text:
What about the aromaticity of cyclopropenylidene?

Here's a compound which we have to tell whether it's aromatic or not.
Drawing pi-orbital for the 3-membered ring, I think there are only 2 pi electrons delocalised all over the ring. The orbital having non-bonding electrons on the carbon having sp2 hybridization(carbon no. 1) is not having the correct orientation with the pi-orbital of the ring so these electrons are not delocalised over the ring. 2 being a Huckel's number, I think that this compound should be aromatic.
But the answer key says that it is ANTI-AROMATIC! Where am I going wrong? Perhaps the ring is not planar?
EDIT: The carbon no. 1 doesn't have a negative charge so it is not an anion. It is neutral.

Are you referring to the cyclopropenyl anion? If so, this has been answered here: https://chemistry.stackexchange.com/questions/51467/is-the-cyclopropenyl-anion-antiaromatic  If you mean cyclopropenylidene, it is aromatic, and you can read about it on the wiki https://en.wikipedia.org/wiki/Cyclopropenylidene

@Blaise Looks like title of my question is edited by someone else. The compound I'm talking about is not correctly mentioned in the title

@Blaise Thanks for that wiki link. I'm talking about cyclopropenylidene (which has a carbene), so is my explanation which says it should be aromatic and those non bonding electrons are not delocalized correct?

A quick google shows a couple papers and a book confirming the singlet carbene to be aromatic. Maybe your book intended to indicate the anion.

This is a problem of counting lone electron pairs as part of the delocalized pi system.  In pyrrole you count the lone pair on the nitrogen, in pyridine you don't.  What's the difference?
In pyrrole you have the lone pair left over after you have given every ring a ligand (the outer hydrogen atoms).  With the sigma bonding requirements thus satisfied, the lone pair goes into the pi system.  In pyridine the nitrogen is missing its ligand, so the lone pair must replace that in the plane of the ring and can't go into the pi system.
Which situation corresponds to your compound depends on what compound is meant, and there seems to be confusion on this point.  Cyclopropenylidene ($\ce{C3H2}$) and cyclopropenyl anion ($\ce{C3H3^-}$) differ only by a carbon-hydrogen bond, and the shorthand method commonly used to render organic structures thus fails to distinguish them.  The problem statement should be checked to ascertain which of these compounds is meant because they give different outcomes, and the answer key is right for only one of them.
If it is cyclopropenylidene
In cyclopropenylidene, the carbon with the lone pair is missing a ligand like pyridine, so the lone pair has to replace the missing ligand in the plane of the ring.  That leaves two pi electrons to be delocalized which makes the ring aromatic.  Cyclopropenylidene is not only aromatic by the electron count, it is a sufficiently preferred three-carbon structure to be seen in interstellar space.
If it is cyclopropenyl anion
Cyclopropenyl anion has all the ligands in place, so the lone pair goes into the pi system like pyrrole.  Unfortunately in this case, adding that lone pair gets you four pi electrons in the ring and that's bad.  In fact it turns out to be really had, even worse than larger (and thus more flexible) formally antiaromatic rings like cyclooctatetraene.
Cyclopropenyl anion goes through some serious contortions to get rid of this antiaromaticity.  See this question.

Sorry I was not asking about Cyclopropenyl anion (someone edited the title of question wrongly). I already know that Cyclopropenyl anion is antiaromatic.


text:
When making a calibration curve, do we have to use the wavelength at maximum absorbance?
If you are making a calibration curve for $\ce{CuSO4}$ where you plot absorbance against concentration, does the wavelength have to be exactly at the maximum of the peak? Or can it be somewhere else along the curve of the peak as long as it is consistent among all the measurements and still get an accurate result?

No it doesn’t. Measurements can in principle be made at any wavelength.
If there are two colored specimens in the same solution, it is actually better to measure at the wavelength where the difference between the spectra is as high as possible.
Measuring at a peak has some advantages: the absorbance is high, giving a good signal/noise ratio, and if the wavelength varies a little it has less effect on the absorbance when measured at a peak than at a slope of the curve.

Yes, it does, and especially for calibration curves [1, p. 439, 455]:

For spectrophotometric analysis, we normally choose the wavelength of maximum absorbance for two reasons:
(1) The sensitivity of the analysis is greatest at maximum absorbance; that is, we get the maximum response for a given concentration of analyte.

Sensitivity $b$ is the slope of your linear calibration curve:
$$y = a + bx$$
where $y$ – signal (absorbance); $a$ – background noise or erroneous values (can be often taken 0 for a dual-beam spectrometer with a blank probe); $x$ – concentration. From differential analysis
$$b = \frac{\mathrm{d}y}{\mathrm{d}x}\approx \frac{Δy}{Δx},$$
so that you are interested in getting the largest signal possible (at the maxima) in order to have the best sensitivity with the set of given concentration, which you cannot increase of decrease infinitely due to limited solubility and device limitations – most spectrophotometers won't register $A > 2$ properly. Further [1, p. 439, 455]:

(2) The curve is relatively flat at the maximum, so there is little variation in absorbance if the monochromator drifts a little or if the width of the transmitted band changes slightly. Beer’s law is obeyed when absorbance is constant across the selected waveband.
...
Measurements should be made at a wavelength of maximum absorbance. Instrument errors tend to be minimized if the absorbance falls in the range $A < 0.3 - 2$.

References

Harris, D. C. Quantitative Chemical Analysis, 9th ed.; W. H. Freeman: New York, 2015. ISBN 978-1-4641-3538-5.

source: Stack Exchange

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

sort: votes


text:
ClO2 molecular orbital diagram?
I was trying to construct a MO diagram for ClO2, and since I'm not too sure about my final result I thought I'd check it from one of the experts here.
ClO2 is of C2v point group, so I just read off the C2v character table and got the following group orbitals [where z axis is that of the principal axis, and the outer atoms are aligned such that the y-axis points towards the centre atom]:
A1 symmetry, two s orbitals
B1 symmetry, two s orbitals
A1 symmetry, two py orbitals
B1 symmetry, two py orbitals
A1 symmetry, two pz orbitals
B1 symmetry, two pz orbitals
A2 symmetry, two px orbitals
B2 symmetry, two px orbitals
Adding those to the s orbital (A1), px orbital (B2), py orbital (B1) and pz orbital (A1) of the chlorine, we get the following (see follow-up post):
Would my prediction of the 12 MOs be correct?
Thanks!

predicted molecular orbitals: https://i.stack.imgur.com/YRsLC.png

Related: Hybridisation of ClO2

Yes, there are 12 orbitals, but they are typically drawn differently from what you have. It is generally assumed that the oxygen $2s$ orbitals lie too low in energy to be involved in bonding, so you have an $a_1$ and $b_1$ orbital at low energy comprised of those two $2s$ orbitals in the same phase ($a_1$) or opposite phase ($b_1$) with no $\ce{Cl}$ orbital involved. You have only included the $b_1$ of this pair. The $a_1$ replaces the $a_1$ orbital you have drawn with oxygen $p_y$ orbitals. Then your $a_1$ pair comprised of $s$ orbitals from all three atoms would instead be comprised of chlorine $3s$ and oxygen $p_y$.
You can read a description of this in Walsh's original paper[1] on AB2 non-hydride molecules.
He also assumes that in bent molecules like $\ce{ClO2}$ there is significant mixing of the $s$ and $p_z$ orbitals on the central atom so that in the $\sigma/\sigma^*$ and $\pi/\pi^*$ $a_1$ groups, the orbital on $\ce{Cl}$ is similar to an $sp$ hybrid in both cases rather than an $s$ orbital in one pair and $p$ in the other.
The relative energy levels of the orbitals for $\ce{ClO2}$ have been reported by Wang and Wang[2].
[1] Walsh, A.D. J. Chem. Soc. (1953) 2266-88.
[2] Wang, X.-B. and Wang, L.-S. J. Chem. Phys. (2000) 113:10928-33.

Would it look more like this (before hybridisation)? https://i.stack.imgur.com/2T1IZ.png

Yes, that's right.

What would the ordering of the MOs look like, if you'd be kind enough to grind through all of that? Thanks a lot!

First comment - The convention is to use $\sigma_v'$ for the plane containing the O ligands and $\sigma_v$ for the plane between them. By that convention, your labels of $B_1$ and $B_2$ are reversed. There is no impact on the $A_1$ and $A_2$.

As for the ordering, I'm not 100% sure of the order. Using your labels, I'd start by assuming the two orbitals from the O2s are lowest, with a1 < b1. Then the two $\sigma$ orbitals, with the a1 < b1. Then b2$\pi$, a1$\pi$, a2 nb (px), b1 nb (pz), a1$\pi^$, b2$\pi^$, b1$\sigma^$, a1$\sigma^$. But the b1$\sigma$ could go up in energy enough to be higher than the $\pi$ orbitals, and the a1$\pi^*$ is lowered in energy, possibly to the point that it's below the b1 nb(pz) (which is raised up). If I recall correctly, that's consistent with the experimental results.

See this reference for details of the energy levels:link. A quick read suggests their results are consistent with a1$\pi^*$ being lower than what you called b1 nb (pz).

Could I inquire what you mean by my labels being incorrect? Thanks!

There's nothing mathematically incorrect about your labels. It's just that the convention (from Mulliken) is for B1 to be -1 with respect to the plane of the molecule and B2 to be -1 with respect to the plane perpendicular to the molecule. You used the reverse. It's only worth mentioning because knowing that might make it easier for you to make sense of literature on the subject. For example, the Walsh paper I linked uses the standard convention, so every orbital that it calls b1 matches to one that you called b2.

Ah, got it. So basically my x/y/z axes labelling was screwed up. Thanks!

Please include human readable citations and use doi links, in order to keep things stable. It would also be great, if you could mark links that lead to pdf (possible via a mirror) as such.

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


text:
What happens to depression when two solutions are mixed?

Equal volumes of $\pu{1.0 M}$ $\ce{KCl (aq)}$ and $\pu{1.0 M}$ $\ce{AgNO3 (aq)}$ solutions are mixed. The depression of freezing point of the resulting solution (with respect to water) will be (assume $K_{f}$ for water is $\pu{1.86 K * kg / mol}$ and molarity ≈ molality).
A) $\pu{1.86 K}$ B) $\pu{3.72 K}$ C) $\pu{0.93 K}$ D) $\pu{7.44 K}$

My answer is.... Since for freezing point the vapour pressure of solute and solution should be same thus adding the mixture would result in same vapour pressure as before... Thus the depression in freezing point should be equal to $\pu{1.86 K}$.
Am I right or wrong? Please help!

Please don't post questions twice! You can edit or delete your own questions.

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

The previous version was correct for multiple reasons.. do not revert a fixed system..

Vapour pressure is not involved in the reasoning.
Freezing-point depression is in fact a colligative property, so it depends on the total number of solute particles and not of their chemical nature, so as you say the freezing-point depression is the same for both solutions ($\ce{KCl}$ and $\ce{AgNO3}$).
However, note that the depression is proportional to the total number of particles - and in the case of strong electrolytes (such as both $\ce{KCl}$ and $\ce{AgNO3}$), we have twice the number of ions in solution than the formal concentration of the salt would indicate.
Given the expression for the freezing-point descent:
$$ \Delta T_{f} = K_{f} · \sum_{i} m_{i} $$
where $\Delta T_{f}$ is the cryoscopic descent, $K_{f}$ is the cryoscopic constant, and $m_{i}$ is the molality of dissolved species $i$.
Since we have strong electrolytes that we can assume are completely dissociated,
$$ m_{\ce{K+}} = m_{\ce{Cl-}} = M_{\ce{KCl}} $$
$$ m_{\ce{Ag+}} = m_{\ce{NO3-}} = M_{\ce{AgNO3}} $$
$$ m_{\ce{KCl}} = m_{\ce{AgNO3}} = 0 $$
where $M$ is the formal molality of the salt. Therefore:
$$ \Delta T_{f} = K_{f} \left( m_{\ce{K+}} + m_{\ce{Cl-}} + m_{\ce{Ag+}} + m_{\ce{NO3-}} \right) = 2 K_{f} \left( M_{\ce{KCl}} + M_{\ce{AgNO3}} \right) $$
Since we're mixing equal volumes (and disregarding any mixing volume change), the formal molalities of both salts in the final solution will be $\pu{0.5 mol / kg}$, and their sum $\pu{1.0 mol / kg}$:
$$ \Delta T_{f} = 2 · \pu{1.86 K * kg / mol} · \pu{1.0 mol / kg} = \pu{3.72 K} $$
So the correct answer would be B) (but see the note below).
Note that you can find an alternative formulation for the expression of the freezing-point depression, using the van't Hoff factor:
$$ \Delta T_{f} = K_{f} \sum_{j} M_{j} · i_{j} $$
where $i_{j}$ is the van't Hoff factor for a salt, i.e. the number of ions into which it dissociates in solution. Again, for our completely dissociated strong electrolytes, $i_{\ce{KCl}} = i_{\ce{AgNO3}} = 2$, which gives exactly the same expression as before.
Note on solubility: the presence of $\ce{Ag+}$ and $\ce{Cl-}$ will result in the formation of a solid precipitate, $\ce{AgCl (s)}$. This will in practice remove $\ce{Ag+}$ and $\ce{Cl-}$ ions from the solution; however, since no value for the solubility product $K_{\rm{s}}$ is given, and no mention is made of this precipitation in the question, I've assumed it is not considered. If in the context of your course or text this is something you'd have to take into account (i.e you are expected to recognise that $\ce{AgCl}$ is insoluble in water and would precipitate in this situation), since you are not provided with a value for $K_{\rm{s}}$, we can only assume that the precipitation is complete, which would remove all $\ce{Ag+}$ and $\ce{Cl-}$ ions from the solution:
$$ \ce{Ag+ + Cl- -> AgCl \downarrow} $$
$$ m_{\ce{Ag+}} = m_{\ce{Cl-}} \approx 0 $$
$$ \Delta T_{f} = K_{f} \left( m_{\ce{K+}} + m_{\ce{NO3-}} \right) \approx \pu{1.86 K} $$
Which makes A) the right answer.

Alright! Thank you bro! But a small question. The molality m(kcl) and m(AgNo3) should be equal to 1 and not zero? Right? And how is M1×i1 + M2×i2 = 2?? Shouldn't it be 4?

Since $\ce{KCl}$ and $\ce{AgNO3}$ are strong electrolytes, they undergo complete dissociation: $\ce{KCl -> K+ + Cl-}$, $\ce{AgNO3 -> Ag+ + NO3-}$, so all the salt is in the form of separate cations and anions, and there is no undissociated $\ce{KCl}$ or $\ce{AgNO3}$ "unit" in solution, hence why there is no "real" concentration of $\ce{KCl}$ or $\ce{AgNO3}$ and why the concentration of the separate ions is equal to the "formal" (theoretical) concentration of the salts. Does that make sense?

On the second part, keep in mind that $\pu{1.0 mol/kg}$ is the molality of the salts prior to mixing. When you mix equal volumes, you are diluting the same amount of $\ce{KCl}$ and $\ce{AgNO3}$ in twice as much volume - the solvent coming from their own solution plus the same amount of solvent coming from the other solution. Hence, once we've mixed them, the formal molality ($M_{i}$) of both salts is not $\pu{1.0 mol/kg}$ but $\pu{0.5 mol/kg}$ - and therefore $\pu{0.5 mol/kg} · 2 + \pu{0.5 mol/kg} · 2 = \pu{2.0 mol/kg}$.

Any teacher should expect that half of his students recognise the insoluble combination of silver and choride ions. OK, a teacher who gives out multiple-choice tests ...


text:
Regarding Analysis for CO2 Absorption in NaOH
Background
Lab experiment, packed glass column (random packing of metal/ceramic material), countercurrent gas absorption, flowrates measured using rotameters, $rt \approx 27^\circ C$.
2 N NaOH flows downwards at 10-30 mL/s; Air + $\ce{CO2}$ (~10% $\ce{CO2}$ (v/v)) flows upwards at total rate of ~350-500 mL/s.
Samples were collected as (I) 10 mL + (II) 10 mL from the bottom of the column. An excess of $\ce{BaCl2}$ was added to (II). Both are titrated against 0.5 N HCl with phenolphthalein ($\ce{HPh}$) as indicator.
Expected Reactions
$$\ce{2NaOH + CO2 \to Na2CO3}$$
$$\ce{Na2CO3 + BaCl2 \to BaCO3 + 2NaCl}$$
Colour changes:
$$\ce{HPh + HO- \to Ph- + H2O}$$
$\ce{HPh}$ is colourless and $\ce{Ph-}$ is pink
The Problem

The first few readings were as expected and no irregularities were observed. The volumes of HCl added were $<$ 15 mL.

Later

On adding indicator and waiting for some time, the colour disappeared
(pink $\to$ colourless) for both (I) and (II). Shaking sped
up the colour change process.
On adding indicator and starting titration immediately, the colour
lightened (did not fade away completely) for a considerable volume of
HCl added, then darkened considerably, and after a few more drops,
suddenly became colourless (white in case of (II)). The volume of HCl added was $>$ 40 mL. Such a great change is not expected.

One Possibility:

$\ce{CO2}$ from the atmosphere reacts with the sample to increase the acidic nature of the solution and causes the disappearance of colour.
This would explain speeding up of colour change by shaking.

But, is the rate of absorption of CO2 into a still sample of $\ce{NaOH + Na2CO3}$ this fast?
Other possibilities

Formation of buffer: $\ce{Na+}$, $\ce{CO3^{2-}}$, $\ce{HO-}$ ions are present in (I). (II) additionally contains $\ce{Ba^{2+}}$, $\ce{Cl-}$. The tap water used might be a source of $\ce{Cl-}$ as well. $\ce{HCO3-}$ formation during titration (?)
The packing material might have reacted into the system somehow. This seems unlikely, and the packings did not show signs of fouling.
The chemicals used contain impurities causing these weird effects, in which case, which type of impurites could cause this?

Note: Do comment if you more information is needed.

What was the concentration of NaOH added to the column, 1 N? How long is it in contact with the 0.1 atm carbondioxide? The pKa of your indicator is 9.4? Can you do a control with 0.001 N NaOH (pH ~ 11) plus indicator to see if the carbondioxide in air is sufficient to give a color change upon shaking?

@KarstenTheis 2N NaOH was added. It's a trickle bed type of arrangement. Normal shaking of NaOH solution + phenolphthalein does not spontaneously go colourless in my experience.

Phenolphthalein becomes colorless again (from pink) in strongly basic medium with time. It is a known phenomenon called the alkaline fading of phenolphthalein. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=fading+of+phenolphthalein+in+alkaline+solution&btnG=
It is not a good choice to use phenolphthalein in the presence of sodium carbonate or bicarbonate. Methyl orange is typically used when NaOH and Na2CO3 are present together. Please consult the classic Vogel's "Quantitative Inorganic Analysis" for more details. Alternatively, you may try a potentiometeric titration (pH vs. volume of titrant). Measure the pH after every 0.5 mL addition of titrant and construct a titration curve.
The reaction of CO2 with NaOH is very fast.


text:
Steric Factors on reaction with grignard reagent

I was asked to find the products X and Y.
For the product X,Here the grignard reagent and even the carbonyl substrate was too sterically hindered to approach the carbonyl center so no reaction would take place, thus the final product will be
I thought the same for Y product .. Here the Carbonyl compound was crowded to approach so no reaction would take place but then I found the answer to be this.

So for the product Y hydrolysis occured due to presence of dilute HCl but it did not occur for the first one. But why? And was I right about the logic above?

https://www.organic-chemistry.org/namedreactions/grignard-reaction.shtm Scroll a bit down. The grignard reagent used for [Y] is most likely to bulky to deprotonate the $\alpha$ hydrogen.

@Soumik Das take a look at the reference Avnish Kabaj linked to. Reduction can occur when there is too much hindrance for nucleophilic attack

@Waylander Hmm. Pretty unusual though.

@AvnishKabaj That means we will get three products in total; and the reduced one is given as a side product though.. That means the product by SET mechanism will be major, isn't it?

@Soumik Das I think that was the point of the question

https://i.stack.imgur.com/Q2G6A.png Looks like the major product will be the reduced form and minor will be the enolized form and surprisingly zero percent addition product.

https://chemistry.stackexchange.com/questions/106987/abnormal-reaction-of-grignard-reagent


text:
How to get $\Gamma_{vib}$ for hexafluoroethane?
An exercise asks me to find $\Gamma_{vib}$ for that molecule.
I know that: $$\Gamma_{cart}=\Gamma_{vib}+\Gamma_{translations}+\Gamma_{rot}$$
Since the molecule I'm considering belongs to the $D_{3d}$ group (in the staggered configuration), from the character table you get:
$$\Gamma_{translations}=A_{2u}+2E_{u}$$
$$\Gamma_{rot}=A_{2g}+2E_{g}$$
and then:
$$\Gamma_{vib}=\Gamma_{cart}-2E_{u}-2E_{g}-A_{2g}-A_{2u}$$
My question is: does exist a fast way to find out $\Gamma_{cart}$?
The only method that I know is to express each operation of the group in terms of matrices using the Cartesian basis, so I should write down 12 matrices 24x24, and then find out the corresponding reducible representation, but I'd take ages. I'm pretty sure that there must exist a faster method.

Possible duplicate of How to systematically find the reducible representation of a molecule? (eg: XeF4)

Unfortunately in that answer is shown only the standard method to get $\Gamma_{cart}$. For water is quite simple since you have just 3 atoms. @Tyberius

The method is in my answer to: https://chemistry.stackexchange.com/questions/70858/how-do-i-determine-the-molecular-vibrations-of-linear-molecules/77474#77474

To transcribe what is given in levineds's linked answer and remove the parts only relevant to linear molecules, here are the steps you can apply to determine the reducible representation of the vibrations.

Determine how the molecule is changed by all symmetry operations. If an atom is moved by a symmetry operation, it contributes it doesn't contribute to $\Gamma_\text{atoms}$. If the atom remains in place, add +1 to the column of that symmetry operation.

For hexafluoroethane, this gives:
$$\small \begin{array}{c|cccccc} \hline
D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6  & 3\sigma_d  & \\ \hline
\Gamma_\text{atoms} & 8 & 2 & 0 & 0 & 0 & 4 &\\
\end{array}$$

Multiply each column of $\Gamma_\text{atoms}$ by the trace of that operation's matrix representation.
$$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$

$$\small \begin{array}{c|cccccc} \hline
D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6  & 3\sigma_d  & \\ \hline
\Gamma_\text{cart} & 24 & 0 & 0 & 0 & 0 & 4 &\\
\end{array}$$

Use reduction formula to determine irreps. $n(i)=\frac{1}{h}\sum_R N\cdot\chi_r(R)\cdot\chi_i(R)$ where $n(i)$ is the number of the $i^{\text{th}}$ irrep, $R$ is a symmetry operation, $h$ is the order of the group, $N$ is the coefficient in front of the operation, and $\chi(R)$ is the character of the operation in the reducible/irreducible representation.

I won't go through the calculation (there are online calculators that you can use to do this), but this gives. $$\Gamma_{xyz}=3A_{1g}+1A_{2g}+4E_g+1A_{1u}+3A_{2u}+4E_u$$
which if we subtract off the translational and rotational irreps you have already obtained gives:
$$\Gamma_{vib}=3A_{1g}+3E_g+1A_{1u}+2A_{2u}+3E_u$$
(You had small error with your rot and trans, the $E$ irreps are degenerate and so count for 2 degrees of freedom. This means in each case, you would only have a single $E$ irrep.)


text:
Concentration of Hydrogen Ions
I am slightly confused about one certain task.

There are 4 different solutions ($\ce{NH3}$, $\ce{HCl}$, $\ce{NaOH}$, $\ce{CH3COOH}$), their concentration is the same $(c = \pu{1 mol L-1}).$ Which solution has pH that is higher than 7, but lower than 14?

It is obvious that $\ce{HCl}$ and organic acid cannot have such concentration of hydrogen ions. Both ammonia and natrium hydroxide have base properties. $\ce{NH3}$ is a weaker base though. I am not sure how to find the concentration of hydroxide in $\ce{NH3}$ and $\ce{NaOH}$  properly.
Thank you for the explanation.

Since $\ce{NaOH}$ is a strong base, the pH can be directly calculated from the analytic concentration (1M):
$\ce{pOH} = -\log\ce{[OH-]} = -\log(1) = 0  $
$\ce{pH} = 14 - \ce{pOH} = 14 - 0 = 14$
For ammonia, you need to consider its reaction in water and the associated $\ce{K_b}$ value,
$\ce{NH3 + H2O -> NH_4+ + OH-}$
The equilibrium concentrations for an unknown $x$ reacted amount (in concentration units) would be
$\ce{[NH3]} = \ce{[NH3]_0} - x$
$\ce{[OH-]} = x$
$\ce{[NH4+]} = x$
$$\ce{K_b = \frac{[NH4+][OH-]}{[NH3]} = \frac{x^2}{[NH3]_0 - x}}$$
By plugging in the proper $\ce{K_b}$ value for ammonia and its analytical concentration ($\ce{[NH3]_0}$), you can calculate $x$, from which
$$\ce{pOH} = -\log(x)$$


text:
Is the equation below for Gibbs energy only applicable under standard conditions?
This is quite confusing to me because it seems as if a lot of people use standard state equations and non standard state ones interchangeably:
$$ΔG^\circ = -nFE^\circ$$

I think the actual difficulty you have is not with the equation for Gibbs energy, but with the fact that you see standard state and non-standard state equations used interchangably. I would suggest you edit your question accordingly.

Please ask your question in the body and not just in the title. The title shall indicate the subject matter of the question (not necessarily the actual question itself) in such a way as to distinguish it from that of other questions. The actual question with all necessary context and details belongs in the body.

The equation as you wrote it refers to the standard state only. However, if you look at a system that is not at standard state, the relevant relationship looks very similar, dropping the standard state from both sides (the assumption then is that both refer to the same non-standard state that you are looking at at the moment):
$$ \Delta G = -n F E $$
The same goes for relating Gibbs energy to enthalpy and entropy:
Both
$$ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ $$
and
$$ \Delta G = \Delta H - T \Delta S $$
are correct. (Again, G, H, and S should refer to the same state in the lower equation.)
However, there are other relationships concerning Gibbs free energy where you have to be at standard state. For example,
$$ \Delta G^\circ = - RT \ln(K) $$
If you wanted a similar relationship for non-standard state, you have to account for it, e.g.
$$ \Delta G = - RT \ln\left(\frac{K}{Q}\right)$$
usually written as
$$ \Delta G = RT \ln\left(\frac{Q}{K}\right)$$

This is quite confusing to me because it seems as if a lot of people use standard state equations and non standard state ones interchangeably.

If you have a correct relationship of Gibbs energy, enthalpy or entropy that is true for any state (i.e. written without the $^\circ$), you can always write the same equation specifically for the standard state by adding the $^\circ$ to all the state functions. The reverse (just dropping the requirement for standard state) is not always possible, as shown in the example above.


text:
Why did J.J. Thomson observe protons in his cathode ray experiment?
In this MIT lecture, at 7:22, the professor says that when J.J. Thomson added a positively charged plate on one side of the cathode ray and a negatively charged plate on the other side, he observed a large deflection towards the positive plate, and a small deflection towards the negative plate (see image below).

This observation implied that hydrogen gas consists of positively charged particles with lots of mass, and negatively charged particles with a small amount of mass.
However, according to Wikipedia, cathode rays are "streams of electrons observed in vacuum tubes." Thus, my question is: in J.J. Thomson's experiment, where are the protons (positively charged particles) coming from?

You might need to reverse the polarity of the high voltage source for that to happen.

I have been wondering the same as well.  There's an anode ray experiment by Goldstein to detect positive charge but the way the professor mentions Thompson's experiment (for the positive charge part) doesn't seem to be correct

Thompson observed more than just cathode rays.
He experimented with different gases and different pressures in his tubes. When there is a significant amount of hydrogen you get ionisation of the gas and more than one type of charged particle can be generated and deflected (though different conditions are needed to see protons than electrons).
Part of the point of this was to distinguish between particles produced by the gas and particles produced when there was little or no gas. The point about cathode rays is that they can still be produced when there is virtually no gas in the tube by thermionic emission from the cathode. They have the same properties as one of the particles produced by ionising hydrogen which established some of the basis for saying that electrons are an important part of atoms.

In the diagram both the ends of battery group is shown as cathode(small line)(?)
Here a perforated anode is used and so only anions-here electrons- originated from cathode can pass through and so beyond anode only stream of electrons would emerge which registers a high deflection towards positive plate in the external electric field, due to their very small mass and so low momentum.
If the applied high DC voltage is reversed in the discharge tube( not of the external electric field), a stream of protons (hydrogen atom minus electron) would emerge out of perforation in the cathode (they originated from anode) would pass though and in the external field of the same strength as in the earlier experiment, would register less deflection due to heaviness of proton(nearly 1836 times that of electron) and so high momentum.


text:
Why are the Covalent Radii of Ruthenium and Osmium So Similar?
Osmium and Ruthenium have covalent radii 144 and 146 respectively (according to Wikipedia).
How are these values so similar when Osmium has 32 more electrons, with another filled p, s, and f orbital beyond Ruthenium?
Is it just because they have similar outer electron configurations?

It has 32 more electrons but also 32 more protons pulling on those electrons. As it turns out, these factors almost cancel out. The cancellation is particularly good in the 4d/5d blocks, because of something called the lanthanide contraction.


text:
Which has a higher lattice energy CaO or NaCl?
I know that as charge increase the lattice energy increases as well. However when the size increases, the lattice energy decreases. Can you please explain this to me? I think it is CaO but i'm not sure. Thanks!

Lattice enthalpy of an ionic solid is the minimum thermal energy required to completely convert one mole of that ionic solid to isolated gaseous ions. Here "isolated ions" in the sense that the distance between the separated ions is sufficient to reduce the electrostatic interaction between them to zero practically.
It is also the energy released for the reverse process of  settling of isolated  ions into the solid lattice.
Greater the charge densities of ions higher the electrostatic interaction and so higher the lattice energy by magnitude. A smaller anion/cation with more charges will have higher charge density. Here calcium ion (114pm)is  not only smaller than sodium ion(116pm) but also having twice the charge as that of sodium ion.
Similarly oxide ion(126pm) is much smaller than chloride ion(167pm). Former is doubly charged where as latter has only single charge. For both these reasons CaO has a much high lattice enthalpy (3414kJ/mol) compared that of NaCl (787kJ/mol)

https://chemistry.stackexchange.com/questions/104805/comparing-the-lattice-energy-of-different-compounds/104864#104864May be of use as well


text:
Is the Gibbs energy of bulk metals in a Daniell cell affected by temperature?
From literature values, it is implied that the change in Gibbs energy of a Cu-Zn voltaic cell is largely due to the difference in Gibbs energy of the bulk metals (solid Cu and solid Zn).
Does temperature affect this?
I was thinking I could use the Nernst equation for this but I don’t know if that would be appropriate.

Use Nernst equation for what exactly? You can expect temperature to affect the Gibbs energy, yes. You can still use the Nernst equation but have to perform the calculations at the temperature of interest.

Your question is a bit unclear, since the Gibbs free energy of formation of both metals is 0 by definition, since they are pure elements in standard form. The $\Delta G^\circ$ for the cell is a function of the difference in free energies of formation of the ions $\ce{Zn^2+}$ and $\ce{Cu^2+}$.
One approach you can use to determine the temperature dependence of the $\Delta G$ of the reaction is to use reported values for $\Delta H^\circ_f$ and $S^\circ$ to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for the overall reaction. If you are dealing with a small temperature range, you can assume that these values do not change appreciably with temperature and calculate $\Delta G^\circ$ at different temperatures and standard concentrations using $\Delta G^\circ_T = \Delta H^\circ - T\Delta S^\circ$. You can adjust for different concentrations with $\Delta G = \Delta G^\circ_T+RT\ln Q$. (I'm using the subscript T to indicate that the temp is different from typical standard state, but concentrations are not.)
Otherwise, you need data on the potential of the cell at different temperatures.


text:
Identifying if reaction is reduction or oxidation. Is there an error in this question?
I am looking at this reaction:

I am trying to determine if it is reduction or oxidation.
In the initial molecule I calculate the overall oxidation numbers as:
ox(Cl)*6 + ox(H)*6 + ox(C)*6 = 0
-6 + 6 +ox(c)*6 = 0
hence, ox(c) = 0

In the subsequent molecule, I calculated the oxidation numbers as:
ox(Cl)*4 + ox(H)*6 + ox(C)*6 = 0
-4 + 6 +ox(c)*6 = 0
hence, ox(c) = 6/2 = 1/3

However, my textbook says the oxidation number of ALL carbon in the second molecule is 2. I am very confused by this, since if all the carbons have an oxidation number of 2 then the molecule would be charged overall? Then I am unsure if my oxidation numbers can be fractional. Is my textbook wrong, and I wrong or are we both wrong?
Overall, what I am really trying to get at is if the molecule is being oxidized or reduced, and how many electrons are being transferred. Is there a better way of doing this?
Thank you kindly

There is nothing to calculate here. Simply seeing double bond between carbon atoms instead of bonds with chlorine should be all you need.

Consider something similar in reverse. If I gave you the product and converted the alkene into a diol, what would that be?

Look at the polarity of vicinal C-Cl bonds. Carbon is positive, chlorine is negative. In the polarized double bond, one carbon is positive the other negative. The net result is a two electron reduction of the carbon framework.

Before the reaction, every carbon is attached to to other carbon atoms, one hydrogen atom and one chlorine atom. So oxidation state of each carbon = 0+0+-1(due to C-H bond) +1(due to C-Cl bond)= 0; average oxidation state of carbon is also zero.
In the product only two carbon atoms suffer a change in bondings; now each of these two carbon atoms is attached to one carbon through double bond, another carbon through single bond and a hydrogen atom; so O.S. of this carbon = 0+0+-1 = -1; so to conclude four carbon atoms remain in zero oxidation states, where as the doubly bonded carbon atoms are in -1 oxidation state. so average oxidation state of carbon = -2/6 = -1/3
P.S.:Fractional oxidation state results only when we take average, and that too in some cases


text:
A confusion regarding gases
A gas molecule is not affected by the gravity and can move anywhere in the confined atmosphere; yet Earth's atmosphere as a whole is retained by the same gravity. To make a simili i can move my hand at my will, but cannot make a hair on my hand fall off by my will. I have a control over the body as a whole, but does not have a control over minute part. Keeping this simili aside, what could be a theoretical explanation on the basis of physical forces, i wonder; help is requested.

A gas molecule is very much affected by the gravity, just like everything else.

Explains atmospheric pressure.

..That being so weak, electrostatic interaction with other molecules give sufficient kinetic energy to move even in opposite direction, right?

Electrostatic or otherwise, that's not important. Imagine a bouncing ball: it is surely affected by gravity, yet it doesn't sit idly on the floor. Same thing here.

But bouncing ball eventually settles where as gas molecule does not...

A bouncing ball settles because its energy dissipates, that is, gets appropriated by the molecules of the surface and the ball itself. A molecule can't lose energy this way. Even if it would, other molecules would share thermal energy with it, so that to make it bounce again.

That makes the explanation perfect. :)

Related: Why don't gases escape Earth's atmosphere?

The correct resolution of your question is that gravity does affect gas molecules but not so much that it matters to a measurable extent when doing laboratory-scale experiments that don't involve weighing the gas.
On a planetary scale the attraction matters as it is what provides an atmosphere. Gas molecules are attracted to the earth and there is a gradient of concentration (which we see as atmospheric pressure) when we travel a large distance vertically. The air at the top of Everest is about 1/3 as dense as it is at sea level. But Everest is nearly 9,000m above sea level. You don't normally have to worry about that distance in a laboratory as few have ceilings >10m above the floor.
It is also easy to see that gravity affects gases by weighing a vessel containing a gas compared to the same vessel in a vacuum. The gas adds to the weight and that is how early chemists worked out how heavy different gases were, an important part of the early understanding of chemical reactions.
If you are doing calculations about gas properties you worry about the pressure and temperature. But, often, pressure is atmospheric which doesn't vary enough across a lab for gravity to be taken into account.

Possibly helpful to note that many of lighter gases have already escaped from our atmosphere.

We do have an effect of gravity.  But, as we will see, it is very small unless you are dealing with heights much larger than you are likely to find in a laboratory.
Say you have an isothermal column of nitrogen gas.  This is not a good model for the actual profiles in our atmosphere, of course, but it sets the scale of heights over which we can expect gravity to have an effect.
The equation for static pressure in the gas is the same as it would be for a liquid:
$\dfrac{dP}{dz}=-\rho g$
$P=\text{ pressure}$
$z=\text{ height}$
$\rho=\text{ density}$
$g=\text{ acceleration of gravity}$
We put in the Ideal Gas Law:
$P=\dfrac{\rho RT}{M}$
$R=\text{ gas constant}$
$T=\text{ absolute temperature}$
$M=\text{ molecular mass}$
Combining these two equations gives a differential equation for the debsity distribution which we can solve:
$\dfrac{d\rho}{dz}=-\dfrac{\rho g M}{RT}$
$\color{blue}{\rho=\rho_0\exp(-\dfrac{gMz}{RT})}$
Now put in numbers.  Remember that everything is to be is SI units, so your molecular weight may seem a little weird:
$g=9.81\text{ m/s}^2$
$M=0.0280 \text{ kg/mol}$
$R=8.31\text{ J/(mol K)}$
$T=298\text{ K}$
Then
$\rho=\rho_0\exp(-1.11×10^{-4}\text{ m}^{-1}z)$
To get the exponential argument equal to 0.05, a 5% decrease in density, you need a height of $z=450\text{ m}$, compared with $416\text{ m}$ for the height of the Willis Tower observation deck in Chicago.  So unless your laboratory workspace is as tall as the Willis Tower, you are not likely to see the effect of gravity versus the forces that drive molecular motion in gases.
To get large gravity effects, your gas column must be so tall that you would see the temperature change in the real atmosphere.  This is no surprise because gravity ultimately drives temperature as well as pressure changes in our atmosphere (at least the troposphere).  Meteorologists have to couple both effects in their models, precisely because you need that Willis Tower level of height, or more, before either temperature or pressure changes noticeably.


text:
Is it correct to say that “some degree of pollution is an inescapable consequence of the Second Law of Thermodynamics”?
The quote above is from ‘Enlightenment Now’ by Steven Pinker, I was just wondering if this was true or if it misrepresented the Second Law.
The full quote below...

...some degree of pollution is an inescapable consequence of the Second Law of Thermodynamics. When people use energy to create a zone of structure in their bodies and homes, they must create entropy elsewhere in the environment in the form of waste, pollution, and other forms of disorder

(Page 123)

Might be seen as obvious, even. It would be strange if we call Law something that doesn't hold.

Yes, this is true. However, the question is where that increase in entropy occurs.
Since the industrial revolution, life on earth is a bit one-sided, with humans controlling the chemistry going on, and a lot of it. When we burn fossil fuels, there is an increase in entropy on earth. Before that, one organism's waste was some other organisms food, and the entropy increase occurred mostly in the sun (giant nuclear reaction slowly going towards equilibrium).
Humans (and all other animals) benefit from the $\ce{O2}$ "pollution" from cyanobacteria and plants. Plants benefit - to a degree - from the $\ce{CO2}$ "pollution" from animals. However, when global concentrations change too fast, some species might benefit a lot, but many other species will go extinct.
If the "plastic ocean" had built up over millions of years instead of 50, some microorganism that uses the polymers as its primary source of food might have evolved, and there would be a cycle instead of a build-up.

It does not have to be true that 'pollution' is a consequence of the 2nd law. All the low entropy, high energy sunshine will ultimately end up as heat (high entropy, low energy), a lot of which is radiated away to allow the planet to reach a steady state. This  would not normally be called 'pollution'.
The quote is of course subjective, as is this answer, however, as a species it is certainly true that we produce waste 'stuff' and do so in enormous quantities, or rather we produce it and are then not prepared to spend money, or equivalently energy, to make further use of it. Perhaps this will change in the future if global warming restricts fossil fuel use.
The quote does have a let out-clause which is the 'other forms of disorder' as described in the first paragraph above.

As they say the devil is in the details.
The definition of "waste" or "polution" requires clarification. We might use the concept of thermodynamic efficiency, which is related to the amount of energy wasted for a specific application (which appears to be Steven Pinker's point), but should also consider the total amount of energy used, and the specific forms of waste we generate. We want to increase our efficiency, but if we insist on living in McMansions, that may not be enough. We may have found ways to produce McMansions very very efficiently, but still produce too many. In addition, our impact extends to the way we produce things (do we generate $CO_2$, do we destroy rainforests, do we polute rivers?). Nuclear power plants are very efficient, but what are we to do with the nuclear waste? Handling waste and environmental degradation also consumes energy. This requires consideration of the entire cycle of production to disposal.

I just googled “McMansions” and I’m surprised there are houses that look like that. But why does a lot of people seem to dislike them and why are they so bad for the environment?

@Alice.Sumarno McMansion is a term used to encapsulate two ideas: that of unfettered consumption, and lack of charm or originality. First, it calls into question whether we consume more than we need to (or more than would suffice to make us reasonably happy). Second, it reflects the nature of modern consumer culture, including its impresively high efficiency in terms of churning out huge amounts of identical consumer goods.


text:
Adiabatic processes and the Ideal Gas Law
in my first year chemistry textbook it wrights that in adiabatic proccess when there is a system of canister filled with gas and a piston above(and wheights on it) if we let the piston move upward by removing some of the weights the temperature of the gas will decrease since the gas has made work.
i understand that in the opposite case when we compress a gas in adiabatic proccess the temperature will increase.
**my question is: isn't it contradicts the ideal gas law which says that the volume of the gas is proportional to the temperature and vice versa
PV=nRT
so when the volume of the gas is increasing, the temperature should increase.
also by intuition: if the volume is increased, there is less pressure , the molecules has now more space and theire kinnetic energy should increase and therefore also the temperature.
how can i look at this?
thanks
 **

I think you should consider the consequences of the word "adiabatic" in more detail and come back to edit your question with that information.

Why is this question classified as off topic?  I see it as on topic

No, it doesn't contradict the ideal gas law at all. When you are saying that rise in volume means increase of temperature, you are inherently assuming that Pressure remains constant. But that is not the case at all in an adiabetic expansion.
In fact, you should realise that it is the product $PV$ which is decreasing. So, Volume might increase, but pressure drops even faster than volume, that's why the product $PV$ overall decreases. If some of the weights are removed, the gas expands slowly according to the equation, $PV^{\gamma} = C$. For most of the gases $\gamma > 1  .(\gamma = \frac{C_p}{C_V})$ Thus by simple observation, you can see if Volume increases $x$ times, Pressure will decrease $x^{\gamma}$ times, which is more dominating in the product.Thus, the product $PV$ will definitely decrease with increase in volume in case of an adiabetic process.
Also, your second argument is also not acceptable, because you already realise that pressure will decrease and by Kinetic theory of ideal gases $PV = \frac{1}{3}m\ n\ v^2_{rms}$. Thus if you understand the decrease in the product $PV$, you will also understand that $v^2_{rms}$ will also decrease, and hence the kinetic energy of gas molecules also.
Also you are thinking in a much more complicated way towards the situation. Think simply. For ideal gases, internal energy ($U$) is a measure of the translational kinetic energy of gas molecules and also $U = \frac{3}{2}RT$ depicting that for ideal gases, kinetic energy is only a function of temperature only. Also, by the first law of thermodynamics, $dU = \delta q + \delta W$ ( $\delta $ here represents inexact differentials). For adiabetic process, $\delta q =0$. So, $dU = \delta W$. Now, $\delta W$ is work done on the system which is negative for an adiabetic expansion, which makes $dU$ also negative, i.e. internal energy decreases and so does temperature.

In an adiabatic expansion, if the volume increases, the pressure decreases, but the pressure decrease (ratio) is more than the volume increase (ratio), so the temperature also decreases to make good on the ideal gas law.  If the volume decreases, the pressure increases, but the pressure increase(ratio) is more than the volume decrease (ratio), so the temperature also increases, to make good on the ideal gas law.

Under adiabatic conditions, heat evolved is zero. Considering first law of thermodynamics, any work performed by the gas or on it would involve expenditure form its internal energy.
If the gas expands, it is doing negative work, hence change of internal energy is negative. So $T(initial)> T(final)$.
Ideal gas law is infact a summation of Boyle's, Charle's and Avogrado's laws which describes the state in which the gas exist. Under adiabatic conditions, pressure changes along with volume and temperature. So with increasing volume, $T/P$ also increases. The extent of decreasing $P$ is greater than decreasing $T$, due to the factor $γ$.
$$P^{1-\gamma}T^\gamma=\text{constant}$$

if the volume is increased, there is less pressure , the molecules has now more space and theire kinnetic energy should increase and therefore also the temperature.

If you have a box divided into two parts, and one part is filled with an ideal gas, the particles will have some average kinetic energy. If you now remove the divider, the particles have more room available to them, but they will not speed up. Instead, the pressure will decrease because less particles will hit a unit area of the wall in some time unit (because there is more space, it will take longer for an individual particle to make a round trip hitting the wall once and then hitting it again).

if we let the piston move upward by removing some of the weights the temperature of the gas will decrease since the gas has made work.

When a particle hits a stationary wall, it bounces off without a change in kinetic energy. If it bounces off a wall that can move, it slows down because some momentum is transferred to move the wall. Try throwing a tennis ball against a concrete wall vs against the side of a cardboard box. (Or take a look how one bowling ball is stopped by another.)

the ideal gas law which says that the volume of the gas is proportional to the temperature and vice versa

The ideal gas law says that V and T are proportional if everything else (n and P) is constant. However, by removing some pebbles from the piston, you change the pressure the piston exerts on the gas, and once the system is back in equilibrium, the pressure inside the canister will also be lower. So V increases as P decreases. In an isothermal process, T would stay constant (particles regain lost kinetic energy from the surrounding) and with it PV. In an adiabatic process, T decreases as well, and you have to look at the specific parameters to calculate how much they change.


text:
which one is the correct statement of "Law of definite proportion"?
according to my genius Teacher ' this law state that "the composition of a compound in terms of atoms (Not Mass) remains constant".
His Proof:
H2O and D2O are similar compounds (because H and D have same atomic number)...Now the story start from here--->
The ratio by mass for H2O= 1:8, and for D2O= 1:4 (both have different ratios, although they are similar compounds- so This law shound not be defined on the basis of mass)
Now, Ratio by atom for H2O= 2:1 and for D2O= 2:1 (both give same results, so this law should be defined on the basis of No.of atoms)...End of Proof
Is the problem lie in similar compounds or my teacher is correct and the world is wrong?

The Law of Definite Proportions was defined before we knew about isotopes. Sure, you can give a modern version of it, but that's not the original version.
If you have no notion of isotopes, then the mass is just a constant multiple (specifically, the atomic mass) of the number of atoms, in which case, the two definitions become equivalent.


text:
What would happen if we had a different kind of metal in our bloodstream?
It is a common fact that we have iron in our blood. But what if we had something different in its place? Like instead of iron what if it was copper, or silver, or what if we had gold in our blood stream instead? Theoretically, what would that do?

If we would use copper to transport oxygen in the blood, we would be more closely related to horseshoe crabs.

Iron in blood is a part of heme group in haemoglobin. Heme is a complex molecule with a porphyrin ring attached to a iron center. Several other such molecules with similar structures are also there, Chlorophyll a has a similar structure with Magnesium at the centre of the Porphyrin ring.
Researching on chlorophyll a will give you an idea.

Instead of one Fe(II) atom like in our heme, the carrier hemocyanin in horseshoe crabs has two bound copper ions to bind to a single dioxygen.

The whole biological system of a human is finely tuned. You can't just willy-nilly substitute some other metal for iron in hemoglobin.

Assuming that you could construct a recombinant protein that was not rejected by the immune system, and used some metal other than iron in its oxygen-binding cofactor, it makes for an interesting if bizarre question. Afaik release of oxygen is passive (ie does not require another protein, even if it's complex) so if you tuned the protein's $CO_2$ and oxygen binding constants just right, why well, I don't know, it might just work. Question is, why don't we have a copper-loaded oxygen carrier? Just because?

A better way to ask this question would be not to assume we could replace iron with something else in our biology but to ask whether alternative systems for carrying oxygen are possible and what the implications would be. Our blood is fine-tuned to use iron, but alternatives are possible.

Don't ask me about the crabs but Cu will be surely toxic somewhere else in our organism. I like the answer by PJR.

This is an interesting question, it is a pity that it is closed. Students in the future will need to tackle questions such as this so they could benefit from some thoughtful answers from people with some knowledge of the area.

To start, transition metal complexes with large porphyrin-like molecules are so electronically complex that standard chemical intuition doesn't apply and you have to either do the experiment or run a (nearly impossible) calculation to determine what effect a non-physiological metal would have on the protein activities. In the case of hemoglobin, your best bet would be to think about metals which have two accessible oxidation states because the interconversion of Fe(II) and Fe(III) is critical to the function of hemoglobin.
To give a broader answer, let's think about the general question of substituting non-native metals into a protein. This is, in fact, an very interesting and open question. On smaller metalloproteins there have been studies probing whether nature 'picked' the best metal for a task or just a convenient/abundant one [1,2]. The focus of those studies is on catalysis, and there are cases where many non-physiological metals can function in the protein and even catalyze the reaction as good or even better than the wild type.
[1] https://pubs.acs.org/doi/10.1021/ar500227u
[2] https://pubs.acs.org/doi/abs/10.1021/acs.jpcb.6b00997


text:
Dioxin synthesis in a rocket stove riser?
I'm building a rocket stove for cooking with a wok at the top of the riser.  At the top of the riser I'm seeing temperatures of 500-700 deg C, and I believe at the bottom it's likely under 1000 deg C ( though others with optimised builds see temperatures closer to 1100 deg C ).  The firebox is comprised of firebricks with the roof and floor lined with ceramic fiber blanket, and the manifold, and riser are all lined with ceramic fibre blanket.

I gather that in the combustion of wood ( biofuels ) you can see de novo synthesis of dioxins and furans in two temperature windows 200-400 deg C, and 600-800 deg C with the latter being much less.
One sees more efficient combustion with the introduction of heated secondary air into the riser, but at these temperatures steel spalls and dies.  Steel pipes need to be replaced every year.
I have placed a copper pipe into the combustion chamber to provide secondary air and threaded it to the riser.  It didn't melt.
So, my question is, does copper help in catalysing the formation of dioxins ( catalytic stoves increase dioxin formation ) and does it matter at temperatures at the bottom of the riser since they should be destroyed?
Edit:

Copper (Cu) is the most potent catalyst for dioxin formation, but Iron (Fe), Zinc (Zn), Potassium (K) and Sodium (Na) have also been found in multiple studies to be correlated with increased dioxin/furan formation. Some studies have also indicated that Manganese (Mn), Magnesium (Mg) and Nickel (Ni) may also serve as catalysts for dioxin formation. Studies have conflicted on whether Aluminum (Al) encourages or inhibits dioxin formation. One study below indicated that Silicon (Si) is negatively correlated with dioxin formation.

https://www.ejnet.org/dioxin/catalysts.html

The formation of dioxins depends a lot of what fuel you are using. Wet wood is the worst (sap contains a lot of chlorine) Dry wood is better. But if you are not using wood but purer fuels like charcoal or LPG there won't be much to worry about.

I use mainly HT pallet wood.


text:
How to distinguish two molecules with same formula but different geometry with vibrational spectroscopy
There is a sample that we know is all made up by only one of these two molecule:

how you can see they have the same formula but two different geometries. The question is if it is possible to know by which of these two molecules the sample is made with an experiment involving vibrational spectroscopy (IR or Raman). The following is a quite wide answer I gave to myself but I'm not sure it is right, so please correct me or confirm what I have written.
We know that if we are considering small oscillations, the vibrational part of the wave function of a molecule with $N$ atoms is the product of $3N-6$ quantum armonic oscillators eigenfunctions, and we also know that in a classic way to proceed the nuclei can move in $3N-6$ differents normal modes. Fixed one of these normal modes, if it is for example IR active, we know that all the $3N-6$ oscillators moving in that normal mode can absorb photons from the external radiation. So if the external radiation has a frequency band enough large to contain all the proper frequencies of the oscillators, we'll notice $3N-6$ absorbtion lines. While if we are considering a not IR active normal mode of course we won't see any absorbtion lines. In general the nuclei of a molecule move on a linear combination of all the normal modes. So we can think that each IR active normal mode of the combination gives the same $3N-6$ lines while each not IR active mode gives zero lines. So I would expect to observe for my real sample 12 absorbtion lines (I am not considering overtones and the noise of the media) which correspond to the proper frequencies of the oscillators. So since the two molecule have exactly the same oscillators (it only changes their position), I should observe the same 12 lines on both cases, so I'd say that it is not possible to distinguish between the two. The same argument can be used for the Raman. However I'm almost sure that I'm wrong and a way there must exists.

see https://www.ideals.illinois.edu/handle/2142/79485

Basically your question is how to identify wich modes of a molecules are IR or Raman active. It seems you are aware. The point is that you wrote "have the same armonic oscillators" which is not true even at a first glance as you have already realised if you have followed the link given above. After that you see how to discern between the two in finer details.

The starting point is to look at the potential point groups of the two molecules and work out if there are common vibrational mode symmetries between IR and Raman spectra, assuming that you are able to obtain both. In the 3rd and 4th columns of the point group are the $x, \,y, \,z$ and $xy, \,xz$ etc operators. The $x, \,y, \,z$ operators transform as does a dipole (so IR transitions) and  act as an alias for the symmetry species.
For example in the $C_{2v}$ point group each  IR transition must be one of $A_1,\,B_1 ,\, B_2$ species and as there are 12 transitions  there must be several of the same symmetry but different frequency.
The Raman transitions correspond to product terms , $x^2,\, xy$ etc. (because Raman spectroscopy depends of the polarisability which is proportional volume and so area in cross-section ) and these have also $A_1,\,B_1 ,\, B_2$ symmetry species as well as $A_2$ species (in $C_{2v}$) so we do expect to see some different transition frequencies between IR and Raman spectra. We do not necessarily know what the symmetry species are for these transitions, just that there is a difference.
In the case of $D_{2h}$ point group as there is a centre of inversion, there are no common transitions between IR and Raman spectra. Look at the point group and you will see this.
In more detailed analysis the polarisation of Raman signals can be used to determine which transition is $A_1$ etc. Spectroscopy books give details of this.
look at Understanding group theory easily and quickly for an explanation of point groups.


text:
Using Hess's law to find molar enthalpy of formation without an initial reaction
I'm having issues relating Hess's law to this question:

The given chemical equation represent the combustion of ammonia and the combustion of hydrogen
$$
\begin{align}
\ce{4 NH3 + 3 O2 &->6 H2O + 2N2} &\qquad &ΔH_1 = \pu{-1516 kJ} \tag{1} \\
\ce{2 H2 + O2 &-> 2 H2O} &\qquad &ΔH_2 = \pu{-572 kJ} \tag{2} \\
\end{align}
$$
What is the molar enthalpy of formation for ammonia?
A. $\pu{-100 kJ}$; B. $\pu{-50 kJ}$; C. $\pu{50 kJ}$; D. $\pu{100 kJ}$.

I'm not sure how I would apply the  product–reactant equation here, or rearrange the equation to get my answer since I don't have an initial reaction to compare it to. How would I go about answering this question?

There is all the info you need. I suggest you start with writing an actual equation for the synthesis of ammonia and then use linear combination of the provided equations.

Hold on a sec, that's for 2 mols of $\ce{NH3}$, I think you forgot to divide it in half (it should be around -50 kJ/mol):)

Are you at least given the heart of combustion?

@user9988, be sure to update the question with your approach to a solution, to make this question less homework-y.

Since you are asked to find the enthalpy of formation for ammonia, it's convenient to write the equation of ammonia synthesis normalized for $\pu{1 mol}$ of $\ce{NH3}$:
$$\ce{0.5 N2 + 1.5 H2 -> NH3}$$
To obtain this equation and apply Hess's law, the following linear combination of the provided equations should be used:
$$
\begin{align}
\ce{4 NH3 + 3 O2 &->6 H2O + 2N2} & ΔH_1 &= \pu{-1516 kJ} &&|\cdot(-0.25) \tag{1} \\
\ce{2 H2 + O2 &-> 2 H2O} & ΔH_2 &= \pu{-572 kJ} &&|\cdot 0.75 \tag{2} \\
\hline
\ce{0.5 N2 + 1.5 H2 &-> NH3} & ΔH_3 &= -0.25ΔH_1 + 0.75ΔH_2 \tag{3}
\end{align}
$$
So the final answer is
$$ΔH_3 = -0.25\cdot(\pu{-1516 kJ}) + 0.75\cdot(\pu{-572 kJ}) = \pu{-50 kJ}$$
The catch here is that you are asked to find molar enthalpy of formation, and calculating for the "convenient" equation with integer coefficients
$$\ce{N2 + 3 H2 -> 2 NH3}$$
you eventually would obtain the doubled value of $\pu{-100 kJ mol-1}$ referred to $\pu{2 mol}$ of ammonia (I almost fell for it, too).


text:
Calculating ΔH from bomb calorimetry
Suppose we carry out a reaction in a bomb calorimeter whose starting temperature is $298.15\ \mathrm K$. Here we assume $\Delta V$ is close enough to zero that we consider the process to be at constant volume.
Based on $\Delta T$ and the calorimeter constant, we determine the internal energy change for the reaction at constant $T$ and $V$, which I'll call $\Delta U_\mathrm r$. (We can assume constant $T$ because we know how much heat flow would be required to return the calorimeter to its starting temperature.)
A typical pchem textbook will then say you can calculate what $\Delta H$ would be if the reaction were instead carried out at the same temperature, but at a constant external pressure of $1\ \mathrm{bar}$, namely $\Delta H_\mathrm r^⦵$, from the $\Delta U$ we determined in the calorimeter, as follows:
$$\Delta H_\mathrm r^⦵=\Delta U_\mathrm r^⦵+\Delta pV=\Delta U_\mathrm r^⦵+p\,\Delta V\approx\Delta U_\mathrm r^⦵+\Delta n_\text{gas}RT$$
But: we didn't determine $\Delta U$ at standard state ($1\ \mathrm{bar}$).  We determined it at constant volume. Thus, strictly speaking, when we do the following calculation using the $\Delta U$ from the calorimeter
$$\Delta H_\mathrm r^⦵\approx\Delta U_\mathrm r^⦵+\Delta nRT$$
aren't we additionally making the approximation that $\Delta U_\mathrm  r=\Delta U_\mathrm r^⦵$? I.e., aren't we ignoring the volume-dependence of $U$? Typically, this is a very good approximation since, for most substances, $U$ is only weakly dependent upon $V$. But it seems it's an approximation nonetheless. If so, it is one that is not (but should be) explicitly acknowledged in these textbooks.
There are other approximations, of course, such as the one shown above, that $p\Delta V\approx\Delta n_\text{gas}RT$ (ignore volume change in liquids and solids, and use ideal gas law to calculate volume change due to change in number of moles of gas), but I'm specifically wondering about this one.

I figured out the answer to my question.  Essentially, the approximation is not significant because it is smaller in magnitude than other approximations that are being made.  I wrote up the details for my class, but since misplaced the notes.  If I find them, or decide to reconstruct them, I'll post it here.

To quote Sime's textbook on Physical Chemistry, experiment 3 on the measurement of heats of combustion of hydrocarbons using bomb calorimetry: "we will assume that $\Delta U = \Delta U_r^o$; that is, we will assume that $\Delta U$ is independent of pressure". There is no further justification of the statement. The Parr Instrument Company provides an introductory guide to bomb calorimetry (document 483M) which has the following to say:

Users interested in high precision calorimetry
will understand that test results obtained in a bomb
calorimeter express calorific values obtained at constant
volume. This differs from the usual conditions
under which fuels are burned at constant pressure
rather than constant volume. If the sample contains
elements which normally would increase in volume
after combustion, a correction must be applied to
account for the heat equivalent of this work. This
value is so small as to be negligible in routine work,
and it usually is disregarded.

The required pressure correction can be computed from thermal expansion and isothermal compressibility coefficients. For instance, for water one estimates that if the bomb pressure rises by $\pu{2 GPa}$ over standard pressure then the energy will increase by $\approx \pu{80 J/mol}$. Compare this correction with the heat of combustion of benzoic acid ($\approx \pu{3221 kJ/mol}$) and (assuming the coefficients for benzoic acid are of the same order of magnitude) it's clear that for condensed substances the correction is (often) negligible.
For gases, based on the assumption of ideal gas behavior, an isothermal PV change for the products will leave $\Delta U$ unchanged, so that in fact $\Delta U_r = \Delta U_r^o$.

I'm afraid that doesn't address my question.  I'm not asking if the assumption that $\text{ΔU}_r=\text{ΔU}_r^⦵$ can be made, I'm asking if it is being made.  In the case of ideal gases, U is always independent of V, so that assumption can always be made.

@theorist I looked my answer over and don't fully understand how this fails to answer your question. The comment you left promising a future answer doesn't seem to go far beyond what I have stated.

An answer requires doing a derivation that formally demonstrates why the  correction is negligible on a general basis, relative to other assumptions that are being made.  That's what I did for my class.


text:
What is the molar enthalpy of reaction
When 0.45 g of Zn is added to 50.0 ml of 0.95 M HCl solution, the solution inside the calorimeter heats up by 12 °C. What is the molar enthalpy of the reaction (in kJ/mol)?
I know how to find the moles using $n=cV$ and $n=m/M$ but I'm not sure I understand why when doing $mc\Delta T$ we use the specific heat capacity of water when the solution is HCl.

Welcome to Chem SE! If you haven't yet, take the [tour] and visit the [help]. A solution of HCl is not liquid HCl, but rather HCl dissolved in water, so for a dilute enough solution you can use the heat capacity of pure water.

Amount of $\ce{Zn}$: $$n_{\ce{Zn}}=\frac{0.45\ \mathrm g}{65\ \mathrm{ g/mol}}=0.0069\ \mathrm{mol}$$
Amount of $\ce{HCl}$: $$n_{\ce{Zn}}=MV=0.95\ \mathrm{mol\ l^{-1}}\times\frac{50\ \mathrm{ml}}{1000\ \mathrm{ml/l}}=0.0475\ \mathrm{mol}$$
The reaction $$\ce{Zn + 2HCl -> ZnCl2 + H2}$$
So $0.0069\ \mathrm{mol}$ zinc requires $2\times0.0069\ \mathrm{mol}=0.0138\ \mathrm{mol}$ HCl; thus $\ce{Zn}$ is the limiting reagent here which decides the extent of the reaction.
Approximating specific heat capacity of solution to be specific heat capacity of water: $c_p=4.2\ \mathrm{J\ ^\circ C^{-1}\ g^{-1}}$.
Net heat released: $$Q=mc_p(T_2-T_1)=50\ \mathrm g\times4.2\ \mathrm{J\ ^\circ C^{-1}\ g^{-1}}\times12\ \mathrm{^\circ C}=2520\ \mathrm J$$
So heat release per amount of zinc: $$Q=\frac{2520\ \mathrm J}{0.0069\ \mathrm{mol}}=365217\ \mathrm{J/mol}=365\ \mathrm{kJ/mol}$$


text:
Percent hydrogen and carbon

A 1 g sample of an organic compound was burned in a combustion analyzer and yielded 1.42 g carbon dioxide and 0.872 g of water. Calculate the percent carbon and hydrogen in the sample. From the percentages, calculate the empirical formula for the compound.

Your initial compound weights 1g, and it is only made of C and H. The products are CO2 and H2O. In other words (the reaction is not balanced, but for our purpose we don't care):
CxHy + O2 --> CO2 + H2O
From the grams of CO2 you can get back to the moles (and therefore grams) of C in the initial compound, while from the grams of water you can get to the moles (and grams) of H.
moles of C = 1.42 g CO2 x $\frac{\text{1 mol CO2}}{\text{44.0 g CO2}}$  x $\frac{\text{1 mol C}}{\text{1 mol CO2}}$ = 0.0323 mol C
You then multiply by the molar mass of carbon, which is 12 g/mol, and you get 0.388 g.
We do the same thing for hydrogen:
moles of C = 0.872 g H2O x $\frac{\text{1 mol H2O}}{\text{18.0 g H2O}}$  x $\frac{\text{2 mol H}}{\text{1 mol H2O}}$  = 0.0969 mol H = 0.0989 g
Note that we did the same thing twice, but since we have two moles of H in water we multiplied by 2 in the second equation.
Now we just have to find the percentage:
You get
$\frac{\text{0.388 g}}{\text{1 g sample}}$ *100 % =  38.8 % C
$\frac{\text{0.0989 g}}{\text{1 g sample}}$ *100 % =  9.89 % H
I hope this helps!

Since the percentages don’t add up to 1, is there oxygen in the compound as well?

Yes, there are some other elements for sure (oxygen, or maybe nitrogen)

Would the empirical formula be CH3?

No, CH3 would be part of the empirical formula, but we don't know the complete one since we don't have all the elements :)

Mass of carbon =(12/44)X1.42=0.39g
So percent mass of carbon= (0.39/1)X100 = 39
Mass of hydrogen = (2/18)X0.872 =0.097g
So percent mass of hydrogen = (0.097/1)X100 = 9.7
P.S.: Please always attempt such simple questions at home; you are sure to get the answer, no?

Here we cannot conclude about the empirical formula since there are possibly one or more other  element(s).

Thank you! That’s what I thought as well. This was a question on my homework. Since the percentages I calculated didn’t add to 1 I thought I was calculating the percentages incorrectly.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Finding percentage of the gas in a binary gaseous mixture of the given density

At STP, the density of a gas in a vessel is $0.9002$. If the gas is a mixture of argon and helium, what percentage of the gas is argon?

I am stuck on this. From what I can gather, the only influencing characters would be moles and grams. This idea is based of off of STP and given constants.
I have a couple equations written down but I can’t seem to the flow going.
R, T, P are all known along with acccording densities. I also want to say $22.4$ is also a known at $V_\mathrm{tot}$ and therefore $n_\mathrm{tot} = 1$.
$$n(\ce{Ar}) + n(\ce{He}) = 1$$
$$\frac{n(\ce{Ar})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{Ar}) + \frac{n(\ce{He})}{n(\ce{Ar}) + n(\ce{He})}d(\ce{He}) = d_\mathrm{tot}$$
$$d = PM/RT$$

Well, I can see all the relations you require knowledge of in the question itself!
Calculate the effective molecular mass from
$$M = \frac{dRT}{P}$$
$R$ is known, $d$ given. $P$ and $T$ are available from the fact that it is at STP.
$M$ comes out to be ${20.176 u}$.
Effective molar mass is easily calculated below:
$$M_\mathrm{eq} = M_1x_1 + M_2x_2$$
where $x_i$ is the mole fraction of each gas. You can replace $x_2$ by $1-x_1$.
Plugging in the values and solving for $x_1$, argon mole fraction comes out to be $0.4493$.
I believe you can now calculate the demanded percentage, be it by mass or by moles.

This seems to be the correct answer. The only thing I'd probably add for clarity is that the density given is the relative density $d = ρ/ρ^\circ$, where $ρ^\circ = ρ(\ce{H2O},\pu{4 °C})$ (usually).

This is correct if I reference the above. This is also another way that I will be analyze do to the time saving nature.

I'd start the solution with this equation:
$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$
where I emphasize that the density of the gas is a molar average value, and rewrite it as
$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$
where $\chi_{Ar} = n_{Ar}/n$ is the mole fraction of argon in the gas mixture and the $d_{i}$ are densities computed assuming all of the gas corresponds to He or Ar, for instance
$d_{Ar} = M_{Ar}n/V  = M_{Ar}/V_{m}$
where $V_{m}$ is the molar volume at STP (22.414 $m^3/kg mol$).
It follows that
$ d_{avg} = \chi_{Ar}M_{Ar}/V_{m} + (1-\chi_{Ar})M_{He}/V_{m} = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})/V_{m} = M_{avg}/V_{m} $
which can be solved for $ \chi_{Ar} $:
$ \chi_{Ar} = ((d_{avg}V_{m})-M_{He})/(M_{Ar}-M_{He})$
Finally, the molar percentage argon is computed as $f_{Ar}=100\chi_{Ar}$.
For your particular problem I get $f_{Ar}$= 45.00%.

Please check your math; I suspect there shouldn't be any quadratic equations. Also, if you check the tabulated values of the gas densities at STP ($ρ(\ce{He}) = \pu{0.179 g L-1}$; $ρ(\ce{Ar}) = \pu{1.784 g L-1}$, it's obvious that they should be in a roughly 1:1 ratio in this mix. In fact, it's about 55% helium and 45% argon.

@andselisk Yeah that quadratic solution didn't make much sense. Hopefully I ironed this out now.

Okay this makes sense. I had the right idea a few times but I got lost in the algebra. The key piece of information in your answer that would have allowed me to figure the problem out is the Vm. Additionally, I didn’t condense the ratio into terms of x. So I had nHe = 1 - nAr plugged into the original equation you started with. That is where I got lost in the algebra and as I had mentioned, the value of Vm that I didn’t realize.

@William R. Ebenezer got the answer right before I did but I thought since I was already on the board to clean up mine and leave it.


text:
Oxidation states of boron
I have just been looking at      https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements and found that boron has a -5 oxidation state. I would like to know which boron compounds form this oxidation state.

As I remember I have read somewhere that $\ce{Al3BC}$ is konwn to have Boron in -5 oxidation state.

you can also reffer to this PDF page no. 139
PDF IN GERMAN

You are correct because (3*3)-5-4=0

It isn't just boron.  The same Wikipedia list gives a -5 oxidation state for all the Group 13 elements except nihonium (which is only very recently known) and, more curiously, aluminum.  For thallium Wikipedia's article on the element gives this reference, where a compound has thallium in various clusters and individual atoms bonded to alkali metals.  The individual atoms at least formally have five negative charges.

Except for Al, seemingly.

Said that, read again.


text:
How is 2-hydroxy-ethoxy named?
2-hydroxy-ethoxy is $\ce{HO-CH2-CH2-O}$.
I have a couple of confusions regarding this substituent.
Ethoxy is $\ce{-HO-CH2CH2}$ or $\ce{-OC2H5}$. So the $\ce{HO}$ in the above formula probably belongs to ethoxy.
Hydroxy (prefix for alcohol group) is $\ce{-OH}$. But if $\ce{HO}$ belongs to ethoxy, from where does hydroxy get its hydrogen?

2-hydroxy-ethoxy is not the name of a molecule but the name of a radical (which usually means there is something else attached to the oxygen as illustrated in nandakumar's answer). Understanding this clears up some of the confusion.

Ethoxy can actually only be $\ce{—O–CH2–CH3}$ and not that impossible $\ce{–HO–CH2CH2–}$ structure (which would require the oxygen to bear a positive charge and the last carbon a negative one).
The "2-hydroxy" derivative is then just replacing one hydrogen atom on $\ce{CH3}$ by a hydroxy ($\ce{OH}$) group, hence $\ce{HO–CH2CH2–O -}$.

This group is called -(2-hydroxyethoxy)
e.g.:


text:
Impact of Pressure on equilibrium of Tin allotropes

The pressure at the bottom of the Mariana Trench in the Pacific Ocean is $1090$ bar. What temperature will the two allotropes of tin be at equilibrium? Assume that the molar volume, energy, and entropy change does not vary with temperature.
Relevant data: Density of white tin: $7.287$ g/mL; Density of grey tin: $5.766$ g/mL
At $1$ bar: The enthalpy change from white tin to grey tin is $-2.016$ kJ/mol and the entropy change is $-7.04$ J/mol K.

My attempt: Since the two species are in equilibrium, I thought first to use the Clausius Clapeyron equation. Assuming that the molar volume is independent with temperature, I calculated the molar volume change to be $4.26\cdot 10^{-6} \frac{\pu{m^3}}{\pu{mol}}$ from white tin to grey tin. Now, I substituted these values into the equation to find that $$\frac{dP}{dT}=\frac{\Delta S}{T\Delta V} =-\frac{1.6\cdot 10^7}{T}$$
upon integration, I find that $$P-P^\circ = 1.6\cdot 10^7 \ln(\frac {T^\circ}{T}).$$
Substituting the temperature of equilibrium at one bar from the data given I find that the temperature is $286.4 \pu{ K},$ thus $T^\circ = 286.4\pu{K}$ and $P^\circ = 10^5\pu{Pa}.$ Thus at $P = 1090\cdot 10^5\pu{Pa},$ $T = 0.317\pu{K}.$ However, the temperature given in the answer is much higher ($220.35$ K).  The given answer was derived from a completely different method not involving the Clausius Clapeyron equation.
Hence I have the following question:
Question: Why is it not valid to use the Clausius Clapeyron equation in this scenario? (Or did I make a mistake in my derivation that caused my values to be off?)

One problem with your approach is that the C-C equation is $\frac{dP}{dt}=\frac{\Delta S}{\Delta V}=\frac{\Delta H}{T\Delta V}$. You've mixed up $\Delta S$ and $\Delta H$.

First (as you already did) you need to determine the equilibrium temperature at $1\ \mathrm{bar}$. This happens to be when  $\Delta G^\circ = 0$, so that $T^\circ_{trans}= \frac{\Delta H^\circ}{\Delta S^\circ}=286.36\ \mathrm K$.
Next you can apply the Clausius equation in integrated form
$$ T = T^\circ_{trans} + (P-P^o) \frac{\Delta V_m^\circ}{\Delta S^\circ} $$
I obtain the following value for $\Delta V_m^\circ$:
$\Delta V_m^\circ = (1/5.766-1/7.287)\times \pu{ 118.71\times10^-6 m3/mol} = \pu{4.2973\times10^-6 m3/mol}$
Plugging this and the values you provide for $\Delta P$ ($1089\ \mathrm{bar}$) and the standard entropy changes $\Delta S_m^\circ$ gives $T=\pu{219.89 K}$. Not exactly what you say is the right answer, but, closer.


text:
Methanol vs water. Acidic strength
Why $CH_3OH$ is more acidic than $H_2O$ while the same is opposite for other alcohols?

https://chemistry.stackexchange.com/questions/57986/is-methanol-really-more-acidic-than-water

The question is based on the assumption, that water has a pKa of 15.7 which is usally not correct.
As a paper mentioned in this thread What is the Real Pka of WATER the pKa of water in water should be 14.It might be different when water is a reactend in organic solvents, for that the paper mentioned by the thread author is helpful.

Like other alcohols Methanol is also a weaker acid than water; sodium methoxide undergo easy hydrolysis with water forming methanol and sodium hydroxide, the equilibrium being more in favour of the products. Since methoxide ion is a stronger base than hydroxide the parent acid methanol must be weaker acid than water
$\ce{CH3O^-Na^+ + HOH<=>CH3OH + Na^+OH^-}$


text:
Is the titration of sodium carbonate with hydrochloric acid temperature dependent?
I am a chemistry teacher in a Cambridge school.
In one of the titrations of sodium carbonate with HCl, there was one question asked:

On heating the sodium carbonate solution before doing the titration, is there any change in the observation/reading?

My opinion is that the rate of reaction will increase due to more collisions among particles due to the temperature change. So the time to completion of the titration should be shortened.
However, the CIE mark scheme suggests that there is no effect, as the concentration remains same.

For all practical purposes, this reaction has no rate. It just occurs instantly.

When in doubt, perform the experiment. Or are you looking for a theoretical explanation of the effect you are asking about?

No because bicarbonate isn't involved in the titration. This is what the question is supposed to check

I think you might have a different understanding of what constitutes a 'change in the observation/reading' from the textbook, resulting in your confusion.
I am assuming that the textbook is only considering the titration graph per se, pH vs volume of acid added.
You are referring to the time to pH equilibrium during each drop of acid added. For the purposes of all titrations I have encountered, this was never a factor that was discussed. As @Ivan Neretin commented, the reaction is assumed to occur instantly.
As the concentration of the species remains the same, the titration graph will look the same. In other words: The temperature has no influence on the equivalence point.

With respect to the reaction rate, the rate-limiting step of a titration is typically the mixing of the bulk solution. This is easy to see if you have an indicator dye in the solution. When you add a drop of the acid solution, there will be an immediate local coloring indicative of low pH (also demonstrating, as others said, that the proton transfer is nearly instantaneous). As the solution mixes, the dye returns to colorless (or to whatever the appropriate color for the bulk solution pH). If you aren't actively stirring the solution, this will take much longer than if you have it stirring.
If the "observation/reading" of the experiment is simply the amount of acid required to neutralize the carbonate, then that outcome is unchanged (and presumably that is what the question referred to). However, if the "observation/reading" includes the pKa of the carbonate, that observation will be temperature dependent, although perhaps not detectable within the limited temperature range you are working, depending on the accuracy with which you can measure pH.
Table 9.1 of this paper gives the temperature dependence of both $K_a$ values for carbonate.

"My opinion is that the rate of reaction will increase due to more collisions among particles due to the temperature change"

Yes, when the temperature rises, the collision fraction which is equal to or greater than the activation energy ($E_\mathrm{a}$) will be increased.This will increase the rate of the reaction.
If the activation energy is very low, close to zero, as for ionic reactions such as precipitation reactions and acid-base neutralization reactions, the fraction of effective collisions does not affect by the temperature rise just with very little value.
The **Arrhenius equation **describes the mathematical relationship between the rate constant $(k)$ and temperature $(T)$ as follows:
$$k = A \mathrm{ e^{-E_\mathrm a/(R T)}}$$
 Where :
$R:$general  gases constant
$e:$ The basis of natural logarithm $~\mathrm {e}=2.718$
$A: $Pre-exponential Factor, the value of this term changes slightly with temperature, but can be considered constant.
The other part of the equation $\mathrm{ e^{-E_\mathrm a/(R T)}} $ is a fraction which represents the ratio of effective collisions and increases with higher temperature.
The value of this fraction is greater for reactions of low activation energies because a greater proportion of collisions have an energy equivalent to activation energy or superiority, Thus, increasing the proportion of effective collisions, and increasing the value of rate constant$ k$.
The value of this fraction $ \mathrm{ e^{-E_\mathrm a/(R T)}}$ is very small and much less than one for most reactions.
 When the activation energy ($E_\mathrm{a}$) approaches zero, the value of this fracture approximates the true one as is the case for ionic reactions such as precipitation reactions and acid-base neutralization reactions. In this case, all oriented collisions lead to reactions, so:$\color{red}{\text{ The rate constant value$~ k$ is very high, and the reaction rate is not affected by increasing the temperature .}}$

I can not comment because I do not have 50 reputation yet so I post this as an answer even thought it is just an estimated guess.
In a titration the reaction should be quite fast so you the result imidiatly. Increasing the temperature would not change much.
I think that the Sodium carbonate in Water will form hydrogen carbonate.
$$\ce{Na2CO3_{(aq)} + H2O_{(l)} <=> 2Na+_{(aq)} + OH-_{(aq)} + HCO3-_{(aq)}}$$
When heating $\ce{HCO3-}$ it seems resonable, that CO2 is evaporating forming another $\ce{OH-}$. Therefore you end up with NaOH in the solution which has a different pKs.
But I do not know if normal heating is enought for that. The thermal combustion of solid $\ce{Na2CO3}$ to $\ce{Na2O}$ and $\ce{CO2}$ needs around 500°C.


text:
Production of ʟ-glucose
What chain(s) of reactions are used to produce ʟ-glucose? I did a search on Google Scholar, but to my surprise nothing came up. My guess would be that the synthesis starts from ᴅ-glucose for which there are abundant natural sources. References would of course be highly appreciated!

Deleting my previous comment and eating my words: http://www.nrcresearchpress.com/doi/pdf/10.1139/v84-112

The Wikipedia page can be a good start: https://en.wikipedia.org/wiki/L-Glucose

Probably easier for synthesis is to use a simpler starting material, more easily derived from feedstocks, e.g., https://pubs.acs.org/doi/abs/10.1021/jo00118a058

related https://chemistry.stackexchange.com/questions/42458/do-l-sugars-enantiomers-of-natural-sugars-have-a-sweet-taste

There is a nice review about synthesis of rare 6-carbon sugars.
They discuss two pathways to L-glucose. The first isomerizes L-fructose to L-glucose:

As you can see from the names of the enzymes, they have different natural substrates, but apparently, L-fructose and L-glucose fit into the active site as well. The reaction creates one additional stereocenter (on carbon 2), and the enzyme is stereospecific.
To make L-fructose, the enzyme L-rhamnulose-1-phosphate aldolase (RhaD) is used to catalyze the reaction of dihydroxyacetone and glyceraldehyde.

Fructose has three stereocenters while dihydroxyacetone is achiral and glyceraldehyde (with one stereocenter) may be used in racemic mixture. The enzyme selects the appropriate enantiomer of glyceraldehyde and is stereospecific in the creation of the new stereocenters (at carbons 3 and 4).
The other pathway starts with sorbitol, a so-called sugar alcohol (i.e. the carbonyl of the sugar has been reduced to a hydroxyl group). In the review cited above, the stereochemistry seems incorrect, and it would make sense to start with L-sorbitol (also known as D-gulitol or L-glucitol). The relationships between different rare (and common) sugars with respect to their stereochemistry is nicely illustrated in this review.

I did a search on Google Scholar, but to my surprise nothing came up.

Searching for "synthesis of L-glucose" brings up several papers. Search for "synthesis of rare sugars review " or "synthesis of rare hexoses review" yields the two references I cited (search done in 2022, not in 2019 when the question was asked).


text:
What is the real pKa of water?
What is the $\text{p}K_\text{a}$ of water? A simple google search yields the value $15.74$, but this site and this paper say it's $14.0$.
According to my understanding, the correct answer should be $14.0$:
$$\text{p}K_{\text{a}}= -\log([\ce{H+}][\ce{OH-}])$$
For $25~\text{°C}$:
$$[\ce{H+}][\ce{OH-}] = 10^{-14} = K_{\text{w}}$$
Thus follows $\text{p}K_{\text{a}}=14$.
Can you tell me which value is correct and why?

Did you read the paper you link to? I glanced through it and it seems to try to explain where this discrepancy comes from.

@tschoppi yup i read the paper but it just confuses me a lot more

Use 14 as this is what all the (thermodynamic) tables of values are based on.  The difference between values is in the 'standard states' used. The normal (thermodynamic) one is to use the pure solvent (water) as the standard state, effectively this means, via the activity, replacing water concentration with 1 in the equilibrium constant equation.  From page 20 in  the paper you quote is a summary of how we  do this.  The 15.74 arises by including the water concentration as 55.3 molar.

Similar to the paper you provide, but perhaps more condensed: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/What_is_the_pKa_of_water%3F

See also this question

The ion product of water is usually expressed as
$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$
At a temperature of $25\ ^\circ\mathrm C$, its value is approximately $K_\mathrm{w}=10^{-14}$, or $\mathrm pK_\mathrm{w}=14$.
However, the ion product of water is not to be confused with the acid dissociation constant of water.
Generally, the dissociation constant for the simplified reaction
$$\ce{HA <=> A- + H+}$$
is defined as
$$K_\mathrm{a}=\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$$
Thus in case of water
$$K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}$$
The concentration of pure water at a temperature of $25\ ^\circ\mathrm C$ is $c=55.345\ \mathrm{mol\ l^{-1}}$.
Therefore,
$$\begin{align}K_\mathrm{a}=\frac{[\ce{OH-}][\ce{H+}]}{[\ce{H2O}]}=\frac{10^{-14}}{55.345}=1.807\times10^{-16}=10^{-15.74}\end{align}$$
or
$$\mathrm pK_\mathrm{a}=15.74$$

hello Sir , I know this calculation but i was asking that which value of the 2 will be used generally for comparision of acidic strength

If you read the full paper, you should see the discussion whether the $\mathrm{p}K_\mathrm{a}$ is $14$ or $15.7$.
From a thermodynamic point your assumption is correct, and $14$ is the "correct" value. In the paper, they mention that

... listed values often are based on conventions which are different from the thermodynamic ones.

So which value is correct for everyday use , beacuse it creates a lot of diffrence one value make it less acidic than methanol and other don't.

As long as you work with your water as a solvent you should use 14. There might be minor cases where water is a reactant where it has a higher pKa but usally it is considered 14.

The situation is the choice of units. When molarity is the chosen unit water must be expressed as a molarity and the pKa = 15.74. The approximation is then made that the concentration of water in dilute solutions is [almost] constant and the equation is modified by multiplying both sides by the concentration of pure water resulting in subtracting 1.74 from 15.74 to give pKw [Not pKa] =14.00 at some T near 298K. This is usually glossed over by saying the activity of pure water is defined as 1..
An activity of 1. is best defined as a mole fraction of 1. now the situation is that the solvent is expressed as a mole fraction when the solutes are expressed as molarities. In comparisons one must examine the units involved. Usually alcohols are expressed as molarities while water is mixed mole fraction [activity] and molarity. Correct units are necessary for evaluations and the correct mole fractions or molarities must be used. This applies to ideal solutions; nonideality brings activity coefficients into consideration.

It is common practice to ignore the change in pH caused by the addition of NaCl to aqueous solutions. This practice is valid for Cl- ions (pKb ≈ 20) because it is a much weaker base than water is. But is this practice valid for the Na+ ion, which has a pKa = 13.9?
If we were to take the pKa of water to be 15.7, the Na+ ion would be over 60 times as strong an acid as water itself is. If that were the case, a 0.1 M solution of NaCl would have a decrease in pH of about 0.5 pH units below the pH of pure water. Ignoring the change in pH caused by the addition of Na+ ions would not be valid in this scenario.
However, if we take the pKa of water to be 14.0, then the Na+ ion is only 1.3 times as strong an acid as water itself is, and the decrease in pH caused by the added Na+ can be ignored in most situations. This is the common experience of everyone who has measured the pH of a solution of NaCl and the expectation of anyone who has made a solution of NaCl.
The pKa of water is 14.0. The molar concentration of water should never be used in equilibrium calculations of aqueous solutions


text:
Grignard Reagent in THF vs in Diethyl ether

I cannot see how Grignard reagent would react differently in an ethereal solvent. While learning about the Grignard reagent I was taught this by my teacher, but I couldn’t justify why it was so.
I need a justification as to why this  happens.

What is a 'resonance stabilised bond'? Never heard of such a thing.

@orthocresol I think what it means is that the carbon-halogen bond has a partial double bond character due to resonance. Of course, it is poorly worded as shown

There are 2 different things when comparing ether and THF: 1) the electron density of oxygen in THF is higher (it is basically ether with its arms attached in its back) so it will stabilise the organomagnesium compound better; 2) the reaction is usually carried out at reflux of the pure ether, so at a higher temperature in THF

@SteffX But according to this formation of a Grignard reagent is a SET mechanism which doesn't involve the solvent. So are you talking about solvation of the organomagnesium compound when you talk of stabilisation?

@YUSUFHASAN — Stabilisation of the product, depending on the conditions, favor the reaction

@orthocresol the resonance stabilised just meant the partial double bond character of that bond, sorry for the poor wording

@SteffX besides electronic effects on stabilising the "Mg", the second point (working at high T) can be addressed by using dimetoxyethane. You can add this and make a nice answer.  For the OP point what is crucial is the higher exposure of the O doublets, that is surely the main point when reacting less reactive C-X bond.

Yes, Indeed it is mentioned in organic synthesis by Micheal B. smith. THF is preferred over diethyl ether if the C-X bond is vinyl or aryl. Though, it is not mentioned that diethyl ether cannot be used, the reaction is sluggish and the yields are poor, when diethyl ether is used as a solvent.

THF has a number of differences with diethyl ether:

Oxygen non-bonding doublets are more available for bonding with magnesium because the two ethyl groups are not freely rotating around (you may think of THF as diethyl ether with its arms tied in its back).
THF boils at 66 °C (ether 34.6 °C). Because the reaction is usually carried out so to maintain a gentle reflux, the higher boiling temperature of THF will speed up the reaction, and solubilize reactants/products better.

As @Alchimista pointed out, THF is just one of the possible solvents you can use if ether fails or gives poor yields. You can also use higher boiling ethers like dimethoxyethane.
Note: this is a collective answer asked by other users.