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What is the vapor pressure of a liquid if it is not in a sealed container(i.e., a river or lake)? Is it just the atmospheric pressure?
From what I understand, the boiling point of a liquid is when its surrounding pressure is equal to its vapor pressure. The vapor pressure is the pressure of the gas form of the liquid in a container when the rate of condensation equals the rate of evaporation. If this is the case, what is the vapor pressure of a liquid that is not placed in a container(i.e., a river or lake)? Is it the same as the pressure of the atmosphere? If so, then how does this affect its boiling point?
(Please correct me if what I know is not true).

Do you mean what the saturation vapor pressure is, or what the dynamic vapor pressure (perhaps as relative humidity) happens to be at a given time?

In such a case, the vapour presure is the pressure of the gas, if all other gases were absent.  In river/lake cases, this presure is variable in space and time, not reaching equilibrium.

From what I understand, the boiling point of a liquid is when its surrounding pressure is equal to its vapor pressure.

This is backwards. The boiling point is when the vapor pressure of the liquid is equal to the atmospheric pressure. So given the average pressure of 1 atmosphere at sea level water would boil at $\pu{100 ^\circ C}$. However on some mountain top water might boil at $\pu{95 ^\circ C}$ since the air pressure is lower.

The vapor pressure is the pressure of the gas form of the liquid in a container when the rate of condensation equals the rate of evaporation.

For a particular temperature, in a sealed container, when the rate of condensation equals the rate of evaporation then the system is at equilibrium. The partial pressure of the gaseous form of the liquid is then the equilibrium vapor pressure of the liquid.
Without being at  equilibrium the partial pressure could be anything.

If this is the case, what is the vapor pressure of a liquid that is not placed in a container(i.e., a river or lake)? Is it the same as the pressure of the atmosphere? If so, then how does this affect its boiling point?

If I have an open beaker of benzene on an outside picnic table then the system isn't at equilibrium and benzene has no defined vapor pressure. There just isn't enough benzene to ever reach an equilibrium with the whole atmosphere of the earth.
This wouldn't effect the boiling point of the benzene. Assuming that the air pressure is 1 atmosphere, benzene would boil at the same temperature whether if it was being boiled  in the open air or refluxed in a lab.
This bring up another point. If we say that the vapor pressure of water is $\pu{0.0231 atm}$ at $\pu{20 ^\circ C}$, we are referring to "local conditions." In other words, that gravity can be assumed to be  a constant within the system. Obviously if I had a cylinder 200 miles high at $\pu{20 ^\circ C}$, the vapor pressure wouldn't be the same at the top and bottom of the cylinder due to the difference in gravity.

For each given pressure $p$ on a liquid there is a temperature at which the liquid will boil and become gaseous.$^1$ This temperature $T_\text{boil}(p)$ is called boiling temperature. The function $T_\text{boil}(p)$  can be calculated by the Clausius–Clapeyron equation.
The pressure on water in a river on earth is $1\ \mathrm{atm}$. The corresponding boiling temperature is $T_\text{boil}(1\ \mathrm{atm})\approx 100\ \mathrm{^\circ C}$.
One can on the other hand invert the relationship and ask for a given temperature $T$ at which pressure the liquid will become gaseous. This pressure is then called vapor pressure.
$^1$ that is assuming that $p$ is smaller or equal than the critical pressure of that fluid.


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how much baking soda and vinegar will you need to fully fill a ziplock bag up without popping it?
EXACTLY how much baking soda and vinegar will you need to FULLY fill a ziplock bag up without popping it or it being deflated?
I am really confused. That is all our teacher told us and he said that we need to be prepared with a procedure for a lab. I know it deals with stoichiometry but don't know where to go from there. We know we can find the volume of the bag by filling it with water and emptying the water into a beaker.
WHAT DO I DO NEXT? help greatly appreciated!
The Ziploc bag is 1 quart.
I got this formula-
HC2H3O2+NaHCO3------>NaC2H3O2+H20+CO2
THANKS IN ADVANCE!!!

There is too little information ,as we do not know the endurance of the bag. Hmm, I see 1 Quart = cca 0.946 L.

Assuming the water doesn't displace much volume in the bag, the volume a mole of CO2 displaces at room temperatre is 22 L.
pV=nRT
p=1 atm
T=298 K
R=0.0821 $\frac{L*atm}{mol K}$
Step 1: find out how much volume the bag is in liters (units need to be consistent)
Step 2: Using pV=nRT solve for the number of moles using that volume of CO2 that you need
Step 3: The ratio of CO2 to acetic acid (HC2H3O2) and sodium bicarbonate (NaHCO3) is 1 so you need that many moles of each
Step 4: Convert moles of HC2H3O2 and NaHCO3 to mass using their molar masses

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Conversion of cis to trans isomer
why in cis and trans isomers the double bond cannot be broken once and again joined to convert them from cis to trans is this type of bond breaking not allowed or is this allowed?

It can.$\mathstrut$

If it could not, you would not see.

The double bond consists of a sigma and pi bond, which is formed between two p orbitals. When the molecule has enough energy, obtained by random collisions in solution, the pi bond can break leaving just the single sigma bond. At this point should the double bond reform it is as if nothing has happened, however, if there is a slight rotation of the end group when only the single bond exists (close to the transition state) then cis-trans conversion can take place and the pi bond reforms after 180 degree rotation of one half of the molecule.
(Recall that a molecule's bonds vibrate and groups have have bending/ rotational  motion also. The actual motion during isomerisation is more complex involving a stretching of the double bond first, which weakens it as the transition state is approached, and then oscillatory motion around the transition state.)

The breaking of double bond requires a sufficently large energy which may also dissociate the compound also first breaking double bond and then again reforming through a reduction reaction is not feasible...pls comment if this helps you

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How to calculate "volume of gas" generated by a given ion ore in a blast furnace?

A blast furnace makes pig iron containing $3.6\% \; \ce{C}$, $1.4\% \; \ce{Si}$, and $95\% \; \ce{Fe}$. The ore is $80\% \; \ce{Fe2O3}$, $12\% \; \ce{SiO2}$, and $\ce{Al2O3}$. The blast furnace gas contains $28\% \; \ce{CO}$ and $12\% \; \ce{CO2}$. Assume that there is no iron loss through the slag. The atomic mass of $\ce{Fe}$, $\ce{Si}$, and $\ce{Ca}$ are $56$, $28$, and $\pu{40 g/mol}$, respectively.

What is the weight of the ore used per ton of pig iron?
What is the volume of the blast furnace gas produced per ton of pig iron?

The first question involved finding the weight of ore used (Exact quote: "The weight of the ore used per ton of pig iron is _"). I was able to do that using algebraic method to generate equations. The method that I applied was very much similar to this. However, a follow up question is asked about the volume of the blast furnace gas (Exact quote: "The volume of the blast furnace gas produced per ton of pig iron is___"). I do not know how to do that. All that I would prefer is to use a similar approach that has been used.

As there is no iron loss through the slag, all your calculations must be made with respect to the iron atom, which is kept constant through all the process. So first, calculate the mass of pure iron in 1 ton pig iron. Then calculate the amount of substance of iron element in this mass. Then divide by two to obtain the amount of substance of iron oxide in the ore. Then calculate the mass of iron oxide in these amounts of iron oxides. Then divide by $0.8$ to get the mass of needed ore. At the end you obtain about $1.7$ tons of ore. Go!
To solve the second part, it is necessary to know the percentage of $\ce{CO}$ and $\ce{CO2}$ in the gas is a percentage in volume (= in amount of substances) or in mass.

What is the weight of the ore used per ton of pig iron?

The ore is $80\% \; \ce{Fe2O3}$, but the pig iron is $95\% \; \ce{Fe}$.
$$\pu{fraction}\ \ce{Fe}\ \pu{in} \; \ce{Fe2O3}= \dfrac{2\cdot56}{2\cdot56 + 3\cdot16}\times 100 = 0.70$$
$$\pu{tons ore} = \dfrac{95\%}{80\%\cdot0.70}= 1.7$$

What is the volume of the blast furnace gas produced per ton of pig iron?

This solution to the second part is a bit more convoluted.
First we were given that "the ore is $80\% \; \ce{Fe2O3}$, $12\% \; \ce{SiO2}$, and [some fraction of] $\ce{Al2O3}$." For the components of the ore to add to $100\%$, there must be $8\% \ \ce{Al2O3}$.
Next the problem states "The blast furnace gas contains $28\% \ \ce{CO}$ and $12\% \ \ce{CO2}$." Here we must just assume that whatever the other 60% of the gas is, that it is inert. Also no temperature is given for the furnace so it is impossible to calculate a volume of gas, only a mass of $\ce{CO}$ can be calculated.
Now a twist here is that the carbon in the pig iron could have come from the $\ce{CO}$ or $\ce{CO2}$. However if the reaction
$$\ce{CO2 -> C + O2}\tag{2.1}$$
happened, then assuming complete combustion
$$\ce{2CO + O2 -> 2CO2}\tag{2.2}$$
Thus all of the carbon and all of the oxide removal is due to reactions with $\ce{CO}$.
$$\ce{Fe2O3 + 3CO <=> 2Fe + 3CO2 }\tag{2.3}$$
$$\ce{SiO2 + 2CO <=> Si + 2CO2}\tag{2.4}$$
$$\ce{Al2O3 + 3CO <=>2Al + 3CO2}\tag{2.5}$$
$$\ce{2CO <=> C + CO2}\tag{2.6}$$
The problem doesn't specify US tons or metric ton, so make it easy and assume metric ton. Now we need to figure out how much CO is needed, component by component.
\begin{array}{|c|c|c|c|}\hline
Element & \%\ in \ pig \ iron & kg & kilo-moles & kilo-mole\ CO \\ \hline
Fe &95\%& 950 & 950/56 = 17.0 & 3/2 \cdot 17.0 = 25.4\\ \hline
Si &1.4\% &14 & 14/28 = 0.5 & 2\cdot 0.5 = 1.0 \\ \hline
Al & - & & \\ \hline
C & 3.6\% & 36 & 36/12 =3.0 &2\cdot 3 = 6\\ \hline
\end{array}
Thus we need 32.4 kilo-moles of $\ce{CO}$ which is $28\cdot32.4 = \pu{907 kg}$, or 0.91 tons of $\ce{CO}$ per ton of pig iron (be they short tons, long tons, Imperial tons, or metric tons!)

The volume of the gas has been asked.

@user586228 - What is the temperature of the gas? As I noted since there is no temperature, it is impossible to calculate a volume.

Assume everything in Standard condition .Take 298K.

298K coming out of a blast furnace?!? // Anyway I've done the heavy lifting, I'll leave the volume calculation to you.Also if you want to do a volume calculation you'll have to assume some particular kind of ton.


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MNDO integrals in the global coordinate system
In MNDO and its descendants (Sewart's PMx models, Austin's AM1, MNDO/d, etc.), the nonvanishing two-center two-electron "coulomb integrals" are evaluated via a classical multipole expansion in a local coordinate system, and then transformed into global coordinates via "standard rotational transformations". How exactly are these transformations carried out?
A similar question would be the methodology behind the resonance integrals in the orthogonalised methods OMx, which are similarly evaluated in a diatomic coordinate system and then transformed to the global coordinate system.

There's a nice open-source C++ implementation of NDDO methods in Sparrow - you can probably hunt around in the code and/or ask questions of the Reiher team.

@GeoffHutchison Does anyone know where to get MNDO99? Walter Thiel was the last known developer but has since passed. I'm interested in using it for OM2/MRCI calculations. If you do, let me know and I'll reward you with a bounty on materials SE (50 pts)!

@CodyAldaz - A Google search turned up a stub of a project on GitHub: https://github.com/andersx/pymndo99 - I'd contact Anders and ask him. Martin Korth and Paolo Drahl both worked with Walter and may have copies. I'd definitely be curious to benchmark OMx methods..

@Cody Aldaz I believe you’re looking for MNDO2005 7.0 (released 2015); I’m looking for it myself. I think your best bet is to contact Thiel’s secretariat (see the Max-Planck-Institüt website)

MNDO2020 (MNDO99) from Prof. Walter Thiel is now free and open source under a special MPI license. Contact: https://www.kofo.mpg.de/en/research/molecular-theory-and-spectroscopy


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Why does copper oxidise if it is considered a "noble metal"?
I'm an electrical engineer and not a chemist, so please excuse me if this is obvious.
When soft-soldering electronic parts, we’re normally taught to clean copper of oxides (like using flux and so on) before soldering. Also I’ve observed that when heating copper to $\ce{\approx 200 ^\circ C}$ copper changes colour and builds an oxide layer. So why is copper considered a noble metal? Shouldn’t it be resistant to oxidation?
If the definition of noble metal is “less oxidation” instead of “no oxidation": Does that mean that gold and silver also build a small oxide layer in the $\pu{25 ^\circ C} - \pu{200  ^\circ C}$ temperature range?

Nobel metal is a very woolly concept without much scientific basis. Silver and mercury are pretty reactive too. I would not give much time to the concept.

Even Pt oxidises, who's Cu?

In context of chemistry of metals, the adjective noble means more or less they do not dissolve in non oxidating acids. But copper slowly dissolves even in vinegar in oxygen presence.

To my understanding "noble metal" meant that those metals frequently found in the metallic form in nature rather than an ionic form. Gold silver platinum and copper were all discovered as they can be readily obtained straight from the ground without smelting an ore. Most modern copper is mined as an ore and smelted to the metallic form, but substantial amounts of the metallic form do naturally occur. Many metals resist bulk oxidation by forming a protective surface oxide layer.

Thanks everybody. I gather that the concept of  "noble metal" is not scientific enough. Regarding my second question: Do gold and silber also build a small oxide layer at their surface under a normal atmosphere in the  25°C – 200 °C temperature range??

Pure gold(24 karat gold) does not tarnish at all but if that is alloyed with other metals, it tends to tarnish(https://www.onecklace.com/tips/does-real-gold-tarnish/). As for silver, it is easily tarnished from oxygen and moisture(https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5111099/) and is common knowledge that silver reacts with sulfur in air to form black $\ce{Ag2S}$. See: https://chemistry.stackexchange.com/questions/78882/abundance-of-sulfur-containing-compounds-in-the-air


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For a copper/copper sulfate half reaction, do sulfate anions move to the anode and lose electrons?
In my (high school) textbook, there's an example on finding the charge on 1 mole of electrons, which involves performing the electrolysis of an aqueous copper (II) sulfate using a copper anode and a copper cathode. I've been given to understand that in electrolysis, the negative ions move to the anode, and lose electrons to the anode.
However, in this example, the copper anode loses mass as the copper goes into solution as copper ions. What I don't understand is, since the sulfate ions in this case would go to the anode, wouldn't it combine with the copper ions to form copper sulfate? Or wouldn't it donate two electrons to the copper ion so it becomes copper metal again? Since the sulfate ion moves to the anode, shouldn't this happen?

That experiment is not usually called electrolysis. https://commons.wikimedia.org/wiki/File:Copper_Raffination.svg Only copper takes part in it, the sulfate ions just stay in solution.

Anode is where oxidation happens: anion to element or element to cation or anything with lower oxidation state to higher oxidation state, in solution next to electrode or stuck to electrode surface.

You are right. The sulfate ions are moving to the anode (positive pole) and attracts $\ce{Cu^{2+}}$ ions. They don't discharge, because SO4 without a charge does not exist. These sulfate ions do not deliver electrons. It is the metal of the anode, copper, that  donates 2 electrons, producing a copper ion $\ce{Cu^{2+}}$ and this ion attracts the sulfate ion arriving at the anode. It makes a solution of copper sulfate. It is the same copper sulfate that has been dissolved in the solution to be electrolyzed.

@Maurice There is no reason for the sulfate to have a net movement. There might be some accumulation or depletion in the double layer, but the charge transport is due to copper ions, which are the species with the net movement, as Karl already stated in his comment.

I've been given to understand that in electrolysis, the negative ions move to the anode, and lose electrons to the anode.

This is one way of achieving charge balance, but there are many other ways.
Charge balance of half-reactions
Half-reactions have electrons either as reactants (reduction half-reaction at the cathode) or as products (oxidation half-reaction at the anode). For the cell in question, the oxidation half-reaction at the anode is:
$$\ce{Cu(s) -> Cu^2+(aq) + 2e-}$$
There is no physical state associated with the electrons, but they are removed through the wire. So if electrons appear in the half-reaction, you need some ions to balance it out. In a oxidation, we produce electrons, so either we need to produce cations at the same time or consume anions (or have ions on both sides of the equation, but of unequal charge). The ions (in solution) are also responsible for the charge transport between anode and cathode.
Oxidation state imbalance of half-reactions
At least one atom (as element, simple ion, or part of a compound or complex ion) has to undergo a change in oxidation state. The species that balance the charges, however, might or might not undergo a change in oxidation state. In the half-reaction in question, copper changes oxidation states, and the copper ions balance out the charge of the electrons so that both sides of the half-reaction have equal charge (zero, in this case).
To give a different example, here is a half-reaction involving lead:
$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + H+(aq) + 2e-}$$
Here, the lead gets oxidized (from element to +II), but remains at the electrode. The charge balance is achieved by hydrogen sulfate ions traveling to the electrode and remaining at the electrode as part of the solid lead(II)sulfate, and the released hydrogen ion.
Direction of ion flow
In the reaction in question, copper cations travel from anode to cathode in the solution to balance the charge transport due to electrons traveling from anode to cathode via the wire. We can write the equation showing this explicitly by combining the half-reactions and keeping anode and cathode species labeled.
$$\ce{Cu(ano) + Cu^2+(cat) + 2e-(cat) -> Cu^2+(ano) + 2e-(ano) + Cu(cat)}$$
Because we can't have electrons or copper ions accumulate in or at the electrodes, we need a transport mechanism for both species (through the wire and through the solution, respectively).
Here are two more examples with different ion transport.
Water electrolysis
$$\ce{4H+(aq) + 4e- -> 2H2(g)}$$
$$\ce{2H2O -> O2 + 4H+(aq) + 4e-}$$
At low pH, the cathode consumes hydrogen ions while the anode produces hydrogen ions. At neutral pH, you would say the cathode consumes hydrogen ions and the anode consumes hydroxide ions produced from auto-dissociation of water (i.e. anions travel to the anode and cations travel to the cathode - the classic case giving anions and cations their name).
Lead-acid battery
$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + 2e- + H+}$$
$$\ce{PbO2(s) + 3H+(aq) + HSO4^-(aq) + 2e- -> PbSO4(s) + 2H2O}$$
The element that changes oxidation state is lead in various solid compounds (elemental, lead(IV) in lead oxide and lead(II) in lead sulfate). The ion transport is due to hydrogen sulfate going from solution to both electrodes, forming lead sulfate, and hydrogen ions going to the cathode to combine with oxygen to form water (and hydrogen ions being released from hydrogen sulfate at the anode). So here, ion transport is accomplished with anions going to both electrodes, but an excess of cations (hydrogen ions) going to the cathode. To emphasize, hydrogen sulfate anions are traveling to the cathode. This is an counter example to "anions travel to the anode".
Cells with salt bridge
None of the cells in my three examples had salt bridges. If half-reactions have separate solutions connected by a salt bridge, ions don't travel the entire way between electrodes; instead, redox-inert ions such as potassium cations and chloride anions negotiate the charge transport between solutions of the half cells.

Karen liked your comments, but did not take any voting action, as your content is a good introductory material, but possibly remiss for those engaging in experimenting with electrolysis.

What is occurring is primarily the electrolysis of water in an electrolyte of CuSO4 with Copper electrodes. The detailed mechanics of the reaction is, however, more complex than is usually discussed.
To begin, I start with the seemingly simple electrolysis of water with a small amount of an electrolyte in acidic conditions as previously presented on StackExchange, to quote:

At the cathode the following mechanism is proposed in acidic media:
  \begin{align}\ce{
 2H+ + 2e- &-> H^{$*$} + H+ + e- \\
          &-> H2
 }\end{align}
  The asterisk superscript denotes, that the formed Hydrogen $\ce{H}^*$ is chemically adsorbed to the active site at the metal surface. They continue that hydrogen is only weakly bound to the metal surface and therefore the hydrogen evolution is much faster than the rate of oxygen production. Hence they focus on the reaction on the anode.

The referred to $\ce{H^{$*$}}$ has also been referred to as 'Hchemisorbed'. To quote from a reference:

The cathodic reaction produces Hchemisorbed by picking up an electron that released in the anodic reaction (H+ + e- = Hchemisorbed ) in Al corrosion in HCl. In such acidic solutions, the Hchemisorbed on the metal surface reacts by combining with other adsorbed Hchemisorbed to form H2 gas molecule, which bubbles from the metal surface. A very small amount of the uncombined Hchemisorbed will remain; however, this amount does not affect the whole process.

Apparently, the Hchemisorbed operational behaves as an (e-,H+) pair, to quote a source (bottom Page 818 here) operating on PbS as follows:
$\ce{ Pb(2+)S(2-) + 2 H^{$*$} -> Pb + H2S (g) }$
Now, with aqueous cupric sulfate, a corresponding postulated reaction:
$\ce{ Cu(2+)SO4(2-) + H^{$*$} -> Cu(+) + HSO4- }$
where the Cu+ concentration is not expected to be stable per this reference and the formation of HSO4- may lower the reaction cell's pH.
Further, in the current context with a cuprous oxide coating on the copper electrode:
$\ce{ Cu(+)O(2-) + 2 H^{$*$} -> Cu + H2O (g) }$
where the above theory corresponds to what is observed, per my experience, the copper anode has been cleaned and also the cathode possibly from increased acidity.
Note: If the Copper electrode is not pure Cu, an alloy, perhaps containing noble Silicon, you may be performing an ascribed path to H2SO4 preparation (see, for example, https://www.instructables.com/id/Make-Sulfuric-Acid-by-Copper-Sulfate-Electrolysis/ ), so they will be a definite rise in pH along with problematic explosive hydrogen generation.


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Is there a correlation between acidity and viscosity of honey?
I was wondering if honey's viscosity has any correlation with its acidity and if not, it would be very helpful if anyone could tell me what variables could affect the viscosity of honey?

Water content will be a big factor

I agree, water is a way to decrease the viscosity of honey but what else?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic in the question, doing at least basic research oneself, writing what he/she found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

another factor can be glucose : fructose ratio, typically 1:1 - 1:1.7 ( remembering from honey keeper books of my father ). The former ratio is more vulnerable to crystalization.

Can you please be more specific? and also what do you think of the main question about the correlation between honey's viscosity and acidity?

More specifically, you have not done your homework in the basic research.

@GeneralMO7 Please keep it civil, I have had to delete your comment. We have a policy about homework questions, which requires you to share your own research. Also, your question title does not match the question body. Even though you have already gotten an answer, it would be nice if you could [edit] your question to make it clearer.

The answer was part of my question because I asked for other variables that could cause a change in viscosity of honey but my main question was that if honey's viscosity could have any correlation with its acidity. So still until now, I do not have an answer to my main question and I don't know how to make it any more clearer.

I have made some changes to the question, hopefully, now it is more clear to everyone.

Please does anyone know if there is a correlation between the viscosity and acidity of honey?

Not much. Given all other variables which influence the viscosity of honey (sugar composition, including polysaccharide content, humidity, degree of crystallization), pH would not be regarded as a key determinant.
Honey has a pH of 3.4-6.1 (average around pH 3.9) according the National Honey Board of America. This translates roughly into a $\ce{H+}$ concentration in the mmolal-$\pu{\mu molal}$ range (because the water content is so low, strictly one should discuss activities). The main organic acid responsible for the low pH of honey is gluconic acid, which looks, as described by the Wikipedia, like the "open chained form of glucose, terminating in a carboxylic acid group". The chemical structure and MW of gluconic acid are so similar to those of glucose, and its concentration in honey is so low, that it does not alter much the viscosity of honey.
As concluded in Ref (1):

Variation  in  viscosity  measures  of  honeys  of  similar  moisture  contents  is  usually attributed  to  non-sugar,  colloidal  material.  Honeys  of  similar  moisture  content  but different  absolute viscosity (Table 1) all obeyed Eq. (3). Thus, any contribution of non-carbohydrate, colloidal material to concentration dependent viscosity was minimal in the honey samples assayed. These findings agree with White,[10] that the types and concentrations of carbohydrates largely determine the gross physical attributes of honey.

References

Brenda Mossel, Bhesh Bhandari, Bruce D'Arcy & Nola Caffin (2003) Determination of Viscosity of Some Australian Honeys Based on Composition, International Journal of Food Properties, 6:1, 87-97, DOI: 10.1081/JFP-120016626

OK, speculation here: time. What?
Well, consider asphalt, as in asphalt concrete. The mass of asphalt (petroleum residue) plus stone is heated in a large mixer, quite hot, then loaded into a truck for delivery to some location for deposit onto a street. It is dropped and rolled to about 2 inch thickness, then rolled again to insure flatness and good compaction, and it rapidly cools down to ambient temperature. (A few hours.) Then we are told to keep off it for a day or so and even then to avoid parking a car on it for a week or so, to prevent forming unsightly depressions that will catch water. The reason is that the asphalt molecules are quickly frozen into some not completely stable configuration as they cool down, and will continue to deform and rearrange until they reach a final structure that then resists further deformation: i.e., has considerable rigidity and strength.
Now, honey. I have noticed, in a qualitative way (having no viscometer at home), that when I take the jar of partially crystallized honey and heat it to get it all redissolved, and tilt the jar, it seems quite fluid , not only when hot, but also when it has cooled (I have thermometers at home, but didn't think it was worth measuring the temperature, till now!), and even an hour or so later... But the next day, the honey seemed considerably more viscous than I remembered, almost as viscous as it should be...And the next day, everything seemed quite normal, so I did not investigate further.
So I wonder if the fructose and glucose (in a concentrated solution: ~85% sugars, 15% H2O) form some metastable hydrogen-bonded polymers with a long-range structure that persists for a long time before sorting out to crystals. The viscosity should then be a function of how long since it had been at a high temperature, and this function should also be influenced by the original ambient temperature viscosity (related to water content), since a lower original viscosity suggests that the original viscosity will be regained faster after heating if the molecules are not tied up so much.
And if the long-range polymer structure is applicable, then perhaps mechanical disruption is also a variable. On the one hand, whipping air into the honey will disturb the viscosity. But perhaps even mild stirring will cause the polymer structure to deform or break down. This could be measured by viscoelasticity (a high elastic component indicates no breakdown of the polymer structure), but if the elastic component of the viscosity diminishes as stirring continues, the polymer structure is being deformed, and the recovery time of the polymer structure could be measured by stopping the stirring and measuring the viscosity periodically.
A very sweet experiment, to be sure!

Yes, honey is a metastable supersaturated solution (when not crystallized that is).

Note also that honey is non-newtonian liquid so it's viscosity is not constant, but is stress dependent.

@Poutnik You're wrong, but I'm worse! I did no literature search, and just speculated - mental mastication. So when I looked it up, I find that honey may indeed be thixotropic (shear thinning), but not very much. Honey is considered Newtonian!  https://www.intechopen.com/books/honey-analysis/rheological-properties-of-honey-in-a-liquid-and-crystallized-state

But thixotropic does not mean Newtonian.

I am sorry everyone for asking one more time, so is there any correlation between acidity and viscosity of honey?

Thixotropic means non-newtonian, but if the thixotropy is small, or found only in certain strains of honey, as reported in the reference, and if it is then claimed to be Newtonian, I'll accept that. You can estimate that honey is Newtonian by the way it drips: it runs like water (if it doesn't dry out) down the jar, just slower. Honeys may vary in viscosity and acidity; both parameters vary only slightly in actuality, and I have not seen any reports about a correlation, and I can't think of a causal relation.

Is this experiment where I have to stir the honey only works with crystallized honey? Can it work with pure honey?

A specimen of clear honey is capable of being subjected to repeatable testing, at least for some period. A specimen with crystals in it is much more difficult to quantify, and is hardly of significance, except as an indicator of impending deterioration and inadequacy. Crystallized honey would hardly ever be used without warming up to dissolve, IMHO.

Excuse my understanding, so what you mean is that I could also stir clear honey and will end up with a lower viscosity? (I can then use Zahn cup and there will be a difference in viscosity?)

Some liquids are shear thinning (technical term: thixotropic). When you stir them, you break weak bonds that we usually ignore, but it can reduce viscosity. E.g., ketchup: shake it and it pours better. A Zahn cup does not input significant shear force vs time, but if you do a Zahn on a sample of honey, then stir the rest of the specimen for 5 minutes, it is likely that the flow will be faster - probably not from specific thixotropy, but rather because the stirring of a high viscosity fluid will raise its temperature. See Joule's paddle wheel calorimeter.


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How to write a wave function for infinite potential well with different width than from 0 to a?
I am working on the perturbation theory problem with the infinite potential well. However, I am not sure how to write a wavefunction for different than the infinite potential well from $0$ to $a$ and in my problem the well is from $0$ to $2a$.
Is it necessary to solve the potential well all over again? Maybe it is possible just to apply the transformation as $a\rightarrow2a$, for the initial wavefunction
$$\psi(x)=\sqrt\frac{2}{a}\sin\left(\frac{n \pi x}{a}\right)$$ ?

Yes, you need to solve a problem all over again, and perturbation theory is a way of solving it (admittedly, not the only one).

@IvanNeretin, yes, but the first-order energy corrections $E^{(1)}=\int \psi^{(0)}_n V(x) \psi^{(0)}_n$ where $\psi^{(0)}$ is the unperturbed Hamiltonian wave function. In this case, I am not sure how to recalculate the unperturbed state wavefunction.

The unperturbed function is the one before the perturbation; you already have it, I suppose. Wait! Now I got it: you have the unperturbed function for a different well. Yes, then just apply... I wouldn't even call that a transformation. You see the equation, where $a$ is any number; you plug any number in place of $a$, and it just works. Oh, and it is not just numbers that you can plug.

@IvanNeretin, yes, I was talking about that! Thank you very much for the help!

Well, yes; the original length $a$ is just a symbol and you can replace it with whatever other quantity you wish. The relevant wavefunctions are thus just
$$\psi_n = \sqrt{\frac{1}{a}}\sin\left(\frac{n\pi x}{2a}\right)$$
You can verify that these wavefunctions are still normalised correctly by explicit integration.


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Why is the transition (0,0,0) -> (1,0,1) observed in a gas phase IR spectrum of CO2?
Let ($v_1$,$v_2$,$v_3$) denote the vibrational state of $CO_2$.
Why is the transition $(0,0,0)\rightarrow (1,0,1)$ observed when the trasition $(0,0,0)\rightarrow(1,0,0)$ (asymmetric stretch) is not observed because it is IR inactive (not allowed)?
Is the following procedure correct for showing that the $(0,0,0)\rightarrow(1,0,1)$ transition is allowed?
The criteria for the transition being allowed is
$$\Gamma_{tot. sym.}\in\Gamma_{initial} \times \Gamma_{x,y,z}\times\Gamma_{final}$$
Where $\Gamma$ is the "irrep" of the group $D_{\infty h}$ and $\Gamma_{tot. sym.}=\Sigma_g^+$. The initial state is $(0,0,0)$ and the irrep is $\Gamma_{initial}=\Gamma_{(0,0,0)}=\Sigma_g^+$
The irreps of the three normal modes $v_1$, $v_2$ and $v_3$ are
$$\Gamma_1=\Sigma_g^+\text{, }\Gamma_2=\Pi_u\text{, }\Gamma_3=\Sigma_u^+$$
From a character table we get that
$$\Gamma_{x,y}=\Pi_u\text{ and }\Gamma_z=\Sigma_u^+$$
Now, this is the step I'm not sure about; we say that the irrep of the state $(1,0,1)$ is $$\Gamma_{(1,0,1)}=\Gamma_1\times\Gamma_3=\Sigma_g^+\times\Sigma_u^+=\Sigma_u^+$$
And finally, using a product table, we get
$$\Gamma_{(0,0,0)}\times\Gamma_{z}\times\Gamma_{(1,0,1)}=\Sigma_g^+\times\Sigma_u^+\times\Sigma_u^+=\Sigma_g^+\ni\Gamma_{tot.sym.}$$
And thus the transition is allowed and we observe it as a band in an IR spectrum.

Everything is allowed in real life, which is what you're seeing. Selection/transition rules are based on approximations.

Your approach seems reasonable

You seem to have answered your own question, (1 0 1) is allowed and observed, (1 0 0) is not and is not.


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Bond strength of carbon compounds
I read online that C=O is more stronger than C=N and the reason behind this was, 'Since bond between C and O is more polar hence it will have a slightly higher ionic nature than C and N. As we know ionic bonds are stronger than covalent bond hence C=O bond is stronger.'
Then I tried to find the bond strenght between C-O and C-N by applying the same logic but the result turned out to be the opposite of what I had calculated. According to factual data C-N is stronger than C-O. Can anyone please explain me what's wrong in my reasoning and whether it's right or not.

The data is right..

That doesn't answer my question ^^"

This page seems to contradict your statement.. http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

@NikhilAnand I gave this statement based on the data given here: https://forums.studentdoctor.net/threads/c-o-c-n-and-ionic-bond-strength.1282556/

And I totally agree with the data given by you cause I referred some other sources also and it matches your data :)

It's not an ionic bond...

@zhe i never said that it's an ionic bond. I was just mentioning about the ionic nature in the bond.

Just because there's charge separation doesn't mean you should start thinking there's anything ionic about the bond at all...


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Use household ammonia damage aluminum?
I'm sorry if this has already been asked but I could not find it on a search here though I also could not determine how to search for the presence of two words ('ammonia' and 'aluminum')
I've read mixed results about whether using household ammonia on aluminum is damaging or not. In my case I want to use it to clean baking trays by putting some (~ 1 tablespoon to 1/4 cup) ammonia on the pan and cover it with a plastic bag and set it aside for hours or a day or so to allow it to soften the baked on grime. I also want to and am using it for a variety of other household and tinkering projects. I read on one site that it is a very bad idea to use ammonia on an aluminum bike frame because it will severally weaken it.
Does household ammonia damage aluminum?

Why the down votes? Please let me know

I don't know the answer, but my guess is that it would be OK. // The gist is that ALL aluminum forms a surface coating of $\ce{Al2O3}$. Concentrated lye solutions (lots of $\ce{OH-}$) will dissolve that layer. So a lye based oven cleaner would be a definite NO. // Chemistry is an experimental science. Try a piece of aluminum foil in the diluted ammonia solution and see how that results.

What exactly is household ammonia?

@Karl Household ammonia is just ammonia water, typically sold as a cleaning aid in grocery stores in the USA. I don’t remember the concentration, but you know when you take the cap off the plastic bottle!

@EdV Found a data sheet for a Walmart product, it´s 2%. But what for? I mean I´m a chemist, and I never looked at some mess in my kitchen and thought "for this I need ammonia". ;)

@Karl I certainly agree! My guess is that it is a holdover from the old days cleaning lore, before folks discovered the magical cleaning power of baking soda, salt, vinegar and lemon juice! ;-)

Yes, aqueous ammonia is likely damaging in time to metal Aluminum in the presence of air and any salts (from outside dust,...) that may further serve as an electrolyte.
Ammonium salts, themselves are good in transporting solvated electrons, and here you are, unintentionally, likely creating a metal-air battery scenario.
Possible half-cell reactions:
$\ce{ Al -> Al(3+) + 3 e- }$
$\ce{ O2 + 2 H2O + 4 e- -> 4 OH- }$
Resulting in a net cell reaction of:
$\ce{ Al(3+) + 3 OH- -> Al(OH)3 }$
The aqueous ammonia may also assist here in the removal of the Aluminum hydroxide by forming a "dissolved/peptised/supersaturated hydrated alumina" (per a paper 'The Precipitation of Aluminium Hydrous Oxide and Its Solubility in Ammonia' by Prideaux and Henness).
In essence, Aluminum is now a more rapidly corroding anode.
Nice experiment, albeit, unintended.

"Ammonium salts, themselves are good in transporting solvated electrons,.." Yes, In liquid ammonia!!

Karl: Check out the electrolyte in some commercial battery cells. See https://chlorine.americanchemistry.com/Science-Center/Chlorine-Compound-of-the-Month-Library/Ammonium-Chloride-Helping-to-Provide-Portable-Electricity/ to quote: "When ammonium chloride is used as an electrolyte in a battery, it is made into a watery paste and placed between the cathode and anode."

I have never heard of a battery in which the electrolyte needs to transport electrons. Except as an overcharge safety mechanism in modern Li-ion cells. Which are definitely and perfectly anhydrous inside. And that webpage you link is an insult to the readers intelligence, sorry.


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Wave function characteristics
In a book, I read that the wave function of a system should not become infinite in a finite region and also should vanish at infinity.
For the first part I think that if wave function is infinite in finite region the normalization constant becomes zero due to which wave function vanishes.
But I am not able to find the reason behind the vanishing of wave function at infinity.
Please clarify this doubt.

It's a boundary condition.

because the wavefunction integral must evaluate to 1, a finite value at infinity cannot sum to 1


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Visible Light Spectrophotometer Recommendations
I’m looking into purchasing a cheap uv-vis spectrophotometer but am not affiliated with an institution at the moment given the whole COVID 19 situation. Are there any good recommendations for cheap visible light spectrophotometers that can analyze wavelengths of light on the order of 370nm and are purchasable directly to consumer?
Edit: Budget: $500-700

yeah, ocean optics are expensive. I've been interested in making a DIY spectrometer for some time but haven't had the chance. It would be significantly cheaper. Unfortunately I can't tell you how easy or reliable they are.

Check the Journal of Chemical Education articles. They have plenty of exercises. Also check Vernier.

Unfortunately, I cannot purchase from Vernier because I’m neither from the US or affiliated with an institute that has a vernier account.

They do sell internationally.

CD fragment as a diffraction grating. Cellphone camera to measure light levels. Yee Ha! :-)

I don't know if you're still looking but you may find this video helpful for building your own light spectrometer using a webcam. I built one and it works very well with a wavelngth accuracey of about +/-3nm. https://www.youtube.com/watch?v=m6zpNSoQTV8

Good spectrometers in "new, unopened" condition are out of your budget. Their prices start at about 2000 USD. This also includes the spectrometers specifically made for education—like the ones from PASCO (example; they are usually based on Ocean Optics components), and from Ocean Optics themselves (example).
One option for your budget is to buy a used spectrometer like Ocean Optics USB2000. There's quite a number of options on Ebay, including one for 700 USD (example; the link will likely rot soon), and some more on Amazon (example). But note: you should make sure you ask the seller about wavelength range and resolution, as Ocean Optics spectrometers of the same model can be configured with different slits, gratings, detectors and filters.
These are the commercial solutions, with a quality reflection-based diffraction grating, focusing mirrors and a CCD detector. Here's the typical optical bench of an Ocean Optics spectrometer (image source; the rubbed pale-reddish line on the left-hand side is the place where the CCD detector is to be placed):

Another option is to get something from Thunder Optics. I have tried "Mini USB spectrometer", which currently costs about 100 USD. But this one not only has much smaller price than the Ocean Optics spectrometers—inside it's basically a DIY spectrometer: a cheap wobbly transmission grating (secured in a plastic holder to avoid too much wobble), a web camera as the detector, and a fixed thick slit in the thick (1.5 mm) aluminium case. The components are fixed in place with some kind of black glue.
This kind of spectrometer isn't useful for anything but a qualitative* look at relatively bright** sources of light.
Here are two photos of my item ("optical bench" and the slit):

* Qualitative, because you can't get raw data from the camera.
** Bright, because the camera has quite low sensitivity, and the optical bench lacking any focusing mirrors reduces light gathering capability.

I was looking into Ocean optics spectrometers and it seems like I need auxiliary materials like patch cords, cuvette holders, and a light source? Is there some seller that includes all of these things together?

@Thunderbolt114 do you mean new ones, not used? In that case I suggest trying PASCO Wireless Spectrometer, which already includes 10 cuvettes, built-in light source, and doesn't need any fiber cables. Too bad they don't list the exact price, but you could request a quote.

For UV-vis I would greatly recommend a Cary 50 UV-vis spectrophotometer from Agilent or a GBC Cintra if you need a double-beam one. Both quite affordable on ebay and also quite reliable. As for cuvette holders, they are certainly not a major price crux, but for a budget version, you could ideally have one 3d-printed.

@flippinpippin that's an interesting direction. I didn't even think of "wavelength-at-a-time" scanning spectrometers. Is this Cary 50 still >2000 USD when sold as new (Agilent's website doesn't list a price)? Or should I edit the first sentence of my answer?

Honestly, I have no idea about the official prices, but the Cary 50 is the most basic model, which should be at a reasonable price. Also, it is nothing new really, measuring spectra in time is a common practice for example in actinometry.

Disclaimer: I'm not affiliated with any company
However,
Ocean Optics cuvette spectrophotometers are a good choice
This model is used in some undergraduate labs and is good because the cuvette holder is built in (i.e. it doesn't require a fiber optic cable).
As long as the optics don’t get jumbled, the prism detector mechanism work well and reliably for many years, although it says to get it checked each year.
If you're on a tight budget I'd recommend looking at a DIY spectrophotometer. In particular the RasberryPI designs seem relatively simple
The Hackaday website has a blog dedicated to DIY spectrometers.

Sigh. When will good spectrometers be affordable... I bought my Amadeus spectrometer (based on OceanOptics components) years ago for roughly the same price.


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Why in nitration of benzene we use concentrated sulfuric and nitric acid and not the dilute version of it?
Is there any different if we use the dilute version? Most experiment I read in book use conc. acid but didn't explain why.

Because dilute acids won't work, as simple as that.

The first reaction happens in the mixture $\ce{HNO_3 + H_2SO_4}$ : $$\ce{HNO_3 + 2 H_2SO_4 -> NO_2^+  +  2 HSO_4^-  +  H_3O^+}$$
Then benzene reacts with $\ce{NO_2^+}$.
But water must be absent, because, if there is the smallest amount of water molecules, the ion $\ce{NO_2^+}$ would immediately be destroyed. It would quickly react according to : $$\ce{NO_2^+  + 2 H_2O  -> HNO_3  +  H_3O^+ }$$ So this would prevent the reaction with benzene.

You're vastly overstating water interference. Even "100 %" H2SO4 has a bit of water due to self condensation, and I mean unprotonated H2O.


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Sulfates and persulfates as accelerants in black powder?
I saw there was a few black powder posts on here so I'd like to pose a question. Would sulfates or persulfates like the oxidixer soduim persulfate be a good replacement for some percentage of the nitrates or usual sulfer content of black powder? Pursulfates are used as an ingredient in flash powder for pyrotechnics and is a strong oxidizer to combustibles, so I assume it would make a good accelerant in some concentration. Would it perform under the pressures and speeds needed for an old fashioned firearm, or would it be too corrosive to be practically used?

Sulphates and persulphates cannot be considered as analogues to chlorates and perchlorates.

Perchlorates are frequently used in professional pyrotechnics, being more reactive than nitrates.
Chlorates are too reactive for safe use.
Sulphates are useless as not acting as oxidant.
Persulphates have limited oxidative oxygen content, compared to nitrates or perchlorates, but as chlorates are too reactive for safe operations.

As a general rule, you must know what you are doing. More reactive mixtures may not only lead to faster wearing of the guns, but also to a gun blowing into your face.
Sulphur cannot be replaced by sulphates/persulphates, as it has the opposite role than oxidants. It would be like if you wanted to replace oxygen for breathing by carbon dioxide, as both contain oxygen.

Gun powder

Chemical reaction
Gunpowder does not burn as a single reaction, so the byproducts are not easily predicted. One study showed that it produced (in order of descending quantities) 55.91% solid products:
   potassium carbonate, potassium sulfate, potassium sulfide, sulfur, potassium nitrate, potassium thiocyanate, carbon, ammonium carbonate and 42.98% gaseous products: carbon dioxide, nitrogen, carbon monoxide, hydrogen sulfide, hydrogen, methane, 1.11% water.
However, simplified equations have been cited.
A simple, commonly cited, chemical equation for the combustion of black powder is
  $$\ce{2 KNO3 + S + 3 C -> +K2S  + N2 + 3 CO2}$$
A balanced, but still simplified, equation is $$\ce{10 KNO3 + 3 S + 8 C -> 2 K2CO3 + 3 K2SO4 + 6  CO2 + 5N2}$$

Black powder is a mixture of three substances : saltpeter $\ce{KNO_3}$, sulfur and charcoal. The reaction taking place with black powder is :$$\ce{2 KNO_3 + 3 C + S ->  K_2S + N_2 + 3 CO_2}$$ It is a rather strange reaction, because two substances are acting as oxidant : Nitrogen from $\ce{KNO_3}$ and Sulfur. These two atoms are reduced, Sulfur from $ 0$ to $ -2$, and Nitrogen from $+5$ to $0$. The only element that is oxidized is Carbon. It also should be mentioned that as soon as $\ce{K_2S}$ is produced, it gets oxidized by $\ce{O_2}$ from the air into $\ce{K_2SO_4}$.

by nitrate as well. As there is not the only one equation describing the gunpowder burn.

Your general comment on persulfates is not likely accurate, in fact, with ammonium persulfate, it may be potentially a very dangerous energetic. See my answer and reference below.

I have written about available oxygen.

Per the Energetic Materials Section in ScienceMadness to quote:

anyone got anything on any high energy fuels for a high explosive flash powder? I know that ammonium persulfate and ferrocerium can be very explosive, and...

Here is a link to Ferrocerium for those interested.
So, please avoid the use of ammonium persulfate in select mixtures due to possible problematic/dangerous energetics per my readings (sorry no personal experience here).


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Is the magnitude of ionisation enthalpy and ionisation energy is same?
In my textbook it is written that ionisation energy and ionisation enthalpy are two different quantities. ionisation energy is the amount of energy provided to extract an electron from the outermost shell of a neutral or ionic gaseous atom and ionisation enthalpy is the change in energy of the 1 mole of gaseous atoms or ions when the outermost electron is removed from the atoms or ions. Since electron is in bounded state in an atom so the net potential energy of the electron nucleus system is negative and to free up the electron from the bounded system we need to change the negative potential energy of the system to zero or positive value by providing an extra amount of energy from our side which is called ionisation energy and since we are PROVIDING the energy mathematically it is written with the positive sign (by definition of ionisation energy), ionisation energy is always positive. But ionisation enthalpy is something different it is the difference in the energy of the state of the system before and after ionisation it can be positive or negative depending on the stability of product with respect to reactant. The question arises here is that are ionisation energy and ionisation enthalpy same in the magnitude.

I also had the same doubt, and IUPAC's Gold Book gives the definition of ionization energy, but not for ionization enthalpy. Searching in journals also didn't give me any hit so this is what I did, I grabbed Physical Chemistry by Atkins, that is a book that talks about both of them.
In table $2.4$ of Chapter $2$ he introduces all the enthalpies of transition. The process of ionization is defined and well known to us
$$ \ce{X(g) -> X^+(g) + e^-(g)} \quad \Delta_\mathrm{ion} H(\ce{X}) \tag{1} $$
In the following page he does that old classic example of calculating lattice enthalpy $\Delta H_\mathrm{L}$ using a Born-Haber cycle. He uses potassium chloride to illustrate. In the path, you need the ionization enthalpy of potassium $\Delta_\mathrm{ion} H(\ce{K)}$. I have the book so I give you the screenshot:

and we clearly see that $\Delta_\mathrm{ion} H(\ce{K)} = 418 \; \pu{kJ mol^-1}$. When we go back to the appendix to search for the ionization enthalpies, he doesn't like that term, and names the table "Ionization Energies". The value listed in Table $9.3$ is $I(\ce{K}) = \pu{418.8 kJ mol^-1}$. Thus, at least for Atkins, they just seem to be the same or $\Delta_\mathrm{ion} X = I(\ce{X}).$
As an additional check, I saw if he did the same for electron affinity. Again, IUPAC's Gold Book defines it, but in the same chapter we are introduced to the electron gain enthalpy as
$$ \ce{X(g) + e^-(g) -> X^-(g) } \quad \Delta_\mathrm{eg} H(\ce{X}) \tag{2}$$
Eq. (2) is the "inverse" of electron affinity.
In the same calculation of the image, Atkins claims that $\Delta_\mathrm{eg}H(\ce{Cl}
) = \pu{-349 kJ mol^-1}$. What happens when we go to table $9.4$ of the appendix? He disregards the term and lists the electron affinities. The value there is $E_\mathrm{ea}(\ce{Cl}) = \pu{+348.7 kJ mol^-1}$ (see the possitive sign). Once more, for Atkins they are just the same $-\Delta_\mathrm{eg}H(\ce{X}) = E_\mathrm{ea}(\ce{X})$ (see the negative sign).
The question as "why multiply entities with no necessity remains". I like more the enthalpy notation of Atkins, though.
References

Atkins, P. and De Paula, J., "Physical Chemistry", 9th ed, W.H. Freeman and Company New York (2016).

Atkins' definition is reasonable. It makes sense to admit that $\pu{−Δ_{eg}H(X)}$ = $\pu{E_{ea}(X)}$

I'm trying to understand this myself, this is what I think.
Enthalpy can be defined as Final Energy of the system - Initial Energy of the system so here Ionisation Enthalpy must mean Energy of system after Ionisation - Energy of system before Ionisation. This is the same as saying how much energy you have to put into or take out of the system to remove an electron which is the definition of Ionisation energy, so both are equivalent terms.
Note:- Ionisation by itself is always endothermic as you have to give energy to the electron to excite it, allowing it to escape the hold of the nucleus so Ionisation Enthalpy is always positive.

@ShrishSrivastava One question I had is for the definition of ionisation enthalpy you used. Is 1 mole used because that is what can be easily measured?

what is the issue with the answer for the downvote?

Ionization enthalpy is defined as the energy required to remove the electron at $0\ \mathrm{K}$, whereas the ionization energy is defined at any temperature.
so, generally, they are not equal quantities.


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Catalytic hydrogenation of polycyclic hydrocarbons
Recently I came across a question-
The compound on the left undergoes catalytic hydrogenation to yield A(not B, according to the answer). My knowledge is limited to the stereospecificity of catalytic hydrogenation, and I don't know if it's regioselective as well or not. Why do we get A as a product and not B? There are 3 moles of hydrogen provided, so how to decide which 3 multiple bonds would be reduced in such a situation? Can someone provide me with a detailed answer?

Consider the ring strain in the [4] rings. The bonds are much less stable and more reactive

The anti-aromatic rings are much more unstable than the aromatic ones. So, the 3 moles of hydrogen gas will prefer to reduce those sites which ultimately lead to the remobal of the AA rings


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Radicals vs atoms having unpaired electrons
In organic chemistry course, I read that a radical is an atom, molecule or ion which have unpaired valence electrons.
So does there any difference between chlorine atom or cholrine radical? Also sodium atom contains one valence electron which is unpaired, then does sodium atom is also considered a radical?
Please clarify it.

There is a difference. Neither sodium, nor chlorine radical needs to be a single atom. The may be who knows how many polyatomic species that can be called like that.

In organic reactions, I studied that Cl-Cl bond breaks homolytically to form two Cl radicals. I am not able to understand your statement. Please explain it more for better clarification.

Many radical cations of chlorine were detected. https://en.wikipedia.org/wiki/Polyhalogen_ions Not to mention monoatomic cation radicals, and excited states.

I upvoted your question because I had to employ one of my recent Question on SE as a source.  I knew the material I provided, although not well received per votes cast, was and may continue to be, a  cited source. Thanks.

Yes, a chlorine atom is very much the same as chlorine radical.
With sodium, the two things are technically similar too, but you'll hardly ever encounter a mention of "sodium radical".
Why is that? Well, I guess there is a difference between Na and Cl. A chlorine radical exists in its own right (if only briefly), and is an important intermediate in many reactions. A sodium radical, on the other hand, is pretty much not a thing. Single sodium atoms just don't wander around alone. You may see reactions involving metallic sodium, or ionic salts of sodium, or maybe (not sure about that) some compounds which can pass for covalent, or (unlikely) even $\rm\bf Na^{\pmb-}$... but that would be all.
So it goes.

Although I get your point, I have seen something like sodium radicals in SET mechanisms like this one for the Wurtz reaction. Doesn't it count for something?

Maybe it does (that's why I said "hardly ever", and not "never"). Or maybe it is just a metaphor of metallic sodium.

\rm\bf Na^{\pmb-} — seriously?!

@andselisk Why not? Alkalides are a thing.

@IvanNeretin I'm referring to the pornographic *TeX syntax:)

@andselisk Er... Sorry about that. But how else would I make it bold?

@IvanNeretin The $\mathrm\LaTeX$ syntax is OK (except for obsolete \rm macro), it's just an overkill to utilize this with MathJax. I'd rather omit bold face at all or go with either Na⁻ (**Na⁻**) or $\mathbf{Na^-}$ ($\mathbf{Na^-}$).

Now I understand. But one more question does nitrogen and oxygen atom is also considered radicals as they contain three and two valence unpaired electrons?

These are in the same basket as Na: technically yes, but rarely called such, because you rarely have them around as single atoms.

To say that the chlorine radical has only a transient life span is NOT technically correct. The .Cl does pair with Cl- to former a rather stable and annoyingly less reactive mixed species, .Cl2- (or Cl2.- depending on one's preference). Source, see for example, https://pubs.acs.org/doi/10.1021/jp036211i

While I essentially agree with the statement that one generally does not differentiate between the chlorine atom and the chlorine radical, this is not the case with hydrogen.
In fact, while the hydrogen atom and hydrogen radical (also called, per a recent thread question of mine, as Hchemisorbed (per my provided answers available here), I also noted in that question's answer that there exists, so-called 'atomic hydrogen', distinct from $\ce{.H}$, which is a fleeting very high energy state of hydrogen created in an electric arc (actually commercially employed in high-temperature welding, see here).
So, in general, depending on the element, the answer actually appears to be mostly yes but sometimes no.

So can we conclude that every atom, or molecule containing unpaired electron is radical but we can use other terminology for them depending on the method they are prepared ( like "atomic Hydrogen" when H radical is formed by electric arc or "Nascent Hydrogen" when H radical is produced in a reaction)?


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Does the absolute value of enthalpy have any physical significance?
I know it is pretty hard to measure the change in volume in open systems. Accordingly, calculating the change in internal energy using the first law of thermodynamics is pretty hard also.
Is this the only reason for introducing enthalpy ($H$) as a new thermodynamic potential to make use of the constant atmospheric pressure, and to calculate the change in enthalpy ($\Delta{H}$) as it is equal to the in/out quantity of heat ($Q$) in this case?
If no, does the absolute value of enthalpy have any physical significance?

You are right. Of course, the enthalpy has been introduced to take care of the change of volume. $\Delta$ H is by definition equal to the quantity of heat Q, without taking the work into account. It is sort of "apparent internal energy". It is a little bit similar to the difference between the weight on an object in a vacuum, and the weight measured in air, which a little bit smaller than in a vacuum, due to Archimede's correction.

The absolute value of enthalpy has direct relation to the mass of the system:
$$H = U + pV = mc^2 + pV$$
with the second term mostly negligible. If the system accepts thermal energy Q, its mass change is $\Delta m = \frac {Q}{c^2}$
For most of enthalpy applications, the absolute $H$ value is irrelevant.
The absolute value interpretation is the total energy stored in the system at the constant pressure, consisting of

the equivalent energy to the mass of contained matter
work done by reaching the system volume against the external pressure.

This energy is formally extractable from the system by annihilation and shrinking to zero volume.


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Understanding chemical reactions
I just got in 10th grade and the first chapter in Chemistry in our school curriculum is Chemical Reactions and Equations
The chapter is littered with a lot of chemical reactions and their equations and most of the teachers I consult say that there is no alternative to revising them but I hate to blindly memorize stuff without knowing why it happens in that manner and not in some other way.
Is it really true that there is no other way than to memorize the equations?
If there is an alternative, please let me know.
And how can I predict the products of a chemical reaction if its reactants are given. Is it done by understanding the types of chemical reactions.
One of my teachers said that it could be done like this : If you're given two elements as reactants and you know that they will react, then you know that it is a combination reaction, you can figure out the chemical formula of the compound that those two elements will form on combination and then balance the equation. This works for simple reactions like Na + Cl $\longrightarrow$ NaCl or for some simple decomposition, displacement or double displacement reactions but what about complex reactions like this :
$2 Pb(NO_ 2)_3$ (s) $\longrightarrow$ $2PbO$ (s) + $4NO_2$ (g) + $O_2$ (g)
I can quite easily visualize a combination reaction as two elements sharing or losing or gaining electrons and combining to form a compound, but how do I visualize these types of reactions
I'm sorry if you find this question silly...

All answers are welcome
Thanks

Nobody will require that you memorize the equation of the decomposition of $\ce{Pb(NO_3)2}$. It is an example of a difficult reaction in the 10th grade. What a teacher can ask is that you find the stoichiometric coefficients in front of the formula. I mean you should find that 2 $\ce{Pb(NO_3)2}$ (and not 1) are required to produce the reaction. And the teacher will tell you that $\ce{PbO}$, $\ce{NO_2}$, and $\ce{O_2}$ are produced. Your job would be to find the coefficient 2, 4 and 1.

Thanks for the edit, first of all, but apparently, in India, that's not the type of questions that are asked in exams. Our teacher gave us an assignment of this chapter and the first question itself happened to be : Write a balanced  chemical reaction describing the thermal decomposition of lead nitrate...

Sindhu. I am surprised of your information. Because, without some supplementary information, nobody can predict tant lead nitrate behave this way. You might as well have proposed an equation like : $\ce{Pb(NO_3)2}$ -> $\ce{PbO_2}$ + $\ce{N_2O_4}$. This last equation is not stupid. Nobody can exclude it without any other information.

And in the next chapter : Acids, Bases and Salts, we have to memorize stuff about every single acid, base and salt given in the book. We have to learn their properties, their reaction with some elements, their formation and what not.

This all is basically memorizing, right? And as I mentioned, I hate to memorize blindly, it ruins all the fun there is in the subject

@Maurice In India, the first thing they'll teach you is : Study for marks. Concepts are valued extremely extremely little in Indian education. If someone tries to understand rather than learn in Inorganic Chemistry, they mostly fail because they don't get any resources to help them for that. Quoting John von Neumann, my role model There is no sense in being precise if you don't know what you're talking about.
I'd rather fail than learn these reactions without knowing what causes them

@Maurice what do you suggest?

Acid-base reactions are systematic. You can easily guess or rediscover how they will react. It is like a machinery that always works with all acids and all bases. No problem ! This not the case of the decomposition of lead nitrate, which produces $\ce{PbO, NO_2 , O_2}$. This cannot be derived by reasoning. For example, with another nitrate, the reaction proceeds another way. With $\ce{ KNO_3}$ for example, it produces $\ce{KNO_2 + O_2}$. And this is impossible to guess, or to deduce in advance. Fortunately the acid-base chapter is absolutely logic.

@Maurice Thank You. So, some chemical reactions are to be memorized and their products were discovered by scientists just by experiments, right? By the way, you were really helpful. :)


text:
Predicting the stability of coordination compounds using Crystal field theory
The question is this:
On the basis of CFT, predict with justification which of the following two complex ions would be more stable [Mn(CN)6]4- or [Fe(CN)6]4-.
Attempt at solution:
Both the complexes are low spin. Filling in the d electrons we get 5 electrons in t2g for Mn2+ and 6 electrons in t2g for Fe2+. There are no electrons in eg for both.
Mn2+ has one unpaired electron while Fe2+ has completely filled t2g level. From here I was in a dilemma because Fe has fully occupied t2g level as well as a higher charge density because the effective ionic size of Fe2+ in low spin is 61 pm while Mn2+ in low spin has size 67 pm. However, due to Jan Teller effect the Mn compound will gain more stability due to the unsymmetrical filling of the t2g level. I know that Jan Teller distortion for t2g is less pronounced, so I was wondering if it was enough to make the complex more stable than Fe2+ complex?

A simpler approach, somewhat ridiculous, I agree. But look : $\ce{Fe^{2+}}$ is {Ar} + 6 electrons. If 6 $\ce{CN^-}$ are approaching, each one brings one doublet. The iron outer shell has now : 6 + 6·2 = 18 electrons. This is like the Krypton atom. Manganese ion cannot do this.

CFT is outdated and deprecated: https://chemistry.stackexchange.com/questions/131780/crystal-field-splitting-of-other-orbitals/131789#comment254617_131789

Horrible question as the word "stability" has little meaning unless you state with respect to what change you are considering some is stable or not. But I suspect https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy may help you give them the answer they want.

@Ian Bush I read that site and drew the above conclusions from there. I just want to know if they are correct. I think stability refers to the complex with the least energy.


text:
Does evaporation occur while boiling as well?
If evaporation were to occur when a liquid was being boiled, then the amount of liquid changing phase would also depend on the surface area of the liquid exposed to the atmosphere (as evaporation rate increases with surface area). However I have read that the surface area doesn't affect the amount of liquid changing phase while the liquid is boiling.
So does evaporation not occur while the liquid is boiling?

Boiling is evaporation.

Evaporation occurs while a random molecule from the bulk collides with the surface molecules and though the upper molecules are less tightly bounded, they gain the energy due to collision and escape from the liquid surface as vapours. But at low temperature this collision frequency is very less. In boiling, we provide heat to the liquid, due to which the collision frequency increases and upper surface molecules more fastly escape. So, boiling includes evaporation. Now for boiling, surface area doesn't matter as the heat which we externally supplied plays a dominant role compared to the surface area of liquid exposed to atmosphere in boiling. The surface area plays negligible role in boiling.


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Is corovin Polypropylene or flat polyester?
I'm wanting to know what material 'corovin' is as described on some duvet covers.  Googling it, some places say it is just another name for 'Polypropylene' while other places say it is 'flat polyester' whatever that means.
Which is correct?  As far as I'm aware Polypropylene and polyester are different types of synthetic fibers.
Thanks

To my understanding, Corovin is a technology using to manufacture Nonwoven fabric. These fabric could either be polypropylens or polyesters, yet technique is first develop for polypropylene (Ref.1). Following is the phrase I borrowed from the book (p. 134):

Companies such as Du Pont, Freudenberg and Corovin began their research and development work on generating spunbonded webs. In Europe, Freudenberg, Terram, and Corovin started production of polypropylene (PP) spunbonded webs at the beginning of the 1970s. Within these technologies, filaments are spun directly out of the polymer melt through small die heads arranged side by side, stretched down to a diameter of $\pu{10-30 \mu m}$ and laid down on a porous lay down and transport belt.

A recent patent has been assigned to Fiberweb Corovin GmbH for manufacture of polyethylene-based, soft nonwoven fabrics as well (Ref.2). The abstract states:

The present invention concerns a nonwoven fabric the fibers of which have polyethylene on at least part of their surface, where the fibers are thermally bonded and the non-woven fabric exhibits an abrasion of less than $\pu{0.5 mg/cm2}$, in particular of less than $\pu{0.4 mg/cm2}$, and an embossed area fraction of less than 35%, in particular of less than 28%. Furthermore a device is made available for the manufacture of a non-woven fabric, using a polyethylene-containing polymer, with a take-down system underneath a spinning plate that causes the take-down of the polyethylene, where the spinning plate has an L/D ratio of between 4 and 9.

References:

H.-G. Geus, "Chapter 5: Developments in manufacturing techniques for technical nonwovens," In Woodhead Publishing Series in Textiles, Number 181: Advances in Technical Nonwovens; George Kellie, Ed.; Woodhead Publishing: Amesterdam, The Netherlands, 2016, pp. 133-153 (ISBN: 978-0-08-100575-0).
Markus Haberer, Henning Röttger, "Polyethylene-based, soft nonwoven fabric," US Patent: 8420557, 2013 (Filed: 2006).


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Is the salt of a weak acid and a weak base also a weak electrolyte?
I'm able to find a lot of information on acids and bases, as well as strong-strong, weak-strong, and strong-weak salts. However all the introductory materials on electrolytes just stop there and say "we won't bother to cover weak-weak salts here". I have also heard blunt statements that all salts are strong electrolytes, but never with satisfactory explanation or even some weak-weak examples.
So are there any salts that are weak electrolytes?
My interest in the subject is mainly directed at finding the van 't Hoff factor, for osmotic effects and tonicity.

Weak-weak salts are typically hydrolyzed all the way to the end, so their solutions are not a thing at all. If they don't do so, they are probably not all that weak.

From wiki, ammonium acetate has a solubility of 143 g per 100 mL of water at 20 deg C. Lots of ions in that solution, that can move in response to a potential difference impressed across a separated pair of inserted electrodes. Seems like a strong electrolyte to me.

Good question, the surprising thing is that you cannot predict conductivity easily. Ammonium acetate, which is a salt of weak acid and weak base, is a comparable conductor as a salt formed by strong acid and strong base (e.g. NaCl). This is a really old table but look at the cases of
a) Sodium chloride: salt of strong acid and strong base
b) Sodium acetate: salt of strong base and weak acid
c) Ammonium chloride: salt of weak base and strong acid
d) Ammonium acetate: weak acid weak base
Not all salts are strong conductors, search the data for organic salts such as potassium citrate - a weaker conductor than many salts.
If you study the table carefully, it shows the strength of acid or base does not matter but what matters is the identity of the ions carrying the current. Also, the concentration of the salt also matters.

Certainly, in the citrate case, arguments regarding the softness of the ion and size / stearic & kinetic hindrance are a relevant. I'd say that given full solubility, the size and "softness" are important factors for conductivity - more so than the strength of the acids / bases (although that hints at solubility)

I don't understand the table, specifically the rows with a concentration of zero. That is pure water - shouldn't it have the same number across the table. I think that problem is that neither concentration nor equivalent conductance are defined, and neither are given with units.

Karsten, this is the equivalent conductance at infinite dilution. So basically,you would plot the conductance with non-zero concentrations and then extrapolate the concentration to zero. The y-value gives the maximum equivalent conductance. Recall the Kohlrausch's laws.

Equivalent conductance is the same as molar conductance, except for the fact the concentration is given in normality = gram equivalents/ Volume in liter. Conventionally, the conductivity is still measured and quoted in gram equivalents.

@M.Farooq Please see my comment under Karsten’ answer. I think the strong and weak labels have not aged so well over the decades: edge cases are not all that rare, e.g., calcium hydroxide. I consider it to be a type of strong base, i.e., a strong electrolyte that is also base, but it is also only sparingly soluble. But not everyone would agree with this.

@EdV, I don't see any problem with your statement. I associate strong and weak with pKa or pKb values. However, conductivity, and I suggest conductivity at infinite dilution should be discussed. As you know conductivity is a function of concentration. The conductivity at infinite dilute gets rid of all solubility problems. This is the true reflection of an ion's properties in water.

When a weak acid such as acetic acid is added to pure water, it does not ionize much, but sufficiently to make the solution acidic. On the other hand, when small amounts of weak acids are added to buffered solution with a pH near neutral, acidic acid dissociates almost quantitatively. So the van't Hoff factor depends on the pH.
The same goes for weak bases used as ammonia - in combination with pure water, you get a basic solution and very little ammonium, but when dissolved in a solution maintaining a neutral pH, ammonia forms the ammonium ion almost quantitatively.

I have also heard blunt statements that all salts are strong electrolytes, but never with satisfactory explanation or even some weak-weak examples.

If we combine ammonia and acetic acid in aqueous solution and remove the water, we get the salt ammonium acetate. If we dissolve ammonium acetate in pure water, the pH will be roughly neutral, and both ions will remain ions. Therefor, this is a strong electrolyte solution.

So are there any salts that are weak electrolytes?

In the pure salt, the components are ions by definition. Water is very good at dissolving ions, so the dissolved ions stay ions. To get a salt that is also a weak electrolyte, you would need an uncharged acid and an uncharged base that  would transfer a proton upon forming a salt, but not while in aqueous solution. I am not aware of any such system.

So even with fairly concentrated solutions of weak-weak salts, the dissolved salt will have complete [or nearly] dissociation of the ions? Just making sure I understand. Would this also mean that precipitation is inevitable when the ions re-associate? It is just difficult for me to see that HA or BOH can readily exist in solution along side -A or +B but BA cannot exist in a significant fraction of the solute. Probably one of those things i just need to memorize for now and will understand later.
(ammonium hydrogen phosphates would be an interesting example, what with the protons and all.)

Yes, when ionic compounds dissolve, each cation is solvated by the solvent, completely separated from solvated anions. And yes, the only time the cations re-associate with the anions is when the ionic compound precipitates. H-A is a covalent bond, so that can exist in solution. BA is an ionic bond, not observed in aqueous solution. The notation BOH is incorrect. For example, NH3 is the base, and NH4+ is the conjugate acid (the ionic form), so there is no involvement of hydroxide ions (those occur in strong bases, e.g. NaOH).

@KarstenTheis Ammonium cyanide is an ionic compound that is very soluble in water. If you make 0.1 M solution, the slightly majority species are aqueous ammonia and aqueous HCN, while ammonium ion and cyanide ion are slightly minority species. But it is still a strong electrolyte and the solution’s electrical conductance is a separate matter: a sufficiently dilute solution of any strong electrolyte is a (like pure water) poor electrical conductor. I think the strong/weak electrolyte notion is separate from the more nuanced electrical conductivity matter. Am I off-track in this?

@EdV, I think the best figure of merit for judging electrical conductivity of a salt is its conductivity at infinite dilution.

@M.Farooq No problem, but I do not see that electrical conductivity has any defining relevance to a classification issue: strong electrolyte or weak electrolyte. As well, there is no defined threshold value for a boundary between strong and weak electrolyte. Seems like apples and oranges to me.

@EdV, Yes I agree. Weak and strong don't have a mathematical meaning.

@M.Farooq I already upvoted Karsten’s answer and will upvote yours right after this. Both answers are good contributions.

@EdV Ammonium cyanide is an interesting case because the pKa of ammonium and hydrogen cyanide are almost the same. In fact, Wikipedia states "Ammonium cyanide decomposes to ammonia and hydrogen cyanide, often forming a black polymer of hydrogen cyanide". But I think this is as close to a counter example as it gets. In terms of van't Hoff factor, you would get the expected number (2), but the ionic strength, conductivity etc. would be surprising if you are unaware of the matching pKa values.

Very interesting! I always thought you could make ammonium cyanide in the gas phase, like ammonium chloride and ammonium acetate can be made, but I did not know it would then (slowly?) decompose. I know HCN can explosively polymerize. Anyway, I ran the numbers here and there is a cross-over situation. At 0.1 M, ammonia and HCN are the majority species. At 0.1 mM, HCN and ammonium ion are the majority species. Around 0.515 mM, the ammonia species are equally concentrated and HCN is the (slightly) highest concentration.


text:
Why is the viscosity of water and ethanol lower than that of a water-ethanol mixture?
We found in an experiment using Ostwalds method that
ethanol is more viscous than water, which
is already surprising as water has stronger intermolecular bonds than ethanol as @MaxW pointed out.
The mixture
of both substances furthermore is significantly more viscous than both individual substances. This
is interesting. We did not come up with a
satisfying explanation for this behaviour.
I know that one shouldn't expect that a mixture behaves the same way or proportional to the concentration of the components, but I would appreciate an explanation of this effect on a molecular level.

Ethanol boils at $\pu{78 ^\circ C}$ and water boils at $\pu{100 ^\circ C}$. I'd say that water has stronger intermolecular bonds.

@MaxW You're right.. I corrected this. Thank you.

This answer just popped up when I searched for viscosity of water-ethanol mixtures, so I just copied the abstract:
Viscosities of Binary and Ternary Mixtures of Water, Alcohol, Acetone, and Hexane
https://www.tandfonline.com/doi/abs/10.1080/01932690802313006?journalCode=ldis20
"This articles studied and determined the viscosities of the binary mixtures of water–methanol, water–ethanol, water–propanol, water–acetone, acetone–ethanol, methanol–ethanol, and acetone–hexane and the ternary mixtures of water–methanol–ethanol and water–ethanol–acetone at 20°C. It is shown that the mixing of water with the alcohols and acetone resulted in a positive deviation of viscosity, which reached the maximum value at the water mole fraction x 1 ∼ 0.7 for water–methanol, x 1 ∼ 0.72 for water–ethanol, x 1 ∼ 0.74 for water–propanol, and x 1 ∼ 0.83 for water–acetone binary mixture. This viscosity deviation can be mainly attributed to the formation of micelles of alcohol or acetone molecules in water because of the hydrophobic attraction between the hydrocarbon chains. The micelle surfaces are surrounded by hydration layers, leading to the positive viscosity deviation in the liquid mixtures because the water in hydration layers has a much higher viscosity than bulk water. Also, the contrary observation was found in the binary mixtures of acetone–ethanol and acetone–hexane, having a negative viscosity deviation."


text:
Why do elements have mass numbers?
I don't really understand why an element has a mass number, I thought that an element has several isotopes, each with a different mass. Is my understanding correct:

The mass number of an element is the mass number of the majority isotope.

Would appreciate anyone willing to clear this out for me. Thanks!

Elements don't have mass numbers. Isotopes do.

hi! apparently my book says that elements ave mass numbers;

Tell your book to shut up.

a row in the chart says
element: Na
atomic number: 11
mass number(A): 23
Atomic mass(u): 22.99

What does it say for an element having more than one isotope? (Cl will do.)

Well, now I 'm intrigued. What does it say for Cl, really?

it says that we take the weighted average of the isotopes (i think this is the atomic mass). So does an element really have no mass number, but it has atomic mass instead?

By the way, they did not say anything about elements having mass numbers. The only place where they hinted at it is when they wrote a table with elements and mass numbers and other stuff.

When in doubt look at IUPAC Gold Book - mass number: Total number of heavy particles (protons and neutrons jointly called nucleons) in the atomic nucleus
. Also called nucleon number
. Symbol m in mass spectrometry
.

The matter on earth has been mixed rather thoroughly through time, expecially in the beginning 4.5 billion years ago. That means that all elements have a rather (some more, some less) constant isotopic mixture across the earth crust, and so you can assign an atomic weight also to each nonmonoisotopic element.
Exceptions prove the rule, like C-14, deuterium, etc. The iridium that came on earth with the impact that killed the dinosaurs has a distictly different composition than all other iridium deposits on earth, as does chromium from the same sediments.
Of course on mars, the numbers would be slightly different.

In the first paragraph "rather" is the keyword. The isotopic variation is why atomic masses for many elements have limited precision. // Don't ever remember that "The iridium that came on earth with the impact that killed the dinosaurs has a distinctly different composition than all other iridium deposits on earth..." To my knowledge iridium was a marker  because it occurred in higher concentrations in particular geological layers. // Atomic weights are known to much better precision than the weighted mass numbers (where weight is isotopic abundance).

Atomic weights might be different on Mars because of isotopic variation, but the mass number of an isotope wouldn't vary.

@MaxW There is a singular iridium concentration in sediments from that time, and it has a different isotopic composition, which more or less proves that it is extraterrestic.

Super Fantastic!!! Learned something new. I poked around a bit and you are right the unearthly isotopic composition was one of the pieces of evidence for the Alvarez hypothesis.


text:
Computational Chemistry Modelling Redox Potentials
How could I accurately determine the Fe3+/Fe2+ redox potential in an aqueous solution? What would be a good computational model(s) for this?

How good? It depends. How good is your voltage meter? see Nernst equation

well this is not an electrochemistry question so im not using a voltmeter. Im using the Guassian 09 program.

You have to calculate the free energy difference of the corresponding redox equation and that is it. You will have a smaller and bigger error, how you treat solvent effects and similar details.

What resources do you have for this? Your personal computer? If so, how powerful is it?

Why is this "off-topic"?

It's a pretty standard DFT calculation (e.g. https://pubs.acs.org/doi/10.1021/jp025853n). The solvent is usually accounted for implicitly.

Here is a non-electrolysis scenario that could be of interest. Add deoxygenated water to FeCl2 in a closed vessel filled with oxygen or air. Record the oxygen consumption.
Chemistry:
$\ce{4 Fe(2+)  + 2H2O + O2 (g) -> 4 Fe(3+) + 4 OH- }$
The wet ferrous chloride should be presented in a fashion to maximize oxygen exposure from air.
[EDIT] I would also place the system (till the reaction is completed) under pressure to limit adjustments owing to the speed of oxygen assimilation.
To be clear, as apparently required by a comment, the purpose of this experiment, per my understanding of the Question, is to serve as a frame for real world validation of a sophisticated theoretical/computational based model.

Question is about computational methods, not wet chemistry

Actually to quote a source: "A new Gaussian MCTDH program: Implementation and validation on the levels of the water and
glycine molecules." So, as I understand it and the Question  (perhaps needing more specificity), there is an underlying frame (aka a redox reaction)  to serve as the basis for computations. Also, if purely theoretical, why is Arum asking non-computational chemists for help? If someone else has other suggested redox involving Fe2+/Fe3+ that are more simple than mine, I would like to review them (as would Arum, I suspect). Also note the word 'validation'!

Could someone sincerely explain how to accomplish validation (via measuring?) of a computational analysis without ever departing from the theoretical realm? My supposition is a 'real' world redox to examine, perhaps even mine?

In general, computational chemistry lags far behind experimental chemistry. There is already a wealth of experimental data regarding the Fe2+/Fe3+ couple. Initial validation of a computational model would simply be showing that it can recapitulate already known experimental results.


text:
Determining mol/L at STP from volume and mol fractions in a CFD program
I'm using some CFD software and the outputs are a bit unusual so I thought I would double check my understanding on how to do these conversions.
The data available per cell is: cell volume, particle volume fraction, average fluid density, mol fraction of gas species, temperature, pressure
I am trying to come up with the mol/L at STP as a way of normalizing for density effects to compare concentration of a gas species throughout the height of a reactor. Temperature and pressure vary wildly, partially because this is a fluidized bed and solids distribution plays a big role in pressure.
So far my process has been convert from mol fraction to mass fraction.
And then multiply (cell volume-cell volume*particle volume fraction) by average fluid density to get gaseous mass in the cell. Then I can use the mass fraction to determine the amount of mass for a given species. Obviously at this point I have the mols of each species.
The problem is then I need the density of that specific species at a given temperature and pressure (currently working at temperatures around 950 C and 110kPa but they very cell by cell obviously) to get back to volume. Is the only way to do this to estimate the density?
The other problem is that let's say I do get the volume by estimating the density. Then I convert to volume at STP by multiplying by 273 K/T and P/101325 Pa or whatever it is. Are the mol fractions I had at non-STP still usable? Or do I just take these volumes and divide by 22.4 and that's the mols I should be using.
Can someone please straighten out my thinking on this. Also are there any good sources for density of a gas species as a function of pressure and temperature. Everything I look at online seems really bad.

Re: "density of a gas species as a function of pressure and temperature" --PV = nRT is pretty much as far as theory goes. However there are formulas with data fitted constants that given better results see Wikipedia article Real Gas. I'm not sure how well the models would work for mixtures.

@MaxW I'm pretty sure these do not count as ideal gases, unfortunately. Really I was looking for empirical data but you're right, I didn't consider that the fact that it's a mixture might make this tough. I'm a bit disappointed this got downvoted, I feel it's a fundamental question though maybe I wrote too much and should've framed it in a simpler way.

Though I just realized that I think I don't need the individual volume contributions and I was being dumb. That simplifies things greatly.

Please avoid acronyms like CFD which not everyone (such as myself) will be familiar with.


text:
What are the alternatives we have to the VEDA 4 vibrational energy distribution analysis software?
Said software is very useful when you need to assign the peaks in vibrational spectra to different vibrational modes found in a hessian calculation for a given molecule, as described in Jamróz, Michał H. “Vibrational Energy Distribution Analysis (VEDA): Scopes and Limitations”. Spectrochimica Acta Part A: Molecular and Biomolecular Spectroscopy, vol. 114, outubro de 2013, p. 220–30. ScienceDirect, doi:10.1016/j.saa.2013.05.096. It's becoming popular, the paper has nearly 450 citations as of april 2020, but it has some downsides for me:

Despite being a freeware that can be downloaded from their site by anyone, it's not free software, or at least open source, and it's windows-only;
Doesn't support people using free tools like Gamess-US or Psi4, only accepting Gaussian output files as its input, a paid and very expensive quantum chemistry software;

So I did some research for alternatives that could avoid these downsides. Among the citations to the VEDA paper, I've found what seems to be a alternative approach to solve the same problem, as described in Teixeira, Filipe, e M. Natália D. S. Cordeiro. “Improving Vibrational Mode Interpretation Using Bayesian Regression”. Journal of Chemical Theory and Computation, vol. 15, no 1, janeiro de 2019, p. 456–70. ACS Publications, doi:10.1021/acs.jctc.8b00439. Their software, called vibAnalysis, is open source, and available from a GitHub repository. It doesn't support Gamess-US or Psi4, but at least supports ORCA, which is free for academics, besides Gaussian.
I wonder if there are other alternatives I have missed in my research, specially if they run on linux and work with Gamess-US or Psi4. If somebody here could provide some guidance, I'd be grateful.

This question would fit well on the new Materials Modeling SE: https://materials.stackexchange.com/

If somebody stumbles in this question, I got some helpful comments in https://materials.stackexchange.com/q/411/243


text:
Reaction of benzene diazonium salts with different compounds

1. $ \ce{ArN2+Cl-}$ gives addition product with $\alpha,\beta$ unsaturated acids, with decarboxylation being integral part of reaction. For example,
  $$\ce{ArN2+Cl- + Ph-CH=CH-COOH -> Ph-CH=CH-Ar +CO2 + HCl}$$
2. With $\ce{SnCl2/HCl, ArN2+Cl-}$ given phenyl hydrazine $\ce{ArNH-NH2.HCl}$

I can not find mechanism of any of these reactions, first reaction seems  similar to Meerwein reaction (without $\ce{Cu^2+}$), but certainly can't follow similar mechanism. Second reaction is completely different from either. What are the reaction mechanisms of each?

The Meerwein Arylation Reaction (item 1) has been reviewed. It requires copper salts. https://onlinelibrary.wiley.com/doi/epdf/10.1002/0471264180.or011.03

As to your first reaction that does not utilize copper, a reference to the source would be useful. Nonetheless, the Meerwein Arylation Reaction has been reviewed in Organic Reactions. The reaction is conducted in the presence of catalytic cuprous ion. Meerwein proposed an aryl cation but a radical mechanism appears plausible. Reduction of the aryldiazonium salt 1 produces aryl radical 2 and cupric ion. Addition of the aryl radical to cinnamic acid 3 to the double bond to give the more stable of two possible radicals, ---namely--- the benzylic radical 4. This radical is oxidized by cupric ion to produce benzylic cation 5 and the catalyst cuprous ion. Loss of CO2 forms the stilbene 6.

Is it E-stilbene? Shouldn't be stereospecific ig..

You mean stereoselective. Probably, that's what I drew.

Didn't you draw stereospecific product, anyway I got your point (+1). What about second reaction..

Not much to say. SnCl2 adds electrons to the diazonium salt and HCl provides protons. The resulting phenyldiimide does the same thing. This was how Emil Fischer made phenylhydrazine with KHSO3. See: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/history/Fischer/fischer.html and https://pubs.acs.org/doi/pdf/10.1021/ed054p295

As to stereoselective vs. stereospecific, see ths ChemSE site: https://chemistry.stackexchange.com/questions/72694/selectivity-in-organic-reactions


text:
Why is the amu of chlorine-35 less than 35?
My book says that a proton weighs 1.0073u, a neutron weighs 1.0087u, and an electron weighs 0.00055u.
Now, why is the mass of chlorine-35 equal to 34.969? Are there not 17 protons, 18 neutrons, and 17 electrons? I calculated it and it sums to around 35.29. Where did I go wrong?

It did not go wrong. The average mass per nucleon decreases toward iron, and then increases toward uranium. Where do you think nuclear fusion and fission take their energy from ? Check masses of 4He and 16O, or Fe isotopes.

hi! my book said that all protons are the same weight, so goes with neutrons and electrons. where did i go wrong when I simply added the weights?

You have not counted mass loss due their bound energy. $\Delta E = \Delta m \cdot c^2$

oh cool! my book did not teach about that yet, so good to know. Thanks for helping me! :)

Related question and answer.

E.g. hydrogen to helium stellar fusion means loss of 0.7% of the mass.

Every proton and neutron is loosing a small part of their masses when they are included in a nucleus.

You need to account for the energy released when nucleons and electrons come together and form a Cl-35 atom. Its called the Binding Energy.
This sort of equation can help to explain :
(Rest Mass Energy of Individual nucleons,electrons*) - (Various Binding Energies) = (Rest Mass Energy of Natural)
*edit

ok thanks! i get it now, since the book did not say anything about binding energy.

Also, I used the term mass-energy. You can interpret it as actual mass, according to the mass-energy equivalence ( the famous E=mc^2 ).

@12345bird As mass of subatomic particles is used to be expressed in equivalent units of energy, usually MeV.

should be:  (Rest Mass Energy of Individual particles, neutons, protons and electrons) - (Various Binding Energies) = (Rest Mass Energy of Natural Nucleus) // The electron binding energies are small compared to the nuclear binding energies, but easily measurable today.

Oh noted, I'm not really up to date on new technologies. I'll make the appropriate edit. Thanks!


text:
Predicting reactions problem

I am not sure where to begin with this problem. I feel Br carries negative charge and is this gonna be no reaction since the alpha carbon is $sp^2$?

On the contrary,the large negative charge density on the vinylic group will be much more reactive(due to the relatively higher instability of the anion),and hence it will be much more prone to be attracted towards the positively charged carbonyl carbon and perform an addition

https://en.wikipedia.org/wiki/Grignard_reagent#Reactions_of_Grignard_reagents will point you in the right direction...

The reagent 1 is a Grignard reagent and your substrate is a ketone. Sounds familiar?

@MathewMahindaratne Ok let me see.