text:
Why isn't pressure filtration popular compared to vacuum filtration?
I want to build a DIY vacuum filtration apparatus with a bike pump but it's very difficult to find one the could be easily converted into a vacuum pump. This prompted an idea to simply build a reverse version where above atmospheric pressure is forcing the liquid to flow through the filter. I notice pressure filtration is rarely mentioned in write-ups, unlike vacuum filtration. Is there a reason for this lack of popularity or a reason why such a DIY apparatus would likely not work?
Could it be that vacuum is necessary for drying the paste that forms at first because it enhances evaporation?

I guess it is because people typically have vacuum lines in the labs, but not high pressure lines.

@IvanNeretin Every lab I've seen has a 5 or 10 bar pressurised air supply, coming from a big compressor in the basement. No? A vacuum line needs a pump usually in the same room, which is high-maintenance. In my old institute, we had only a few rooms with vacuum lines, mostly for the inorganic first-year lab courses.

@Karl Then I guess we were in different kinds of labs.

Not to answer your question, but perhaps to solve your problem: You can buy a plastic Venturi Pump online for less than $20; you attach a hose, screw it into your faucet, let the water flow, and air is sucked through the hose (via the Venturi Effect). I used one myself for at-home vacuum filtration, and it seemed to achieve quite low pressures. Just make sure you don't turn the water on so high that pressure builds up in your water pipes and they leak.

@IvanNeretin I think so! ;-) But what are you using them for? Anything except water chemistry ( or other high-boiling solvents) would be a lot of hassle with waste disposal, and slightly dangerous, wouldn't it?

Because regular glass parts structurally much stronger when low pressure in - high pressure out due to their cylindric/spherical shape, but much weaker structurally to hold high internal pressure. Pumping any glass cylinder can produce millions of small shrapners, especially if you cannot control the pressure inside. Vacuum pumping can be still dangerous, but the chances that you have more than 1 atm between the two sides of the glass wall is pretty low.

@Karl: both vacuum or pressurized air may be used to drive tools. Vacuum in particular is good e.g. for lifting pads and fixation systems. In research labs while many applications may need "vacuum" the particular requirements may differ widely so that locally producing the "right" kind of underpressure is better: e.g. evacuating dewars or IR detectors needs extremely low pressure but almost no flow whereas filtration needs less underpressure but higher flow and the ability to deal with a contaminated gas stream. The mechanical vaccum tools need high flow, no extremely low pressure and ...

... no particular care about contaminated air (the appropriate system for this could directly be used as pre-vacuum for the detector evacuation pump).

@electronpusher In some European universities these water venturis are banned already for multiple years, and were collected from both teaching as well as research labs, too.  From outside the labs (law) because of their water consumption, as well as for the water pollution; from inside the lab since their power varies too much if multiple room on a floor engage them.

@cbeleites Evacuating dewars or IR detectors needs a dedicated turbo pump, costing dozens of k€, that's not what we're talking about here. Agreed about a (20-50 mbar) vacuum line being potentially useful, but I've rarely seen a modern instrument using it. I'd be interested to hear about actual, current uses. The ordinary lab today has a portable membrane pump, which is used a few times a week, for filtration purposes. Schlenk lines, of course, but that's a different story again.

@Karl: the vast majority of house-wide pressurized air and vacuum systems I've seen were rather for mechanical (including stabilized tables for analytical instrumentation - but arguably that was optical instrumentation as well, just used to analyze chemical samples) than chemical purposes. And pressurized air orders of magnitude more common than vacuum lines. I've seen vacuum lifting pads in action in the chemical industry for lifting/handling bars of precious metals. My prediction (and dream as spectroscopist working with biological samples) for the not too distant future is, though, that..

... the automatization technology that is getting smaller now like pick & place machines (they work with vacuum suction, not sure about pressure, though) will become available for us in the form of autosamplers for solid pieces just like pipetting robots are around already. I still guess they'll come with their own vacuum pump. And I totally agree that for most chemical labs, the membrane pumps are the better solution for filtration because you only need to worry what happens in your lab (i.e. what spill/contamination/gas gets into the pump).

It's much easier to assemble a system that's safe under external pressure than one that's safe under internal pressure. Convex flasks and tubes resist external pressure better than internal pressure. If something leaks or breaks, you have air rushing into the system, rather than reactants and products blasting out. For that matter, if you've put things together with ground glass joints or rubber stoppers or corks, internal vacuum (external pressure) works to keep them together, while pressurizing the system's interior tends to force them apart.
Also note that a vacuum system will never be under more than one atmosphere of external pressure. With a pressurized system, you need a way to limit the pressure to safe levels, and you need to make sure you know what those levels are.

Also, it is easier to apply more sample to the filter if it is not under pressure, or wash it with some fresh solution. You do have to turn off the pump when you want to remove the filtrate. Flash chromatography does work with pressure on the column bed rather than vacuum applied to the outlet of the column.

Pressure filtration is commonly used in industrial applications outside of the lab. In this case vessels and process lines are usually made from metal or plastic and can be easily designed to operate at a given pressure. From an engineering standpoint, pressure filtration offers the ability to have a higher driving force and allows you to limit the required filter area. With vacuum filtration the driving force across the filter is limited to 1 atmosphere. Pressure filters can be safely designed to operate at much higher pressures.

I'd like to expand on some points that have been mentioned in the comments by @IvanNeretin, @Karl and @electronpusher, but IMHO deserve some more attention:

As @electronpusher commented, aspirator (Venturi) pumps are cheap and easily available.
They are also super robust: each pump is usually made from a single material, and a range of materials (brass, PP, glass) is available. This makes it easy to get one for pretty much any kind of solvent - no worries about pump membranes etc.. Also, no moving parts, no maintenance required other than making sure it doesn't get clogged.
In addition, with good water supply, you can get quite close to vapor pressure (either of the water driving the pump or of your solvent/eluent/washing medium).
Yes, the exhaust gas is released and gets mixed with the water. However, you can take care of both locally: collect the water for appropriate treatment and work under a hood that is properly equipped to deal with the substances in question.

In contrast, a house-wide vacuum system means that you really have to make sure that no exhaust gas enters that vacuum system where it may condense and/or mix with other exhaust gas or air from other outlets to form corrosive or dangerous (combustion!) mixtures.
Any mistake here can be very costly and dangerous.
(Side note: always make sure a vacuum outlet is meant for "chemical" use and not only for workshop/manufacturing/handling purposes such as pneumatic systems with lifting pads or fixation pads)

@jeffb and @KarstenTheis have already explained the safety and practical advantages of working with vacuum in lab settings rather than with positive pressure.
In addition, pressurized air lines often deliver oiled air: it is never safe to assume pressurized air is "just" air.
This is good for pressure tools (who otherwise need an extra oiler) and also cheaper to produce. There exist oil-free compressors and de-oiling systems that produce oil-free air for the whole house but it is usually more economic to go for oiled pressurized air in general and de-oil only the part of pressurized air that really needs to be oil-free and to the degree needed. Particularly if that pressurized air system is not exclusively for chemical lab use but also used to drive dampers for optical tables, pneumatic system, and all kinds of tools in the mechanical workshop.
Bonus point: for this reason, be very careful what pressurized air you use for cleaning (not only in terms of chemistry, also optical parts). It's perfectly appropriate to buy expensive bottled "cleaning air" or have an oil-free compressor when there's a pressurized air outlet in the very same lab!

Positive pressure filtration is extremely common, probably even more common than vacuum filtration.

source: https://andyjconnelly.wordpress.com/2016/09/28/syringe-filters/

I worked as a grad student in organic chemistry labs and both vaccuum lines and compressed air are very common.
The reason most filtering setups uses vaccuum (especially for gross extraction after recristallization) is that glassware does not tolerate pressure very well.
We do use air pressure to speed up purification with manually-packed silica gel chromatography columns because installing a vaccuum apparatus for each tube would be a major assle. But even then we use very little pressure as this can break the column. There is specialized glassware for this - it has a crude teflon pressure regulator with a rubber O-Ring (that has to be changed regularly as it dries).
Low-grade vaccuum setups are relatively easy to improvise - with an appropriate adaptor, you can use your home sink tap as a vaccuum source. This setup is often used for rotavapors in organic chemistry labs.


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Is it possible to have only liquid in a closed container?
I recently read an answer by Aman Rusia for this question: Is a liquid in a container always in equilibrium with its vapour?
In it, they say that

You can't have a container filled with only liquid. Think about that
  for a moment. Similarly you can't have a container filled with only
  ice. Some water, or water vapour, or even combination of the two will
  appear alongside.

However, my textbook says that

Imagine that you have a sample of water in a cylinder fitted with a piston at low pressure. Suppose that the temperature is held constant at 50 °C, and that weights are placed on a piston to exert a pressure of 1.0 atm. Only liquid water is present. The piston presses on the surface of the liquid. Now gradually reduce the pressure by removing some of the weights. At first, nothing seems to happen. The high pressure is keeping all the water molecules in the liquid state, and the volume of a liquid changes very little with pressure. However, when so many weights have been removed that the pressure has fallen to 0.12 atm (the vapor pressure of water at 50 °C), vapor begins to appear. The sample is now at the vapor-liquid boundary on the phase diagram. The pressure remains constant so long as the liquid and vapor phases are both present at equilibrium and the temperature remains constant. You are free to pull up the piston by an arbitrary extent, but enough water will evaporate to maintain the pressure at 0.12 atm. When you pull the piston out far enough, the liquid phase disappears; you are now free to modify the pressure of the vapor at will.

Atkins, Jones, and Laverman, Chemical Principles
So, is it true that you can have a container filled with only liquid?

Of course you can. Dive to the bottom of a lake (short of a lake, a bucket will do), and open your container there. Will some water magically boil and turn into vapor inside it? I don't think so. Maybe it will when you close the lid? Still unlikely.

Considering my textbook's information, what if gaseous nitrogen with a partial pressure of 1 atm was introduced to the piston? At what total pressure would the water vapor start to form, and at what total pressure would the liquid water completely vaporize?

You can't have just liquid if its volume is smaller than that of the container. Is the wird "filled" that makes the difference, and in this sense the answer you mention is not accurate. The P of the liquid must balance vapour pressure. See @Karl answer

Note that even your question can be interpreted.  I would perhaps say "no" to the title question but answer "yes" to the last sentence.

In a tight, solid vessel, you can of course have a liquid (or any condensed phase) completely fill it.
There are basically three possibilities

the inside is at (potentially very) high pressure. no problem, water is compressible, just not very much, and also the vessel has a finite $E$ modulus
the inside is just at the vapour pressure of its content, but (because of the different expansion coefficients of vessel and content) the temperature must be kept perfectly constant at exactly the level where this is the case
the inside is below vapour pressure, but no gas bubble has evolved yet, that's principally an unstable condition, but if the deviation isn't large, it can stay practically forever.

Because of surface tension, there is even a small pressure range where the last case is actually stable.
I say your statement is basically right: You can have a completely filled vessel, but you can't do anything useful in it before it either bursts or bubbles.

For case 1, imagine a balloon filled with a liquid such as gallium (negligible vapor pressure at STP). Or a bellows, for that matter.

@DrMoishePippik So? The vessel excerts a certain pressure on the liquid, which tries to push back with its vapour pressure. If one of them wins, you are at points one or three, if they are identical, point two.

Just citing an example...

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Calculating ideal chemical work using extent of reaction and chemical potential
I have a electrochemical reactor assumed isothermal and isobaric with 4 reactions and I am trying to calculate the "ideal chemical work' exerted by each reaction using the extent of reactions for each reaction (which I have already calculated).
I am calculating the 'chemical work' $w$ by multiplying the extent of reaction by the Gibbs energy of the respective reaction $k$:
$$w_k = \mathrm{d}G_k = \Delta_\mathrm{r} G_k \cdot \mathrm{d}\xi _k $$
I'm calculating the reaction Gibbs energy $\Delta_\mathrm{r}G$ as basically the stoichiometric sum of the chemical potentials
$$\Delta_\mathrm{r} G = \sum_i(\nu_i \cdot \mu_i)$$
However, I am not sure which conditions to use for the calculation for the reaction Gibbs energy as the reactions are reversible and for 2 of the reactions I have taken the forward reaction as the opposite of the expected (i.e. the forward reaction is the fuel cell reaction but I am operating in electrolyser mode so the extent of reaction for these reactions is negative).
Do I use the conditions at the reactor outlet? Average of the inlet and outlet or am I going about this completely wrong?

If the concentrations of reactants and products are changing over the course of the reaction, you have to take that into account and integrate over the Gibbs free energy of reaction with the extent of reaction as integration variable (or over the electrochemical potential with the charge through the wire as integration variable). See also https://chemistry.stackexchange.com/questions/6076/express-the-maximum-work-from-a-voltaic-cell/112313

Thanks for responding! Okay so if I integrate over the course of the reaction, with respect to the extent of reaction, $\xi$, where at the beginning of the reaction $\xi = 0$ and at the end of reaction, the extent is known, then the solution of the integral would simply mean the change in Gibbs energy (work) is the product of Gibbs energy of reaction (second reaction in OP) using the final conditions and the final/overall extent of reaction. When I use these conditions I get similar results to when I did an overall Gibbs energy change but not the exact value which is why I'm so confused.

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Bond dissociation energy calculation
I am looking for methods to determine the dissociation free energy in a precise way, ideally from a simulations. My hope is to get the most precise values possible that are specific to the structure of the chemical element, rather than the more general energies I have found that completely ignore the context of the molecule (like in "Bond Dissociation Energies of Organic Molecules", ACCOUNTS OF CHEMICAL RESEARCH 2003). A machine learning program that can assess the bond energy of an input molecule using parameters learned from a database would be ideal.
Alternatively, I would like to know what are the most accurate and reliable source for such values, as I have been having trouble finding sourced values.
The goal is to use these values to calibrate other properties of a system based on thermodynamic cycles. Since this calibration is for molecular simulations, I need accuracy but most importantly, consistency between methods. I can live with not so accurate parameters, as long as the overall force field is balanced, meaning the differences are balanced.

NIST is a precious source of physical data of molecules. You might be interested in: https://srd.nist.gov/NSRDS/NSRDS-NBS31.pdf as well as the whole database itself
https://webbook.nist.gov/chemistry/form-ser/

Thanks, I did not know of this resource, will definitely help.


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Why does silver (I) have a larger lattice enthalpy and hydration enthalpy than sodium?
The ionic radius of the $\ce{Ag^+}$ ion is $129$ pm, and that of the $\ce{Na^+}$ ion is $116$ pm.
Since the sodium ion is smaller than the silver ion, it makes sense that it has a stronger polarizing power than the silver ion. Due to the higher charge density, the lattice energy of $\ce{NaX}$ compounds is predicted to be larger than those of $\ce{AgX}$ compounds. However, that is not the case, as shown here:

Furthermore, the enthalpy of hydration of $\ce{Na^+}$ is expected to be more exothermic than that of $\ce{Ag^+}$, because $\ce{Na^+}$ is more polarizing than $\ce{Ag^+}$. However, that is not the case:

Why are the lattice enthalpies of silver compounds larger than those of sodium compounds, and why are hydration enthalpies of silver compounds more exothermic than those of sodium compounds, despite the silver ion being larger than the sodium ion?
Sources:
R. D. Shannon (1976). "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides". Acta Crystallogr A. 32: 751–767. Bibcode:1976AcCrA..32..751S. doi:10.1107/S0567739476001551.
Atkin's Physical Chemistry

Partly at least because ionic radii are not as uniquely or strictly defined as you would seem to believe. And, partly because lattice enthalpies vs ionic radii is a very tenuous connection.

The silver ion is larger and can have more interactions at the same time.

... and because it's relatively larger, it forms a different crystal structure, and then quantitative comparison of molar properties becomes very questionable

Crystal structure shouldn't affect enthalpy of hydration, right?

not "crystal", but (statistically) more water molecules fit around a larger ion

The electron configuration of sodium ion is that of the inert gas neon (1s^2 2s^2 2p^6), a spherical shell with a positive charge at the nucleus. There are 10 electrons inside this shell.
The electron configuration of silver ion is that of krypton plus 4d^10, a spherical shell with a positive charge at the center. There are 46 electrons inside this shell.
While the slightly smaller sodium ion might have slightly more polarizing power than silver ion, silver ion has far more polarizability than sodium ion. Polarizability is the distortion of the electron cloud as a result of a point charge (or dipole). Dispersion forces are a form of polarizability where two polarizable entities polarize each other. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Polarizability
The effects of dispersion forces due to the polarizability of the anion can be seen by comparing the enthalpies. The crystal structures for the fluorides, chlorides and bromides and NaI are FCC, rock salt. AgI is wurtzite. The difference in enthalpies for the fluorides is 43, but as the polarizability of the chloride anion increases, the attraction of dispersion forces increases the difference to 125, and for bromide, 148. Even tho the crystal structure of AgI is different, it still seems to have a polarizablity-polarizability effect (difference = 181).
This is not what you would call a predictable effect, but does a good job of explaining the trend. It is not a first-order attraction, nor a second-order polarizability, but a third-order polarizability-induced polarization effect.
And very perspicacious of you to have observed this trend.


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Can someone please explain to me where the flaw is in my proposed synthesis mechanism?

For this I have to convert benzene into the product in the box using the reagents listed. The order of reagents that makes sense to me is IDGBEH. However, this is not the correct answer. Somewhere I am missing 2 steps to the synthesis and I am at a complete loss. I understand all of the basic and complex concepts of organic chemistry we've covered in my class thus far, so I'm confused as to how my proposed mechanism has an issue. If it's any help I know the mechanism is supposed to be 8 steps.
In my mind IDGBEH makes complete sense and should give me the product that I want, unless I am missing something or I misunderstood my professor.

You forgot protection/deprotection of amino group.

I am sure you can come up with a specific title.

The steps for conversion of Benzene to 1,3-dichloro-5-(propan-2-yl)benzene are shown below (Scheme).

A possible improvement would be to use a large protecting group like pivaloyl instead of acetyl. This should more effectively prevent ortho alkylation. Then again, it's probably also more annoying to deprotect. There's probably a good intermediate size group.

You have to protect the amino group because the amino group is so activating that it is hard to stop multiple substitutions from occurring. By reacting it with the anhydride, the lone pair on the amino group gets delocalized due to conjugation and is therefore less available to the ring, making it easier for one addition. However, it seems that you want the chlorines ortho to the amino group, since it can't go para due to the isopropyl. So multiple substitutions shouldn't matter.


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What is the pH of a solution of iron acetate?
What is the pH of a $\ce{0.1 M}$ solution of $\ce{Fe(CH_3COO)_3}$? I'm not sure how to do this, as it would involve complicated mass balance equations, but I think I can apply the approximation $\ce{pH = \frac{1}{2}*(pKa(Fe^{3+}) + pKa(CH_3COOH)) = \frac{1}{2}*(4.76 + 2.22) = 3.48}$.
Is this approximation valid in this situation? If not, how would I correctly calculate the pH of the solution without solving complex mass balance equations?

Well the acetate will protonate making the pH go up. The iron can form hydroxides making the pH go down. Assume salt completely ionizes, and first assume that iron can be ignored. If iron is now above its solubility limit then you have to solve iron and acetate simultaneously.

Does your approximation completely ignore the concentration of the solution?

PS - Iron has no pKa.

@MaxW Why not? The iron ion is a weak acid in water

Compare ammonium acetate, https://chemistry.stackexchange.com/questions/75047/ph-of-ammonium-acetate-solution

Well, not iron but its aquo complex to be precise...

Like $\ce{ [Fe(H2O)6]^3+ + H2O <=> [Fe(H2O)5(OH)]^2+ + H3O+}$

Ah.. yes. Thanks for the correction. Normally I'd just work the math for the "total" reacton, $\ce{Fe^{3+} + 3OH- -> Fe(OH)_3}$, but the iron complex would deprotonate in steps. Still bothers me to think of that being an "acid" since you can't have $\ce{Fe(H2O)_5(OH)Cl}$ as a salt.

Won't the iron precipitate as $\ce{FeOH3}$? You need to consider the $K_{sp}$ of the hydroxide as well.


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Will the volume occupied by a mixture of nitrogen and water be same as the volume occupied by them individually?
I am currently studying gaseous state and I encountered the following problem.

A container with a volume 3 liter holds $\ce{N2(g)}$ and $\ce{H2O(l)}$ at $\pu{29 ^\circ C}$. The pressure is found to be $\pu{1 atm}$. The water is then split into hydrogen and oxygen by electrolysis according to the reaction
  $$\ce{H2O (l) -> H2 (g) + 1/2 O2 (g)}$$ After the reaction is complete, the pressure is $\pu{1.86 atm}$. What mass of water was present in the container? The aqueous tension of water at $\pu{29 ^\circ C}$ is $\pu{0.04 atm}$.

This is one of the steps of the solution given along with the question (edited to show the units):

Amount of substance of nitrogen:
  $$n(\ce{N2}) = \frac{\pu{0.96 atm} \cdot \pu{3 L}}{\pu{0.0821 atm L mol-1 K-1}\cdot \pu{302 K}} = \pu{0.116 mol}$$
  Amount of substance of water:
  $$n(\ce{H2O}) = \frac{\pu{0.04 atm} \cdot \pu{3 L}}{\pu{0.0821 atm L mol-1 K-1}\cdot \pu{302 K}} = \pu{0.00484 mol}$$

The thing I don't understand here is that how can the volume of nitrogen and water be taken as $\pu{3 l}$? Both of them would occupy some volume not the whole volume will not be occupied by one of them.

Hello and welcome to our site! We have a chat room specifically for JEE Chemistry. You're welcome to join it, but you'll have to cross 20 reputation first to speak there.

You asked: "how can the volume of nitrogen and water be taken as 3 L?"  The answer is that the water is primarily in the liquid phase (with some water vapor, at a pressure equal to the vapor pressure of water at 29C), while the nitrogen is in the gas phase.  Thus you have liquid water at the bottom of the container, and gaseous nitrogen and gaseous water filling the remainder of the 3 L container, giving a total volume for all species of 3 L.

@theorist That is where I am confused. After the water is filled, the volume left will not be 3 L. Instead, it will be less than 3 as some volume will be occupied by the water. Thus your statement "remainder of the 3 L container", doesn't really makes any semce.

Not sure about the solution, but after the reaction is complete, you've got a 3L container that contains a mixture of N2(g), H2(g), and O2(g).  The volume occupied by each is 3L, since each gas has full access to the entire container.  Consider how you would calculate n for each gas with the ideal gas law:  n=PV/(RT).  For each gas, you would use P = the partial pressure of the gas, and V = the volume of the container (3L).  [Or, equivalently, you could use P = total pressure in the container (1.86 atm), and V = partial molar volume of the gas; but I'm guessing the solution used the former.]

@theorist No reaction takes place between nitrogen and water. Nitrogen just dissolves in water. I might have agreed with you if you said $N_2$ will occupy a volume of 3L. But, $H_2O$ is present in a limited region only and thus how it can occupy 3L of volume?

The question clearly says the liquid water is converted to H2(g) and O2(g), by electrolysis.  I.e., after the reaction is complete, no liquid water remains. Consistent with this, I wrote:  "after the reaction is complete, you've got a 3L container that contains a mixture of N2(g), H2(g), and O2(g)."

The idea behind the question is that you know the partial pressure of N2 is 0.96 atm (1-0.04) atm.  And you know that the partial pressure of all the H2 that can be produced from the water is thus (1.86 - 0.96)* 2/3 = 0.6 atm. [Ignoring the volume of liquid water, which is only ~1/1000 that of the gas.] From this you can determine the number of moles of H2, which is equal to the number of moles of water =0.6 *3/(302 * 0.0821).  Note the number of moles of water given in your answer isn't the total no. of moles of water, it's the number of moles of water vapor prior to the electrolysis.

@theorist I think I got it. Thanks for help.

@theorist Please write an answer and avoid answering the question in comments itself! Have a great day.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

The idea behind the question is that you know the partial pressure of N$_2$ is 0.96 atm (= 1 atm - 0.04 atm). And you know that the partial pressure of all the H$_2$ that can be produced from the water is thus (1.86 atm - 0.96 atm)* 2/3 = 0.6 atm. [Ignoring the volume of liquid water, which is only ~1/1000 that of the gas.] From this you can determine the number of moles of H$_2$, which is equal to the number of moles of water = 0.6 atm *3 L/(302 K * 0.0821 atm L / mol K).
Note the number of moles of water given in your answer isn't the total no. of moles of water, it's the number of moles of water vapor prior to the electrolysis.


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Reference on Pople basis set
I need to write reference on 6-311++g(d,p) basis set in my work where I used it on atoms of I-II periods. Does anyone know where it was first published?

If there is no particular reason to use Pople-type (Pople only developed double zeta quality) basis sets, you probably better switch to a more recent and more efficient one.

You should look on Basis set exchange
After you select your basis set of interest, on the bottom left you click "Get References for selected basis", and in this case you obtain (I did it for elements C,H,O):

References for the basis set
H
## VTZ Valence Triple Zeta: 3 Funct.'s/Valence AO for H,He
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## 1P Polarization functions associated with 6-311G for H,He
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## Diffuse functions asociated with 31G Hydrogen
Clark, Timothy, Chandrasekhar, Jayaraman, Spitznagel, Günther W.,
        Schleyer, Paul Von Ragué
Efficient diffuse function-augmented basis sets for anion
        calculations. III. The 3-21+G basis set for first-row
        elements, Li-F
J. Comput. Chem., 4, 294-301 (1983)
10.1002/jcc.540040303

C,O
## VTZ Valence Triple Zeta: 3 Funct.'s/Valence AO
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## 1P Polarization functions associated with 6-311G
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## Diffuse functions asociated with 6-31G
Clark, Timothy, Chandrasekhar, Jayaraman, Spitznagel, Günther W.,
        Schleyer, Paul Von Ragué
Efficient diffuse function-augmented basis sets for anion
        calculations. III. The 3-21+G basis set for first-row
        elements, Li-F
J. Comput. Chem., 4, 294-301 (1983)
10.1002/jcc.540040303


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What is the cause of decrease in volume of gaseous mixture in the following problem?
I am studying Gaseous State.
The problem says: A mixture of gas contains $\ce{N2}$ and $\ce{C2H2}$. $\pu{20 \\mL}$ of this mixture is added to $\pu{70\\mL}$ of $\ce{O2}$ and combustion is allowed to take place over mercury. After cooling the volume of gas was found to be $\pu{68.5\\mL}$. When the resultant gas mixture is passed through a $\ce{KOH}$ solution, the volume of residual gas is $\pu{38.5\\mL}$. Then the percentage composition of $\ce{N2}$ is:
In the solution to the question there is this following step which I am not getting:
:Volume of $\ce{CO2} = \pu{68.5\\mL} - \pu{38.5\\mL}$.         How come?
Also please clarify why nitrogen does not undergo combustion.
Please help.

Wait until you'll be studying Chemistry, and all this will start to make sense. There is no shortcut.

But  I need a bit help.

Chemistry is more than just a bit. It is a vast body of knowledge. There are whole textbooks on it, some of them thick enough to kill a man.

For starters, nitrogen is pretty unreactive... It's chemistry alright, but it's also general knowledge. Never heard of aircraft tires filled with nitrogen?

Could please come up with a solution to the problem. I need it.

If you are wondering why you didn't get any answers here, please read through the help section on the homework problem policy. Also, begging rarely receives a positive response (i.e. "Could please come up with a solution to the problem. I need it."). Regardless, welcome to chemistry.se and better luck with your questions in the future.

I think the fact you are missing is that after combustion, water vapors will be given away and the volume lost during combustion will be equal to the amount of water lost. $\ce{KOH}$ will absorb $\ce{CO2}$ and the volume absorbed by $\ce{KOH}$ will be equal to volume of $\ce{CO2}$ in the solution. Thus, you will have two equations and you will be able to proceed with the solution.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also note that capitalization plays significant role in STEM sciences, for example, "KOh" makes no sense at all.


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Effects of gamma irradiation on pH buffer
Does anybody know if it is possible to gamma-irradiate a bottle of pH buffer to sterilise it? Are there any known effects (short-term or long-term) of the irradition process on the pH qualities of the buffer? This would be for a pH4 and pH7 buffer.
The other option I am considering is autoclaving, but I am told this will definitely affect the pH of the buffer solution, which I don't want.

I think it's virtually guaranteed that you'll reach sterilising doses of gamma radiation far before you create a bulk chemical change in the solutions, especially if the buffers are made with simple inorganic oxyhydrides (phosphate buffers?). Life is chemically far more fragile than a buffer solution. Also, the fact that gamma irradiation is a safe and proven method for sterilising food for human consumption is an additional argument in your favour. That said, gamma irradiation seems overkill. I'm not sure that autoclaving would irreversibly alter the pH, again depending on your buffer.

Wonder how an autoclave would work with a solution. If the pressure was created by steam, then the container shouldn't be at risk from the start of the cycle till 121 C is reached. However once the steam stops, the pressure inside the autoclave would start to drop. However the bottle of liquid would still be near 121 C, a high enough temperature to boil at atmospheric pressure.


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What's the definition of multi phase materials? Does it mean the different phases represent different materials?
I'm from a CS background and just trying to understand the real meaning behind multi phase materials. Following is the experiment which drove me towards this question.
With the help of DFT and machine learning, we identified that the material LaBiO3 could display interesting dielectric properties. Hence, a team here mixed the oxides in stoichiometric ratio in a ball mill and observed the XRD patterns of the formed composite powder.
The XRD data suggest that LaBiO3 is unstable and form Bi1.4La0.6O3 R-3m phase and Bi2La4O9 R3m phases.
So my question is, is this a composite that has nothing to do with LaBiO3 and they are just two different materials? Or could this be categorized as a metastable phase of LaBiO3?
Could anyone explain this with the definitions?

La2O3 and Bi2O3 have different crystal structures. This suggests fairly strongly that a (BiLa)O3 phase will not be stable.

The definition of a substance that is "multi-phase" is that there is more than one distinct compound in it and the compounds form distinct regions in the substance with different properties.
Examples make this clearer. Salt dissolved in water forms a uniform mixture: a salt solution. There is more than one compound there but they are perfectly mixed.
Fatty organic compounds are not soluble in water but can form small dispersed fatty particles. We call the result milk (I'm simplifying a little as there are other things in there as well). The result has more than one phase: a microscope would reveal very small, dispersed particles of fat floating in water. Milk looks like a single uniform substance, but isn't when you look more closely.
Fibreglass is, essentially, a bunch of small fibres made of glass embedded in a matrix of some polymer. There are two distinct phases with different compositions in the substance.
The characteristic of multi-phase compounds is that different regions of the substance contain different compounds (these regions can be very small as in some metal alloys). In your case the reaction seems to produce two distinct compounds which may be finely mixed together. The XRD pattern shows two different crystal structures. This may result because this specific reaction does not create the compound you wanted under these conditions. Which could be because the compound you want is not stable or could be because the reaction conditions allow more than one product to form. Either way the reaction is giving a mixture not a single compound.

Thanks @matt_black for the reply, it does simplify the term "multi-phase". So just to make sure I'm getting this right, salt solution you mention is single phase, result milk is multi-phase but looks like a uniform substance (clarification asked below) and fibreglass is multi-phase with visible signs to the outside? If so, does this result milk show uniform physical properties as well? like uniform density, viscosity and all? If it does we can clearly differentiate these from fibreglass kind of compositions.

@avin "Uniform" depends on what scale you measure on. Milk looks uniform in bulk, but not when you look closely enough. Then you see droplets of oil which have different density and viscosity to the water than surrounds them.

This is actually a super important concept in chemistry. Without belaboring it.  There is a difference between having a mixture of different substances (e.g., carbon and oxygen, graphite and $\ce{O2}$ molecules) and having a new compound, the carbon and oxygen reacted to make $\ce{CO2}$.
In solid state compounds, this can sometimes be confusing, because there is no big change evident (usually) in terms of a gas being generated or the like.  So, for instance (true story), I found $\ce{ZnSnO3}$ reported as a "compound" in the powder diffraction database.  But when I looked at the spectrum--and I was intimately familiar with this phase space--I saw that the spectrum of "$\ce{ZnSnO3}$" exactly matched a mixture of $\ce{ZnO}$ and $\ce{Zn2SnO4}$ (known compounds).  An email to the JCPDF database got the spurious $\ce{ZnSnO3}$ deleted.
The reason I am telling this story is it is directly analogous to what your guys are going through.  And it is not common, but also not vanishingly rare, in solid state chemistry to get spurious new compound reports based purely on powder x-rays (even when just looking at that would tell you, you have a mixture). This is especially the case when people are "trying" to get a certain new compound. Lesson is....watch out for mixtures.


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Br- and I- ions give layer test, but Cl- ion doesn't. Why?
In a paragraph about "non-metal displacement", my book talks about chlorine displacing bromide and iodide ions in an aqueous solution. This results in the evolution of bromine/iodine gas depending upon the reaction. When the evolved gas is dissolved in Carbon tetrachloride, the resulting colour change is used to determine the presence of Bromide and Chloride ions in the laboratory. This test is named as 'Layer test' in my textbook.
My question is, if chlorine gas is dissolved in Carbon tetrachloride will it induce a colour change? If not, why?
One of the questions I found in an NCERT book. Interestingly I could not find the answer anywhere on the internet. Hence, stack exchange.

Welcome to ChemSE.  Please substantiate your question by addition of futher information providing (chemical) context + at least the start of your argumentation to solve the question.  Layer of what?  Since NCERT is an organisation, would it be a question about organic chemistry or inorganic chemistry or something else?

@Buttonwood The organization NCERT writes textbooks for Indian schools which are colloquially called NCERT books, and the chemistry textbook contains organic, physical and inorganic chemistry together. And BTW I haven't come across this "Layer test" ever in it, but again I never read the book seriously.

@Soham can you type out how the "Layer test" described in the book? Some chemical context is mandatory. I can't find anything by that name anywhere.

This (Layer test) is an ancient chemical method. It's as outdated as the theory that everything is made of five elements. If you want to study real chemistry don't read NCERT. It's a really difficult read and hasn't been reeditioned in decades.

The so called layer test is a qualitative analysis of bromide and iodide salts (old school chemistry). The salts are oxidized in water and then chloroform or carbon tetrachloride is added. This forms a layer. Since the partition coefficient of iodine or bromine is much higher for the organic solvent, a colored layer is formed sitting in the bottom of the test tube (hence the name layer test). Bromine shows an orange color and iodine shows a beautiful violet color. Chlorine is almost colorless, it is a gas, it will not be visible in these chlorinated solvents. And above all, very few oxidizing agents can form chlorine with chloride.

Organic Layer test is a test involving redox reactions among halides. This test is used to detect in the presence of bromide ions and iodide ions in a solution (e.g., Lassaigne’s test solution).  Layer test is carried out by adding organic solvent such as carbon disulphide or chloroform in the presence of nitric acid and chlorine water to the given solution. For example, Lassaigne’s extract is acidified with dilute nitric acid and chlorine water is added to it. When organic solvent is added to this aqueous solution and shaken the mixture vigorously but carefully, the liberated bromine or iodine liberated by redox reaction of bromide or iodide ions with chlorine) would be traveled to organic layer to give distinguished color. In case bromine is present, it will give an orange layer while it will give a violet layer in case of presence of Iodine. Relevant redox reactions are given below:
$$\ce{Br- + Cl2 -> Br2 + Cl-}$$
$$\ce{I- + Cl2 -> I2 + Cl-}$$
The key is bromine and iodine are much soluble in organic solvents than in water. Since we are using chlorine as a oxidant, chloride ions are negated here.


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Glass joint stuck in flask
I am doing distillation using system of glassware including a flask and a glass joint. Recently I've faced a problem: the glass joint stuck in the flask. I tried to remove it but it broke and hurted my finger. I want to ask how to prevent this from happening again: can I use a paper masking tape or continue using grease? I am thinking to try the adhesive tape do you think it deserves trying?

Silicon grease on the ground glass joint. Use a heatgun on seized joints

Safety note: as you've apparently figured out, you should not apply heavy torque to a stuck glass joint with unprotected hands. Something like heavy leather gloves should be used. Also, welcome to chemistry.se.

Brings back the good old days in high school circa 1970. Used a lot of rubber stoppers and rubber tubing on glass tubes. Had to carry a pocket knife to cut them off lest you break the glass. can you imagine carrying a pocket knife into a school these days?

The next time try to use a wood handle of some sort or a wood rod to beat the joint. You won't risk injuries and very often it works. You can also envelope the joint/glassware in a cloth for maximum safety and if "energic" beating is required.

I found, lying around my lab, a staple remover with the handle perfectly fitting between two jointed pieces of glassware. When two pieces get stuck, I press the handle, and they get apart with no effort. Rather than knocking the glass on a surface "the usual way", I would try to get creative and find something like that :D

@The_Vinz is it possible without breaking the joint?!! Do you use it when hot or could do it after cooling and even if it's stuck for too long?? You are so creative!!

Yes, you just use it as a lever to force the joint vertically without putting stress on the delicate parts of the piece

You have multiple options.  All of them assume that the distillation setup is empty, at ambient pressure.

Of course, like mentioned in the comment by @Waylander, put a little of (silicon) grease on the glass joints prior to the distillation, just enough that they become transparent.  There are multiple kinds of grease, suitable for different temperatures / pressures during the application.
If grease is not an option, there are sleeves made of PFTE / teflon to be put on the joints.  They are available in different diameters, and forms, e.g.

(source, source)

Especially for vacuum distillations, but not limited to this: the best remedy for this issue is to dismount the distillation right after the recovery of the product fraction(s), while the parts (e.g., column, head) are still considerably warm.  Otherwise, once cooled to rt, the setup then often will stuck much better than ever intended.  Depending on size of the distillation and remaining temperature of these pieces, wearing gloves (to protect your hands against the heat) may be helpful.  Then, it is better to allow the warm pieces to cool to rt while being put e.g. on cork rings.
«Forgotton» and already stuck glass joints may be detached from each other if you immerse them in the basket of a ultrasound bath.  Addition of some of detergent to the water bath helps while occasionally gently pulling the pieces apart.
If you have the dexterity (some develop this over time / by experience), wear both gloves to protect your hands against heat and goggles: you may heat the circumference of the outer wall of the glass joint with a heat blower or even a heat gun / pocket burner.  The aim here is that the outer part of the joint dilates quicker than the inner part.  It is the most risky variant because uneven heating may shatter your distillation into pieces; especially if the parts joined together are made of different types of glass with different expansion coefficients.  It this is not for you, just ask your local glass blower for assistance.

Dear Buttonwood, thank you so much for answering my question. I was just asking if a normal masking tape will tolerate such a high temperature and vapors during distillation but I think it will not and it will losses its adhesion and maybe will affect the joint this is what I want to make sure of. What do you think?

Bad idea. This is not what the masking tape is designed for. You will also get the adhesive in your distillate

@ShafikSaad I second Waylander.  Masking tape (by your comment) or adhesive tape (by your question) contain multiple compounds which the distillate might solute.  At elevated temperatures, they may melt and drop into your flask; if they solidify they might glue your joints together, too.  Each of these perspectives are counter-intuitive to the aim to purify (i.e., to separate) your target compound from other compounds.  And even at room temperature, mechanically speaking, adhesive tape is not an alternative to a clamp to keep your pieces of a distillation joined.

@Buttonwood Many thanks for your help, I've bought a new flask and joint and I am using grease now and I am removing the joint immediately after distillation even if hot so I think I will not face this problem again, also I have a sonicator here and I've been told that a sonic bath with soap could solve it so I will give it a try... many thanks for your help

Silicon grease is the one we used normally. But, your post indicates that grease hasn't worked for you. Any adhesive tape wouldn't work because it will increase the risk of leaking (specially when you are refluxing). However, we have used PTFE sleeves between joints, which have worked wonders. These sleeves are not cheap but very durable and reusable. Following are the websites you can order them:
https://www.thomassci.com/Laboratory-Supplies/Stoppers/_/Sleeves-PTFE?q=Joint%20Ptfe%20Sleeves
https://www.thomassci.com/Laboratory-Supplies/Glass-Joints/_/TEFLON-SLEEVE-GASKETS?q=Joint%20Ptfe%20Sleeves
https://www.thomassci.com/Laboratory-Supplies/Glass-Joints/_/SYNTHWARE-PTFE-Joint-Sleeves?q=Joint%20Ptfe%20Sleeves

Dear Mathew, thank you so much for answering my question. I was just asking if a normal masking tape will tolerate such a high temperature and vapors during distillation but I think it will not and it will losses its adhesion and maybe will affect the joint this is what I want to make sure of. What do you think?

I would say the same. It won't tolerate masking tapes.


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Which metal ions precipitate in NH4OH?
From Wikipedia-1, and from an answer here on Stack-Exchange, I can list at least the following elements which will dissolve in aqueous $\ce{NH4OH}$:

$\ce{Cr(III), Co(III), Ni(II), Cu(I, II), Zn(II), Ag(I),}$ along with  "several platinum group metals"

From reading Wikipedia-2, I also suspect the following elements to dissolve in liquid ammonia:

$\ce{Mg, Ca, Sr, Ba, Eu, Yb, Li, Na, K}$ and $\ce{Rb}$

As a matter of fact, according to Wikipedia "almost all metal ions bind ammonia as a ligand." I thought it would be interesting, and potentially very useful, to find out which metal ions are NOT soluble in $\ce{NH4OH}$.
I can list two for now:

$\ce{Fe(OH)3}$
$\ce{Mg(OH)2}$ with a solubility product of only $5.61×10^{−12}$ (credit to Oscar Lanzi)

There is a huge difference in pure liquid $\ce{NH3}$  and an aqueous solution of $\ce{NH4OH}$. I can't imagine any of the metals Mg, Ca, Sr, Ba, Li, Na, K or Rb forming amine complexes.

For the case of magnesium ion, at least, it seems a stretch to say it dissolves in ammonia solution.  Assume that the ammonia solution has $\mathrm{pH}$ of $11$ and magnesium hydroxide has a solubility product of $5.61×10^{-12}$. We infer that the magnesium ion solubility at equilibrium with the hydroxide in ammonia solution is only $\pu{5.6μM}$ which is not really all that soluble.
Magnesium ion is appreciably soluble not in an ammonia solution but in a buffered solution of ammonia and an ammonium salt.


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Wave-particle duality
Can I get an answer stating some analogies on how electrons behave like a wave, and what does it even mean for an electron to behave like a wave?
Also, if going around the nucleus, is electron a real physical wave or a probability distribution wave? And when in motion otherwise, (like an electron shot towards something) does it travel like a wave? I mean like up and down and up down that sort? Or is this wave different in some kind?
I'm really having a very hard time understanding this stuff, please help
PS- can you suggest a book which explains me this in kinda fun and intuitive way rather than mathematical balderdash?

That mathematical balderdash is what called science...

Fun and intuitive only gets you a certain part of the way. Would you like a pilot, a surgeon or nuclear engineer learn their craft in a "kinda fun and intuitive way"? There is a benefit to a quantitative treatment.

@KarstenTheis But that builds a base, at least.

@user231094 Sometimes it builds a strong base and sometimes it builds something that you have to remove again before you can continue (like a misconception). But I guess that's why you are asking for a recommendation.

One major complication of intuition is there is no guarantee that your intuition has anything to do with my intuition. If I tell you to think of the electron as a blue, fuzzy sphere with orange sparks flying off of it, you'd think I'm crazy. Well, I just might be, but it kind of works (well, the sparks aren't orange, but you get the picture).

Electrons and photons behave similarly regarding diffraction and interference.

First consider a wave going through a single slit, as the amplitude of the waveform and its diffracted component interfere, creating a high central peak and weaker side-lobes. This follows Fraunhofer's formula, which applies to sinusoidal waves.
When a photon or an electron beam passes through a double-slit, it forms an interference pattern combined with the diffraction pattern, as the peaks and troughs of the waves sum or cancel.

In both the case of the electron and photon, detection relies on the behavior of a particle, as if all its mass/energy is at a point. A single electron can be observed as a scintillation in a phosphor (e.g. cathode ray tube, CRT). A single photon has the energy to activate one particle of a silver halide emulsion.
If the flux of electrons (or photons) is reduced so that only a single "particle" is going through a double slit at a time, the double-slit interference pattern is not disturbed at all! This is one of the mind-boggling challenges of physics -- the electron or photon clearly is detected at only a single point, i.e. as a particle, yet behaves as a wave by interfering with itself. Which slit did the single particle go though? Both at once and the same time!

See this NYU lecture slide on wave-particle duality. This video and this one might help, too.


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Zero-filling Fourier Transform in NMR
I understand zero-filling increases the resolution of an NMR signal in the frequency domain, but I fail to comprehend how. Why do the artifical points appear between the real points in the frequency domain? Is it a byproduct of the fast fourier transform algorithm? Can someone explain this to me? Thanks!


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When boiling water, how is evaporation rate affected by the surface area of the container?
Imagine two pots boiling on a stove. One is tall, while one is wide. Both contain the same volume of water.
What will be the difference in the rate of evaporation between the two containers?

A precise modelling isn't easy at all. For comparable size and shape and the same stove flame you could expect that water evaporates faster (absolute volume reduction) from the wider one.

Boiling water
When you boil water on a stove, the temperature on the bottom of the pot is higher than the boiling point. Steam bubbles are generated, and they help to heat the water at the top of the pot (convection also helps, and conduction of the pot itself). Once you have a rolling boil, the rate is determined by how quickly you can supply more heat (almost all of it now goes into the phase transition). If you use the same burner, and mostly heat the pot (so no tiny pot on a big burner), at that stage the rate of boiling off should be the same.
Evaporating water
When the temperature of the water is below boiling, the mechanism is different. Rather than steam displacing the air above the water, single water molecules break away from the liquid and mix with the air. Evaporation rates depend on the partial pressure of water just above the surface (this influences condensation, i.e. the reverse reaction), the surface area, and the vapor pressure of water in the liquid state. Anything that lowers the partial pressure of water (such as "wind" bringing in dry air), increases the surface area (different pot geometry) or increases the vapor pressure (temperature) will increase net evaporation rates.
Beaker vs Erlenmeyer
Tall vs. wide pot changes the contact area with the stove top and the area of the liquid:gas interface at the same time. If you compare a beaker with an Erlenmeyer (same bottom area), it should show similar boiling rates (maybe a big mess with overboiling for the Erlenmeyer) but very different evaporation rates.

See this answer for an example how phase transitions help in heat transfer.

There are multiple effects which differ in both cases.
Firstly, convection (e.g. "wind") in the air will be much more efficient in the wide container. The tall container will limit the quantity of air flowing just above the surface. This, in turns, means that the water that does evaporate will remain for longer just above the surface of the water, and has more chances of returning to solution.
Secondly, the container itself will act as condenser. As the metal is typically cooler than the water vapour, the metal can cool down the vapour and turning it back into water, which conveniently flows back into the pot. The more exposed surface you have around the boiling water, the more important this effect will be.
You might also consider the power of the stove itself. This is hard to account for, but briefly, there might be some differences in the heat diffusion in the water.

Not clear if water is boiling or just evaporation is considered. You mention metal at room T than stove, etc.

Boiling is fast evaporation. Feel free to edit my answer if you want to be more precise.
Good point about the metal, I edited this.

See my comment. It is really tricky and engineering oriented. Water in a baking tin might even not boil, and in a very extended ideal one can fast evaporate. It is really complicated, at least in my opinion.  Obviously there are parts of truth in your answer, at least listing the parameters, or some of the parameters, involved.


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How can vinyl chloride arise from treating 1,2-dichloroethane with a solution of potassium hydroxide in ethanol?
Liebig apparently created it that way (https://en.wikipedia.org/wiki/Vinyl_chloride#Production). However, googling I can not find chemical equation for such reaction, I only see ethene as result of such elimination reaction:
https://www.quora.com/What-is-the-product-when-1-2-dichloroethane-reacts-with-alcoholic-KOH

The Quora answer is wrong. Vinyl chloride arises from the elimination of one eq of HCl from dichloroethane. Push it harder and you get ethyne.

I get it, thanks. Is the result then Vinil chloride + potassium chloride + water?

Correct. Vinyl chloride, water and KCl


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How to calculate current density in an Electrolysis Process?
I have an anode & cathode, both of $100 \times \pu{100\\mm}$ size. The process requires $\pu{450\\A/m2}$ current density. How do I reverse calculate this? Do I use the total surface area of both Cathode and Anode or just Anode?
Please help.
Thanks

You use the cross-section, just like it were a wire.

Cross-section of both Anode and Cathode or just Anode?

Just one. Like in a wire, you would measure the cross-sectional area once, not cut the wire into many piece and add up all the areas. See https://qph.fs.quoracdn.net/main-qimg-aabadab6ad3379dffe9a3ecb89f466ee.webp

Got it now. Thank you very much! Appreciate your help.

Do I use the total surface area of both Cathode and Anode or just Anode?

You use the cross-section, just like it were a wire.

Cross-section of both Anode and Cathode or just Anode?

Just one. Like in a wire, you would measure the cross-sectional area once, not cut the wire into many piece and add up all the areas.

What about if anode & cathode are not of same dimensions? If I have an Aluminium sheet hardly 0.5mm thick and a Lead anode that is 3 mm thick, which cross section should I refer?

If the electrodes have different areas exposed to the electrolyte, then the electrode with smaller area would have higher current density because the current density at an electrode is simply the current divided by the electrode’s area in contact with the electrolyte.


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How do I view a DSN6 file as text?
I am trying to isolate certain pieces of information within a dsn6 file. Specifically, I want to find the electron density around two atoms within a protein. I want to then run a python code to determine if there is positive electron density between the two of them. In order to do so, I believe I need to convert the dsn6 file, which is stored in a very compressed format, into a format I can understand.
I have searched around found ways to view dsn6 files (such as through PyMOL or COOT in CCP4) as a model. However, I do not know a way to view this decompressed file as text. I have tried, but have been completely unsuccessful in trying to figure out how these programs decompress/read the files.
This is what a piece of a dsn6 file currently looks like:
˘>˝‰‡>™†≠>I14>)B=*À¢<™ÅK=i§=‰≤Ñ=ê8ºO∑ææÑ¸ΩﬁZ
Ω2…=wÇ=œjÖ=ÿm=◊6=â(:ÿ◊ôΩl≥æ=sËΩ,vzºËm
>dg>ÄàG>>hº&>è,>SªÂ=

I hope that the decompressed dsn6 file will look similar to what a pdb file looks like. For example,
ATOM    228  OG1 THR A  58       7.843  45.672  59.760  1.00 15.56   O
ATOM    229  CG2 THR A  58       6.672  44.132  61.223  1.00 14.90   C
ATOM    230  N   PRO A  59       9.292  42.373  62.754  1.00 13.16   N
ATOM    231  CA  PRO A  59       9.409  40.982  63.160  1.00 13.07   C

I asked this question in Stack Overflow, but thought that it might be better suited here. Thank you for the help!

I commented under your SO question: https://stackoverflow.com/questions/56987505/how-do-i-decompress-a-dsn6-file-into-a-readable-format

It looks like your file is saved in a binary file file format. It is possible to decode is, particularly if you have an idea of what is in the file, but this is a challenge. Can you save the file in any other format?

I tried to save it as a .pdb file, but this was unsuccessful. I am unfamiliar with the binary file format. If you want to download one of the files to see, here is a link: https://edmaps.rcsb.org/maps/5x22_2fofc.dsn6

There is a  software called  xtal.js ,  referred to at github.com/uglymol/uglymol/wiki/ccp4-dsn6-mtz , that may do what you need

can you provide some metadata of the measurment that you shared e.g. x start, y start, sampling step,  listed in marcin comment maybe we can create a script for opening the files

There is a program called mapman that can convert between different formats. The .xplor format is a text-based one, the others are all binary, so that's the one you need. The program runs on linux and mac platforms.
Other crystallographic software suites (ccp4, etc) might have similar capabilities, but this one might be the easiest to install (standalone program).

I want to find the electron density around two atoms within a protein.

This is much easier with a molecular browser. You could just set the contour levels to the ones your are interested in, and take a look. If you need to do this programmatically for thousands of structures though, that will not help you.
It will be kind of tricky to find the right spot in the density, though, because they use fractional coordinates (makes generation of symmetry much easier) and your coordinates are in orthogonal coordinates. If this is a problem, just ask another question.
Here is the description of how data is organized in an .xplor map:

Following the map header, the density matrix is then written section-by-section with c moving slowest (in z-sections). Each section of the density map is preceded by a section number.
Thus, for the formatted map file each section of the density map is written using FORTRAN statements of the type
WRITE(OUNIT,'(I8)') KSECT
WRITE(OUNIT,'(6E12.5)') ((SECTON(I,J),I=1,ISECT),J=1,JSECT)
and the resulting map is written with six pixels on each line. The binary format is identical except the format statements are missing, so that each line that is written contains the entire length of map along the `fast' (a-axis) direction.

Indeed, I have thousands of proteins I need to run through. What you've given here is a great boost though, and I will make much use of it. Thank you!

@Ben - If you are programming in python anyway, you could try the Computational Crystallography Toolbox. It calls https://github.com/cctbx/ccp4io/blob/master/libccp4/ccp4/cmap_data.c to read the binary map files.

Here are the docs: https://cci.lbl.gov/cctbx_docs/index.html


text:
Non-chlorine bleach vs. hydrogen peroxide 3% soln from pharmacy: Disinfecting gravity water filter system
I'm seeking to disinfect a Platypus gravity water filter made with hollow fibers.  My main concern is to address any mold / mildew that may have developed inside the hollow fiber filter due to long term storage.
The manufacturer suggests
"To disinfect your Platypus filtration system, fill it with a solution of unscented household bleach and water, using a concentration of 2-5 drops per liter of water. Close and swirl for 10 seconds. Leave filled overnight.
"Filter solution through any filters and/or hoses, then rinse system (including filters and hoses) three times with hot water. Let air dry.
"Note: for an eco-friendlier option, use a chlorine-free, hydrogen peroxide-based bleach alternative.
I called the customer hotline and asked if I could just pour in hydrogren peroxide from the pharmacy, but they said I should just stick to the recommended guidelines.  The agent also reminded me that if I use chlorine bleach, it's best to use a fresh bottle.  Do you think the same is true for hydrogen peroxide bleach?
Could I dilute 2-5 drops of the pharmacy 3% hydrogen peroxide with one L of water?
I looked at the Seventh Generation non-chlorine bleach MSDS, and it lists hydrogen peroxide at 3-5% concentration:
https://www.seventhgeneration.com/sites/default/files/msds/en/sds-chlorine-free-bleach-en-20160925.pdf
Any insights would be welcome.  Thank you!

The manufacturer gives specific instructions, confirmed by customer service. What advantage could you gain by making up your own procedure? Bleach and H2O2 are both good, but not equivalent. The H2O2-based bleach is a powder. But I am not rigid - I might tolerate 6-7 drops of chlorine bleach from a year-old bottle. Chlorine is a great bactericide and cleaner-up. And it is not eco-unfriendly, just aggressive.

@james gaidis actually the questioner was correct - they are now selling good ol' liquid H2O2 (3-5% on SDS) in the laundry care isle of the supermarket.


text:
Is an acid a salt or not?
In our office, we discussed today what the exact definition of a salt is and whether an acid itself can be classified as a salt.
Our first problem was that we couldn't get a definitive definition for what exactly a salt is. Wikipedia states:

In chemistry, a salt is an ionic compound that can be formed by the neutralization reaction of an acid and a base.

Let's take, for example, carbonic acid: it is a molecule comprised of the ions $\ce{H+}$ and $\ce{HCO3-}$. We could say that this molecule forms due to the reaction of a base ($\ce{HCO3-}$) and an acid ($\ce{H_3O+}$):
$$\ce{H3O+} + \ce{HCO3-} \rightarrow \ce{H2CO3} + \ce{H2O}$$
Thus, it seems that the acid fulfills the definition above. We are pretty sure that acids are not considered salts. So where is the error in our reasoning?

They use a different definition of acid.

There's not satisfying  def. of "salt" and I don't see it coming. Abandoning such concept would be somewhat reasonable.

One might argue that "formed by the neutralization reaction of an acid and a base" is a necessary, but not sufficient, condition for classifying a compound as a salt. You can say that a dog is an animal, but it doesn't follow that all animals are dogs.

It is a semantics question with an open ended discussion. If you recall the old problem "How many angels can dance on a pinhead? Medieval problem, this issue here is similar.
Basically in chemistry, all the terminology is controlled or endorsed by the International Union of Pure and Applied Chemistry (IUPAC). They have a very open ended definition of a salt: "A chemical compound consisting of an assembly of cations and anions."
If we follow some really old chemistry texts (>1850s), yes an acid was defined as a salt of hydrogen ion, i.e., an acid is a salt whose electropositive element is hydrogen Definition of an acid.
Did this definition by a few chemists become popular? The simple answer is no. Not in a general sense.
The modern interpretation of an acid, again from IUPAC is, "A molecular entity or chemical species capable of donating a hydron (proton) or capable of forming a covalent bond with an electron pair."
If you have heard of ionic liquids, they consist of a large organic cation and an anion, this is a salt as well.
Then you have so-called acid salts, such as sodium hydrogen sulfate e.g.,  $\ce{NaHSO4}$. Their pH is very acidic and these "salts" will neutralize bases like a typical acid.
Thus one may loosely say, that all acids may be considered as salts but not all salts are acids.

Thanks for the historic perspective, and kind of sad that the high school texts (and wikipedia) can't move on. Oh, I guess I could edit the wikipedia definition... help them move on.

I am very much disappointed by modern general chemistry texts. It turns away brightest students from chemistry. Some radical changes must be adapted.

Wikipedia has the IUPAC definition: https://en.wikipedia.org/wiki/Salt_(chemistry) Let's see if it sticks.

+1 for this answer. IUPAC is the gold standard for definitions. However to me that definition leaves out two important characteristics. First a salt is a solid. Second I would add that a salt should have an appreciable solubility in water. I wouldn't expect $\ce{CuS}$ for example to be classified as a "salt." Yes it has a cation and an anion, but its solubility is next to nothing. If you leave out solubility then every solid inorganic compound is a salt. // Going back to the OP's question, I would consider oxalic acid as being a salt and an acid.

The point I'm trying to make about solubility is that a salt dissolves in water to give cations and anions. No molecular species, or at least none to any extent. So I would not consider phenol a salt. Yes you can remove a hydrogen in a basic enough solution, but the molecule doesn't "completely" dissociate into ions when added to water, so it isn't a salt even though it is a solid under ordinary conditions.

How will you view ionic liquids? They are "organic" low melting point salts and they don't dissolve in water. They are very hydrophobic.

We are pretty sure that acids are not considered salts

Definitions in chemistry don't have to be exclusive (that is, if X is an acid, it cannot be anything else). It's not very useful to call acids "salts of hydrogen", but it's not fundamentally wrong.
The entire DHMO joke is based on giving water unusual but nevertheless technically correct names.

You write:

[..] carbonic acid. It is a compound of the ions $\ce{H+}$ and $\ce{HCO^3-}$

This is the error in your reasoning. There is a covalent bond between the hydrogen and the rest of the molecule. When it reacts as an acid, this bond breaks.
I like definition d) of the online MW:

any of various compounds that result from replacement of part or all of the acid hydrogen of an acid by a metal or a group acting like a metal : an ionic crystalline compound

So starting with the acid $\ce{H2CO3}$ (carbonic acid), you would let it react as an acid (in water, perhaps), and replace the lost hydrogen ion by a sodium ion, giving you $\ce{NaHCO3}$, sodium bicarbonate. If you isolate that as a pure solid, you get a typical salt (ionic crystals, dissolve more or less in water).
If you want to get back to your definition, you would add carbonic acid and sodium hydroxide in water. Carbonic acid would act as an acid, sodium hydroxide would act as a base, and depending on the mixing ration (stoichiometry), you would get sodium carbonate or sodium bicarbonate. Removing water would give you the ionic compounds.

I wouldn't go to a general-purpose dictionary for precise definitions of scientific terms.

II edited the Wikipedia article to have the IUPAC definition, I  can’t edit MW, but I consider their definition mostly harmless. When you’re discussing chemistry “in the office”, those are the types of references you would check first. @DavidRicherby

It's great that you are discussing chemistry in your office! Now you can go back to discussing how to [https://stackoverflow.com/questions/25078285/replacing-a-32-bit-loop-counter-with-64-bit-introduces-crazy-performance-deviati])popcount large arrays)...


text:
Why does the SCF not converge in a structure optimization with GAMESS of the thiourea molecule at the PM6 level of theory?
I'm using GAMESS to run a geometry optimization on a thiourea molecule, using PM6. However, the output file keeps displaying the error message, "NO FORCE FIELD, SCF DOES NOT CONVERGE." Is there an error in my input code?
 $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE MAXIT=100 MULT=1 COORD=UNIQUE
    NOSYM=1 $END
 $BASIS GBASIS=PM6 $END
 $DATA
THIOUREA
C1
C          6  -0.000001   -0.216235   -0.000006
N          7  -1.149850   -0.964256   -0.062205
N          7   1.149844   -0.964254    0.062165
S         16   0.000003    1.474558   -0.000003
C          6  -2.468377   -0.327323    0.016251
C          6   2.468378   -0.327328   -0.016221
H          1  -3.255938   -1.092284   -0.077195
H          1  -2.599975    0.405207   -0.806499
H          1  -2.609097    0.216356    0.970441
H          1   3.255932   -1.092302    0.077168
H          1   2.599974    0.405143    0.806581
H          1   2.609108    0.216419   -0.970372
H          1  -1.133325   -1.963288    0.067490
H          1   1.133319   -1.963297   -0.067428
 $END

In these cases, the first thing to do might be to increase the number of SCF cycles, so try MAXIT=150, and see what happens.

I am unfamiliar with GAMESS, but make sure you have specified the method correctly. Also, check your input geometry visually and consider switching to another method. Maybe the C-S bond is badly described in PM6.

I couldn't find any indication that pm6 is actually available in GAMESS. Maybe try pm3, and see whether it works. I think pm6 would do fine for the molecule, pm3 probably too. If you require pm6 you might have to try a different ESS code.

I spent some time yesterday trying it on WebMo (I have no local GAMESS installations on the OS I am working on right now), and even if I am not familiar with GAMESS, it looks like for some weird reason PM6 doesn't like that molecule (increased optimization and scf cycles, modified convergence criteria). If I recall correctly, anyway, it worked with PM3.

I believe the problem has to lie within GAMESS. I just ran an optimisation with Gaussian 16 B.01 and pm6, which looks like you'd would expect it. ( cc @The_Vinz ) Could you please specify the version you are using. Have you tried other molecules; did they work? Maybe good choices to localise the problem would be urea, CS2/CS, maybe the problem lies there. I'll see if I can get GAMESS to work, then I might try again.

TL;DR: PM6 only works for elements up to fluoride in GAMESS.
I have played around with the GAMESS version installed at our cluster, the ancient 5 DEC 2014 (R1) and I came to the conclusion that there must be a 'bug' in the implementation of PM6 with GAMESS. (I reproduced the problem with 20 APR 2017 (R1); I have also downloaded 30 Jun 2019 (R1), but I won't try it as I think the result is the same.) When I tried it with Gaussian 16 Rev. B.01 on the PM6 level of theory it converged fine.
Actually, I have looked at the documentation, and it seems to be officially not supported (from INTRO.DOC):

Capabilities

   [...]
   3. Calculates semi-empirical MNDO, AM1, or PM3 models
      using RHF, UHF, ROHF, or GVB wavefunctions.
   [...]

Or from INPUT.DOC:

        * * * semiempirical basis sets * * *

GBASIS = MNDO - selects MNDO model Hamiltonian
       = AM1  - selects AM1 model Hamiltonian
       = PM3  - selects PM3 model Hamiltonian
       = RM1  - selects RM1 model Hamiltonian
       = DFTB - selects tight binding Hamiltonian

I am pretty sure the error is associated with sulfur. I have tried a few molecules, among them $\ce{H3C-NH2}$, $\ce{(H3C)2NH}$, $\ce{(H3C)2CO}$ (see input below), urea, which converged fine. Whenever I tried a molecule containing sulfur the SCF seemed to be oscillating. I have tested $\ce{CS2}$, $\ce{SO2}$, $\ce{(H3C)2CS}$, and (obviously) thiourea. (With PM3 everything worked fine.)
My best guess is that they simply didn't finish implementing it. You could issue a support ticket on GitHub: issues. I found an article by (the here known) Jan Jensen from 2013: Project status: PM6 and PCM in GAMESS. And then from 2014: PM6, DFTB, GAMESS, enzyme design, protein-ligand binding, and protein structure validation. I stopped there.
I'd recommend using a different program, the obvious choice would be MOPAC (http://openmopac.net/). Gaussian has it implemented and a couple others probably too. Or you switch to a completely different method like xtb, HF-3c (in ORCA), DFTB (e.g. DFTB+), etc.. Depending on your system size, you can probably also get DFT based calculations done.
Here is also something related: GAMESS crash course?

 $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE MAXIT=100 MULT=1 COORD=UNIQUE
    NOSYM=1 $END
 $BASIS GBASIS=PM6 $END
 $DATA
 C3H6O
C1
CARBON      6.0      0.156332162        -0.059231571         0.417752882
OXYGEN      8.0      1.200569512        -0.535604279         0.831319560
CARBON      6.0     -0.037914597         1.466530485         0.340822109
HYDROGEN    1.0      0.279389785         1.860652999        -0.631337255
HYDROGEN    1.0     -1.089141584         1.742544046         0.481697509
HYDROGEN    1.0      0.545018653         1.982387227         1.112307247
CARBON      6.0     -1.053925162        -0.882111362        -0.061565447
HYDROGEN    1.0     -0.987243509        -1.102851552        -1.132983096
HYDROGEN    1.0     -1.121341433        -1.838731138         0.468944868
HYDROGEN    1.0     -1.993938204        -0.344384630         0.106761952
 $END


text:
I can't find reaction enthalpy ∆H for water formation reaction
$$\ce{1/2 O2(g) + H2(g) → H2O (l)}$$
The tabulated value is $\Delta H = \pu{−285.8 kJ/mol}$
I'm using the bond energies values for each molecule and for some reason, I am getting a different result, not even close to $\pu{-285 kJ/mol}$.
Bond dissociation energies:
$\ce{H-H} = \pu{436 kJ/mol}$
$\ce{O=O} = \pu{499 kJ/mol}$
$\ce{O-H} = \pu{464 kJ/mol}$
My attempt to calculate from bond dissociation energies:
(500.1/2) + 436 → 464.2
250 + 436 → 928
($∆H$ found is positive)

Well, i wanted to see someone doing the correct way, so i could see my mistakes and fix them

Bond formation is exothermic and bond breaking is endothermic.

i know that, i want to know how to get the result using the bond energies.

@BrunoMachado Thanks for showing your attempt. Zhe guessed correctly that you had trouble with plus and minus. I guessed incorrectly that you where concerned about the difference between -285 and -245 kJ/mol. Now we know.

Here is the tally of bonds:

One $\ce{H-H}$ bond broken
One half $\ce{O=O}$ double bond broken
Two $\ce{O-H}$ bonds made

Breaking bonds costs (positive sign), making bonds gains (negative sign). The result will be an estimate of turning reactants to products in the gas phase. It is an estimate because the strength of an $\ce{O-H}$ bond depends on what else is attached to the oxygen. For H-H and O=O, this is no problem because there is only one molecule, respectively, with that kind of bond.
$$\Delta H = \pu{436 kJ/mol} + \pu{249.5 kJ/mol} - \pu{928 kJ/mol} =  \pu{-242.5 kJ/mol}$$
We got the right sign and are not too far off in magnitude.
Why is the physical state important?
In solids and liquids, there are intermolecular interactions that contribute to the enthalpy. This is why the enthalpy of sublimation and boiling is positive - it takes energy to break the intermolecular forces. Here are the values for enthalpy of formation for water in the two states at room temperature:
$$\Delta H_\mathrm{f}(\ce{H2O(l)}) = \pu{-285.8 kJ/mol}$$
$$\Delta H_\mathrm{f}(\ce{H2O(g)}) = \pu{-241.8 kJ/mol}$$
And here is the enthalpy of vaporization
$$\Delta H_\mathrm{vap}(\ce{H2O(g)}) = \pu{43.99 kJ/mol}$$

But ∆H is products - reagents, right?

why you did reagents - products?

Products - reactants is for calculating reaction enthalpies from $\Delta H_f$ values. Bonds broken - bonds made (i.e. bonds dissociation energies for the reactants - bond dissociation energies for the products) is what you need for this kind of exercise.

It is a common confusion. First, sort out what type of problem it is, then make sure the pluses and minuses are correct.

but i still can't understand how water has more energy than the sum of reactants, considering that this is an exorthermic reaction

Let's consider a simpler reaction: two hydrogen atoms forming the dihydrogen molecule. This is an exothermic reaction. the dihydrogen molecule has less energy that the two separate atoms (it just gave away that energy in an exothermic reaction). It would take energy (endothermic reaction) to separate the molecule into atoms again. In general, molecules with stronger bonds have the lower energy, molecules with weak bonds have higher energy.

hypotetically speaking, what would be the values for h and h2, just to exemplificate?

$\Delta H_f$ for H2 is zero by definition. $\Delta H_f$ for the H-atom would be positive, 218 kJ/mol. If you do products - reactants for the enthalpies of formation, you get a negative value - the expected exothermic reaction. Or you could simply say you are making an H-H bond and not breaking anything. With the bond dissociation energy of +436 kJ/mol, you would come to the same conclusion (bond formation: exothermic: negative change in enthalpy).

Here is the logic for the formulas: In one case (using enthalpy of formation) we go from reactants via elements to products. Enthalpy of formation is elements to compounds, so reactants to elements is the wrong direction - change of sign. In the other case (bond dissociation energies) we go from reactants to atoms (or fragments) to products. We are making, not dissociating bonds of the product, so products to atoms is the wrong direction - change of sign.

@KarstenTheis Sorry, I don't think I made it clear that that comment was to Bruno... :( Probably best for me just to delete it at this point...

@Zhe no problem, I thought I should spell out the rationale. I could have put it in the answer, but I didn’t know what the misconception was when I wrote it.

Karsten Theis' answer is a perfect one for your question. He also did his best to explain why is he used the bond-making bond-breaking sign conversion but you are still in mind set of product-reactant explanation. So I decided to explain it in different direction.
Remember the law of conservation of energy, which states that the total energy of an isolated system remains constant? So it applies here as well. As Karsten Theis has explained, for the reaction, $\ce{H2 (g) + 1/2 O2 (g) -> H2O (g)}$:
$$\text{energies of $\ce{H2}$ and $\ce{1/2 O2}$ } = \text{energy of $\ce{H2O}$ + energy absorbed or released} $$
Thus, $$\pu{436 kJ/mol} + \pu{\frac{499}{2} kJ/mol} = \pu{928 kJ/mol} + \Delta H $$
$$\therefore \;  \Delta H = \pu{436 kJ/mol} + \pu{\frac{499}{2} kJ/mol} - \pu{928 kJ/mol} = \pu{−242.5 kJ/mol}$$
Rest of the explanation to get $\Delta H_f (\ce{H2O(l)})$ is exactly same as Karsten Theis'. I hope it is clear to you now.


text:
What do the horizontal lines in a P-V phase diagram mean?
I have been studying phase diagrams of pure substances, and for a substance such as carbon dioxide for example, it's diagram looks something like this:

The horizontal lines in the liquid phase puzzle me. It seems if we shrink the specific volume by a certain amount, the pressure change is much much less. This doesnt seem to be what is observed in reality (liquids have a very high bulk moduli and are generally considered incompressible), which would require a large pressure change for a less change in specific volume. The solid phase has vertical lines, which seems to be true, but the liquid portion baffles me.
Can someone please help me understand what I'm missing here?

the graph is trying to show three things (pressure, density and temperature)  in 2D.

The yellow portion shows the super critical fluid of $\ce{CO2}$.
The blue portion liquid $\ce{CO2}$
The light green portion shows the gas phase for $\ce{CO2}$.
The horizontal lines are called "tie lines." The two phases at the ends of the tie lines are in equilibrium.
If you follow the line for say for 12 $^\circ$C, from the degree notation there is at first one phases, a gas phase.
Now when you reach the dotted curve there are two phases a gas phase and a liquid phase which are joined by a horizontal tie line.
The whole thing is a lot easier to figure out with a more traditional phase diagram.

Found above on page:
https://www.engineeringtoolbox.com/CO2-carbon-dioxide-properties-d_2017.html

Thanks for your answer, looks like I've been reading it wrong the whole time. Btw, which portion would be solid $\ce{CO2}$?

Furthermore, does this mean that the tie-lines are just a diagrammatic representation, and not the actual relation between pressure and specific volume (reducing volume doesnt increase pressure by alot)

No  solid Co2 on your diagram. Look at the lower one. You have to get to about -55 C to get solid CO2. // Quote "The horizontal lines are called "tie lines." The two phases at the ends of the tie lines are in equilibrium."

@PrittBalagopal If I understand correctly, along a tie line, you can increase and decrease specific volume without changing the pressure, because these parts of the diagram represent mixtures of liquid and gas. Increasing the specific volume will cause some of the liquid to evaporate, and decreasing the specific volume will cause some of the gas to condense; in either case, the equilibrium pressure remains the same.

@MaxW Is my understanding in my last comment correct? Do you want to include any of that in your answer?

@TannerSwett Ahh, so basically, as I move along left in the tie line, I'm increasing the percentage of liquid in the mixture, am I correct?

@PrittBalagopal I think so. The left end of the tie line is where all of the gas has condensed, so you're left with only liquid, and the right end is where all of the liquid has evaporated, and you're left with only gas. That's my understanding; hopefully someone else can confirm.

@TannerSwett - You can't change the specific volume. It is what it is. Think of an ideal gas in the green area. P and T fix V. You can't just change V without changing P or T. // The gas and liquid at opposite ends of the tie line are in equilibrium. // Think of water. At 100 C it is all gas. Drop the temperature to 90 C. Now some of the water condenses. You can calculate how much water is a gas from Vapor pressure of water at 90 C and knowing the volume. The rest is liquid water. If you want to change how much water is liquid, then you need to change the temperature.

The horizontal lines represent a combination of liquid and vapor at the same temperature and pressure.  At the left of the horizontal line is the liquid specific volume and on the right side is the saturated vapor specific volume.  Locations between the two ends represent the specific volume of the combination, which is proportional to the amount of each.

You're right. The tie lines join the gas and liquid phases. I corrected my answer based on yours. Thank you for noting the error.

The horizontal lines are tie-lines, as explained in another answer, but between gas and liquid $\ce{CO_2}$. Note the curves labeled with temperatures lie below the critical temperature and above the triple point. The horizontal lines therefore represent regions where gas and liquid $\ce{CO_2}$ coexist. The single point at the cusp of the dotted line is the critical point. At points on a tie-line you can use the lever rule to determine the fraction of liquid and gas from the corresponding distances to the dotted lines on the two ends of the tie-line. The volume at points on the tie line is an average over the two phases. The fact that the pressure (and temperature) are constant on a tie-line is consistent with out understanding of a first order phase transition.
I recommend searching online for another clearer version of the phase diagram.

You're right. The tie lines join the gas and liquid phases. I corrected my answer based on yours. Thank you for noting the error.


text:
Why isn't hydrogen gas an element according to the definition of an element?
I read the definition of element:

Element consists of only one kind of atom and cannot be broken down into a simpler type of matter by either physical or chemical means.

Then as per definition, $\ce{H2}$ is also an element as it is made of up one kind of atom but it can be broken down as in case of $\ce{H2 + Cl2 -> 2HCl}$. Here hydrogen get broken down in order to bond with chlorine.
Hence my question is,
Is $\ce{H2}$ an element or not? If not, why?

Instead of your definition, which can be misleading if you are a beginner, I would use the Wikipedia definition instead. You see that an element is better defined using the atomic number, i.e. the number of protons in the nucleus: we define elements as those entities which have the same number of protons in the nucleus. That's why we can distinguish between hydrogen and oxygen, or between carbon and iron. It is implied that you cannot break down an element into simpler terms, otherwise we would be doing nuclear physics, not chemistry, since we would be dividing it into neutrons, protons and electrons.
In the case of $\ce{H2}$, we see that we have two hydrogen atoms bonded together. Therefore, you see that $\ce{H2}$ can actually be broken down into simpler terms, i.e. into the two $\ce{H}$ atoms, because $\ce{H2}$ is a molecule (see the Wikipedia definitions for atom, and Wikipedia definitions for molecule). I hope this helps!

I'll add that elements have a standard state. So often when we're talking about "gold" (e.g Gold reacts with ...) the implication would be gold in its standard state. // For metals the standard state is typically the solid metal. But a chunk of gold isn't really an atom. The gold atoms are bonded in a 3D matrix.  For hydrogen the standard state is the molecular gas $\ce{H2}$.

@Pier Could we say that $\ce{H2}$ is an allotrope of hydrogen?


text:
Test for unsaturation

Ethene and ethyne can be distinguished by employing the test
a. $\ce{Br2}$ in organic solvent
b. Baeyer's test
c. Phenyl hydrazine
d. Tollen's reagant

According to me it must be both a and b, but there's no such option.

None of these techniques will distinguish between alkene and alkyne. Both will react with a and b, neither will react with c and d. See http://academics.wellesley.edu/Chemistry/chem211lab/Orgo_Lab_Manual/Appendix/ClassificationTests/unsaturation.html

But it's a mcq of a test

And I've to choose one answer as a right one

It's like that Higher the unsaturation greater would be the extent of decolorization in Bromination..So can I take "a" as a right option on the basis of extent of decolorization

Terminal alkynes could react with tollens reagent.The following  link from stack exchange could help you https://chemistry.stackexchange.com/questions/33854/color-of-precipitate-formed-on-reaction-of-acetylene-with-tollens-reagent&ved=2ahUKEwj5p7WF6K7jAhWSeisKHaHhDBYQjjgwA3oECAoQAg&usg=AOvVaw1RWjIx1ZvYmoHKTEqEUJPz

Please choose a or b,just forget c and d

@ChakravarthyKalyan  good spot. You should make this an answer

@MaryamMumtaz  I think you should answer d to this question. See the line just below the equation for the reaction with aldehydes in the Wiki articl:  https://en.wikipedia.org/wiki/Tollens%27_reagent

Okay...but key to this test has marked "a" as the right option to this question,so please give your five minutes to this question and think that why it's "a" and not "b"...there must be some reason which we all are neglecting...So please do think about it.

@MaryamMumtaz Given answers can be wrong. Don't always trust answer keys.

I have two guide books  by two different authors,and both have same answer(option a) to this mcq.But you're also right.

Neither of the authors has adequately researched the question - possibly one copied the other. Don't worry about it, nobody in the real world uses these tests, I had to look up Baeyer's test as it is not something I'm familiar with.

I see two possible answers. Not (A) and (B) but rather (A) and (D).
Chakravarthy Kalyan is responsible for reminding us that (D) can be an answer too (refer to comments).
For option (D):
$\ce{HC ≡ CH + 2AgNO3 + 2NH4OH→ AgC ≡ CAg↓ + 2NH4NO3 + 2H2O}$
The greyish precipitate of silver acetylide will readily form when ethyne is passed through freshly prepared tollen's reagent.
You may read more from this answer by @ron: Does acetylene give a positive Tollens' test? If not, why is this so?.
For option (A):
While $\ce{Br2}$ looks like a seemingly unsuitable candidate, it can actually be used as a crude way to differentiate them.
Take the same volumes of equal strength of bromine water in two vessels and allow equal amounts of ethyne in one and ethane in the other to pass through them (this shouldn't be difficult, just passing equal volumes at STP will do the trick).
Carefully observing which vessel has the more intense color will help you identify which one is ethene, which would consume half the amount of bromine consumed by ethyne.
If you're not convinced, put the final solutions in a cuvette each and measure the absorption or transmittance in a spectrophotometer at 530 nm.
P.S. Make sure that you have a sufficient amount of bromine water, or you'll end up comparing two colorless solutions.
P.P.S. You could also use titration to find which one reaches the end point first, which would be ethene.


text:
How to distinguish mass concentration and density?
I am writing a proof where I have to mention both with density and of a material and mass concentration (mass/volume) of a material.
I am using the letter $\rho$ to represent density, and I didn't want to abuse the same symbol for concentration because it will make the whole proof much harder to follow. I would also prefer to avoid subscripts as the only distinguishable element.
What other symbol or syntax should I use to distinguish density from mass concentration of a material?

Will $c\mathstrut$ do?

@IvanNeretin If that's a accepeted convention, yes... but I actually have other constants named c :P

According to IUPAC's Green Book [1], both $γ$ and $ρ$ can be employed to denote mass concentration. Fragment of a table from the section 2.10 GENERAL CHEMISTRY [1, p. 48]:

$$
\begin{array}{llll}
\hline
\text{Name} & \text{Symbol} & \text{Definition} & \text{SI unit} \\
\hline
\ldots \\
\text{mass concentration, (mass density)} & γ, ρ & γ_\ce{B} = m_\ce{B}/V & \pu{kg m-3} \\
\ldots \\
\hline
\end{array}
$$
[…]  In polymer science the word “concentration” and the symbol $c$ is normally used for mass concentration.

Also, from the SI Brochure, [2, p. 139]:

$$
\begin{array}{lll}
\hline
\text{Derived quantity} & \text{Typical symbol} & \text{Derived unit expressed} \\
  & \text{of quantity} & \text{in terms of base units} \\
\hline
\ldots \\
\text{mass concentration} & γ, ρ & \pu{kg m-3} \\
\ldots \\
\hline
\end{array}
$$

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 9th ed.; Bureau international des poids et mesures: Sèvres, 2019.

Thanks. I never saw it before, but in IUPAC I'll trust and I shall use $\gamma$.


text:
Phase Diagram-Fe-C, and TTT: Hardness Scale
So far I know that the Hardess scale of these materials is this:
Martensite> Bainite> Perlite > Cementite> Ferrite
Where is Austenite along this scale? I know that Cementite is harder than Austenite.

Google, Wikipedia, and your closest copy of a metals handbook would be good starting points...

This is much too general statement to have an answer. "Hard "martensite is softer then cementite . Depending on the heat-treat condition one could rewrite the listing several ways.


text:
Inductive effect of alkyl ammonium groups

Why $\ce{-NR3+}$ has more -I effect than $\ce{-NH3+}?$
And what about order of $\ce{-NHR2+}$ and $\ce{-NH2R+}?$

My attempts
I got idea about electronegativity of carbon atom in $\ce{R}$ is more than hydrogen atom, hence positive charge gets more concentrated on nitrogen atom and it exerts more electronic attraction. But this didn't worked out in latter case when -I effect of $\ce{-NHR2+}$ is less than $\ce{-NH2R+}.$

The inductive effect is both distance-dependent and χ-dependent. N-H bonds are 1.01 Å, yet the χ of H is only 2.1. N-C bonds are 1.42 Å, yet the χ of C is 2.5. At the extremes, where there are only N-C or N-H bonds, the ordering of -I is easy to rationalize.
However, the two, short N-H bonds in -NH2R+ outweigh the higher χ of N-C bonds, giving -NH2R+ a higher -I than -NHR2+.

What that X means dipole moment

@user49564 It's not "X", it's a Greek letter χ ("chi") . Guessing from its numerical values and the fact it's dimensionless, my bet is it's electronegativity. Mark, it would be nice to add clarification to the answer.

Oh yeah, it's electronegativity. Thanks :)

Can anyone tell me why are not we considering the +I effect shown by alkyl group and hydrogen atom to determine the order of -I effect in $NR_3+$ and $NH_3+$?


text:
Current reversal in galvanic cell according to Nernst equation
In a galvanic cell with zinc, zinc sulphate solution and copper, copper sulphate solution, if at equilibrium concentration of zinc sulphate solution increased/copper sulphate solution decreased, will current flow in the opposite direction (from copper rod to zinc rod)?
(The value of cell potential according to Nernst equation will be negative in this case, and to re-establish equilibrium, current should flow from copper rod to zinc rod. )

If I understand you correctly, the answer is yes, at least in ideal circumstances. As a practical matter, it would be more likely with a pair of metals and their salt solutions that were closer in reduction potentials, e.g., tin and lead or perhaps nickel and copper. And you are correct about the Nernst equation being involved.