text:
Why fossil fuel is too dangerous?
This seems stupid question, but from many years I am studying that burning/combustion fossil fuel cause pollution because it releases $\ce{CO2}$.
I have recently read about $\ce{CFC}$ here, which is more dangerous than $\ce{CO2}$ and seems it is not release by burning of fossil fuel then why are scientist more concerned about fossil fuel instead of $\ce{CFC}$?
I want proper explanation Why fossil fuel is too dangerous?

Fossil fuels are burned at the rate of billion of barrels a day. The production of the more hazardous CFC's has been banned. See http://en.wikipedia.org/wiki/Chlorofluorocarbon

Why fossil fuel is too dangerous?

Combustion of fossil fuel does not only release $\ce{CO2}$ but many other harmful substances(pollutants) like oxides of sulphur ($\ce{SO_{x}}$), oxides of nitrogen($\ce{NO_{x}}$), Carbon Monoxide($\ce{CO}$), etc...
oxides of sulphur ($\ce{SO_{x}}$):
When sulphur containing fossil fuel is burnt, then oxides of sulphur is produced. Sulphur Dioxide($\ce{SO2}$) is generally produced.
Low concentration of $\ce{SO2}$ causes diseases like asthma, swelling and irritation of respiratory tract.
High concentration $\ce{SO2}$ leads to stiffness of flower buds, which eventually fall off!
Oxides of Nitrogen($\ce{NO_{x}}$):
When in automobile engine, fossil fuel is burnt at high temperature significant amount of nitric oxide($\ce{NO}$) and nitrogen dioxide($\ce{NO2}$) is released.
$\ce{NO2}$ is lung irritant. it is highly toxic to living tissue.
$\ce{NO2}$ decrease the rate of photosynthesis, cause respiratory diseases in children, etc.
Carbon monoxide($\ce{CO}$):
$\ce{CO}$ is produced due to incomplete combustion of carbon. It is major constituent in smoke of vehicles.
$\ce{CO}$ binds with haemoglobin and form carboxyhemoglobin, which is 300 times more stable than usual oxygen-haemoglobin! If concentration of carboxyhemoglobin reaches 3 to 4% then oxygen carrying capacity of blood reduces.
Few more!
$\ce{SO2}$ with smoke and fog is responsible for classical smog(London smog)
$\ce{NO2}$ and $\ce{NO}$ is responsible foe photochemical smog(Los-Angeles smog)
Update:
I was reading more about environmental chemistry and i discovered that $\ce{NO_{x}}$ is responsible for 40% depletion  of concentration of Ozone layer!

why are scientist more concerned about fossil fuel instead of CFC?

As commented by LDC3 The production of the more hazardous CFC's has been banned.
There are 95 derivatives of chloro and bromo fluorocarbon which are listed as ODS(Ozone Depleting Substances).
On 16th September, 1987 93 countries sign Montreal Protocol and accept not to use ODS.

Reference: GSEB textbook pdf

Don't forget that even though other gases may be better greenhouse gases, their concentrations are so low compared to that of water and CO2, they're not significant contributors to the greenhouse effect.

@MichaelD.M.Dryden Yes are not in significant amount. But I think OP wanted to know about other harmful substances.

Could you cite some of the sources for your numbers that you use?

@jonsca I m afraid, this is from my textbook. But I will try to find references.

@Freddy Including the title and author and year of publication is fine.  Ideally we should use the ACS format for citations, but it's not absolutely necessary

@jonsca Sorry for delay. But i have added pdf link of book. :)


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Is the condenser for a Soxhlet extraction left open on top
My question is about the condenser used in a Soxhlet extractions. It's unusual looking compared to the rest of condensers, used in distillation for example, I've noticed it has a thin and short tube on top and my question is, do one leave it open when performing a Soxhlet extraction or should it be stopped with a stopper or something?

The condenser is open at the top. That is why it is important not to leave the apparatus unwatched. There have been many fires started from the ether overflowing and being lit from the heater.

First thing, you should NEVER perform any heated reaction with a stoppered condenser. You shouldn't even be allowed to work independently in the lab if this rule is not coming automatic to you. I am not sure what you mean by a special condenser being used in Soxhlet extractions. You can use any condenser that fits. Since you don't do a distillation you can use a shorter condenser so your whole Soxhlet set-up has a reasonable height.

Very true, one thing I have noticed with soxhlet extractions is that when the liquid drops into the flask at the bottom it can suddenly cool down the flask at the bottom. This will make the vapour / gas volume inside the glassware greatly contract. So the volume of gas which enters and leaves via the top of the condensor can be very large compared to the amount which goes in and out of a condensor being used to reflux something in a round bottomed flask. As a result it is even more important not to heat a sealed soxhlet extraction than a "reflux".

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Will this beta-branched secondary alkyl halide undergo SN1 or SN2?

What is the correct mechanism here? Would SN1 occur here? I chose SN2 because we have a secondary alkyl halide with a weak base $\ce{CN-}$.
However, the solution says SN1 is major. Can you explain why?

My friend got full credit for this:

Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.
Full mechanism:

I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.
I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.

Jerepierre, How could I have known SN1 would have won? Solvent effet and branching at beta makes the situation a bit ambiguous and when we see CN-, we are tempted to do SN2?

You make a hydride shift from the tertiary carbon.

Murpy, TJ, J. Chem. Ed. 2009, 86, 519-524 shows that secondary alkyl halides do not undergo SN1 reactions.
This is counter to what we have been (and still are in many cases) teaching.

Perhaps you could give some more details about the paper in question and ideally provide a link to it. At the moment this seems more like a comment than an answer.

It is in $2^\circ$ position, so both, $\ce{S_{N}1}$ and $\ce{S_{N}2}$ can happen.
$\ce{Br}$ is a good leaving group probably the third best in the groups very common(depends on opinion, but it is a good one anyways though):
$\ce{OTs>I>Br>Cl>...}$ but LG accelerates both reactions
$\ce{CN-}$ is a very good nucleophile in the list of common ones(depends on opinion, but it is a good one anyways though): $\ce{PhS- > CN- > I- > ...}$, the second best so $\ce{S_{N}2}$ is favourable.
The solvent is water, which is polar protic which highly favours $\ce{S_{N}1}$
The temperature is normal so there's no elimination
Finally  fourth point overthrows thirds and moreover water reduces nucleophilicity of $\ce{CN^-}$

I think the cyanide won't be a good nucleophile because it will react with the water molecules to form a hydrogen cyanide. this will make the cyanide weak so there is no way the cyanide can display the bromide. therefore a strong nucleophile is required

Although HCN is a weak acid, in water, $\ce CN^-$ will stay as is because if  it takes $\ce H^+$ from water, the net equilibrium constant of the reaction will be $\ce  {K_w}/{K_a} $ = $\ce 10^{10} * 10^{-16}$ = $\ce 10^{-6}$

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Predict the Product(s) of a reaction of tertiary benzylic alkyl halide

I could find two products:
The first one where is $\ce{Cl}$ is replaced by $\ce{CN}$.
AND the second one where $\ce{CH3CH2O}$ replaces $\ce{Cl}$.
Did I make a mistake or miss anyone?

What about elimination?

Ok, so there would be another two that are alkenes.

OUPS, three others.

$\ce{CN-}$ is a significantly stronger nucleophile than $\ce{EtOH}$ and will therefore primarily displace the chloride. The facts that substitution takes place at a tertiary benzylic carbon, and that a polar protic solvent (ethanol) is used indicate that the reaction follows the $\ce{S_{N}1}$ mechanism (sources: 1, 2, 3). Due to the achiral intermediate (carbocation), the nitrile will be obtained as a racemic mixture, i.e. 2 products.
Elimination is also a possible reaction pathway which needs to be considered, since cyanide is a good nucleophile but a weak base, which favors an E2 reaction unless an excess of $\ce{CN-}$ is present. In this case, the product with the more substituted double bond should be favored.

nucleophilicity $CN->I->EtO-$ which implies cyanide attacks, no heating means no elimination generally


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Inductive effect and hyperconjugation - one elephant, different parts?
Are the inductive effect and hyperconjugation both different ways of looking at the same phenomenon - i.e. methyl groups donate negative charge for example. Inductively we can argue that carbon is of greater electronegativity than hydrogen, and so carbon withdraws negative charge from hydrogen, and thus negative charge is "donated" to whatever that the methyl is attached to.
From a hyperconjugative effect we can argue that there is some interaction between the C-H orbitals and some antibonding orbital ... how is this supposed to be stabilizing if it involves an antibonding orbital?

Related: Does hyperconjugation and induction always have the same effect?

Are the inductive effect and hyperconjugation both different ways of
  looking at the same phenomenon

I think so, both inductive and resonance (that's what hyperconjugation is after all) effects move electrons around.  I tend to separate the two effects as follows:

inductive effects are associated with electron movement through sigma bonds due to electronegativity differences
resonance effects are associated with electron movement through p orbitals due to electronegativity differences

For example, we often say that a methyl group stabilizes a double bond (carbon-carbon or carbonyl) or a carbocation.  The methyl group is roughly $\ce{sp^3}$ hybridized while the double bond carbon or carbocation carbon is roughly $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character.  Therefore electrons will generally prefer to flow from $\ce{sp^3}$ to $\ce{sp^2}$ orbitals.  The direction of this electron flow also indicates  that $\ce{sp^2}$ orbitals are more electronegative than $\ce{sp^3}$ orbitals.  How can the $\ce{sp^3}$ methyl group shift electrons to the lower energy, more electronegative $\ce{sp^2}$ carbon?  It can do it inductively (through sigma bonds) or through resonance (through p orbitals).
I've added a picture showing resonance structures involving hyperconjugation, just to be clear as to what hyperconjugation involves.

why would increased electron density in a double bond stabilize it?

see this link

Does this mean that the inductive effect of $-CH_3$ won't take place if there's no chance for hyperconjugation to happen?

@AumkaarPranav  No, as I note above inductive effects are transmitted through the sigma skeleton and resonance effects (including hyperconjugation) are transmitted through the pi framework. So inductive effects can still occur when resonance effects cannot.  Resonance and inductive effects both move electrons around - that is their similarity, but through different types of orbitals (pi and sigma respectively).  Hope this helps.

Inductive effect and hyperconjugation are two distinct phenomena.
Inductive effects are purely through sigma bonds. For example, in an organofluorine compound, the fluorine withdraws electrons from carbon by not sharing the electrons of the sigma bond equally. We explain this using the rationale that fluorine is more electronegative than carbon.
Similarly, in a carbocation, the electron-deficient carbon is more electronegative than any substituent carbons. Thus, the carbocation will withdraw electrons from substituents through the sigma bonds.
Hyperconjugation is electron donation of a filled bonding orbital to a nearby low-lying, unfilled orbital. Considering t-butyl cation as an example, a filled C-H bonding orbital can donate electrons to the empty p-orbital on the electron-deficient carbon. From a molecular orbital view, this essentially serves to lower the energy of the C-H sigma orbital and raise the energy of the p-orbital. With only two electrons in this system, the effect is stabilizing.

Notice that this means there are two effects that explain why greater substitution increases carbocation stability: inductive effect and hyperconjugation.
Hyperconjugation can also be invoked for the substitution of alkenes. In this case, the filled C-H sigma orbital overlaps with the empty C-C pi-antibonding orbital.

A good rule of thumb is that a filled orbital overlapping with an unfilled orbital is a stabilizing interaction.

How does overlapping with an anti bonding orbital lend something to stability?

@Dissenter When two orbitals (of any type) are combined, two new orbitals are generated, one of lower energy and one of higher energy. If there are only two electrons available to fill the two new orbitals, they will go into an orbital that is lower in energy than either of the two originals. So if we overlap an empty anti-bonding orbital with a filled orbital,a higher energy anti-bonding orbital will result. But that orbital is empty anyway. The important part is that the electrons can be placed in a lower energy orbital.

@Dissenter Perhaps you are thinking of a situation where the end result places electrons in an anti-bonding orbital. That would be destabilizing.

@Dissenter Keep in mind that many reactions require that a filled orbital (i.e. nucleophile) overlap with an unfilled orbital (i.e. electrophile) which is often an anti-bonding orbital.

Right ... The c-c anti bonding orbital is only antibonding with respect to the alkene.

And being able to delocalize electrons from one atom to another atom is inherently stabilizing ...


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Carbocation Rearrangements in SN1 reactions

Would one get 1-(1-chloroethyl)-1-methylcyclobutane?
I am not sure if the driving force identification is correct? Is my going from a 2° to a 3° carbocation correct? Does anyone have any tips for knowing when to look out for rearrangements?

I think the solvolysis would involve rearrangement to the cyclopentyl carbocation as indicated in the figure below.  The driving force for the rearrangement is two-fold

without rearrangement, the cyclobutyl compound would generate a secondary carbocation; with rearrangement a more stable tertiary carbocation is generated instead
release of strain as the 4-membered ring opens to a 5-membered ring

The tertiary cyclopentyl carbocation ultimately captures a solvent molecule as indicated.

the + charge goes onto the cyclic ring. Does that destabilize the ring more than having a 4-membered ring?

An $\ce{sp^2}$ carbon would like to be around 120 degrees; that angle is not destabilizing in 5-membered or larger rings.


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Is resonance always stabilizing
Does resonance always stabilize a molecule or can resonance effects destabilize molecules?
The only example I can think of in which resonance (more accurately, conjugation) destabilizes molecules is in the cases of anti-aromatic compounds. Is this a valid example, and are there any more such examples?

If resonance is not stabilizing, then why would it occur?

Couldn't you say that about a lot of things? If anti-aromatic compounds were unstable, why do they exist? I think the answer to your question lies with meta-stability.

What I was trying to point out is stated by Ron, Resonance only occurs when the result is beneficial. Anti-aromatic compounds exist since the molecular bonding releases energy from the state of individual atoms.

Resonance only occurs when the result is beneficial, that is when a more-stabilized molecule will result. If resonance (delocalization) occurred in a compound such as cyclobutadiene to produce a square molecule with equivalent bond lengths, an antiaromatic molecule would be the result.  Therefore, resonance does not occur and a square molecule with equivalent bond lengths is not produced.

As you say that a square molecule with equivalent bond length will not result therefore the rectangular molecule will be called anti aromatic or non aromatic?

The double bonds will be localized, the molecule will be non-aromatic.

Yes, resonance can definitely destabilize.
One example you can think of is destabilization of carbanions, which is something we often consider when trying to determine the relative acidities of some given species.
Consider the relative acidities of o-Methoxytoluene, m-Methoxytoluene and p-Methoxytoluene. Destabilization of their respective conjugate bases due to resonance plays an important role here, as mentioned in the renowned book Solomons|Frhyle|Snyder.

As clear in the attached figure, +R effect, where a substituent increases the electron density in a ring, destabilizes the conjugate base. Hence, the order of relative acidities here is: B > A > C
I encourage the reader to try and draw the resonance canonical forms for the three species mentioned, and realize why this happens for themselves.

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SN2 vs SN1 in Acidic Environments for Secondary Alkyl Halides
I have researched three books: Brown's, Solomons' and Wade's. I could not find enough information about this.
My prof. says that secondary alkly halides undergo SN1 only under acidic environments.
We had this quiz. I got a terrible mark on one question.

I think these are the missing structures.

Now, I am wondering, did Sn2 occur here, if yes, at what rate? Was it more important than Sn1? What was the major pathway?
What is the major pathway for all secondary alkyl halides with acidic environment and with NONacidic environments?
Could anyone also just check my missing structures?

It is an SN1 reaction. Your major error was forgetting the double bond in the product.

Jerepierre, yes I forgot the double bonds on my exam.
BUT, there is also the fact that there can be carbocation rearrangement. The double bond moves away. (I actually made a mistake on the uploaded images too. The double bond must be on the adjacent carbon.)

Note that the starting material is drawn as a chiral molecule.  Solvolysis of this starting material will generate a symmetric, achiral, allylic carbocation.  Since a stabilized allylic carbocation is generated, the mechanism likely proceeds through an $\ce{S_{N}1}$ mechanism.
Both ends (there is positive charge delocalized at both ends of an allylic carbocation - draw the resonance structures and convince yourself)  and both faces (top and bottom) of the allylic carbocation can capture the nucleophilic solvent molecule to restore neutrality as shown below. This results in the formation of two diastereomers of the 3,5-disubstituted cyclohexene - the cis and tran isomers. For each diastereomer, both enantiomers are also produced, as they must be, having passed through an achiral intermediate.

Ron, could you please elaborate as to why SN2 is not major?

Normally solvolysis of a secondary halide can proceed through SN1 and\or SN2.  Stabilizing the incipient ion further, as in this case through conjugation, shifts the balance towards SN1.  Even if an SN2 component remains, it would be accompanied by an SN2' pathway as well.  So again all 4 compounds would be formed, they just might have an excess of one enantiomer over the other.


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How do chemical reactions taking place in an alkaline battery produce electrons?
I'd like to understand the chemical reaction in a typical alkaline battery a little better and how that produces a flow of electrons.  I read wiki, how stuff works, and some more advanced articles.  They all seemed to either stop short of explaining the nitty gritty details, or went straight to chemical formulas.
I understand that a chemical reaction occurs between the cathode and the anode and that the electrolyte helps make that happen.  I don't really understand the flow of electrons though, and how the anode/cathode are consumed.  I read a bit about ions but not sure I grasped how that was moving electrons about.
Also, does this reaction start right away or only after a circuit is completed and electrons can flow?  Or has the reaction already started, and there is an excess of electrons on one plate that only start flowing after the circuit is completed?
What forces the potential to stop at 1.5V?

An alkaline battery is a specific type of galvanic cell.  As I am sure you found in your research, it involves an oxidation-reduction reaction between zinc($\ce{Zn}$), and manganese(IV) oxide ($\ce{MnO2}$.)  They are called alkaline batteries because these chemicals are surrounded by an alkaline (basic) solution of potassium hydroxide, $\ce{KOH}$.
Your question is really one that is more about galvanic cells in general.  When an oxidation-reduction (a.k.a. redox) reaction occurs, electrons are transferred between species.  This sort of reaction is exceedingly common.  That's all that happens in a battery, it's just that people have figured out how to catch the electrons on their way from species A to species B and use their energy to do work.
If I took a piece of zinc and placed it in a slurry of $\ce{MnO2}$ and $\ce{KOH}$, it would react according to the following balanced chemical equation and produce a lot of heat.
$\ce{Zn + 2MnO2 <-> ZnO + Mn2O3}$
Suppose that I separated the two pieces of the reaction into their own jars.  A piece of zinc placed in a solution of $\ce{KOH}$ in one jar and some $\ce{MnO2}$ mixed with $\ce{KOH}$ in another.  The reactants are all present, but they have no way of interacting, so nothing would happen - there's no way for electrons to move from the zinc to the $\ce{MnO2}$.
Now I attach a piece of wire to the zinc on one end and into the $\ce{MnO2 - KOH}$ slurry at the other. Now electrons can move from one jar to the other.  If I have a voltmeter that can measure a very brief signal, I can actually see a quick pulse of electrons flowing through the wire, from the zinc to the $\ce{MnO2}$.  The pulse ends almost immediately though, because the movement of electrons away from the zinc and toward the $\ce{MnO2}$ creates a charge separation.  The former becomes positively charged and the latter negatively charged.  Continuing to move electrons against this charge separation is progressively more difficult, and it doesn't take long for the electrostatic force of the charge separation to balance out the electrochemical potential of the reaction and the entire thing just stops.
To fix this problem, I need a way for the charges to balance back out without allowing the reactants to mix directly.  This is done with some sort of salt bridge.  A salt bridge allows positive and negative ions to move between the jars, but it still keeps the jars separate from one another.  This could be as fancy as an actual salt bridge (a glass tube filled with a gel containing some sort of salt) or as simple as a piece of paper soaked in the $\ce{KOH}$ solution.  Now when I connect the wire, electrons flow from the zinc to the $\ce{MnO2}$, negatively charged $\ce{OH-}$ ions move through the salt bridge toward the zinc, and positively charged $\ce{K+}$ ions move through the salt bridge toward the $\ce{MnO2}$.  The charge stays balanced in both jars, so there is no electrostatic force preventing electrons from moving, and they flow freely through the wire.  If I put a motor in the middle of the wire, it turns.
Specifics of Alkaline Batteries
That all was regarding galvanic cells in general.  A battery has additional requirements.  It needs to be compact and safe, in particular.  The construction of an alkaline battery involves a zinc core with a paste of manganese(IV) oxide and potassium hydroxide wrapped around it.  The paste and zinc are separated by a paper sheet (the salt bridge in this galvanic cell.)  Because all of the reactants and products are solid, there is no concern that they will diffuse through the cardboard; it just keeps the reactants in their places.
The reaction does not start until the battery is put into a device and the device is turned on.  Flipping that switch gives the electrons from the zinc a pathway to reach the manganese(IV) oxide.  Until the circuit is complete, there is no buildup of electrons at one end of the battery.
Electrical potential (voltage) is the ratio of the energy carried by an object to that object's charge.  One electron in this system has a potential of 1.43V, two electrons carry twice as much energy, but they also have twice the charge, so they also have a potential of 1.43V.  Electrical potential is sometimes called electromotive force, and that idea, that we're measuring how hard the electrons are being pushed through the wire, may help you understand the battery's behavior better.
Why 1.43V though?  It's because of the properties of the chemical reactions the battery is based on.  You probably know that if you burn some amount of propane it will produce a given amount of heat, and if you burn an equal amount of acetylene, it will produce a different amount of heat.  The same thing is true here.  The energy associated with the half reactions in each part of the battery depends on the chemicals involved.  Rather than list a total amount of energy available, batteries list the energy per charge, voltage.  This battery's half reactions are:
$\ce{Zn + 2OH- -> ZnO + H2O + 2e-}$  E°=1.28V
and
$\ce{2MnO2 + H2O + 2e- -> Mn2O3 + 2OH-}$ E°=0.15V
Each half reaction gives a specific amount of energy to the electrons involved in it.  Since the energy:charge ratio (E° above, called a reduction or oxidation potential) is constant, we can use that to find the energy:charge ratio of the entire battery just by adding them together.  Why doesn't it go above 1.43V?  For the same reason that a fire doesn't burn hotter.  The chemicals involved only release so much energy per reaction.  Note: The potential can be changed in some systems by playing around with the concentrations of the chemicals in the cell.
As the battery is used up, the zinc slowly but surely turns into zinc oxide and the manganese(IV) oxide turns into manganese(III) oxide.  As their amounts fall to zero, the battery's potential falls and it quits working.


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Am I correctly showing a hydride shift on the alkyl halide?
I did this quiz and I forgot the hydride shift.

With the hydride shift, would these be the right answer?

Huh? Check your images again--they both have 9 carbons, but you're adding an EtOH (2 carbons) to an 8-carbon molecule. I think once you make the numbers add up, you'll see the right answer.

Instead of the H, we should see a CH3 group.

Excellent! Now if you draw the molecules with CH3 instead of hydrogens, what happens?

We get the right amount of carbons? :P

When you solvolyze the starting bromide a secondary carbocation would be generated.  Since a secondary carbocation is not particularly stable, the molecule will explore other pathways such as hydrogen shift, alkyl shift, etc., in order to produce a more stable carbocation.  The figure below, explores these options for your starting bromide.  Realize that although I've drawn a discreet secondary carbocation that then undergoes rearrangement, in reality cation generation and alkyl or hydrogen shift may occur in a more or less concerted fashion.

In this case there are 3 options (actually there are 4 options, but the fourth one would produce a primary carbocation, so I left it out - can you find it?), a, b and c. Pathways a and c produce secondary carbocations; however pathway b, involving hydrogen shift, produces a more stable tertiary carbocation.  Capture of the tertiary carbocation from pathway c by solvent should produce the expected product.

Very clear. Thanks ron. I guess I just wrongly wrote H instead of CH3 in my attempt to correct my answer.


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Does the amount of acid change when I add water to vinegar?
I have a question that asks me to describe why adding distilled water to my vinegar sample would not affect my calculated Molarity (determining amount of acid in the vinegar be titration with a base).
At first thought, it wouldn't, because the same amount of acid is still there, we'll still have the same amount of base reacting. So you'll still get the same amount by measuring how much base was needed.
But that got me thinking. There is a lot of diluting acids with water, but isn't water amphiprotic? Why doesn't it react with the acid, or react with it enough to matter at least? What information am I missing here?

First off, your thinking about the titration volume of a dilute solution vs the concentrate from which it was made is correct.
Why we don't need to worry about water acting as an acid or a base in this context:  Consider the reaction that occurs between a weak acid, HA, and water (a very weak base.)
$\ce{HA + H2O <-> A- + H3O+}$
The products of the reaction are a weak base and a strong acid (hydronium, the strongest acid that can exist in aqueous solution.)  If I add a base to this mixture, it doesn't matter whether that base reacts with an undissociated HA molecule or with a hydronium ion, in either case it has removed a particle of one acidic species from the solution, leaving behind A-, the conjugate base of HA.  The total number of such events that has to occur before the titration is done is a fixed value, regardless of how many of the events involved hydronium or HA.

That's interesting. Glad I thought to ask. Thanks!


text:
Combustion of paraffin
Find the heat of combustion for 1 mol of $\ce {C21H44}$ in the reaction where some wax is oxidized to elemental C (soot) and water vapor. The Heat of Formation of $\ce {C21H44}$ is -476 kJ/mol.
When it says some, does that necessarily change the molar ratio?
I thought the balanced equation would be:
$\ce {C21H44 + 11O2 -> 21C (graphite) + 22H2O}$
$\ce {O2}$ and elemental carbon are zero for heat of formation
22(-241.818 kJ/mol) - (-476 kJ/mol) = -4843.996 kJ/mol
Is this right?
Thanks

The question is ambiguous, unfortunately.  There are two possibilities that I can see.  Either the question:

is asking about a reaction in which some quantity of wax undergoes the reaction you give, in which case you are 100% correct.
is asking about a reaction in which part of the paraffin is being oxidized to C and part is being oxidized to CO2, the usual reaction that occurs.  Though that is more likely to occur in real life, in my opinion, it also is impossible to solve without further information.  The best you could do would be to set a lower limit (the value you've already found) and an upper limit, the heat of reaction for $\ce{C21H44 + 32 O2 -> 21 CO2 + 22H2O}$.  (I guess that would be an upper limit and a lower, numerically, since the values are both negative, but this reaction would release far more heat.)

OR you could write a pair of equations, one describing the reaction giving carbon, the other carbon dioxide. You could then derive an answer where the heat of combustion depended on the proportion of the reaction going down either path.


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"Orthogonality" of chemical elements
Chemical elements are said to be immutable at normal energies, but not at the energies that are orders of magnitude higher.
Given that nucleus at normal energies gives us the Periodic Table and at higher energies becomes unstable, can we say that (with fusion) we will be able to transform an arbitrary chemical element into some other chemical element (if we have enough energy)?

Certainly nuclear fusion and nuclear fission can convert elements into other elements. Indeed, this is how different elements form in the first place.
The one issue I have is with your "arbitrary chemical element." You can use fusion to put together lighter nuclei or elements into heavier ones - up to iron, or use fission to divide more massive nuclei into lighter ones.. up to about iron.
The limit here is the binding energy of the nucleus which maximizes with $\ce{^{56}Fe}$.
In general, though, your point is a good one. Given enough energy (e.g., in a supernova), you can create lots of different elements.


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Why can't saliva clean "stickiness"?
Why is it that when you consume something that is sugary-sticky (like candy, chocolate, fruit juice, etc.) and get it on your skin, when you try to clean it off with saliva (particularly, licking/sucking it off) it doesn't work?  What is it about the makeup of saliva that doesn't dissolve the sugar?

I actually don't think this actually happens...we'll see though. I just slathered my left arm in syrup and am waiting for it to dry. I'll report back either once I've had a chance to test it out or once the ants have chewed my arm down to the bone.

So I could have sworn I've cleaned stuff off with saliva before, but I wanted to make sure, so I ran a quick impromptu experiment.
I slathered the underside of my left arm with some maple syrup I had lying around. I waited about 10 minutes for it to dry, then I mentally sectioned off my arm into three parts and cleaned each part off in different ways:
The section closest to my hand was cleaned by putting my other hand under running water and wiping the syrup off (basically, getting my right hand wet and making wiping motions across the arm, then returning the hand to the water, until I felt like I was no longer making any progress).
The middle section was left untouched.
The section closest to my elbow was cleaned by licking my hand, then using it to wipe that section down (essentially the same as the water, except I used my mouth as the source of liquid).
After the cleaning, I let everything dry again for 3 minutes. Then I placed some toilet paper over my entire arm and attempted to remove it. Here are the results:
(Sorry for the horrid image quality)
Unfortunately, I didn't have any skin marking tools on me at the time (I use pretty much only water-soluble inks), so I had to add the lines afterwards.
As we can see, the section of my arm that was cleaned with water and with saliva are pretty much TP free, aside from a small section near my elbow that I think was just me missing some spots. The section cleaned with water was similarly TP-free. The middle section is pretty heavily TP-ed. What you're seeing is what got left behind when I attempted to rip the TP free.
So it seems like saliva actually can clean off stickiness pretty well! This is backed up by the comment that Michiel made, which points out that saliva is 99.5% water.
That being said, running this experiment gave me a possible explanation for why it might seem like it doesn't: it takes a LOT of saliva to do this!
With the water, I was easily able to wash off all of the syrup off of my hand and get a new coating of water on my fingers. Unfortunately, I can't generate that much saliva at once, and so I couldn't wash off the sugar that was already on my fingers that effectively. That made the saliva cleaning take a LOT longer than washing with water, both in terms of how many wipes it took to clean off most of the stuff and how much time it took overall (sadly, I cannot drool at the same rate a bathroom sink outputs water).
There's another possible explanation that springs to mind: when you have syrupy stuff stuck to you, you probably just ate some of the syrupy stuff. That means your mouth probably still has a good bit of sugar in it, which would also make it harder to clean off additional sugar.
Still, based off of this experiment, I feel fairly confident in saying that saliva cleans off sugar stickiness almost as well as regular old water, as long as you have enough of it lying around.
NOTE: I initially tried this with a flavored whiskey, hoping the alcohol would help it dry faster. Terrible mistake! Not only was there not enough sugar in the whiskey to get any sort of real stickiness, most of it basically ran off of my arm right away, so I ended up losing a lot of it. Save your flavored sprits for drinking, use a syrup!

Nice hands on approach and discussion of the reason for the observation of the OP. I agree with your assertion that saliva should also work because human saliva is 99.5% water. So essentially: it is the same thing (in terms of dissolving sugar). That also means that you indeed need as much saliva as water, and I agree that saliva doesn't form like water from a tap does (luckily, otherwise we would all be drooling like crazy)


text:
Electronegativity in krypton and xenon?
Why do krypton and xenon have high electronegativity? Noble gases are supposed to be "happy" with the amount of electrons they have, because they have 8 valence electrons (thus, most noble gases have no electronegativity). So why do krypton and xenon have electronegativity? Why do they tend to "want" one more electron? It makes no sense... their outer shells are already full!
I'm going to do a research paper on this question, so I would really like some good resources that could help me answer this question for myself. Know any?

I think you misunderstand the definition of electronegativity. Have you read the Wikipedia article?

For Pauling electroneg. bonds are needed, but they are no problem - there plenty of compounds of Xe and quite a bit for Kr. For other definition's you don't even need that.

Why do Krypton and Xenon have high electronegativity?

As you point out, krypton ($\ce{Kr}$) and xenon ($\ce{Xe}$) are members of the Noble gas family.  They are generally unreactive (noble) because all of their occupied orbitals are filled with electrons - they really don't want to gain or lose an electron.  However, in the 1960's researchers found that $\ce{Kr}$ and $\ce{Xe}$ did react with extremely electronegative elements such as fluorine and oxygen to form new molecules.  In the reaction process, $\ce{Kr}$ and $\ce{Xe}$ basically gave up an electron to the very electronegative elements they reacted with.
$$\ce{Xe + 3F2 → XeF6}~~(1)$$
$$\ce{XeF6 + 3H2O → XeO3 + 6HF}~~(2)$$
The most common definition of electronegativity is based on Pauling's work and is given by the equation
$${\chi_{\rm A} - \chi_{\rm B} = ({\rm eV})^{-1/2} \sqrt{E_{\rm d}({\rm AB}) - [E_{\rm d}({\rm AA}) + E_{\rm d}({\rm BB})]/2}}$$
where ${\chi_{\mathrm{A}}}$ and ${\chi_{\mathrm{B}}}$ are the electronegativities of atoms $\ce{A}$ and $\ce{B}$ and ${E_{d}(AB)}$ represents the bond dissociation energy of molecule $\ce{A-B}$.
Using $\ce{HBr}$ as an example, if we know the bond dissociation energies of $\ce{HBr,~ H2}$ and $\ce{Br2}$, then we can calculate the difference in electronegativity between $\ce{H}$ and $\ce{Br}$.  From this, if we know the electronegativity of hydrogen, then we can determine the electronegativity of bromine.
In essence, for any element that can react and form a molecule, we can calculate its electronegativity.  Since Kr and Xe do react to form molecules, we can use the above methodology to calculate its electronegativity.
Note: In the case of atoms like Kr and Xe that do not form a diatomic species like $\ce{Xe2}$, we need an extra step to find their electronegativity.  Since there is no $\ce{E_{d}(Xe-Xe)}$, we must use Pauling's equation twice, once with reaction (1) and once with reaction (2).  We now have 2 equations and two unknowns (${\chi_{\mathrm{Xe}}}$ and ${E_{d}(Xe-Xe)}$) so we can solve for ${\chi_{\mathrm{Xe}}}$, the electronegativity of $\ce{Xe}$.


text:
NaCl dissociation not spontaneous?
It's a well-known fact that the dissociation of $\text{NaCl}$ can be represented by the series of equations $$ \begin{eqnarray*} \text{Na} &\rightarrow& \text{Na}^+ + \text{e}^- \qquad \Delta H = 495 \, \text{kJ} \\ \text{e}^- + \text{Cl} &\rightarrow& \text{Cl}^- \qquad \qquad \Delta H = -348 \, \text{kJ} \end {eqnarray*} $$
Adding gives $ \text{Na} + \text {Cl} \rightarrow \text{Na}^+ + \text{Cl}^- $ with $ \Delta H = 147 \, \text{kJ} $, which is positive so it is not spontaneous. What explains this?

The reactions of elements to give ions is irrelevant to the dissolution of NaCl. NaCl is already ionic and the dissolution is dependent on the interaction of those ions with water molecules. The initial formation of NaCl from its elements is irrelevant.

That's not a well-known fact because it's not a fact...

Why atomic chlorine?

You might be confusing $\ce{\Delta G}$ and $\ce{\Delta H}$.
$$\ce{\Delta G = \Delta H - T \Delta S}$$
$\ce{\Delta G}$, free energy, measures the "usefulness" of a system. If $\ce{\Delta G \gt 0}$, then the reaction is not thermodynamically favored - it's NOT SPONTANEOUS.
$\ce{\Delta H}$ being positive means that the reaction is endothermic, or makes the surroundings cooler. While this would CONTRIBUTE to a positive value of $\ce{\Delta G}$, it DOES NOT GUARANTEE it.
In the case of your example above, I would assume the $\ce{T \Delta S}$ factor would make $\ce{\Delta G \lt 0}$ if that reaction really is SPONTANEOUS (I don't have an entropy table with me).

What is this?? The equation $\ce{Na + Cl -> Na+ + Cl- }$ does not represent the dissolution of NaCl.

Sodium chloride does not spontaneously dissociate into ions; table salt pretty much remains as it is. ($\Delta G$ is positive even allowing for entropy $RTln(2)$). Perhaps you are thinking about aqueous solution? In aqueous solution the molecule does dissociate, and this is due to the stabilisation given to the ions by the highly polar water. The Coulomb energy between two ions of charge $z_1, z_2$ is $w=z_1z_2e^2/(4\pi\epsilon_0\epsilon r)$ and is reduced by $1\epsilon$ the dielectric constant. In water $\epsilon= 78$  compared to approx 2 for hexane and 1 for a vacuum. Thus in water, ions can exist separately, as their electric field is 'quenched' and their interaction is weak vs thermal energy (random collisional disruptive motion) and so dissolve, whereas in a non-polar solvent the ions can interact with one another  even when well separated (since their electric field spreads out significantly) and so recombine more easily, and with greater stability, and do not dissolve


text:
How does a redox reaction between NaClO and KBr take place? will bromide be oxidised to bromine?
How does a redox reaction between $\ce{NaClO}$ and $\ce{KBr}$ take place?
will bromide be oxidised to bromine?
$\ce{2Br^{−} + ClO^{−} + H2O ⟶ Cl^{−} + 2 OH^{−} + Br2}$
or will bromide be oxidised to $\ce{BrO^{−}}$?
$\ce{Br^{−} + ClO^{−} ⟶ BrO^{−} + Cl^{−}}$


text:
Which substance has the higher oxidation strength?
I have to decide which one has a higher oxidation strength $\ce{ClO-}$ or $\ce{Cl2}$
$\ce{ClO- + H2O + 2e- -> Cl- + 2OH-}$
$\ce{Cl2 + 2e- -> 2Cl-}$
Will this reaction occur?
$\ce{ClO- + H2O + 2Cl- -> Cl- + 2OH- + Cl2}$
In my opinion not, because otherwise the $\ce{ClO-}$ is reduced to $\ce{Cl-}$. If the reaction works, the $\ce{Cl-}$ could react as reduction agent again. Therefore if I would add $\ce{Br2}$ to $\ce{ClO-}$:
$\ce{2 ClO- + 2 H2O + 4Br- -> \mathbf{2 Cl- }  + 4OH- + 2Br2 }$
$\ce{ClO- + H2O + 2Br- + \mathbf{2 Cl- } -> \mathbf{Cl- } + 2OH- + Br2  + \mathbf{Cl2}}$
So therefore chlorine has a higher oxidation strength than $\ce{ClO-}$?

Adam, I have used mathjax in combination with our chemical equations typesetting on your post to make it a bit more readable. Please check the syntax so you can do it yourself next time.

Chlorine dissolves in alkaline solutions forming mixture of chloride and hypochlorite. However, in acidic solutions the reaction is reversed. For this reason acidification of bleach solutions is not advised. The reason is that hypochlorite half-reaction includes hydroxide/proton, so its potential is susceptible to change depending on solution pH

Look right here ... .
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html
Hypochlorite ion (+0.90 V) has a lower reduction potential in basic solution than elemental chlorine (+1.36 V), so the hypochlorite ion is less strongly oxidizing and more stable in basic solution (which is what you write the reaction for).
Household "chlorine" bleach, which really uses hypochlorite, is made alkaline as one way to stabilize the hypochlorite ion and prevent evolution of chlorine gas.


text:
Gas phase chiral molecules
Are there any molecules that are chiral that are also gases at room temperature?
Failing that, what chiral molecules have a high vapor pressure? Propylene oxide has a boiling point of only 34 C, so that seems like a good one. Are there others?
EDIT: I appreciate all of the great suggestions! I should have mentioned before, but I am looking for something that can be purchased in both R and S enantiomers.

Chances look slim. Bromochlorofluoromethane has a bp of 41.2 deg C according to Chemspider. Looking at the HCFC page of Wikipedia doesn't hit paydirt.

The molecule has to be small. trans-1,2-difluorocyclopropane is certainly chiral, and its predicted boiling point is ~ 5 deg C.  trans-1,1,2,3-tetrafluorocyclopropane is another likely candidate.  This one will be easier to prepare (from perfluoropropylene oxide and trans-1,2-difluoroethylene).
Separation of enantiomers is the hard part.

Sorry, you lost me, what is "trans-1,1,2,3-tetrafluoroethylene"

It's tetrafluorocyclopropane.  Note to self: don't post without coffee.

1-Chloro-1,2,2,2-tetrafluoroethane has bp -12 Celsium. For comparison, 1-chloro-1-fluoroethane has bp +16 Celsium. In most cases, replacing good share of hydrogen in an organic molecule with fluorine leads to significant decrease of bp. perfluoropropylene oxide  has bp -27 Celsium, comparing to +34 of vanilla propylene oxide you noted.

Considering molecules that have been prepared and resolved:
1,3-dimethylallene (2,3-pentadiene) boils a bit above room temperature at 48.5 C (see  page 11 of this reference for the experimental preparation), but likely has a reasonable vapor pressure since it has a lower molecular weight than pentane.  The advantage to this molecule is that it has already been (at least) partially resolved (see here and here) in a one-step hydroboration.
Another simple hydrocarbon to consider is trans-cyclooctene.  It has a higher boiling point (143 C) but a significant vapor pressure, it flashes at 25 C.  The molecule is chiral and has also been resolved.

If you use deuterium and tritium, you can make stereocenters from methyl groups. For example, 1-deutero-1-tritioethane is be chiral.


text:
Is there a pH range for "neutral" detergents?
There exist many home made pH scales like these

But do any of you know if there is an "official" scale that defines if a neutral detergent is 7 sharp or within a range _______?

Strictly speaking, anything other than pH 7 is not neutral, because it has an imbalance in the ratio of $\ce{H+}$ to $\ce{OH-}$.

I know that chemically this is the definition :-)

Well, I would argue that chemically$=$officially

I'm more thinking like public affairs or expert panel that is generally accepted as giving guidelines. Like some EU group og FDA declaration. What I could find was GSA http://www.gsa.gov/portal/content/113006 but not exactly hitting the spot

I have dissected the definition of 'neutral' in this question which might help.

Neutral detergents are not neccesarily at pH 7, and I know of no legal restriction (at least in the EU) on labelling detergents as neutral based on pH. For example, this patent claims protection for a set of neutral detergent formulae with pH of "from about 5.5 to about 8.5".
An example of a commercial product is Proctor and Gamble's "Tide", which has a pH of 8.1-8.5 according to their safety data. This is advertised as "100%" neutral !

Thank you for your answer. Listing commerical products is not advertisement in my oppion, its just showing that there are no clear standard and companies are allowed to interpret neutral as they like.

I want to add some imformation to J.LS answer. According to U.S. General Services Administration, neutral detergents are produced with strong alkalis and weak acids. They are the most common soapless detergent and are usually alkylbenzenesulfonates. So, neutral detergent is not that neutral after all.


text:
What does oxidizing strength mean?
What does oxidizing strength mean?
Does it mean the strength of a oxidation agent that is needed to oxidize the reduction agent, or does it just mean the tendency of a substance to lose electrons (oxidate)?

If X has a higher oxidising strength than Y, then X would probably be able to oxidise Y (even if X/Y are both oxidizing agents).

See also Comparing Strengths of Oxidants and Reductants.

The strengths of oxidizing and reducing agents are indicated by their standard electrode potentials. Let's consider standard potentials of some redox couples:
$$
\begin{align}
E^\circ(\ce{Li^+/Li}) &= \pu{-3.04 V} \\
E^\circ(\ce{K^+/K}) &= \pu{-2.92 V} \\
E^\circ(\ce{Cu^2+/Cu}) &= \pu{+0.34 V} \\
E^\circ(\ce{F_2/F^-}) &= \pu{+2.87 V}
\end{align}
$$
The values above are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.


text:
Acidity of substituted phenols
Phenol has a $\mathrm pK_\mathrm a$ approximately equal to $9.9$.
When one studies the acidity of chlorophenols, one notices the following:

First of all, chlorophenols are more acidic than phenol, due the negative inductive effect (−I) of chlorine, that reduces the negative charge, located on the oxygen of the phenolate anion.
Among the different chlorophenols, the observed trend is explained by the fact that the stabilizing −I effect of chlorine decreases with the distance from oxygen.
Now focusing on another class of substituted phenols, such as methylphenols (= cresols), the $\mathrm pK_\mathrm a$ varies as follows:

In this case, the methyl group yields a positive inductive effect (+I), thus increasing the negative charge on phenolate oxygen. That's why cresols are less acidic than phenol. One can further realize that the meta compound is more acidic, and this is due to the resonance structures which show a negative charge in ortho and para, but not in meta.
I would appreciate to learn whether such a reasoning, for these two type of substituted phenols, is plausible or whether further more solid arguments should be invoked.
In particular, I would appreciate to understand why the difference in acidity between 2- and 4-chlorophenols is much more pronounced than the same difference between o- and p-cresols.

I would appreciate learning whether such reasoning, for these two
  type of substituted phenols is plausible or whether furthermore
  solid arguments should be invoked.

Your reasoning seems generally well thought out.  You clearly have a good understanding of inductive effects and you mention resonance effects in the cresol series.  Structural effects don't play a role with these compounds, but let me mention them for completeness.  If we have a compound such as I with a carbonyl ortho to the hydroxyl substituent, then in addition to inductive (stabilizes the anion) and resonance (stabilizes the anion) effects, we should also consider hydrogen bonding; which, in this case, would stabilize the phenol making it less acidic.  On the other hand, in compound 2, the bulky t-butyl group would force the ester carbonyl to twist out of the plane of the benzene ring.  In II, the inductive effect of the carbonyl would still operate, but the resonance and hydrogen bonding effects on acidity would be diminished.

I would appreciate learning why the difference in acidity between
  2- and 4-chlorophenols is much more pronounced than the same
  difference between o- and p-cresols

This link discusses electrophilic aromatic substitution, where again both inductive and resonance effects need to be considered.  About two-thirds of the way down the page we see that chlorobenzene undergoes electrophilic nitration about 30 times slower than benzene, but is still ortho-para directing.  We also see that toluene reacts about 25 times faster than benzene and is also ortho-para directing.  From these and other examples, we learn that chlorine donates electrons weakly by resonance and withdraws electrons strongly through its inductive effect; and that a methyl group is a mild electron donor through resonance (hyperconjugation), while its inductive effect is also electron-donating, but is small.
Let's apply these concepts from electrophilic aromatic substitution to your two phenol series.  In the chlorophenols, we would expect the chloro substituent to influence the series primarily through a strong electron-withdrawing inductive effect.  Indeed, as the chlorine moves closer to the hydroxyl substituent, the inductive electron withdrawal increases in power and the $\mathrm pK_\mathrm a$ increases dramatically.  On the other hand, in the methyl series we expect mild resonance (hyperconjugation) effects to dominate (even though the inductive effect is in the same direction), therefore in the ortho- and para-cresols we expect the anion to be weakly destabilized and consequently a bit less acidic than phenol itself; while the meta isomer, with minimal resonance and inductive effects affecting the anion, should have an acidity close to phenol.

Thanks for the awesome explanation. But what is the order of ortho meta para floro phenols ? ... I understand that ortho floro phenols would possess h bonding and thus less acidic so acc to me answer should be meta>para>ortho but why does ortho comes in first place in reality ?

@ChiragLathi I'm glad the explanation helped.  Maybe take a look at this question, it might be helpful.

which is more acidic? I or II?

@ibuprofen II is more acidic.

@ron Can you please explain why carbolic acid is more acidic than nitrophenols even though there is strong electron withdrawing nitro group which stabilises the conjugate bases?


text:
Ca energy release when gaining an electron
Why does $\ce{Ca}$ have a negative electron affinity, i.e. energy is released when it gets an electron (ca. $\pu{-2 kJ/mol}$), when $\ce{Be}$ and $\ce{Mg}$ have positive values?
I know that the electron will be settled in the $\ce{3d}$ orbital which has a lower principal quantum number than $\ce{4s}$, but still, why would energy be released?

Two sign conventions: (1) the more common one states that a positive
  electron affinity value represents energy release when an electron is
  added to an atom; (2) the other states that a negative electron
  affinity represents a release of energy.

http://www.rose-hulman.edu/~brandt/Chem251/GenChem_Review_v3.pdf

Recommended values for these electron affinities, in the units
  commonly employed in introductory texts and with the sign convention
  used here, are 2.37, 5.03, and 13.95 kJ/mol for Ca, Sr, and Ba,
  respectively. The endothermic electron affinities often quoted for Be
  and Mg are also too large and should be reported simply as ">0". An
  argument for a return to the original sign convention for the electron
  affinity is presented in this paper.

http://pubs.acs.org/doi/abs/10.1021/ed074p123

OP: I know that the electron will be settled in the 3d orbital

Actually, as explained in the above article, the electron occupies 4p.

OP: but still, why would energy be released?

Very little energy is released and the lifetime of Ca- is on the microsecond scale in practical experiments.  The reference Contributions to the electron affinity of calcium and scandium may be the best source for theoretical models that do and don't predict a slight energy release.  According to Atomic negative ions: structure, dynamics and collisions, core electrons need to be considered in addition to the valence electrons in calculating the stability of Ca-.  Early calculations that considered only valence electrons predicted that Ca- was unstable.  Configuration-interaction study of differential correlation energies in Ca-, Ca, and Ca+ presents detailed calculations of energy levels.

Can't access it.

can you access http://www.sciencedirect.com/science/article/pii/S0370157304000316 ?

I can, but I couldn't find a satisfying explanation (probably there isn't one).

try this one: http://journals.aps.org/pra/abstract/10.1103/PhysRevA.62.052518

I read that they relativistic effects i.e. electrons have bigger speed, lower mass, the orbital contracts, the electrons spend more time near the nucleus and have lower energies. Did I figure it out correctly, or is the explanation different?

Basically, they are saying that only through accurate calculations of the interaction of electrons with each other in the atoms and ions, including relativistic effects and core electrons as well as valence electrons, can the energy levels be calculated well enough to conclude that energy should be released upon Ca bind and electron.  I don't think there is an answer beyond that.

Please verify your data.
$\ce{Be}$ has negative electron affinity (precisely -2.4 eV).
Also calcium has positive electron affinity (precisely 0.0245 eV).
Check the Wikipedia page Electron Affinity Values
I assume you got confused between electron affinity and gain enthalpy. Electron gain enthalpy is negative of electron affinity.
EDIT 1: DAVE PhD, you misunderstood me. Electron affinity has absolute definite values. Only one convention is used. Electron affinity can assume both positive and negative values. But beryllium has negative electron affinity, by any convention you can't assume it to be positive. You can even check the Wikipedia page.

There are two conventions. I used the one where negative electron affinities represent stabilization. In your case, the positive electron affinities represent stabilization of the system.

Nit picking, but those values can't be "precise" and exact because they're not definitions.

authors: محمد الحربية


text:
Why does electron velocity increase with the increase of atomic number?
I know that for atoms with big values of Z, relativistic effects have to be applied. Why does electron velocity increase with the increase of atomic number?

You can easily understand that based on the particle in the box model. Large Z corresponds to large electrostatic field around the nucleus which act on the electrons. This leads to a smaller "box", ie. the core electrons of heavy elements are moving on orbitals with radius orbitals than electrons of light elements.
Since the energy of the electron is (if we use the simple 1D version) $$E = \frac{n^2h^2}{8mL^2}$$
where $L$ corresponds to the box size and the energy ($E$) is purely kinetic energy. You can see that smaller box corresponds to larger kinetic energies.

Why does electron velocity increase with the increase of atomic
  number?

Newton's second law tells us that for an electron moving in a stable circular orbit around a nucleus the centripetal force pushing out must equal the electrostatic attraction pulling in, or
$${\frac{m_e~ v^2}{r}} = {\frac{Z~k_e ~e^2}{r^2}}~~(1)$$
where $m_e$ is the electron's mass, e is the electron charge, $k_e$ is Coulomb's constant and $Z$ is the atom's atomic number.  Rearranging equation (1) yields
$$v = \sqrt{ Z~k_e~ e^2 \over m_\mathrm{e} r}~~(2) $$
From equation (2) we can see that the electron velocity increases as the atomic number, $Z$, increases.
Edit: Response to OP's comment

I am expecting some quantum mechanical explanation

As the electron velocity given by equation (2) becomes comparable to the speed of light, the mass of the electron increases due to relativistic effects.
$$m_{rel}=\frac{m_{e}}{\sqrt{1-(v_e/c)^2}}$$
As the electron mass increases, the s (and somewhat for p) orbital radius contracts.  So relativistic effects affect the electron mass and the orbital radius.  In terms of electron velocity, these 2 relativistic effects offset one another and the electron velocity is not affected; equation (2) is still valid.

I am expecting some quantum mechanical explanation.

The quantum mechanical response is similar. The potential energy of the electron includes the electrostatic interaction between the nuclear charge Z and the electron. Even ignoring the "stable circular orbit" idea, there is a balance between the angular momentum and the electrostatic attraction and the remaining equations are similar.


text:
Finding the Ka of a weak acid after addition of a strong base
When 50.00 mL of an unknown 0.2461 M weak acid are titrated with 0.1968 M NaOH, it was found after the addition of exactly 16.00 mL of base that the pH of the solution was 3.79. From this information, calculate the Ka of the acid.

I have an idea of how to do this question, but looking in my book, I seem to be arriving at a wrong answer.
$$50 \text{mL of }0.2461\text{M weak acid} = 0.2461 \cdot 0.05 = 0.012305\text{ mol weak acid}$$
$$16 \text{mL of }0.1968\text{M NaOH} = 0.2461 \cdot 0.013 = 0.0031488\text{ mol NaOH}$$
$$\begin{array}{|c|cccc|}\hline
   & \ce{HA} & \ce{NaOH} & \ce{A^{-}} & \ce{H_{2}O} \\ \hline
  \mathrm I & 0.0123 & 0.0032 & 0 & -\\
  \mathrm C & -0.0032 & -0.0032 & +0.0032 & - \\
  \mathrm E & 0.0091 & 0 & 0.0032 & -\\ \hline
 \end{array}$$
$$\text{molarity } \ce{A^-} = \frac{0.0032~\textrm{mol}}{0.066~\textrm L} = 0.0485~\textrm M$$
from here, I work on the assumption that the weak conjugate base will associate in water.
$$\mathrm p\ce{OH} = 14 - \mathrm p\ce H = 14 - 3.79 = 10.21$$
$$[\ce{OH^-}] = 10^{-\mathrm p\ce{OH}} = 10^{-10.21} = 6.17\times 10^{-11}~\mathrm M$$
$$\begin{array}{|c|cccc|}\hline
   & \ce{A^-} & \ce{H_2O} & \ce{HA} & \ce{OH^-} \\ \hline
  \mathrm I & 0.0485 & - & 0 & 0\\
  \mathrm C & -6.17\times 10^{-11} & - & +6.17\times 10^{-11} & +6.17\times 10^{-11} \\
  \mathrm E & 0.0485 \text{ (negligible change)} & - & 6.17\times 10^{-11} & 6.17\times 10^{-11}\\
 \hline\end{array}$$
\begin{align}K_\mathrm {b} &= \frac{(6.17\times 10^{-11})^{2}}{0.0485} = 7.85\times 10^{-20}\\ & = \frac{K_\mathrm W}{K_\mathrm b}\\ & = \frac{1.0\times 10^{-14}}{7.85\times 10^{-20}} = 127388.54 \leftarrow\text{This is obviously wrong}\;.\end{align}
Where did I go wrong? Any help is much appreciated!

Ka of an acid doesn't change with addition of a strong base.

When you used the ICE method the second time, you forgot that HA does not have an initial concentration of 0. Instead, the initial concentration of HA is 0.0091 as you calculated previously. For simplicity's sake, you can just use Henderson-Hasselbalch equation after your first ICE method to determine the answer.

I had been trying to get the proper answer ($5.61\cdot10^{-5}$) and for whatever reason I kept getting the wrong answer. Maybe I was using the moles of $HA$ and $A^-$ instead of their molarities in 66 mL of solution. Thanks a lot though! Can you think of situations where you would need to set up a second ICE table for its association in water: situations where doing just one ICE table isn't enough?

I guess the issue is not so much about whether you use one or two or even three ICE tables, but more so on looking at the net change in concentration of each specie in solution.

Here is a different approach:
Initially we have: 50,00 *0,2461 = 12,305 mmole of acid A.
This amount of the acid is being titrated with 16,00*0,1968 M = 3,1488 mmole of NaOH.
By the titration, 3,1488 mmole of the acid A will be converted to its corresponding base B.
pH = pka + log [B]/[A]
=>
3,79 = pka + log [3,1488/(12,305-3,1488)] = pka - 0,46357 => pka = 4,25
=> ka = 10-4,25 = 5,62*10-5


text:
Is there any free GAMESS input file maker for Mac OS?
My main purpose is to simulate excited state calculations on a molecule with Time Dependent DFT.
I am looking for free software that makes input files for GAMESS. I have tried Avogadro and it works fine (in Extensions> GAMESS > Input Generator > Advanced tab) but it can not make TD-DFT input codes.
I also checked the recommended software by the Gordon Group (here) and tried installing some of them but couldn't install Gabedit at all (I was unable to compile the binary, maybe because I am using Mac OS X Yosemite).  I also tried WebMO but it is available just on iTunes for iOS.
What I need is a software that makes the input code interactively to avoid possible errors.

personally, I export .xyz from Avogadro and use primitive bash script

If you can give me the requisite keywords for TD-DFT in GAMESS, it's not hard to add that to Avogadro
I've never done TDDFT in GAMESS because I use other programs, but given a link to the docs, it's easy enough to enable more drop-down menus.
That said, you can interactively edit the text in the Avogadro input for any quantum program (not just GAMESS).

If you don't want to use Avogadro, the "native" program for GAMESS is MacMolPlt which was developed in the Gordon group itself. I haven't checked, but I'd assume it supports TDDFT options, since it enables pretty much everything else in GAMESS.

@Aug I suggest you jump on Geoff's generous offer as soon as possible.  He can fix you up with an Avagadro update that may meet what you require.

Wow! I just noticed your profile, you are developer of the Avogadro and I think I've found a gold mine! Here is what I am looking for:I want to simulate reaction of Methylene Blue in Excited state( MB+ *) with some amines.I need to set up TDDFT but it is a little confusing as TDDFT=EXCITE needs DFTTYP= functional and I think it must be in RHF/UHF.Case becomes more complicated if I want to use Transitional states or Analytic nuclear gradients available in $TDDFT.I know that I can type the input commands in Avogadro but it would be awesome if it was available as some menus ( to avoid mistakes )

If you want to use Gabedit, it's in Homebrew. Just brew install gabedit after you install Homebrew. I just installed it without issue on Yosemite and it was pretty fast. It seems to have a lot of options for GAMESS, but it's been a long time since I've done computational chem, so I can't tell you whether it has all the TD-DFT options. I didn't get any weird errors or anything, so it seems to run fine.

Thanks!  yes I have used Homebrew and it said installation is complete but the problem is I can not find the installed software !   It is not in Application folder and a spotlight search brings up nothing!  How do you run it?

Just type gabedit into the terminal; it doesn't make an application bundle.


text:
Elastic scattering of x rays in 3D: the principle for XRD analysis
I have some trouble visualising scattering of x-rays in 3D, since all textbooks and examples present it in 2D. Under I will share some reflections (no pun intended) about the elastic scattering of x rays, and the diffraction effects observed, which is the basis of XRD analysis.

Here you see a figure of two crystal planes, separated by the distance d. Two parallel and in-phase x-rays hit an atom in the planes. The scattering cones are drawn. The darker area represents those scattering directions going into the crystal itself, and thus not detectable and uninteresting for XRD analysis. The light area are those scattered rays exiting the crystal.
Thinking in three dimensions is different from thinking in one dimension. To achieve constructive interference, the rays must travel in the same direction and also be superimposed on each other. Even though the scattered rays ' angle is, say, 45 degrees, they may not travel in the same direction.
However, if Bragg's law is valid for 45 degrees, all rays of 45 degrees scattering angle will interfere constructively. So we would get a diffraction cone of increased intensity. However, we cannot detect the whole diffraction cone, partly because some of the rays travel into the crystal, and partly because the detector surface area only covers parts of the diffraction cone; we would measure a curved segment of the diffraction cone. This is called the Debye-Scherrer ring, yes?
Though we get constructive interference in all directions of 45 degrees scattering angle, the scattering angle itself is random. This means there are scattered rays which do not undergo constructive interference in all directions, regardless of angle (though smaller angles are more probable). So, in an ocean of random scattering, we have diffraction cones of much greater intensity, since Bragg's law is fulfilled.
Is this a correct assessment of the XRD situation?

In 3D you will see spots not cones from a single crystal.  If you Google for "XRD spot pattern" images, you will probably find yourself enlightened about how many different shapes, and forms diffraction patterns can take.
Debye-Scherrer rings are one special case which occurs when you have a diffraction pattern from a very fine powder.  In such a powder you are summing the diffraction patterns from thousands to millions of individual crystallites.  In this case, you can imagine a large number of spot patterns rotated randomly around the beam axis so each spot projects into a cone.  Each individual crystallite, however will produce a spot pattern, and there is never anything but a spot pattern from a single crystallite.  (Sometimes spots can be smeared when there is strain in the crystal, or when the crystal is too small to produce high quality diffraction, but these are really cases of imperfect crystals.)
You can see diffractions from rays transmitted through the crystal.  However, X-rays tend to be absorbed within a few microns to millimeters in most materials so that limits the size of crystal you can use for transmission X-ray diffraction.
Finally, I noticed you said "Two parallel and in-phase x-rays..."  I just want to note that diffraction occurs with a single photon.  A single photon can scatter, and produce constructive/destructive interference with ITSELF.  You need to do this with many photons to see a full interference pattern, however, and this is usually the case in experimental situations.  If this is new to you, then take a look at Young's double slit experiment.
Hopefully this clarifies things a bit.

authors: By แป้งมันส์


text:
What is the pOH of a 0.0100 F solution of dipotassium phtalate?
What is the pOH of a 0.0100 F solution of dipotassium phtalate? pKa1 = 2.950 and pKa2 = 5.408 for phtalic acid.
I did it for phthalic acid by calculating pH and then just doing pOH = 14.00 - pH, which gave me pOH = 11.60. Not sure if I did it correctly, I'm only just (re)learning this. I just need a clear and basic explanation, if possible. The two pKa's are what confuse me... which one do I use? I've been searching on google and through my textbook for 2 days now but I'm just confusing myself even more at this point. I know it's probably really easy stuff, but I haven't done chem in 4 years... trying to refresh my knowledge.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

So, we know that dipotassium phtalate is a conjugate base of a diprotic acid, so the addition of it into water will remove some $\ce{[H+]}$s from $\ce{[H2O]}$ and increase the level of $\ce{[OH^{-}]}$ with means a lower pOH.
We calculate the pOH by first converting to $\mathrm{K_a}$
$$\mathrm{pK_{a2}=10^{-5.408}=3.908*10^{-06}}$$
$$\mathrm{K_{b1}=\frac{K_w}{K_{a2}}= \frac{1*10^{-7}}{3.908*10^{-06}}=2.559*10^{-2}}$$
$$\ce{\frac{[HA][OH^-]}{[A^-]}=\mathrm{K_{b1}}}$$
$$\mathrm{\frac{x^2}{F-x}=2.559*10^{-2}\rightarrow 1.60*10^{-2}}$$
We did that last one assuming that x was negligibly small.
We now know
$$\ce{[HA]}=\mathrm{x=1.60*10^{-2}}$$
$$\ce{[H+]}=\mathrm{\frac{K_w}{\ce{[OH^-]}}=\frac{K_w}{x}=6.25*10^{-6} \rightarrow pH=5.20_4}$$
$$\mathrm{pOH=14-pH \rightarrow 14-5.20_4 = 8.79_6}$$


text:
Skin Adhesive Emulsion
I'm working on upgrading my current skin adhesive emulsion. It is a roll-on skin adhesive used to keep clothing attached to the skin. Ingredients are; Alcohol Denae,Ethyl Ester of PM/MA Copolymer,Water, Butylene Glycol, and Aminomethyl Proponal.
What ingredient(s) could be used to:

Improve formula so that it is more pliable(as far as the skin moving with fabric).
Still maintain a strong bond/hold(possibly even stronger then current formula).
Have the overall formula be slightly less wet/runny (as it sometimes absorbs through clothing).
And when the formula sets, remains fairly matte/dry to the touch.


text:
Make a silver zinc battery
I have a shaver that runs off a rechargeable battery that is dying. Would it be feasible to make a silver zinc battery and use it to replace the existing battery? Cost is not an object, anything less than $2500 I would consider doable. I have a machine shop and a small chemistry laboratory with the standard equipment and glassware, including high vacuum capability, a centrifuge and simple glass blowing capability.
I found a book on silver-oxide zinc chemistry and battery design, but it is $500 and I don't want to spend that if the information can be obtained just as easily elsewhere.
I have tried to find commercial options, but had no luck. Most silver zinc batteries seem to be just for large (multi-million dollar) military or satellite applications. Sony makes a line of silver oxide primary cells for hearing aids, but these are not rechargeable. There is a company called Ultralife that makes medical and military batteries and might have something viable, but before I call them I wanted to check out the opinion of the experts here.

Would you consider just replacing the shaver battery with a similar rechargeable battery? It's tough enough to find a commercial battery of the right size and voltage and clip and resolder the tiny wires. I looked at doing the same thing with a rechargeable toothbrush - then just got a new one (it's better!).

This turns out to be very difficult to do for two reasons. One problem is that the voltage of a silver cell is different from that of Ni-Cd cell, so it would require a specialized, multi-cell configuration to emulate the voltage characteristics of the Ni-Cd rechargeables.
The other problem is that silver cells generally have a sophisticated frame inside of them that is produced by an intricate high-temperature welding process. To duplicate this process and produce a suitable frame would require a significant amount of experimentation and work.


text:
Naming system -"ylene"
There are various compounds which have common names that end in "ylene", like acetylene and ethylene. However, how come acetylene is an alkyne whereas ethylene is an alkene. Both of these have the same "ylene" ending though. What is the cause for this discrepancy?

Related Q: Nomenclature of common naming of alkynes vs alkenes; both -ene?

These are common names, and predate our knowledge about bonds. My guess is that the "ylene" comes from reduction -- reducing ethyl alcohol gives ethylene, reducing acetic acid  gives acetylene.


text:
Is the ring or the chain chosen as the parent structure in octan-2-ylbenzene?

How would you name the above compound according to IUPAC rules? I see some sources say it is 2-phenyloctane, but ChemDraw and a few other sources say it is 2-octanylbenzene.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
(exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): (octan-2-yl)benzene (preferred IUPAC name; ring preferred to chain)
Method (2): 2-phenyloctane (chain has greater number of skeletal atoms)

If both the methods are written in IUPAC book , then how is (1) more preferred than (2) ?

Well basically there are 8 carbons in the long chain
The Phenyl Group is actually considered to be kind of a Functional Group
Hence the name should be:

"octan-2-ylbenzene"
Although some sources say that the name should be 2-phenyloctane , They do it on the basis that :the unsaturated acyclic structure is recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure, which is kind of not used and considered to be wrong by many according to IUPAC rules.


text:
What is the non-ideal equivalent of mass action kinetics?
Starting from the equations of mass action kinetics, one may show that (for suitably chosen numbers $A_i$), the function $\sum_i x_i(A_i + \log x_i)$ is a Lyapunov function, i.e. it decreases over time.
Apart from an irrelevant factor of $RT$, this is the expression for the free energy of an ideal solution (or ideal gas, etc). So it seems that mass action kinetics carries with it an assumption that the reactions are taking place in an ideal solution.
I'm interested in how to think about kinetics in non-ideal solutions, but I haven't been able to find any information about it. Is there a standard set of kinetic equations that remain valid when the chemical potential is given not by $\mu_i^0 + RT\log x_i$ but by $\mu_i^0 + RT\log \gamma_i x_i$, for activity coefficients $\gamma_i$ that depend on the concentrations?
There are several alternatives to mass action kinetics that I know about, such as Michaelis Menton kinetics for enzymes - but these are actually derived from mass action kinetics applied to elementary steps, so they don't seem to be able to escape from the assumption of ideality.

I think you probably mean "non-unity activity coefficients" where you currently say "nonzero".  Also, keep in mind that what makes the problem difficult is that $\gamma_i$ is a function of all of the $x_i$ and $x_j$.  If the $\gamma_i$  where non-unity but constant, the problem easily reduces to mass action kinetics.  The devil is in the details of the activity coefficient model you use to relate the $\gamma_i$ to the $x_i$ and $x_j$.

@CurtF. thanks, I've updated that part of the question.

According to Catalytic reaction rates in thermodynamically non-ideal systems Journal of Molecular Catalysis A: Chemical 163 (2000) 189–204 (citing to kinetics textbooks) non-ideal kinetics needs to consider the activity of the transition state in addition to the activities of the reactants:
For $\ce{A + B ->}$
rate $=\frac{k_BT}{h}K^{\ddagger}\frac{a_Aa_B}{\gamma_\ddagger}=k_0\frac{\gamma_A\gamma_B}{\gamma_\ddagger}C_AC_B $
Where $K^{\ddagger}$ is the equilibrium constant for the formation of the transition state, $a$ is activity, $\gamma$ is activity coefficient, $C$ is concentration, $k_B$ is Blotzmann constant, $T$ is temperature, $h$ is Planck's constant and $k_0$ is the thermodynamically ideal rate constant.
Another reference to look at for more info is The status of transition-state theory in non-ideal solutions and application of Kirkwood–Buff theory to the transition state J. Chem. Soc., Faraday Trans. 2, 1986,82, 1297-1303.

Interesting, thank you. This makes a lot of sense. If you take the first equality and lump all the constants into a single constant $k$, it amounts to a mass action type equation, but using the activities in place of the concentrations, which is conceptually very striaghtfoward.

You can formulate the kinetics using extent. The extent can be related to the chemical potentials (through the Gibbs free energy). The activity coefficients might have a concentration dependencies, and the changes of extent is describing the changes of concentration.
Doing so for ideal gasses, you will derive mass-action kinetics.

I am familiar with the extent of reaction, but usually when formulating kinetics in terms of extent, we start by making a mass action assumption, no? (There is nothing on the wikipedia page you linked to that relates to kinetics.)


text:
Thermodynamics: Is Isolated system possible?
I have just started learning thermodynamics. I have learned that The system which is not able to exchange energy and matter on its own with surrounding is called isolated system.
I wanted to know that practically is it possible to make 100% isolated system?

You could argue that the universe as a whole is such an isolated system.

@Philipp ah interesting! Please put that as answer.

There are no 100% isolated systems - even the universe itself may not be truly isolated (depending on how you define universe).
However, in practice, many systems are "isolated enough" that we can treat them as conserving mass and energy, especially over short time frames or small distances.
For example, we often treat a well-insulated and closed reaction vessel as isolated when doing thermodynamic calculations, and this approximation works very well for fast reactions.
Another example where this works is an adiabatic process. Many processes can be successfully modeled as adiabatic even though they happen in poorly-insulated containers - for example, the rapid expansion or compression of a gas.
Another example is the use of small sub-domains of a non-isolated system, which over short time scales behave as isolated systems themselves. For example, in finite element methods (FEM), a system is broken up into very small sub-domains. Each one acts as a small system in which mass and energy can transfer across the boundaries. Locally, an element is an open system, and over long time scales the entire system is open, but on short time scales, the sets of domains that share boundaries are treated as isolated. If energy leaves one domain, it enters the neighboring domain - it doesn't completely leave the system. In effect, you can model an open system as a large collection of isolated systems, with the outermost having open boundaries.
As you can see, even though the assumption of an isolated system is not really true, you can find many, many places where it is "true enough" that we can use it practically. In fact, it would be very hard to do anything in thermodynamics without it!

It is an idealized situation, an extreme, just as well all the other thermodynamic models.

An isolated system does not exchange energy or matter with its surroundings. For example, if soup is poured into an insulated container and closed, there is no exchange of heat or matter. The fact that, in reality, a thermos is not perfect in keeping things warm/cold illustrates the difficulty in creating an truly isolated system. In fact, there are a few, if any, systems that exist in this world that are completely isolated systems.
The universe of course is an another approximation of an isolated system.
So,  practically  it is impossible to make 100% isolated system.

"The universe of course is an another approximation of an isolated system." Do we know that? Has it been proven or are we just assuming here?

We have conservation of energy in the universe (It's a physical principle). And we have the principle of the equivalence mass-energy . So, it's not an assumption.

So is the conservation of energy shown or simply postulated? We also have principles in quantum chemistry but some of those are simply postulates.

Complete isolation also assumes constant volume. The Universe is expanding, ie. has no constant volume.


text:
How to find the empirical formula for an nitrogen oxygen compound from the given mass of nitrogen and the product?

$5.00~\mathrm{g}$ of nitrogen is completely converted into an oxide of nitrogen. The mass of the oxide formed is $19.3~\mathrm{g}$.
  The empirical formula of the oxide would be?

My working:
$$\ce{N2 + O_2 ->  N$_x$O$_y$}$$

$m(\ce{N2}) = 5~\mathrm{g}$
$n(\ce{N2}) = 5/28~\mathrm{mol}$
$n(\ce{N$_x$O$_y$}) = 19.3~\mathrm{g}$

How do I approach this?

$\ce{x/2 N2 +y/2 O2 \rightarrow N$_x$O$_y$}$
We have two unknowns $x$ and $y$. So, to determine them we need two independent  equations:
1 - The conservation of nitrogen mass: $5=14x$. So,  $x= 0.35714$.
2- According to the stoichiometry of the above balanced chemical equation: The number of moles of  $\ce{ N$_x$O$_y$}$ equals $n= \frac{5}{28}\times\frac{2}{x}$.
On the other hand, the number of moles of  $\ce{ N$_x$O$_y$}$ equals   $n= \frac{19.3}{14x+16y}=\frac{10}{28x} $.
By substituting $x= 0.35714 $ into the last equation, we find $y=0.89375$.
We find $\ce{ N_{0.35714}O_{0.89375}}$. We divide by $ 0.35714$, We obtain  $\ce{ NO_{2.5}}$.i.e. the formula of nitrogen oxide is $\ce{ N_2O_5}$.

why did you write the equations with x/2 and y/2
I love the algebraic approach though - using 14x + 16y=19.3

We have to balance the chemical equation.

If the question specifies that only one NxOy is formed (so not also NpOq in some ratio) , you need to calculate how many atoms of Oxygen you will have added. Then, you can deduce the number of Oxygen atoms per Nitrogen atom and find x and y. If you find decimal values, adjust the stoichiometry accordingly.

$$\ce{N2 +O2 -> N_{$x$}O_{$y$}}$$
Now using POAC (Principle of Atomic conservation),
$$\frac5{28}\cdot2=\frac{19.3}{14x+16y}\cdot x\implies \frac{y}{x}\approx2.5025\sim2.5\implies y'=5,x'=2\implies \ce{N2O5}$$

Subtract the mass of $\ce{N2}$ from the mass of NXOY to get 14.3g.
This means there is 14.3g is from the $\ce{O2}$ so you can find the moles of $\ce{O2}$ and use the coefficients to solve for the ratio between $\ce{N2}$ and $\ce{O2}$ which will give you X and Y.


text:
Using sodium peroxide to scrub carbon dioxide emissions
I am a student from a school and recently I came across the reaction
$\ce{2Na2O2 + 2CO2 -> 2Na2CO3 + O2}$
Can't we use this reaction to reduce $\ce{CO2}$ in the atmosphere by placing some $\ce{Na2O2}$ filters in places where there is high level $\ce{CO2}$ emission? I then realized that the filter would have to be loaded with lot of $\ce{Na2O2}$ to achieve this. So why not adsorb as much $\ce{CO2}$ as much as possible using a good adsorbent for $\ce{CO2}$. I am not sure if there are other methods better than these but I would like to know if there is any?

Please revise the question to show the correct chemistry. The equation you gave does not balance. Is it supposed to be 2 Na2O2 + 2 CO2 -> 2 Na2CO3 + O2?

In any case, the amount of energy (i.e. carbon footprint) it takes to make sodium peroxide is likely greater than the carbon footprint of the CO2 you would capture.

I think the sodium peroxide chemistry is used to get oxygen/remove CO2 in spacecraft, but I've never heard of it being used for large-scale carbon capture. This sort of thing seems more suited, perhaps?

The problem with carbon capture is not so much that we lack the technology to do it, it's more that we lack the technology to do it in a way that is economically feasible.
Take your example - according to the US Energy Information Adminstration, it takes about 1.09 lbs of coal to generate 1 kWh of electricity. A typical coal-fired plant might produce an average of 500 MW throughout each day (this is a rough estimate), which works out to roughly 12,000 MWh, or 12,000,000 kWh. So that's something like 12 million pounds of coal each day.
If we assume that coal is 100% carbon (close but not quite), then it's about 37.8 mole C per pound of coal. That's 453.6 million moles of carbon each day being converted to $\ce{CO2}$, and since the combustion of carbon follows:
$\ce{C + O2 -> CO2}$
that means a single medium sized-coal plant produces roughly 454 million moles  of $\ce{CO2}$ each day. By now you are probably getting a sense of the scale of this problem. Since sodium peroxide and carbon dioxide have a 1:1 ratio in the absorption equation:
$\ce{2Na2O2 + 2CO2 -> 2Na2CO3 + O2}$
you will need an equal number of moles of sodium peroxide. That works out to about 78 million pounds per day.
Alibaba lists sodium peroxide at around \$400-500 per ton, so that means it would cost you (at a minimum) \$15.6 million dollars per day to scrub the $\ce{CO2}$ emissions from one medium-sized power plant. The real costs would be much higher because of the additional equipment and operating expenses.
Electricity currently sells for about \$0.10 to \$0.20 per kWh in the US, and around half of that is usually for production, with the rest being delivery. That means our 500 MW coal plant is going to be bringing in something like (at the high side) \$1.2 M each day. As you can see, the costs for scrubbing with sodium peroxide far exceed the potential revenue for the plant.
Now, there are other options:
One is to reuse the sodium peroxide, although the costs of regenerating it might exceed just buying more, and you would still need to find a place to store the $\ce{CO2}$.
Another is to use a cheaper/more effective absorbent - according to Wikipedia, amine-gas scrubbing is the most promising technology so far, although I am sure it faces the same kinds of cost scaling problems.
Another more popular approach is carbon capture, where carbon dioxide is basically just pumped into underground reservoirs in the hopes that it will stay there. While this is probably going to turn out to be the most economical approach, it has the risk of releasing large amounts of $\ce{CO2}$ accidentally, as well as having unknown effects on the local geological formations.
Yet another option is to use photosynthesis to scrub carbon dioxide. This has the advantage of being very cheap in comparison to other methods, but the downside is that the rate at which plants and algae can consume $\ce{CO2}$ is much slower than the rate at which we produce it.
It's a big problem, and all of the proposed solutions so far have been very expensive to implement - which is the main reason the fossil fuels industry opposes them.

This is a great analysis. Presumably there are energy costs associated with producing the 78 million pounds of sodium peroxide per day, energy that comes from a coal plant...

@jerepierre - yes, definitely, and that would have to be considered if the economics weren't the limiting factor. There are three important considerations: 1) what is the net energy change for the overall process; 2) what is the net cost per unit of energy; and 3) what is the net change in CO2.


text:
Synthesis of Pseudopelletierine, Mannich reaction base catalysed
The synthesis of Pseudopelletierine starts from glutardialdehyde, which reacts with methylamine and acetonedicarboxylic acid in a two-fold Mannich reaction.
This is the reaction:

I need to write the mechanism for this reaction. It needs to be under basic conditions also.
.
MY ANSWER SO FAR:
There will be a Mannich reaction followed by a second intramolecular Mannich reaction in order to form a ring by joining al three reagents together. Finally there will be a decarboxylation to remove the carboxylic acid groups.
This is the mechanism I have done:

Is this right? Any help would be appreciated.

I was going to post essentially the same answer as @jerepierre but he beat me to it. Don't forget that you have a N-Me group missing in the structures in the second row from the end. You can also compare your mechanism with this

Thanks for telling me about the missing N-Me group. That link you gave me is an acid catalysed reaction and the problem i am having is that mine is base catalysed.

You are very close. The only issue I see is that the Mannich reaction takes place between an enol(ate) and an imine (iminium). Your mechanism goes to the hemiaminal, and then the enolate must displace a hydroxide leaving group. Essentially that's an SN2 with a hydroxide as a leaving group, which does not occur. Instead, from the hemiaminal, the nitrogen kicks out the hydroxide to form an iminium, which is a much more reactive electrophile. The enolate attacks readily. Repeat that process, decarboxylate & tautomerize twice, and you're there.

on the second reaction arrow, where it says "H+ transfer",  how would you draw curly arrows to show that? Is one of the H's from the N+ being put onto the O-? Also where would I then be using the Na2HPO4 in my reaction for the synthesis of Pseudopelletierine?

@user5181 It sort of depends on your instructor. The most correct way would be for the anionic oxygen to take a proton from HPO4- and for hydroxide to take a proton from the cationic nitrogen in two steps. I'm a lazy guy, so I usually draw what you suggest, with the anionic oxygen deprotonating the cationic nitrogen in a single step. Some instructors don't like that, so I tend to not draw that on this site. Since the proton transfer steps are fast and we don't always know the exact species taking or giving the proton (or the order, like in this case), some people just write "H+ transfer".

@user5181 The mixture of Na2HPO4 with hydroxide gives a phosphate buffer that you can use to shuttle protons around. HPO4- will be the strongest acid and PO4(2-) will be the strongest base in the reaction mixture, so those species can play the acid and base roles in your mechanism.

so can i actually draw a mechanism for the Na2HPO4 acting as the buffer or is it just not included in the mechanism. Also can i just check, can the H+ transfer still occur in basic conditions? I'm just worried as there isn't supposed to be any H+ in the reaction.

@user5181 yes, proton transfer still occurs under basic conditions. There just can't be any free protons or strong acids generated.

In your answer, you mentioned "Repeat that process, decarboxylate & tautomerize twice, and you're there." However, that does not seem to explain the formation of the NMe bridge. How does the bridge actually form? It may seem as if one of the amines attacks the carbon bonded to the other N but such a nucleophilic substitution is not feasible as we would generate a nitrogen anion in solution.


text:
How does one analytically quantify whether or not a compound is a SIRT1 activator?
So there has been a lot of noise over whether or not certain molecules activate SIRT1. Resveratrol was thought to be a specific SIRT1 activator, but many have argued against this claim.
Similarly, this has also happened to SRT1720 too.
My question is - why is it so hard to quantify whether or not a molecule is an SIRT1 activator? How would one go about doing this?

It's important to distinguish SIRT1, which is a gene, from the protein it encodes, sirtuin 1 (some papers use SIRT1 for the protein too, but to avoid confusion, I'll use Sirt1). When looking at compounds that affect gene expression, one method of measuring the effects is measuring the change in mRNA levels for that gene when the compound is administered, usually with northern blotting or qPCR. In this case, because the gene encodes a protein, quantifying the protein levels also tells us about the expression of the gene.
The SIRT1 gene doesn't have any regulatory role itself, so in this case, people are more interested in the Sirt1 protein. While we can measure protein concentrations, Sirt1 is an enzyme that deacetylates proteins and resveratrol has been proposed to directly alter the activity of the enzyme (it's also known to increase SIRT1 expression, but this isn't the controversial aspect), so it's important to examine the effects of resveratrol on Sirt1's ability to deacetylate certain proteins. The way this is done is to expose various proteins/peptides to Sirt1 and look at whether they deacetylate. One method of measuring this is to use a special peptide as the substrate to which a fluorophore can be attached if the acetyl group is cleaved.
The reason resveratrol's effects have been debated is that the early studies that showed large activation effects were later found to only occur because of the fluorophore on the substrate and not resveratrol alone. Some work since then using other detection methods have shown activation effects: This paper tried many different substrates on a microarray using fluorescently labelled antibodies as well as kinetic measurements by mass spectrometry and found that resveratrol increased Sirt1's activity for some substrates, decreased it for others, and left many the same, though they didn't offer much in the way of structural insights that might be responsible. I haven't done a thorough literature search, but it seems like the debate is less contentious now that people are using other assays to examine activity.
Just to summarize your questions, it's not hard to quantify whether a molecule increases the expression of SIRT1, but because Sirt1 is an enzyme, any kind of label you add to measure activity could have its own effects on that activity. Any detection method that can measure the deacetylation of substrates could potentially be used for studying Sirt1 activity, and there are many, but label-free techniques like mass spectrometry are especially useful because they reduce complications in simulating biological conditions.