text:
How do we figure out whether reactions are endothermic or exothermic?
Are the following reactions exothermic or endothermic:
$$\ce{H2 + I2 -> 2HI}$$
$$\ce{N2 + O2 -> 2NO}$$
So I know the second reaction is endothermic as we need to break triple bonds between nitrogen molecules. I am not sure about the first one. Actually, I want to know if there is some way we can decide whether a reaction is endothermic or exothermic just by looking at the equation?
I got a test coming up in a few months so this was one of the few topics I was having trouble with. Any help would be appreciated.

Without looking up the thermodynamic data, or without background chemical knowledge, it cannot be said just from looking at the equation.

Question titles should be like book titles. They distinguish the book from
other books, give a hint about content, but do not tell the full story of
the book. The content then should elaborate the topic to full depth the
author is able to do.

@Poutnik , i am sorry, this was my first question here and this usually how we ask questions on MSE so i thought CSE would be similar sorry. Do you know if the first reaction is endothermic or not?

The first reaction is definitely endothermic and your reasoning is quite correct, Aditya.

If you search for "standard enthalpy of formation hydrogen iodide", you get positive value $\pu{26.5 kJ/mol}$, what means the reaction is endothermic, as by convention, the values for elements in their standard state is zero.
The other one is strongly endothermic, what you realize, if you search the same for nitric oxide.
For a general reaction, search for standard enthalpy of formation of all reactants and products, multiply the values by stoichiometric coefficients and subtract the reactant values from the product values. If the result is negative, the reaction is exothermic. For the positive value, the reaction is endothermic.
For starters: enthalpy is defined as $H = U + p . V$, where $U$ is system internal energy, $p$ is pressure, $V$ is volume. Change of system enthalpy in constant pressure condition is change of system internal energy + volume work done by system: $\Delta H = \Delta U + p . \Delta V$. By other words, $\Delta H$ is the energy you have to provide to a system at constant pressure to increase its internal energy by $\Delta U$, if the system changes its volume by $\Delta V$.
For guessing the formation enthalpy patterns, there is no easy shortcut. For simple inorganic compounds, a lot of help in noticing patterns is the periodic element table.
All comes with experience and learnt expirical knowledge and patterns. Chemistry is in large extent empirical science, where many things cannot be derived. Or can, but it is much easier to just remember, unless one wants to jump in quantum chemistry. Even then not everything can be calculated.

Thanks this is super helpful. I just got one more doubt though. I know that in exothermic reactions, energy on reactant side > energy on product side. So is this standard enthalpy of formation similar to energy levels? ( I haven't studied thermodynamics yet), Also is there some certain trend with which I can know the standard enthalpy formation value for compounds. If not then i'll happily learn those values but I think understanding it would be beneficial. Thank you so much though.

See the answer update.

Thank you. This is really useful. After reading your answer i read a little bit about Thermodynamics and it seems really interesting, and apparently it's a really important topic too (where I live). Am going to have fun studying it, also thank you (this was my first question on the site)

You are more than welcome.


text:
Uncertainity principle of a particle on a ring in $m=0$ state
When a particle is on a ring, we know that the commutator $[L_z,x]=i\hbar y$ and $[L_z,y]=-i\hbar x$. So we can't measure $L_z$ and $x$ & $L_z$ and $y$ simultaneously with infinite accuracy.
When $m=0$, then the energy of the particle is 0. So we can say that $L_z$(angular momentum in z-direction) is 0 as we consider potential energy to be 0 on the ring. This also implies $p_z$ is also zero as $L_z=rp_z$.
My question is that if the particle is in $m=0$ state on the ring, then if we find its position $x$ and $y$ simultaneously {as $[x,y]=0$}, we are confident that $L_z=0$ on the ring.
Then doesn't the commutator relation for $[L_z,x]=[L_z,y]=0=[p_z,z]$ (as $z=0$ is considered on ring) for $m=0$?
Here we see that commutator relation and uncertainty principle is violated. I know that it can't be the case as they are generalized. Please tell me what is wrong in my reasoning.

Your definition of $L_z = rp_z$ isn't quite right.
$$\vec{L} = \vec{r} \times \vec{p} \implies \pmatrix{L_x \ L_y \ L_z} = \pmatrix{x \ y \ z} \times \pmatrix{p_x \ p_y \ p_z} = \ldots?$$

Oh, I see. But the total energy is 0 in m=0, doesn't this imply that energy by virtue of any translation and rotational motion is 0 and L and p=0 because potential energy is considered to be 0 on ring?

The commutators you need are $[\hat L_x,\hat L_y]=i\hat L_z$ and cyclic variants on $x,y,z$. When the particle has zero energy its orientation is unknowable meaning that it can be anywhere on the ring which is accord with the uncertainty principle.

Ok, now I have understood

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text:
Problem encountered during high school unsaturated fats experiment
To get straight to the point, my chemistry experiment was looking at the amount of unsaturated fats in avocado oil with respect to temperature. I did this by testing for the $\ce{C=O}$ bonds using bromine water. The full procedure is as follows, followed by the actual theory behind the experiment:

Pipette out approximately 5 mL of the avocado oil into a boiling tube
labelled ‘r.t.’ (room temperature) and place the boiling tube on a
test tube rack.

Measure and record the room temperature.

Prepare a blank sample in a flask labelled ‘Blank’ by pipetting out
10mL of bromine water into the flask.

Pipette out another 5mL of the oil into another boiling tube
labelled '$50˚C$’ and heat the boiling tube over a Bunsen burner until
the oil reaches a temperature of $50˚C$. Place the boiling tube on the
test tube rack, and repeat this step for the following temperatures
(˚C): $150, 250, 350.$

From the boiling tube labelled ‘r.t.’, measure out approximately
$0.2g$ of the oil into an iodination flask labelled ‘Test’. Note that the oil does not need to be exactly $0.2g$ as the results will be
standardized when processing the data, but should still be within
$0.05g$ of $0.2g$.

Close all the curtains and blinds in the room to ensure the room is
dark. This is due to the fact that the $\ce{Br2}$ in the bromine water is
susceptible to photo-decomposition, meaning it can be broken down
into bromide ions by light. Since carbon-carbon double bonds can
only be broken down by $\ce{Br2}$ molecules and not bromide ions, this
photo-decomposition will causes inaccuracies in the data.

Measure the temperature of the oil in the flask to make sure that it
is at room temperature value that was recorded in step 2. The oil
should be at room temperature but if it is higher, wait until the
temperature lowers to room temperature. If it is lower than room
temperature by more than $1˚C$, gently heat the flask over a Bunsen
burner until it reaches room temperature before immediately take the
flask off heat.

Pipette out 10mL of bromine water into the flask and, after
stoppering the flask, thoroughly shake it to mix the contents.

Into the ‘Test’ flask, pipette out 5mL of the 10% $\ce{KI}$ solution before
stoppering the flask and thoroughly again shaking the solution. The
colour of the solution should change to a dark brown-red colour. As
explained in the prior calculations, when the $\ce{KI}$ is added, the
$\ce{KI}$ dissociates and the $\ce{I-}$ ions oxidise to $\ce{I2}$ while
reducing the remaining $\ce{Br2}$ molecules to $\ce{Br-}$. The produced
$\ce{I2}$ will now be titrated against $\ce{Na2S2O3}$. We perform this back
titration with KI as there is no known direct titration for bromine.

Rinse the stopper and the inside of the ‘Test’ flask with 30mL of
distilled water so that the distilled water flows into the flask.
This is done in order to ensure that all traces of the solution
that could be on the stopper or on the inside of the flask, caused
by the swirling of the flask, enter the solution at the bottom of
the flask.

The solution inside the ‘Test’ flask is then titrated against the
0.1N $\ce{Na2S2O3}$ until the solution fades to a pale straw-yellow colour
(take a photo at this point). At this point, most of the $\ce{I2}$ would
have reacted with the $\ce{Na2S2O3}$ to reduce the $\ce{I2}$ back to iodine
ions, which are colourless in solution and hence, the colour fades.
Remember to shake thoroughly.

After the solution has faded to a straw-yellow colour, pipette out
1mL of the starch indicator solution. The colour of the solution
should now change to a purple-blue colour. Continue titrating with
the $\ce{Na2S2O3}$ until the colour of the solution turns colourless.
Record this disappearance of the blue colour as the endpoint of the
titration. The starch indicator combines with $\ce{I2}$ to form the
purple-blue colour, hence indicating the presence of $\ce{I2}$. However,
the $\ce{I2}$ will react preferentially with the $\ce{Na2S2O3}$ which will
cause the $\ce{I2}$ to be taken from the starch to the $\ce{Na2S2O3}$. The
starch solution is only added after the solution reaches a straw-yellow
colour, when most of the $\ce{I2}$ has already reacted, due to the fact
that if left too long, the $\ce{I2}$ might get permanently bound to the
starch, which would cause inaccuracies in the data.

Empty the flask and stopper and rinse both three times with
distilled water to eliminate any trace of the solution.

Repeat steps 9 - 13 for the ‘Blank’ flask and record the endpoint
of the titration.

Repeat steps 5 - 13 for the oil samples from the remaining seven
boiling tubes on the test tube rack.

Some theory on what occurs during the reaction
This reaction is supposed to happen due to the fact that, even though it is initially non-polar, $\ce{Br2}$ is a very polarizable molecule. When the $\ce{Br2}$ molecule gets close enough to the electrons in the double bond of the alkene, the bonding electrons in the $\ce{Br2}$ molecule repel away from the bond, hence ‘inducing’ a dipole in the bromine molecule. This is shown in Figure 2(a) where the electrons in red repel away from the electrons in blue, causing a dipole moment. This causes the red electrons in the bond to act as a nucleophile and attaches itself to the positively charged bromine atom, which is acting as an electrophile as it wants electrons. At the same time the single bond in the Br2 molecule breaks with the blue electrons going to the negatively charged $\ce{Br}$ atom while a long pair of electrons on the positively charged $\ce{Br}$ atom, shown in purple, bond to one of the carbon atoms in the former double bond. The result, known as a cyclic bromonium ion, is shown in Figure 2(b).

Note that there is a dipole moment as bromine (electronegativity of 2.96) is slightly more electronegative than carbon (electronegativity of 2.55). This causes the negatively charged bromide ion to attach itself, from the bottom, to one of the carbon atoms, which is acting as an electrophile due to the dipole moment and wants the electrons of the bromide ion. Consequently, on the side that the bromide ion attaches itself, the single bond that is already attached to the bromine at the top breaks away and the bonding electrons go to the bromine atom. Figure 3 shows the full halogenation reaction at the site of the carbon-carbon double bond.

So what went wrong?
After conducting this experiment, the results I got were:

This shows that, as you increase the temperature, more bromine reacts with the oil, implying that there are more double bonds.
Both my chemistry teacher and I could not figure out what is happening or how the method could be flawed.
Any insights or thoughts would be much appreciated.

What was the rationale behind the experiment? As iodine or bromine number is a standard parameter in oils - edible or not -, characterization, there should be plenty of informations available, at least to the teacher. Naively I would have expected the opposite. Were the titration steps (all of them) done at room temperature? Also you may think on what would possibly link your results to oxidation. It is possible that oxidation of the fat (that would be peroxide value) is linked to bromine/iodine value but I must think of that as I have no clear picture in a glance.

The experiment was conducted to gain an experience of conducting an experiment and analysing the results gained. All titration steps were done at room temperature and the oil was left to cool to room temperature before being used. Will try and approach this problem from the oxidation viewpoint, thank you very much for your thoughts @Alchimista.

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text:
Do we consider the spatial orientation of a molecular orbital while assigning plane of symmetry?
I am following the video lectures by Neeraj Saini sir and in this video$^{\dagger}$ he said that $\ce{CO2}$ has infinite planes of symmetry.
But I think that if we take into account the orientation of $π$ bonds (which I don't know if we have to or not) then $\ce {CO2}$ molecule must  have $2$ planes of symmetry, as per the following argument:
A $\ce {CO2}$ molecule has similar $\pi$ bond orientation as that of an allene.
Now if we consider the orientation of pi bonds then $\ce {CO2}$ must have only 2 planes of symmetry.
So, to check whether infinite planes of symmetry statement is correct or not, I have the following question:

Do we consider the spatial orientation of molecular orbitals while assigning plane of symmetry? And if not then why not?

$\dagger$ Note that he uses Hindi language for the purpose of conversation.

We don't.$\mathstrut$

@IvanNeretin: I agree that we don't. But why don't we? We learn about symmetry of molecular orbitals and molecules as a whole. But in a molecule, we don't consider the orientation of orbitals. Is this simply a convention?

In a way, orbitals themselves are a convention.

You can use symmetry to examine orbitals, for example those in benzene or butadiene, but to determine a point group the atoms are all that is needed. The allene belongs to d$_{2d}$ for example and the orbitals / bonds can be remove and the conclusion is the same.

The reason that orbitals are not considered in the symmetry of a molecule is that the orbitals cannot decrease the symmetry of the molecule. If we look at the canonical delocalized molecular orbitals of any molecule, everyone of them is symmetric or antisymmetric with respect to a symmetry operation of the molecule, with those symmetry operations determined only by the positions of the atoms, not the electron density.
Looking specifically at carbon dioxide, the simplistic picture of $\ce{CO2}$ as being like allene with lone pairs instead of the C-H bonds is not an accurate view of the electron density. If one looks at the delocalized pi-type orbitals, both span all three atoms in either the xz or yz planes (using the standard convention of the z axis being aligned with the molecular axis). Thus, both orbitals are symmetric with respect to the xy plane of symmetry of the molecule (which passes through the C).
Similarly, the lone pairs do not occupy separate sp2 orbitals that have specific alignment in a plane perpendicular to the p orbital involved in pi bonding. Instead, the create a combined electron density that can be deconvoluted into an sp-type orbital along the z-axis and partial contributions from the px and py orbitals, such that the total has no bias along the x or y axis.
If we were to look at the total electron density of $\ce{CO2}$, we would therefore see only a pill shape that does not vary as one rotates around the z-axis.
To visualize this, I recommend looking at all of the calculated delocalized molecular orbitals of $\ce{CO2}$. You can do this on the molcalc.org website if you do not have other software for that purpose.


text:
Dipole Moment of p-dibromobenzene

What is the dipole moment of p-dibromobenzene?

I know that p-dichlorobenzene has zero dipole moment, but my teacher said that this is not the case for bromine. I am still confused and feel its dipole moment as zero. My teacher said that it does have some dipole moment (supported through experimental data).

Who is "Sir" ? A teacher ? Why should p-dibromobenzene have a dipole moment ?

Hmmm. This source mentions a $\mu$ of 0.00 but PubChem mentions a $\mu$ of 1.43 in gas state.

PubChem gives dipole moment ($\mu$) of 1,4-dibromobenzene (p-dibromobenzene) as $\pu{1.43 D}$ in gas phase and $\pu{1.87 D}$ in liquid phase at $\pu{20 ^\circ C}$. It has given 1987 version of Handbook of Organic Chemistry (J. A. Dean, Ed.) as the reference. PubChem also gave $7.77$ as dielectric constant at $\pu{10 ^\circ C}$ and $6.7$ at $\pu{40 ^\circ C}$ with the same reference. However, I afraid to tell you that all of these values are incorrect. The 1999 version of Handbook of Organic Chemistry (same reference in different edition) listed followings for 1,4-dibromobenzene and its isomers (also included 1,4-diiodobenzene and its isomers for comparison):
$$
\begin{array}{c|ccc}
\hline
\text{Compound} & \text{Dipole moment, }\mu & \text{Dielectric constant, }\epsilon  & \text{reference page}\\
\hline
\color{blue}{\text{1,4-Dibrmobenzene}} & \color{blue}{\pu{0.0 D}} \ (-) & 2.57 \ (\pu{95 ^\circ C}) & \text{p. 5.111}\\
\text{1,3-Dibrmobenzene} & \pu{1.5 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 4.21 \ (\pu{20 ^\circ C}) & \text{p. 5.111}\\
\text{1,2-Dibrmobenzene} & \pu{2.13 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 7.86 \ (\pu{20 ^\circ C}) & \text{p. 5.111}\\
\color{red}{\text{1,4-Diiodobenzene}} & \color{red}{\pu{0.19 D}  \ (\pu{20 ^\circ C}\text{ in benzene})} & 2.88 \ (\pu{120 ^\circ C}) & \text{p. 5.113}\\
\text{1,3-Diiodobenzene} & \pu{1.22 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 4.11 \ (\pu{50 ^\circ C}) & \text{p. 5.113}\\
\text{1,2-Diiodobenzene} & \pu{1.70 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 5.41 \ (\pu{50 ^\circ C}) & \text{p. 5.113}\\
\hline
\end{array}
$$
Accordingly, 1,4-diiodobenzene actually has a dipole moment, but, every other 1,4-dihalobenzenes do not have a dipole moment. The values given in PubChem must be corrected during 15th edition of the book.
Reference:
John A. Dean, Editor, Lange's Handbook of Organic Chemistry, Fifteenth Edition; McGraw-Hill, Inc.: New York, NY, 1999, p. 5.111-5.113 (ISBN: 0-07-016384-7).

Could you provide an explanation for the non-zero dipole moment in 1,4-diiodobenzene? On the basis of symmetry, I'd expect the bond moments to cancel each other out, similar to other 1,4-dihalobenzenes. Also will it be possible to determine the direction of the dipole moment vector?

@Guru Vishnu: 1,4-diiodobenzene evidently has shown some value. I guess it's due to high polarizability of large atom. But, you should ask theoretical chemist to have an explanation.

Thanks for your reply! I've asked the same question separately - Why does 1,4-diiodobenzene have a non-zero dipole moment? as I don't know any theoretical chemists other than users here.


text:
CFSE Calculation
I have $\ce{[ML6]}$ & $\ce{[MD6]}$ as two complexes. I know the $\Delta_0$ & hybridisation of each of them. How do I calculate the CFSE for $\ce{[ML4D2]}$ ?
This is my way of thinking :
I can calculate the CFSE of $\ce{[ML6]}$ & $\ce{[MD6]}$ in terms of Pairing Energy and then do some manipulations to cancel the pairing energy to be left with an equation of the form :
$$a\times(\text{CFSE of }\ce{[ML6]}) + b\times(\text{CFSE of }\ce{[MD6]}) = c$$
where $a$, $b$ and $c$ are known. How do I proceed to get the CFSE for $\ce{[ML4D2]}$?

In the future, please format your posts using $\LaTeX$


text:
Leaving group ability and solvent's effect
Regarding leaving groups and solvent effects (SN1 and SN2 reactions), I would like to ask some questions.
As the nucleophilic character is also influenced by the solvents, there is some indication (to me at least) that the leaving group can be influenced as well. Unfortunately, there is not much information found. Let us consider the halides, the quality of the leaving group is $\ce{I-} > \ce{Br-} > \ce{Cl-} > \ce{F-}$ as the larger ions are more polarizable, more stable and less basic.
So imagining a polar protic solvent, wouldn't the solvation stabilize $\ce{F-}$ via hydrogen bonds? I totally understand, that fluorine forms tighter bonds with carbons, so it naturally is a bad leaving group.
Considering the Finkelstein reaction, the driving force of the reaction is the formation of an insoluble salt, so basically the solvent influences the leaving group ability.
So, can I generally say that the solvent can theoretically influence the leaving group ability? Is there some general thumb rule I can use?
Until now, I haven't seen any reaction, where $\ce{Cl-}$ is preferred over $\ce{I-}$ as a leaving group except for the Finkelstein reaction.

The solvent generally doesn't affect the leaving group ability of an anion. But coming to your Frinkelstein case, it a special one cause here some precipitate is being formed (yes, that's correct that it is due to lower solubility product of say $\ce{NaCl}$ in acetone). As soon as some ppt. is formed, those atoms lose their ionic identity, settle at the bottom and don't interfere in the solvent as solute and so their concentration decreases in the solvent. Due to Le Chatelier's principle, the reaction goes forward to a great extent. This is also the case for other super leaving groups like gases (for eg. $\ce{N2}$, $\ce{H2}$).


text:
Casimir effect and van der Waals forces
The Casimir effect makes two close metal plates in a vacuum attract.
I was pondering and figured out that two close molecules should experience the same effect, as they are close and between molecules there is nothing. Maybe even stronger as they are usually way closer than metal plates can be.
So, is there any theoretical work on this? Is the Casimir "force" considered one of the van der Waals forces? Am I talking nonsense?
Also, could the attraction between the plates result from van der Waals interactions and not from the quantum field stuff? How is it known that this is not the case?

You are correct in your ideas. The effect is important in the van-der-waals interaction and appears as an additional  $1/r^7$ term compared to the usual $1/r^6$ term. (This is sometimes called a retardation effect). It is present only at larger distances (a few nm) and arises only in induced-dipole interactions (dispersion interaction) because of the time it takes for the interaction to propagate between molecules. During this finite time (governed by the speed of light in the medium) the first dipole's direction has changed during the time it takes to cover distance $2r$ so that the inducing and induced dipoles become less correlated the farther the two dipoles are apart. The equation describing this effect is called the Casimir-Polder eqn. $W(r)=-23\alpha_0^3hc/(8\pi^2(4\pi\epsilon_0)^2r^7)$ where $\alpha_0 $ is the polarisability and $\epsilon_0$ the permittivity of free space.


text:
Why isn't formation of benzyne a 1-step process?
According to "Organic Chemistry as a secondary language", during elimination-addition, formation of benzyne involves 2 steps (as shown in the bottom photo). I wonder during 2nd step, it is just electron transfer and 'kick out' the Cl as leaving group.
Can the 2 step process be simplified as 1 step? (as shown in the upper photo) The only difference is that I used the electrons from C-H bond to form the 3rd C-C bond, where the book suggest a deprotonated anion is first formed then C-C is subsequently formed. I wonder what causes the difference? Thank you.

The loss of HX from chloro- and bromobenzene is concerted with amide ion while fluorobenzene loss of HF is stepwise. J. D. Roberts, et al., J. Am. Chem. Soc., 78, 611 (1956).

But I would really like to know why it must be a 2- instead of 1-step process, is the (central) anion more stable? What caused the electron to stay at 1 point but not immediately transferred to form 3rd C-C bond? Thanks

look carefully at user55119's comment and you will realize that it is satisfying the criteria you have yourself stated. The anion formed in the fluorobenzene case will be quite a bit more stable than that in chlorobenzene and bromobenzene, hence theformer goes through a stepwise process, while the other halobenenes adopt a concerted path

OK thanks. So in this case, forming benzyne from chlorobenzene is actually a concerted process rather than 2-step process as the book says? (since the anion is much less stable)

Yes, benzyne to chlorobenzene is concerted


text:
Can someone explain this question in detail?
An infusion stated to contain 0.95% w/v NaCl was diluted so that its Na content  could be determined by flame photometry. The following dilutions were carried out:

10 ml of the sample was diluted to 250 ml with water.
10 ml of the diluted sample was diluted to 200 ml with water.
The sample was found to contain 0.74 mg/100 ml of Na.
Atomic weights: Na = 23 Cl = 35.5
Calculate:
The % w/v of NaCl present in the infusion
The % of stated content of NaCl.
The concentration  in mmoles/litre of NaCl present in the infusion.

We will not make all the calculations without you trying to enter the problem. So try and start. Tell us where you find a difficulty. We will help you go ahead.

It's an example question that I have the solution to and I just want it explained better.

Better explaining without knowledge what has to be explained better is guesswork. Say explicitly in the question what you need to explain.

It was this bit of the answer: Conc. of NaCl in the injection = 370 x (58.5/23) = 941 mg/100 ml

Since you have an answer, would you answer my question?: Is infusion contains $0.94% (w/v)$?

I didn't realize you have the answer in comment. So it is 941 mg/100 mL =0.94% and my answer is correct.

I'm just trying to understand the relationship between the atomic weight and the concentration

In flame photometry, we need to dilute the NaCl sample because the calibration is not linear at high NaCl concentration.
It is a classical dilution problem. I find it useful to remember
$$C_iV_i=C_fV_f$$
where $i$ indicates initial concentration (before dilution) and volume and $f$ indicates final concentration and volume (after dilution).
Now note that your sample concentration is in 0.95% w/v NaCl. The flame photometer only responds to Na ions in the solution. So your flame photometer results tells you how much Na is there, not NaCl.
Try to use mole ratios between
grams of Na => moles of Na
What is the relationship between moles of Na and moles of NaCl?
Once you sort this out, edit your question with your attempt.

Thanks  it's an example so I have the answer and calculations but just don't really understand the whole process in my notes.

@Amy, do you understand it now? What is specifically confusing you?

In my notes this bit Conc. of NaCl in the injection = 370 x (58.5/23) = 941 mg/100 ml

They are arriving at the answer too quickly. What I do is that I draw block diagrams of dilution on a piece of paper and label the initial concentrations and final concentration to keep track of how much volumes were transferred.

I'm not from a chemistry background so all this is very new to me.

There are plenty of videos on dilution problems. It takes a little bit practice even with chem students. https://www.youtube.com/results?search_query=Dilution+formula


text:
I am having problem with this question including topics of redox reactions and charges
What is the oxidation number of $\ce{Fe}$ in $\ce{K4[Fe(CN)6]}$?
This seems to be a pretty easy problem. Just input the oxidation numbers of everything and equate for 'x'. But the problem is that I don't know the charge on the Cyanide ion right there. Now, I knew from doing many questions that $\ce{Fe}$ usually has +2 or +3 so i guessed +2. But I don't want to be guessing on the actual day.
I essentially need to know the best way to find charges on polyatomic ions because there's just loads of them and I don't think I can remember over 300 oxidation states. I know the charge can be calculated by adding up Oxidation states but most elements have variable Oxidation states too so I am really confused.
Even if you know a certain website that may be useful, please link it. Any help is appreciated.

Maybe read your textbook? Cyanide ligand is clearly mentioned to be $\ce{CN-}$ in your text (which I'm guessing is NCERT). You have to remember the basic oxidation states of the most common ligands.

@AniruddhaDeb that isn't really very helpful as that's not what I was asking. But allright, thanks

@AniruddhaDeb Hello, do you know which class's NCERT contains those polyatomic ion charges. I'll buy that book so I can read that too. Thanks

Know how to draw cyanide (it’s not hard … only two atoms) and then the linked dupe will help you.

As the formula of the substance is known, you should first state the formula of the ions produced when the substance is dissolved into water. Here, $\ce{K_4Fe(CN)_6}$ gets dissolved in water and produced $4$ ions $\ce{K^+}$ so that the $4$ corresponding negative charges  must be fixed on the remaining anion, which has the formula $\ce{[Fe(CN)_6]^{4-}}$ with $4$ negative charges. Now you have to think how to synthesize this complex anion from $\ce{1 Fe^x+}$ ion and $\ce{6 CN^-}$ ions. The equation is $$\ce{Fe^{x+} + 6 CN^-  -> [Fe(CN)_6]^{4-}}$$ Counting the charges gives x positive charges from $\ce{Fe^{x+}}$, 6 negative charges form the six cyanide ions, and this makes a total of $\ce{4-}$. So the initial $\ce{Fe}$ ion must have been charged $+2$
The nature of the ion $\ce{CN^-}$ remains to be explained : Well ! The carbon atom has 4 independent valence electrons. Nitrogen has 5 valence electrons, but 2 of them are taken in a doublet. So there is only 3 available electrons in N atoms for bonding. When 1 C is attached to 1 N, this makes 3 covalent bonds between C and N. As a consequence one electron is still available on the carbon atom. This is not a stable compound. The carbon atom needs one more electron to get 8 electrons in its outer layer. This electron can be offered by one H atom, which would make a covalent molecule HCN, with one covalence between H and C, and three covalences between C and N. But this last electron can also be given by a sodium Na or a potassium atom K. This would produce the compound NaCN or KCN, where  Na (or K) and the carbon atom are bound in a ionic bond between $\ce{Na^+}$ (or $\ce{K^+}$) and the ion $\ce{CN^-}$. This is the origin of the ion cyanide $\ce{CN^-}$.

Thanks, but I was asking if there was any way to find out that CN has a negative charge apart from just knowing it. Thanks for your help though

If you try to join one C atom and one N atom, you should see that there is a triple bond between C and N. This being done, a doublet is remaining on the N atom, and this N atom has now an octet around it, which is supposed to be good. But a free electron remains on the Carbon atom, and the outer valence shell of C has only 7 electrons. A good way to stabilize this C atom is to approach an atom like sodium which can easily loose and give an electron, making CN negative. This being done, C has now 8 electrons in its outer shell, and is stable : this is the CN- ion.

thanks so much.... I understand it now


text:
How much NaOH I have to add to increase pH?
I'm stuck in solving this problem. I have two solution of $\pu{5 M}$ $\ce{H2SO4}$ that needs two consequential $\mathrm{pH}$ increasing: the first from the natural $\mathrm{pH}$ of $\pu{5 M}$ $\ce{H2SO4}$ (around zero) to $\mathrm{pH}$ 2 and then from $\mathrm{pH}$ 2 to $\mathrm{pH}$ 7. The second solution from the natural $\mathrm{pH}$ of $\pu{5 M}$ $\ce{H2SO4}$ to $\mathrm{pH}$ 2 and then from $\mathrm{pH}$ 2 to $\mathrm{pH}$ 7. I would like to understand how can I calculate the volume of $\ce{NaOH}$ (let's say $\pu{5 M}$) I have to add for each step.
at the beginning, to calculate the Volume from $\mathrm{pH}$ 0 (as you said) to 2 I was considering this formula:
$$\ce{[H3O+]} \ \text{(after adding $\ce{NaOH}$)} =(V(\ce{H2SO4}) \times \ce{[H2SO4]} \ \text{(initial)} − V(\ce{NaOH}) \\ \text{added} \times [\ce{NaOH}])/V(\ce{H2SO4}) + V(\ce{NaOH}) \ \text{added}$$
But somebody told me this calculation is wrong because I have to consider that $\ce{H2SO4}$ is diprotic.
The value I used are:
$\ce{[H3O+]}$ (after adding $\ce{NaOH}$) $= 10^{-2}$;
$V(\ce{H2SO4})= \pu{1L}$; and
$\ce{[H2SO4]} \ \text{(initial)} = \pu{5 M}$

@Blade: I appreciate your hard work on editing. However, keep in mind that we do not approve MathJax formatting on the title as a policy.

@MathewMahindaratne Got it! Thanks for noting that.

Can't quite follow your equations but your friend is right. It takes twice as much $\ce{NaOH}$ as $\ce{H2SO4}$. In other words for every $\pu{1L}$ of $\pu{5M}$ $\ce{H2SO4}$, you need $\pu{2L}$ of $\pu{5M}$ $\ce{NaOH}$.
The reactions are:
$$\ce{2H+ + 2OH- -> 2H2O}$$
$$\ce{2Na+ + SO4^2- -> Na2SO4}$$
OVERALL:
$$\ce{2NaOH + H2SO4 -> 2H2O + Na2SO4}$$

Numbers don't match in your first equation. 6 H on LHS and 4 H on RHS. Also, what are .=. and =. ?

@Blade: Use $\pu{}$ for units, e.g.., $\pu{1 M}$ conveted to $\pu{1 M}$.

@MathewMahindaratne Thanks! What about constants? same or \mathrm?

@Blade: Check This and this. Good luck! :-)


text:
If acid releases proton then how can a proton react with another proton to form hydrogen gas?
As we know that acid releases proton (H+ ion) when dipped in water hydrogen has a proton and electron only . To form H+ hydrogen releases 1 electron and it becomes proton. But when Acid reacts with metals hydrogen ion reacts with another hydrogen ion ( which is nothing but proton ) to form hydrogen gas then how can it be possible that one proton reacts with another proton to form hydrogen gas?

They can't, of course. One proton is positively charged, and so is another. There must be something else to it.

A proton does not react with another proton, as you state. It reacts with a metal. And this metal gives one electron to one proton. This makes a Hydrogen atom, then a Hydrogen molecule. Finally the metal becomes a positively charged cation, which passes into solution

Usually there is some form of a redox reaction happening there, for example like
$$\ce{M(s) + 2H+(aq) \to M^{2+} (aq) + H2(g)}$$
where $M$ is just some metal (like copper, iron, etc...) reducing hydrogen to its elemental/gaseous form.


text:
Sketching lattice points, primitive axes, primitive cell, and basis of atoms
I am currently studying Introduction to Solid State Physics, 8th edition, by Kittel. Chapter 1 provides the following figure and accompanying explanations:

It also says the following:

An ideal crystal is constructed by the infinite repetition of identical groups of atoms. A group is called the basis. The set of mathematical points to which the basis is attached is called the lattice. The lattice in three dimensions may be defined by three translation vectors $\mathbf{a}_1$, $\mathbf{a}_2$, $\mathbf{a}_3$, such that the arrangement of atoms in the crystal looks the same when viewed from the point $\mathbf{r}$ as when viewed from every point $\mathbf{r}^\prime$ translated by an integral multiple of the $\mathbf{a}$'s:
$$\mathbf{r}^\prime = \mathbf{r} + u_1 \mathbf{a}_1 + u_2 \mathbf{a}_2 + u_3 \mathbf{a}_3. \tag{1}$$
Here $u_1, u_2, u_3$ are arbitrary integers The set of points $\mathbf{r}^\prime$ defined by (1) for all $u_1, u_2, u_3$ defines the lattice.
The lattice is said to be primitive if any two points from which the atomic arrangement looks the same always satisfy (1) with a suitable choice of the integers $u_1$.

Here are some lattice points I drew:

It doesn't seem to me that any of these lattice points are primitive; that is, I don't think that any of these lattice points satisfy $\mathbf{r}^\prime = \mathbf{r} + u_1 \mathbf{a}_1 + u_2 \mathbf{a}_2 + u_3\mathbf{a}_3$. The irregular pattern present in this space lattice makes it difficult to see how this could be done.
So, in order to satisfy figure 3c, what are the lattice points of interest?
I would greatly appreciate it if people would please take the time to explain this.
EDIT

How about a cell larger than your 1st try but smaller than the last one?

@marcin Is that the cell that you were referring to? But isn't there still a gap?

A cell, above all other definitions, is the thing that repeats. Does your first tentative cell repeat to the right and to the left?

The cell after EDIT looks good to me. There won't be a gap (unless I overlooked something)

@marcin shouldn’t the cells be making contact, so that where one ends another begins?

I think the cell after EDIT is correct. Here I added a few neighboring cells:

Admittedly, the atom in the right bottom corner is missing. Apparently the author didn't consider it important.

You have to take care about the definition of lattice and crystal structure. Quoting your text "An ideal crystal is constructed by the infinite repetition of identical groups of atoms. A group is called the basis. The set of mathematical points to which the basis is attached is called the lattice". So the crystal structure is constituted of a basis (an atom or a group of atoms) attached to a lattice.
By definition a lattice is an infinite array of mathematical points in space, in which each point has identical surroundings to all others. your Fig. 3a represents a lattice. Conversely your fig 3c does not represent a lattice; in fact in the figure caption it is stated that the points represent atoms. In this case case each pair of points represent your basis, not lattice points. You're asked to find the lattice points. The cell after EDIT is correct; in that case lattice points are at the same position of one of the two points forming the pair.


text:
Quinones - Radical Scavengers
In the textbook of "A Guidebook to Mechanism in Organic Chemistry" by Peter Sykes, it is mentioned that peroxides are radical sources and quinones are radical absorbers or scavengers, but it had no explanation, and quick search gave results about HALS, but not about quinones. Also I understand that peroxides can act as radical sources due to homolytic cleavage and its 'O-O' bond energy being very low, around 140 kJmol^-1. But can you explain how quinones are good radical scavengers? Also its mechanism, if any, would be appreciated.

Look up the redox chemistry of these compounds, that should be covered in many basic chemistry books.

Might I ask the reason for downvote?

@Martin, Also, I searched in L.G.Wade before asking, and there was no mention

HALS: https://en.wikipedia.org/wiki/Hindered_amine_light_stabilizers?

I am not a particular fan of libre texts, but the following has quite a bit of what is needed to understand this: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/26%3A_More_on_Aromatic_Compounds/26.02%3A_Quinones


text:
Deriving the absorption coefficient from state transitions
This source states that the absorption coefficient $\gamma$ is proportional to the difference in particle number density $\Delta n$ per unit volume between the initial and the final rotational state and also proportional to the dipole moment $\mu_T$
$$\gamma \propto \Delta n\cdot \mu_T$$
According to the Beer-Lambert Law the absorption coefficient is also equal to:
$$\gamma=\log_{10}\bigg(\frac{I_0}{I}\bigg)\cdot \frac{1}{c\cdot l}$$
Where $I_0$ and $I$ are the incidental and transmitted intensity, $c$ is the concentration of the solution and $l$ the traveled length through the solution by the incidental beam.
From this Beer-Lambert law I would deduce that the absorption coefficient is in terms of per 1 unit length and per 1 unit concentration.
I would assume that the difference in number density $\Delta n$ is analogous to the concentration of a solution.
But how is $\gamma$ then proportional to the $\Delta n$ if it should be per 1 unit of $\Delta n$ just as it is per 1 unit of sample concentration?

$\Delta n$ is the population difference between the two rotational states of the same molecule per unit volume and is mostly a function of temperature as dictated by the Boltzmann distribution.

@Paul In that case, I would assume that the total number density of particles $N$ is the concentration $c$ instead and that the absorption coefficent $\gamma$ should be per 1 unit of $N$, correct? Such that the absorptivity $\log_{10}\bigg(\frac{I_0}{I}\bigg)$ is proportional to: $$\log_{10}\bigg(\frac{I_0}{I}\bigg)\propto \Delta n \cdot \mu_T \cdot N \cdot l$$ Is this correct?

When the population difference is zero there is no net absorption because of stimulated emission, i.e 'absorption' from the upper state to the lower state plus a photon. This is common in rotational spectroscopy where levels have very similar populations at room temperature, unlike in the uv -vis and near ir. The $\gamma$ in the second equation is the extinction coefficient, a property of the molecule and proportional to the transition dipole,  and not the same as in your first eqn, it seems that $\Delta n \mu\equiv \gamma c l$ but you should check.

@porphyrin The source mentions right beneath Example 1 that it can also be the absorption coefficient instead of the extinction coefficient. How is that possible?

These are different words for the same thing, it's best to look at how it is defined and then decide what it is. Not all authors use the same definition. Extinction coefficient is still very widely used in the scientific literature and textbooks.

@porphyrin In your derived equation $\Delta n \mu\equiv \gamma c$, is $\gamma$ the extinction or the absorption coefficient? Perhaps it's better to name them $\epsilon$ and $\alpha$ respectively. Based on the relationship between Transmittance $T$ and Absorbance $A$, I have deduced that $$\epsilon=\frac{A}{\ln(10^A)}\cdot \frac{\alpha}{c} = 0,434 \cdot \frac{\alpha}{c}$$ ..but I can't seem to match it with your derived equation.

Starting from scratch; $I=I_0e^{-\epsilon cl}$ is the Beer Lambert law where $I$ is transmitted intensity, $I_0$ initial amount, $\epsilon$ is the extinction coefficient , $c,l$ concentration and path length. $\epsilon c l$ is dimensionless. (in practice power 10 is used instead of e, it makes no odds). In your case you need $\Delta c$ not $c$ as there is population in both levels. $\epsilon$ is proportional to the transition dipole, thus $\epsilon \Delta c l \sim \mu \Delta c l $.

@porphyrin This has cleared it up for me, thanks! One last question, the source says that $I=I_0e^{-\alpha \cdot l}$. Combining it with your mentioned $I=I_0e^{-\epsilon cl}$ means that $\alpha = \epsilon \cdot c$. But shouldn't this be $\alpha = \ln(10)\cdot \epsilon \cdot c$?

there are two definitions, one with powers of 10 , used in practice in spectrophotometers etc , and one base on powers of e , it does'n't matter which one is used as long as it is defined Normally the decadic extinction coefficient is quoted.


text:
Crystmet or alternative metals/alloys/intermetallics powder diffraction database
Does anyone know what happened to Crystmet, the metals structure database? The website seems to have disappeared.
Is there any way to access the database or is there a recommended powder diffraction database for metals/alloys/intermetallics that goes beyond PDF or ICDD? I would like to have access to CIF files for relatively uncommon non-stoichiometric intermetallics.

Our lab would use the Fink Index to search XRD patterns , intermetallics, corrosion products,etc. . It was a very large collection  of patterns in  print, not computer.

NIST offers a metals/alloys/intermetallics database but it has been offline and not available for sale for months due to updating. I have tried contacting NIST many times about updates with no reply.


text:
Why ester reaction with one equivalent of Grignard reagent produces ketone?
While reviewing my old materials before an external exam which also covers these topic I stopped by one of the reactions - Grignard w/esters:

From what I remember, and what my notes are also telling me, Grignard is performed by creating Grignard reagent containing alkane & MgX. Then, such complex attacks the carbonyl leaving oxygen with a negative charge, which easily deprotonates water creating 2° alcohol.
However, when done with ester, which contains a good leaving group the reaction produces ketone. Is this only because no aqueous workup is performed after the first round?

Your picture is correct. The second paragraph is incorrect. Even with one equivalent of ester and Grignard reagent there is overreaction because the ketone that is formed in situ is more reactive than the ester.

So if I understand correctly ketone while being an intermediate cannot be readily isolated and we should assume the major product of ester+G.reagent will be a 3° alcohol? Unfortunately I didn't have a chance to perform this rx in practice.

If you want to make a ketone from a carboxylate group don't use an ester. The standard reagent is a Weinreb amide.

True, you get the alcohol. Follow @Waylander's lead. Alternatively, the carboxylic acid, six-carbon or greater, will react with 2 equiv. of RLi to give the lithium dialkoxide of the the ketone which, upon aqueous workup, gives the ketone.


text:
Mechanism for simultaneous reduction of nitro group (into aniline) and aromatic substitution (into benzonitrile) by Ethyl cyanoacetate?
I stumbled upon this paper when I'm trying to look at established protocol to synthesize one of the compounds (compound IIk) mentioned in the paper.
Studies on Aromatic Nitro Compounds. V. A Simple One-Pot Preparation of o-Aminoaroylnitriles from Some Aromatic Nitro Compounds

The paper described transformation of aromatic nitro compounds into o-aminoaroylnitriles by ethyl cyanoacetate in the presence of strong base. While the general scheme showed fused aromatics, this reaction also works for any electron-withdrawing (in addition to the nitro) containing single aromatics.
I have replicate the reaction for compound IIk and I was abled to purified the compound with yield consistent with what was described in the paper.
I have never seen this kind of transformation before, especially with ethyl cyanoacetate. The paper itself mentioned that the mechanism is unknown and I haven't found any subsequent papers that try to elucidate the mechanism.
Anyone has any idea how this reaction works?

As a noobie, realize that you are expected to show some effort toward an answer. Then answers may ensue.

That's the problem. None of the typical organic chemistry knowledge applies here.
Reduction of nitrile for example requires a reductant. Obviously ethyl cyanoacetate act as a kind of reductant here, but no known reaction (aside from this) use ethyl cyanoacetate as reductant.
Ethyl cyanoacetate typically is used in the presence of base for aldol-type reaction. But I fail to see how the generated enolate interact with nitro group in the aromatic ring here,
Aromatic substitution too, especially with C-H activation, typically requires metal catalyst. This reaction has none.

you are correct about ethyl cyanoacetate being a reductant. Note that the initial products contain oxalate units that are formed from the cyano ester.

This sure is interesting. I wager reaction may still be similar to aldol, with further ring closure and subsequent reopening.

This reaction was conducted using a 3- to 5-fold excess of ethyl cyanoacetate (ECA) and KOH over the arylnitro compounds in dimethylformamide (DMF). It is also important to realize that most KOH is 15% water, which allows for a source of protons in this basic medium. Steps 1 --> 3 (Scheme 1) illustrate a possible route to nitrone 4. In so doing the ECA moiety is being oxidized as the arylnitro compound is being reduced. Addition of hydroxide to nitrone 4 forms 5, which can collapse in two ways. One pathway (another will be mentioned later) leads to reactive ethyl oxocyanoacetate 6 and an aryl hydroxylamine. Base deprotonates the hydroxylamine which adds to 6 to afford species 7 and liberates cyanide anion. Addition of cyanide to the aromatic ring of 7 affords arylamine 8 with loss of the half ester of oxalic acid. A second equivalent of 6 reacts with amine 8 to produce target compound 9.

Condensation (Scheme 2) of the anion 2 of ECA with ethyl oxocyanoacetate 6 provides ketodiester 10 which condenses with arylamine 8 to form the second targeted product 11. Scheme 3 provides another mode of collapse of intermediate 5 which leads to 9. The hydrolysis of 9 and 11 is routine. There are undoubtedly other variants of these Schemes that may apply.

Update: The reference provided by the OP was not the original paper in the series and the picture provided referred to reference 5. The original[1] reactions with 6-nitroquinoline used KCN as the base providing yields of 9 (IIa) in the 50% range and trace amounts of 11 (IIIa) in reaction times up to 5 hours.  After 24 hours the yield of 9 is greatly reduced with the formation of the oxamic acid derived from ester 9 by an unspecified mechanism of formation. The appearance of 16.5% of 11 should got be construed that 11 is derived from 9. Significantly, the use of KOH as a base produces 9 in 66.5% yield without the formation of 11.

Y. Tomioka, A. Mochiike, J. Himeno and M. Yamazaki, Chem. Pharm. Bull., 1981, 29(5), 1286.

Interesting and certainly possible. Given the conditions of excess cyanoacetate, why would we not see only the products of cyanide addition to the ring and not cyanoacetate anion addition?

@Waylander: Good question! There are a lot of questions about this reaction. The yields are moderate. Why isn't structure 10 just deprotonated given its acidity? If the yields of 9 and 11 were high, then every time 6 is generated an arylnitro cmpd is required. But then again, the yields are not quantitative. I'd be interested in your thoughts. Even the authors didn't venture a mechanism. I'm still working on it.

Why is 6 not hydrolysed immediately in an environment of excess OH-? Its reactivity is similar to ethyl oxalylchloride. Wouldn't 5 be more likely to expel CN-?

I buy that. The loss of cyanide from 5 is in Scheme 3 which explains the formation of 9 but not 11 unless 9 is saponified to 8. Does not appeal to me. As I noted earlier, there are issues here.

@Waylander: See the update.


text:
Is there any actual reaction between ferric chloride and CO2 gas?
I want to make Ferrous carbonates with Ferric chloride using $\ce{CO2}$. Does it even react with $\ce{CO2}$? how to know if it reacts to form $\ce{FeCO3}$?

The way to know it is to learn chemistry. Short of that, you will be left to wonder in the dark.

Does CO2 have reducing and alkaline properties ?

Per Atomistry on Ferrous carbonate, to quote:

In the laboratory ferrous carbonate may be obtained as a white precipitate by adding sodium carbonate to a solution of ferrous sulphate in the absence of air. If air is present, the precipitate rapidly turns green in consequence of oxidation, carbon dioxide is evolved, and ultimately the mass consists almost entirely of ferric hydroxide.

As such to prepare FeCO3, I would recommend adding your FeCl2 to aqueous Washing Soda (Na2CO3):
$\ce{FeCl2 + Na2CO3 (aq) -> 2 NaCl (aq) + FeCO3 (s) }$
Controlling air/oxygen exposure perhaps with a blanket of CO2 by performing the above reaction in a larger vessel filled with carbon dioxide may work.
So, you do still get to employ your CO2 in the preparation, but as a direct reaction path, it apparently is not.
[EDIT] Apparently, one can employ FeCl3 to create a transient  Ferric carbonate in the above scheme, as per a reference, to quote:

Physical properties of iron (III) has not been evaluated due to its instability and quickly decomposition. It is hygroscopic and decomposes in water.

So, employing an aqueous ferric salt, or allowing the oxidation of the ferrous, does not lend itself to the formation of a stable iron carbonate salt here.
Note, the action of Fe on FeCl3 absence oxygen exposure is a possible path to FeCl2:
$\ce{Fe + 2 FeCl3 <-> 3 FeCl2}$

Hmm, but the OP considers ferric chloride FeCl3, not more suitable ferrous chloride FeCl2, unless there is an error.

@poutnic Ferric or ferrous, anything that'll do, I just want to use CO2 gas I have with something to make something strong ;) such as ca(oh)2 + co2 --> caco3

By way of an [EDIT], I have addressed the reason for the use of the ferrous here.


text:
Difference between Soap and Detergent
In most articles and texts that I've read, soaps and detergents seem to be talked about collectively. I know that they are both surfactants and that soaps are usually natural, in comparison to detergent which is artificial. However, what are the differences in terms of chemical structure and properties (between soap and detergent)?
NOTE: not the same as Is there an important difference between soap and detergent?, as I'm asking about chemical structure and properties rather than the effects in everyday life.

There is no restriction that soap be natural or unnatural. Soap are salts and detergents are not necessarily salts. The latter is a very broad term. Soap has a narrow meaning.
IUPAC clearly defines soap as

"A salt of a fatty acid, saturated or unsaturated, containing at least
eight carbon atoms or a mixture of such salts."

Just note that not all soaps are used for cleaning nor there is any condition that they should be soluble in water. Insoluble soaps of metallic salts serve as lubricants.
Similarly,
IUPAC defines a detergent as

"A surfactant (or a mixture containing one or more surfactants) having
cleaning properties in dilute solutions (soaps are surfactants and
detergents)"

Just like banana is a fruit but not all fruits are bananas, in the same way, all water soluble soaps are detergents and surfactants. Not all detergents are soaps.
Visit the IUPAC Gold Book and search the entries detergents and soap.

Anything chemical structure wise? Like are there differences in the chemical structure of the molecules

@M.Farooq I suppose you are trying to quote IUPAC's web version of "Gold Book". Keep in mind you can include an URL in a format [<text>](<link>), and use > for quoting third-party sources. Currently the quoted definitions are not referenced.

@andselisk, You might have noticed that I do not provide links to papers but write the textual form form of the reference. The intention is to promote students to search themselves. In my humble opinion, they should be curious enough to search the quote and land at the right website within a few seconds. You are right, I should provide the main page link to IUPAC Gold book so that they at least scroll down and look up the relevant entry.

@An0n1m1ty, I have given you enough description to begin the journey of search. Now it is your duty as future curious scientist to search the web. Search what how do fatty acids salts look like. Also read the Wikipedia entry on detergents and start from there.

@M.Farooq While I can clearly see you mean well, the problem is that SE sites are not exactly educational in a sense there is little to no dialogue in Q&A section, and, as a consequence, all answers must be complete and self-contained. Hints, unfinished posts and general recommendations are not answers. Yours, however, is, but it lacks referencing, which if good in a pedagogical sense for one person (OP), but drastically lowers the quality of your answer for all other users.

The main difference between soap and detergent is the effects they produce when dissolved into pure  water or tap water. Usually, tap water contains calcium carbonate, or at least calcium ions. When dissolved into tap water, soap reacts with these calcium ions, producing an insoluble substance which could be called "calcium soap". This substance stays usually in emulsion in water, making it cloudy or turbid. In the rotating washing machines, this deposit remains in the clothes, which become rough when dry. This does not happen with detergents, which do not react with calcium ions, and wash without producing this turbidity.


text:
Why does CCl4 not undergo hydrolysis while other chlorine-derivatives of methane do?
In one of my textbooks, I came across a question as to why does $\ce{CCl4}$ not undergo hydrolysis. The answer given went along the lines of carbon lacking a vacant d-orbital to accept the lone pair of oxygen from water.
However I was curious to find out that other chlorine-derivatives of methane i.e. $\ce{CH3Cl}$, $\ce{CH2Cl2}$ and $\ce{CHCl3}$ (to some extent) do undergo hydrolysis, even though carbon lacks the vacant d-orbital.(Mechanism involved I figured would be Lone pair of oxygen attacking the vacant anti-bonding molecular orbital of carbon)
What is the actual reason that carbon tetra-chloride doesn't undergo hydrolysis while its other chlorine-derivatives seem to undergo hydrolysis effectively?
My only logical approach was the large size of chlorine provides stearic hindrance towards the attacking lone pair of water, But I didn't find any substantiation for this.
A detailed explanation would be appreciated! thanks in advance!

"Does not undergo hydrolysis" is quite a bit of an overstatement. It does, albeit very slowly. Then again, other derivatives don't react fast either.

But the rate with CCl4 is negligible compared to the rate with other derivatives, As similar to the SF6 example, where no hydrolysis takes place and the reason is stearic hindrance contrary to SF4.

Plus in the case of carbontetrachloride , "for all purposes, we can assume no reaction takes place" right

Maybe this could help: Hydrolysis of $\ce{CCL4}$.


text:
Phase equilibrium in mixtures
How should I determine the total number of phases that are at equilibrium with each other in a multi-component mixture?
For example, this problem is from the 2007 GRE published by ETS (Problem 24):

What is the maximum number of phases that can be at equilibrium with each other in a three component mixture?

I initially thought the answer should be 3 since no matter what kind of mixture it is there should always be solid, liquid, and gas. But then I understood that there can be situations like immiscible liquids which increases the total number of phases. The actual answer is 5. What does this number entail?

In equilibrium,  solid quantity produced no effect on equilibrium.

Does this answer your question? melting and boiling ...really equilibrium?

The linked post (supposed dupe) does not address the question in this post so I vote to reopen. In addition the linked post has also been closed.

The solution is a direct application of the phase rule
$$F=C+2-P$$
with the number of components C=3 and F the number of degrees of freedom (the number of intensive variables that can be varied while phases continue to coexist). The phase rule can be derived by considering all the conditions of balance (chemical, pressure, temperature) that must be satisfied by the phases in mutual equilibrium.
Solving for P (the number of phases).
$$P=5-F$$
The maximum number of coexisting phases is observed when there are no degrees of freedom left (a coexistence point), that is when
$$P=5$$


text:
Mesomeric effect in crotonaldehyde
I tried making structures using lone pair of oxygen but I am unable to achieve a structure which doesn't violate the octet rule.

You should shift the pi elctrons in -C=O on O.

Well, yes, the mesomeric effect is shown by the carbonyl group. It's an electron withdrawing group.
What happens here is that since the oxygen is a highly electronegative atom, it will tend to pull the $\pi$ bond electrons towards itself. Thus, the carbon becomes electron deficient i.e. it has it's electrons drawn towards the oxygen.
So, we can safely say that a $\ce{-C=O}$ i.e. the carbonyl group is an electron withdrawing group. So, it has a -R/-M effect.


text:
Does K2CO3 react in minute amounts with glassware at higher temperatures?
At a temperature of $\pu{750 ^\circ C}$, $\ce{K2CO3}$ and $\ce{SiO2}$ react with each other in substantial proportions, forming $\ce{K2SiO3}$ and $\ce{CO2}$ in the process$\ce{^{[1]}}$.
We all know as well that potassium hydroxide reacts with glass. So, I was wondering if $\ce{K2CO3}$ was reacting with pyrex and/or normal glass in minute amounts, either in solution at say >$\pu{150 ^\circ C}$ or when being dried under vacuum in an oven.
One putative indirect mechanism in my wild imagination would be if $\ce{CO2}$ could leave the solution (maybe only at higher temperatures) and if potassium hydroxide at all existed in solution, upon the dissolution of $\ce{K2CO3}$ in water?
References:

Shikha Agarwal, In Engineering Chemistry: Fundamentals and Applications;  Cambridge University Press: Delhi, India, 2016 (ISBN: 978-1-107-47641-7).

No clue at all about possible mechanisms, but some minor level of reaction seems plausible. After all, Gorilla glass involves reactions of glass with molten potassium salts.

Even at lower temperatures (300 to 373 K) and dissolved in water, alkali metal carbonates attack glass slowly and will etch it. Ground glass fittings are difficult to take apart in equipment that has held the carbonates for a long while. Of course, some glasses are more resistant than others; see ISO rating: https://www.iso.org/standard/4895.html

In agriculture, potassium silicate ($\ce{K2SiO3}$) is used as foliar fertilizer ($\ce{K2SiO3}$ is water soluble). The major synthetic method of the preparation of $\ce{K2SiO3}$ is heating $\ce{SiO2}$ and $\ce{K2CO3}$ at high temperatures $(\pu{600-850 ^\circ C})$ in various mole ratios (e.g., Ref.1):
$$\ce{SiO2 + K2CO3 ->[\Delta] K2SiO3 + CO2} \tag1$$
According to Ref.1, alkali metal carbonates such as $\ce{Na2CO3, K2SiO3, Rb2CO3,}$ and $\ce{Cs2CO3}$, all give corresponding metal silicates in this reaction (e.g., silica to metal carbonate mole ratios used in these preparations are $1:2$ to $3:1$).
Keep in mind that the temperatures used in these preparations are below the melting points of both reactants, $\ce{SiO2}$ ($\pu{1713 ^\circ C}$) and $\ce{K2CO3}$ ($\pu{891 ^\circ C}$), thus, the reaction is essentially a solid state one.
Potassium silicate fertilizer grades were successfully produced by direct fusion of silica ($\ce{SiO2}$) and potassium salt ($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of the mixtures (melt temperatures are as high as $\pu{1350 ^\circ C}$). The range of the weight ratio of silicon dioxide/potasium solid used in this method is $1:1$ to $5:1$. The reaction between $\ce{SiO2}$ and $\ce{KOH}$ is being:
$$\ce{SiO2 + 2KOH ->[\Delta] K2SiO3 + H2O} \tag2$$
According to the Ref.1, the other salts (e.g., halides and sulfates) of the metal carbonate catalyze the reaction. For example, in preparing sodium silicate, 6.12 parts by weight sodium carbonate, 3.47 parts by weight silica of minus 325 mesh particle size, and 0.41 part by weight sodium sulfate were ground together (which is $\ce{SiO2 + Na2CO3 + 0.05Na2SO4}$). The ground mixture was heated at a temperature of approximately $\pu{700 ^\circ C}$. for $\pu{4.5 h}$, after which time, the reaction was found to be 51% complete (c.f., without catalyst under same condition, completion was ~$23\%$).
Based on these date, it is safe to say that when silica made contact with an alkali carbonate or hydroxide at elevated temperatures, we can expect a reaction between two chemicals, regardless of they are been in solid form or melt. Once washed, the formed metal silicate would be washed away by water, leaving silica with damaged surface. According to Wikipedia:

The most familiar, and historically the oldest, types of manufactured glass are "silicate glasses" based on the chemical compound silica (silicon dioxide, or quartz), the primary constituent of sand. Soda-lime glass, containing around 70% silica, account for around 90% of manufactured glass.

Therefore, it is no wonder when you applied alkali hydroxide solution (e.g., $\ce{NaOH}$ or $\ce{KOH}$) solution to glass surface for a long time and found the surface become roughed. The alkali carbonated would do the same in higher temperatures.
Note: $\ce{Li2SiO3}$ is only partially soluble in water. However, its $\ce{Li4SiO4}$ version has an advantage of being $\ce{CO2}$ absorber (Ref.3) according to the reaction:
$$\ce{Li4SiO4 + CO2 ->[\Delta] Li2SiO3 + Li2CO3} \tag3$$
Unlike other carbonates, two $\ce{Li2CO3}$ molecules react with one $\ce{SiO32}$ molecule to give $\ce{Li4SiO4}$:
$$\ce{2Li2CO3 + SiO2 ->[\Delta] Li4SiO4 + 2CO2} \tag4$$
References:

Isadore Mockrin, “Production of Silicates,” US Patent 1958, 2,823,098 (PDF).
Srie Muljani, Bambang Wahyudi, Ketut Sumada, Suprihatin, “Potassium silicate foliar fertilizer grade from geothermal sludge and pyrophylite,” MATEC Web of Conferences 2016, 58, 01021 (DOI: 10.1051/matecconf/20165801021) (Proceedings of the 3rd Bali International Seminar on Science & Technology)(PDF).
Xianyao Yan, Yingjie Li, Xiaotong Ma, Jianli Zhao, Zeyan Wang, “Performance of $\ce{Li4SiO4}$ Material for CO2 Capture: A Review,” Int. J. Mol. Sci. 2019, 20, 928 (22 pages) (DOI: 10.3390/ijms20040928).

Thank you mathew. Do you know anything about what happens in the sub 200C in aqueous medium?

I'm sorry I dont know. But, we all know that $\ce{NaOH}$ in any concentration during titration experiments, we keep the solution in plastic bottles because $\ce{NaOH}$ reacts with glass and concentrations get reduced giving us wrong determinations.

Yes it does.
Some years ago I have made an experiment to extract potash K2CO3 from oak ashes.
I have evaporated the filtered leach liquid in an Pyrex dish while having the liquid close to the boiling point.
This whole operation made the Pyrex dish milky looking – respectively the K2CO3 did attack the glass.

What was the concentration and temperature? Did you measure how much loss? I expect it to be milder than for the hydroxides but I'd love to know how much less (if at all).

I have evaporated to dryness. So the concentration was at a maximum at some point.
I dont know about the loss... I can only guess about it. If I have to guess, its just a few mg, but enough to make the glass milky.

About what concentration do you think you were at when the solution started to be milky?

the solution did not turn milky - the glass turned milky as it got etched.

at some point crystals of K2CO3 formed in the solution, thats where the maximum concentration was reached.

And the glass had turned milky after the saturation was reached? I know the reaction of hydroxides with glass has a very exponential curve vs temperature, and I suspect the same is the case here. I'm just wondering if there's a similar relationship with the carbonate concentration (i.e. if lower concentrations of K2CO3 wouldn't react with glass at all vs. near saturation).


text:
Why does a sealed plastic bag of dish soap, water and air expand when shaken?
I've noticed that when I put dish soap, water and a little air into a sealed plastic bag, and then shake the mixture, it forms soap suds and the bag expands significantly in volume. Why does it expand?
Update:
After being prompted by the comments, I tried to re-create and measure this phenomenon under more controlled conditions, but the bag does not expand.
The less-controlled conditions, where the bag expanded in the past, were:

A plastic bag, slightly dirty from crumbs of cooked ground corn, and canola oil (the bag held arepas)
A little squirt of Ultra Ajax Triple Action Orange dish soap (not automatic dishwasher soap). The ingredients (and purpose) are:
Water (Consistency)
Ammonium C12-15 Pareth Sulfate (Cleaning and Foaming Agent)
Lauramidopropylamine Oxide (Cleaning and Foaming Agent)
SD Alcohol 3-A (Controls Thickness and Clarity)
Sodium Chloride (Controls Thickness)
Poloxamer 124 (Controls Thickness)
Fragrance (Pleasant Scent)
Methylisothiazolinone Benziosothiazolinone (Preservation)
Pentasodium Pentetate (Maintains Product Stability)
Sodium Bisulfite (Maintains Product Stability)
Dyes (Color)
Chlorinated tap water
A little air
Everything at around 30 C (a little warmer than room temperature)
Seal the bag and shake. Bag volume increased approximately 2x, with no noticeable change in temperature.

I could observe a similar phenomenon in the past by sealing my palm over the open end of a dirty drinking glass containing dish soap water and air, shaking it, and feeling outward pressure on my palm. I don't recall what the glass was dirty from.
Today, under more controlled conditions, I used a clean bag, and a clean drinking glass, and I could not create any expansion. So I think either the "dirt" in the vessel contributed to the phenomenon, or the tap water was more chlorinated on the days where I observed the phenomenon, and that somehow contributed to expansion. In today's experiments, the water doesn't smell chlorinated.
If I can re-create the expansion later, I will note the conditions as well as I can.
Update 2:
At the request of a commenter, "under more controlled conditions" as described above (and notably where the water does not smell chlorinated), I added table salt, and it did not cause expansion. When the local water authority chlorinates the water again, I will retry the experiments.
Update 3
A similar question has been posted at https://physics.stackexchange.com/questions/617113/why-does-pressure-in-a-thermos-increase-after-shaking-up-hot-water-and-soap

There is no gas, it is trapped air. Point to ponder, would a detergent form foam or suds in vacuum?

Never noticed/thought about. But if the V increases the pressure inside does ultimately originates by mechanical forces linked to surface tension. A bubble must form and so does lifting the bag wall. It principle the air pressure inside has to decrease.

Maybe shaking the solution helps releasing gas that was originally dissolved in water. You could trying either shaking water alone before the experiment, or boiling/cooling it before

What typ of dish soap is it? For manual dish washing or automated? Does the label reveal anything about its content?

May you add some numbers about «expands significantly in volume»?  And, what about the temperature of the bag / the bag's content prior / during / past the experiment recording these numbers?

Could you try the controlled experiment again with one change: add a tablespoon or two of ordinary table salt?

Sorry my half-baked hunch failed! Thanks for doing the experiment and I just upvoted. This is an unexpectedly perplexing little problem.

None of these components releases any kind of gases when (further) diluted or in contact with ordinary dirt. Thermal expansion of the air trapped in the bag and release of gasses dissolved in the tap water may add a little to the volume, but would hardly result in a 2x increase.


text:
Why does p-nitrophenol give sodium bicarbonate test, but o-nitrophenol does not?
Both ortho- and para-nitrophenol are less acidic than carbonic acid. But still paranitrophenol reacts with sodium bicarbonate to give carbonic acid and sodium paranitrophenoxide.
However, this is not true for ortho-nitrophenol. Any explanations?

Hint: consider H-bonding in o-nitrophenol

Thanks for your reply. All it shows is that orthonitrophenol is less acidic but my point is that why does paranitrophenol even react with sodium bicarbonate despite being less acidic than carbonic acid. Try checking the reaction and pka values to satisfy yourself

p-Nitrophenol does have a higher $\mathrm{p}K_\mathrm{a}$ than carbonic acid (7.15 vs 6.3), the two logarithms differ by less than one unit, if we render the acidic species in carbonic acid as $\ce{CO2}$ rather than $\ce{H2CO3}$.  So the equilibrium constant in the reaction
$$\ce{p-C6H4(NO2)OH + HCO3- <=>  p-C6H4(NO2)O- + CO2 + H2O}$$
is close to $1$ and a large percentage of the p-nitrophenol could react with bicarbonate.
o-Nitrophenol actually has $\mathrm{p}K_\mathrm{a} = 7.23,$ about the same as the para isomer.  It should, in fact, react similarly with sodium bicarbonate.  It is the meta isomer that has higher $\mathrm{p}K_\mathrm{a}$ and may not visibly react.

Thanks Oscar, I too had the same views but the textbook i referred says that ortho nitrophenol does not give sodium bicarbonate test. Should I consider that wrong?

Likely the book missed.  Based on duissociation constants the meta isomer is the one that drags its feet.  The proton is indeed hydrogeb-bonded in the ortho isomer, but the closer proximity of the nitro group to the hydroxyl group in that isomer.

@OscarLanzi Can you please explain why p-nitrophenol is less acidic than carbolic acid? Despite that the conjugate base of p-nitrophenol is more stable due to electron withdrawing nature?

Carbonic acid, $\ce{H2CO3}$, is not to be confused with carbolic acid which is another name for phenol $\ce{C6H5OH}$.

@OscarLanzi Oops, I am extremely sorry.

o-Nitrophenol won't react with $\ce{NaHCO3}$ because when $\ce{-NO2}$ group is attached at ortho-position, hydrogen bonding is observed between $\ce{-H}$ of $\ce{-OH}$ and $\ce{-O}$ of $\ce{-NO2},$ due to which acidic strength of o-nitrophenol decreases.
Since o-nitrophenol becomes less acidic than $\ce{H2CO3},$ the equilibrium shifts in backward direction. In other words, it doesn't react with $\ce{NaHCO3}.$


text:
How do scientists know they have discovered an element?
I have a background in math not chemistry and perhaps this question is inappropriate but I will try. I read that after pitchblende was discovered Martin Klaproth in 1879 a latter date discovered that the pitchblende contained another element which he called uranium.
How did he know the uranium was an element and not just another compound that could not be broken down further?  I read Mendeleeve published the periodic table in 1869 but how about before this date? Somehow the first few elements had to be discovered without the benefit of that publication and scientists had to know the substance discoverd and in question was not a compound but an element.

The question in the title is likely a duplicate of How were the elements discovered before modern technology?. If your question is specifically about uranium, Wikipedia has a brief description of the chemistry allowing to determine that the new element was indeed new. And oh, it's 1789, not 1879. Would be nice if Klaproth lived a bit longer though:)

Mistakes were not uncommon. Compounds were mistaken for elements and vice versa.

History tends to remember those who got it right.

@Karl Except, of course, for Stigler’s Law of Eponymy. ;)

From @Buttonwood again, here is a bonanza of periodic tables: https://www.meta-synthesis.com/webbook/35_pt/pt_database.php?Button=1950-1999+Formulations. Have fun!

@Ed V nice link  that will come in handy !!!


text:
Ortho-para directing and meta directing groups: Activating or deactivating the aromatic ring?

Determine the reactivity of the above compounds (activating or deactivating; ortho-para or meta directing). Explain your answers. You may use resonance structures in your explanation. Which molecule will have the highest rate of reaction?
What are the products formed when the above compounds are reacted with $\ce{Br_2}$ and $\ce{FeBr_3}$. Predict the major and minor compound, if there is any.

My answers:
For Question (1):
E is an activating group because it has a nitrogen with a lone pair. F is a deactivating group because $\ce{C=O}$ is directly attached to the benzene ring. In G, $\ce{N}$ is an activating group but $\ce{C=O}$ is deactivating, so I'm not sure if G is an activating or deactivating compound. Also, How can resonance structures help to explain the reactivity of the compound? I'm not sure how to draw the resonance structures. How do we draw resonance structures that has two substituents such as G. For F, does the lone pair on $\ce{N}$ affect the resonance?
For Question (2):
E: $\ce{Br}$ will attach at ortho and para, so it has two products.
F: $\ce{Br}$ will attach at meta with 93 % (major) and ortho 7 % (minor).
G: I think $\ce{Br}$ will attach para to $\ce{N}$ because $\ce{N}$ is a strong activating group and it will not go to between the two substituents because it is crowded, is this true?

I'm not sure if G is an activating or deactivating compound.

You may want to look at the relative order of activation/deactivation of some substituents,
(source: www.chem.ucalgary.ca)

As per the image, $\ce{-NR2}$ is a strongly activating group, while $\ce{-CONH2}$ is a moderately deactivating group (bcoz, it lies somewhere in between $\ce{-COOH}$ and $\ce{-COCl}$). Hence, the net effect would be activating.

How can resonance structures help to explain the reactivity of the compound?

Resonance structures can tell whether the substituent is o,p -directing or m -directing. It can't say anything about rate of reaction.

How do we draw resonance structures that has two substituents such as G?

Just like you've drawn for a single substituent. The only difference is that the other substituent has replaced one of the $\ce{H}$ from the resonance structure drawn with single one. So, the other may participate in delocalization.

For F, does the lone pair on N affect the resonance?

Obviously, but not significantly, as it will participate in resonance too.

I think Br will attach para to N because N is a strong activating group and it will not go to between the two substituents because it is crowded, is this true?

Yes, but it is partially true.
para product (wrt. $\ce{-NR2}$) will be major, however, ortho product (wrt. $\ce{-NR2}$) will be also present as the minor one.


text:
Hydrophobic interactions and Gibbs Free Energy
I want to understand the basis of the hydrophobic effect.
In biochemistry, protein structures are often explained in terms of hydrophilic and hydrophobic effects.  The hydrophilic interaction is understandable in terms of electrostatic interactions but I don't understand the nature of the hydrophobic effect.
I have seen explanations in terms of the Gibbs free energy of a system. For example, if I take an oil and water mixture we know the mixture spontaneously separates, but what are the forces involved?
I get it in terms of Gibbs Free Energy:
$\Delta G = \Delta H -T \Delta S$
If the oil and water mixture is shaken so that there is an emulsion and then it's left to settle, hydrogen bonds are broken and form to the value of $\Delta H$.  I assume that more hydrogen bonds are formed as water and oil separate but I believe that the main driver is entropy.  Overall the entropy change is positive because the ordered structure of water bonds around each individual fat droplet are broken down and entropy increases, and hence $\Delta G$ is negative. OK, but to me that does not explain why individual fat particles to move.  If there's a change in inertia, then there's a force?  What is the force?
I can accept this explanation on an energetic level but what forces are acting on the oil droplets to effect this?  I can't see any Coulomb forces here.  Whilst I can accept that energetically it might make sense for oil and water to separate, I feel I need a force to explain why matter moves.  Could someone explain what the nature of the hydrophobic interaction is because I don't understand at present.
Thank you.
My explanation was taken from this link: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Hydrophobic_Interactions#:~:text=Hydrophobic%20interactions%20describe%20the%20relations,example%20of%20this%20particular%20interaction.
Edit:  I note someone asked the exact same question.  I didn't see it before but I don't think it's been answered.
How does the entropic force arise within the hydrophobic effect?

Not very clear. Which motion do you refer too? All molecules are in motion in your sample. If you mean why fat and water layer, then it is basically density. If the two phases would have very similar density but the same energetics would be involved, I suppose the emulsion would be equally unstable but it can take very long to layer.

Before Walter Kauzmann "discovered" hydrophobicity (?? see the libretext), Einstein explained theoretically Brownian (diffusional) motion.

Alchimista. I am asking what the hydrophobic effects is between water and non-polar substances such as hydrocarbons.  Molecules with long hydrocarbon chains are described as hydrophobic.  What is the basis for that?

The question you link to does have an answer. However your question appears to be "I feel I need a force to explain why matter moves". That's thermal energy

I knew about that explanation.  I refer to it in my question.  I can see how the hydration entropy would lead to a lower energy state for for aggregating non-polar molecules in water but as Buck says I struggle to see how a thermodynamics sum can propel a hydrocarbon molecule to move away from water molecules.  I am probably not grasping something but it does not make sense to me.

So it is energetically favourable for hydrophobic molecules to interact with each other because by doing so the entropy of water molecules increases when they are no longer highly ordered when surrounding small particles of hydrophobic molecules.
OK.  I concede.  In effect I can see that an input of energy would be required to cover the loss of entropy when water bonds to itself around hydrophobic molecules, and in practical terms I can see that work would have to be done.  I don't think a hydrophobic interaction is really a very good description of this phenomena.

You appear to mix a few different topics so I'll try addressing only two. First, equilibrium thermodynamics deals with macroscopic bodies or with ensembles, which is to say, it describes average static (time-independent) behavior, it does not deal with forces or instantaneous values of the potentials that are typically used in describing dynamics. Non-equilibrium thermodynamics does deal with systems that are dynamic (away from equilibrium), and maybe that is a more useful tool to describe what you are interested in.

You may be right, perhaps hydrophobicity is in fact not a good term to describe what is happening when water and oil mix, because it is repulsion of oil by water that accounts for most of the work in separating the two components. Perhaps it should be called "oleophobicity". The relevant forces would be hydrogen bonding interactions between the water molecules.

Thanks Buck.  If I take two molecules, a water molecule and a lycine molecule, and I reduce their separation from infinity, I believe they would describe the Lennard Jones potential.  Initially attractive forces would be generated by induced dipoles and a net repulsion would only be exhibited when electron clouds start to interact.  I think the hydrophobic effect is a macroscopic effect due to the affinity of water to hydrogen bond with other water molecules rather than by any interaction between polar and non-polar molecules.

The hydrophobic effect does not require a macroscopic system to be evident, and it is ultimately explained in terms of entropy (the answer to the linked post explains how this works). You might think of its like this: waters like to H-bond to each other. When an apolar compound is solubilized, waters tend to form shells around the solute while continuing to H-bond with other waters. Waters about the solute have less H-bonding choices and therefore lower entropy, which is unfavorable.


text:
Vibrational state dependency of the reaction cross section
I'm reading p.1238 of Mcquarrie's Physical Chemistry: A Molecular Approach textbook.
In the last line of the page, the author explains that the value of the reaction cross section in the constant total energy depends strongly on the vibrational state of the reactant. Of course, it means that not only translational kinetic energy but also internal energy(such as vibrational state) contributes to the value of the reaction cross section values.

However, the book puts a limit on the 'constant total energy'. If both translational kinetic energy and vibrational energy contribute to the reaction cross section value, does the explanation in the book imply that vibrational energy contributes more to the reaction cross section value than translational kinetic energy?
If that's true, is there a way to explain the difference in contribution qualitatively? Or is it just a matter of quantitative differences?

Vibrations are very important for reactions. In the particular example of $\ce{H2+(g) + He}$

vibrational levels 0-3 have an energy less than E0 (minimum energy for reaction), so additional transitional energy is needed to induce reaction. That’s why we see a threshold.
vibrational levels > 4 already have more than E0 energy so they react even with no additional energy

The greater the vibrational energy level of $\ce{H2+}$ the easier it will be for the substitution reaction, and this makes sense considering the transition state diagram

You may also be interested in the "harpoon" reaction which demonstrates how vibrations are important 1.
This reaction is known as a harpoon reaction because an electron is first transferred from M in $\ce{M...XR}$ or $\ce{[M...XR]^*}$ to $\ce{RX}$ and thereby forms the ionic transition state $\ce{M+...RX-}$. The negatively charged $\ce{RX-}$ is unstable and can dissociate, forming the products $\ce{M+X-}$ and $\ce{R}$.
These type of reactions are, in general, valuable for studying how rovibrational states affect the reaction probabilities because they can be triggered by light, and are small enough to model with extremely accurate theories.
In particular, the reaction probability has notable "resonances" spaced precisely 10.9 $cm^{-1}$ apart. These resonances are most likely due to low frequency bending mode of the T-shaped complex.

References:

Skowronek, S., Jiménez, J. B. & González Ureña, A. Resonances in the Ba...FCH3+hv→BaF+CH3 reaction probability. J. Chem. Phys. 111, 460–463 (1999).


text:
Why does barium sulphate not dissolve in HCl while barium sulphite does?
Intuition tells me that the ionic bond in the sulphate must be stronger than that in the sulphite.
Is this true?
If it is then why? The charge on the barium ion is 2+ in both cases and the charge on the sulphate and sulphite ions are both 2-.
The sulphate has an extra oxygen atom. It's presence is somehow making the ionic bond stronger. But how/why?

Do you know about the reaction of sulfites and acid?

The bond is pretty much the same. That's not where the explanation lies.

Consider sulphurous acid ( rather SO2(aq)) acidity constants pKa1=1.857, pKa2=7.172. As consequence, the BaSO3 solubility in HCl is high, as SO3^2- concentration is low.

Strongly related: https://chemistry.stackexchange.com/questions/115827/is-barium-sulfate-soluble-in-dilute-mineral-acid

and https://chemistry.stackexchange.com/questions/17271/how-to-recognize-barium-sulphite

@EdV No , what can you tell me about it ?

@IvanNeretin Where does the explanation lie ?

Follow the second link provided by @NilayGhosh. Look at the answer there.

$\ce{BaSO_3}$ has the same property as $\ce{BaCO_3}$ :
When reacting with an acidic solution, these two substances produce a gas which can get out of the solution, and displaces the equilibrium to the right  : $$\ce{BaCO_3 + 2 H^+ ->  Ba^{2+} + CO_2(g) + H_2O}$$
$$\ce{BaSO_3 + 2 H^+ -> Ba^{2+} + H_2O + SO_2(g)}$$ This allows these two substances to be soluble in $\ce{HCl}$ solutions. As a contrary $\ce{BaSO_4}$ has not this possibility to produce a gas in an acidic solution : it is and remains insoluble in $\ce{HCl}$ solutions.


text:
How to convert an XYZ file to Z-matrix?
I would like to generate a Z-matrix from the following XYZ file (C2H6dimer.xyz):
16
C2H6dimer
C    0.539202   -0.539202    1.933833
C   -0.539202    0.539202    1.933833
H    1.536064   -0.098299    1.932264
H    0.459532   -1.174264    2.816008
H    0.456405   -1.179449    1.055941
H   -1.536064    0.098299    1.932264
H   -0.459532    1.174264    2.816008
H   -0.456405    1.179449    1.055941
C    0.539202    0.539202   -1.933833
C   -0.539202   -0.539202   -1.933833
H    1.179449    0.456405   -1.055941
H    1.174264    0.459532   -2.816008
H    0.098299    1.536064   -1.932264
H   -1.174264   -0.459532   -2.816008
H   -0.098299   -1.536064   -1.932264
H   -1.179449   -0.456405   -1.055941

However I can't do it with the Gaussian utility called newzmat, because after typing newzmat -ixyz C2H6dimer.xyz -ozmat, I receive the following error message:
 Error in cartesian coordinates.
  WANTED A FLOATING POINT NUMBER AS INPUT.
  FOUND AN END-OF-LINE FOR INPUT.
 16
   ?
 Error termination via Lnk1e at Sun Jun 21 01:09:30 2020.
Segmentation fault

Could you please help me to figure out what the problem is?

Please include anything that is a prerequisite to create the problem. Am I right to assume that the program newzmat is a utility of Gaussian (which version)?

Can you try to remove the last (empty) line? Gaussian can be very strict about empty lines in places it does not expect them. Maybe newzmat behaves similar.

According to the manual of the newzmat utility of Gaussian 16, it accepts unadorned Cartesian coordinates (G16 online manual). The format you have posted is actually (simple) xmol. This means the first line denotes the number of atoms, the second line denotes a comment, and both need to be skipped.
Therefore your file should look like this:
C    0.539202   -0.539202    1.933833
C   -0.539202    0.539202    1.933833
H    1.536064   -0.098299    1.932264
H    0.459532   -1.174264    2.816008
H    0.456405   -1.179449    1.055941
H   -1.536064    0.098299    1.932264
H   -0.459532    1.174264    2.816008
H   -0.456405    1.179449    1.055941
C    0.539202    0.539202   -1.933833
C   -0.539202   -0.539202   -1.933833
H    1.179449    0.456405   -1.055941
H    1.174264    0.459532   -2.816008
H    0.098299    1.536064   -1.932264
H   -1.174264   -0.459532   -2.816008
H   -0.098299   -1.536064   -1.932264
H   -1.179449   -0.456405   -1.055941

In Gaussian terms, a block of cartesian coordinates is also a z-matrix. Therefore you also need to tell Gaussian to rebuild it (-rebuildzmat). This will likely fail on your input with an error:

Input z-matrix variables are not compatible with final structure.

If you allow rounding (-round) it will work. The complete command is:
newzmat -ixyz -ozmat -round -rebuildzmat <input.xyz> <output.com>

The following file is produced (it is a valid Gaussian input file):
# HF/6-31G* Test

-- No Title Specified --

 0,1
 C
 C,1,R2
 H,1,R3,2,A3
 H,1,R4,2,A4,3,D4,0
 H,1,R5,2,A5,3,D5,0
 H,2,R6,1,A6,3,D6,0
 H,2,R7,1,A7,6,D7,0
 H,2,R8,1,A8,6,D8,0
 C,3,R9,1,A9,2,D9,0
 C,9,R10,3,A10,1,D10,0
 H,9,R11,3,A11,10,D11,0
 H,9,R12,3,A12,10,D12,0
 H,9,R13,3,A13,10,D13,0
 H,10,R14,9,A14,3,D14,0
 H,10,R15,9,A15,14,D15,0
 H,10,R16,9,A16,14,D16,0
      Variables:
 R2=1.53
 R3=1.09
 R4=1.09
 R5=1.09
 R6=1.09
 R7=1.09
 R8=1.09
 R9=4.04
 R10=1.53
 R11=1.09
 R12=1.09
 R13=1.09
 R14=1.09
 R15=1.09
 R16=1.09
 A3=111.
 A4=111.
 A5=111.
 A6=111.
 A7=111.
 A8=111.
 A9=81.
 A10=94.
 A11=22.
 A12=128.
 A13=104.
 A14=111.
 A15=111.
 A16=111.
 D4=120.
 D5=-120.
 D6=180.
 D7=120.
 D8=-120.
 D9=-76.
 D10=-21.
 D11=-144.
 D12=-120.
 D13=113.
 D14=167.
 D15=120.
 D16=-120.

 1 2 1.000 3 1.000 4 1.000 5 1.000
 2 1 1.000 6 1.000 7 1.000 8 1.000
 3 1 1.000
 4 1 1.000
 5 1 1.000
 6 2 1.000
 7 2 1.000
 8 2 1.000
 9 10 1.000 11 1.000 12 1.000 13 1.000
 10 9 1.000 14 1.000 15 1.000 16 1.000
 11 9 1.000
 12 9 1.000
 13 9 1.000
 14 10 1.000
 15 10 1.000
 16 10 1.000

Whether or not this z-matrix does actually fits the purpose you are needing it for, is a different question. For the most tasks, where z-matrices are the superior input, I recommend building it by hand.

I copied and pasted your XYZ into this and got the Z-matrix for your XYZ coordiantes (see below). Geoff Hutchison's comment points out that this uses OpenBabel. Here is a ZMAT for your XYZ coordinates:
16
C   1
C   1 1.525
H   1 1.090  2 111.141
H   1 1.090  2 111.120  3 119.9
H   1 1.090  2 111.206  3 239.9
H   2 1.090  1 111.141  3 179.8
H   2 1.090  1 111.120  3 299.7
H   2 1.090  1 111.206  3  59.7
C   8 3.216  2 130.964  1  32.0
C   9 1.525  8  85.518  2  56.3
H   9 1.090  8  56.495  2 297.1
H   9 1.090  8 161.358  2 262.3
H   9 1.090  8  72.021  2 170.2
H  10 1.090  9 111.120  8 129.0
H  10 1.090  9 111.141  8 248.9
H  10 1.090  9 111.206  8   9.1

In my experience, this is the fastest and easiest way to to convert an XYZ file to Z-matrix format, except maybe in extremely difficult cases.

Incidentally, this uses Open Babel behind the scenes.

There is a way to do this in Avogadro Molecular Editor and Visualizer that you can download freely.

Open your .xyz file in Avogadro. Click on the extensions in the top.

In the extensions, select Gaussian. This opens a guassian input editor like below.

Change the format to Z-matrix or Z-matrix (compact), whichever one you like.

This editor has a nice interface that you can use to edit your Gaussian input.

Incidentally, this also uses Open Babel behind the scenes.


text:
TKN: Why are they called Missouri tablets?
When analyzing for Total Kjeldahl Nitrogen (TKN), you can make a copper sulfate digestion solution in the lab or purchase a variety of pre-prepared digesting tablets for example on VWR. My question is, why are some formulations called Missouri tablets?
As far as I can tell, the state of Missouri does not specifically require this formulation, and I haven't seen any digestion tablets named after any other states. Were they perhaps invented in Missouri?

Comments are not for extended discussion; this conversation has been moved to chat.

As suggested in the comments, the predecessor of the method that uses "Missouri" tablets was developed at the University of Missouri (Columbia, Missouri, USA). I believe the first report was from 1972 and named the method as the "Missouri Automated Nitrogen Method (MANM)". They also refer to it as the "Missouri Technicon" method because it uses a Technicon AutoAnalyzer instrument. Many variations of this method have been described, and the specific formulation of Missouri tablets is for one of them.

An automated spectrophotometric method, utilizing Technicon AutoAnalyzer modules, has been developed to determine total nitrogen in fertilizers containing only ammoniacal, nitrate, and urea nitrogen. This colorimetric system employs the Berthelot ammoniaphenate-hypochlorite reaction. A homogeneous chromous/titanous reduction system for the automatic reduction of nitrates was interfaced with the digestion unit and color development manifold, resulting in a total nitrogen system for fertilizers with an effective analysis rate of $\pu{20-25 samples/hr}$. A new sample retrieval system removes digested samples from the helix. The totally automated system was optimized to obtain the highest sensitivity for nitrogen. Data were compared for $\pu{458 samples}$ analyzed by the automated method and the comprehensive nitrogen method (CNM), 2.053–2.054. The average difference between the 2 methods (MANM – CNM) was +0.04% nitrogen. The average relative difference was 1.12%. The average value for $\pu{140 samples}$ of $\ce{KNO3}$ primary standard analyzed by the MANM was 13.84% nitrogen (theoretical content 13.85%). The results of these standards ranged from 13.58 to 14.14% nitrogen and the standard deviation was 0.11%. The relative standard deviation was 0.80%. Of fertilizer samples received for analysis, less than 1% contained insoluble organic nitrogenous material which was unsuitable for analysis by the MANM. These few samples were not included in the statistical evaluation. The MANM is rapid, accurate, and precise and is a convenient and dependable automated method for the determination of total nitrogen in fertilizers containing only ammoniacal, nitrate, and urea nitrogen.
 Source: Charles W. Gehrke, John S. Killingley, Larry L. Wall, Sr. (1972) Missouri Automated Nitrogen Method (MANM) for Fertilizers. Journal of Association of Official Analytical Chemists 55:467-480. https://doi.org/10.1093/jaoac/55.3.467

A rapid automated method for total protein nitrogen has been developed, using the Technicon AutoAnalyzer with block digestor. Ammonia is determined by the automated ammoniasalicylate reaction at a rate of $\pu{40 samples/hr}$. A number of variations in digestion parameters have been evaluated. Hydrogen peroxide was used as a digestion accelerator. A salt-acid ratio of 1:1 and a block temperature of $\pu{425 ^\circ C}$ were chosen. The catalysts evaluated included $\ce{HgO, CuSO4, SeO2,}$ and $\ce{TiO2/CuSO4}$. Maximum nitrogen recovery in the shortest time (30 mill) was achieved with HgO as the catalyst. The results from multiple analyses of 10 experimental samples of different refractoriness with the block digestor method, using $\ce{HgO}$ or $\ce{CuSO4}$ as a catalyst, compared well with the results by the official AOAC Kjeldahl method. The accuracy, precision, economy, and saving of space offered by the Missouri-Technicon block digestor method make this an attractive alternative to classical Kjeldahl analysis for large numbers of total protein determinations in samples of different refractoriness. The AutoAnalyzer cartridge manifold is simple and reproducible in its performance, and a large dilution of the sample is made on dialysis which eliminates the matrix effects from the sample digests. At least $\pu{250 samples}$ can be analyzed in $\pu{9 hr}$ with 1 instrumentation setup.
Source: Larry L. Wall, Sr., Charles W. Gehrke (1975) An Automated Total Protein Nitrogen Method. Journal of Association of Official Analytical Chemists 58:1221-1226. https://doi.org/10.1093/jaoac/58.6.1221

I think it's worth adding that nowadays "Missouri" refers to one specific type of Kjerdahl tablets (among many others) containing 0.3% Cu and is essentially a trademark given to the most widely used composition in tribute to the extensive agricultural chemical research performed at Missouri's Agricultural Experiment Station Chemical Laboratories.

I edited to address your point. Any better now? I don't think it's quite accurate to say that the name was given in tribute, since the Missouri group had already established the name of the method as Missouri, so my impression is that the choice of "Missouri tablets" for the product is simply a way of saying "tablets for use in the Missouri method"