text:
Why is zinc sulfate used in daniell cell
Why is zinc sulfate needed in the construction of a daniell cell.  It seems that because the zinc is getting oxidized and not reduced there is no need to have a solution of zinc sulfate as it will produced when the zinc is reduced.  The sulfate ions would simply cross the salt bridge and balance out the charges.

Zinc sulfate is not needed. With any other neutral salt in its place, your cell will work just fine.
Without any salt at all, though, it will not work, since pure water is a poor conductor of electric current.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post.

I do not actually see how I could provide an answer (in the strict sense) to the question, when the question itself relies on a wrong presumption. Why does a cat bark, really?

If you have doubts about the premise of the question, it is better to ask for clarification in a comment.

I do have doubts about many other questions, but this one is crystal clear. Also, if we clarify it, wouldn't the question vanish into nothing?
Anyway, I'll follow your advice the next time.


text:
Identification of types of alcohols

An unknown compound was found to react with sodium hydride with the evolution of hydrogen. The compound could not be acetylated under the normal conditions for acetylation. The compound resisted oxidation under mild conditions, and under vigorous conditions yielded only products of molecular weight much smaller than the unknown. On the basis of these facts the compound is:

a primary alcohol
an aldehyde
a secondary alcohol
a tertiary alcohol
a secondary amine

The correct answer is (4).
From the first statement, it is clear that this compound must be an alcohol (since alcohols react with sodium hydride and produce hydrogen gas in this reaction).  I could guess that the answer was a tertiary alcohol as secondary and primary alcohols would easily oxidize, but I'm not sure what the products of the oxidized tertiary alcohol would be?
Also, why wouldn't a tertiary alcohol undergo acetylation?  I don't see any reason why it shouldn't?

Reaction of tertiary alcohol with NaH
As all types of alcohol can easily reacts with NaH to produce hydrogen gas and sodium salt of alkoxide ion or alcohol. This shows alcohols have acidic hydrogen. It is type substitution reaction. The general reaction is as follows :

When tertiary alcohol say $\ce{(CH3)3COH}$ reacts with $\ce{NaH}$, hydrogen gas is released as follows :
$$ \ce{(CH3)3COH + NaH  ->  (CH3)3CONa + H2}$$
No acetylation of tertiary alcohol occurs under normal conditions
The acetylation means addition of acetyl group to a substrate. In the case of primary alcohol, acetylation is done under normal condition. Reactivity of secondary alcohol is somehow decrease for acetylation. In the case of tertiary alcohol, it is possible in extreme conditions. There is main reason behind.
As we know acetylation is nucleophilic substitution reaction. Here, the attacking nucleophile for acetylation of alcohol is alkoxide ion. More is nucleophilicity, more be reactivity of nucleophile to attack at carbon of carbonyl group of acetyl chloride.
Nucleophilicity order : $\ce{CH3O- > CH3CH2O- > (CH3)2CHO- > (CH3)3CO-}$
This is reason that tertiary alcohols are least reactive for acetylation. You can more understand by the following mechanism :

Oxidation of tertiary alcohol
Tertiary alcohols Undergo oxidation with oxidizing agents such as chromium tri oxide, potassium permanganate, potassium di chromate or nitric acid in neutral, acidic or alkaline medium to give the products. The oxidation may be partial or complete. Tertiary alcohols does not get oxidized by mild oxidizing agents either in aqueous, alkaline or under neutral conditions. They undergo oxidation under acidic condition.In this process of oxidation tertiary alcohols first undergo dehydration to form alkenes. The alkene thus formed then undergoes further oxidation and forms a ketone with one carbon atom less than the alkene. On further oxidation the ketone forms carboxylic acid with one more carbon atom less than the ketone. Thus the oxidation of tertiary alcohol gives a product with less number of carbon atoms than that of the original alcohol taken.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Can a tertiary enamine tautomerize to an imine?

Which of the following compounds does not show tautomerism?

The correct answer is (a), and I'm curious if this is because to show tautomerism, there can't be any net charge.
For instance, ketones can tautomerize to enols, and more importantly, enamines can tautomerize to imines.  In (a), we would get an imine-like thing, but the nitrogen would be positively charged.  Does this still count as tautomerization?
EDIT: Thank you everyone for your helpful clarifications, but I'm still a bit confused as to why, mechanistically, we need a hydrogen on the nitrogen.
When I Google the mechanism for imine tautomerization, what if find is this:

Where the nitrogen lone pair pushes down and the double bond reaches out to pick up a proton.  There isn't a transfer of hydrogen from the nitrogen to the solvent in the case shown below, yet the molecule still tautomerizes.

As this is a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you. Additionally, all I am seeing a black box where the image should be.

Hi, sorry about that.  I had just posted it to see if the black box that I was seeing would correct itself, but obviously it hadn't.  I wasn't actually done writing my post when you found it, but I appreciate your dedication to answer questions.

@coloratura Sometimes you have to use the HTML workaround to resize images when it fails via imgur. Have a look at the source when you want to know how it is done.

Thank you for fixing the image and adding a considerable amount of your thinking - close vote and downvote retracted.

What's you're talking about is simply another mesomeric structure...

Tautomerization is a chemical reaction whereby constitutional isomers interconvert. The interconversion process involves the shift of a hydrogen atom accompanied by the shift of a double bond.  Here are a few examples.

In molecule (a) pictured above, there is no hydrogen attached to a nitrogen (it has been replaced by a methyl group), so there is no hydrogen atom available to shift.
Response to OP's edit:
The picture you posted in your edit is incorrectly labeled.  The charged molecule on the left is an iminium ion, not an imine.  So this is not an example of an imine-eneamine equilibrium.  You are correct that in this equilibrium (the one pictured in your edit) there is no hydrogen transferred to nitrogen, but again, this is not an example of an imine-eneamine tautomerization.  I think this error is the source of your confusion.
Here is a picture of an imine-eneamine tautomerization.

As you can see, a hydrogen is transferred between nitrogen and carbon.  This hydrogen transfer can occur with or without solvent molecules playing a role.  That is to say, the hydrogen can be transferred directly between the carbon and nitrogen atoms, or it can be transferred from one of the atoms to a solvent molecule which can then transfer it back to the other atom involved in the tautomerization. The image source also provides a nice explanation of tautomerization.

Does the hydrogen necessarily need to be transferred from one nitrogen to the other nitrogen?  In mechanisms, it's portrayed that the hydrogen comes from solvent.

Tautomerization can also occur in solvents that can't supply a hydrogen.  In those cases the reaction can be intramolecular with the hydrogen transferring from one nitrogen to the other.

I see, so in this problem, we must assume that the solvent can't donate a hydrogen?

No, the solvent doesn't really matter in your question.  Even if the solvent is involved, the first step would involve transfer of the hydrogen on the nitrogen to solvent and then from solvent it would eventually return to the other nitrogen.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Distribution of acid in Organic and Aqueous Phases
The value for the distribution constant for an organic acid, HA, between chloroform and water is $10^2$.  The dissociation constant of the acid in water is $10^{-5}$, and it can be assumed that it does not dissociate or dimerize in the chloroform phase.  Increasing the pH of the aqueous phase from 4 to 9 has what effect on the fraction of HA extracted into the chloroform phase?
(a) The fraction is more than doubled.
(b) The fraction is increased but not doubled.
(c) The fraction is unaffected.
(d) The fraction is decreased but is not halved.
(e) The fraction is reduced to less than half.
The correct answer is (d).  I understand why the concentration of HA in chloroform should decrease, but I'm not sure I'm completely confident as to why it decreases to less than half.
As the acid in the organic phase is in equilibrium with the acid in the aqueous phase, obviously, increasing the pH past the pK$_a$ of the acid will drive it to completely dissociate, causing the HA in the organic layer to go into the aqueous layer.  How can I quantitatively know that this process will decrease the $\frac{[HA]_{CHCl_3}}{[HA]_{water}}$ to less than 50?
EDIT: So, I've taken your suggestion, but I seem to be getting stuck.
I'll use the notation HA$_{org}$ to denote the acid in the chloroform layer and HA$_{aq}$ to denote the acid in the water layer.
H$^+$ + A$^-$ $\rightleftharpoons$ HA$_{aq}$
HA$_{aq}$ $\rightleftharpoons$ HA$_{org}$
K$_a$ = $10^{-5}$ and the equilibrium constant for the bottom reaction is K$_D$.
We know that:
Our wanted ratio, D = $\frac {[HA]_{org total}}{[HA]_{aq total}}$ = $\frac {[HA]_{org}}{[HA]_{aq} + [A^-]} =  \frac {[HA]_{org}}{[HA]_{aq} + 10^{-5}[HA]_{aq}/[H^+]} = \frac{K_D [H^+]}{[H^+] + 10^{-5}}$
But, I don't know what K$_D$ is and it would also change with pH.

KD is the distribution coefficient for HA between the organic phase and the aqueous phase = [ha]o/[ha]aq with equal phase volumes.


text:
Calculating Enthalpy Changes at Different Temperatures

To calculate $\Delta H$, the change in enthalpy at $\mathrm{100^\circ C}$ for the reaction below, one needs what addition information?
$$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$$
$\Delta H^\circ = \pu{-92.0 kJ}$ at $\mathrm{25^\circ C}$
(a) The equilibrium constant for the reaction at $\mathrm{100^\circ
 C}$
(b) The molar heat capacities of the reaction and the products as a
   function of temperature
(c) $\Delta E^\circ$, the standard internal energy change for the
   reaction
(d) The partial pressures of the reactants and products at
   $\mathrm{100^\circ C}$
(e) The entropies of formation for the reactants and products at
   $\mathrm{100^\circ C}$

The correct answer is (b), but I'm not entirely sure why.
I know it can't be (a) because I think we need $\Delta S$ as well.  Also, (c) won't be helpful because it is at standard temperature, and we are wanting information at $\mathrm{100^\circ C}$.  Finally, (e), seems correct but they are actually talking about entropy instead of enthalpy.  However, I'm not sure how to rule out (d) or justify (b).

@Chet already gave the qualitative answer but the actual equation it is based on is:
$$
\Delta H^\circ(T)=\Delta H{^\circ}_{T_1} + \int_{T_1}^{T} \Delta C_p{^\circ}(T^\prime ) \,\mathrm{d}T^\prime$$
If you know the enthalpy at a reference temperature and the heat capacity as a function of temperature you can calculate the enthalpy change at any other temperature.
Actually d) and e) together are not wrong. We all know these equations:
$$\Delta G= - RT\ln (K) \qquad \Delta G= \Delta H - T\Delta S $$
With d) you can calculate $\Delta G$ and with e) you can calculate $\Delta H$.
Lets look deeper into: $\Delta G^\circ$, $\Delta G$, $K$ and $Q$.
To answer your second question completely I would recommend to look again into a textbook of your choice. Especially the definitions of activity and chemical potential.
If you know the chemical potential $\mu^\circ$ of a pure substance you can calculate the chemical potential of mixtures if you know the activities:
$$\mu= \mu^\circ + RT \ln(a)$$
Now we make two assumptions:

We have constant temperature and pressure. ($\mu = \Delta G$)
We can calculate the activities simply over the partial pressures or concentrations.

This leads then to your equation:
$$
\Delta G= \Delta G^\circ + RT \ln(Q)
$$
where $Q$ is the reaction quotient.
Since we assume equilibrium,  we know that $\Delta G=0$ and $Q = K$. This leads to:
$$ 0= \Delta G^\circ + RT \ln(K) \Rightarrow \Delta G^\circ = -RT \ln(K) $$
The last equation is the handy one you use for calculating equilibrium constants from energies and vice versa.
So to be really precise one has to write:
$$ \Delta G^\circ= \Delta H^\circ - T\Delta S^\circ $$
But one usually skips the superscript $^\circ$.
At equilibrium $Q$ and $K$ are essentially the same.

Forgive if I'm wrong, but isn't ∆G˚ = -RTln(K) and ∆G = ∆G˚ + RTln(Q)?  So, how can one get ∆G if we only have K?

I edited my answer to adress this question. Perhaps you edit your question.

Since $\Delta H$ is a function of state, you can get $\Delta H$ at 100 C by taking the reactants at 100 C, cooling them down to 25 C, then carrying out the reaction at 25 C, and then heating the products up to 100 C.  This procedure makes use of Hess' Law.  Getting the amount of heat removed to take the reactants from 100 C to 25 C requires using the molar heat capacities.  Same for heating the products back up to 100 C.

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


text:
Hydrolysis of Radiolabeled Esters
Hydrolysis of C$_6$H$_5$COO$^{18}$CH$_3$ with aqueous NaOH produces what final product.
(a) CH$_3$$^{18}$OH
(b) CH$_3$CO$^{18}$ONa
(c) CH$_3$C$^{18}$OONa
(d) CH$_3$$^{18}$ONa
(e) CH$_3$OH
The correct answer is (a), which is puzzling to me.  I think it should be (d), as when the radiolabeled methoxy leaving group is pushed out by the reformation of the carbonyl, it will be negatively charged.  The only cation that is floating around is Na$^+$.
Where, oh were, did this magical proton appear in basic conditions?
EDIT:  Since I think my comments will get longer I'll just post my question here.  I agree that the methoxy is in equilibrium with water:
CH$_3$O$^-$ + H$_2$O $\rightleftharpoons$ CH$_3$OH + OH$^-$
But the pK$_a$ of methanol is 15.54, and the pK$_a$ of water is 15.74.  Thus this equilibrium will favor the methoxy/water side.  Thus, the slight majority of product should be a methoxy salt.

You are surely right in that the labeled oxygen will stay with the alcohol and not with the acid. Now is the time to recall what do you know about the comparative acidity of alcohol and... wait, what does the word "aqueous" stand for, really?

Ah, I see.  The labeled methoxy will be protonated by the water in the system.

But, the equilibrium CH$_3$O$^-$ + H$_2$O $\rightleftharpoons$ CH$_3$OH + OH$^-$ should favor the left hand side, as methoxy is the weaker base.  So, wouldn't the majority of the hydrolysis product be in methoxy form?

You are overthinking it; your first answer was right.

It would be more proper to take proton from transient acid. Also methoxide is stronger base.

The product of the hydrolysis is benzoic acid: $$\ce{C6H5COOH}$ which is promptly deprotonated. As the reaction proceeds, it gets less basic.

@Mithoron I looked up the pK$_a$s of methanol and water and methanol is a slightly stronger acid, which means that methoxy is the weaker base, thus the equilibrium favors left hand side.

To understand why the answer a is correct, we need to understand two things.

As Ivan has pointed out water is a stronger acid than methanol, so a solution of sodium methoxide and water will be mostly converted into sodium hydroxide and methanol.
The which of the single bonds CH3-O and C(=O)-O will break during the reaction, I am sure that the CH3-O bond will stay intact.

Lastly why do you assume that oxygen-18 is radioactive,
Oxygen has three stable isotopes, 16, 17 and 18.
Oxygen-14 has a half life of about 1 minute while oxygen-15 has a half life of 2 minutes. I think that it would be very hard to make an organic compound with a radioactive oxygen lable and then to do a chemistry experiment on it. I think that making an ester and doing the experiment within 20 minutes of getting the oxygen-15 enriched water from a cyclotron would be the test of all tests in terms of working very quickly.

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


text:
What does my textbook mean by saying this about petroleum?
It reads:

Petroleum consists of crude oil and natural gas and contains a mixture of up to 300 hydrocarbons, including sulfur and nitrogen compounds. It is formed over thousands of years from the remains of fossilised marine organisms.

Is petroleum a single entity that is simultaneously a gas and a liquid oil compounds? How does that work? It is also a mixture of up to 300 hydrocarbons, so it is a liquid and gas in one? I don't understand? Do they mean that petroleum can be either gas OR it can be either liquid? What about the 'mixture of 300 hydrocarbons bit'. Does that mean you will never extract petroleum of a single hydrocarbon eg pure $\ce{C6H14}$ etc? I haven't revised this in a long time so I forgot..

Petroleum is a mixture of many different compounds.
Some of those compounds are gas.  Some are liquid.  Some are solid.
Many of them are hydrocarbons, and there are many different hydrocarbons included in the mix (e.g. different numbers on the C and H).
Petroleum is generally refined into more specific products before use, at a refinery.
Another example to consider might be "mud" - that's a mixture of some liquids, maybe some dissolved gases, and a bunch of different solids, usually including clay, maybe a bit of gravel, and some organic materials formed from decaying plants and animals.
If you take mud and compress it underground for thousands of years, then you'll have petroleum.
The second sentence seems clear enough.

How can something be both a gas and a liquid? Or is petroleum used to refer to a collective group of hydrocarbons, some of which are either gases or liquids?

The latter.  Have a soda with some ice - there in your cup you'll have solid, gas (which makes the soda fizzy), and liquid.  Then have some ice water- you'll find the same substance in solid, liquid, and (a small amount) gas forms all in the same place at the same time (stable only at 0C).

To understand what petroleum is you need to understand where it is usually found.
Deposits of crude oil are usually found deep in the earth in porous rocks under high pressure (the organic material that forms crude oil needs pressure and heat to convert it to oil). Often the rocks that store oil and not where the oil was originally formed, but, being daily mobile, it tends to migrate to the highest place in the rock formation that is still porous (and is often trapped by a higher layer of a non-porous rock).
The mix of components in oil varies a lot. There will usually be a range of hydrocarbons from methane to higher hydrocarbons with dozens of carbons or perhaps even hundreds. There will also be sulphur and nitrogen compounds, sometimes in large quantities. Sometimes the gaseous components will separate out into a layer in the deposit, but even then there will be a lot of gas dissolved in the liquid parts (oil is usually under a lot of pressure so there can be a lot of gas in solution).
When you drill into an oil formation one of the things that happens is that you release the pressure. As the pressure lowers much of the gas in the solution will be released even if there isn't a separate layer of gas in the deposit. The oil driller than has to handle the separation at the wellhead in a controlled way. But the major point is that what comes out of the ground is often a messy mixture of gas and liquid and your description is reasonably accurate there. This, however, would not normally be called petroleum which is usually reserved for certain particular refined mixtures.

Petroleum is a mixture of many hydrocarbons and other things as you said. Due to the pressure and temperature,  some light fractions will dissappear and your petroleum will be heavier. Let's think about the condensate. Condensate is mainly composed of C8 C11 fractions and if you put it in a open bottle in atmospheric conditions, light fractions will be gaseous and leave the bottle. The light color will be darker and the amount will be less. This means that your crude oil contains gas and liquid substances. The conditions are important. In reservoir, due to the high pressure and temperature,  light fractions are soluable in it.
Hope it helps you to understand the situation.


text:
How to light a candle with small amounts of safe to use chemicals?
Already saw some youtube video where they lit a camp fire with glycerol and potassium permanganate and tried those but the amount of the stuff you need to create enough heat is just too much.
So is there a way to light a candle with small amount of something and something, like having the candle wick dipped/powdered with something and then just letting it get into contact with something else?

Well, you could coat the head of the candle with phosphorous and then scrape it against a rough surface...but that would basically be a match.

What is wrong with matches? A small amount of chemicals applied to a small stick to create fire in a safe way...

Safety and ability to light a fire, as far as I understand, directly contradict each other. What starts a fire on your wish, would occasionally start a fire against your wish. The glycerol/sulfuric acid/permanganate recipe, like it or not, is probably the safest of all.
Others include:

White phosphorus, not only flammable but also severely toxic by itself.
Butyllithium and many other organometallic compounds that would ignite instantly on air; that's not quite what you want, anyway.
Hydrogen against platinum black (danger of explosion).

I definitely would not try any of these at home.


text:
Why is there a subtractive volume correction in the van der Waals' equation for real gases?
Why there is a subtractive volume correction in van der Waals' equation for real gases, both real and ideal gases occupy volume of container in which they are kept, so there shouldn't be any correction?

Your post is rather unclear at the moment: please edit it for grammar and clarity and be sure to add your thoughts concerning the answer. If this is homework, please be sure to read the site policy on such.

This is a wholly different kind of volume. Sure, any gas occupies the volume of container, but here we are talking about the volume of molecules themselves. Ideal gas has none; real gas has some.

I don't understand whose volume is talked about of in ideal gas and real gas

Is the volume is of container or gas molecules or something else

You have two volumes here. The volume of the container V and the volume of the gas molecule, b. Ideal gases are thought of as being points ie zero volume, b=0. Therefore they move about the total volume of the container. Real gases have a volume, b>0. Each gas molecule takes up some space in the  container, meaning there is less effective volume for them to move about in. Effective volume is V-nb where n is no of molecules ( or moles depending on units of b).

Ok in case of gases we take the  volume in which they are free to move, tell me if I am wrong.

Exactly, so since we are now assuming that they have a non-negligible volume, it means that they have less space to move about. Therefore $V \rightarrow V - nb$.

Also see https://chemistry.stackexchange.com/questions/126495/why-is-the-pressure-correction-added-in-the-van-der-waals-equation

@Leeser In that sense, the pressure also reduces in case of real gas and effective pressure should be P - something?

@orthocresol In that sense, pressure is also reducing for real gas. Therefore, $P \ce{->} P-an^2/V^2$ ?

The van der Waals equation is an attempt to explain the pressure-volume relationship for real gases. For example, what will the pressure be of a given amount of gas in a vessel of a fixed volume.
In ideal gases the size and interactions between the atoms or molecules of the gas don't matter at all. We think of them as point-like entities that don't interact except to bounce elastically off each other and off the walls of the vessel. Their behaviour is described by the ideal gas equation and doesn't depend at all on the nature of the particles making up the gas.
Real molecules and atoms deviate from this ideal behaviour because the particles of the gas both have finite volume and attractive interactions which mean they don't behave exactly as the ideal gas equation describes. The van der Waals equation is a simple attempt to explain those deviations from ideality. The volume correction is one of those factors. It takes into account the fact that real gases are made from particles that actually have a finite size and are not mathematical points. The factor is about the volume of the particles of the gas not about the vessel the gas is in. The result of the correction is that the pressure of a given amount of gas in a vessel of fixed size is slightly different than would be expected from the ideal gas equation.

The factor nb is about the particle of gas but the V in PV=nRT is volume of the vessel. So, the correction is on the volume of the vessel (gas)?

One thing that this discussion missed is that the volume occupied by a gas is the volume available for molecular motion. So the volume of a real gas can be given as:
Volume of gas = Volume of the container - Volume not available for molecular motion
While studying ideal gas the volume not available for molecular motion turns out to be zero (due to the assumption of negligible molecular volume).
But for real gasses that is not the case and hence you apply that correction term.

We often frame ideal gas interactions as "particles do not attract or repel each other" and "particles have no volume." Often, it is more useful to call an ideal gas a gas where particles pass through each other. This framing implies both principles. Real gases are any gases where this behavior is not exhibited (i.e. all actual gases).
Let's say that particles in a closed container do not interact except they collide with each other and bounce off elastically like billiard balls. The volume term in the ideal gas law $V$ stands for the volume of the entire container; however, if two molecules collide elastically with each other, then for every molecule added to the container, there is a molecule's volume less space for every other molecule to be. So, while we still see the volume of the container as $V$, from the molecule's perspective there isn't a full $V$ volume worth of space to go, because other molecules are in the way.
For an analogy, if you walk into an empty room, you can navigate every square foot of the room easily. However, if you invite a bunch of friends over, suddenly there is much less space to navigate even though the room did not literally shrink in size—you effectively have less space to move. We'll call the amount of space that the molecule can actually move "effective volume" or $V_\text{eff}$.
We expect for every additional molecule we add that the volume $V$ stays the same, but the effective volume $V_\text{eff}$ decreases a little bit. So, our effective volume should be 1) linearly related to the number of molecules proportional to the molecule's volume and 2) smaller than the original volume. We can use this to make an equation, where $b$ depends on the volume of the molecule:
$$V_\text{eff} = V - nb$$
The properties of the ideal gas law come from conditions in the particle's perspective. This is because gas laws are derivable from the behaviors of individual particles through the application of statistical mechanics, a set of mathematical methods that can connect particle states to bulk-system states. In the previous analogy, each person represents a particle. Your behavior in a large room, like how fast you can run around in it, is a function of the room size and of other people in the room. The ideal gas law does not take this latter consideration into account; it assumes each person is a ghost that can phase through any other ghost in the room. In reality, you are limited to the unoccupied space available to you. This is also the case for every other person in the room—your individual behaviors are all linked to the room size and the number of people present.


text:
Why is sodium chloride edible whereas both sodium and chlorine are highly reactive?
Sodium chloride is edible. But elemental sodium and chlorine are highly reactive and poisonous. What is the difference?
Also, is there any difference in eating sodium chloride in powder form and after dissolving in water? It gets ionized, correct?

An atom and its ion are not the same thing.

Sodium chloride is edible. But elemental sodium and chlorine are highly reactive and poisonous. What is the difference?

Solid sodium chloride is an ionic compound. This means it consists of positively charged sodium ions ($\ce{Na+}$) and negatively charged chloride ions ($\ce{Cl-}$) arranged in a cubic lattice.

On the other hand, sodium metal and chlorine gas are elements. $\ce{Na}$ has an electron configuration of $\mathrm{[Ne]~3s^1}$ and so it really wants to lose that single electron to form a full outer shell as a $\ce{Na+}$ ion. This makes it highly reactive towards species which can easily accept an electron such as $\ce{Cl2}$, $\ce{H2O}$ etc. $\ce{Na+}$ has a full outer shell, being isoelectronic with neon, and so it is quite happy to stay as an ion and not react. Chlorine, with an electron configuration of $\mathrm{[Ne]~3s^2~3p^5}$, really wants to gain a single electron to get a full outer shell, isoelectronic with argon, as the $\ce{Cl-}$ ion. This makes it highly reactive towards species which can easily donate an electron, such as $\ce{Na}$.

is there any difference in eating sodium chloride in powder form and after dissolving in water?

Not really. Sodium chloride dissolved in water consists of hydrated $\ce{Na+}$ and $\ce{Cl-}$ ions. If you eat solid sodium chloride then it rapidly dissolves in the water in your body, so the effect is the same. Interestingly though, solid salt tastes quite nice, but salty water tends to taste horrible.


text:
How to find the number of atomic microstates for a given electronic configuration?
Given some electron configuration, I know the following formula can be used to determine the number of microstates:
$$\text{# mircostates} = \frac{(\text{# electron positions})!}{((\text{# electrons})!(\text{# positions} - \text{# electrons})!)}$$
So for 2 electrons in a $\mathrm{p}$ orbital it would be $6!/(2!4!) = 15$, or for two electrons in a $\mathrm{d}$ orbital it would be $10!/(2!8!) = 45$.
I'm a little confused how to apply this for a situation where all the electrons are not in the same $l$ level.
For a configuration like one electron in $\mathrm{s}$ and one in $\mathrm{d}$ would the number of microstates be the product of the above formula for each $l$ level?
$$\frac{2!}{1!1!}\cdot \frac{10!}{1!9!} = 20$$
Or for one electron in $\mathrm{s}$ and one in $\mathrm{f}$:
$$\frac{2!}{1!1!}\cdot \frac{14!}{1!13!} = 28$$
I assumed it would work like this because the formula reminds me of multiplicity, so it would make sense for it to be a product. But it also occurred to me that the 2 $l$ levels could be combined for the formula, so for a $\mathrm{(s^1)(d^1)}$ electron configuration, the number of positions would be $$\frac{(10+2)!}{2!(10+2-2)!} = \frac{12!}{2!10!}= 66,$$ which is must bigger than what I had before.
Can anyone clear up how to apply this formula to a case like this?
For clarification, the exact specification in the problem was:

Imagine a $\mathrm{s^1\, f^1}$ electron system (that is, 1 s electron and 1 f electron; no p or d electrons).  How many total microstates exist for this system?

I think this means that the one $\mathrm{s}$ and one $\mathrm{f}$ electrons are each confined to their $l$ level, which is why I thought the multiplicity of microstates would be the product of the above formula for each, but I'm not sure.

You really have to calculate out things exactly, and particularly carefully if electrons are in the same orbital, e.g. $p^2$, when the Pauli principle has to be checked. Its a bit tedious but it can be very quick with a bit of practice especially if electrons are in different orbitals.
You start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number s = 1/2 and magnetic quantum number $m_s$ = +-1/2. Orbitals have angular momentum of s=0, p=1, d=2, f=3 etc units.
The total spin angular momentum is the series of values $S=|s1+s2|...|s1-s2| $ where s1 and s2 =+1/2 are the quantum numbers of each electron. The total angular momentum is   $L=|l1+l2|...|l1-l2| $  and total angular momentum $J=|L+S|...|L-S| $
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by 1. Note that sometimes there is only one value and all values are positive. e.g. J=5/3, 3/2, 1/2 or J could be 0 or 3 etc.
The term symbol has the form $^{2S+1}L_J$.
The super-prefix is the spin multiplicity, for spin angular momentum $S$ this is  $2S+1$ or in general $2X+1$ for angular momentum X.
 In your question $s^1f^1$ the total spin using the series above is  S = 0 and 1. The spin multiplicity is thus 1 and 3, singlets and triplets terms only. The total orbital angular momentum L = 3 since the s orbital has zero angular momentum. Thus the terms so far are
$^1F$ and $^3F$ The number of microstates (multiplicity) in each term is $(2S+1)(2L+1)$. This is 7 for the $^1F$ and 3.7=21 for the $^3F$ making 28 in total. These states are also degenerate in energy. You should calculate that $s^1d^1$ has 20 microstates.
The total angular momentum J series associated with S=1 is $J=|3+1|.... |3-1|$ or 4, 3 and 2 and with S=0 , J=3.  Thus only triplet states have J = 4 and 2 and both singlet and triplets J=3. The terms symbols for the levels are
$^1F_3$, $^3F_4$, $^3F_3$,  $^3F_2$
 The total multiplicity from the J values is is 7+9+7+5 = 28 the same as from terms.
When the electrons are in the same orbital then the only secure way of calculating terms is to list all the individual quantum numbers in a table, with columns $m_{l1},m_{l2},m_{s1},m_{s2}$, S and L and remove those for which the Pauli principle removes a row. (See table in Steinfeld 'Molecules & Radiation' chapter 2 and Engel & Reid, 'Physical Chemistry' chapter 22). In particular you have to look out for when two rows have the same set of quantum numbers but in a different order; for example $m_{s1}$=0, $m_{s2}$=1 in one row and $m_{s1}$=1 and $m_{s2}$=0 in another but the $m_l$ values are the same, and vice versa. Doing this you should find, for example that $p^2$ has 15 terms, as $^1D$, $^3P$ and $^1S$ Hope this rather long answer helps you understand things.

Your equation (1) is correct. The assumption about the degeneracy of orbitals as well. But you miss another important point - electrons are indistinguishable. You do not know which one is where, only how many of them are in a given "slot". So for your question, you have 2 electrons in $(1+7)*2=16$ positions.
In addition, as the system is $s^1f^1$ it means, the orbitals are near in energy, otherwise the electrons will be paired. As a side note, the microstates are Slater determinants and for 2 electrons you will get both singlet and triplet states.

What do you mean the microstates are Slater determinants and therefore only singlet/triplet states? I have only seen slater determinants in reference to Hartree Fock and other computational methods.


text:
Sodium phenoxide bonding
Benzene has O and Na attached. Oxygen with a negative charge and sodium with positive , so they write it as
$\ce{benzene-O- Na+}$
Why couldn't it simply be like oxygen bonded to benzene and sodium there wouldn't be excessive charges.
I think sodium bonds with benzene as well as oxygen ($\ce{benzene-Na-O}$)
Am I right?
Why is it like this? I usually get stuff on my own this seemed really odd for some reason

Because this is an ionic compound. You must have heard of such in the inorganics course.

In your second diagram, did you mean to put the sodium between the oxygen and benzene or on the end?

The reason people do this is to highlight the fact that the bond between the oxygen and sodium atom is of a more ionic character than of a covalent character.

That's not what the question is asking about. If you read it carefully you will see that it is asking about the order in which the atoms are bonded, not the type of bond.

Actually I see that this might be a mistake on the part of the OP. I will retract my downvote for now until there is clarification.

Yes, since he refers to it twice, I just assumed it was a mistake since the title is Sodium Phenoxide and he writes the structure correctly once. I have no privileges to do anything so I can't really comment like I'd prefer to do.

My hypothesis is a very simple one you know, and its that elements want to be in the lowest state of energy generally by the means of completing a shell or shedding electrons down till they achieve a complete valent shell.
In this situation its better for sodium to shed a electron and form a ionic bond to achieve the aforementioned than to form a covalent bond with oxygen.


text:
Which of the following have the same molecular formula as the molecule shown in the model?

This model has 3 hydrogens connected the the first and last carbon and 2 hydrogens connected to the other carbons
Finding the same molecular formula as the alkane model I have to check all that apply

I know the last two options are not correct, because they don't have the same amount of carbons, but when it comes to the first two I'm confused because there is only two carbons attached to 3 hydrogens yet in these options there are 3 different carbons attached to 3 hydrogens.
Would they both be correct or does it matter if it's connected to the second bond instead of the third?
Im assuming since they are isomers that they are equal to eachother.

Please see the site's homework policy. You'll need to share your thoughts as you work towards an answer, otherwise your question might get closed.

As you correctly observed, the model has two $\ce{CH_3}$ groups, while each of the options has 3 of them. Would you expect 2 to turn into 3 if you rotate picture left, or right, or maybe upside down, or do anything else with it?

Remember they are asking for the molecular formulas which are the same, not the structural formulas. The structural formulas are shown in the question, but for example the first option, if you count up the number of carbons and hydrogens you get the molecular formula $\ce{C7H16}$, so you'll need to check if the molecule shown has the same formula.
So for molecular formulas, it doesn't matter how the atoms are arranged; all isomers have the same molecular formula. Only in structural formulas do you get information on the arrangement of atoms in the molecule.
Note there is also condensed structural formulas which look similar to the molecular formula but is still a structural formula. For the first option the condensed structural formula would be $\ce{CH3CH(CH3)CH2CH2CH2CH3}$.


text:
Is heating dilute HCl dangerous?
Say dilute $\ce{HCl}$ is used to convert some alkaloid to salt form, then the alkaloid is dried with heat, would gaseous $\ce{HCl}$ be released in the process?  Is there a threshold concentration of the acid, below which,  this would be acceptable from a safety point of view?

Why would anyone want to do that outside of a chemical lab with proper safety equipment (fume hoods, etc.)?

Say dilute HCl is used to convert some alkaloid to salt form, then the alkaloid is dried with heat, would gaseous HCl be released in the process?

Likely yes, unless it's the limiting reagent in the reaction in question. Probably best to assume some of it will come off as gas - that's the safest approach, in my opinion.

Is there a threshold concentration of the acid, below which, this would be acceptable from a safety point of view?

Yes, there are several metrics used by various authorities to determine what "safe" means. You'll have to decide which one (or ones) are relevant to your situation.
Here in the U.S., the Occupational Safety and Health Administration (OSHA) - which is under the Department of Labor - has published this information on hydrochloric acid exposure limits and hazards.
In that linked document, several other agencies such as the National Institute for Occupational Safety and Health (NIOSH) and the American Conference of Governmental Industrial Hygienists (ACGIH) list permissible exposure limits ranging from 3 to 7 mg/m$^{3}$; the Environmental Protection Agency (EPA) given an inhalation reference concentration (RfC) of 2 mg/m$^{3}$, detailed here as the concentration above which (I'm paraphrasing) bad things start to happen.
Good luck on your project and stay safe.


text:
How to esterify phenol with oxalic acid?
Is it possible to fully esterify phenol with oxalic acid in order to get diphenyl oxalate? What catalyst should I use and is the synthesis possible to perform at home with hobby equipment?

I don't know for sure, but it could work provided some conditions be fulfilled

use of oxalyl chloride
use low temperature, in order to try to suppress fragmentation of the intermediate (semi-ester) into the simple monoacyl chloride (by CO loss): in such cases $\ce{(COCl)2}$ behaves much like phosgene ($\ce{COCl2}$.
use the phenol as a dissociated salt (sodium, potassium) in polar solvents such as sulfolane or acetonitrile, this is optional and possibly risky, due to enhanced competitive "C-acylation" if ortho/para positions are free.
use Piridine-based solvents as reactivity enhancers (they convert acyl chloride to acyl immonium salts). Optionally, even stronger catalysts such as DMAP (p-dimethylamino pyridine) could work

Anyway, to sum up, the major risks stem from:

"C-acylation" (which is generally kinetically slower but irreversible and thermodynamically favored over "O-acylation"), if some suitable position is avail.
formation of diaryl carbonates (on extrusion of CO).
I guess this last drawback coul be at least reduced by working at low temperature.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Diphenyl oxalate can be obtained by transesterification of dimethyl oxalate with phenol.


text:
Does oxygenating tap water remove the tap water flavor?
I know flavor is subjective and not exactly quantifiable, but I think there's merit to this question.
We've all experienced that tap water often tastes a lot worse than bottled water or filtered water. I've always felt, however, that cold water actually tastes much purer than warmer water.
For instance, I was recently in a house where I knew the water was so called "hard water" and I could taste that it had the typical mineral flavor straight from the tap.
So, I took a glass of this tap water and refrigerated it for about a half hour and then drank it and I couldn't taste the typical tap water flavor at all.
I'd encourage people to try this just to make sure I'm not crazy.
Additionally, I have family that lives in Ohio and if anyone has ever tasted Ohio's tap water after living in a place like Washington (where I live) which has extremely fresh water, you know that Ohio's water is disgusting. I have found, however, that it tastes much better after being refrigerated for a while.
So, there are a couple of questions here.
Is it even true that refrigerating water (and thus cooling it) would cause a significant amount of oxygen to be dissolved in the water?
If so, would the amount of time the water is being refrigerated (even at a constant temperature) continue to dissolve more oxygen into the water? In other words, how long might it take for oxygen to reach it's equilibrium level of dissolution?
Those last two are definitely answerable chemistry questions. This next one is a little hairier.
Is there any reason, related to chemistry, that oxygen rich water would taste better than water not saturated with oxygen? Meaning, does the oxygen actually have a well-defined effect on the flavor of water, or is this simply answered by the fact that most people like the feeling of cold water more than luke-warm water and hence I identify cold water as "tasting better"?

To my recollection, any bunch of water which is allowed to stand for a while will lose its "chlorine taste", with or without refrigeration. It has nothing to do with oxygenation, but it should be more pleasant to drink. And given the number of questions about water tonight, I assume that it was all planned, so I (and others) should give a global answer: 1. Boiling the water once or twice will not create any new contaminant, nor will it remove chemical contaminants. It will only kill bacteria. 2. The difference between mineral/spring/tap water depends on your country's laws. Please refer

... (con't from above) to them especially for toddlers.
There is no such thing as a "miracle" water that everybody should drink, but it is clear that too much of highly mineralized water is unhealthy (once again especially for toddlers).
It is known that levels of nitrates and fluorides (not an exhaustive list) should be of particular concern, depending on the age.

Many public water systems spray water into the air or trickle over rocks (aeration) to let some dissolved gases bubble out (e.g. odorous $\ce{H2S}$). This also may make the water less acid and thereby precipitate dissolved manganese and iron.
When tap water treated with chlorine, bromine, ozone or other disinfectant is allowed to stand, open, then those chemicals also can escape.
However, I doubt anyone actually tastes dissolved oxygen (the human body can sense $\ce{CO2}$, but not oxygen, which is why miners kept canaries. In addition, one can taste the acid of $\ce{CO2}$ in soda water, and feel the bubbles bursting.)
So the practice of letting water stand and refrigerating is useful in improving flavor, but $\ce{O2}$ has little to do with it. Chilling water does increase the solubility of gases, though.


text:
What is difference between H⁺ and proton?
What is difference between $\ce{H+}$ and a proton?

Do you think there should be any difference?

There is no chemical difference, only a psychological one: how do you think about it. They are both the same thing, but many people associate $\ce{H+}$ ions with chemical reactions and protons with particle physics.
A hydrogen atom has one electron and a proton, no neutron. Therefore $\ce{H+}$ is just a proton.
That is why acids are sometimes referred as proton donors as they donate $\ce{H+}$ to a base (also known as proton acceptor).

Yes, and also H+ makes me think more about the solvated proton.  For instance, in chemical reactions where what shows up in equations as H+ is really H3O+, etc.  Using the word proton emphasizes the physics over chemistry, thus it would be odd to refer to proton NMR as H+ NMR, again, that would be 'correct' but confusing.

This answer is good and valid in the narrow (but common) case where $\ce{H}$ refers to $_1^{1}\ce{H}$ only (with $\ce{D}$ equalling $^2_1\ce{H}$ and $\ce{T}$ for $^3_1\ce{H}$). This means that in terms of molar mass, there is a difference between the case just outlined and the concept of $\ce{H}$ for hydrogen as an element with different isotopes.

In gaseous/plasmatic phase, there is no difference - $\mathrm{p}$ and $\ce{H+}$ are synonyms for a proton.
The former ( p, proton ) is more often used by physicists in subatomic particle context, the latter by chemists in hydrogen properties/behaviour context.
In polar solvents like water or liquid ammonia,  "naked" protons nor electrons cannot exist, but they are solvated, forming molecular ion or reacting with solution components.
$\ce{H+}$, frequently used in electrochemistry and ion equations, is the implied shortcut for hydrated proton aka hydronium $\ce{H3O+}$.
Strong enough acids form solid hydronium salts like e.g. hydronium perchlorate $\ce{[H3O+][ClO4-]}$.
There are consider also larger scale hydronium hydration structures
$\ce{H9O4+ = H3O+ \cdot 3 H2O}$
$\ce{H3O+ } \cdot  6\ \ce{H2O}$
$\ce{H3O+ } \cdot  20\ \ce{H2O}$,

I have not noticed until now the Q and 1st A are quite old, being among active question.:-)


text:
Assuming ideal gas behavior, how can I tell when the change in enthalpy is equal to the change in the inner energy?
If ideal gas behavior is assumed, for which of the following reactions does $\Delta H = \Delta U$?
\begin{align}
\tag{a} \ce{N2O4               &-> 2 NO2(g)}\\
\tag{b} \ce{CH4(g) + 2 O2(g)   &-> CO2(g) + 2 H2O (l)}\\
\tag{c} \ce{SO2(g) + 1/2 O2(g) &-> SO3(g)}\\
\tag{d} \ce{Br2(l) + 3 Cl2(g)  &-> 2 BrCl3(g)}\\
\tag{e} \ce{Cl2(g) + F2(g)     &-> 2 ClF (g)}\\
\end{align}
The correct answer is (e), and I just want to make sure I understand why.  For $\Delta U = \Delta H$, no work can be done as $\Delta U = H - W$ (where $W$ is work done by the gas).
This means that the total number of molecules can't change, otherwise a change in volume will occur, causing work to be done upon molecules or the molecules to do work.  Likewise, phase changes will also require work to be done on the molecules.  Thus, we rule out (a), (b), (c), and (d)?
I'm not sure if this is too simplistic or the correct way of thinking about this problem.

You got it right, and there is really nothing to add. Same number of molecules means no change in volume, hence no work is done, hence $\Delta H=\Delta U$.
On a side note, I find it hardly believable that burning of methane (b) will result in liquid water, but if the problem statement was given to you exactly like this, well, so be it.

The heat of reaction, by convention, is defined in terms of the amount of heat that needs to be added in bringing about the transition at constant temperature and pressure from a thermodynamic equilibrium state of pure products to a thermodynamic equilibrium state of pure reactants.  If the initial and final states happen to be at room temperature, then specifying liquid water for the products is perfectly reasonable.


text:
What happens when butane and methane both absorb 100 J of heat?

Two cylinders, one containing 1 mole of $\ce{C4H10}$ gas at $\pu{1 atm}$ and the other containing 1 mole of $\ce{CH4}$ gas at $\pu{1 atm}$, are at $\pu{288 K}$.  If each gas absorbs $\pu{100 J}$ of heat under conditions of constant volume, which of the following is true?
  (a) The temperature of the $\ce{CH4}$ increases more than the temperature of the temperature of the $\ce{C4H10}$.
  (b) The internal energy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
  (c) The heat capacity of the $\ce{C4H10}$ is less than the heat capacity of the $\ce{CH4}$
  (d) The entropy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
  (e) The heat transferred to the $\ce{C4H10}$ is greater than the heat transferred to the $\ce{CH4}$.

The correct answer is (a). I am trying to eliminate answers, but I seem to get stuck.
We know that (b) is clearly false as we've added heat and no work is done, therefore the internal energy for both gases must increase. Likewise, (d) is false as the entropy should increase as the molecules have more heat and become more disordered. Finally, (e) is incorrect as we transferred $\pu{100 J}$ of heat to both systems.
But here is where I get stuck. I can't seem to eliminate (c) from just the information given.

Well, c) is opposite to a) so you need to understand heat capacity.

@Mithoron Yes, (c) is opposite to (a), but short of looking up those values, I'm not sure how to rationalize which heat capacity is larger than the other.

First, consider this from a nice write-up on heat capacity:

The amount of energy needed to raise the temperature of a substance by one degree in temperature will depend upon how many different ways the atom or molecule can store the energy in the form of these thermal motions and how much energy each kind of thermal motion involves.In general, for molecules with similar bonding the bigger the molecule the more heat has to be added to raise its temperature by one degree.

Now, examine the degrees of freedom for each molecule: there are more ways to store energy in the larger butane molecule (with 36 vibrational modes in addition to the translational and rotational modes) than there are in methane (9 vibrational modes plus translational and rotational ones).
Therefore, we'd expect a greater increase in temperature for the smaller molecule than the larger one for an equivalent amount of heat added to the system.

It may be worth mentioning that at room temperature many of these vibrational modes are "frozen", so they don't really contribute to the heat capacity.

Not according to the equipartition theory and my experience modeling energy flow between vibrational modes in non-linear hydrocarbons (low-frequency bends and torsions come to mind as modes that will be populated at room temperature). While the mass difference manifests in varying translational and rotational sinks, there are plenty of vibrational modes that are populated above zero-point, not to mention Coriolis coupling (roto-vibrational mixing modes), which is certainly a non-negligible effect.

Well, I never said that all of them are frozen.

@IvanNeretin - I appreciate the comment, thanks!

To substantiate the existing qualitative answer, I made the corresponding numerical calculations using REFPROP – NIST Standard Reference Database 23, Version 9.0, which provide the following quantitative results.
$$\begin{array}{lll}\hline\text{Quantity}&\text{Unit}&\text{Methane}&&\text{Butane}\\&&\text{Initial}&\text{Final}&\text{Initial}&\text{Final}\\\hline\text{Temperature}\ T&\mathrm K&288.00&291.70&288.00&289.13\\\text{Pressure}\ p&\mathrm{atm}&1.0000&1.0129&1.0000&1.0043\\\text{Volume}\ V&\mathrm l&23.586&23.586& 22.831&22.831\\\text{Amount of substance}\ n&\mathrm{mol}&1.0000&1.0000&1.0000&1.0000\\\text{Internal energy}\ U&\mathrm{kJ}&11.847&11.947&33.150&33.250\\\text{Heat capacity}\ C_V&\mathrm{J/K}&27.050&27.178&88.677&88.944\\\hline\end{array}$$

A is the correct answer.
C is incorrect because butane is larger than methane. This means butane has more London Dispersion Forces than methane, and so needs to absorb more heat to gain the same amount of temperature.
A is correct for the same reason: given that both absorbed the same amount of energy, methane is going to have more temperature than butane because it has fewer intermolecular forces holding it together.

Sorry but your reasoning is wrong - not interactions but degrees of freedom are important here.


text:
Copper and hydrochloric acid
When hydrochloric acid is added to a sample of copper why does it look pink at first then form a pale blue solution ?

Well copper is pink...

The copper was not pure but contaminated with another substance.
Likely your sweaty oily hands.

If pure hydrochloric acid is added to a sample of pure copper, there is essentially no reaction; the solution should not turn blue.
At a guess, you're seeing the copper turn pink because it has a significant oxide layer that is reacting with the acid, forming copper (II) chloride and leaving bare the pinkish metal.  The solution could be pale blue or pale green, and the time delay of the color would be a result of the amount of time required for the reaction to occur.  Note that if this is the case, you would not expect to see any bubbling on the copper, as copper (II) oxide will react with hydrochloric acid to form copper (II) chloride and water.
As Ivan Neretin pointed out in comments, large pieces of copper will react with hydrochloric acid over a period of several hours to days in the presence of air.  Permeakra added that a fine copper powder will react in minutes with a solution of hydrochloric acid in the presence of air.  In both cases it is the addition of oxygen that causes the copper to react, and in both cases the solution will be a greenish yellow color due to the excess of hydrochloric acid required.
As an example, a student of mine recently asked about this with regard to a penny.  We cut open a modern US penny (95% zinc, 5% copper by mass) to expose the zinc core, and placed it in 200mL of 6M HCl.  The oxidation layer on the penny was gone almost immediatley after it entered the HCl solution.  The zinc reacted quickly and by the next day the penny had been converted into a hollow copper shell.  The shell sat in the remaining HCl solution (still 5.7M) over the weekend with no visible reaction but the solution slowly turned green and the copper dissolved completely by the following Friday.
If your solution is turning pink, then I'm not entirely sure what's going on.

In fact, copper would slowly dissolve in acid in the presence of air, even if initially there were no oxide layer at all.

@IvanNeretin Indeed, but this takes a loooong time before the reaction is noticeable, at least with the acid concentrations I've used for this (several days.)  Though kay_student doesn't specify a timeline, I assumed it was faster than that.  I thought about mentioning the oxidizing acid aspect of HCl and air, but decided against it for simplicity's sake.

@JasonPatterson This depends on the copper sample. Fine powder in excess of acid dissolves within minutes.

@permeakra Good point.  Edited to include additional information from previous comments.


text:
Why CCl4 does not dissolve in water while SiCl4 does?
Why $\ce{CCl4}$ does not dissolve in water while $\ce{SiCl4}$ does?
Does it have to do with d-orbitals? If yes, then how are they related?

$\ce{SiCl_4}$ does not quite dissolve in water; rather, it reacts with water. So does $\ce{CCl_4}$, albeit extremely slowly, so for most purposes one may safely assume it doesn't. The difference is mostly due to the atomic size of C and Si. Smaller С atom is completely blocked off by four bulky chlorine atoms, so the water molecule can't reach it. Larger Si is easier to attack.
Another factor is that silicon is a less electronegative element than carbon, so that the Si-Cl bonds will have more polar character.
Whether or not the d-orbitals of silicon contribute significantly to the transition state, is (IMHO) a question of secondary importance.

Similarly, while methane will burn in air, silane is pyrophoric, that is it will spontaneously burst in to flame (and then there is disilane!).

A silicon tetrachloride reacts with water, while carbon tetrachloride does not. This is due to the fact that the carbon does not have d-orbitals to accept lone pair of electron from water, while silicon has vacant d-orbitals to accept lone pair of electron from water.


text:
Master equation for catalyst and substrate binding?
I recently asked this on Biology.SE, and need to start with the master equation for the reaction:
$$\ce{E + S -> ES}$$
Where $\ce{E}$ is our enzyme and $\ce{S}$ the substrate to get a probability density function for the binding time of a substrate with one specific enzyme to be:
$$f{\left(t\right)}=k\left[\ce{S}\right]\mathrm e^{-k\left[\ce{S}\right]t}$$
I have read up on master equations, but really have no idea where to start in this derivation, any help would be appreciated?

It is not clear what system you want to discuss and what is the question.

The way to approach this is not how you have mentioned in your answer.
The representation of the master equation is correct and the generalized form of the chemical master equation (CME) [1] would be:
$$\frac{\partial P(x,t)}{\partial t}=\displaystyle\sum _{\mu=1}^M a_\mu(x-v_\mu)\ P(x-v_\mu,t) - a_\mu(x)\ P(x,t)$$
Where:

$x$ denotes the state of the chemical system (number of molecules of each species)
$\mu$ denotes a type of reaction
$a_\mu$ denotes the probability of happening of reaction-$\mu$ in a small interval (also called reaction propensity)
$v_\mu$ denotes the change in the state due to the reaction-$\mu$
$M$ denotes the total number of possible reactions

As far as I can understand (also from your question on Biology.SE), you are interested in knowing the distribution of the time interval when no reaction happens. A chemical reaction can be modelled as a Poisson process which is based on 3 main postulates:

$\lim\limits_{h\to0+}\dfrac{P(N_h=1)}h=\lambda$  i.e. the probability of occurrence of one event in a very small interval of time is equal to the macroscopic rate or intensity ($\lambda\,$).
$P(N_h\geqslant2)=o(h)$ i.e. the probability of occurrence of more than one events in an infinitesimal interval is essentially zero.
Consecutive events are independent of each other.

A chemical reaction seems to follow all these postulates. The detailed derivation of the CME can be found in [1]; I am trying to present a short derivation based on Poisson postulates.

The probability that system is in state $x$ at time $t+\Delta t$ would
  be the probability that system is in state $x-v_\mu$ at time $t$ times
  the probability that reaction $\mu$ happens in the interval $\Delta t$, or the probability that system is in state $x$ at time $t$ times probability that no reaction happens in the interval $\Delta t$.

$$P(x,t+\Delta t)=\sum_\mu P(x-v_\mu,t)\ P(\mu) + P(x,t)\left(1-\sum_\mu P(\mu) \right)$$
when you rearrange this equation, divide by $\Delta t$ and take the limit of $\Delta t \to 0$, you will end up with the CME. Note that, as per the first postulate: $$\lim\limits_{\Delta t\to 0}\frac{P(\mu)}{\Delta t}=a_\mu$$
Now, the probability that $k$ reactions would have happened in a time interval $\tau$ follows the Poisson distribution (see [2] for the derivation).
$$P(N=k)=\frac{(\lambda\tau)^k e^{-\lambda \tau}}{k!}$$
This is a well known property of the Poisson distribution that the time interval between reactions will follow exponential distribution. The probability that no reaction has happened in time interval $\tau$ would be:
$$ P(N=0)=\frac{(\lambda\tau)^0 e^{-\lambda \tau}}{0!}=e^{-\lambda\tau}$$
in this case, $\lambda=\displaystyle\sum _{\mu=1}^M a_\mu$ i.e. probability of any reaction.
Conversely, the probability that some reaction would happen in time $\tau$ would be $P(r\leq \tau) = 1-e^{-\lambda\tau}$. Where $r$ denotes the first reaction time. This is the cumulative distribution function for the exponential distribution.
In other words, the reaction time between intervals would follow an exponential distribution such that the probability that no reaction would happen will exponentially reduce with the size of the interval $\tau$.
In your specific case there are two reactions — association and dissociation, the propensities of whose will be:
$$a_f=\frac{k_f}{V}n_E.n_S\\[1em] a_r=k_r n_{ES}=k_r(n_{E_0}-n_E)$$
Where $k_f$ is the rate constant of association, $k_r$ is the rate constant for dissociation (as per mass action kinetics), $n_i$ denotes the number of molecules of species-$i$ and $V$ is the volume (of the reaction vessel). See [3] for details. Note that the amounts of the species change with time, and so will the reaction propensities. You can calculate the exact reaction time distribution when the system is at the deterministic steady state (using the steady state concentrations).
I think this answers your question on [biology.se] as well.

References:

Gillespie, Daniel T. "A rigorous derivation of the chemical master equation". Physica A: Statistical Mechanics and its Applications, 188 (1992): 404-425.
Hogg, Robert V., and Allen T. Craig. Introduction to mathematical statistics. New York: Macmillan, 1978.
Gillespie, Daniel T. "A general method for numerically simulating the stochastic time evolution of coupled chemical reactions." Journal of computational physics 22.4 (1976): 403-434.

You can use \displaystyle to force the "normal" behaviour in inline math, hence the formula will be bigger and you don't need explicit \limits - useful for any lists.

(Note I am the OP)
Master equation
The master equation can be written as:
$$\frac{\mathrm d p_n{\left(t\right)}}{\mathrm dt}=\sum_{n'}\left\{W_{nn'}p_{n'}{\left(t\right)}-W_{n'n}p_n{\left(t\right)}\right\}$$
Where $p_n$ denotes the probability that the system is in state $n$ at time $t$. And $W_{nn'}\Delta t$ denotes the probability of the system changing from state $n'$ to the state $n$ in the time interval $\Delta t$.
Specific case
Let us say we have $n=1$ free enzyme, then:
$$ \frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=W_{01}p_1{\left(t\right)}$$
So what is $W_{01}$? This is the probability that the one enzyme is used up, let us call this $\mu$.
$$\frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=\mu p_1{\left(t\right)}$$
The average number of 'free enzymes' is given by:
$$\left\langle n \right\rangle=p_1{\left(t\right)}$$
Differentiating this gives:
$$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=\frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}$$
But in our case,
$$ \frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}=-\mu p_1{\left(t\right)}$$
Thus
$$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu p_1{\left(t\right)}$$
$$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu\left\langle n \right\rangle$$
From here with the assumption that $\mu\propto [\ce{S}]$ you can follow through my derivation on Biology.SE
The reason for the assumption $\mu\propto [\ce{S}]$ can be explained by the fact that $\mu$ is related to the rate of the reaction, which increases (approximately) linearly with substrate concentration.  If this assumption is not valid, please can you comment.
Sources
1.'Stochastic processes in physics and chemistry' by N.G. Van Kampen chapter V

http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf


text:
Total pressure in a chamber calculation
I have a chamber containing two different gases and a liquid which needs to be heated at a certain temperature to bring it to vapor state.  I have worked out that we can calculate the total pressure of a mixture of gas using Dalton's law using partial pressure of each gas. To calculate total vapor pressure we have Raoult's law. If I have both gases and vapors in the medium can we calculate the total pressure by adding them both, as in a "collecting gases over water experiment"?

Okay, let me put it this way. If I know the partial pressure of 2 gases and vapor pressure of liquid turned vapor, can I directly add them to get total pressure in the chamber?

This not my homework question. I am an electronics engineer working on an interdisciplinary project wherein I need some information on mixture of gases. I have worked out that we can calculate the total pressure of a mixture of gas using Dalton's law using partial pressure of each gas. To calculate total vapor pressure we have Raoult's law. If I have both gases and vapors in the medium I think we can calculate the total pressure by adding the both like 'collecting gases over water experiment'. I read these things long ago so I wanted to confirm it.

Thank you for the clarification, could you please add this detail to the question

Dalton's Law is only true for ideal compounds, which theoretically have no intermolecular forces between their molecules. Those forces become more and more significant when increasing the concentration of any compound, because all molecules become packed up closer.
This will lower the pressure, because if any molecule is about to hit the container wall, it gets pulled backwards $-$ and thus slowed $-$ due to the attraction of the other molecules behind it. But for high temperature this effect is negligible, because they can't be slowed down so easily.
Now lets come to the calculation using Kay's rule:
Since you only have one liquid component, you don't need Roult's Law to calculate the vapor pressure for a mixture of liquids.
1.) You need to find the values for the critical temperatures and critical pressures for your components (gases and water vapor).
2.) You need to know the relative amounts of each substance:
$$y_i = \frac{n_i}{n_m}$$
$n_i$... amount of a specific component [mol]
$n_m$... total amount of substance of the mixture [mol]
3.) Calculate the pseudo-critical temperature and pressure:
$$T_c' = y_1 \times T_{c1} + y_2 \times T_{c2} + y_3 \times T_{c3} $$
$$P_c' = y_1\times P_{c1} +y_2\times P_{c2} +y_3\times P_{c3}$$
$T_c$... critical temperature of a specific component [K]
$P_c$... critical pressure of a specific component [Pa]
4.) calculate the pseudo-reduced values for T and the volume:
$$T_r = T / T_c'$$
$$v_r' = V / (R \times T_c' / P_c')$$
$T$... temperature in the container
$V$... volume of the gas phase [m$^3$]
$R$... universal gas constant = 8.314 J/(K mol)
5.) Now you need to look up $v_r$ and $T_r$ in a compressibility chart to get the value for a specific $Z$ factor, where their lines intersect.
6.) And finally calculate the pressure: $$P = ZRT / V$$
This will be more accurate than if you would use the standard way for ideal gases, especially for low temperatures
References:
real gases
calculation
compressibility chart examples


text:
How to separate sodium iodide dissolved in water?
Am fairly new to chemistry and apologize if the question and the description are too vague.If you think I lack basic background knowledge to deal with something like this, please point me to a resource, a book or anything that can help me in this regard. So in case this question is off-topic, please suggest where I should begin- at least starting point for my research will be helpful. However, in that case, if possible, please be specific, cause its too much of an effort to read entire chemistry textbooks to find out the answer to this one question.
Would like to know how to seperate NaI dissolved in water. Please suggest a way to do so without heating and evaporation. Like, what compound should I add to the solution so that the NaI can react with that resulting in solid products, maybe in the form of lumps/signicantly sized suspended particles that can be easily filtered out.
Additionally I would like to know how to find out, given a solvent (or any liquid compound), whether NaI dissolves in it or not. For eg: say that I have methanol, will NaI dissolve in it? For methanol i know the answer and its a yes. But how do i test solubility for any other compound. Of course chemists should not have to carry out practical experiments to find out this(it must probably be part of a known set of facts(about a particular compound) to chemists or they will be able to deduce it from some other known facts. How will a chemist answer this question?)
And finally, once I have any solution with NaI dissolved, how to separate it without evaporation(but by some chemical reaction). For eg: how to separate NaI from methanol? how to figure out the reactants I'll need in case the solvent is known?

Why are you so set against evaporation? It's by far the easiest method.

evaporation can be risky if the solvent is flammable or toxic. hence the issue

alright, if someone can describe that evaporation is safe, then am open to that idea.

You were talking of water solution. Water is neither flammable nor toxic. Will you take my word for it, or should I find the proper reference?

in the last part of the description, i have pondered upon what's to be done in case the solvent is methanol for instance. is it safe enough to carry out in a lab, if its methanol?

So you're basically asking about precipitation with another solvent added?

I summarize some answers given in comments.
Evaporation is really one of the standard methods in any combination of solid (cristalline, amorph, polar, unpolar) and solvent (polar, unpolar, flammable...).
But you must not think of it as the pot heated on a flame.
In modern synthesis you often already work with a nitrogen or noble gas atmosphere in a closed apparature and use varying pressure for controlled destillation.
So it is no problem to just reduce the pressure and evaporate your (organic) solvent.
In your special case you even can think of a pot heated on a flame, there is no risk and disadvantage. It is simply an effective method.
Do you know the concept of polarity? A good rule of thumb is that polar substances mix with polar ones but not with unpolar ones.
So water dissolves salt but not paraffin wax.
Benzene on the other side dissolves parrafin wax but no salt and water and benzene do not mix with themselves.
=================Edit (Addition of Information, because of question in comment)
So let's see how to regain the solvent.
First of all there is not only one boiling temperature. The boiling temperature is always a function of pressure. Approximately it can be described by the Clausius-Clapeyron-Equation.
$$
T_2(p_2)=(\ln(\frac{p_1}{p_2})k+\frac{1}{T_1})^{-1}
$$
Here $k$ is some constant depending on the specific solvent and $T_1$ and $p_1$ is a corresponding pair of known boiling temperature and pressure. For example $100^\circ$C and 1bar for water.
In general the boiling temperature goes down when pressure goes down. (Water boils faster in the Himalaya)
So if you have a typical organic solvent like aceton with a boiling temperature of $56^\circ$C under 1bar it evaporates already at $20^\circ$C when applying a pressure of e.g. 200mbar. (it is not the exact number)
But it also condenses again when cooling down. Using this you can make it liquid again and regain it. Compare this image of a typical distill.
http://www.ssc.education.ed.ac.uk/BSL/pictures/distill.jpg
Summarizing: you use a pump to get into your desired range of boiling temperature and cool the gas in the condenser again to a temperature below the boiling temperature. Depending on your system you have boiling temperatures below $0^\circ$C and cool with liquid nitrogen in the condenser or normal temperature range and cool with water or...

a little bit of an advanced answer for me at the moment. will need to learn more. one very important thing i forgot to mention is that its the solvent that am intending to recover and keep. not the NaI

May I ask if you are in school and what class. I can write more, but I do not want to be boring explaining things you already know  ;)

nope. am a working professional . software engineer. Wish i had paid more attention to my chem lectures when i was in school :-)

And I wish I will be interested into other subjects this way, when starting to work. ;)
Unfortanetly I only know good introductory textbooks in German. But you could look for something like chemistry for biology/medical... students. I think that you will learn polarity and other basic concepts more efficiently with such a textbook instead of forums. You will get a overall picture to ask the right questions.

So I will assume school knowledge in chemistry and advanced knowledge in maths for editing the answer.


text:
Why must both the critical temperature and pressure be exceeded to achieve the supercritical phase?
The characteristics and properties of supercritical fluids and the associated vapor-liquid critical point are well established.  One thing that I've always been curious about, though: what is the physical explanation for why it's necessary for both the critical temperature and critical pressure both to be exceeded simultaneously in order for this phase transformation to occur?
ChemWiki's description ("Critical Point") provides some measure of an explanation:

[A]bove the critical temperature, it is impossible to condense a gas into a liquid just by increasing the pressure. All you get is a highly compressed gas. The particles have too much energy for the intermolecular attractions to hold them together as a liquid. The critical temperature obviously varies from substance to substance and depends on the strength of the attractions between the particles. The stronger the intermolecular attractions, the higher the critical temperature.

Is there more to the physics than just "[t]he particles have too much energy for the intermolecular attractions to hold them together as a liquid"?
Per an early comment by Mithoron, the fact that increasing pressure on a supercritical fluid can eventually drive it to a solid phase would seem to imply that it's not quite that simple.  See, for example, the phase diagram of carbon dioxide(source):

Text about having too much energy is rather bad as supercritical water can be turned into ice with very high pressure.

There are no underlying physical reasons. This border is entirely arbitrary, much like the border between Nevada and Utah.

@IvanNeretin So, in effect, it's a definition of convenience? (If so, it explains why both the supercritical phase boundary and the Utah-Nevada border are both straight lines....)

Yes, I think it is.

The critical point is a phase transition of phase transitions.  It's a different beast than your everyday run-of-the-mill phase transition like water boiling to vapor or ice melting to water.
The diagram you included is very misleading in its use of color.  The "yellow" supercritical regime doesn't really exist.  Beyond the critical point, gas is indistinguishable from liquid.  So instead of having a special yellow region, the supercritical region should be some gradient of green (liquid) and pink/orange (gas).  The vapor pressure curve between the triple point and critical point is the curve along which liquid is in equilibrium with gas, i.e. those phases have the same molar free energy $G$.  However, they obviously differ in many other properties, including molar enthalpy $H$, molar entropy $S$, molar volume $V$, etc.  Moving along a trajectory in $PT$ space that crosses the vapor pressure curve results in a discontinuity of those values.  Thus, such a phase transition is a first-order phase transition, as first-order derivatives of $G$ (such as $S=\left (\frac{\partial G}{\partial T}\right )_{P}$ and $V=\left (\frac{\partial G}{\partial P}\right )_{T}$) are discontinuous through the transition.
As you move along the vapor pressure curve closer to the critical point, the size of the discontinuous jump in $S$, $H$, $V$, etc. gets smaller and smaller.  Finally, at the critical point, the discontinuity disappears.  So the existence of a first-order phase transition goes away, and no phase transition at all takes its place.
Say you were to start with liquid $\ce{CO2}$ at a $T$ slightly below and $P$ slightly above the critical point, and move through $PT$ space in a counterclockwise cyclic orbit around the critical point.  As you lowered $P$, there would be a distinct first-order phase transition to gas -- the $\ce{CO2}$ would boil -- but then as you reached the bottom of the circle and started increasing $P$ and $T$, moving past the critical point, there would be no phase transition at all as you went to the yellow region in your chart.  There would similarly be no phase transition as you reached the top of the circle and moved back into "green" territory.  You'd get liquid carbon dioxide again, and when you go again through the circular path through $PT$ space, it would boil -- a distinct first-order phase transition -- again.  But it would never condense, in the sense of a sudden change from gas to liquid.  The change back to liquid $\ce{CO2}$ would be gradual, with no discontinuities.
So to specifically answer your questions:

what is the physical explanation for why it's necessary for both the critical temperature and critical pressure both to be exceeded simultaneously in order for this phase transformation to occur?

The pseudo-fallacy in the question is using the phrase "phase transformation" to describe critical phenomena.  Passing the critical point is not a first-order phase transition like boiling water under normal conditions.  It's a much higher-order phase transition.  Here's a good chapter with more info.

The fact that increasing pressure on a supercritical fluid can eventually drive it to a solid phase would seem to imply that it's not quite that simple. See, for example, the phase diagram of carbon dioxide.

Think in terms of free energy landscapes.  Solid, liquid, and gas phases are at the bottom of potential wells or pits in this $G$ surface.  Past the critical point, the liquid "well" and the gas "well" are no longer distinct minima, but are the same.  However that doesn't change the fact that if you move "up" in the free energy landscape by increasing $P$, you will eventually find the still-distinct pit (free energy minimum) for solid.

Figured you'd have thoughts, Curt. :-)  Thank you. I think if I read this enough times, I might actually get it. Casting it in terms of free energy landscapes I think will really help.

Why must both the critical temperature and pressure be exceeded to achieve the supercritical phase?

Consider the four paths through the phase diagram.

Comparing (A) and (C), we are below the temperature of the critical point in (A), so an increase in pressure will liquify the gas. In (C), we are above, so increase the pressure does not lead to a gas/liquid two-phase system. Comparing (B) and (D),  we are below the pressure of the critical point in (B), so an increase in temperature will vaporize the liquid. In (D), we are above, so increase the pressure does not lead to a gas/liquid two-phase system.
You can imagine why we call the yellow region that of the supercritical fluid. We can make this state coming from a liquid in process (D) or coming from a gas in process (C), but we never see a second phase. So is it a liquid or a gas?

what is the physical explanation for why it's necessary for both the critical temperature and critical pressure both to be exceeded simultaneously in order for this phase transformation to occur?

As Curt F. stated in another answer, there is no phase transformation, it is just an area in the phase diagram that is not delineated by a phase change as you move into it or out of it unless you look at the phase boundary with the solid.


text:
Butler-Volmer equation with concentration dependence?
For a simple elementary reaction on electrode
\begin{equation}
O+e^-\rightleftharpoons R
\end{equation}
We can derive the Butler-Volmer equation. But it seems that the formula found on John Newman's Electrochemical System seems to be different from the one found on Allen Bard's Electrochemical Methods, both of which are considered classics. Below is my understanding, if not incorrect.
In the most general case, Newman writes
\begin{equation}
i = i_0\biggl[\exp\biggl(\frac{(1-\beta)nF}{RT}\eta_s\biggr)-\exp\biggl(-\frac{\beta nF}{RT}\eta_s\biggr) \biggr]
\end{equation}
where $\eta_s=V-U$. $U$ is the equilibrium which depends on the surface concentration, and so do $i_0$, the exchange current.
Bard, on the other hand, writes, in the most general case, the current-overpotential equation
\begin{equation}
i_n = i_0\biggl[\frac{C_O}{C_O^*}\exp\biggl(-\frac{\alpha nF}{RT}\eta_s\biggr)-\frac{C_R}{C_R^*}\exp\biggl(\frac{(1-\alpha) nF}{RT}\eta_s\biggr) \biggr]
\end{equation}
which only leads to Butler-Volmer equation if mass-transfer is not concerned. Here $\eta_s=E-E_{eq}$ (with its notation). $E_{eq}$ is the equilibrium potential established on bulk concentration, which is a constant taken to be initial condition. $i_0$ is dependent on the bulk concentration.
Aside from the anodic, cathodic sign convention different, what Bard has written explicitly depends on the surface concentration, which in a sense suggest that even if the over-potential is negative (which, in Bard's convention, leads to positive cathodic current), anodic reaction can still be established if $C_R$ is sufficiently high.
The more widely known equation (also found in Bockris's Modern Electrochemistry vol.2) is the one Newman wrote. But the equation will only lead to (in Newman's convention) positive anodic current if the over-potential $\eta_s$ is positive, even if $C_O$ dominates (which leads to cathodic current in Bard's formula).
It seems that the derivation differ on the reference potential: Newman sets equilibrium potential $U$ to be dependent on the surface concentration while Bard let it be referenced to the open-circuit potential with bulk concentration. What's right and wrong?

I just figure it out that both are right (of course!). It's just that the definition of over-potential $\eta_s$ is different. In the case of Newman's, positive $\eta_s$ will guarantee positive (anodic) current. Here $\eta_s=V-U=\phi_{metal}-\phi_{solution}-U$ is positive in the sense that $V>U$, where $U$ is the equilibrium potential corresponding to the surface concentration. Whereas in the case of Bard's, sometimes positive $\eta_s$ will lead to positive (cathodic) current if $C_O$ is sufficiently large (possibly with convention or precipitation). But here $\eta_s=V-E_{eq}$, and $E_{eq}$ is defined as equilibrium potential corresponding to bulk (initial) condition. So the value of $V$ here will result in negative $\eta_s$ in the case of Newman's
In short, it’s just that the sign of over-potential is different with their corresponding definition of equilibrium potential.

The difference between the approaches is also in the definition of the reaction rate.
Bard assumed the bulk concentration is the same as the surface concentration in equilibrium (Nernst equation). Newman did not use this assumption.

I think Bard's is more general than Newman's, Newman's assumes Cx=Cx* where Bard does not. Bard says the current is proportional to the forward rate, and if there are metal ions near the electrode, the forward reaction removes electrons from the electrode to reduce the ions. That means the current flows into the electrode. Bard then says the applied potential is E(electrode)-E(electrolyte) which would cause current to flow out of the electrode. Bard is not inconsistent, just kind of wierd.


text:
Is there any difference between de-Broglie wavelength and the wavelength of normal waves?
Is de-Broglie wavelength different from wavelength of waves such as electromagnetic waves?
If yes, they differ in what respect?
And can we use the formula of wavelength of light ($c=v\lambda$) to calculate the de-Broglie wavelength?

de-broglie wavelength is the wavelength associated with a particle with mass. h/p=de-broglie wavelength, where p is momentum, h is planck's constant.

duplicate of http://chemistry.stackexchange.com/questions/33460/why-replace-c-with-v-in-broglie-hypothesis ?

Is de-Broglie wavelength different from wavelength of waves such as
  electromagnetic waves?

I would atually say no, there is no difference, since the wavelength is purely mathematical characteristic of a wave, the distance over which the wave's shape repeats. There is, however, an important difference between de-Broglie waves on the one side and usual physical waves (mechanical & electromagnetic) on the other:

The later consist of periodic oscillations of a physical quantity, such as, for instance, the pressure for sound waves or electrical and magnetic fields for electromagnetic waves.
The former consist of periodic oscillations of a mathematical quantity, the probability of finding a particle at some location.


text:
What is the known cheapest chemical solution to extract CO2 from air on industrial scale?
As far as I know, the cheapest non-chemical solution is: grow grass, and burn it. The resulting $\ce{CO2}$ is from the air. But I am interested on chemical industrial solutions now.
As far as I know, the main difficulty of the process is that the ratio of the $\ce{CO2}$ in the air is very low. Thus, extracting it means a large entropy difference.
My layman's intuition is that maybe some substance exists - probably an organic compound - which binds especially strongly to $\ce{CO2}$ and it is easily regenerated. One would pump air through it, and then regenerate it to get the $\ce{CO2}$ back. If such a compound exists, what is it?
The process should be powered by electrical energy only. Thus, solutions such as "buy gas, burn it, and you get much cheaper $\ce{CO2}$" are not acceptable.
On "cheapest" I understand, what is cheapest in industrial size, not in a laboratory? (for example, in a large chemical factory)

Surprisingly, the cheapest solution might be refrigeration. Water vapor will freeze out at $0\ ^\circ\mathrm{C}$ and the next major component of air to freeze out will be carbon dioxide at $-78\ ^\circ\mathrm{C}$.

@BenNorris The ratio of CO2 in air is 0.035%, what is the vapor pressure of -78% CO2? Ok, maybe there is an optimum which depends on the cost of the freezing (and the heat exchanger), where the extracted CO2/$ is maximal.

Your intuition is correct, and such compounds are plenty. The cheapest are probably $\ce{NaOH}$, $\ce{Ca(OH)_2}$ and the like, but these are not easily regenerable. Then there are organic amines, like monoethanolamine, which can do the same in a reversible manner. Indeed, they are used in industry.

This is right. Organic amines are what I've heard the most about for recovering CO2 from gas streams.


text:
Crystal field splitting pattern with strong field and weak field ligands
What will be crystal field splitting pattern for octahedral complexes if some of the ligands are strong and some are weak field ligands?

Technically this will no longer be an octahedral splitting pattern, making the results much more complex.

@Jonathon I think it would be valuable to make this an answer.

Well, the split would depend on:
The identity of the metal ion as well as the identity of the ligands themselves. You'd have to consult a spectrochemical series to determine the magnitude of the split caused by the ligands.
Alternatively, you could beam white light through the complex and using the light that appears, determine the wavelength of the light absorbed (the complementary colour) to promote electron(s) from the lower split d orbital to a higher one... This would give you more information as to the magnitude of the split (sometimes, the experimental value will provide an answer that theoretically cannot be determined).


text:
What does the acid do in this reaction?

  Which of the following is the product of the series of reactions shown above?

The correct answer is (C).
So the first step of the first reaction meant to create an enolate (deprotonating the hydrogen sandwiched between the two carbonyls), which upon addition of the substrate, forms:
.
I think the second step, potassium hydroxide is meant to hydrolyze the ester to a carboxylic acid.
Finally, the third step is meant to decarboxylate the acid by emitting $\ce{CO2}$.
However, I'm unsure as to what role the acid plays in the last step.  The only thing I could think of is that the acid helps the enol (the actual product from decarboxylation) tautomerize, but I don't think acid needs to be present for tautomerization.

My guess is that the acid is just there to protonate the carboxylate salt and allow decarboxylation. Hydrolysis of the ester with $\ce{KOH}$ generates the potassium carboxylate salt:
Since the mechanism for thermal decarboxylation involves a cyclic transition state including the carboxyl $\ce{-OH}$ group, I expect that the carboxylate salt must be protonated before decarboxylation can take place, hence the addition of aqueous acid.


text:
what makes an acid a strong acid or weak acid?
When my teacher referenced nitric acid as a strong acid, I thought it was strong because it donates more than one H+ per molecule, but then I saw the equation is $\ce{HNO_3}$ and it only donates one. Acetic acid is also monoprotic, but it is considered a weak acid. What makes an acid strong or weak? If an acid donates more than one proton, does it make it stronger? Feel free to make any formatting changes to formulas as well.

I recall at least 2 definitions from school: (1) the dividing line between weak vs. strong acids is at p$K_{\rm a}$ = 0, with weak ones having p$K_{\rm a}$ > 0 and strong ones with p$K_{\rm a}$ < 0; and (2) an arbitrary cutoff defined by $\ce{HF}$ as the weak acid, and anything with a p$K_{\rm a}$ < the p$K_{\rm a}$ of $\ce{HF}$ being a strong acid. I have no idea where (2) came from but I recall it.

I feel it would be appropriate to advice the topic starter to read something (there are quite a few reliable sources out there, Wiki is as good as any other), and then come back and ask what's left unclear.

related http://chemistry.stackexchange.com/questions/19799/strong-and-weak-acid, http://chemistry.stackexchange.com/questions/5155/what-is-a-strong-acid


text:
Structure of KC8

Graphite reacts with potassium to produce a compound with the empirical formula $\ce{KC8}$. Of the following, which is the best description of this compound's structure?
  (a) $\ce{K+}$ ions close packed with polyhedral $\ce{C8-}$ ions;
  (b) $\ce{K-}$ ions close packed with polyhedral $\ce{C8+}$ ions;
  (c) $\ce{K+}$ ions packed with $\ce{C2^2-}$ ions;
  (d) Negatively charged hexagonal carbon layers with intercalated $\ce{K+}$ ions between them;
  (e) An expanded diamond lattice with $\ce{K+}$ ions in tetrahedral holes.

The correct answer is (d). I'm not very strong in inorganic chemistry, so I'm unsure if there is a way to deduce this answer.
The only thing I'm sure of is that it's not (b), since it seems very unlikely that potassium would have a negative charge. But I don't know why (a), (c), or (e) are all false (and why (d) is true).
Any help would be greatly appreciated.

Do you know the structure of graphite? Knowing this it would be rather obvious

This question is kind of "illogical", in that the answer can't quite be derived by logic alone; you either know it, or you don't. Well, of course you are right about (b). Also, it's not (c) either, because if we balance the charges to get a neutral compound, it would be $\ce{K_2C_2}$, and that's not what we want (though this thing does exist, too). Then what?
Graphite is known for its ability to intercalate pretty much anything, as its layers are held together only weakly. Diamond is not like that. Also, changing from graphite to anything else would require some rearrangement of covalent bonds, and that's not going to be easy (think of the harsh conditions needed to make diamond from graphite). So (d) is quite likely, while (a) and (e) are not. That's just about it.
All that being said, carbon is a surprisingly versatile element (think of organic chemistry), so I'd rather avoid making too general conclusions. Suppose someone brings you a sample; you analyze it and see that it contains just K and C, in molar ratio 1:10, so you may say it's $\ce{KC}_{10}$. What's the structure of it? The problem is almost similar to your original question, except they don't tell you they obtained it from graphite, but it looks kinda like graphite anyway (a fine black powder), so shouldn't the logic be pretty much the same? What would you expect? Another graphite intercalation compound like your (d), only having slightly less potassium? Yeah, sure.

Graphite is the most stable form of carbon, and forms as sheets of hexagonal carbon networks, with sheets that are held together by vanderwaal forces.  One set of electrons from each carbon is delocalized to the plane between sheets.  Because of this delocalization, it is able to stabilize a negative charge relatively easily (apparently an extra electron for every 8 carbons).
The others are incorrect, because they are forcing the carbon to adopt a less stable form, which is energetically unfavorable.

Nearly all organic chemistry is energetically unfavorable in the same sense, so what? Diamond, for that matter, is indeed less stable; does that make it non-existent?


text:
Does adding lemon juice to water chemically alter its nutrients and thereby negating any of its health benefits?
An acquaintance who apparently has some chemistry understanding told me that eating lemon juice freshly squeezed to drinking water negates all of its health benefits because the water chemically alters the vitamins minerals and enzymes. I am very dubious because a lemon is 82% water and a human body is 75% water and of course lemon juice must be to a very great extent just water.
Also lemon juice is renowned as being a great morning drink because of its health benefits.
My acquaintance however emphatically stands by his statement.
Please clarify this situation for me.
Does water really chemically alter the nutrients in lemon juice rendering a lemon drink of no health benefit whatsoever other than the obvious water base?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

It seems this guy is making a fool of you, only thing I can think of is oxidating wit. C by chlorine in water.

This question is suitable for [skeptics.se] or [health.se]

Does water really chemically alter the nutrients in lemon juice rendering a lemon drink of no health benefit whatsoever other than the obvious water base?

No, of course not. Otherwise, humanity would have succumbed to scurvy a long time ago.
I encourage you and your friend to visit PubMed where you can peruse hundreds of refereed abstracts of studies that illustrate that nutrients from citrus fruits are bioavailable to humans and other species and necessarily require administration with water.
Here is a one-pager on lemon juice from the Cleveland Clinic, which is probably a reliable enough source in this case if you want to skip the more technical items at PubMed.