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Why does an increase in Kinetic Energy increase the likelihood of successful collisions?
As far as I understand bond formation, bonds must be broken between reactant particles to allow reactant particles to form new bonds and create product particles. This involves increasing the potential energy of the reactant particles, such that the attractive forces between say, reactant A and reactant B > the attractive forces between reactant A particles.
My textbook states that reactant particles must have sufficient kinetic energy so the collisions are energetic enough to break reactant bonds. What I don't understand is how the KE of the particles is related to breaking bonds. KE describes the energy needed to shift a body from rest to a given velocity, not the magnitude of a distance between particles (potential energy). So my question is: how is KE related to collisions, potential energy and product formation?

You buy 2000 iPhones and start throwing them against a brick wall at different speed, just out of curiosity. You notice that at low speed most of them are but slightly cracked or visibly unharmed at all. At higher speed (and consequently, higher kinetic energy) the picture is different: they start smashing into pieces. Well, molecules are like that.

So has it got something to do with Newton's third law of motion?

@TomBrooks an intuitive answer: (1) particles are not billiard balls -- they don't "collide" physically, they repel each other electrostatically when they are close. (2) molecules are held together by electrostatic attraction, (3) this electrostatic attraction can be cancelled by electrostatic repulsion of a "colliding" particle nearby, (4) high-speed particles will tend to get really close to our molecule.

Consider you have only two molecules, A and B, enclosed in a box and that the temperature is low. Furthermore, we suppose that A and B will react forming AB, only if (1): A and B collide and if (2): A and B have the right kinetic energy. A and B are moving randomly in the box. If the temperature is low, the kinetic energy of A and B is low. It will then take a long time before A and B will collide. This is because A and B are moving randomly quite short distances per unit time, i.e. the likelihood they will collide is low.
Now, we increase the temperature => the kinetic energy of the molecules will increase.
A and B are now moving at a much higher velocity, i.e. they move much longer distances per unit time. The probability that they will collide has increased per unit time.
A and B will now collide once and a while, but no AB is formed.
We therefore increase the temperature even more. By that, A and B will collide frequently, resulting in the forming of AB. A and B has now the kinetic energy required for colliding within a reasonable time, as well as, the required energy to react with each other.

The kinetic energy of a particle is an alternative way of measuring how fast it moves (the mass of the particle also matters but for a given type of particle of a given mass the only thing that matters is how fast it is moving).
Potential energy has nothing to do with how far particles are apart (unless they are very close) and is pretty irrelevant in this situation.
In a gas or liquid, molecules are moving around a lot and banging into each other a lot. The speeds of the molecules are distributed statistically but their average speed is related to temperature. Whether a reaction happens when they collide depends on whether the molecules hit each other with enough energy to cause a reaction to happen (the orientation of the molecules might also matter but there are usually enough collisions in many different orientations that we don't think about this explicitly other than as part of the rate of the reaction).
In general, more energy in the collisions will lead to more reactions as more molecules will have enough energy to make the reaction happen (rather than just bounce off each other). More energy means more molecules have faster speeds. So, in general, more kinetic energy leads to a higher probability that any given collision will lead to a reaction.

Consider a simple reaction of the following type:
$$\ce{A-A(g)->2A(g)}$$
In a concentrated enough sample, the molecules are colliding all the time.
We can then invoke the equipartition theorem to say that the kinetic energy in the system is equally distributed between all of the vibrational, rotational, and translational modes of the system. Now, if you vibrate the central single bond in $\ce{A2}$ hard enough, the molecule will fragment and the reaction will occur.
This idea also applies to how reactions work in general. Reactants must reach a high energy transition state before transforming into product. Having the energy to deform to this high energy state allows the reaction to proceed, but with enough molecular collisions, higher kinetic energy in the system translates to stronger deformations and the ability to reach the transition state.


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What causes KNO3 to decompose into KNO2 and Oxygen in gunpowder?
I've been studying fireworks and from what I understand:
$\ce{KNO3}$  decomposes to form $\ce{KNO2}$ and $\ce{O2}$ gas, and this free oxygen reacts with carbon to react with carbon and sulfur to form $\ce{CO2}$ and $\ce{SO2}$ gases in exothermic reactions.
What I don't understand is: What causes the $\ce{KNO3}$ to decompose, what is the purpose of using$\ce{KNO3}$  in gunpowder?

Nitrogen's oxidation state in $\ce{KNO3}$ is +5,which is the maximum it can exhibit, so it tends to act as an oxidising agent, by releasing oxygen and getting reduced to a +3 state

KNO3 (Potassium Nitrate) is a molecule that is used with pyrotechnics and is typically mixed with Carbon and Sulfur as fuels.
When Potassium Nitrate is thermally decomposed it forms Potassium Nitrite and Oxygen.
$$\ce{2KNO3 -> 2KNO2 + O2}$$
Potassium Nitrate is known as an oxidizer for this reason (it provides Oxygen to the reaction). This free Oxygen then reacts with the fuel sources in combustion to create an exothermic reaction. Carbon is typically the main fuel source and Sulfur is added to reduce the ignition temperature required to begin combustion of the Carbon-based fuel.
If you had a mixture of Potassium Nitrate and Sucrose (table sugar), you would see something like this:
$$\ce{5C12H22O11 + 48KNO3 -> 24K2CO3 + 36CO2 + 55H2O +24N2}$$
Where you can see that you have a combustion of the two reactants (after the Potassium Nitrate thermally decomposes, it provides the necessary Oxygen to begin combustion) resulting in the products of combustion, Carbon Dioxide and water, along with the byproducts of Nitrogen gas and Potassium Carbonate.

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PVC reacting or self-decomposing?
I have a PVC inflatable device (a custom made inflatable pillow) which is, according to the spec sheet, made out of PVC. The PVC material itself, however, is not the one you'd find on usual swimming rings or such, but something closer to that of the PVC pipes (matte texture, white color).
The problem I encountered is, after some time of storage (deflated of course) in a dark and dry room, some kind of an oily fluid appears on the surface of the device.
Initially I thought that, as it is a pillow, it might have accumulated some of the body sweat over time, and after some time it came out as the material was returning to its original size after being stretched due to inflation. So I thoroughly washed it, but after a couple of days of storage, oily stains can be seen again on it's surface. I didn't kinda think it through, but it might have been helpful to collect the accumulated fluid before I washed it off to run a pH test on it (can't do much other tests at home).
So, may that be a sign of PVC degradation of some sort, and if so, what is that? This occurs whether it's stored in a polyethylene bag or just plain put on a wooden shelf with nothing in contact.
I was looking in various sources but most only specify that PVC degrades either under higher temperatures (above 60ºC) or in contact with quite agressive fluids.
Picture of plain material texture:

Picture of material with stains that appear (reflective):

What you are seeing is most likely PVC plasticiser.
Pure PVC is a hard, inflexible solid. It doesn't make for good flexible objects. To make it flexible, plasticisers are usually added. These plasticisers are often compounds like diisononyl phthalate (DINP), diisodecyl phthalate (DIDP) and di-2-ethylhexyl phthalate. These are, when pure, oily liquids. They often slowly migrate from the PVC object and can do so faster in certain circumstances (eg they will be pulled out when the PVC is in contact with hydrocarbon solvents).
So, most likely, you are seeing small amounts of plasticiser migrate from the PVC and not any degradation of the polymer itself. Eventually, of course, if enough plasticiser leaves the PVC object it will become brittle and inflexible. But it may be quite some time before that happens.

Hm, so I thought it was just my memory tricking me into thinking that it was more flexible a couple years ago. But as it turns out it isn't. Is that plasticiser dangerous?

Also, can I somehow restore the flexibility, probably even by obtaining that kind of plasticiser somewhere? And can I somehow prevent it from escaping? Also, why doesn't this happen to all sorts of PVC insulated flex cables I have lying around, the oldest ones dated mid-80s, probably there was an error in the production process that now makes the plasticiser escape?

@VladislavKorotnev Plasticisers are usually very safe. I don't know whether it is easy to restore PVC plasticiser but I guess it probably isn't. Cable probably uses a different plasticiser so may not show the same behaviour but may be much more expensive due to electrical standards.

does the plasticiser have to be added while the PVC is liquid? If so, that's probably beyond restoration at home...

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Chemical name : Melamine
melamine is a tri-azine with 3 amine groups how would you name it,
as 1,3,5-triamino-2,4,6-triazine
or
2,4,6-triamino-1,3,5-triazine ?
please explain the reason too if possible

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), cyclic parent structures containing one or more heteroatoms with no more than ten ring members are named by using the extended Hantzsch-Widman system. The locant ‘1’ is always is given to a heteroatom. The corresponding rule for cycles with more than one of the same heteroatom reads as follows.

P-22.2.2.1.2 A multiplicity of the same heteroatom is indicated by a multiplying prefix ‘di’, ‘tri’, ‘tetra’, etc., placed before the appropriate ‘a’ term. The final letter ‘a’ of a multiplying prefix is elided before a vowel, e.g., tetrazole, not tetraazole. Lowest possible locants are assigned to heteroatoms, locant ‘1’ being assigned to one of the heteroatoms. Locants are cited at the front of the name, i.e., before the skeletal replacement (‘a’) term and any preceding numerical prefixes.

Therefore, the systematic name of the parent structure in melamine is 1,3,5-triazine.

This numbering is also used in substituted heteromonocyclic compounds. A heteroatom keeps the locant ‘1’; then low locants are given to any substituents.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(a) fixed numbering in chains, rings, or ring systems, i.e., when the numbering of a system is fixed, for example in purine, anthracene, and phenanthrene, this numbering must be used, both in PINs and in general nomenclature;
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)

Therefore, the nitrogen atoms of the heterocycle in melamine have the lowest locant set ‘1,3,5’ according to (a), and then the lowest possible locant set ‘2,4,6’ is given to the amine groups according to (c). Thus, the preferred IUPAC name is 1,3,5-triazine-2,4,6-triamine.


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Why does slaked lime water lose alkalinity over time?
When I throw 3 handfuls of white lime $\ce{Ca(OH)_2}$ into 60 gallons of tap water, the initial pH value is 11.4, but unknowingly it changes to pH 7.
What is the chemistry behind it?
How long does it take to completely change to neutral pH at open space?
I put lime in the water to pasteurize straw for mushrooms, and I need to know how long and how to make neutral pH faster so that I can pour the neutral water in the soil, so it won't ruin the soil.

If you dissolve slaked lime in water it will absorb carbon dioxide from the atmosphere over time.
Carbon dioxide will react with hydroxide ions from the slaked lime forming mainly hydrocarbonate ions. Hydrocarbonate ions will equilibrate with water and at about pH < 8 carbon dioxide will slowly be liberated from the solution. After some time, no significant amounts of hydroxide ions will be left from the slaked lime and the pH of the solution will close into neutral pH.
Well, what will happen to the calcium ions from the slaked lime? You will probably find them as an insoluble deposit at the end. This is because carbonate ions will also be formed during the process, forming calcium carbonate of very low water solubility.

I have never heard the term "hydrocarbonate", though it's clear that it's a reference to what is commonly called bicarbonate, or according to IUPAC "hydrogen carbonate".

You are right. It should read hydrogen carbonate. However, I do not think this a site for linguistic issues. Misprint happens. I´m sorry for that.

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What makes certain zeolites better at ion exchange?
I would have thought that the bigger the pores in the zeolite, the more ions it can contain, and therefore the better the use of the zeolite for ion exchange.
However, zeolite A has small pores but its main use is as an ion exchanger, as opposed to many other zeolites which have larger pores but are not used as ion exchangers.
Therefore, why are some zeolites better than others at ion exchange? Or, what influences the ion exchange capabilities of zeolites beyond pore size?

Zeolites are selective ion exchangers.
What is usually required for a good zeolite is that it selectively absorbs the ion you care about rather than any old ions in the mixture.
That selectivity is a product of the shape of the holes in the 3D structure. Bigger holes will be less selective than smaller holes. If all the ions in the mix fit in the hole the zeolite won't reject the ones you don't want. So, ideally, the holes will be big enough for one ion but too small for another one. This allows selective exchange.
That is about it. It isn't about general equilibration of ions; it is about selective exchange of ions.

Can some other porous substance like MCM-41 (https://en.wikipedia.org/wiki/MCM-41) be used for ion-exchange?

The ion exchange mechanism will not simply fill the inner cavities up to the rim with cations; rather than substitute already present ions by others.
For water softening, if the the free / accessible diameter of the pores of zeolite(s) is at least 3 Å, more relevant properties for ion exchange for a potential application are, for example,

the theoretical number of sites where ions may be exchanged per formula unit (exchange capacitity)
the strength the ions to be released (in exchange to the incoming ones) are bound to the framework
the strength the newly bond ions are bond to the frame (often you want to regenerate / recycle the zeolite)
if there is selectivity in binding ions
the exchange rate between incoming and released ions
resilience of the framework under the parameters of use (temperature, pressure, acidity of the medium; mechanical stress by weight / fluid flux; repeated regeneration)
the absence of toxic / harmful products relased during its deployment
cost in production and waste disposal of this material

The use of synthetic zeolites of different diameters accessible (3, 4, 5 Å, to mention just a few) often aims to benefit from their reactivity as Brønsted / Lewis acid as well as their selectivity due to the geometrical constrain imposed.  Hence a 3 Å-zeolite may be used to trap selectively remaining small quantities of water out of ethanol used as biofuel, for example.


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Does the temperature of liquid ammonia stored at high pressure, room temperature suddenly drop to −33.35 °C when poured into a bowl?
Boiling point of ammonia is $-33.35\ \mathrm{^{\circ}C}$ and its critical temperature $132.4\ \mathrm{^{\circ}C}$ is well above the room temperature. So we can liquify the ammonia gas by compressing it into a bottle, and store it at room temperature for ever.
(Even though the bottle is kept at room temperature, the ammonia will stay in liquid state due to high pressure. )
Hope I'm correct with the above thinking. Next, I open the bottle and pour some liquid ammonia into a bowl. Since this bowl is at 1 atm, I expect something to happen. Does the temperature suddenly drop to $-33.35\ \mathrm{^{\circ}C}$? If so, why?
I'm trying to relate this to adiabatic expansion of gas – the gas does work on the surrounding as it expands and loses its kinetic energy.
But here ammonia is in liquid state as its temperature drops to $-33.35\ \mathrm{^{\circ}C}$. I don't see any work getting done by the ammonia, so I don't see how the temperature drops. I feel I'm missing some important concept.

When the pressure drops the ammonia wants to evaporate. but evaporation costs energy which can only come from the environment (which consists of the bowl and the ammonia). Now do you see what happens?

At 1 atm and room temperature, the ammonia in the bowl is well above its boiling point. So yeah it does want to evaporate. Does the energy required for the evaporation come from the liquid ammonia itself ? Thus, cooling the liquid ! Wow! Thank you (:

Hey just a small question. Can the liquid ammonia cool below its boiling point due to evaporation ? I know that the temperature stays constant during state change. But it seems here the situation is different, I don't see why it cannot go below the boiling point..

The temperature can go pretty cold. The situation is not in equilibrium so the simple idea that the temperature stays the same in a state change doesn't apply.

@Hiiii liquids want to evaporate because the vapor pressure in the surrounding air is less than the saturation vapor pressure, not because  of some relation to boiling temperature. The boiling temperature is just the case where saturation vapor pressure equals the ambient atmospheric pressure.

The key to understanding what happens is that evaporation costs energy.
Changing the state of a liquid to a gas requires the input of energy: this is basic thermodynamics. This is in addition to any effects related to a change in temperature which also costs energy or releases energy.
When liquid ammonia evaporates to gaseous ammonia, energy is required from somewhere. If liquid ammonia at room temperature is poured into a bowl at room temperature it will want to evaporate as it is above its boiling point. But that evaporation requires an input of energy. That energy will come from the environment which consists of the bowl and the rest of the liquid. The only way to provide that energy is to cool the rest of the liquid and the bowl. This happens quickly and the temperature will often fall locally to below the boiling point of ammonia as there isn't enough time to equilibrate the temperature of the liquid with the temperature of the environment surrounding the bowl.
The net effect is that you will end up with some liquid ammonia at some temperature below its boiling point and a lot of gaseous ammonia (don't do this outside a good fume hood).
Forced evaporation of volatile liquids used to be used to freeze water via a similar process. For example, place a small bowl of water in a larger bowl of ether. Blow air onto the ether, which will evaporate, drawing energy from its environment (including the remaining ether and the bowl of water). The water will freeze if you used enough ether. I don't recommend doing this as ether is very flammable, but it was once a common demonstration of this process in chemistry classes.

Correction: the liquid ammonia will boil until it reaches its boiling point and no further. It will not drop below its boiling point. Also, the process will happen essentially instantaneously, i.e. in the valve where the pressure is released.

@LevelRiverSt You are only right if the mixture is at equilibrium. Otherwise it is easy to get colder than the boiling point.

Why do I miss here any consideration of the Joule-Thomson effect here?  With the hint in the original question of a critical temperature of Ammonia of 132 degC, and the rule that the Joule-Thomson inversion temperature of a van der Waals-like Gas of 6.75 times the critical temperature (in K), any sample of ammonia gas initially stored at room temperature and under pressure will cool already by the mere expansion.  If the sample is just large enough, and the system is kept adiabatic...

@Buttonwood It would be interesting to see if that is a significant contributor to cooling in this circumstance. It might not be as the expansion is not happening in an insulated system with no heat transfer.

@Buttonwood sounds like a separate answer there...? Please post it, it sounds interesting

@matt_black I agree it were interesting because the reasoning assumes the sample were already a gas, and not a liquid.  To point to Joule-Thomson effect simply came to my mind as recently I read about ammonia used as a coolant for a bobsleigh run, that would allow to use the track even if the ambient temperature would settle around 20 degC (45 t of ammonia in the circuit and four 250 kW aggregates...)

@Tim Let me reflect on this for some time, prior to shooting this as a new answer.  My reasoning was based on a sample already in the gaseous state; the original question however mentions liquid ammonia (like in the Birch reduction) under pressure.

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Equation for collision frequency of molecules/ions in aqueous solution?
There are equations that describe collision frequency in the gas phase, but I am wondering if this collision theory has been extended to describe the collisions of molecules and/or ions that are dissolved in water. I would imagine this equation would incorporate the aqueous diffusion coefficients of the species in some way and also not have the gas constant (R) in the expression.

I believe this page from the University of Malta goes into at least some of what you are looking for. In particular, it describes how the diffusion coefficients influence the rate of reaction in solution and discuss limiting cases, such as diffusion controlled and activation controlled, for a generic aqueous system.


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How to show that when the equilibrium constant is larger than one, some real-life equilibria are still reactant favoured?
My instructor assigned us with this question:

When real examples of chemical equilibrium are used, the conclusions about reactant- or product-favoured versus the value of $K$ must be modified (e.g. real equilibria can have $K > 1$ but still be reactant-favoured). Discuss this idea, using examples of real chemical equilibria and assumed values of equilibrium concentration.

I was taught that when $K > 1$, it will be product favoured, assuming it’s already at equilibrium ($Q = K$).
How would I go about answering this question? Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

This question can't be answered without a more precise definition of "product-favored" and "reactant-favored."

@Zhe Doesn't "reactant-favored" imply $k_\mathrm{b} > k_\mathrm{f}$? Or are there other definitions?

@GaurangTandon In a case of $\ce{A->B}$, sure, but what about an arbitrarily complex reaction? What if the system is very concentrated versus very dilute?

The terms reactant-favored and product-favored
As discussed in the comments and in YB609's answer, these are not technical terms with precise definitions.
First possible definition
One straight-forward definition would be to call reactant-favored reactions those for which $K$ is smaller than one, and product-favored those for which $K$ is larger than one. You would have to specify the reaction (with phase information), the temperature and the standard state you are using. With this definition, you would be correct to say that such a reaction, if initiated at standard state, would go in the net forward direction to reach equilibrium if $K$ is greater one (and therefor, according to this definition, it is product-favored). Of course, if you take this definition, the question does not make any sense.
Second possible definition You could make a definition based on whether there is "more" reactant or product at equilibrium. This is problematic for multiple reasons. First, you could have a lot of one product, little of another product, and something in between for the reactant. So is there more reactant or more product? Second, if you use amounts of substance (moles) as the criteria for "more" and have a heterogeneous reaction, the relative amounts depend on the volumes (compare oxygen in equilibrium between hemoglobin and the dioxygen in a test tube vs in the entire atmosphere).
Third possible definition If you look at the concentrations to define "more", it still is confusing under some circumstances, even for a homogeneous reaction (everything in the same phase). If the sum of coefficients of reactants is not equal to that of products, diluting the system will disturb an equilibrium, so reactant: product ratios will depend on the dilution of the system.
Answering the question

How would I go about answering this question?

You would start by asking for the definition of reactant-favored and product-favored.

Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

Once the terms are clearly defined, it would be sufficient to look at equilibrium states (and you need to know the definition of the standard state, otherwise the value of $K$ has no meaning).

Actually you should specify correctly what exactly product favoured and reactant favoured mean according to you. Some text use product favoured or reactant favoured term to notate the direction of reaction and some text use product favoured or reactant favoured term to notate the domination of concentration.

If product favoured term is used to notate direction of reaction:
If $K < 1$ that means reactant concentration is greater than product concentration. Now at equilibrium concentration of reactant can be greater than that of product or vice-versa. Say for a reaction $\ce{A -> B}$, we get $K = \frac{\ce{[B]}}{\ce{[A]}}$ and let's say this $K$ is less than 1. Now this ratio will always remain constant. Now if you add some more reactant, then it will get consumed and yield product so as to keep $K = \frac{\ce{[B]}}{\ce{[A]}}$ constant. Now adding reactant made the reaction move towards right i.e product favoured (i.e reaction moves towards the product) but still $K < 1$ is being maintained.

If product favoured is used to notate domination of concentration in the reaction i.e. if by the term product favoured you mean to say at equilibrium concentration dominates product concentration, then according to this notation:
Again taking the example of $\ce{A -> B}$, we get $K = \frac{\ce{[B]}}{\ce{[A]}}$ and let's say this K is less than 1. That means at equilibrium reactant concentration is greater than product concentration. So according to the notation we are using now, we should say it to be reactant favoured. Now $K < 1$ will always be reactant favoured and can never be product favour if you are using the term reactant favoured or product favour to notate domination of concentration in the reaction.

If you are using the terms product favoured and reactant favoured to notate the direction of reaction, then saying K < 1 and saying product being favoured makes sense. According to your professor I think product favoured and reactant favoured mean the direction of reaction. So, it depends how you use those terms.

By only discussing $\ce{A -> B}$ examples, you miss a lot of the subtleties.


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Why are complex basis functions not used in electronic structure calculations?
I've noticed while reading about Hartree-Fock in Modern Quantum Chemistry (Szabo and Ostlund) that most basis sets use only real basis functions. What I'm wondering is why aren't complex basis functions used? I might be mistaken, but from my experience it seems as though integration on the complex plane would have a lot of convenient properties that would make computations easier. Put another way, are complex basis functions not used because of some difficultly in implementation or is it because of difficulty in interpretation of properties like the molecular orbitals?

I have found a recent article that discusses using complex basis functions that might offer some insight, but I currently don't have access to it. Even if there are some cases where it is used, I'm interested in why it wouldn't be the standard approach.

"(...) integration on the complex plane would have a lot of convenient properties that would make computations easier." Could you give us some examples?

@pentavalent I guess I was thinking about it the wrong way in my explanation. It in most cases wouldn't make things easier to compute, but it does open up a wider array of functions to describe an orbital, which might allow fewer basis functions to be used to obtain an accurate description. I'm not certain this would necessarily be useful, but one could also potentially take advantage of path independence of complex integration to solve certain integral. But I admit, its purely speculation on my part whether these properties would be of any use in practice.

Also, I read a bit of that article and found this book which discusses using complex basis sets for a specific type of problem. It seems in both cases that complex wavefunctions are used because they allow one to obtain eigenvalues of an analytically extended Hamiltonian and that these eigenvalues are useful in the context of resonance scattering.

The basic answer is rather simple: In the non-relativistic limit it is just not necessary. There is no complex operator and hence nothing which could change the complex component of your basis functions. Using complex basis functions would only double the required work (i.e., computer time). There are very efficient schemes in order to calculate the standard integrals (McMurchie-Davidson, Obara-Saika, ...). I don't think that the representability of MOs is a reason for using real basis functions, since, strictly speaking, you can interprete only the square of the MOs and not their amplitude.
However, there are situations, where your MOs can be complex. This can happen when you have a complex operator in your equation. An example are the Pauli spin matrices of which $\sigma_y$ has only an imaginary compound. They occur, e.g., the Dirac equation (see here one formulation where they explicitly state the incorporation of the Pauli matrices). Complex MOs can be obtained either by using complex basis functions and real MO coefficients or by using real basis functions with complex coefficients. Starting from a non-relativistic implementation it might be simpler just to keep your basis functions real and to choose the MO coefficients complex.

I haven't had much experience with the particular algorithms used to compute the integrals in these calculations so thank you for introducing them. I'm curious about one point in your answer: why would using complex basis functions "double" the work required? Does it eliminate certain symmetries that allow less integrals to be computed or do these efficient algorithms either not work or work less effectively for complex functions?

@Tyberius : If you multiply two real numbers, you basically have one operation: $AB$. If you instead multiply two complex numbers $(A+iC)(B+iD)$ you end up with $AB - CD + i ( AD +  BC)$. In this case you would increase thenumber of operations by a factor of 6. So, the "double" was maybe not correct. Indeed, for GHF calculations, as they pop up in relativistic QC, one has to drop the spin symmetry. I.e., the orbitals are spinors which contain $\alpha$, as well as $\beta$ components. Hence you loose the blockdiagonal nature of many operators' matrix representation.


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Comparing magnesium ion sensitivity in ribosomes to sensitivity in double-stranded DNA
I was doing some reading on this, and $\ce{Mg^2+}$ has a much greater sensitivity to ribosomes than to dsDNA (double-stranded DNA). It seems the best answer I can think of it that $\ce{Mg^2+}$ provides greater stability and complexation to ribosomes, while it is not as crucial to dsDNA.
Does anyone have a better biochemical explanation why?

cross-posted http://biology.stackexchange.com/questions/57428/magnesium-has-a-greater-sensitivity-to-ribosomes-than-to-dsdna-why


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Optical rotation of mixture

The specific rotation of optically pure adrenaline is $-53$. A mixture of (R)- and (S)-adrenaline was found to have a specific rotation of $-45$. Calculate the % ee of the mixture (enter numeric value only, without % sign, rounded to the nearest whole percent).

This should be $\%~\text{ee}= \text{specific rotation sample}/\text{specific rotation enantiomer}$
But $100\times (-53/{-45})=117.78~\%$
The fraction the other way is $84.9~\%$ but doesn't stick to the formula.

Enantiomeric excess (ee) is defined as:
$$\text{ee} = \frac{\alpha(\text{sample})}{\alpha(\text{pure enantiomer})}\tag{1}$$
Here, we are given the specific rotation of our sample ($-45^\circ$) and that of a pure enantiomer ($-53^\circ$). To calculate the ee, all you need to do is to plug these two values into equation $(1)$ and work out how to turn a fractional number into a value in $\%$.

 The answer is $85~\%~\text{ee}$.

"Optically pure" means pure enantiomer?

@user41987 Yes.


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How does a hydroboration mechanism actually work?
Let us say that you have 1-propene reacting with $\ce{BH3-THF}$ and then later with hydrogen peroxide and sodium hydroxide. I eventually get 1-propanol. The way it was explained to was that after reacting with $\ce{BH3-THF}$  we have $\ce{CH3CH2CH2BH2}$ and after the second set of reagents we get our product.
But, I never understood the mechanism, the explanation seems forced. $\ce{BH2}$  goes to the less hindered carbon which explains why its at the terminal. And then for some reason, $\ce{BH2}$  gets replaced by $\ce{OH}$.

I had to write the complete mechanism for you so that you can understand it better. See the image below. I have explained it in the simplest form. Some of the points I would like to discuss before you look at the mechanism are:

In BH3, Boron is less electronegative than H (i.e. hydrogen) , hence in the syn addition I have shown, hydrogen ends up being attached to a more substituted carbon atom.
For every bonding orbital we have an anti-bonding orbital which lies exactly opposite to the bonding orbital and each bonding-antibonding orbital pair can contain a maximum of 2 electrons in them.

Ok now lets head towards the mechanism which I wrote exclusively for you:

If you have any further doubts in the mechanism then you can ask in the comment section below my answer. I hope it helps.

B(OH)3 or H3BO3 ionises as H+ & [B(OH)4]- in alkaline aqueous media, so the ionization step seems suspect. It is also a very weak monoprotic acid. Now if you changed the final byproduct to 3NaB(OH)4, the reaction would make more sense.

Ya I understand, I even had this as my doubt when I used to study this, but later on in various texts I noticed the above kind of ionisation i.e. Na3BO3 one .

Can you mention the texts in which you found the above? Anyway, even with the standard ionization the overall reaction doesn't change, just the stoichiometry of the last step and the boron byproduct.


text:
How does equilibrium change when pressure is increased?
What happens to the equilibrium when an increase in pressure is applied to a system with the same number of moles of gas on both sides of the reaction, according to Le Chatelier's Principle?
For example:
$$
\ce{2 HBr (g) <--> H2 (g) + Br2 (g)}
$$

According to Le Chatelier's Principle, when there is an increase in pressure then the equilibrium will shift towards the side of the reaction which contains fewer moles of gas and when there is decrease in pressure, the reaction favors the side with more number of moles.
In your question, as both the side of the reaction contains there same number of moles of gas i.e. 2 moles of gas on the left and 2 moles of gas on the right side of the reaction, hence any change in pressure will have no effect on the system.
Therefore, there is no effect of any change in pressure.

There will be no change even if the pressure is increased. Because in both sides, the number of molecules are same.And by Le Chateliers principle, the equilibrium should move in the direction of lower pressure, but here, there is no such possibility as the number of molecules on either side is the same ie $\Delta$n = $0$


text:
Calculating Surface Area of Compound
Is there any way to calculate the surface area of compounds such as 2,5-dimethylhexane or 3-methylpentane? If not, is there a database that has information on surface area of these compounds (not polar surface area). I'm doing an in-depth research between the correlation of surface area and boiling and melting point in branched chain isomers and really need to calculate the surface area of these branched chain isomers.

if it's through a program, please specify if the program is free to use

This article looks promising. I will try to give an explanation once I have access to it.

I found the data at [https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwj255Hwz-PSAhVE6SYKHXLgAnEQFggoMAE&url=http%3A%2F%2Fwww.academia.edu%2F25385484%2FMolecular_surface_calculations_on_organic_compounds&usg=AFQjCNGS9L27LsHmJ1dfpBADGaTlwLrAAA&sig2=8FMJVht6UwAhhq8XcCQG0A.]

Cool, so it looks as though 3-methylpentane is included in the data set. You could presumably find the value for 2,5-dimethylhexane using the formula in their paper in reverse by experimentally measuring solubility. If this doesn't work this book has more info on MSDOT.


text:
A non-life process making glucose from carbon dioxide and water using electricity or light?
update (Sept. 2018): Here is a bit of news that someone might be inclined to cite in a "No, not yet, but..." answer.

NASA has announced a CO2 conversion challenge!

Get out of your element! Convert CO2 into molecules to power bio-manufacturing in space.
Convert CO2 to Glucose:
Help us discover ways to develop novel synthesis technologies that use carbon dioxide (CO2) as the sole carbon source to generate molecules that can be used to manufacture a variety of products, including "substrates" for use in microbial bioreactors.
Because CO2 is readily abundant within the Martian atmosphere, such technologies will translate into in-situ manufacturing of products to enable humans to live and thrive on the planet, and also be implemented on Earth by using both waste and atmospheric CO2 as a resource.

It looks like this will be a multi-year project.

original question (Mar. 2017):
A large mass fraction of foodstuff a person would need on a long trip (for example, in space or otherwise not on Earth) would come from its caloric (energy) value. Scenarios have been explored for decades to use solar photovoltaic (or other electric) power to generate light to grow plants locally for food. Light provides energy for photosynthesis which combines $\ce{CO2}$ and $\ce{H2O}$ and makes glucose, (later converted to and stored as starch) which then provides caloric value to food.
$$\ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)}$$
I'd like to know if a purely non-life based self-contained chemical process has been developed or is at least close to continuously converting $\ce{CO2}$ and $\ce{H2O}$ to glucose, using electricity directly, or electrically generated light.
By "non-life based" I mean it's not a bioreactor based on living cells. It could certainly use processes that are similar to those used in organisms.
The process would have to be potentially able to run stand-alone with minimum consumables, in the same sense of the term "process" as is mean for the Haber Process.
I am not asking if you think it could be done, but if there is a way that it could be done, based on well-documented research towards implementing complete Artificial Photosynthesis using light or electricity, not just a demonstration where a laboratory technician regularly adds separately prepared reagents and precursors for each step.

related http://chemistry.stackexchange.com/questions/32059/what-are-the-steps-to-creating-synthetic-glucose

also http://chemistry.stackexchange.com/questions/696/energy-efficiency-aside-what-are-the-chemical-constraints-on-co%e2%82%82-capture-and-me

@Mithoron No, those are not really related at all. I've specified starting with $\ce{CO2}$ and $\ce{H2O}$, not cellulose, and ending up with glucose, not methane.

I am not certain, but there may be something of use in the research discussed in Phys.org's New dual-atom catalyst shows promise to yield clean energy by artificial photosynthesis which refers to the recent paper in PNAS Stable iridium dinuclear heterogeneous catalysts supported on metal-oxide substrate for solar water oxidation. I've reduced the constraints of the original question a bit, because I think the answer would have been clearly no.

It is clearly possible that some chemical process can do this (after all plants do it) but the complexity of the processes used in photosynthesis suggests it is very hard to do it. Possibly alternative routes might be easier but using elements not commonly available to plants (heavier, rare metal catalysts, for example).

If I remember the figure quoted to me by Dan Nocera, plants rebuild photosystem II roughly once every 20 minutes. If I remembered that right, it gives a sense for how messy this process is and therefore, complex from a chemical engineering perspective.

possibly helpful; Phys.org's Water-splitting module a source of perpetual energy

I was thinking of this too!  I think this is a good candidate for both X Prizes - making CO2 extraction valuable, and removing CO2 from the atmosphere!

@Jonathan I don't have any real money to contribute towards an X Prize, but I'll add a nice bounty to an answer that solves these problems :-)

I was talking about the X Prizes that Elon Musk is putting out there, 100 million!

Popular research these days for CO2 fixation involves the following enzymes with associated conversion rate:

RuBisCO - 10 molecules/sec
ECR(Enoxyl-CoA carboxylase/reductase) - 80
molecules/sec

The major noncarbohydrate precursors are lactate, amino acids, and glycerol.
Here are some links to current research:

TiO2-catalyzed synthesis of sugars from formaldehyde
CO2 reduction -> Hydrocarbon eg. Formaldehydes/acids (CH2O).
Then Formose/Butlerov reaction (catalyst:base+Ca/Mg) -> sugar
Volume 1 of Comprehensive Glycoscience
CO2 to Glucose Whitepaper

Some topics for research:

Gluconeogenesis
Formose reaction
Calvin cycle (absence of light)

Thanks! This is the type of answer that I was hoping for. The whole synthetic process (CO2 to sugar via electricity or light) hasn't been developed, but you have highlighted work on several key steps along the way.

Turns out I was born a bit late and most of the ice sheet has melted; am working on a similar project  with a different mission than providing sustenance on space missions


text:
Home science optical absorption test for approximate caffeine quantity in coffee?
I started with How can you measure the caffeine content of a liquid at home? in the cooking stackexchange, but the test strips were only qualitative (yes/no) and don't seem to exist anymore, and the part about rubbing dichloromethane on ones gums frightened me off.
The 2014 NY Daily News article Scientists working on at-home caffeine detection test links to Validation of Caffeine Dehydrogenase from Pseudomonas sp. Strain CBB1 as a Suitable Enzyme for a Rapid Caffeine Detection and Potential Diagnostic Test (Mohanty, S. K., Yu, C. L., Gopishetty, S. and Subramanian, M., J. Agric. Food Chem., 2014, 62 (31), pp 7939–7946) but that's not for home use. I found this nice experiment but that is still serious analytical chemistry.
However the article does mention the use of UV spectrophotometry, and that's something I could imagine playing with at home personally, using UV LEDs or other light sources, a Raspberry Pi camera or some photodiodes, a piece of diffraction grating film, basic physics, duct tape, etc. I'm OK with the optics and electronics, it's the chemistry I need the most help with.
The 1994 paper A Simple Method for Determination of Caffein Content in Tea Samples shows a correlation of caffeine assay from various types of brewed tea with absorption at 440 nm. Strong coffee would contain perhaps 5X more caffein, but being coffee, would probably be 100X or more higher in absorption in blue or UV.
This and this discuss UV absorption of caffeine in a solvent, but I am not sure if there are any solvents that I can find or use at home. Further, while I might be able to find a 275nm LED, it probably limits me to glass optics and expensive detectors, most plastics probably absorb too much there, and most silicon detectors have thin inactive layers that absorb this wavelength of UV; backside-thinned detectors are usually necessary. However, possibly the longer wavelength tail in the direction of 300nm might be useful.
My Chemistry Question: Are there any ways to separate the caffeine in coffee from the organics that make it so dark, using "household chemistry", so that I could try to check  blue or UV optical absorption?
I'm only looking for a way to tell if brewed coffee has a big or a small caffeine "kick" to it, so even something as bad as +/-20% repeatability might do in a pinch. I don't need an absolute measurement - just something that roughly tracks the caffeine content.
Possibly helpful reddit, What wavelength should I measure the absorbance of my coffee to get an idea of how strong it is?, which links to an article I can't read right now, as I am drinking coffee on the wrong side of the paywall: Measurement of caffeine in coffee beans with UV/vis spectrometer.

above: An old Certificate of Analysis from a Caffeine Standards Kit, from Agilent.

above x2: Caffeine absorbance in water, from here (top) and here (bottom).

A thoughtful discussion of the analytical chemistry of, as well as the sociology and effects of caffeine at Periodic Videos. (Ethyl Acetate used in this example.)

Home science optical absorption test for approximate caffeine quantity in coffee?

Yes..., except that for "...in coffee?" it's No.
AAI Solutions' Measuring Caffeine Concentration; Applied Analytics Application Note No. AN-019 front artwork certainly looks hopeful...

(click for larger)
After all an Optical Multichannel Analyzer (OMA) in this case, to a home scientist, looks like some light source, a grating, three photodiodes and some tedious calibration, this is really for extracted caffeine in supercritical CO2.
However, from Background in

As different literatures indicated spectrophotometric determination of caffeine is also reported as preferred method of determination such as UV–Vis spectrophotometry because of its relatively low cost, rapidity, high accuracy and reproducibility. But UV–Vis spectrophotometric method cannot be used directly for determination of caffeine in coffee beans extracted with water owing to the matrix effect of UV–Vis absorbing substances in the sample matrix [9]. In aqueous solution of coffee beans it was observed that there is spectral interference from caffeine and chlorogenic acid in the wavelength regions of 200–500 nm. Yet this method requires the extraction of caffeine from the aqueous solution of coffee beans using dichloromethane for the spectroscopic determination. This is necessary since the caffeine spectrum is overlapped with other compounds found in coffee. Hence, the use of dichloromethane limits the wider application of UV–Vis method.
[9] Genetic diversity and correlation of bean caffeine content with cup quality and green bean physical characteristics in coffee (Coffea arabica L.)

In simple language: Coffee is full of all kinds of molecules besides caffein, so there will likely be many broad spectral features in the UV interfering with any simple way to measure one particular molecule's concentration.
You're going to have to resort to some kind of specific extraction. If not supercritical CO2 mentioned above, then something like this:
SPECTROPHOTOMETERS: How can I measure the caffeine content in drinks? with 16 items of apparatus, two reagents including somewhat dangerous to "home scientists" Dichloromethane
or like this: Jenway's 73 SERIES SPECTROPHOTOMETER Application note: A09-010A; The quantitative determination of caffeine in beverages and soft drinks using UV wavelength spectroscopy again with the dichloromethane!
Bottom line
Try not to drink so much coffee that you need to worry about all of this. Listen to "Sir Professor" Martyn Poliakoff's and switch to tea. (note, I'm also the OP)

Dichloromethane is relatively nontoxic. In an Agilent HPLC application note they don't bother with any preextraction (although they add MgO prior to filtration of the sample and injection), and running phase is relatively benign water/methanol. But it's HPLC, and methanol. You might be able to run gravity column chromatography with ethanol instead, or other available organic solvent.

I believe the MgO is for metal removal which might poison the column?

@BuckThorn if you think it can fit into a "Home science" setting, please feel free to add an answer!

Depends what kind of resources you have, I suppose. I'm still looking for a "home science" guide that would explain how to do the various steps (filtration, purification, separation, detection). If I do maybe an answer is in order.


text:
What is the most polar covalent bond?
Which polar covalent bond of the following:
Cl-F
S-O
P-N
C-Cl
It's trivia - I guess. And I got it wrong.
Electronegativity list was given for the following:
F, O, Cl, N, S, C, H, P
My wrong reasoning was:
why would two highly electronegative atoms bond with each other. Even they did, given their high electronegativity as both pull the electron cloud to each side (tug of war, weak dipole moments) with much strength by forming a weak bond. I refused to consider the difference between electronegativity of the species. (Even I did, S-O would be the highest).
I applied the same to S-O and P-N as well.
C-Cl was the pick and bad!
C-Cl has a acceptable dipole moment given C's less en 2.55 and Cl's high en. 3.16. The size of atoms wise, Cl-F is shorter compared to C-Cl. Which is questionable.
I checked few other related questions and answers, such as this peculiar one and this.
As I read ron's answer (the latter), the concept is not about a compound like CCl4 which negates the polarity between C-Cl bonds in the tetrahedral geometry. Thus C-Cl is a polar covalent bond. (isn't it?)
Where did I go wrong in reasoning, what did I miss?
PS: Not homework but some theory revision questions. Trying to reinstate/expand the understanding of concepts.

This is really difficult to answer. (1) There are multiple tables of electronegativity values. So which were you given? (2) Off the top of my head I'd assume that since you were given a table of electronegativity values that the pair with the greatest difference would be the most polar. (3) you say that it isn't S-O or C-Cl. So what was the right answer? What are we trying to rationalize for you?

Cl-F was the answer. I have no clue how it came about. Care to discuss and shoot an answer?

@MaxW revisiting this question again, and looking at your comment, why do we have multiple tables of electronegativity values for the same bonds from single periodic table?

There are various electronegativity scales.
Electronegativity vales from Wikipedia
$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Element} & \text{Pauling} & \text{Allen}& \\
\hline
\text{H}  & 2.20  & 2.30 \\
\text{C}  & 2.55  & 2.544 \\
\text{N}  & 3.04  & 3.066 \\
\text{O}  & 3.44  & 3.610 \\
\text{F}  & 3.98  & 4.193 \\
\text{P}  & 2.19 & 2.253 \\
\text{S}  & 2.58 & 2.589\\
\text{Cl} & 3.16 & 2.896 \\
\hline
\end{array}
$$
The compounds are all binaries, so the polarity, $\mu$, is equal to the difference in charge, $\delta$, times the distance, $d$, between ions ($\mu = \delta \cdot d$). Only given a table of electronegativities, the first level of analysis would be to assume that all the bond lengths were equal and that that the polarity would simply be a function of the difference in electronegativity. The differences are thus:
$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Element} & \text{Pauling} & \text{Allen}& \\
\hline
\Delta\text{Cl-F}  & 0.82  & 1.297 \\
\Delta\text{S-O}  & 0.86  & 1.021 \\
\Delta\text{P-N}  & 0.85  & 0.813 \\
\Delta\text{C-Cl}  & 0.61  & 0.352 \\
\hline
\end{array}
$$
So on the Pauling scale S-O should be the most polar, but on the Allen scale Cl-F should be the most polar.

Are you implying the diff between electronegavity of the bond was the only reason? That feels too naive to conclude... because the values given were similar to Pauling scale and therefore answer can not be Cl-F. That's where my doubts began and to post this question.

I'm not saying that the difference in electronegativity is the only reason. The dipole moment is given by the different in charge times the bond length. However only given the tables of electronegativities, I'd assume the bond lengths for the four compounds were equal and choose an answer based on the electronegativity difference. Don't really know how else you're supposed to analyze the problem. // I edited the problem to reflect the reasoning better...

I need sometime to reply to you. I will have a bit more sub questions to discuss.


text:
Are wrinkles\folds in paper reversible by heat, pressure or something else?
I ran into this question on Lifehacks.SE, regarding how to remove wrinkles from a crumpled piece of paper. Suggestions included ironing the paper or placing it under some heavy books for a few days.
I am wondering about the chemistry of these processes and whether they are actually effective. This Reddit thread gives some general background about the molecular structure of paper and what happens when you fold paper with enough force to permanently create folds, i.e. to break bonds (as opposed to rolling it within the "elastic boundaries", as it says there). See also this excellent infographic from Compoundchem.
So, can ironing\applying pressure really straighten out a folded paper back to factory level? If not, is there any other way to unwrinkle a paper?
I'd appreciate explanations in the molecular level.

So, did you try by yourself if wrinkles in paper may be ironed out?  By the way, modern paper for newspapers is chemically different from the cotton paper used for banknotes; so the latter could behave similar to a shirt / your hair and their intermolecular H and S bonds.

Once you fold a paper or crush a paper, then it was found experimentally that the internal geometry of paper developed ridges throughout inside it and these ridges inside the folded or crushed paper holds the energy you imparted by folding the paper. This energy imparted by you by folding the paper gets stored in the ridges of the paper in form of bending energy (and this bending energy is also the reason that a crumpled paper ball cannot be further compressed beyond around 90% air).
The crumpling of a piece of paper leaves localised permanent marks. This shows that, by some mechanism, the crumpling has focused the stress at some points, exceeding there the limit yield of the material and leading thus to irreversible deformations.
Indeed, a piece of crumpled paper does not usually return to its planar shape even if you iron it or apply pressure to it because along the folds, the material is often strained beyond the level of reversible elasticity.


text:
Free radical bromonation of 1-ethyl-4-nitrobenzene

In the above reaction , I am confused between . Among them which is major product and why .

The key is the relative stability of each alkyl radical formed as an intermediate in the reaction. This is explained rather well in Bruice's Organic Chemistry, 7th edition, section 13.4, pp. 561-563.

@AdityaSher but there will be more ellect of NO2 (-I , -M)

Oops sorry I just missed that! Let me work on it

This is would be a free radical bromination reaction. Bromine is very selective. It is going to replace the hydrogen from the highest substituted carbon. So according to me second one should be the answer i.e. Br attached to secondary carbon.
Hope it helps

but there will be more effect of NO2 (-I , -M)


text:
Help me understand electrochemistry and batteries
I have a brief understanding of how batteries work, but there are still some things that I can't put together. Therefore I would like to ask for help in understanding how batteries work. Every time I try to learn about batteries, I end up seeing the Zinc and Copper cell, but I can't connect with any of these materials. Therefore I would like to ask what would happen if instead of Zinc and Copper, one would make a battery made out of... Steel and Aluminum, for example. Or any materials that are randomly or wrongly chosen.

I have chosen a random example because I think I can understand better if a system doesn't work rather than having a system that works. So now the questions:

Would such a system work if the electrolyte would be a solution that would allow the transfer of ions?
If not, why would this not work?
What can one do/change to make such a system work?
What would happen at an atomic level with both electrodes?

By work, I mean store energy. Numbers such as electrode potential and cell potentials would also be useful.

You can't "connect" with Zn/Cu? You'll need to be more clear, I don't understand what you mean. (You seem to be implying that you understand Al and steel in a more comprehensive way than zinc or copper, but that doesn't seem plausible to me.)
The zinc half cell reaction is Zn°(s) + H2O + H+ → Zn++(aq) + 2e- + H2O + H+ (pH << 7 ) or eliminating the "by-standers" Zn°(s)→ Zn++(aq) + 2e-. The chemical reaction at the copper electrode (and it would be the same reaction for many electrically conductive and inert electrodes, such as Pt (platinum), C (graphite), Au (gold), Aq (silver), and many more) is 2H+(aq) + 2e- → H2↑ (pH << 7) where I've 'canceled' the by-standers again. There are two chemical reactions going on, one at each electrode. The simple analogy is two cups with a tube between them. If the cups and the tube are all filled with water, then water will flow from one to the other if one is higher (and of course one lower) than the other. The water flows because the gravitational potential energy is greater for the water in the higher cup. Well, energy flows if the half-cell reaction (the reaction on one electrode) is more energetic than the other. Please refer to http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm or http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html or https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page) for tables of just a few of the possible redox reactions (note that most of the half-reactions listed are inorganic, with a couple of exceptions. There is an enormous number of organic redox reactions which aren't generally tabulated.) The hyperphysics page links to a further discussion. Aluminum is always (at least, on this planet's surface) covered by a insulating layer of Al2O3. I don't know what would happen if you connected a piece of Al to a piece of "steel". First because I don't know what your electrolyte is, second because commercially available aluminum is, just like steel, an alloy of several elements. In other words, neither is a pure (or nearly pure) element, unlike the copper or zinc you can get. But, there's no reason why you couldn't substitute a piece of stainless steel for the Cu electrode, or even carefully open up a zinc-carbon battery's carbon electrode (if you can find one).=============== The important facts are that each half-cell reaction has a "potential" and the larger the difference is between the two half-reactions, the greater the current flow and voltage will be between electrodes. Since many of the half-cell reactions won't work in the same electrolyte, salt bridges (see Wikipedia) can be used to connect the fluids. (BTW, there's another reason why the use of steel can be a problem, there can be various reactions on the surface of the steel (corrosion) in which one part of the surface acts as an anode, and another part acts like a cathode. If an electrode is "too reactive" in a particular electrolyte, these side reactions can decrease the amount of electricity flowing to the other electrode. Cu/Zn is popular because all you need is a copper wire, a lemon (or even some vinegar (and salt)), and something that's either made of zinc, or has been galvanized (like a galvanized nail, which is coated with a layer of (mostly) zinc). While it doesn't seem like the hydrogen half-reaction involves the copper, actually the voltage (potential) of that reaction depends on the nature of the surface of the metal, sanding the surface until it is 'bright' will increase the amount of H2 produced. So, to sum it up, first you need two different oxidation/reduction half-cell reaction components, then you need to connect them in an electrolytic system so that the circuit is complete. An important fact, but sometimes overlooked in beginning discussions of half-reactions, is that the reaction should be considered reversible with the arrow → reversed ← as well as changing the sign of the potential (multiplying it by -1, so if it is negative it becomes positive, and vice versa). Also note that many tables assume an aqueous environment, and often an acidic or basic pH. The tendency of any "randomly chosen" electrodes to work, depends on the system they create (including your choice of electrolyte and pH). In a sense, half-reactions are a clever way of separating chemical reactions, like putting a paddle wheel into a stream of flowing water, to use some of the energy being release by the chemical reaction. If you didn't keep the two reactions separate, all that would happen is the chemical reactions would generate heat instead of electricity (and heat). In other words, putting zinc metal into acid will generate hydrogen gas (as long as a surface, maybe even the zinc) exists for the gas to form on), and won't generate any electricity (because the electrons aren't being forced through the circuit). This answer is a little disorganized, sorry.

Please consider reformatting your answer.  To render this more accessible and legible to the general audience, structure your contribution by paragraphs.  The chemical equations deployed here are better set not in-line, too.  As a side-effect, your answer may be a shorter one, too.


text:
how to understand to what subshell is related basis function in the Eigenvectors block (GAMESS output)?
So, each subshell of some atom is described by one or combination of few BF's. But I can't find correlation between representation of subshells in eigenvectors block and, say, basis set or something else.
How it looks:
eigenvectors block (tiny part of it), actual DFT calculation of BF2H:
                           1           2           3    ---> MO number
   atom   shell type -24.7137    -24.7137     -6.8636   ---> MO energy (a.u.)
BFs | atom num |          B2          A1          A1    ---> orbital symmetry
1   B    1     S           0      1e-006    0.496374    --->
2   B    1     S           0   -1.6e-005    0.572289    --->
3   B    1     S           0   -4.3e-005    0.097616    --->    eigen
4   B    1     S           0   -0.003145   -0.089222    --->   vectors
5   B    1     S           0    4.6e-005    0.003663    --->
6   B    1     X           0           0           0    --->

At MO 3 we can see 1s subshell of B distributed mostly on BF 1 and 2. But for another atom, shell (2s, 3s, 4s etc.), subshell type (px, py, pz, dxy etc.) the distribution is very different. I can understand there what is, but I nedd to look by my own eyes on the numbers, but for systems with 30-50 and more atoms it's totally unacceptable.
That's why I use the power of C++ to parse megabytes of eigenvectors (I'm very new to programming (5 or 6 days) and I write very silly code, but... ).
So, I'll be happy if somebody help to understand on that numbers in the output to "look" to correctly associate basis funtions with shell number (1, 2, 3...).


text:
Sigma Complex vs Resonance Structure
What is the difference between a sigma complex state of a Bezene and the resonance structure?

When a positive charge is conjugated in the ring,it is called sigma complex.
The resonance structure comprises of no such excess charge, only pi electron clouds

Please check out how I edited the MathJax of this post. Please do not add \ce{...} into every mathematical expression like a robot.

I'm sorry, will take care from next time


text:
Atomic mass unit based on other atoms
Question:

Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?

My attempt:
Current standard is $1/12$th the mass of a C-12 atom. New standard is $1/9$th the mass of a Be-9 atom.
Now I understand that the only the basic unit of measurement of masses is changing. The absolute masses of the different atoms in the universe remains the same. That means the absolute mass of an He-atom in the old system and in the new system must be the same.
Now, absolute mass (of an atom in any system) = no. of "amu"s of that atom * mass of 1 amu (in that system)
We are supposed to determine the relation between the mass of 1 amu in the old and in the new system. Obviously, the one having more "amu"s would have a smaller mass of 1 amu.
And now I am stuck. I can't proceed further than this much explanation. I've looked on Wikipedia and some other questions but can't find similarities. Hints are appreciated!
UPDATE: although not explicitly mentioned, the book does NOT expect readers to look up data tables to find the answer. It expects a possible reason as to why is the Be-9/C-12 scale larger.

It might be worth looking into the original definition of amu, which was based on the mass of $\ce{O^{16}}$ rather than carbon. http://chemistry.oregonstate.edu/courses/ch121-3s/ch121/Answers%20to%20interesting%20questions/history_of_the_atomic_mass_unit.htm

Not oxygen-16.  The earlier standard was developed before it was realized there were isotopes, so just plain oxygen (which, we know now, is mostly oxygen-16 in nature) was 16.

@OscarLanzi Physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. Anyway, the amu became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$.

Question:

Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?

Answer
It is easy enough to look up the weights of $\ce{^{12}C}$ (12.00000000...) and $\ce{^9Be}$ (9.0121822) based on $\ce{^{12}C}$ scale and make a decision from that data. Thus a mass scale based on $\ce{^9Be} =9.00...$ would be heavier than an mass scale based on $\ce{^{12}C} = 12.00...$.
"Justification"
The OP has asserted that the problem is not supposed to be just a "lookup problem" but that some reasoning should be offered.
I content that the question simply asks which is larger. It doesn't ask why. Furthermore I'd assert that there isn't any "simple" reasoning by which the problem can be analyzed other than lookup.
Neutron to proton ratio reasoning
In his answer QuarkyLittleThing's reasoned that a neutron is heavier than a proton and since $\ce{^9Be}$ has a higher neutron to proton ratio (5:4) than $\ce{^{12}C}$ (6:6), it ought to be heavier.
Using the neutron to proton ratio reasoning, $\ce{^9B}$ only has 4 neutrons but 5 protons, so a $\ce{^9B}$ scale should be lighter than the $\ce{^{12}C}$ scale. But the mass of $\ce{^9B}$ is 9.0133288 u, so the neuron to proton ratio isn't entirely sound, and a mass scale based on $\ce{^9B} =9.00...$ would in fact be heavier than a mass scale based on $\ce{^{12}C} = 12.00...$.
Lower nuclear binding energy reasoning
Mithoron's made a comment on QuarkyLittleThing's answer asserting that the reason was "lower nuclear binding energy."
To properly analyze Mithoron's comment one must note that the nuclear binding energy is based on the energy differences of the actual fission/fusion of nuclei. Thus as NicolauSakerNeto pointed out on an earlier version of this answer, this is really not a "reason", but a tautology due to $\text{E}=mc^2$. So a decrease in the mass of the nucleus is proportional to an increase in the nuclear binding energy for that nucleus.

The current system, takes 1 amu as $1/12th$ that of a $\ce{C^12 atom}$.
The $\ce{C^12 atom}$ consists of 6 protons and 6 neutrons, so the average mass would be the average of a proton and a neutron.
Now if a $\ce{Be^9 atom}$ is considered, it has 4 protons and 5 neutrons, making the average tilted toward the neutron side, therby giving the value of 1 amu as slightly more than that given by the $\ce{C^12 atom}$
Hence the one based on the $\ce{Be^9 atom}$ is larger.
Hope this helps

It would be bigger, but because of lower nuclear binding energy

It is easy enough to look up the weights of C-12 (12.00000000...) and Be-9 (9.0121822) based on C-12 scale and make a decision from that data. Trying to rationalize why is apt to get you into trouble.


text:
Role of Hg2+ as catalyst
in the reaction of addition of hydrogen halides to alkynes $\ce{Hg^2+}$ can be used as a catalyst. But what is its role? How does it increase the rate of reaction.
I think that the vacant orbitals in Hg would help in the process of ionisation of hydrogen halides. Is this concept correct? please help

The main use of $\ce{Hg}$$^{\text{2+}}$ is to polarize the electron cloud of the alkyne.
Else the attacking specie wont be able to approach the bond

You know you can comment under questions? If you wanna say one sentence from the top of your head, comment is a way to go.

Im sorry, but i aint got a commentable amount of reputation, like you. Seemingly want to maximize earnings dude @Mithoron

@QuarkyLittleThing you mean that cyclic intermediate that $\ce{Hg^2+}$ forms with triple bond

Then make better answers. Also it's not like rep is a big deal.


text:
Why pyranosides do not mutarotate?

For mutarotation, at least one hemiacetal group must be present in the
  sugar. Therefore, non-reducing sugars do not show mutarotation.

Why?
Consider the following reaction:
$$\ce{\beta-D{-glucose} + CH3OH + HCl -> \alpha-D{-glucopyranoside} + \beta-{D-glucopyranoside}}$$
Even though the pyranoside has an acetal group, the reaction is still reversible? Therefore, shouldn't pyranoside show mutarotation? Isn't the mechanism for the reversible reaction for acetal same as the one for hemiacetal?

In principle, the acetalisation reaction is indeed reversible. Typical conditions to reverse it would be to use $\ce{TsOH}$ or $\ce{PPTS}$ and water.
However, both full acetalisation and liberation of the aldehyde require acidic conditions. Hemiacetalisation — the prime mechanism of mutarotation — proceeds without acidic catalysis, i.e. the inherent (low) acidity of water is sufficient. And to observe mutarotation, you typically dissolve the pure sample in a pure solvent (with no acid added).
If you dissolve e.g. methyl α-ᴅ-glucopyranoside in water and add catalytic amounts of acetic acid, you will observe mutarotation but not if you simply dissolve the O,O-acetal in water.


text:
Which is most likely the acidic proton in ascorbyl palmitate?
The structure of ascrobyl palmitate contains a vinyl group, so I assume the red proton would be:

What about the other OH group on the other side of the double bond?

The other H (hydrogen) will be more acidic. The ascorbate anion is stabilized by electron delocalization, as I have show below in terms of resonance between two canonical forms.


text:
How do we perform the isomerisation of 1-butyne to 2-butyne using NaNH2?
How do we perform the isomerisation of 1-butyne to 2-butyne using NaNH2? As a first step, I think that $\ce{NH2^-}$ would remove $\ce{H^+}$ from the third carbon to give an allene intermediate. But I don't know how to proceed after that. Please explain the other steps.

This transformation isn't likely to happen, afaik isomerisation of alkynes with NaNH2 gives the terminal alkyne (because that can be deprotonated).

@orthocresol This is a popular question. The answer below has shown deprotonation of a non-terminal alkyne (apparently contrary to what you said "because that can be protonated"). And has an in fact converted a 2-alkyne to a 3-alkyne. Also, there's this question: Does a non-terminal alkyne react with sodamide? So, what is correct?

@GaurangTandon answer below is wrong

Here in the following image I have explained the mechanism for transformation of 2-alkyne to 3-alkyne. Isomerisation from 1-butyne/ 1-alkyne to 2-butyne / 2-alkyne is not possible.

Hope it helps.


text:
What gives brown butter its color?
Brown butter is butter that has been heated until the water content has boiled off and the remaining liquid has turned brown. It is often called for in cooking and baking, for example in recipes for financiers.
Discussions of the process typically explain the brown color as browning of the milk solids in the butter due to the Maillard reaction. However, even when these milk solids are removed, as is done in most recipes by filtering or careful pouring, the brown color remains.
I believe I understand how the flavor components of the brown butter survive filtering - the volatile products of the Maillard reactions remain dissolved in the liquid fat - but is there a similar explanation for the brown color? Or is it down to the coarseness of the filters allowing small suspended particles through with the liquid?
Any details on the particular compounds responsible for the color, or the chemical processes involved in creating them, would be much appreciated.

Although at first glance, this would appear to belong on the cooking.stackexchange.com, I feel this is a more appropriate forum as the question is asking about the chemistry underlying this culinary process.

I have several more links to illustrate various aspects of this question, but cannot post them due to lack of reputation. I'd appreciate it if a more veteran user could edit these into my question.

Financiers: https://en.wikipedia.org/wiki/Financier_(cake))

Discussions of the browning process: https://www.chefsteps.com/activities/brown-butter-solids http://eviltwin.kitchen/how-to-make-brown-butter/ http://www.thechoppingblock.com/blog/2014/09/brown-butter.html http://joythebaker.com/2013/09/on-browned-butter/

Recipe calling for filtering to remove the milk solids: http://www.epicurious.com/recipes/food/views/brown-butter-383967, and one calling for careful pouring: https://www.budgetbytes.com/2016/11/make-brown-butter/

The Maillard reaction: https://en.wikipedia.org/wiki/Maillard_reaction

The Maillard reaction is also called the non-enzymatic browning reaction. The reaction occurs when there are reducing sugars and amino acids present. The reaction is controlled by solvent choice, time and temperature considerations.
When browning butter, certainly, the Maillard reaction is occurring. There may also be other reactions happening such as Strecker degradation and caramelization.
As noted, browned butter is made by essentially heating butter until the water boils off at 100C and then the temperature ramps upward. So, what is causing the browning once the browned particulates are strained out? Some are chemicals that are formed from the reaction.
The color comes about due to production of melanoidins. As the Maillard reaction produces a range of chemical compounds, some of the components may have a lot of color on their own, such as furfural and its derivatives.
Professor Chi Tang Ho at Rutgers University has done extension work on Maillard reactions in model systems. Heath & Reineccius in their book Flavor Chemistry and Technology write about the Maillard reaction in more general terms that includes more on color production as part of the process.

Your first link (Brown Butter) states that heating allows for separation of the butterfat from the milk solids, which sink to the bottome of the pan and turn brown.
According to this Wikipedia page about a popular drink in rural Russia called baked milk:

In rural areas baked milk has been produced by leaving a jug of boiled
  milk in an oven for a day or for a night until it is coated with a
  brown crust. Prolonged exposure to heat causes reactions between the
  milk's amino acids and sugars, resulting in the formation of
  melanoidin compounds that give it a creamy color and caramel flavor.

While the color of baked milk is described as "creamy", the picture below (from the same Wikipedia page) shows that it has a distinct brown color. I suspect the reactions described above that give baked milk it's color also give the milk solids in brown butter it's distinctive brown color.


text:
Why is phenol a solid at room temperature?
The commonly cited reason is inter-molecular hydrogen bonding. But won't the hydrogen bonding in phenol be weaker than that of most alcohols because of resonance-effect? And yet, alcohols with stronger hydrogen bonding (such as ethanol, propanol, etc.) aren't solids at room temperature.
Also, if the reason behind its state is it's molecular weight, then why is benzyl alcohol a liquid at room temperature? The hydrogen bonding in benzyl alcohol is stronger (less resonance-effect) and its molecular weight is also higher.

It can also be due to $\pi$-stacking

Benzene freezes at +5°C, cyclohexane at +7. It's a matter of efficient packing in the crystal, plus a little hydrogen bonding.


text:
Calculate thr unit cell length of perovskite
In a ideal unit cell of perovskite $\ce{KMnF3}$ (closed packed with purely ionic bondings),  calculate the unit cell length:
radius:  $\ce{K}=1.7~$  $\ce{F}=1.2~$   $\ce{Mn}=0.8$
The structure

Green $\ce{K}$, red $\ce{F}$, blue $\ce{Mn}$. (ignore the sizes in the picture)
Am I going right by using the $\ce{F-K-F}$ diagonal length to calculate the cell length  using pytagroas, as the Mn diameter is to small and therefore the $\ce{Mn-F-Mn}$ will have some spacing on it and are not fully adjacent.
I would get unit cell length =4.101.   can somebody confirm this?
Just a bit confused as I have the radius of $\ce{Mn}$ but don't use it for calculation
Many Thanks

Presumably the radius of Mn is there for you to figure out which set of ions is close-packed.

Experimental X-ray crystallography data yields a value of 4.182 for the lattice constant for $\ce{KMnF3}$. See this reference for further details. Given that the paper was published in 1961, there are likely more recent refinements experimentally. That said, your number is within the error of observed values.
Reference: K. Knox. Perovskite-like Fluorides. I. Structures of $\ce{KMnF3}$, $\ce{KFeF3}$, $\ce{KCoF3}$, $\ce{KNiF3}$, and $\ce{KZnF3}$. Crystal Field Effects in the Series and in $\ce{KCrF3}$ and $\ce{KCuF3}$. $\it{Acta\,Cryst.}$ (1961) $\bf{14}$, 583.


text:
Formal potential
I've got the following question and I'm having doubts with it:

$$\ce{NAD+ + 2e- + H+ -> NADH}$$
Calculate ${E^\circ}'$ for the half cell $\ce{Pt | NADH, NAD+, H+}$ at pH 7, given that $E^\circ = \pu{-0.358 V}$ at $\pu{298 K}$.

It wants the formal potential. So I write the Nernst equation:
$$E = E^\circ - \frac{RT}{nF}\ln\left(\frac{[\ce{NADH}]}{[\ce{H+}][\ce{NAD+}]}\right)$$
so $$E = E^\circ - \frac {RT} {nF} \ln\left(\frac {1} {[\ce{H+}]}\right) - \frac {RT} {nF}  \ln\left(\frac{[\ce{NADH}]}{[\ce{NAD+}]}\right)$$
Now if I understand correctly, $$E^\circ - \frac {RT}{nF} \ln\left(\frac {1} {[\ce{H+}]}\right)$$ is the formal potential, I calculated this and got $\pu{-0.56 V}$. But the literature values are different. Can you help?

Instead of deleting your old question, you could edit it - the edit will push it into a queue for reopening, and I would probably have reopened it too. Anyway, it's OK. By the way, you can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details. I've done up a bit and will leave the rest for you.

With the standard electrode potential assumed to be $\pu{-0.358 V}$, I am getting the same answer as you but I googled the standard electrode potential of the above reaction and I found it to be $\pu{-0.320 V}$. On substituting $\pu{-0.320 V}$ as the standard electrode potential, I got $\pu{-0.529 V}$.


text:
How can I remove water vapor from an airstream without a cold trap?
I'm trying to perform vacuum distillation at home with a rotary vane vacuum pump.
In order to not ruin my vacuum pump, solvent vapor (in my case, water) should be removed from the air before it reaches the pump inlet. The only method that I've encountered for accomplishing this, is using a cold trap to chill the air below $\mathrm{-50^\circ C}$ to deposite the water. To this end, the cold trap is filled with a cryogenic liquid such as liquid nitrogen.
Unfortunately I do not have access to liquid nitrogen, dry ice, etc. Is there any other affordable way to remove water vapor from the air before it reaches the pump?
Could I use a container filled with crushed silica beads or a liquid to absorb the water in the air? Or would using a very hydrophobic membrane work?

Hello and welcom to Chemistry.SE. If you have any questions about the site you can take the short [tour] or visit the [help]. Regarding your question, drying the air to a dew point of -50C without cooling might be challenging, but hopefully someone here has a good idea (and you probably don't need the air to be that dry). Best of luck!

I guess it depends on where you live, but it seems like dry ice would be rather cheap, easy, effective and readily available compared to things like dessicants, membrane technologies, etc.

The products of the Drierite company may interest you.  www.drierite.com

Thanks, airhuff. I live in Europe, dry ice is not readily available. Fishmongers and such use regular ice, not dry ice, and I know of no other business that (may) use dry ice. Twelve pounds of dry ice costs close to 40 dollars. And it's not like you can really store it for any reasonable amount of time.
Waylander, thank you, their products seem to somewhat confirm the validity of my idea that a column with a dessicant might work. I think I could construct something like that myself.

You won't easily be able to use some other liquid to pull the water vapor out of the air. I think the cheapest option is a hydrophobic membrane or filter that you feel comfortable using. A commercial desiccant system will be the highest performance next to cryogenic cooling but may not be cost effective (depends on the scale you are working at). You can also try salt in place of silica beads for absorption drying versus adsorption drying; I imagine the silica will be more efficient but salt will be cheaper. When using things like silica, salt, or commercial desiccants you may run into issues of too much pressure drop as the gas travels through the packed column. You will have to figure out how often the desiccant media needs to be regenerated/replaced to make sure your drying is adequate.

To remove water vapor from an airstream without a cold trap, consider using a desiccant dryer or a moisture-absorbing filter. These alternatives help reduce humidity and prevent water vapor from reaching downstream components.


text:
How to predict (know) the products of the reaction SnO+NaOH?
I would like to know how to complete the reaction $\ce{SnO}+\ce{NaOH ->}$, how can I form a salt and water from this?
I've seen that the complete reaction is:
$$
\ce{SnO + 2NaOH ->Na2SnO2 + H2O}
$$
How to know what will be formed?

I've discovered the answer, at least I guess...
The given reaction is
$$
\ce{SnO + 2NaOH -> Sn^{2+} + O^{2-} + 2Na+ + 2OH-}
$$
The hidroxy breaks in $\ce{H2O + O^{2-}}$. This released oxigen will get together with the oxigen from the $\ce{SnO}$, to form $\ce{O2^{2-}}$.
$$
\ce{SnO + 2NaOH -> Sn^{2+} + \underbrace{O2^{-2}}_{\textrm{nox = -4}} + 2Na+ + H2O}
$$
The oxidation state of this molecule is -4. Performing the double replacement reaction we get
$$
\ce{SnO + 2NaOH -> SnO2^{-2} + 2Na+ + H2O}
$$
And finally,
$$
\ce{SnO + 2NaOH -> Na2SnO2 + H2O}
$$
What is already balanced.


text:
Just to Clarify for the Isomers of C5H12O
Are these the isomers to $\ce{C5H12O}$? If so what makes them so different from each other?

pentanol
2-pentanol
3-pentanol
3-methyl-2-butanol
2-methylbutanol
methyl butyl ether
methyl isobutyl ether
methyl-t-butyl ehter
methyl-s-butyl ether
ethylpropyl ether
ethylisopropyl ether

Corresponding to $\ce{C5H12O}$  there will be a total of 14 structural isomers.
8 isomers with alcohol functional group:
(1) $\ce{CH3CH2CH2CH2CH2OH}$  : n-pentanol
(2) $\ce{CH3CH2CH2CH(OH)CH3}$  : 2-pentanol
(3) $\ce{CH3CH2CH(OH)CH2CH3}$  : 3-pentanol
(4) $\ce{CH3CH2C(CH3)(OH)CH3}$  : 2-methyl-2-butanol
(5) $\ce{CH3CH2CH(CH3)CH2OH}$  : 2-methyl-1-butanol
(6) $\ce{CH3CH(OH)CH(CH3)CH3}$  : 3-methyl-2-butanol
(7) $\ce{CH3CH(CH3)CH2CH2OH}$  : 3-methyl-1-butanol
(8) $\ce{CH3C(CH3)2CH2OH}$  : 2,2-dimethyl-1-propanol
6 isomers with ether functional group:
(1) $\ce{CH3CH2CH2CH2-O-CH3}$  : methyl butyl ether
(2) $\ce{CH3CH2CH2-O-CH2CH3}$  : ethyl propyl ether
(3) $\ce{CH3CH(CH3)CH2-O-CH3}$  : methyl isobutyl ether
(4) $\ce{CH3CH2-O-CH(CH3)CH3}$  : ethyl isopropyl ether
(5)  $\ce{CH3CH2CH(CH3)-O-CH3}$ :methyl sec-butyl ether
(6) $\ce{(CH3)3-C-O-CH3}$ : methyl tert-butyl ether