text:
Hydro-Bromination Product 1,4-Benzoquinone

What will the products on addition reaction of 1,4-benzoquinone with HBr ?

This is the mechanistic approach :

The Dissociation of HBr into H+ and Br- is implicit!

Thus final product is ortho bromo, 1,4 diphenol

Why don't you add the info in your comments to your question? And, do you have a better image, please? This one hurts eyes.

I agree, the picture is barely readable, your comments also belong into the post. Just [edit] it. The post was flagged low-quality because of these issues. For now I am not voting to delete, but these kinds of posts should not be the standard.

I apologize, the last twonsteps are tautomerism to give enol form which is aromatic, hence stable!


text:
Why is NaCl3 possible?
There. And there. Almost a year ago, a group of scientists claimed to have reached compounds of $\ce{Na}$ and $\ce{Cl}$ with weird stochiometries ($\ce{NaCl3, Na3Cl, NaCl7, Na3Cl2}$ and $\ce{Na2Cl}$).
What one of the articles says:

These compounds are thermodynamically stable and, once made, remain indefinitely; nothing will make them fall apart.
For example, $\ce{NaCl7, NaCl3, Na3Cl2}$, and $\ce{Na2Cl}$ are metals (that explains the apparent violation of electroneutrality since charge balance rules are inapplicable to metals), and only one semiconducting phase of $\ce{NaCl3}$ is stable in the pressure range between 250 and 480 thousand atmospheres.

At first, I thought this whole thing was bogus. But then, I faced another, apparently convincing, article and even one from RSC. Then the questions popped up:

Why does this happen? (Why is such a compound more thermodynamically stable at high pressure?)
How can the bonding in these be explained?

(Source)

Please have a look at the web site of Artem R. Organov and read the 2013 paper in Science, where he outlines the synthesis of $\ce{NaCl3}$ from $\ce{NaCl}$ in a laser-heated diamond anvil cell in the presence of excess sodium and chlorine. He suggests that under these conditions, rather unusual $\ce{[Cl3]-}$ anions are formed.

I only read the article superficially and I'm bit on a schedule ;) If you like to read the article more careful than I did, please feel free to post the answer yourself :)

related https://chemistry.stackexchange.com/a/29610/9961

Yes NaCl3 is possible.
Think of the more common iodine analog.
$\ce{KI + I2 -> KI3}$
For a comprehensive analysis see:
Unexpected stable stoichiometries of sodium chlorides

At 25-48 GPa, NaCl3 is stable in the Pnma structure, which has 4 formula units in the
  unit cell. Unlike all the other new phases predicted here (which are metallic), this phase is
  a semiconductor. Its structure (Fig. 2(B)) contains almost linear asymmetric Cl3 groups.
  Bader analysis (19) shows that the middle atom in the Cl3-group is nearly neutral, with
  most negative charge on the side atoms (Table 1) and the total charge of this anion group
  being ~-0.8. This result is reminiscent of well-known triiodides and azides with I3
  - and
  N3
  - ions, and of recent calculations on hypothetical H3
  - ions (20), which were found to
  have charge configuration [H-0.81H+0.72H-0.81]-0.9. Pnma-NaCl3 still can be viewed as an
  octet compound, with covalent bonding within [Cl3]--groups and ionic bonding between
  Na+ and [Cl3]-. At 48 GPa, with the formation of a new phase of NaCl3, the octet rule
  breaks down.

This research group compressed chlorine gas between NaCl plates, while heating with lasers and observed formation of $\ce{NaCl3}$.

Aside: the "new" higher pressure NaCl3 is valence isoelectronic with the known intermetallic compound Cr3Si (NaCl3: 1+(3×7)=22 valence electrons, Cr3Si: 6+(3×4)=22). So the trichloride ions polymerize to make in effect an intermetallic compound!

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text:
Singlet oxygen production
Why is in the reaction of hydrogen peroxide and sodium hypochlorite, singlet oxygen obtained? Why isn't triplet oxygen produced instead, as it is thermodynamically more stable?

Isn't the other question a duplicate of this one?

You are starting with spin-paired starting materials (an even number of electrons, with half in the up state and half in the down state), and they react (exothermically) to create an oxygen molecule. Initially, spin is conserved so that the oxygen molecule also has to be spin paired (singlet), despite the fact that the triplet state of oxygen is thermodynamically more stable. It will then react with a substrate or eventually decay over minutes to hours (most likely thermally) to the triplet state.

What else starting materials could be used?

(1) Irradiate oxygen in the presence of a photosensitizing dye.

(2) You can also use a trialkyl phosphite, treat it with ozone to make a phosphite ozonide, and then use a basic catalyst (pyridine) to decompose the ozonide to give singlet oxygen.

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text:
Can I use my extech pH 60 meter to measure soil pH if so how?
Can I use my extech pH60 meter to measure soil pH if so how?
I included a link to the meter.
Extech pH 60

Take a soil sample and mix it with distilled water. The most common ratio used for soil-water pH is 1:1 soil:water.
Stir the soil and solution and wait 15 minutes to 1 hour to set the slurry.
Measure the pH (check the temperature first).

In the video below it says to do 1:2 (1 part soil 2 parts distilled water) which is correct? https://youtu.be/int0FevLw8I

There is more sampling method. Ratio 1:2 is often used for soils containg high amount of organic material. Also electrolyte solutions (instead of DI water) are used. For detailed info please see https://www.academia.edu/3326561/Models_relating_soil_pH_measurements_in_water_and_calcium_chloride_that_incorporate_electrolyte_concentration

Here's a great video showing how to do this
In the video below it says to do 1:2 (1 part soil 2 parts distilled water)  https://youtu.be/int0FevLw8I


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Geometrical Isomerization
Can you explain to me:

What are Cis-Trans, E-Z and Syn-Anti Isomerization?
What is the difference between them?
How many functional groups is cis-trans isomerism applicable to?

I  know mostly about cis trans and E-Z but have no idea about syn- anti isomers. Correct me if needed: I think cis and trans isomers respectively occur when identical atoms or groups are on the same or opposite side of the double bond respectively. E and Z isomers are for trans and cis in complex cases where more functional groups are involved.
Also, I couldn't follow the Wikipedia Link.

Syn and anti are used more commonly in case of addition to an unsaturated compounds not to denote the actual compounds.

see this earlier answer

cis and trans are older notations that express roughly same concept of E and Z.
Explicitly, E and Z have absolute meaning and are assigned based on precedent, where the Z is assigned to the case where the highest priority substituents are on the same side of the alkene/double bond and E being opposite.  These notations explicitly refer to a pair wise relationship.  It is not normally said '... the group E to the ...'
cis and trans refer to substituents of context, with cis being substituents on the same side of the alkene/double bond, and trans being opposite.  Frequently these refer to the most important nomenclaturally, but not always.  Additionally cis and trans are used in discussions such as '...the group trans to the ...' describing a relation of one object relative to the other.  cis and trans are no longer considered IUPAC-correct nomenclature, but still have some usage for descriptive discussion.

what is syn anti that is my main question

Perhaps then, those terms should be first in your list of 'difference between' items.  Ron's comment contains a post to an answer which covers syn and anti.


text:
Effect of pressure on solubility of salts

The volume of certain saturated solution is greater than the sum of the volumes of the water and salt from which it is made. On increasing pressure solubility of this salt
(a) Increases
(b) Decreases
(c) Remains unaffected
(d) Can't be predicted

The solution says

Density of product < Density of reactant [mass of reactant = mass of product]
So on increasing pressure equilibrium shift toward higher density solid or liquid.

According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?
Is my interpretation of the Principle wrong or is the solution of the answer wrong?

You're wrong, maybe you mixed density and volume?

The answer is "(b) Decreases."  The reason is that the saturated solution occupies more volume than the starting reagents.  At higher pressures, creating this extra volume requires more $P dV$ work be done by the dissolving reagents.

According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?

I find it easier to think in terms of volumes, not densities.  High pressure should favor low volumes.  Imagine a balloon filled with saturated solution.  If we squeeze the balloon by raising the pressure on the outside of it, the balloon will "want to" shrink.  In the example given, the balloon can shrink only by saturated solution converting back into solid salt and liquid water, so that is what would be thermodynamically favored.
Update: Based on discussion in the comments, I wanted to add a point about compressibility and thermodynamic stability.  The degree to which a system opposes a change in pressure is called the compressibility and is related to dV/dP for a given system, which you can find from the equation of state. In general, systems in thermodynamic equilibrium cannot have a positive dV/dP value. That is, the volume must shrink as pressure rises for mechanically stable thermodynamic systems.

"High pressure should favor low volumes." According to me, high pressure  creates low volumes and the system should oppose such a change and try to regain its earlier higher volume. What's wrong with this interpretation?

The degree to which the system opposes such a change is callled the compressibility and is related to dV/dP for a given system, which you can find from the equation of state.  In general, systems in thermodynamic equilibrium cannot have a positive dV/dP value.  That is, the volume must shrink as pressure rises for mechanically stable thermodynamic system.  Your interpretation implies the opposite.

Please edit your post to include this point. I feel this make the answer more complete.

It shifts towards high density and not low.

You are right to suggest that when applying Le Chatelier's Principle, it's often useful to write and equation. We are told:
  graphite <--> diamond
  We are also given that diamond has a greater density meaning that it occupies less volume than the same mass of graphite, so the balanced equation could be written as:
  graphite <--> diamond + volume
  Now think of it this way: if you were to add volume, you would shift to the left according to Le Chatelier's Principle. Therefore, an increase in pressure (= decrease in volume) must shift to the right.

Source:
(http://www.mcat-prep.com/forum/gs-1-physical-sciences-f8/topic2040.html)

There was another answer here which got removed because it just contained a link. The link was also useful to me in addition to Curt's answer. I'll quote the relevant portion here.
The topic in discussion was over the conversion of graphite to diamond but the concept applied was the same as in this question.

You are right to suggest that when applying Le Chatelier's Principle, it's often useful to write and equation. We are told:
$\ce{graphite <=> diamond}$
We are also given that diamond has a greater density meaning that it occupies less volume than the same mass of graphite, so the balanced equation could be written as:
$\ce{graphite <=> diamond + volume}$
Now think of it this way: if you were to add volume, you would shift to the left according to Le Chatelier's Principle. Therefore, an increase in pressure (= decrease in volume) must shift to the right.


text:
Are methyl n-propyl ether and methyl iso-propyl ether metamers?
Are methyl n-propyl ether and methyl iso-propyl ether metamers?
It says so here.
So the difference must be in number of carbon atoms or the distribution?
Definition(http://dictionary.reference.com/browse/metamerism)

http://en.wikipedia.org/wiki/Metamerism I agree with @lighthart. Metamerism seem to be a limited definition of isomerism. As so it's chosen to be obsolete, you mustn't worry about it.

The term metamer is equivalent to isomer.  Isomers differ in connectivites of atoms between molecules with the identical number and type of atoms.
Methyl n-propoyl Ether and methyl iso-propyl ether have the same numbers of carbon, oxygen and hydrogen atoms.
As an aside, I have never heard the term metamer before, I expect it is archaic.

Did you check that link? It has a definition.

Neither have I. It isn't in the Gold Book either.

http://dictionary.reference.com/browse/metamerism @KlausWarzecha

@ChemExchange Irrelevant! No IUPAC, no cigar.

@KlausWarzecha Okay

I think metamer has an interesting specification: it has to have the same molecular formula and the same functional group. While we could just say "isomeric ethers" or something, I think it's a cool concept, but I'll probably never use it, and it doesn't make much difference as far as this question goes. Now if the question had been about 1-methoxypropane and 2-methyl-1-propanol, that would come into play.

Compounds having the same molecular formula but different number of carbon atoms ( alkyl groups) on either side of functional group ( i.e., $\ce{-O-,-S-, -NH-, -C(=O)-}$) are called metamers and the phenomenon is called metamerism. metamerism occurs amongst the members of the same homologus family.
For example:

$\ce{CH3CH2-O-CH2CH3}$ is a metamer of $\ce{CH3-O-CH2CH2CH3}$ OR $\ce{CH3-O-CH(CH3)2}$
$\ce{CH3CH2COCH2CH3}$ is a metamer of $\ce{CH3COCH2CH2CH3}$ OR $\ce{CH3COCH(CH3)2}$

It may be noted that metamers may also be position isomers, e.g., pentan-2-one and pentan-3-one may be regarded as position isomers as well as metamers.

Quoting from Wikipedia without mentioning the source: https://en.wikipedia.org/wiki/Metamerism

It is a type of positional isomerism in which functional group changes its position in the backbone of parent chain,methyl propyl ether and methylisoptopyl ether are metamers from Above def.


text:
Amount of substance of a molecule in a solute the same as amount of substance of constituent elements?
I’ve done absolutely no chemistry since high school.  Now, 16 years later, I’ve been asked to look at some chemical analyses and I’m having trouble!
Context: I’ve got a series of spectroscopy results giving the concentrations (in mg/l) of chlorine and $\ce{SO4}$ in a sample.  I want to find the molar and mass ratios of $\ce{S}/\ce{Cl}$.
What I’ve done: I convert the mg to g, then (mass) concentration into a molar concentration by dividing by the molar mass (35.5 g/mol for $\ce{Cl}$ and 96 g/mol for $\ce{SO4}$).
Question: am I right in thinking that the molar concentration of $\ce{S}$ will be the same as the molar concentration of $\ce{SO4}$?  My reasoning is that 1 mol $\ce{SO4}$ contains 1 mol $\ce{S}$ and 2 mol $\ce{O2}$.
Furthermore, to get the mass concentration of $\ce{S}$ alone, can I just multiply my concentration of $\ce{SO4}$ by the molar mass ratio, $\frac{M(\ce{S})}{M(\ce{SO4})}$?

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Your approach seems solid.

You are right thinking that the molar concentration of S will be the same as the molar concentration of SO4, unless there is no other molecule whith S in the sample.
If there is 1 mole of molecules of sulphate (SO4), there will be exactly 1 mole of atoms of sulfur (S), and 4 moles of atoms of oxygen (O).
On the other hand, though "theoretically" it is correct to say there will be 2 moles of molecules of O2, it is really not correct, because there is absolutely no molecule of O2 in your sample. The amount of oxygen atoms is the same in both cases, but how they are organized is not.
If you want to get the mass concentration of S, say in g/l, you would just multiply the concentration of SO4 in g/l by the mass ratium of sulfur within sulphate, (sulfur mass over sulphate mass, 32/96=0.33333). Since for every sulphate gram there is 1/3 sulfur mass and 2/3 oxygen mass, 1/3 of your sulphate mass is due to sulfur.
Another strategy would be to get the concentration of sulphate, which is exactly the same as that of sulfur (1 mol of atoms of S for every mol of molecules of SO4), and multiply that by the molar mas of the sulfur atom (32).
In brief, the molar ratio between S and SO4 is 1, and the mass ratio is 1/3 grams of sulfur for every gram of sulphate

Just a small correction at the end: the molar ratio of $\ce{S}$ and $\ce{SO_4^{2-}}$ is exactly 1, but the molar mass ratio is approximately $\frac{1}{3}$, as you mention above

I just realised it too, now that I looked back at the answer. Thanks, I will edit it right now!


text:
What causes passivity of specific metals in conc. nitric acid?
I know  that specific metals like $\ce{Pb}, \ce{Sn}$ and $\ce{Al}$ become passive in conc. nitric acid. My questions are:

Why does this happen only with conc. nitric acid, and not other conc. acids?
Why is the passivity specific to these metals? Do they have something in common or are there individual reasons?

Aluminium
This link states that:

Aluminium metal dissolves readily in dilute sulphuric acid to form solutions containing the hydrated Al(III) ion together with hydrogen gas, $\ce{H2}$. The corresponding reactions with dilute hydrochloric acid also give the hydrated Al(III) ion. Concentrated nitric acid passivates aluminium metal (due to formation of thin protective oxide layer, $\ce{Al2O3}$ on surface of metal which cuts off further reaction. )

$$\ce{2Al(s) + 3H2SO4(aq) → 2Al^3+(aq) + 2SO4^2-(aq) + 3H2(g)}$$
$$\ce{2Al(s) + 6HCl(aq) → 2Al^3+(aq) + 6Cl-(aq) + 3H2(g)}$$
Lead
This link states that:

The surface of metallic lead is protected by a thin layer of lead oxide, $\ce{PbO}$. This renders the lead essentially insoluble in sulphuric acid. Lead reacts slowly with hydrochloric acid and nitric acid. In the latter case, nitrogen oxides are formed together with lead(II) nitrate, $\ce{Pb(NO3)2}$.

$$\ce{Pb + HCl -> H2[PbCl4] + H2}$$
$$\ce{Pb + 4 HNO3 → Pb(NO3)2 + 2 NO2 + 2 H2O}$$
Tin
This link states that:

Tin is attacked only slowly by dilute acids such as hydrochloric acid and sulfuric acid. [...]Tin dissolves easily in concentrated acids.

$$\ce{Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O}$$
$$\ce{Sn + 3HCl → H[SnCl3] + H2}$$
$$\ce{Sn + 4HNO3 → H2SnO3 + 4NO2 + H2O}$$
Summarizing as a table,

Do they have something in common or are there individual reasons?

The three metals do not have common properties and reacts with different acids differently as we can see in the table.

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text:
Reactions of alkynes with sodium/ammonia vs sodium amide
What happens when alkynes react with sodium in ammonia ($\ce{Na/NH3}$), or when they react with sodium amide ($\ce{NaNH2}$)?
Are these two sets of conditions different?

Sodium in ammonia ($\ce{Na/NH3}$) and sodium amide ($\ce{NaNH2}$) are very different sets of conditions, despite superficially looking similar!

Sodium metal itself, $\ce{Na}$, is a one-electron reducing agent – it is oxidised to $\ce{Na+}$ in the process. When it reduces neutral hydrocarbons, negatively charged species are formed, which pick up a proton from the ammonia solvent. Overall, $\mathbf{Na/NH_3}$ is a way of reducing some organic compounds (including alkynes).
Internal alkynes $\ce{R-C#C-R}$ can be reduced to trans-alkenes using $\ce{Na/NH3}$. Specifically, solvated electrons are created which add to the triple bond as shown. Ammonia acts as a proton source to protonate the carbanions formed. The vinyl anion intermediate 4 can interconvert between a cis and trans geometry, and the trans geometry is preferred as this minimises steric repulsion between $\ce{R^1}$ and $\ce{R^2}$. When this anion is protonated, the trans-alkene is formed. This is a nice synthetic procedure for the preparation of trans-alkenes, complementary to Lindlar hydrogenation which produces cis-alkenes from alkynes.

On the other hand, in sodium amide $\ce{NaNH2}$, sodium is already in the +1 oxidation state and is no longer a reducing agent. However, the amide ion $\ce{NH2-}$ is a very strong base (the $\mathrm pK_\mathrm a$ of ammonia, $\ce{NH3}$, is 38). So $\mathbf{NaNH_2}$ is a way of deprotonating some organic compounds (including terminal alkynes).
The proton attached to the terminal carbon in a terminal alkyne is acidic enough ($\mathrm pK_\mathrm a \sim 25$) to react with sodium amide, forming the corresponding carbanion 7. Such carbanions are good nucleophiles and can be used in many synthetic procedures involving nucleophiles, such as addition to a carbonyl compound, as illustrated here.

In the first case hydrogenation occurs(anti addition) whereas in the second case, acid base neutralization occurs. So they aren't identical.
According to Wikipedia,

$$\ce{2 Na + 2 NH3 -> 2 NaNH2 + H2}$$

This shouldn't occur "usually" in reactions, unless it is specified at very low temperature.
However, using FeCl3 as a catalyst one can produce sodamide and hydrogen from sodium metal and ammonia in appreciable amounts.

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how do you enable result summary in gaussview
On G2:M1:V1 window, result summary is grayed out. Does anybody know how to enable so that I can see my calculation summary?

What file do you open into GaussView?


text:
Is it correct to say that dehydration isn't a subset of condensation?
To my understanding, dehydration is defined as a subset of a condensation reaction. A condensation defined as being two molecules joining together (in the case of loss of water, this is referred to as a dehydration).
I also thought however, if a molecule of water leaves without a condensation necessarily occurring, that this is also a dehydration. In this case, how could the dehydration be a subset of condensation?
I'm not familiar with any such molecule off the top of my head, but some easily protonated alcohol that is resonance stabilized when water leaves? Would this leaving group of water not be a dehydration, without a condensation, since no addition would occur thereafter?

Alkanols are indeed a good example. Think in the E2 and E1 reactions of primary and tertiary alkanols  in the presence of sulfuric acid at elevated temperatures, yielding alkenes.


text:
How to calculate valence factor (n-factor) for a element?
Suppose I am given some reaction in which
$$\ce{C6H12O6 -> CO2}$$
and I want to calculate n-factor for this reaction to ultimately calculate equivalent weight of carbon for this reaction. Since $$E=\frac{M}{n}$$
I know that n-factor is no of electron gained/lost by one atom of a compound.
So Initial Oxidation state of carbon is $$6x + 12-12=0$$ $$x=0$$ meaning
and final oxidation state of carbon is $$-4$$ Hence, n-factor of this reaction should be $4$ but it's given to be $28$. How is this possible? Can someone explain how to correctly calculate n-factor with few more examples?

How did you get the 2 of 2x+... ?

Sorry for typo .

Are you sure its 28 and not 24?

Not sure , I could be wrong.

You say it's given to be 28. Is it given 28 on some website/text?

It was 28 in my notes , I may have made mistake .

answer is 24. 46=24.the ans provided by ur notes is wrong
the oxidation no. of carbon on product side is 4 and the no. of carbon atom on reactant side is 6.so n-factor of C 64=24

while balancing, you should take oxidation number on both sides, not the oxidation states. On the right, you will have 6CO2, so, ON=6*4=24. On the right, you have zero. So, n factor=24

Hint :
n-factor of a molecule/compound is defined as the change in oxidation state per molecule.
You have correctly calculated the change of one carbon atom as 4.
But how many carbon atoms are there in the glucose molecule?
Note:

The average oxidation state of carbon in glucose is zero while in reality the different carbons have different OS. (Reference)
n-factor of a reaction is not defined but it is defined for a single species participating in that reaction.
n-factor of glucose is not equal to the n-factor of the product formed in that reaction, that is, $\ce{CO_2}$.

And I differ from the answer provided by your notes. If I am wrong somewhere, feel free to correct me.

Okay what if we had 2 carbon atom on product side?

Hypothetically let's suppose c2o4

@Heisenberg The OS of Carbon in $C_2O_4$ is the same as in $CO_2$. Hence the n-factor with respect to glucose shall remain the same.

atomicity of only reactant? is multiplied ?

Yes @Heisenberg. I have edited the answer.

Your reaction, $\ce{C6H12O6 -> CO2}$ is a redox half-reaction. The n-factor of a molecule/compound in a redox reaction is defined as the change in the oxidation state per molecule (as defined in Del Pate's answer). The easy to visualize definition is the amount of electrons (in $\pu{mol}$) donate or accepted per $\pu{1 mol}$ of the compound in the given redox half-reaction.

Can someone explain how to correctly calculate n-factor with few more examples?

Let's look at following redox half-reactions and find the n-factor by these two different definitions:
$$\text{Example 1:} \qquad \qquad  \ce{MnO4- ->[acid] Mn^2+ }$$
The oxidation number of $\ce{Mn}$ of the reactant is $+7$ while that of the product is $+2$. The change in the oxidation state per $\ce{MnO4-}$ is $+5$. Thus, the n-factor of $\ce{MnO4-}$ is $5$ in acid medium.
If we balance this redox half-reaction, we get:
$$\ce{MnO4- + 8 H+ + 5 e- -> Mn^2+ + 4 H2O}$$
Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{MnO4-}$ in the given redox half-reaction is $5$. Hence, the n-factor of $\ce{MnO4-}$ is $5$ in acid medium.
$$\ce{MnO4- ->[base/neutral] Mn^4+ \ (\text{as} MnO2)}$$
The oxidation number of $\ce{Mn}$ of the reactant is $+7$ while that of the product is $+4$. The change in the oxidation state per $\ce{MnO4-}$ is $+3$. Thus, the n-factor of $\ce{MnO4-}$ is $3$ in base/neutral medium.
$$\ce{MnO4- + 2 H2O + 3 e- -> MnO2 + 4 OH-}$$
Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{MnO4-}$ in the given redox half-reaction is $3$. Hence, the n-factor of $\ce{MnO4-}$ is $3$ in base/neutral medium.
$$\text{Example 2:} \qquad \qquad \ce{CuS -> Cu^2+ + SO2 }$$
The oxidation number of $\ce{S}$ of the reactant is $-2$ while that of the product is $+4$. The change in the oxidation state per $\ce{CuS}$ is $+6$. Thus, the n-factor of $\ce{CuS}$ is $5$ in this redox reaction.
If we balance this redox half-reaction, we get:
$$\ce{CuS + 2 H2O -> Cu^2+ + SO2 + 4 H+ + 6 e- }$$
Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{CuS}$ in the given redox half-reaction is $6$. Hence, the n-factor of $\ce{CuS}$ is $6$ in this redox half-reaction.
Now, let's look at the given reaction in the question:
$$\ce{C6H12O6 -> CO2}$$
Suppose we don't know the oxidation number of $\ce{C}$ of glucose $(\ce{C6H12O6})$ since there are $\ce{6C}$s in the molecule. Thus, we can approach to solve the problem by the second method. If we balance this redox half-reaction, we get:
$$\ce{C6H12O6 + 6 H2O -> 6CO2 + 24 H+ + 24 e- }$$
Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{C6H12O6}$ in the given redox half-reaction is $24$. Hence, the n-factor of $\ce{C6H12O6}$ is $24$ in this redox half-reaction. This demonstration reveals that even without knowing the change in oxidation number during the reaction, you can find the n-factor.
Note: This finding shows that the given answer for the question is incorrect.


text:
Abnormal termination of GAMESS in frequencies calculation
I want to optimize geometry and to analyse frequencies for transition metals ($\ce{Cu}$, $\ce{Mn}$, etc.) compounds using GAMESS. I use Avogadro for preparing input files and I have a problem: one of my calculations terminates abnormally with the following message in the output file.

NO FORCE FIELD, SCF DOES NOT CONVERGE AT VIB0 POINT
  EXECUTION OF GAMESS TERMINATED -ABNORMALLY- AT Thu Mar 26

What is the actual problem and how can I overcome it?

Consider adding at least the input file so we can see what can be changed.

Looks like there are some problems with the SCF convergence. First, try a bigger number of SCF iterations by setting MAXIT to, say, 100 in $CONTRL group.

In Avogadro GAMESS Input Generator choose "Advanced Setup" tab and set "Max SCF Iterations" to 100.

Welcome to chemistry.SE! I want to second @Wildcat's request. Adding the input file could really help.


text:
Interaction between metalic nanoparticles and aggregation
I can't find any publication to deal with aggregation of colloidal metalic nanoparticles. What type of interaction(s) is/are there between aqeous metalic nanoparticles which give(s) rise to aggregation?

Metallic nanoparticles, considered as colloids, are attracted to each other by van der Waals forces. To prevent their aggregation, we normally use dispersant using electrostatic repulsion or steric hindrance to stabilize the colloids.


text:
What's a good resource/book on pH measurements?
Most resources on pH and the electrochemistry behind it that I've found are IMHO too superficial. For example, they explain that an electrode with less than 80% slope should be discarded, but not why the decrease in slope happens, or why it's a problem if you can just calibrate it. Or they explain the potential generated at the Ag/AgCl eletrode and the glass membrane potential, but not the interaction between the two - does the H+ combine with the e- from the electrode? And what happens if the current is reversed? After reading for a few days I still feel like I don't understand any of it.
So far I've read a couple PDFs on nico2000.net (the 30 page Introduction to pH measurements), the 160 page book pH Measurements by C. Westcott (1978), a few manufacturer's guides to pH measurement theory and practice (by Radiohm, Emerson, Mettler) and the relevant chapters in the Analytical Chemistry books that my library had.
All help appreciated!

Looks like you understand pH measurement, but you are trying to go deeper into the details of potentiometry and open circuit potential measurements.
There I would recommend basic electrochemistry texts. Here is one simple book that is cheap, easy to read and very informational:
Electrode Potentials; Richard G. Compton and Giles H. W. Sanders


text:
Stabilisation of phenoxide ion

Which one of the following substituents at para-position is least effective in stabilizing the phenoxide ion $\ce{PhO-}$
(a)$\ce{-CH3}$
(b)$\ce{-OCH3}$
(c)$\ce{-COCH3}$
(d)$\ce{-CH2OH}$

According to me it should be $\ce{-OCH3}$ because of its mesomeric effect which will counter the mesomeric effect of $\ce{O-}$.
$\ce{-CH3}$ has a positive inductive effect but, the mesomeric effect of $\ce{-OCH3}$ should be stronger and destabilize the phenoxide ion to a greater extent

Seems to me the answer has already been given by Ron here.

@NicolauSakerNeto Thanks for pointing that out. I had completely forgotten about that question I had asked.

Which one of the following substituents at para-position is least
  effective in stabilizing the phenoxide ion $\ce{PhO^{−}}$

Yes, the p-methoxy group is the least stabilizing of the choices. I'm not quite sure what you mean by

its mesomeric effect which will counter the mesomeric effect of $\ce{O^{−}}$
  .

so let me add a few words of explanation for the choice.
In the figure below, I've drawn a few of the resonance structures possible for p-methoxyphenoxide (A-C).  The methoxy group donates electrons through resonance and resonance structure B illustrates this effect.  However, note that in resonance structure B we have placed two negative charges close to each other - this is very destabilizing.  Therefore, as the methoxy group donates electrons through resonance it destabilizes the system. In resonance structure C, the electron withdrawing inductive effect of the methoxy group is stabilizing.  However the resonance effect outweighs the inductive effect and overall the p-methoxy group is strongly destabilizing in the case of phenoxide.
Resonance structure D illustrates the p-methyl case.  Methyl is very mildly electron releasing both through resonance and inductively, so it also destabilizes the phenoxide anion, but since its effect is weak, it is not as destabilizing as the p-methoxy substituent.
Finally, notice in the p-acetyl case we can draw resonance structure E which actually stabilizes the phenoxide anion via resonance.  The p-acetyl group is also inductively stabilizing in the phenoxide example.

I meant that $\ce{O-}$ will try to donate electrons to the ring and to the methoxy group (and will be very successful in doing so).The methoxy group will try counter the effects of donation of electrons by $\ce{O-}$ by its mesomeric effect.

In this answer, you explain why methoxy at para position will stabilize phenoxide more than methyl at para position. Doesn't this contradict the above answer?

I would agree Ron, however the pKa's then seem counter intuitive, since pKa of p-cresol = 10.26 and pKa of p-Methoxyphenol = 10.16. (http://pubchem.ncbi.nlm.nih.gov/compound/p-cresol#section=Surface-Tension and http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf). Or am I not seeing something? There differences are very small anyhow.

Thanks for clarifying.  That is correct, the phenoxide oxygen does donate to the ring, but it can't donate to the methoxy oxygen.  Try to draw a resonance structure to show this and you wind up with 10 electrons around the methoxy oxygen.

No, I explain why methoxy destabilizes.  Resonance structures like B, where the methoxy donates electron density into the ring, place 2 negative charges close to one another - destabilizing.

@Jori  This question is specific to phenoxide anion.  Pka's involve comparison of substituent effects on the anion and the neutral phenol (e.g. you must compare both reactant and product, here we're just looking at the "product").

I agree that methoxy destabilizes, but the question was to compare the amount of destabilization. In this answer, you explain the reason why p-methoxyphenol is more acidic than p-cresol. Wouldn't that imply that methoxy stabilizes phenoxide more than a methyl group?

In the earlier question,  I failed to include an analysis of the starting phenols.  I will rework the earlier answer.

What exactly do you mean by "starting phenols"?

When you are comparing the acidity of compounds you must compare the effect of the substituent on both the neutral (reactant) phenol and the ionized phenol (product).

So this answer is correct and the other one is wrong?

Yes, the other answer is incomplete.  I cannot delete it unless you remove the acceptance.  If you do that, I will delete it.

Let us continue this discussion in chat.


text:
Why are so many wave functions associated with hydrogen?
According to Wikipedia, there's an infinite set of possible wavefunctions (orbitals) for the hydrogen atom:
$$\psi_{n\ell m}(r,\theta,\phi) = \sqrt {{\left (  \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{- r/na_0} \left(\frac{2r}{na_0}\right)^{\ell} L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right) \cdot Y_{\ell}^{m}(\theta, \phi )$$

Would an unperturbed electron ever go outside the lowest orbital?
At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?

"Probability densities for the first few hydrogen atom orbitals" (source)

1.Would an unperturbed electron ever go outside the lowest orbital?

First, there is no such thing as "outside the lowest orbital" because the lowest orbital has probability density at all points in space (no nodes).
But yes, at a given temperature electrons will be distributed amoung the energy levels in accordance with a Boltzmann distribution.

2.At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?

There are infinite solutions none of which exceed the ionization energy (the ionization energy is only approached as n approaches infinity).
Think of a series like 0, 3/4, 8/9, 15/16, etc. where there are infinite members without reaching 1 (1 is approached in the limit of infinity).

so if I have only one electron, which can be at all points in space via the lowest orbital, what additional information I get from all the other orbitals? especially since they completely change (hybridize etc.) when around other atoms.

@Sparkler although in the ground state the electron can be anywhere is space, each point has a probability density.  If the electron enters a higher energy state, the probability density will change.  For example, if the atom is in a 2s state instead of a 1s state, the electron has a greater probability of being further from the proton.

and is there any "use" of superposition of orbitals when discussing a single isolated atom?

"There are infinite solutions none of which exceed the ionisation energy" is not the whole story, strictly speaking. Technically, the solutions are not limited to such bound states which give rise to discrete energy spectrum, the are also infinite number of solutions all of which exceed the ionisation energy. These solutions correspond to unbound states and give rise to continuous part of energy spectrum.

See, for example, here.

@Wildcat, right it's not the whole story, in addition to infinite discrete bound solutions there is a continuum of infinite unbound solutions; the whole story would need to include relativity, nuclear spin, electron spin etc. too.

@Sparkler a superposition of equal-energy wavefunctions will also be a solution to the Schrodinger equation.

i know superpositions can be used to approximate helium for example, but do these superpositions actually needed to represent an electron behaviour in an isolated hydrogen?


text:
Molecular weight determination from freezing point depression

Which is the better solvent for molar mass determinations by freezing point depression, benzene or cyclohexane?
Which is the better method for molar mass determination, freezing point depression or boiling point elevation?

What is the correct approach for solving these?

In my opinion, resonance structures might be the key concept, but I don't know how it would help or hurt.
I think boiling has a decomposing effect that we must avoid, meaning freezing the compound(s) is preferable.

But, both of these statements are inadequate.

Molar freezing point depression
Benzene      5.12 °C/m
Cyclohexane  20.0 °C/m

Molar boiling point elevation
Benzene      2.53 °C/m
Cyclohexane  2.79 °C/m

I would base my argument on the constants, instead of waving hands.

What about the second question?

Which is better, 20.0 °C/m or 2.79 °C/m?

I don't really understand the "better" concept here.

Better in the sense of determining the molecular mass of the compound from using either the boiling point or the freezing point. If you took a solution and divided in 2. On one you determine the freezing point depression and with the other you determine the boiling point elevation. If your thermometer is off by 0.2°C, which method would give you the closer value?

what is m in °C/m stands for? minute or molar

@Eka It would be moles/kilogram of solvent.


text:
Reduction of acids with borane
What is the mechanism of the reduction of organic acids with borane? Borane reductions don't work well with electron deficient carbonyl groups. Why is that so, and how isn't the carboxyl group an electron deficient group?

What is the mechanism of the reduction of organic acids with borane?

Source:
Henry Rzepa. Chemistry with a twist. Mechanism of the reduction of a carboxylic acid by borane: revisited and revised. October 16th, 2011. http://www.ch.imperial.ac.uk/rzepa/blog/?p=5114
DOI: 10.15200/winn.142795.57029 provided by The Winnower, a DIY scholarly publishing platform.

but I see an ester in your answer.

I think quoting that R=H is a possibility from the original source might clear up the confusion.


text:
Raoult's Law explained with thermodynamics/ free energy
I got some confusion going on with this explanation of Raoult's  Law.
This is what my book say:

Because the ideal solution has a higher entropy than the pure solvent, the energy of a vapor increases as its molecules expand into a larger volume, that is, if they are at a lower pressure. Thus, solutes in a solution reduce the vapor pressure of the solvent.

Also, I should mention both solutions have the same $\Delta S$ based on the same $\frac{\Delta H}{T}$.
The point is I am confused about the last part of this explanation about how molecules expand. First off, this is all in a closed container, otherwise there's no vapor pressure. If so, the volume should be the same there is no larger or smaller volume.

It is possible that adding more paragraph the idea turn to be more clear. I don't get the author point (at least with this info.) He/she should be talking about molar quantities (or some fixed amount of substance). The volume shouldn't be important, because of the necessity stated in previous sentence.
The grosse idea should be. There are two cases. Pure and mixed solution in liquid phase, both in contact with respective gas phases. Both cases need to have the same molar Gibbs energy. Comparing the Gibbs free energy of the solution and pure state, it should be predictable the change in

pressure due to the constrain of same molar Gibbs energy of the two phases.
But I think that you should add more data to your question or read another text.

Basically a material in a mixture will have a lower vapor pressure than if it were pure.  We know that the vapor pressure of a liquid can be determined by:
$$\Delta G = \Delta G_v^o + RT \ln(P_x)$$
Given that $\Delta G$ is $0$ at equilibrium, we get:
$$\Delta G_v^o = \Delta H_v^o -T\Delta S_v^o = -RT\ln(P_x)$$
Being an ideal mixture $\Delta H$ is $0$ and the equilibrium vapor pressure is determined by:
$$-T\Delta S_v^o = -RT\ln(P_x)$$
$$\Delta S_v^o = R\ln(P_x)$$
Meaning that a higher difference in molar entropy ($\Delta S_g^o$) between the phases means there will be a higher equilibrium vapor pressure ($P_x$). Since gases have higher molar entropy ($S_g^o$) the process of vaporization creates an increase in entropy (positive $\Delta S$).  However since a mixture has a higher molar entropy ($S_m^o$) in the initial state than the pure material ($S_p^o$) the phase change to vapor has a lower change in molar entropy:
$$S_g^o>S_m^o>S_p^o$$
and thus:
$$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$
meaning the $\Delta S_g$ will be higher for a pure material than a mixture.
Q.E.D.

Can you tell what the physical implication of vapour pressure being related to entropy is?


text:
What is the highest possible expanded octet?
Often called "hypervalent", chemicals like phosphorous pentachloride and sulfur hexafluoride are possible due to the fact that their central atoms form covalent bonds with more than four other atoms, giving rise to uncommon arrangements.
In the case of $\ce{PCl5}$, the phosphorous atom forms 5 bonds, giving rise to trigonal bipyramidal shape according to the VSEPR model. With $\ce{SF6}$, the Sulfur atom forms 6 bonds, giving rise to octahedral shape.
My question is, which central atom can achieve the highest number of bonds through the use of an expanded octet, and what is its shape?
The greatest I can think of is iodine heptafluoride, whose shape is pentagonal bipyramidal due to its fluorine atom of 7 bonds.

What you really want to know is what is the highest possible coordination number. It's possible to go quite high; take a look at this Wikipedia article. Also, though it's not completely related, I used to have a great link with all the (theoretical) lowest energy coordination geometries up to massive coordination numbers like 30. I'll see if I can find it.

Interesting, I never knew about those. It would seem strange to have one equal to 30, as it would seem that such an arrangement would require the use of electrons from lower principle energy levels, if I'm reading that article correctly, that is.

Oh the thing I said about a coordination number of 30 is purely theoretical, just a mathematical investigation for the fun of it! The highest I know is actually 16 though, from uranocene. Though perhaps one should establish a difference between coordination number for single-atom ligands (i.e. hapticity 1) and for multiple-atom ligands (hapticity greater than 1).

as for doubtful octet exp. concept: http://chemistry.stackexchange.com/questions/13949/how-can-sulphur-and-phosphorus-expand-their-octet, http://chemistry.stackexchange.com/questions/19433/how-can-iodine-bond-with-5-fluorines-in-iodine-pentafluoride

also http://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit

Hapticity is a new idea to me, though I guess it makes sense even if the required hybrid orbital theory is pretty advanced. When I was asking the question, single-atom ligands were all I knew were possible, though now I'm interested in both and may have to ask a separate question.

Thanks, Mithoron, I saw two of those posts that challenge the octet rule already. Even if it isn't a rule, my question still remains as to how great a number of single-atom ligands are possible around a central atom.

I took an interest in this question because it's something I recently wondered myself. First of all, I should clarify that while you mention hypervalency, what you seem interested in is hypercoordination, or even more generally, just compounds with high coordination numbers (hypercoordination is used specifically when the number of ligands in a compound is larger than "normal"). Hypercoordination and high coordination numbers are entirely independent of hypervalency or VSEPR theory altogether. Regardless of the precise electronic structure in a compound, coordination numbers can often be determined far more directly, especially when the compounds can create crystal structures for x-ray crystallography or neutron diffraction.
Usually there is little focus given to compounds with more than six ligands, as the vast majority of compounds will have atoms surrounded by six or less ligands. However, there are some representatives for higher coordination numbers. Some books will mention iodine heptafluoride, $\ce{IF7}$, as a seven-coordinate compound, with its pentagonal bipyramidal structure. It is still possible to add a fluoride and obtain an example of coordination number 8 in the octafluoroiodate(VII) anion, $\ce{IF8^{-}}$, which has an interesting square antiprismatic geometry (take a cube and twist a face by 45°).
For coordination number 9, representatives can be found in the transition metal hydrides, such as the nonahydridorhenate anion, $\ce{ReH_9^{2-}}$, and the lowest energy configuration in this case is a curious tricapped triagonal prism.
Is it possible to go higher than coordination number 9 in coordination compounds? While there are examples, at the present it seems that none of them contain solely monodentate ligands, that is, individual ligands which only bind to the centre once. This is because either there would have to be a lot of crowding over the central atom, creating repulsions between the ligands, or because to allow enough space the ligands would have to stay relatively far from the central atom, making their bonds weak.
However, if a single ligand is allowed to bond to the centre through more than atom simultaneously, then the coordination number can keep increasing without requiring the presence of too many ligands. Actinides have a very rich coordination chemistry and are capable of generating several impressive compounds, such as uranocene with coordination number 16, but it seems the current record is held by actinide elements surrounded by four cyclopentadienyl rings, each with five carbon atoms, reaching an amazing coordination number of 20 in tetrakis(cyclopentadienyl)thorium(IV) ($\ce{Th}\mathrm{(\eta ^5-}\ce{C5H5}\mathrm{)_4}$ or its uranium analogue.
To finalize, it's interesting to note that though there are no examples yet of compounds with coordination number 10 or above containing only monodentate ligands, their expected geometries can be calculated even for much higher coordination numbers (under certain assumptions). For very high coordination numbers, calculations and physical reality will likely diverge, but perhaps a few more coordination numbers with the expected geometry will be unlocked by the study of ultraheavy element chemistry. Of course, the calculations linked here don't help much to study the geometry of coordination compounds with polydentate ligands, as they rely significantly on the geometry of the ligand itself
Edit: This very relevant article raises some interesting points. For example, endohedral fullerene compounds could be thought of as a central atom surrounded by a single ligand in the shape of a cage, so one could possibly make a case for structures with coordination numbers of 60, 70, 80 or even more. The article also calculates the possible existence of a compound containing 15 monodentate ligands, the cation $\ce{PbHe_{15}^{2+}}$, though it would be very weakly bound (as you might expect from a helium compound) and possibly restricted to the gas phase.

Additional content very related to this answer (and partially superceding it) can be found in this more recent question. In particular, the coordination number for cyclopentadienide and other  cyclopolyene ions arguably is different from what I wrote here.


text:
Why do many hypervalant chemicals have non-constant bond angles?
According to the VSEPR model, when a molecule in formed in the shape of $\ce{AB_{x}}$,

like $\ce{IF7}$, $\ce{SCl6}$, and $\ce{PCl5}$,

its structure is found by reducing the repulsions between peripheral atoms. This gives rise to many shapes, most of which have constant bond angles. Examples of those with constant bond angles include

$\ce{CH4}$ with 109.5° bond angles
$\ce{CO2}$ with 180° bond angles.

My question is then, why do some have non-constant bond angles?
The following do not

$\ce{IF7}$ with 72° and 90° bond angles
$\ce{PCl5}$ with 120° and 90° bond angles.

The $\ce{IF7}$ and $\ce{PCl5}$ molecules have 7 and 5 ligands, and pentagonal and trigonal bipyramidal geometries, respectively. These geometries happen

simply because
  there is no geometrical arrangement which can result in five equally
  sized bond angles in three dimensions.

(Wikipedia)
There is no arrangement in which all the angles would be the same, except in a plane, where the repulsion would obviously be a lot higher than in the pyramids.


text:
Is there a difference in the reactions of a dilute acid and a concentrated acid?
All of the exams that I've sat so far, and the homework questions I've had, always ask for the reaction between a substance and a dilute acid. Is there a difference in the reaction between a dilute acid and a concentrated acid when each is applied to a substance? If so, what is it, and how dilute is dilute?

Can you provide an example? I might have an answer, but it depends on the context.

Example:  https://drive.google.com/file/d/0B_LohPswqrHqXzRFOERhR0xPUzg/view?usp=sharing

I presume you're a high school student, in which case specifying your educational board would yield more relevant answers.

While it would be easier with examples, I leave here some ideas.
This could be because the idea of concentration is an approximation of the concept of thermodynamic activity, which is frequently referred to as effective concentration. Depending on the question being asked, it might be important that the solute behaves as if it was in an infinite concentration, where there's no interaction between solute molecules/ions and the concentration equals the activity.
The mention of dilution might also be important if the reaction involves water as a reagent. Even though water usually does not receive much attention when discussing reactions, since it's the solvent with "constant" concentration, it is important to note that several reactions wouldn't happen in its absence. That would be the case if the acid is available in a purified form, such as a gas (like HCl) or as solid (such as benzoic acid). It's also worth mentioning that the concept of acidity is inherent to solutions, so the two above aren't acids unless diluted.
The above also highlights something that is noteworthy. While diluted is the opposite of concentrated, it may also be said as the opposite of "not-in-solution", meaning not mixtured with the solvent. This could also be a reason to mention dilute acids, to make it clear they are in solution instead of any other form.
Edit:
Given the example, I do believe it's because the reactions happen in the presence of water, but honestly, I'm not sure.
Edit 2:
Another possible reason to mention dilute acids would be in case it's necessary to consider the concentration of a medium-strength acid. In dilute solutions, those can be considered completely dissociated, which makes calculations a lot easier, if that's the case.

Acidity isn't inherent to solutions - acid-base reactions can occur also in solid or gaseus phases.

It depends to some extent on the acid. The most important example would be sulphuric acid because in concentrated form ~98% there is not enough water present to accept the protons available so it becomes a weak acid if concentrated enough. This is not the case for conc. nitric acid (~65%) or conc. hydrochloric acid (~36%). In addition conc. H2SO4 is a powerful dehydrating agent (eg. with sucrose). It is also a strong oxidising agent eg. gives SO2 rather than hydrogen with metals. It can also be simply that the reaction could be inconveniently vigorous with too high a concentration. I hope this helps.

It seems rather common misconception to think that lack of water could weaken acid stronger than water because, it seems, people are stuck in pH reasoning which breaks down in very concentrated acids - see http://chemistry.stackexchange.com/questions/27905/how-concentrated-can-an-acid-be


text:
Sodium thiosulfate formation
I have a problem solivng this chemical problem. Any help would be highly appreciative.
Sodium Thiosulfate can be made by the reaction
$\ce{Na_2CO_3 + 2Na_2S + 4SO_2 ->  3Na_2S_2O_3 + CO_2}$
Question: How many grams of $\ce{Na_2S}$ are required to react with 25.0 g of $\ce{Na_2CO_3}$ if $\ce{SO_2}$ is present in excess?
My attempt:
$25.0 \text{ g } \ce{Na_2CO_3}\frac{1 \text{ mol } \ce{Na_2CO_3}}{105.99}\frac{2}{1}\frac{78.05}{1}$
The answer from the book states: 25.9g$\ce{Na_2S}$

Welcome to Chem.SE! Its good to see proper formatting of chemical equations in one's first post. Are you missing an equality sign under My attempt ?

You are completely correct in your dimensional analysis and stoichiometry.

36.8g is the correct answer.

Molar mass $\ce{Na2CO3 = 106g/mol }$
${25g \rightarrow 25/106 = 0.236}\text{ mol } \ce{Na2CO3}$
This will react with ${0.236*2 = 0.472} \text{ mol } \ce{Na2S}$
Molar mass $\ce{Na2S} = 78.0 \text{ g/mol}$
$0.472 \text{ mol } = 0.472*78.0 = 36.8 \text{ g } \ce{Na2S}$ will be needed.
I will now prove your book wrong
25.9g $\ce{Na2S}$ = 0.332mol $\ce{Na2S}$
0.332mol $\ce{Na2S}$ reacts with 0.166mol $\ce{Na2CO3}$
0.166mol $\ce{Na2CO3}$ = 17.5g of $\ce{Na2CO3}$, which is different from our starting quantity of 25.0g.

The answer from the book says 25.9g Na_2(S)

I'm certain your book is wrong. We both solved in different layouts and got the same answer. I'll check and see what may be the issue.


text:
The concentration is unity! Why?
I want to know why concentration of solids and liquids are taken unity in expression of equilibrium constant( heterogenous equilibrium). It is usually mentioned that the density of solids and liquids( ideally) does not change and hence concentration does not change and so rate does not depend on solids and liquids. I accept that it is true. But will not it change the actual value of $ K_c$? We use $K_c$ everywhere like calculating extent of reaction, in thermodynamics- calculation of standard gibbs free energy and so on...
How the assumption of unit concentration in expression  is justified?
I want to give you an example:
 Suppose in expression of $K_c$ the concentration of water(liquid) is involved  then the concentration wil be $(1000 g/L)/(18g/mol)$ ( density/ gram molar mass) Now if we take the concentration of water(liquid) unity then definitely value of $K_c$  will change and this change will be reflected. for example: when we want to predict extent of reaction $OR$ put value of $K_c$ in different equations relating it like $G^○=-RTlnK_c$

Check this: http://dlrgenchem.com/LECTURES/omitsolids.htm

This doesn't look like a duplicate. The linked question talks about constant concentration while this is about taking unity as the concentration of solids and liquids in calculations.


text:
How hydrate isomerism results difference in colours?
The standard example is $\ce{CrCl3.6H2O}$, which can have three distinctly different
crystalline compounds now known as
$\ce{[Cr(H2O)6]Cl3}$ (violet),
$\ce{[CrCl(H2O)5]Cl .H2O}$
(blue-green), and
$\ce{[CrCl2(H2O)4]Cl.2H2O}$ (dark green).
How does this difference of ligands in octahedral system affect the colour of compound? I think it may be due to the change of splitting energy but if that so then how does it actually work?

Your idea appears to be correct.
Initially, the structure of $\ce{CrCl3.6H2O}$ corresponds to the dark green compound $\ce{[CrCl3(H2O)3].3H2O}$. This complex shows hydrate isomerism; i.e. the chlorido ligands are step by step replaced by water molecules:
$$\ce{$\underset{\text{dark green}}{\ce{[CrCl3(H2O)3].3H2O}}$ <=> $\underset{\text{dark green}}{\ce{[CrCl2(H2O)4]Cl.2H2O}}$ <=> $\underset{\text{blue green}}{\ce{[CrCl(H2O)5]Cl2.H2O}}$ <=> $\underset{\text{violet}}{\ce{[Cr(H2O)6]Cl3}}$}$$
Thus, the actual complexes are:
$\ce{[CrCl3(H2O)3]}$ (dark green)
$\ce{[CrCl2(H2O)4]+}$ (dark green)
$\ce{[CrCl(H2O)5]^2+}$ (blue green)
$\ce{[Cr(H2O)6]^3+}$ (violet)
Each of these complexes has octahedral geometry. Cr(III) has a $\mathrm{d^3}$ electronic configuration. In the ground state, three electrons occupy the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$). The two $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2{-}y^2}$ and $\mathrm{d}_{z^2}$) are empty.
Three spin allowed d–d transitions are expected for Cr(III) complexes. The corresponding absorption maxima of $\ce{[Cr(H2O)6]^3+}$ are:
$$\begin{alignat}{2}
&\mathrm{^4A_{2g}\to{^4T_{2g}}}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(F)}}\quad&&\tilde\nu_2=24\,500\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(P)}}\quad&&\tilde\nu_3=38\,600\ \mathrm{cm^{-1}}
\end{alignat}$$
The first transition $\mathrm{^4A_{2g}\to{^4T_{2g}}}$ corresponds to promoting one electron from the $\mathrm{t_{2g}}$ level to the $\mathrm{e_g}$ level. Thus, it directly gives the value of $10\ \mathrm{Dq}$ as $17\,400\ \mathrm{cm^{-1}}$.
The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the low-field ligand $\ce{Cl-}$, the value of $10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are lowered:
$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\\
&\vdots \\
&\ce{[CrCl6]^3-}\quad&&\tilde\nu_1=13\,200\ \mathrm{cm^{-1}}
\end{alignat}$$
By way of comparison, if the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{CN-}$, the wavenumber of the corresponding absorption maximum is increased ($\tilde\nu_1=26\,700\ \mathrm{cm^{-1}}$ for $\ce{[CrCN6]^3-}$).
The wavenumbers can be converted to the wavelength of the absorbed light:
$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\quad\lambda=575\ \mathrm{nm}\ \text{(yellow)}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\quad\lambda=610\ \mathrm{nm}\ \text{(orange)}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\quad\lambda=640\ \mathrm{nm}\ \text{(red)}\\
\end{alignat}$$
Since the colour of the absorbed light in this series changes from yellow to orange to red, the corresponding colour of the remaining visible light changes from violet to blue green to green.


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What are the products of the haloform reaction of acetylacetone?
I searched a lot to find the products of the haloform reaction of acetylacetone (pentane-2,4-dione),  but I didn't find any results.
What are the products of the haloform reaction of acetylacetone?

Not sure about this but it looks like malonic acid should be formed.

Because acetylacetone is a 1,3-dione, the carbonyls are very enolizable.  This makes the hydrogens attached to carbon 3 on acetylacetone very acidic.  When introducing the conditions for a haloform reaction, base is added.  The base will take carbon 3's hydrogens, in the presence of some halogen, they will be replaced, this happens twice.  Now, with the product 3,3-dihalopentane-2,4-dione, the reaction will end.
If you insist on a haloform reaction occurring, a much stronger base will be needed and the reaction vessel would probably need to be heated.  It will proceed as usual and create first 2,2-dihalo-3-oxobutanoic acid (2,2-dihaloacetoacetic acid). The reaction will probably stop at this point. Upon heating though, this compound may decarboxylate to give 1,1-dihaloacetone and carbon dioxide. If you wanted to continue with this, you could do another haloform reaction to first generate perhaloacetone, and if you wanted to, this may be able to become trihaloethanoic acid and haloform.
But realistically, the reaction would end at the 2,2-disubstituted dihalo acetoacetic acid (2,2-dihalo-3-oxobutanoic acid)

This is kinda hard to follow without chemical structure diagrams, it would be nice to see what's happening visual.

In 1933 Hurd and Thomas1 investigated the action of bleaching powder (calcium hypochorite) on ethyl acetoacetate (1). They isolated dichloroacetic acid (2) in 60% yield without the production of chloroform. Their conclusion was that chorination did not occur on the methyl group (no chloroform) and that dichloro acetoacetic ester 2 was the intermediate. Reasonably, hydroxide attacks at the ketone, as is typical with disubstituted β-ketoesters.2 Acetic acid was not isolated but is implied by the mechanistic considerations.
In a similar fashion acetylacetone (4) would be expected to afford dichloride 5. Cleavage by base (red arrows addition to the carbonyl; blue arrows for cleavage) yields the anion of dichloroacetone (6), which is in equilibrium with its protonated form 11. Further chlorination of anion 6 yields trichloroacetone 10. A final haloform reaction provides chloroform 7 and calcium acetate 9.

1) Hurd, C. D. and Thomas, C. L., J. Am. Chem. Soc., 1933, 55, 1646.
2) The hydroxide cleavage of ethyl 1-methyl-2-oxo-cyclopentanecarboxylate affords α-methyladipic acid.

authors: คุณากร เพิ่มพูน


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Confused about strong/weak dibasic and tribasic acids
I know that sulfuric acid is considered a strong acid, and the dissociation of its first $\ce{H+}$ ion is complete and of its second $\ce{H+}$ ion is partial.
Phosphoric acid is weak and all three of its $\ce{H+}$ ions dissociate partially.
I just wanted to clarify: are dibasic and tribasic acids considered strong if their first dissociation is complete (and subsequent ones will naturally be partial)?

I mean whenever you have a polyprotic acid/base and they say that if it is strong or weak you can only assume the first dissociation is strong or weak.
Say like you have $\ce{H2SO4}$ and we all know that it is a strong acid. This only means that the first dissociation is strong.
$$\ce{H2SO4 -> HSO4- + H3O+}$$
So in this case above only the first dissociation is strong. And then you cannot and do not know if $\ce{HSO4}$ is strong as well.
So overall in any case we cannot assume and the data must be provided.
Good Question.

If you think about each dissociate state separately, than it's easy to answer your question. $\ce{H2SO4}$ is a very strong acid $(\mathrm{p}K_\mathrm{a}\approx -3.0),$ but $\ce{HSO4-}$ is not so strong $(\mathrm{p}K_\mathrm{a}\approx 2.0).$
Phosphoric acid starts with $\mathrm{p}K_\mathrm{a}\approx 2.14,$ which as as you said is not so strong, and $\ce{H2PO4-}$ and $\ce{HPO4^2-}$ have $\mathrm{p}K_\mathrm{a}\approx 7.2$ and $\mathrm{p}K_\mathrm{a}\approx 12.3,$ so definetly weak.
So yes, we say sulfuric acid is strong because we're talking about $\ce{H2SO4},$ and what comes after should be mentioned separately.


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Can we prolong life?
This is rather a bunch of questions that I decided to post on Chemistry StackExchange since I thought the chemists would have the most knowledge about the chemical processes of life.
So recently I have came across stem cells and a lot of revolutionary medical products that can rebuild tissue and even organs.
And then I thought that can we use these products to transplant our organs and in a sense, rebuild our own body. I have even came across making a heart and lungs as well.
If so is there any thing that stands between having a longer life?
What else do we need to have in order to prolong and in a sense immortalize ourselves.
I know that it would be impossible to do this for everyone since our planet cannot sustain more life, but theoretically is it possible to do this?
What are you views into this?

It's rather a question for biology, related: http://biology.stackexchange.com/questions/30298/do-all-organisms-have-to-die, http://biology.stackexchange.com/questions/2879/are-human-bodies-programmed-to-die?rq=1, http://biology.stackexchange.com/questions/1495/what-evolutionary-explanations-are-there-for-death?lq=1, http://biology.stackexchange.com/questions/17077/why-does-evolution-not-make-our-life-longer?lq=1, http://biology.stackexchange.com/questions/30298/do-all-organisms-have-to-die

@Mithoron  IMO, I think this is a fair question for chemists too.  That's why a lot of (bio)chemists are being paid to work in labs and find treatments for this disease.

Even then it's still too broad. Also about telomers: http://biology.stackexchange.com/questions/13841/is-solving-cancer-required-in-order-to-avoid-aging

I am pretty sure there is a reason for the tag: biochemistry, medicinal chemistry, drugs. Of course there might be other stackexchange sites that can answer this question as well but there isn't anything wrong for it being on Chemistry. After all chemistry defines us.

Ageing is a disease.  Just as treatments have been found for other diseases, ageing can be slowed or reversed as well.
Not that long ago, scientists noticed that at the end of each DNA strand there were thousands of nucleotides.  Nucleotides are used for genetic coding.  That is to say, certain sequences of nucleotides instruct the body to synthesize  certain proteins when the genetic code in the DNA is being read.  DNA is really like a massive software program.
In any case, when scientists noticed all of this code at the end of a DNA strand, at first they couldn't figure out what is was for, so they thought it must be gibberish.  Later they noticed that everytime a cell divided (e.g. everytime the DNA was read) a piece of this gibberish was lopped off.  Eventually it became understood that this "gibberish" was the basis of aging.
Each lopped off piece was termed a telomere.  Each telomere contains the same nucleotide sequence (code).  There are about 15,000 telomere nucleotides at the end of the DNA strands in an embryo.  This has been shortened to about 10,000 when you are born and when it reaches a length around 5,000, the body dies.  It turns out that the telomeres play a role in stabilizing DNA strands during the copying process, when their length gets down to around 5,000 nucleotides the telomere length becomes to short to provide adequate stabilization and errors (genetic mutations) occur in the copying process and/or the cells become senescent (become inactive, go to sleep).  Errors or sleep, either way, the body stops functioning efficiently.
This telomere shortening occurs during cell division in all human cells except for sperm and egg cells.  Sperm and egg cells are immortal.  There is an enzyme called telomerase, it is expressed only in the sperm and egg cells.  After these cells divide a telomere is lopped off, but telomerase re-adds a telomere to the DNA in sperm and egg cells.  Hence, their telomere length never shortens - they live forever.
All cells contain the code to make telomerase, but that code  (or signaling) is turned off in all cells except for sperm and egg cells.  A specific protein combines with a certain portion of DNA and that is all that is required to prevent telomerase expression in a cell.
If scientists could find a way to prevent that (repressor) protein from complexing with DNA, then telomerase expression could be turned on and that cell would then be immortal.
As you can imagine, this is a big area of research.  Here is a link to a company that has been active in this area for a long time, and here is a link to a basic tutorial on the subject of telomeres.  Poke around the web site, the videos and other tutorials are extremely interesting.  Even aspects such as the social consequences of curing aging are discussed.
Enough progress has been made that you can buy products in the over-the-counter market that reduce telomere shortening, but they're very expensive at the present time.

+1. But how come that the body comes equipped with this repressor protein? So say that someone has a genetic mutation and the protein you talked about above is not present, then is there a chance that the person can live forever assuming that he/she doesn't die because of environmental causes?

That's an excellent question.  They touch on it in the basic tutorial.  They point out that the first unicellular organisms on earth were immortal - they still exist, our sperm and egg cells can be traced back to them (germ line).  They go on to say that as multicellular organisms evolved, the cells "became afflicted with a disease" and "the cause of this disease is still speculative".

-1:  I'm no expert on microbiology, but I believe there's a lot more than just telomere shortening that causes aging.  There's a biologist, Aubrey de Grey, who identifies 7 causes of aging of which only the first will benefit from a fix to telomere shortening.

@Garrett wrote, "I believe there's a lot more than just telomere shortening that causes aging. ".  There probably is.  Like all diseases there are probably multiple approaches to delay, eradicate and/or reverse the effects of ageing.  Take a look at this link where a number of experiments are described that provide a "proof of concept" for the significant involvement of telomeres on ageing.  Again, I agree that there may well be other approaches, but a lot of money and effort appears to have been focused on this one.

There's oodles of evidence that teleomere shortening is not the primary cause of aging, ranging from little to no correlation between telomerase activity and lifespan through to an almost complete lack of significant lifespan extension in model organisms from modifications to telomeres (save for early death if they get really short really fast--but that's true of everything).

Also, sperm and egg cells are absolutely not immortal.  In fact, they age faster than many other cell types (e.g. human women's eggs lower in quality dramatically by the 40s, when many other cellular processes are still humming along fine).  Instead, it is that they are a step in the immortally-privileged process of the lineage of the species; exactly what step in that process confers immortality (or age-reversal) is not fully clear, but it is clear that it is not the lack of aging of sperm and egg cells themselves (or their progenitors).

@RexKerr  I agree that sperm and egg cells are not immortal, it was the DNA contained in these cells I was speaking about.  Let's agree to disagree.

@ron - This isn't a matter of opinion.  Your post, right now, literally says that sperm and egg cells are immortal.  They're not.  It's wrong.

@RexKerr  Have you had a chance to poke around the Sierra Science website? As they point out, an embryo develops from parental sperm and egg DNA.  If this DNA is telomerically shortened, why aren't babies born with 30 year old (or whatever the parents age is) DNA.  The answer lies in sperm and egg DNA not being telomerically shortened;  said differenty such DNA does not age, e,g it is immortal.

If that was how it happened, your post would still be wrong because that's not what it says.  Also, that's not how it happens; oocytes can have short telomeres.  They just fix it after fertilization.  http://www.nature.com/stemcells/2007/0711/071129/full/stemcells.2007.123.html

There is no chemical reason why life couldn't be prolonged at least somewhat.  There are a huge variety of undesirable biochemical reactions (oxidation of proteins for instance) that cells must deal with as part of being alive, and given that lineages are immortal (as the process of growing up and creating germ cells and then fusing them into a new being somehow resets the aging process) we can see that chemical immortality is entirely possible.
In fact, that's what DNA is: the chemically immortal molecule.
But there are all sorts of practical reasons why it may be difficult.  There is extensive research in a variety of organisms implicating all sorts of fundamental biochemistry in lifespan: insulin signaling, the EGF pathway, TOR, the unfolded protein response, and so on.  Ultimately it is not the chemistry which is the barrier: life already knows how to select and grow its way out of any undesirable side-reactions.  It is the biology: can you actually regulate yourself to do the biochemistry you need to do to live longer?
So the answer to this chemistry question is: chemically speaking, it's a biology question.

To provide a more abstract answer not particularly founded in science, have you considered that we have a finite lifetime in order to encourage us to reproduce and therefore create/promote genetic variability.
Thus providing a far greater chance for our DNA to live on forever (albeit in a different form). This seems to already offer us immortality in some way. Unless of course global disaster strikes causing total extinction.
To me it seems this is the only way to achieve true immortality as something as simple as a disease mutating could wipe us all out, variation is what allows DNA to continue to live.
Ultimately what would living forever allow us to achieve, there would be no goal in life and instead the potential for our DNA's death would increase as other reproductive organisms evolve and adapt.


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Structure of a metal in gas phase
I'm going to use aluminium in my example, but this is applicable to all metals.
Aluminium has delocalised electrons. It has aluminium cation -electron attraction. It's easier to visualize the atomic structure of aluminium in solid form.
But if it's in the gas phase, will delocalised electrons return to the atom itself, or is there going to be an electron and cation mixture (electrons floating around)?
So can aluminium (or any other metal) conduct electricity as a vapour, and what would its conductance be compared to its solid form?

You need to be more specific: do you mean an isolated, electrically neutral Al atom? An Al ion of some kind?  A cluster of Al atoms? A cluster of variously ionized Al atoms?  A dispersed gas of Al atoms? A dispersed gas of Al ions? Etc. The answer depends somewhat on which specific system you're asking about.

no,i am talking about cluster of atoms.

Metal vapors do conduct electricity, as in the mercury vapor lamp (also used in fluorescent lamps), but, since the atoms are isolated or in small (molecular) clusters, conduction is ionic, not metallic. Since electrons are far less massive (and therefore more mobile) than the heavy ions, they are responsible for the majority of the conducted current.
Electrons are forcibly ripped away from the mercury atom (well, raised to higher levels) -- the first ionization potential is 10.39V, and its second is 18.65V, so in a mercury vapor lamp or mercury vapor rectifier, the potential drop of at least 20 volts implies more than one of the outer electrons are lost from many of the atoms.
Note that ionized gases are much less conductive than solid or liquid metals. Consider a 70 watt sodium vapor lamp, with an arc length of perhaps 10 cm and a potential drop of ~18 volts. The finest commercially available wire, 40 gauge, would have a potential drop of just ~1.3 volts for metallic copper wire, or ~4 volts for sodium, which has 1/3 the conductivity of metallic copper (if they made fine sodium wire), at the same length and current (~3 A).
See the useful chart explaining sodium-vapor ionization .


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Can we picture metallic bonding as an equilibrium between electrons and cations?
Can we picture metallic bonding as an equilibrium between electrons and cations?
Suppose:

$$\ce{Al^3+ + 3e- <=> Al}$$

I mean isn't $\ce{Al^3+}$ stable, why would it need the extra electrons since its outer shell is stable?

Related although not a direct duplicate. http://chemistry.stackexchange.com/questions/25003/clarification-for-metallic-bond?rq=1

In metals, electrons are non-localized, forming a "sea" of electrons, rather than having them localized, as in the $\ce{Na+Cl-}$ lattice of crystalline salt. See Metallic bonding for a more complete description.
It is, of course, a matter of degree, as covalent, ionic and metallic bonding can "blend" from one to the other. A bond can be considered partially ionic and covalent, for example; see these helpful graphics


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How to measure band gap of powder sample with using diffuse reflectance method
We have UV SHİMADZU 2550 and its diffuse reflectance attachment. I want to measure band gap of single crystal samples. What should I do? Which formulas should I use to calculate band gap? I don't have much sample...

It's not too hard to determine band gap.  When the energy of a wavelength of light is higher than the energy of the band gap the light will be absorbed, any wavelengths that are lower in energy will reflect or transmit. Simply make the spectrum, there should be a cliff-looking cut-off in transmittance, determine the wavelength and corresponding energy and that is your band gap.


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How is the pKa of extremely weak acids determined?
According to Wikipedia, the conjugate acid of LDA has a pKa of about 35 - which is in line with figures I've seen elsewhere.
How then can such a weak acid's pKa be determined? If the determination is made in water, how is that even possible? Water would be a stronger acid by far and couldn't it easily mask the acidity of LDA's conjugate acid?
Is LDA's pKa value even determined in water?
Similarly, how are extremely strong acids' pKa values determined? Wouldn't the leveling effect come into play as one introduces, say, tosic acid into water? How then can we determine the strength of an acid stronger than hydronium ion in water?

There are a number of different spectroscopic techniques that lend themselves to measuring pKa's.  One approach involves ion cyclotron resonance.  It can accurately assess gas phase equibria.  Starting with a molecule with known pKa, add a second molecule with unknown (but expected to be within a few orders of magnitude) pKa and measure the equilibrium constant; work your way up to methane.  Other studies have shown a high correlation between gas phase and solution pKa's, see for example pp. 97-98

in this link

http://www.sirius-analytical.com/science/pka/methods-measuring-pka ? This kind of method is often used for pharmaceutical compounds, which are weakly acidic/basic to a certain extent. Shows good correspondence with lit. values for pKa of common chemicals too.

Sure enough, such extreme pKa values can't be determined in water for the very reason you've specified (the so-called leveling effect). Now, pretty much every textbook containing the phrase "leveling effect" would also provide a word of explanation as to how can we move beyond the common pH range of water. Even the link above does that; please, read it till the words "discrimination window", at least.
Long story short, we abandon water altogether and move to another solvent (more acidic, like acetic acid, if we are interested in super-strong acids, or more basic, like liquid ammonia, if we want to deal with super-weak acids and super-strong bases). We take our compound in question which is too acidic (or too basic) to be measured in water, pit it against some known compound which is less acidic (or basic) so that its pKa (or pKb) is known from the measurements in water, measure the constant of equilibrium between them and thus derive the unknown pKa. If our compound is too strong for that solvent, we step further to another solvent. Thus we build a kind of ladder, much like the ladder of years in dendrochronology or the ladder of distances in astronomy. And that's how the wild values of pKa like -20 or +40 can be reached.

Standard pKa measurements
pKa is, as you mentioned, commonly measured in water. This worked well initially, but as chemists got more ambitious in their measurements, they began to come up against the levelling effect, whereby the acidity/basicity of the compound being measured was limited by the acid/base properties of the solvent being used to measure it.
To a certain extent, this was mitigated by moving from water to a solvent that is unchanged over a wider range of pKa (the range of pKa is known as the discrimination window), DMSO being the example chosen. This change to DMSO allowed the pKa of far more compounds to be measured. DMSO has been used to measure the pKa of LDA, the example you gave.
Even in cases where DMSO is unsuitable pKa may be measured in other solvents. There is however an issue with using different solvents to measure pKa in that the values are not directly comparable since different solvents affect the ease of protonation/deprotonation. The discrimination windows of several solvents are given below (Source: Wikipedia).

Standard pKa measurement is carried out by dissolving the compound in the solvent and titrating in acid/base slowly, whilst measuring the pH change, which can be done in many ways. This gives a curve from which the pKa may be read off (pKa is the point at which the compound is 50% ionised).

Measuring pKa of weak acids
Whilst pKa has been determined for many common chemicals, there are instances where pKa will be extremely high/low. In these situations, we can abandon the direct measurement as described above, and look for some other way of doing it.
-A brief interlude- To give an analogy that may (or may not) be more familiar. In medicinal chemistry, logP is an incredibly useful property. It was traditionally measured by partitioning the drug molecule in-between octanol and water and seeing how much of the compound ended up in each layer. This was laborious, and so medicinal chemists moved from directly measuring logP to carrying out other analysis from which logP could be derived. The common way of doing this is to use a column (similar to chromatography, but specially adapted). The chemist first runs a set of samples with known logP through the column, followed by the compound being studied. By looking at retention times and then comparing and doing some maths, a calculated logP can be obtained which is generally fairly reliable.
With pKa measurement, the situation is a similar in that we can indirectly measure the pKa and extrapolate.
There are, however, two separate issues here. A first problem is that pKa measurements at the extreme are just difficult (i.e. very small or very large) due to the titration curve becoming very difficult to measure accuratly, it isn't impossible however. This is really the main issue.
The other problem is related to levelling, but as you can see from the table above , we do have solvents available to measure over pretty much as wide a range as you could ever want (note:  if you're considering measuring a pKa in hydrofluoric acid, you should probably reconsider why you need to measure it in the first place).
The following paper (free to view) details the most common direct and indirect methods of determination.
From my experience, there are three methods I think are worth discussing here as alternatives to direct pH change measurement. . .

HPLC measurement in which the pH of the mobile phase is varied. As the pH changes, the compound may ionise, which changes its retention time on the column. This may be used to extrapolate pKa by comparison to samples of known pKa. The HPLC columns used now are able to tolerate the use of highly acidic/basic mobile phases for pKa determination.
NMR measurement in which we 'titrate' in the NMR tube, adding acid/base, and watching as the chemical shift changes. This has allowed some super accurate pKa's to be measured.
Comparative measurement in which a compound with a known pKa is reacted with our compound for analysis, and we see the extent to which they protonate/deprotonate one another. By looking at this equilibrium constant we essentially can derive the pKa.

Last, but not least...
Computational determination of pKa
Although not a 'measurement' of pKa, in silico computational methods are becoming increasingly accurate. This is especially true within a normal pKa range, but has also found applications with highly acidic/basic molecules. Some common packages for this are ACDpKa, Epik and Marvin.


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When and why do hot material take heat from cold material?
This is what I read on Chemwiki:

Rudolf Clausius was a German physicist, and he developed the Clausius statement, which says "Heat generally cannot flow spontaneously from a material at a lower temperature to a material at a higher temperature." However, there is one exception to this statement and that is sometimes a hot material will take heat from the cold material, making the hot material warmer and the cold material colder.

Chemwiki does not provide a reference for the exception about heat transfer from cold material to hot material. How is this possible? What is an example of this happening?

Hmm, good question. I suppose the exception is when there is some work done on the system.

Fridge or example :)

I admit that was confused about that sentence too and agree it didn't make much sense. I have erased it from the ChemWiki. We accept all suggestions to improve the site graciously and appreciatively.

The statement is wrong and/or poorly formulated. Heat cannot spontaneously flow from cold areas to hot areas, there are no exceptions known. It contradicts Clausius' statement of the second law and hence also the more general observation/statement that the entropy of an isolated system is increasing.
Of course you can drain heat from cold regions and transport it to hotter regions, but this should always go paired with some other (larger) increase in entropy somewhere else, e.g. by doing work like in a Linde refrigeration. An other example is that some salts have a very high entropy of solvation and can cool water to ~ -15 °C in a lab at room temperature. These salts can be/are used in practical chemistry for cooling a reaction mixture.


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Why heat of reaction under neither constant pressure or constant volume is considered (in my textbook, or probably somewhere else)?
So my textbook gives me this random equation:
 $\Delta U = q + w$. Then consider two conditions, under constant pressure and under constant volume. Because $w = -P\Delta V$, constant volume will make $\Delta U = q$. And constant pressure equation is basically the same.
But why is isn't heat under neither condition considered? Because it is of significance?

The constant pressure and constant volume scenarios are presented when you first study thermodynamics because they present an easy way to simplify the calculation of energy changes ($\Delta U,\ \Delta H,$ etc.) based on the change of one easy to measure phenomenon.

Internal Energy $U$

So, the change in internal energy must equal the sum of the work done $w$ and the heat transferred $q$. This is not random, it is a consequence of the Law of Conservation of Energy. If the internal energy of a system has changed then energy must be exchanged with the surroundings. There are only a few ways this can happen (heat, work, electromagnetic radiation, ...), and heat and work are the most important for most thermodynamic scenarios.
Thus, $\Delta U = q + w$. We need to be able to relate $q$ and $w$ to things we can measure. We calculate $q$ based on the temperature change of the system $\Delta T$ and the heat capacity of the system $c$. Thus, $q=c\Delta T$. Work is defined (in one way) as the change in volume of the system times the pressure. The negative sign is present to make sure that an expanding system is doing work on the surroundings (and thus losing energy): $w=-P\Delta V$.
Now we have $\Delta U = c\Delta T -P\Delta V$

Constant Volume

If volume is constant $\Delta V = 0$, which means that $w=0$, regardless of pressure. Thus, $\Delta U = q = c\Delta T$.

Constant Pressure

This one is trickier, but we can do it for a gaseous system.
We do not need to measure volume, since $PV=nRT$, but it is easier to measure volume than $n$, which may also be changing. If $n$ is constant, then, the equation for change in internal energy becomes:
$$\Delta U = c\Delta T - nR\Delta T$$
If $n$ is not constant, then we need to measure that change in volume and temperature.
If the process is in another state, like in solution, $\Delta V$ might be very challenging to measure. Let's hope we can measure $\Delta n$.

Enthalpy $H$

Because of the challenge of measuring volume changes in solution, we invented a new thermodynamic potential $H = U + PV$. Total differentiation and whatnot gives us:
$$\Delta H = \Delta U +P\Delta V + V\Delta P = c\Delta T - p\Delta V + P\Delta V + V\Delta P$$
$$\Delta H = c\Delta T + V\Delta P$$
Thus, at constant pressure, $\Delta H = q = c\Delta T$, which is easy to measure. Enthalpy is similarly challenging at constant volume as internal energy is at constant pressure.

Constant pressure and volume.

Under these conditions, $\Delta U = \Delta H = q = c\Delta T$.

Pressure and volume are not constant

Okay, so if $w=-P\Delta V$, how do we calculate work? If pressure is changing, which value of $P$ do we use? It turns out we use all of them and we need some calculus to do it. This case is often left out because many introductory textbooks consider the likelihood that many students beginning their study of chemistry may not have taken calculus.
Here we go. We need to replace the total change in $V$ with the infinitessimal change in $V$ at each moment $\Delta V \rightarrow  dV$. $\Delta V$ is the sum of all of the infitinessimal $dV$ values. Now, we can multiply each $dV$ with the pressure at that moment and add them up to get work (which involves integration):
$$w = \sum_{j=1}^n P_j dV_j=\int_{V_i}^{V_f}PdV$$
However, we cannot leave well enough alone, because $P$ is changing, so we need to substitute for $P=\frac{nRT}{V}$.
$$w=\int_{V_i}^{V_f} \dfrac{nRT}{V}dV=nRT\dfrac{\ln V_f}{\ln V_i}$$
Now the above maths only work if temperature is constant. What if it's not?


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Electrophilicities of carboxylic acid derivatives

I would like to ask about the rate of hydrolysis for 4-hydroxyphenyl acetate (A), N-(4-hydroxyphenyl)acetamide (B), 4-hydroxyphenyl chloroacetate (C), and methyl 4-hydroxybenzoate (D). I know that hydrolysis depends on the positive charge of the carbonyl carbon and also the substituent around it (e.g. electron withdrawing).
Therefore, from the picture above I can see that compound C is the most reactive since it has a phenol and a chlorine as its substituents. Compound B is the least reactive due to nitrogen being the weakest electronegatively as compared to oxygen (I think?).
I am confused between compounds A and D because they are isomers of each other. The oxygen and phenol group are on the same substituent in compound A and are in two different substituent in compound D. Does this affect anything? or can I say that the strength of the electron withdrawing groups are additive, meaning compound A and D are similar in terms of reactivity?

A and D are not similar.  When the methyl is attached to the carbonyl it shifts electron density towards the carbonyl carbon.  What does the hydroxyl-substituted benzene do when attached to the carbonyl carbon?  Draw some resonance structures for D where you use the OH oxygen lone pair.  What effect does that have on the carbonyl carbon?

hi there, thanks for the reply. I have drawn the resonance structure of the hydroxyl-substituted benzene and realised that the negative charge is right beside the carbonyl carbon (stabilizing it? less reactive?). but still this doesnt help me understand whether compound A or D is more prone to hydrolysis. Both the methoxy group in compound D and the phenol + oxygen group in compound A are electron withdrawing. therefore both compound A and D have 1 group that stabilize the carbonyl carbon and 1 group that makes it more reactive? what does that mean in terms of reactivity?

Yes, stabilizing it, less reactive. Methoxy and phenoxy are roughly the same; methyl is less electron donating to the carbonyl carbon than the hydroxyphenyl.

Thank you, that really helped me out and i understand it now. Therefore the answer to this question is C>A>D>B from the most reactive to the least.

That is how I see it.

Background
When a carbonyl compound reacts with a nucleophile we form a tetrahedral intermediate.  Since the carbonyl carbon becomes $\ce{sp^3}$ hybridized in the tetrahedral intermediate there is no significant resonance stabilization - the various tetrahedral intermediates have (approximately) the same energy.  However, strong resonance effects can occur in the starting carbonyl compound, potentially stabilizing the carbonyl compound.
In other words, while a substituent attached to the carbonyl carbon can stabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.
If we place the carbonyl compound and the tetrahedral intermediate on a reaction coordinate we can see that anything that stabilizes the carbonyl compound will increase the activation energy for reaction, slowing down the reaction rate.

Below are the resonance structures we can draw for our carbonyl compound.

If structures II and  III contribute significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structures II and III contribute less.
Answer
Three of your compounds are esters, one is an amide.  We know that resonance structure III is more important in an amide than in an ester (due to electronegativity differences, positive charge on nitrogen is better than positive charge on oxygen).  Therefore the amide (B) will react the slowest.
Now to separate the three esters.  You are right that ester C will be the most reactive.  This is because the inductively electron withdrawing chlorine substituent destabilizes resonance structure II.
In compound D we can draw stabilizing resonance structures like the following;  the para-hydroxy substituted aromatic ring is very effective at moving electron density towards the electron deficient carbonyl carbon thereby stabilizing it.

The methyl group in compound A also donates electron density to the carbonyl carbon, but not nearly as effectively as the aromatic ring.
Therefore our overall order of carbonyl reactivity is C > A > D > B


text:
Why does an increase in temperature decrease the electrode potential?
According to the Nernst equation, the electrode potential $E$ is
$$E = E^\circ - \frac{RT}{nF}\ln\frac{[\ce{M}]}{[\ce{M^{$n$+}}]}$$
Question is, why? Assuming a simple galvanic cell, we know that in general if the temperature is increased, there should be more collisions of the solution with the electrode, resulting in a higher accumulation of the metal ions on the electrode. It should also make it easier for the metal atoms in the electrode to leave the electrode and go into the solution.
Also, if temperature reaches zero, the reaction should stop altogether as there would be no more collisions and no more potential difference should be developed on the electrodes. Yet according to the Nernst equation if temperature reaches zero, maximum standard electrode potential would be reached instead. Where am I going wrong?


text:
Concentration cell lower half cell concentration from measured cell potential
I am trying to find the Ksp of silver thiocyanate:
AgSCN(s) --> Ag+ (aq)SCN-(aq)
but I need to first prove that the lower [Ag+] = 1.27-12 M, but don't know how to find it.
The data I am given:
Volume of 10-4 M Ag+ solution used: 10.00ml
Volume of 1.0 M KSCN added: 50.00ml
Measured cell potential when an Ag(s) electrode was placed into the above mixture, and then joined to the Ag+(0.1M)/Ag(s) half cell: 0.645V
Results:
Calculated lower [Ag+] from cell potential: 1.27-12 M
I am suppose to use this equation for the concentration cell:

where n is the number of electrons.
So really my question is how I would go about finding the higher concentration of Ag+ and then use that to determine that lower [Ag+] = 1.27-12M

Can you define what lower and higher mean? A sketch of your cell could be useful.

This is a concentration cell, which has the same components in each half cell, but the concentrations are not standard (1mol/L) and the difference in concentration results in an electrode potential. By higher I mean the half cell with a higher concentration of Ag+ which is the cathode half cell, and the lower concentration half cell is the anode. So the net cell reaction is Ag+(higher)-->Ag+(lower) and thus the Nernst equation is link The idea is that the concentration of SCN- is much larger than the lower concentration of Ag+