different manner. The gist is they are all happening in parallel and producing more educts for each other.

This question https://chemistry.stackexchange.com/questions/14704/reaction-mechanism-of-combustion-of-hydrogen?answertab=active#tab-top has a nice answer on hydrogen combustion, which adds a few liteature reads.

No alkanes have the formula $\ce{C_nH_{2n}}$ as given in the question. Alkanes may be $\ce{C_nH_{2n+2}}$, but not $\ce{C_nH_{2n}}$, which is more probably an alkene, or a cycloalkane. The theory developed by Karl is OK, but it used $\pu{R}$ to avoid this difficulty, where $\pu{R}$ is all the rest of the alkane molecule when $\ce{CH3}$ has been removed


text:
Buffer made of salts of polyprotic acid: reactions
I was trying to make a buffer with $\ce{K2HPO4}$ to use it in a bacterial culture.
If I use $\ce{K2HPO4}$ I think I have to add $\ce{KH2PO4}$. I am trying to think it writing the equations of the process but I have troubles.
I think the buffer's reactions will be:
Salt dissociation in water:
$$\ce{KH2PO4_{(aq)} <=> K+_{(aq)} + {H_2PO4^-}_{(aq)}} $$
$$\ce{K2HPO4_{(aq)} <=> 2K+_{(aq)} + {HPO4^{-2}}_{(aq)}}$$
Anion hydrolysis:
$$\ce{HPO4^2-_{(aq)} + H2O_{(l)} <=> H2PO4^-_{(aq)} + OH-_{(aq)}}$$
$$ \ce{H2PO4-_{(aq)} + H2O_{(l)} <=> HPO4^2-_{(aq)} + H3O+_{(aq)}}$$
If I add acid:
$$\ce{K2HPO4_{(aq)} + H3O+_{(aq)} <=> H2O_{(l)} + H2PO4-_{(aq)}} $$
If I add base:
$$\ce{H2PO4-_{(aq)} + OH-_{(aq)} <=> H2O_{(l)} + HPO4^2-_{(aq)}}$$
Is it correct? And if so, is it possible to add only $\ce{K2HPO4}$ and water to make the buffer?

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu{}$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

For use in a bacterial culture, you no doubt want a buffer solution with a $\mathrm{pH = 7.20}$, which corresponds to the $pK_{a2}$ for phosphoric acid. At that pH $\ce{[H2PO4-] = [HPO4^2-]}$. The $\ce{K+}$ ion is basically a spectator ion and doesn't influence the pH directly. So typically you'd start with a solution which has equal molarities of $\ce{KH2PO4}$ and $\ce{K2HPO4}$ and then adjust the pH with $\ce{H3PO4}$ and $\ce{KOH}$ using a pH meter to get $7.20$ "exactly."

Ok. Thanks. I have understood. So... Are the reactions I wrote fine, @MaxW ? (Because I have to write it, too). @MaxW

Well, you went a little overboard. All you need is $$\ce{K+H2PO4- <=> K+ + HPO4^2- + H+}$$


text:
Is cyclohexanol or hexanol more acidic?
If we had a hexane ring with an $\ce{OH}$ substituent vs. a hexane chain with an $\ce{OH}$ substituent, which of the two, if any, would result in a weaker conjugate base?
My confusion: The ring appears to be a bulky shape, but the chain would have one more hydrogen in it...
What do you think and do you have a source reference? There isn't anything in the books I have on structures and their influence on acidity.

It's not a "forum" but Q&A site (yes there's a difference ;) As far as your question is concerned, you need to specify which hexanol isomer you want to consider. Another thing is does it matter much which will be more acidic? I think that difference should be negligible for most purposes.

It's just for something silly like ranking acidity. Generally, in the text books i'm reading, there is not a way to distinguish from two very similar alkane molecules. I was more curious if there was a way I wasn't exposed to. (Probably no practicality, i'm just being picky?)
I guess structure is more important, but I was hoping to find a generic "rule of thumb" similar to the "ARIO" or "CARDIO" pneumonic. Since Charge, Resonance, Induction, and Orbitals are all similar, the only thing I could think that would affect acidity would be its geometry?... In a normal case I would just check pka.

Well, simple picking the isomer and editing your question would do more good. BTW pKa values of such weak acids are quite uncertain.

My guess was that the electron donating character of the neighbouring methyl groups will play a deciding role. The more electrondonating group are attached to the hydroxyl carbon, the less stable the resulting alkoxide anion (the oxygen would become more negatively charged), and the less acidic.
So then a secondary alcohol would be less acidic than a primary alcohol, and cyclohexanol would be less acidic than n-hexanol. In comparison with e.g. hexane-3-ol the difference will be very subtle. One explanation I found on quora (https://www.quora.com/What-is-the-order-of-acidity-among-primary-secondary-and-tertiary-alcohols) points out the importance of steric factors. Indeed, the negative charge will be further stabilized by surrounding water molecules. The potential steric hindrance caused by neighbouring alkylgroups would appear to be less for the cyclohexanol anion than for hexane-3-ol anion which should lead to greater stabilization of the cyclohexanol anion. The latter would probably be slightly more acidic then. Keep in mind that this is all handwaving to some extent, definitely since these differences are so subtle and small unforeseen effects can easily change the ranking of acidity.


text:
Why do gases coexist as liquid and vapour?
I have having difficulty in understanding the liquefaction of gases.In the graph
of isotherms of $\ce{CO2}$ I cannot understand how can gas coexist as liquid and  vapour under area of $\mathrm{XCEBY}$.

I have read the following statements in my textbook:

At $\pu{21.5 ^\circ C}$, $\ce{CO2}$ remains a gas up to point $\mathrm B$. At point B a liquid of specific volume appears.In fact,the gas and liquid  $\ce{CO2}$ coexist along the horizontal line $\mathrm{BC}$. The decrease of volume represents condensation of more and more $\ce{CO2}$ gas till point $\mathrm C$ is reached.

A careful study of isotherms reveal that it is possible to change gas to liquid by following a path in which a single phase is present throughout the process.For example,move vertically from $\mathrm{A}$ to $\mathrm{F}$ by increasing the temperature of the gas.Now move from point $\mathrm{F}$ to point $\mathrm{G}$ by increasing pressure at constant temperature.Move from point $\mathrm{G}$ to point $\mathrm{H}$ by lowering the temperature.As soon as we cross point $\mathrm H$ we get a liquid.

(i)Why $\ce{CO2}$ coexist as liquid and vapour in the area of $\mathrm{XCEBY}$?(instead of converting directly into liquid by condensation of all the molecules simultaneously)
(ii)Why under some range of pressure and temperature $\ce{CO2}$ converts directly into liquid without going through the two-phase region?

In order to change from vapor to liquid at constant temperature and pressure, you need to remove heat.  The amount of liquid formed depends on the amount of heat removed.  If you only remove a tiny amount of heat, you can't suddenly change all the vapor to liquid.

@Chester Miller :I am talking about liquefaction when pressure is not constant.

But the pressure is constant (p2) along the 21.5 C isotherm in the 2 phase region.

https://chemistry.stackexchange.com/questions/101900/interpreting-straight-lines-in-a-graph-of-isotherms-of-carbon-dioxide

I have a feeling, you're getting too far, not knowing most elementary things. 1) If you put some water into pressure cooker, it won't suddenly boil off. 2) There's no boiling at all in this region, because you have supercritical fluid.

@Mithoron Do you really think water in a pressure cooker is supercritical (i.e., above the critical temperature)?

@ChesterMiller I put those numbers for a reason - addressing two OP's questions.

@ChesterMiller So why do we call it as liquid- vapour equillibirum and not simply show as fraction of gas liquiefied?

Actually, many plots such as the one you have shown also include contour lines showing the fraction of gas liquified.

Vapor-liquid coexistence.

This is possible because changing the system's molar volume $V_\mathrm m$ does not affect the Gibbs free energy of either phase, so equality of the Gibbs free energy (that is, phase coexistence,) can be maintained for a range of $V_\mathrm m$; this is the coexistence region $\mathrm{XCEBY}$.
On the other hand, changing $p$ and $T$ will affect the Gibbs free energy, and if instead we plot this phase diagram on the $p,T$-plane, the entire coexistence region $\mathrm{XCEBY}$ will reduce to a single curve, the liquid-vapor coexistence curve.
In other words: Fix $T$. Only one value of $p$ results in phase coexistence; phase coexistence is a curve on the $p,T$-plane. A range of $V_\mathrm m$ results in phase coexistence; phase coexistence is a region on the $p,V_\mathrm m$-plane.

Phase transitions without coexistence.

This is made possible because the liquid and vapor phases share continuous translational symmetry. In contrast, a phase transition from the liquid phase to a solid, crystalline phase requires breaking this continuous symmetry, resulting instead in the discrete translational symmetry possessed by a crystalline lattice. This shared symmetry permits the possibility of continuously going from one phase to the other without an explicit phase transition.
This possibility is realized past the critical temperature, the temperature represented by the critical point $\mathrm E$. Past the critical isotherm identified by line segment $\mathrm{HE}$ is the supercritical fluid "phase", which shares properties of both liquid and vapor. Isotherms like $\mathrm{GF}$ do not represent a liquid-vapor transition, because the species remains as a supercritical fluid throughout. A liquid-vapor transition that bypasses the coexistence region, like $\mathrm{AFGHD}$, is a continuous transformation from liquid to supercritical fluid to vapor, rather than a discrete phase transition.

It seems beyond my scope of understanding from what I have studied upto now,but thanks for the answer.

@pranjalverma, I'm happy to help, and might be able to answer any questions you have regarding my answer. In general, however, I would recommend that you wait longer before accepting an answer, especially if it's not easy for you to understand, because someone else might provide a more accessible answer.

Your answer was indeed useful as I will understand all the stuff after reading carefully about thermodynamics.


text:
If a single electron fills a degenerate energy level, does it fill all degenerate orbitals at once?
For example, if a single electron fills the 2p orbitals. Since all orbitals are degenerate, there is no reason why there would be any preference for px, py, pz. Is the electron hence in a superposition of these 3 orbitals?
And what if we consider hydrogen for n=2, where 2s, 2px, 2py and 2pz are degenerate?

It is worse than that. While the orbitals are degenerate, you have no way of knowing which of them should be called $p_x$, etc.

@Ivan, yeah that makes sense, but is it then true that the electron probability is symmetric around the core, i.e. it exists in a superposition of all possible degenerate orbitals?

Yes, it is symmetric all right.

If one would like to describe this mathematically, choosing a configuration like $2s^2 p_x^1$ will give a wrong (to high) energy. To fix this you have to mix in the other 2 configurations ($2s^2 p_y^1$ and $2s^2 p_z^1$). See https://chemistry.stackexchange.com/questions/82656/what-constraints-are-imposed-on-a-wavefunction-by-the-symmetry-of-the-system/82682#82682 for details.

Thanks! And another follow-up question: what happens if a second electron fills the singly occupied p-shell? Even if they cannot be placed in the same orbital there would still be degeneracy right? How does that work then?

You would mix three different states as well ($\ce{2s^2p^1_xp^1_y}$, $\ce{2s^2p^1_yp^1_z}$ and $\ce{2s^2p^1_zp^1_x}$).


text:
Calculate Rate constant, k, for a consecutive reaction where k1<<k2
I have a consecutive reaction where the first step is the rate determining step so $k_1<<k_2$:
$$\ce{A ->[$k_1$] B ->[$k_2$] C}$$
The rate law can be shown to be:
$$[\ce{C}] = {1 - \exp(-k_1t)} [\ce{A}]_0\tag{eq. 1}$$
My reaction is pseudo first order since the first step is a hydrolysis. If I take the natural log of eq. 1 will I be able to convert it into the following form?:
$$\ln[\ce{C}] = -k_1t + \ln[\ce{A}]_0\tag{eq. 2}$$
... because If I could get it to this form, I will be able to plot $\ln[\ce{C}]$ vs $t$ to get $k$ from the slope.
If I have made a mistake with the natural logarithm (or if it is not possible to convert it to the form of eq. 2), what function do I need to use to find $k$?
My question is specifically: What do I need to plot to determine my rate constant considering I have done a time course NMR experiment?

Your eq. 1 is wrong since 1 is dimensionless an [C] has the dimension of a concentration. Also keep in mind that [A] + [B] + [C] is constant over time and [B] can be neglected.

In a consecutive reaction A-B-C, A decays as $A=A_0\exp(-k_1t)$ with $A_0$ the initial amount (initial B and C are assumed to be zero) and B rate expression $\displaystyle \frac{dB}{dt}=k_1A-k_2B$  which when substituting for $A$ and integrating gives $\displaystyle B=A_0\frac{k_1}{k_2-k_1}\left(\exp(-k_1t) -\exp(-k_2t  \right)$. C is found as $A_0-A-B$
then $\displaystyle C= A_0\left(1+\frac{k_2\exp(-k_1t)-k_1\exp(-k_2t)}{k_1-k_2}  \right) $
In your case $k_1<<k_2$ then $\displaystyle C= A_0\left( 1 - \exp(-k_1t)  \right) $ which is almost your equation.
Taking logs gives $\displaystyle\ln\left(\frac{C}{A_0}\right)=\ln\left(1-\exp(-k_1t)\right)$ and this is what you should use.
[To go further you will have to expand the series for log and exponential but this will lead to a polynomial and will only be an approximation anyway.  The first few terms are $\log(k_1t)  - k_1t/2 + (k_1t)^2/24\cdots$; very messy.]

I still don't understand what I need to plot to determine the observed rate constant for my reaction

You should then plot $ C/A_0$ vs $t$ because $ C/A_0=\left( 1 - \exp(-k_1t)  \right) $. The curve rises then becomes constant. Then fit the curve by guessing values of $k_1$ until you get a match. You should use a computer to do this. By expanding the exponential, at small times the slope is $+k_1$ so this will give you a starting value.

yes, you can plot $\ln[\ce{C}]$ vs $t$ to get $k$ from the slope. // If you have made a mistake with the natural logarithm, it is impossible to predict what happens with the data.


text:
What makes Citrate or Citric Acid an acid?
As far as I know, an acid is something that gives off a proton or hydrogen ion $H^{+}$.
But when I look at Citric Acid,

There are three $COO^{-}$, which I think is a carboxyl group, that lacks $H^{+}$.
They seem to rather demand a hydrogen ion from its environment than donate one, which make it basic and not acidic. But the name is still citrate or citric acid. Can someone explain?

That's not citric acid, that's the citrate anion. Replace those negative charges with hydrogen atons to get citric acid. The citrate anion is a base, and does demand protons from the environment.

my mistake. It is from biology text book and citrate appears in Kreb cycle. That was why I posted this question here. And also somehow the text book says that is citric acid. And that is from Kreb cycle.

It depends on the pH of the solution.  At higher pHs the carboxyl groups will be dissociated (as drawn).  At lower pHs the equilibrium will shift and most of the COO- will become protonated.
In general, carboxylic acids are relatively weak compared to stronger acids that dissociate almost completely when added to an aqueous solution.  For example, acetic acid, or vinegar, will not burn a hole in cotton clothing, whereas sulfuric acid (the kind in an automobile battery) will easily burn a hole in cotton cloth.
Also, don’t confuse carboxylic acids with their carboxylate salts.  Sodium citrate, where the three protons are replaced with Na+, will act as a base when added to water, and raise the pH.

The crux of the matter is that in biochemistry the names of conjugate acid-base pairs (in this case, citric acid and citrate) are used interchangeably. Thus, the picture you have shows citrate (which, as you correctly say, is not an acid), but books may still refer to it as citric acid. See also: Why are lactate and lactic acid used synonymously in biochemistry?

There is a nomenclature confusion. The word citrate means 'a citric acid who already donated protons'. The reason why your books shows the picture with COO- and not COOH is that because at physiological pH the citric acid immediately gives away protons.
The same things applies for amminoacids, who can both be written with NH4+ and COO- groups OR you can use NH3 and COOH. It means the same thing, just a different molecular state.


text:
How does chlorine in water evaporate?
Drinking water (e.g. from your home's tap) is spiked with chlorine, for an amount of just some ppm.
Purpose of this operation is to kill bacteria, for our health.

Background: chlorine in water is potentially dangerous for plants, or better, their soil.

A common solution to get rid of chlorine in tap water is leave it in an open pot all-night-long. In this way chlorine evaporates.
Now, in my mind, chlorine in dissolved in water as chloride ion ($\ce{Cl-}$). Just like salt in water.
And just salt in water, if you leave a pot of that mix in a room, after some time water evaporates completely and salt remains on the bottom of the pot.
Question is, how can chlorine evaporate from water?

And  why do you think it shouldn't?

Because it has a higher vapor pressure than the water it's dissolved in?

I updated OP in the hope I well clarify question

The chlorine in chlorinated water is NOT as chloride, it as molecular chlorine - a different oxidation state to chloride.

@Waylander You have highlighted the key point. But the actual chemicals in the water will be a mixture of things like hypochlorite as the chlorine reacts with hydroxide in a complicated equilibrium. But the key part is that it isn't chloride.

Chlorine and Free Chlorine
Chlorine is the term commonly used to describe the addition of chlorine gas, sodium hypochlorite (i.e., bleach), or calcium hypochlorite (i.e., bleach powder) to water to form free chlorine. Free Chlorine is the sum of hypochlorous acid ($\ce{HOCl}$) and hypochlorite ($\ce{OCl-}$); a monoprotic acid with a $\ce{pK_a} = 7.54$.
Chlorine gas ($\ce{Cl2}$) does not exist in water (except highly acidic) as it hydrolyzes to $\ce{HOCl}$. Chloride ($\ce{Cl-}$) forms from the redox reaction of $\ce{HOCl}$/$\ce{OCl-}$, but it is not reactive.
Both $\ce{HOCl}$ and $\ce{OCl-}$ are volatile, with $\ce{HOCl}$ being $\approx\!5\!\times$ greater (Henry's solubility constant being $H = 0.076$ and $0.013$, respectively).$^{[\text{citation needed}]}$ When you open a pitcher of water containing free chlorine to the atmosphere, the free chlorine comes into equilibrium with the atmosphere and thus it volatilizes. As such, you do not lose "$100\%$ or all" of the free chlorine, but enough is removed that the concentrations drops below your taste threshold.

Side Note
If the water contains dissolved organic carbon and organic nitrogen (e.g., natural organic matter, pathogens), then the free chlorine could be reacting while it is sitting and thus you could be losing free chlorine this way as well. Further you can also add combined chlorine (e.g., monochloramine) to water in place of free chlorine. This also forms from free chlorine reaction with nitrogen compounds in water (e.g., urine) and is the common "pool smell" observed in swimming pools that are not properly chlorinated.


text:
How do we know cyclopentadienylmolybdenum tricarbonyl occurs as a dimer?
Cyclopentadienylmolybdenum tricarbonyl, $\ce{MoC8H5O3}$, dimerizes to form the cyclopentadienylmolybednum tricarbonyl dimer, $\ce{(MoC8H5O3)2}$.
$\hspace{3cm}$
How was it discovered that it does this?
If you had an sample of a pure substance, but didn't know if it had dimerized or polymerized, what information would you need to find out if it had?

Asking in general, makes it broad-ish  - there are various crystallographic, spectroscopic and "traditional" methods, to check out the structure.

$\ce{^{95}Mo}$ NMR?

The suggested answers from the textbook are to look at the output of a mass spectrometer, or to observe the macroscopic crystal structure.
However, NMR spectroscopy is also a perfectly acceptable way to go about it.

For reference: http://scripts.iucr.org/cgi-bin/paper?S0567740868002773


text:
Reaction of soluble salts
If we were to mix two soluble salts together, for instance sodium nitrate and potassium chloride together, and their products are also soluble, will any reaction still occur? Say if we were to boil the solution, will we obtain only sodium chloride and potassium nitrate? Or will it be a mixture of all 4 possible salts.

Unless a precipitate forms from a double replacement reaction (basically cat$_1^+$an$_1^-$ + cat$^+$an$^-$ --> products), no reaction will occur from the dissolution of two salts. The precipitate can even be water, or a gas (like carbon dioxide, which will form from the decomposition of carbonic acid, alongside water).
e.g. H2SO4 + BaCl2 --> 2HCl + BaSO4 occurs because barium sulfate precipitates.
e.g. 2HNO3 + BaCl2 --> NO REACTION because neither hydrochloric acid nor barium nitrate precipitates. They just hang in the water as free ions (hence ionic equations).

There are other reactions besides precipitation, for example redox reactions.

What if you remove the water? Do you get your original salts back (sodium nitrate and potassium chloride) or new salts (sodium nitrate, sodium chloride, potassium nitrate and potassium chloride)

@LuoZeyuan I'm not quite sure. I know that with one dissolved salt, it will reform once the water has boiled out. Not all cation-anion combos do this though, for example vinegar (acetic acid HC2HOOH or HC2H3O2). But when you have two salts dissolved... maybe if the ionic bonds are strong enough to reform, but that goes deep into chemistry.

You will get a mixture of all 4 possible salts if you evaporate to dryness

Some salts are easily separated by crystallization. In salt ponds, this is used to separate highly soluble salts, such as LiCl, from less soluble, e.g. CaCO3. See http://www.hydrochemistry.eu/exmpls/sea_evap.html


text:
Is there a detailed explanation as to why polarity of a molecule affects the absorption intensity of IR radiation?
Polar molecules have an electric field due to there being a net dipole moment. The electric field will be stronger with greater polarity. The stronger electric field in the polar molecule will interfere with IR radiation stronger than a weaker electric field. But how does the interference between the molecule's electric field and the IR radiation have to do with intensity of IR radiation absorbtion of the molecule? Is there a causal relationship between the two or is there simply a correlation?

Huh, I'd be interested why it was "favourited" twice. This question doesn't seem to make any sense... Why? False premise.

To answer your question 'yes' there is a relationship. The process is not quite as you put it however. If a bond has atoms of different electronegativity then obviously there is a dipole formed and as the molecule vibrates about its mid point the dipole also oscillates  since the nuclei move. The electric field of the radiation can interact with the molecules's oscillating dipole and only if the frequency is very, very close to that of the vibration energy can be transferred from the radiation to the molecule. This 'resonance' results in an absorption band at the vibrational frequency, and a molecule with one extra quanta of vibrational energy.
This is a phenomenological description, in a quantum description the probability of absorbing energy and moving from one vibrational level (stationary state) to another is proportional to the square of the strength of the radiation's electric field $\epsilon$ and the square of the transition moment $M$ of the vibration. Thus probability of absorption $\sim  \epsilon^2M^2$. The electric field squared is proportional to the light intensity; no light no absorption, obviously.
The transition moment is proportional to the dipole $\mu$ as $M = \int \psi_i^*\mu \psi_f dx= q\int x\psi_i^* \psi_f dx$ where $\psi$ are the vibrational wavefunctions for the initial and final states in the transition, say v = $0,1$ and $q$ the charge on the electron. This shows that if the dipole is zero then there is no transition, just as in the case of homonuclear diatomic molecules. Additionally this shows that even if $\mu$  is large if the wavefunctions do not overlap (their product is zero) then the integral can be zero. This leads to the selection rules for transitions, which means that not all potential transitions occur.


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Solve for wavelength of photon given threshold frequency
The threshold frequency of a potassium metal plate is $\pu{4.35\times10^14 Hz}$. Determine the wavelength of a photon required to reach $\pu{2.50 eV}$ of the maximum kinetic energy of an emitted electron.
Options:
A. $\ \ \ \pu{2.88\times10^2 nm}$
B. $\pu{-4.96\times10^2 nm}$
C. $\ \ \ \pu{9.50\times10^-21 nm}$
D. $\ \ \ \pu{6.69\times10^2 nm}$
E. $\ \ \ \pu{2.02\times10^15 nm}$
From what I understand, I used the equation $KE = h\nu + h\nu_0$ and replaced $\nu$ with $\frac cA$ because wavelength and frequency are inversely proportional. Then I plugged in the given components, $\nu_0$ and a fraction of $KE$ to solve; however, I don't understand how to approach this as I'm only given $2.5 \pu{eV} (4.005 \times 10^{-19} \pu J)$ of maximum $KE$, not an actual value for $KE$. The answer I'm getting is $\pu{1.14 \times 10^{-12} nm}$ if I just use $\pu{2.5 eV} = KE$ max, which isn't an answer choice, and the given answer is A. $2.88 \times 10^2 \pu{nm}$. Please help.

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu{}$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

Please do not remove formatting markups without cause. They make the question more presentable.

Always work in energy not wavelength. You are given $\nu_0$ in Hz ($\mathrm{s}^{-1}$), multiply this by $h$ to convert it to Joules; ($h$ has units J s). You have the kinetic energy in eV, use the conversion  $1\mathrm{eV} = 1.602\cdot 10^{-19}$ J to get into joules. Now you have only to find energy $h\nu$ by subtraction. Finally convert energy to frequency using $E=h\nu$ and then to wavelength using $c=\lambda \nu$. Make sure $c$ is in m/s and as you answer will be in meters then convert to nm from metres.


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Strengthening a clay rocket nozzle with sodium silicate and calcium chloride
"I strengthen the bentonite clay nozzle by dipping the rammed nozzle in a $50\%$
 sodium silicate solution and allow to dry in warm oven. Then dip in a strong solution of Calcium Chloride, again dry. Then another dip in $50 - 70\%$ sodium silicate and again air dryed in warm oven. A third silicate treatment can be done. This makes for a strong nozzle that doesn't erode unless the motor thrust is over about $\pu{150kN}$." My question is: What is going on here chemically? Is it crosslinking?

The Na-silicate solution would seep into pores in the clay and polymerize into silica. Not sure why the dips in calcium chloride solution.

a google search shows that calcium chloride and sodium silicate react in water to make calcium silicate.  I am not exactly sure why that would strengthen/prevent erosion in a rocket nozzle, but further research shows that calcium silicate melts at 3,870 °F.  So I guess that's a clue.


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Electrochemical cells. Why is a salt bridge needed at all when the system is open?
I get that the purpose of salt bridges is maintaining the amount of negative and positive charges in the solutions of two connected half-cells when electrons flow from one to the other. But when the system is open, as it is when using a high impedance voltmeter to meassure the potential of the cell. There is no flow of electrons to counter. Does a salt bridge have any purpose then?

When using a voltmeter there is a very very small current flow. There must be some current to get a voltage, remember that $V=iR$.

But isn't that current flow negligible? Is it enough to need a salt bridge?

With no salt bridge there is no current flow at all. $V = iR$ now let $i=0$, so $V = 0\times R = 0$

Now why would there be no current flow without the salt bridge? Electric current travels through the wire connecting both half-cells (in this case through a voltmeter that allows a very tiny amount of current through). The salt bridge only exhanges ions with the solutions to maintain the amount of charges in both solutions.

Below is an image of the Daniel cell from Wikipedia which I modified a bit.

Again from Wikipedia...
In the Daniell cell, copper and zinc electrodes are immersed in a solution of copper(II) sulfate and zinc sulfate respectively. At the anode (on the left), zinc is oxidized per the following half reaction:
$$\ce{Zn(s) → Zn^2+(aq) + 2e−} \tag{EMF -0.7618 V}$$
At the cathode (on the right), copper is reduced per the following reaction:
$$\ce{Cu^2+(aq) + 2e^− → Cu(s)}\tag{EMF +0.340 V}$$
The total reaction being:
$\quad\quad\quad\quad\ce{Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s)}\quad\quad\quad\text{(Open Circuit voltage 1.1018 V)}$
Now from the zinc half cell electrons to the copper half cell via the wire. Considering what would happen if the flow of electrons could happen continuously without a salt bridge:

The zinc half cell would become positively charged, since there would be more $\ce{Zn^2+}$ ions than $\ce{SO4^2-}$ ions.
The the copper half cell would become negatively charged since there would be less $\ce{Cu^2+}$ ions than $\ce{SO4^2-}$ ions.

You can't generate a continuous current like this since it would create  "charged solutions", so sulfate ions must flow, from the copper half cell to the zinc half cell, through the salt bridge to balance the charges and create neutral solutions.

Now what does the phrase "Open Circuit voltage 1.1018 V" mean?
You can think of a battery being like this:

So the battery has an internal resistance $R_s$. Now we add an external load (be it voltmeter or resistor...) on the battery so that current flows.

Now the current, $i$ is flowing through both the internal resistance of the battery, $R_s$ as well as the external load, $R_l$. So the Voltage measured across the load of the Daniel Cell is $V_{out} = iR_l$, but for the whole cell:
$$ V_s = 1.1018 V = iR_s + iR_l$$
Rearranging a bit we get $V_{out}$ as related to the "open circuit" voltage of 1.1018 V.
$$ V_{out} = 1.1018 V - iR_s $$
So if any current flows in the cell then the load voltage, $V_{out}$ will be less than 1.1018 V. However if the current is small so that $iR_s < 0.0001 V$  , then for all practical purposes we can ignore the internal voltage drop of the Daniel cell and claim that the load voltage measured is the "open circuit voltage."

Thanks for the detailed answer, but I'm afraid the question remains, if the current is small enough that the reaction could be practically considered to not be happening at all, then no half cell will become positively or negatively charged and the salt bridge will no longer have a purpose, right?

"You can't generate a continuous current like this since it would create 'charged solutions', so sulfate ions must flow from the copper half cell to the zinc half cell to balance the charges and create neutral solutions." // I quit. I don't know how to explain this any better.

But how would the solutions get "charged" if there is practically no current going on? I'm not trying to generate a continuous current, Im just meassuring the potential of a cell to create one, without it practially happening. I'm sorry for being so dense, but I can't wrap my head around it, you tried, man, thanks a lot.

I'll try again...  (1) With no salt bridge there is no continuous current flow so the solutions don't get "charged." The point that I was trying to make is that if a continuous current did in fact flow then the zinc solution would become more and more positively charged and  the copper solution would become more and more negatively charged.  (2) You got to think about the numbers involved here. 1.00 microamp of current for 1 second is $1\times10^{-6}$ coulombs. A coulomb is $6.24\times10^{18}$ electrons, so that would be $6.24\times10^{12}$ electrons. That is a lot of electrons!

(3) A real voltmeter can't measure a "theoretical voltage." It works by measuring a small current flow. The overall notion is that the internal resistance of the voltmeter is much much greater than the resistance of the load. So practically you can ignore the current flowing though the voltmeter. That isn't to say that there is 0.00... amps flowing through the voltmeter.

(4) I quibbled and wrote about a continuous current flow. With no salt bridge there would in fact be a itsy bitsy teeny tiny  current pulse as the electrodes became polarized in the solution. Think of the electrode-solution interface as sort of capacitor. But that would be a current pulse, not a continuous current flow that a typical voltmeter could measure. I was hesitant to mention this because I knew it would just confuse the issue more.

I think I got it now, you have some patience!

@MaxW Excellent answer (+1)! Especially nice to see the cell resistance, discussion of the loading and your comments!


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Precipitation titration with Volhard method
In the Vollhard method Chloride is precipitated from a solution using escess silver ions.
$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)}}$$
After precipitation the solution is filtered and ammonium thiocyanate is added to the silver supernate with a small amount of ferric ammonium sulfate as an indicator where thiocyanate reacts preferentially with the silver:
$$\ce{Ag+_{(aq)} + SCN-_{(aq)} -> AgSCN_{(aq)}}$$
Once the all of the silver has reacted with thiocyanate the $\ce{Fe^3+}$ will then react to form $\ce{Fe[SCN]^2+}$ which produces a dark red color and indicates an equivalent point. and the chloride concentration is calculated by difference.
$$\ce{Fe^3+_{(aq)} + SCN-_{(aq)} -> Fe[SCN]^2+_{(aq)}}$$
Since the silver in the precipitate, why will you have to filter $\ce{AgCl}$ before you can start back titration instead of treating the solution? Would filtering also necessary for $\ce{AgBr}$ and $\ce{AgI}$?

Welcome to Chemistry.SE!  Good question, but please note the edits made and try to demonstrate and understanding of the underlying principles in future questions. We are a community happy to help, but not act as instructors. Requiring users to understand the underlying principles first helps maintain quality on this site.

Yes filtering is necessary as thiocyanate can complex the solid silver chloride per:
$$\ce{AgX_{(s)} + SCN-_{(aq)} -> Ag[SCN]X-_{(aq)}\tag{X = Cl, Br, I}}$$
This additional reaction requires more thiocyanate for back titration and thus gives a lower chloride concentration than is true.


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How&why does change in concentration affect cell potential

I'm a bit confused about why (in 1c) if the air pressure is lower, and there is a lower concentration of oxygen, why the cell potential decreases.
It checks out fine with the Nernst Equation, but conceptually, voltage is the driving force or energy of the current. Wouldn't there just be less electrons, not a lower voltage?
Like how the standard reduction potential isn't based on moles of reactant?

I'm a bit confused about why (in 1c) if the air pressure is lower, and there is a lower concentration of oxygen, why the cell potential decreases.

Think about it this way. Oxygen will diffuse into the cathode regardless if current is flowing or not. Obviously an infinite amount of oxygen can't flow into the cathode, or else all the oxygen would be sucked out of earth's atmosphere. So there has to be some distribution coefficient, D, that gives the limit of how much oxygen the cathode can absorb. The notion here is that if 100% oxygen allows an amount $x$ of oxygen to diffuse into the electrode, the the atmosphere with roughly 20% oxygen and 80% nitrogen would allow $0.20x$ amount of oxygen to diffuse into the electrode. Thus much akin to the Nernst Equation for liquids applies and the cell voltage with only 20% oxygen is lower than if the cell is in 100% oxygen.


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Formation of caffeidine in Noon Chai
Noon Chai (as described in this question) from Kashmir is prepared by adding common salt and sodium bicarbonate to green tea and cooking it for quite some time (~ 30 minutes).
Noon Chai is also associated with gastric cancer; the mechanism is assumed to be endogenous nitrosation of caffeidine by which then carcinogenic nitrosamines (dinitrosocaffeidine) are produced.
The question is: how does Noon Chai get its high amount of caffeidine in the first place?
Somehow the caffeine

is converted to caffeidine

by the special way to prepare the tea.
What is the reaction that happens there?

Basic hydrolysis - the sodium bicarbonate generates a basic solution aqueous solution which is then heated.


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what is superimposability of mirror images of this molecule?
I don't get why (a) and (d) are achiral when we cant superimpose their respective mirror images?Or can we superimpose them?

Why, we surely can.

how exactly can we do that? i am unable to.

These are like easiest examples among these...

It depends on where you put the mirror, in the case of a a horizontal mirror is required, or rotate the molecules 90 degrees before using a mirror to the left.
For d it is best if you rotate the C-C bond between the six memebered ring and the amide group before turning the molecule 60 degrees clockwise, then hold it beside the imaginary mirror to the left of it.


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What is the predominant enol tautomer of 1-phenylbutane-1,3-dione?
In 1-phenylbutane-1,3-dione (1), which enol tautomer will be predominant?

Both carbonyl groups possess α-hydrogens, so can undergo keto-enol tautomerism. In particular, the carbonyl group on the right can enolise in two ways since there are two unique sets of α-hydrogens.
The answer is supposed to be enol 2. Why can't it be either 3 or 4?

You need to retain a conjugated carbonyl group which is more stable than two enols. [Acetone exists as a ketone with little enol.] The real question is why the given answer without considering the tautomeric form of the second structure.

The question editor has modified the question, and maybe assumes that he has thought of every possible enol structure, but he didn't.

@mykhal Hmm, this is kind of two years old, but I merely copied the structures from the previous version of the question. It's in the revision history... Yes, there are missing enol forms as user55119 also suggests, but the original question did not have those either.

Two enol tautomers are possible for 1-phenylbutane-1,3-dione. Both are shown in the figure. Enol tautomer 1 is predominant due to more extended conjugation than Enol tautomer 2 which leads to more resonance structures give extra stability to 1.

The compound is highly stabilised by higher Conjugation and is thus favored over the other enolic form. Another factor is intramolecular hydrogen bonding which stabilises the enolic form even more.
It can also be explained on the basis of acidity of H in keto from.

Due to presence of 2 Electron withdrawing groups, the H atom is highly acidic and will favor the enolic form.
The reason (3) is not formed is because after the formation of (2), the H marked in the 2nd picture will not be acidic anymore, making the equillibrim highly towards (2) form. Also, formation of the allene leads to leads to loss of comjugation, since the pi orbitals are now perpendicular to each other, which makes the equillibrim even backwards (towards 2nd form).

I think this is not really answering the question. The diagrams in the question were not very clear, but I have edited it. There is a more fundamental reason why the allene 3 does not form, namely the loss of conjugation (as the pi orbitals are orthogonal). And you also don't explain why the terminal methyl group doesn't enolise, e.g. in 4. To me, it feels like you are just explaining why 1,3-dicarbonyls tend to enolise, without explicitly considering the alternative structures which OP asked about.

@orthocresol I have added explanation for (3). If the answer still doesn't answer the question, please inform me and i will delete it because i cannot answer the question any further than that.


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What is the actual density of CAS 7647-14-5 (NaCl, 99+%)?
I'm not a chemist by profession, so please forgive any syntactical or vernacular oversights on my part.  Basically, I'm getting mixed messages from different sites about the density of sodium chloride. Wikipedia and some other sites list the density as $2.16$ or $\pu{2.17g/cm^3}$, while other sites list it as $\pu{1.199g/ml}$.  Paradoxically, this one lists it as both:
https://www.bio-world.com/productinfo/4_847_63_473/9221/Sodium-chloride.html
Would someone be kind enough to point out what I am missing, please?  I have reagent-grade NaCl that I need to measure accurately in order to obtain a correct weight/volume percentage.  Thanks

The CRC Handbook of Chemistry and Physics (75 ed) gives values as $2.1634\pm 0.002$ measured by x-ray crystallography and 2.1654 in another table with unspecified method so these values rounded give your first two values.  The third value (1.19 ) seems to be wrong.

If $\pu{1.199 g cm^−3}$ would be the correct density if sodium chloride, it would float in a saturated solution of itself in water at $\pu{20 °C}$. Since this is not the case this value must be wrong.

Do you need the value for a specific purpose? Because density can vary for a lot of reasons, temperature being one, measurement method another (XRD ignores any amorphous part), size of particles, impurities, etc. If it has to be really precise, the question is what is the right value for your application?

There is a simple explanation: one density is the density of the pure substance; the other is the density of the container with imperfectly packed crystals of the substance (so the overall density contains a lot of air because the crystals don't pack tightly). The second matters to shippers; the first to chemists.

Oh, and my explanation would be the reason why you should only measure solids by mass not volume as the volume depends on how well the particles or crystals pack and will vary a lot.

I'm almost certain the value of 1.199 g/mL refers to the density of a saturated aqueous solution of NaCl. See for example this data sheet. This kind of thing happens a lot in sites which just scrape whatever property values they can find and don't do any sanity checking.

Dead Sea water has an density of $\pu{1.24 g cm^-3}$. Also no reports about saltbergs from there. ;-)

Thank you everyone for your input and leads, and thanks for the saturation chart, Nicolau.  I'll seek experimental confirmation for this and the points others mentioned.


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Which chemical could be used to decrease water boiling point to a minimum possible level without heating it
I aim to decrease boiling point of water that will be used to generate steam. I wonder maybe there is an opportunity to add a substance that will help to decrease boiling point of water without harming environment with toxic byproducts. Do you know any solutes or substances?

No, that can't be done.

Why do you want steam? If it is about driving an energy source or for cooking then the temperature matters (both depend on the amount of energy which depends on the temperature so won't work as well if the temperature is lower even if you can achieve that goal chemically). You can get lower temperature boiling with lower pressure, but the same issues apply.

If you add another low-boiling chemical that forms an azeotrope, e.g. ethanol, the b.p. is lowered... however, you are no longer boiling water but a mixture. In this example, it would be 95.6% ethanol and 4.4% water. Ethanol is comparatively inexpensive and environmentally "friendly", but would need to replace most of the water. Check a table of azeotropic mixtures for other compounds that could be used.
Perhaps an ammonia-water working fluid would be more suitable for your use, or just pure $\ce{NH3}$. Ocean thermal energy conversion (OTEC) often uses ammonia as the working fluid. Ammonia-water is also used in absorption refrigerators, the converse of a motor. Ammonia is also inexpensive, and safe if fully dispersed into the environment (i.e. it's biodegradable), but is quite toxic in close quarters.


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What is the correct name of 3-isopropyl-4-methylhexane?

In the branched alkane shown above, I've been told that 3-isopropyl-4-methylhexane is supposedly the correct name. However, in the alternative numbering scheme on the right, the parent chain is still of the same length (six carbon atoms) and has more substituents, so shouldn't that be the preferred name?

Both the names you have given are not the preferred name for the molecule above. You are correct that the preferred name maximises the number of substituents when there is a choice: see Alkane nomenclature - choosing the parent chain - maximize or minimize substituents? So, a name of the form a-ethyl-(b,c)-dimethylhexane (where a, b, and c are appropriately chosen locant numbers) is preferred over a name of the form a-isopropyl-b-methylhexane, as it has three substituents as opposed to two.
However, the numbering on the right is not correct either. If you number the carbon chain the other way around instead, you would obtain the name 3-ethyl-2,4-dimethylhexane:

This is preferred over your proposed name 4-ethyl-3,5-dimethylhexane, because the locant set (2,3,4) in 3-ethyl-2,4-dimethylhexane is lower than the locant set (3,4,5) in 4-ethyl-3,5-dimethylhexane.
Note also that according to current IUPAC nomenclature recommendations (P-57.1.1.1 of the 2013 Blue Book), the preferred prefix for $\ce{-CH(CH3)2}$ is "propan-2-yl", not "isopropyl".


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Formation of (Gamma) Methylparaconic acid
I came across this question recently,

I tried to devise a mechanism but I'm struck here,
My try -

Moreover, I think my mechanism is somewhat inefficient because of formation of some byproducts, which we don't want.
In short in my mechanism,
First we convert succinic acid to its ester (by esterification using ethanol).
Then, we use this ester to react with acetaldehyde in basic medium (dilute NaOH), intramolecular magic occurs to yield an ESTER of the desired product.
Now, here's where i'm struck, how do I convert ester of the desired product into acid, Can I do Hydrolysis, but I think then the ring might get cleaved!
Moreover, I think during this reaction, some acetaldehyde may undergo self condensation to yield a byproduct, this may be true for ester of succinic acid.
Any help will be appreciated!

The mechanism looks OK. Though catalytic NaOEt would be a better base than NaOH as the EtO- is regenerated as the reaction proceeds.
The opening of the lactone under basic conditions is reversible but the reformation of the lactone is favoured. A mild basic hydrolysis - 1eq LiOH in 19:1 THF/H2O - should give you your target.
Self condensation is always a possibility with acetaldehyde - this is why few reactions go with 100% yield in real life

Thanks but any better way to obtain the same product?

Because I want to get as pure product as possible and therefore avoid self condensation.

One step condensation of Succinic anhydride with acetaldehyde:-  generate the anion at low temp with non- nucleophilic base such as NaH, add acetaldehyde, allow to warm to rt, work up with NHCl soln.

If you want clean product from your route, do column chromatography on the ester/lactone before the hydrolysis.

Whoa! Wait I don't think I fully understood your method, can you please explain it in detail as an answer (that I will accept) (I guess I'm a chem novice). As I said I'm much more interesting in finding a better method. Thanks in advance!

Succinic anhydride (structure here: https://en.wikipedia.org/wiki/Succinic_anhydride) will deprotonate next to one of the carbonyl groups with strong base - something that will not attack at the reactive anhydride.  That anion will condense with acetaldehyde and the sodium alkoxide of the intermediate will react intramolecularly with the anhydride carbonyl to give the 5-membered lactone and the sodium salt of the acid.

WOW I GOT IT! THAT WAS A COOL METHOD! Thanks a lot!


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What bonding allows chloroaluminate (AlCl4-) to form?
How is $\ce{AlCl4-}$ formed when $\ce{Al}$ only has $3$ electrons in it's outershell. It would be able to give $3$ electrons away with $\ce{AlCl3}$ for it to have a full outershell but how does it give another one away to the 4th $\ce{Cl}$? Wouldn't that cause it to be unstable as it wouldn't have a full outershell anymore? Also why would the final product $\ce{AlCl4}$ be negative? Wouldn't it be positive as it is giving electrons away?

Better to think of it as a covalent compound instead of an ionic compound.

If it was a covalent compound the 3 electrons would again be used to form the 3 bonds and so it wouldn't have any other electrons to donate to the 4th Cl?

See: Dative Bonds

Of course, but after forming three bonds, Al would only have six electrons around it, so there's room for two more from a chloride ion.

So chlorine forms a dative bond with the Al by donating 2 electrons to the covalent bond? But then the Cl wouldn't have a full outer shell as it requires 1 electron from the Al but the Al doesn't have one to give.

Chloride doesn't donate outright, it shares the electrons in a covalent bond with the aluminum. Yes, the aluminum has a formal charge of 3+ but it can covalently bond with 4 chlorides, each sharing 2 electrons to have a full octet of electrons in the n=3 shell. Boron does the same thing for the n=2 shell.

In order for the covalent bond to work wouldn't the Al need an electron? It has 3 in its outer shell and 1 is shared in each bond within ALCl3. It wouldn't have another electron to share with the Cl in the 4th Cl covalent bond?

Also the Cl would share 2 electrons with the Al but would only recieve 1 in return from the Al for it to have a full outershell that wouldn't work would it as both have to donate the same amount to the bond unless its dative? Also I thought covalent bonds were only between 2 non metals?

Think of it as $\ce{Al^{3+} + 4 Cl- -> [AlCl4]-}$.

So the 4th chlorine doesn't gain an electron from the Al and instead shares 2 of its own?

Correct. The same pattern is present in the aluminium trichloride dimer and in solid aluminium trichloride. See https://upload.wikimedia.org/wikipedia/commons/f/fa/Aluminium-trichloride-dimer-3D-balls.png and https://upload.wikimedia.org/wikipedia/commons/2/24/Aluminium-trichloride-crystal-3D-balls.png

Thank you for your help. 1 Last question, if it is dative covalent bonding as the lone pair from the Cl is being shared with the Al how come it isn't drawn as dative covalent bonding?

@James There are quite a number of ways to look at this situation. We can see it as: an aluminium ion in the centre with 4 chloride ligands surrounding it. Each chloride ligand donates a lone pair of electrons. The net result is an aluminion atom covalently bonded to an aluminium atom with a -1 formal charge. Most of the time, this would be indicated outside of the square brackets to indicate that the charge is delocalised over the entire complex ion. Why do we represent it without the dative bonds? Well... Dative covalent bonds and covalent bonds with formal charges are the same.

In this case, we would rather use formal charges because it would be weird to show each chloride ion having a negative charge and the aluminium ion still having the +3 charge in the overall structure


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How to loosen up Tris HCl?
I have this bottle of Tris HCl that I need to use and apparently it's all hard. I don't have time to get a new bottle since this is needed for a lab for tomorrow and this is all I was given. Is there an easy way of loosening it up to be free flowing short of throwing the bottle around?

You could try warming the bottle gently in an oven or hot water bath (not on a dry hot plate!) - I would try 50 C for starters. If it is in a plastic bottle I would avoid getting it too hot. Maybe it will soften, like brown sugar, and then you could scoop it out with a spatula.
Alternatively, you could weigh the entire bottle, use the contents to prepare a stock solution of known volume, measure the tare weight of the now-empty bottle, calculate the concentration of the stock solution, and use it to prepare buffers of known TRIS concentration.


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Calling Diamond an element?
Is it scientifically correct to call Diamond an element? Carbon has a wide range of allotropes, so would it be correct to actually just call graphite, diamond etc an element. For that matter, could we call any molecule containing only one type of atom, an element?

Phosphor has (among others) an allotrope called "red phosphor" and one called "white phosphor". Does that anwer your question?

Can we have that idiotic IUPAC definition https://goldbook.iupac.org/html/C/C01022.html that probably raised this question again? It was cited in an answer that has been deleted. They really should stick to defining nomenclature and not write dictionaries.

It might be scientifically correct but it is linguistically misleading
The sentence "diamond is an element" can be seen to be misleading when compared to the sentence "diamond is an allotrope of the element carbon". Or even "diamond consists of the element carbon".
The issue is that clear language should distinguish between the form and the composition of an element with more than one allotrope. Saying "diamond is an element" confuses the fact that diamond is made from carbon with the fact that it is made from an infinite tetrahedral array of carbon atoms. It also subtly excludes the fact that there are other allotropes.
For clarity in language a specific form of an element (an allotrope) should not be referred to as "an element" but should be referred to as one form of that element. As in "diamond is one form of the element carbon, buckminsterfullerene is another form."

Reminds me of https://xkcd.com/169/ ;-)

Then what is the point giving that definition of "element"? I mean it would be better to say periodic table of atoms than elements.

@adosar The periodic table provides more than just a list of atoms. It also tells us the normal form in which those elements appear. But there can be several common forms. The definition of element is that there is only one type of atom. But this is incomplete as we also need to know the typical forms (diamond, graphite, buckyballs...). The periodic table gives both.

Technically Yes, but Scientifically No.
Diamond is made of pure Carbon, so yes. However, it is not on the periodic table nor is it classified as an element by most scientists, so on that case, no. It really depends what you think.

The IUPAC definition of element is as follows:

A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. IUPAC Goldbook definition of element

This seems to allow diamond to be called an element. But it's not a good idea, scientifically or linguistically, since it doesn't make sense.
Carbon is an element.  Diamond is a form that the element carbon can take.
Let me give you a car analogy:
Toyota is a brand of car.  The Camry is one example of a type of Toyota car. I.e., it is one form a Toyota can take. That does not make "Camry" a brand of car, nor would it make sense to call a Camry a car brand.
In summary, referring to diamond as an element, rather than a form an element can take, is like referring to Camry as a car brand, rather than as one of the brand's models.

Diamond is a form of pure carbon (which is an element). Logic may lead you to believe that diamond is, therefore, an element. However, this is a bit circular because if you say that diamond is an element, then every allotrope of every element also falls under the same consideration.
Also, in order for diamond to be a unique element, the molecules that make it up should have a different number of protons than any other element on the periodic table. Since a diamond is made of carbon, it is really just another form of carbon.

Welcome to Chemistry.se! I would say this does not answer the question, but it only restates it. I agree the reasoning is circular, but according to some definitions available, a more rigorous answer would be preferable.

I felt that I had provided that with my second paragraph where I touched on proton counts, but I can try and make my answer more detailed.

No. Diamond is not an element.
It is a name for a  gemstone, a particular occurence of an allotropic form of carbon that may or may not contain other elements in addition to carbon.
And while "pure" diamond is an allotropic form of the element carbon, diamonds that are actually pure carbon are usually synthetically made, as most naturally occuring diamonds have impurities/inclusions in the matrix.