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Structure and Vibrational Spectroscopy
What are the irreducible representations of the C-O stretching modes in free CO, $\ce{[Fe(CO)5][Fe(CO)4]^2-}$?
I think Free CO irreducible representation is A2.
For $\ce{[Fe(CO)5]}$ irreducible representation is 2A1'+E'+A2'' but I am really not sure about $\ce{[Fe(CO)4]^2-}$.

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

You skipped a lot of steps - how did you derive the irreps for Fe(CO)5? Exactly what problems are you facing in applying the same steps to Fe(CO)4(2-)?


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2 electrons with the same spin in the same orbital
When two electrons are in the same orbital and they have the same spin do we say that the electronic configuration is impossible or that it refers to an excited state?

As your question is written: "Impossible."

https://en.wikipedia.org/wiki/Pauli_exclusion_principle

this question can have a good explanation in physics!

It is impossible. Having two electrons with the same spin in the same orbital is a violation of the Pauli exclusion principle.

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What happens at a molecular level when you cook meat?
When we cook food how does good smell produces? and which thing converts into taste.
Please ignore my English grammar mistakes.

"Millard reaction" (google it...) which denatures proteins. http://www.exploratorium.edu/cooking/meat/INT-what-makes-flavor.html

^ Maillard reaction

I also want to point out that he Maillard reaction does not denature proteins (heat does this) - it is a reaction between free amino acids and saccharides.

How do proteins denature?

Thanks for the corrections. // About proteins denaturing... Think of cooking an egg. Cooking creates other "unnatural" bonds (such an egg is no longer fertile of course) in the egg which cause it to solidify.  Cooking an egg is a great example of an irreversible reaction - there is no way to uncook an egg! https://en.wikipedia.org/wiki/Denaturation_%28biochemistry%29

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Strength of HClO4 and resonance
Why is $\ce{HClO4}$ the strongest oxyacid?
My effort: The structure of $\ce{HClO4}$ should be tetrahedral. So when there is a negative charge on oxygen, it should not be in resonance with other oxygen atoms as a condition for resonance is that the atoms taking part in resonance should lie in a plane.

What about inductive effects?

Actually $\ce{ClO4^{-}}$ is tetrahedral. I forget, to which oxygen does the hydrogen attach?!?

Related: Halogen Oxyacid Trend (acid strength), Relative strength of strong acids, and How to determine which substance is the strongest base / acid?

Geez, stop thinking with primitive mesomeric models. All oxygens in anion are single bonded and identical. That's it.

Well, they are all identical. And yes, they're probably essentially single bonds. So is induction the only cause of lower pKa? Because then we have to go back to the question on why $\ce{H3PO2}$ has a lower pKa$_1$ than does $\ce{H3PO4}$.

@DGS possibly pi back-donation from oxygen to P-H antibonding orbitals is working on hypohosphorous acid (in a sense passing some positive formal charge from P to O).  P-H bonds are weak so the antibonding orbitals are at low enough energy for pi acceptance.

Assume an acid $\ce{HA}$. An acid is strong if

protons are donated easily, i.e., the $\ce{H-A}$ bond is readily broken;
the conjugate base $\ce{A-}$ is stable, i.e., $\ce{A-}$ is reluctant to combine back to $\ce{HA}$.

When are protons easily donated?
First parameter to consider is the polarity of the $\ce{H-A}$ bond. More polar bonds mean less electron density on the bond. Thus the dissociation equilibrium
$$\ce{{{HA_{(aq)}}+H2O_{(l)}}<=>H3O_{(aq)}+ + A_{(aq)}-}$$
is tilted strongly towards the right. For this reason, $\ce{HF}$ is a stronger acid than $\ce{NH3}$.
In addition, the size of $\ce{A}$ is important. If $\ce{A}$ is very big, essentially it is harder to hold on to the hydrogen. Thus the $\ce{H-A}$ bond is weaker. This is why $\ce{HI}$ is a better acid than $\ce{HF}$.
Stability of the conjugate base
Charge delocalisation is key here. If all atoms bonded to a central atom are equal, then the hydrogen ion does not have a preferred location at which to bond. A large charge in a small volume would also translate into an unstable ion.
Oxoacids

Look at the number of oxygens bonded to the central atoms. More oxygens -> stronger acid, e.g., $\ce{HNO3}$ is stronger than $\ce{HNO2}$.
If the number of oxygens is equal, consider the electronegativity of the central atom. Higher electronegativity -> stronger acid. For instance, $\ce{HBrO3}$ is stronger than $\ce{HIO3}$.
When electronegativity is similar as well ($|\Delta EN|\le 0.1$), it sometimes helps to look at the size of the central atom. For example, $\ce{H2SeO4}$ is stronger than $\ce{H2SO4}$.

The case at hand
Polarity of the $\ce{O-H}$ bond in $\ce{HClO4}$ is guaranteed via

the electronegativity of the $\ce{O}$ atom itself;
the electronegativity of the central atom $\ce{Cl}$;
the electronegativity of the three additional $\ce{O}$ atoms.

Thus, the electron density is pulled from the  $\ce{O-H}$ bond and the proton can be easily donated.
Here is a picture where you can compare the charge delocalisation of various chlorine oxoanions.

Clearly, the perchlorate ion $\ce{ClO_4-}$ is the most stable. Every valence orbital and valence electron of chlorine takes part in the bond formation.
Extra
This reply discussed relative acid strength with respect to Brønsted–Lowry acid–base theory (in water). Do not forget that these are pointers; there will be other parameters to consider and exceptions to take into account.
$\ce{HOF}$ is the single known fluorine oxoaxid. The analog $\ce{HFO4}$ is simply not possible. It would violate the octet rule.
EDIT: Answer to comment

You asked: "When you say charge delocalisation in ClO4- anion is it due to resonance? Is resonance possible?"
From IUPAC Gold Book: "[Resonance] refers to the representation of the electronic structure of a molecular entity in terms of contributing structures. Resonance among contributing structures means that the wavefunction is represented by 'mixing' the wavefunctions of the contributing structures."

Therefore, the answer is yes. The main resonance structures are given in the picture below.

The resulting ion is a resonance hybrid. In classical valence bond theory, charge delocalisation is not due to resonance; rather, resonance is delocalisation.
Furthermore, coplanarity is not a requirement. It is fair to say that there is (almost) always some delocalisation (again, see the definition given). Still, having atoms that lie in the same plane can be advantageous. Qualitative models of aromaticity and conjugation may even give coplanarity as a necessary condition. This is not always the case, however. Quantitative models are used to more rigorously predict aromaticity and antiaromaticity (see NICS, QTAIM models).
P.S. Resonance energy is important. High, positive resonance energy indicates increased stability; slightly positive or even negative values point to unstable particles. For instance,  antiaromatic molecules generally have low resonance energies. And even though planarity is a prerequisite of qualitative antiaromaticity, too, it is not actually a rule, instead a rough indicator. Example: 2,3,4,5-tetraphenylcyclopenta-2,4-dienone here and there.

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Is there a systematic way of remembering the names and formula of Oxoacids of Phosphorus?
My question is exactly what the title says: Is there a systematic way of remembering the names and formula of Oxoacids of Phosphorus ?
I am asking this because I find it very daunting to memorize their names and chemical formula and also a significant number of questions are asked regarding them.

You mean what really? IUPAC naming? And "P, S, Cl etc" looks like joke. Do you have idea how many oxoanions sulfur alone makes?

http://chemistry.stackexchange.com/questions/40506/what-does-the-prefix-meta-mean-in-inorganic-chemistry http://chemistry.stackexchange.com/questions/43826/problem-in-understanding-nomenclature-of-mineral-acids

Chemistry is daunting?!? Who doesn't like playing with LEGOs - The periodic table is the ultimate LEGO set.


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is ph at eq point dependent on acid concentration at all
Know for a weak acid strong base titration that the ph will be above 7, but does the concentration of the acid affect the ph at the eq point at all?

Of course it does

ok thank you thought yes though wanted to be sure

The concentration of both acid and base solution matters - or rather the ratio of concentrations matters. Let's look at some simple examples titrating acetic acid, which has a $K_a = 1.74 \times 10^{-5}$ with sodium hydroxide.
$$\ce{HC2H3O2 + NaOH -> H2O + NaC2H3O2}$$

Example 1 - 20.0 mL of 0.100 M acetic acid titrated with 0.100 M NaOH

In this example, since the concentrations are the same and the stoichiometric ratio is 1:1, we will need 10.0 mL of NaOH solution to reach the end point. The new volume is 40.0 mL and the concentration of acetate anion is 0.0500 M (volume doubled, concentration halved). Some of that acetate ion reacts with water in the following equilibrium:
$$\ce{C2H3O2- +H2O <=> HC2H3O2 + OH-}\\ K_b=\dfrac{[\ce{HC2H3O2}][\ce{OH-}]}{[\ce{C2H3O2-}]}=5.76\times 10^{-10}$$
To figure out how much hydroxide is formed (and thus eventually the pH), we need to do an ICE problem:
$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0500\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0500-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$
Then we plug in the concentrations into the law of mass action:
$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0500-x)}$$
Solving for $x$ (since I am not making any approximations, I am letting Wolphram Alpha do the grind:
$$x=5.32\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 5.42 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.85 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.734$$

Example 2 - 20.0 mL of 0.100 M acetic acid titrated with 0.200 M NaOH

In this example, the NaOH solution is twice as concentrated, so it only takes 10.0 mL to titrate the acid. The concentration of acetate anion is 0.0667 M.
$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0667\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0667-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$
Then we plug in the concentrations into the law of mass action:
$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0667-x)}$$
$$x=6.15\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 6.25 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.60 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.796$$

Example 3 - 20.0 mL of 0.200 M acetic acid titrated with 0.100 M NaOH

In this example, the NaOH solution is half as concentrated as the acetic acid solution, so it takes 40.0 mL to titrate the acid. The concentration of acetate anion is 0.0333 M.
$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0333\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0333-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$
Then we plug in the concentrations into the law of mass action:
$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0333-x)}$$
$$x=4.33\times 10^{-6})\ \text{M} \\
[\ce{OH-}] = 4.43 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 2.26 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.646$$

A really nice and complete answer!


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Thermodynamics : Work Done by a Gas
Suppose in a piston in which there is a gas, the gas exerts pressure P on surroundings, whereas the surrounding exerts a pressure $P[ext]$ on the gas.
In order for the gas to expand, P must be greater than P[ext]. So far I understand. Suppose initially that $P>P[ext]$ and the gas expands until $ P=P[ext]$ This is what I don't understand is that when you calculate the work done by the gas, you use the external pressure and not the gas pressure. This makes no sense to me. After all, if you calculate the work done on something, you should use the force or pressure applied on that object, or not the other way around, correct? So why do we use external pressure instead of internal pressure when calculating the work done by the gas?

Assuming an adiabatic wall between gas and surroundings then the expansion of the gas will do work on the surroundings.  What is that work?  It is the pdV work using the external pressure acting against the piston from the surroundings.   You are asking why not use the pressure of the gas.  You could, but only if the process was reversible.  That is, the gas is at equilibrium throughout the expansion process.
But what if the gas is not at equilibrium throughout the expansion?   The pressure may not be uniform inside the gas throughout the expansion.  If the gas is expanding irreversibly then the pressure just behind the piston will be lower than the other parts of the gas.   It's easier to calculate the work on the surroundings since you know the external pressure is uniformly applied.

To expand on this, if the expansion is irreversible, there are also viscous stresses present within the gas that also affect the pressure at the piston, such that the force per unit area exerted by the gas on the piston face depends not only on the volume of gas but also on the rate of change of volume.  So the ideal gas law couldn't be used even if the pressure and temperature were uniform (which they are not, as pjg pointed out).  For a massless frictionless piston, the gas pressure at the piston face always matches $p_{ext}$, but the average gas pressure in the cylinder is higher.


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what would be p-v diagram for irreversible isothermal expansion
We say that area under the curve gives the work done. So for an irreversible iso thermal expansion, we have work done= -P external( V final- V initial) signifying a rectangular graph. But we know it's isothermal so it would be along an isotherm? Or no. I believe if we have external pressure at the y-axis, graph would be rectangular and if we take pressure of gas at y-axis we would get a curve, same as isothermal reversible but would end at a lower volume than the former. So what do we take for the p-v diagram.

If you have constant external pressure $P_{ext}$, the work on the PV diagram will be minus the area of a rectangle that is $P_{ext}$ high and $V_{final} - V_{initial}$ long.

When you want to show work as an area, you must plot external pressure on the y-axis of a PV diagram, not gas pressure. In the special case of a reversible expansion or compression, the gas pressure $P$ is equal to the external pressure $P_{ext}$ at every point during the process, so you are still plotting external pressure if you're relating work to the area under the curve.
Why does the external pressure matter in the calculation of expansion work in irreversible processes, and not the gas pressure?
The reasoning goes like this:
expansion work = opposing force x distance moved
opposing force = external pressure x area
volume change = area x distance moved
so
expansion work  = -(external pressure) x (volume change)
The minus sign comes from our decision to make work negative when energy leaves the system in an expansion, and positive when energy enters the system from the surroundings in a compression. This is the convention used in chemistry; you'll find the opposite convention used in physics!

That would be plotting the external pressure with volume of the gas. What if we plotted pressure and volume of the gas?

You could do that, but the area under the curve would NOT correspond to the work in that case.

What would that graph look like?

Is external pressure greater than internal pressure in irreversible expansion

The most extreme form of an irreversible isothermal expansion is the adiabatic isothermal expansion of an ideal gas into vacuum. In this case no pressure-volume work is done, so in the PV diagram pressure first drops to zero, then volume increases to the final volume, and the pressure increases to the final value. The area under the curve is zero.

A more common irreversible isothermal process can be seen as a combination of this totally irreversible process with the reversible process.

Why does the pressure drop to 0? There is some pressure inside the gas, right? Pressure would drop to 0 if we are plotting external pressure and volume of gas. If we plot pressure of gas, then? Which one is more commonly used?

When expanding into vacuum there is no pressure to work against. The pressure in the diagram is external pressure since the work we are interested in is energy transferred by the system to its surroundings.

@aventurin The pressure of the gas won’t suddenly drop to zero. And the external pressure is zero throughout.


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How many total absorptions would appear on proton -NMR spectrum for this molecule?
I count 7 equivalent absorptions total, but answer says 8. I don't see where this eighth absorption comes from?
Heres how I'm counting starting on right side of molecule:
1 is Carbox acid H
2 is the 1 H just left of the double bonded Oxygen
3 is the 3 H's of the methyl group next to the lone H
4 and 5 are the 2 sets of equivalent H's in the ring structure
6 is is the 2 H's to left of ring
7 is the 2 methlys on the isoproplyl (these are equivalent so same signal)
How do they get 8?

8 is the H on the central carbon of the isopropyl group

thanks for the reply ron. ..I actually already accounted for the H at that location. The isoproply has 2 absorptions: the 2 equivalent methls + the H you just mentioned.  So it still sums to 7

No you didn't...

oh wait.. ..you are 100% correct.. I totally looked over that.. Shouldve noticd the only 3 bonds there hence an H..

It's actually ibuprofen. And technically it's an isobutyl group not isopropyl.

Noooooo. The methyl groups on the isobutyl group are diastereotopic because of the presence of a stereogenic center. They are not equivalent.

To answer this question properly one must understand that the field of the instrument is important. At 400 MHz, ibuprofen displays 7 distinct carbon-bound proton signals. Below is the spectrum from the SDBS website. Although the two methyls of the isobutyl chain are recorded [G] as a "singlet" at δ 0.889, they are diastereotopic and inherently different no matter how far they are removed from the asymmetry on the other side of the ring. If this spectrum were recorded on a 1,020 MHz 1H NMR, perhaps separation of the two methyl signals would occur. That the two methyls are different, imagine that one of them is a perdeutero methyl group. This compound is a diastereomer of the compound having the other methyl group as the perdeutero group. So there are actually 9 total distinct types of hydrogens (including the carboxyl hydrogen) that, given the right spectrometer, could appear in a spectrum of ibuprofen. It is a matter of resolution. [As a case in point, see the 1H NMR of cholesterol at the SDBS website and note that the two methyl groups at the end of the side chain have different chemical shifts even though the are separated from the nearest center of asymmetry by three methylene groups.]
Acknowledgment: http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi

There are 8 non-equivalent protons as shown.

adianadiadi how did you number the protons? i mean what software? I was trying to do this in chemdoodle but no success. i was totally missing #6 proton for some reason, but ths is pretty clear now thanks

Numbering done manually.


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Which is more basic, piperidine or pyrrolidine and why?
Pyrrolidine seems to be less basic as it is a 5-member ring and +charge on N atom won't be stable due to steric factors as compared to piperidine but the results are opposite.

Aren't their pKa's about the same (~11.2)?

Yup about the same, but I've read at a lot of different places calling either of them more acidic, and I'm confused.

I think I may have a possible insight into the small difference. Wikipedia gets the pKa from J. Am. Chem. Soc. 1957, 79 (20), 5441–5444 and lists pyrrolidine at 11.27 and piperidine at 11.22.
Calculating the equilibrium structures (B3LYP 6-31g*) for both species in their neutral and protonated states, I found something interesting.

For piperidine

Neutral and protonated species both achieve the optimal staggered conformation about all bonds.

For pyrrolidine

For the neutral species, the C-C bonds have their hydrogens in an eclipsed conformation. In the protonated species, the ring gets a slight twist, which allows the C-C bonds to have their hydrogens in a staggered conformation.

So...

The hydrogens being moved into a staggered conformation upon protonation of pyrrolidine could be the source of the extra stabilisation of the protonated form.
I'm also thinking if the cation form is favoured slightly more in pyrrolidine because its aliphatic portion is less, thus leading to a bigger release of solvation energy upon protonation (calculated to be 1 kJ/mol more favourable for pyrrolidine, but this could be within the error of the method).
If I had to pose arguments for this small difference in basicity, these would be the ones I'd choose.

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How to remember/write molecular formulas?
For example, when given the reactants and asked to write a complete formula including their products for nitrogen and oxygen...:
$\ce{N2 + O2 -> 2NO}$ (this is the correct one right)
but what naturally comes to mind is that I know what will be formed in this case as in:
$\ce{N2 + O2 -> NO}$ ( how am I suppose to know that there is 1 "O"s or that there is only  1 N!!)
Ok example two!!
The desolation of nitrogen dioxide in water(acid rain and stuff):
$\ce{NO2 + H2O -> H+ + NO3- + HNO2}$ ( correct version)
alright right here, I know that an acidic oxide + water gets you oxyacid the $\ce{HNO2}$right. But how in the world was I suppose to know $\ce{H+}$ and $\ce{NO3-}$ would be there??
So basically, is there like no way to find what the products will be without doubt with only knowing some basic rules(such as the products of an acidic oxide and water and others) or with electron configuration in example one.
Or is the only way, just to remember the whole formula.

There are definitely some rules. Read a book on inorganic chemistry :-) I think the one by Housecroft and Sharpe is the standard at the moment, but would love to hear other suggestions. I personally found organic chemistry more easy to swallow, but honestly would recommend a good foundation of inorganic chemistry as well.

Chemistry doesn't have the same sort of predictive powers as physics or math, but there sure are some rules. Experience helps too. I'd suggest JD Lee..it is good for inorganic chemistry(especially chemical bonding). As @Brian mentioned, organic chemistry is a bit more logically convincing; LG Wade or Morrison-Boyd are good books if you want to have a go.


text:
Creating ethane through electrolysis of vinegar
Some
 sources that I studied say that it is possible to produce ethane
through the electrolysis of ethanoic acid. Would this work with vinegar
(5-10% acid)? Also, other sources say that ethanoate salts should be used.
 I can easily obtain such salts (sodium hydrogen carbonate (baking soda)
 + ethanoic acid (vinegar)), and vinegar as well, so which one is better
for creating ethane? I don't care about the amount of carbon dioxide,
oxygen, hydrogen or any other gas produced.

Ethane is formed at the anode through $\ce{CH3COO- -> CH3. + CO2}$ and then $\ce{2 CH3. -> C2H6}$.
Since acetate solutions have a higher concentration of acetate anions than acetic acid I would use an acetate solution to get a higher concentration of acetate at the anode in favor of the formation of ethane.

Indeed that is quite possible. The reaction was first described by Hermann Kolbe and is hence called "Kolbe's Electrolysis"
It is a decarboxylative dimerisation of two carboxylate ions (carboxylate salts--either sodium, or potassium are used instead of carboxylic acids for precisely this reason as they disassociate more readily to furnish carboxylate ions).
The general reaction scheme is is a two-stage radical process: first, electrochemical decarboxylation gives a radical intermediate; then two such intermediates combine to form a covalent bond.
For sodium acetate (again would furnish acetate ions more readily than acetic acid) one would see the following reaction:
$\ce{2CH_3COO^-}\ce{Na^+} \rightleftharpoons  \ce{2CH_3COO^-}+ \ce{Na^+}$
At the anode: a) $\ce{2CH_3COO^-} \stackrel{-2e-}\rightarrow \ce{2CH_3COO.}$
Electrons shuffle around a bit in the acetate radical and we get
b)$ \ce{2CH_3COO.} \rightarrow \ce{2CH_3.} + \ce{CO_2} $
c)$ \ce{2CH_3.} \rightarrow \ce{CH_3CH_3} $
At the cathode: a) $\ce{H_2O} + \ce{e-}\rightarrow \ce{OH^-} + \ce{H.}$
and, finally $ 2\ce{H.} \rightarrow \ce{H_2}$
Now, here we observe that the solution is becomes more basic as the reaction proceeds since hydroxide ions are being produced. If we make the conditions mildly acidic, and draw on Le Chatelier's principle, we can shift the reaction to the right.
So, it would make sense to use sodium acetate, in the presence of ethanoic acid to perform this reaction.

So essentially if I were to do this entirely at home, I could neutralise some baking soda with excess of vinegar, and then electrolyse using carbon electrodes, collect all gas, and burn? (CO2 may be a problem though.)

Not neutralise. What I meant was simply try electrolysing a slightly acidified sodium acetate solution. I've never performed this synthesis myself so I can't be sure; If you try it out, let me know how it goes.

However, I would produce the sodium acetate using neutralisation of baking soda and vinegar, so it would work...? Ethanoic acid == acetic acid.

Ah yes, indeed. sorry i misinterpreted what you just said.


text:
Colligative properties
I have some problems with these properties.
1) Lowering vapor pressure and boiling point.
It happens because the solute molecules take space in the surface of the liquid so a lower number of the liquid molecules change to the gas form. Also the vapor pressure increases when you increase the temperature of the liquid because more molecules get to escape into a form of gas.
So the real question here, What makes boiling point related to vapor pressure?
I have read a lot of articles about that.. The definition of boiling point is the temperature at which the vapor pressure of a liquid is equal to the pressure of the atmosphere on the liquid. Which will rapidly increase the rate of forming gas but why does that happen? It says because of the vapor pressure balanced the atmospheric pressure which makes bubbles form. How does it exactly balance out? Isnt pressure in all directions? How can you be sure that it balances out? Also, Bubbles are created at the bottom of the liquid, How does atmospheric pressure affect it?
2) Same applies to Freezing points.
I need a microscopic explanation. Also without some advanced chemistry complexity.

Here's an answer without complexity which I read in my book. When you heat a liquid in open vessel, its molecules turn to vapours, but the atmospheric pressure above is pushing these vapours down. Some vapours would have enough energy to tackle this atmospheric pressure and would get away from the liquid leading to evaporation. Only some particles at the surface would have enough energy but when vapour pressure reaches atmospheric pressure, all liquid molecules have enough energy to counteract the pressure on them and thus liquid can form vapours from the whole bulk of the liquid leading to boiling.

Thanks! but you still didnt tell me how vapor pressure cancel out the force that the atmospheric pressure exerts on the liquid.
Also another point is the bubbles, How does atmospheric pressure affect them? Like it is at the bottom of the liquid

Pressure exerted by a gas is the force exerted by the the collisions of gas particles per unit area. So basically air particles above a liquid are constantly colliding with the vapour particles and pushing it down, kind of. Surface molecules have the highest energy and that's why they are the only liquid particles that could escape the pressure of air above them. Liquid particles at the bottom don't have enough energy to be converted to vapour. At boiling temperature all the liquid particles have enough energy, including those at bootom. As vapour is a gas, we see bubbles.

So I can just assume that the vapor pressure is pushing upward and the atmospheric pressure is pushing downward. So they cancel out at the Boiling  point. Now most of the molecules will have enough energy to escape the intermolecular force and there wont be a problem with atmospheric pressure, Right?
Also another question if you don't, What happens if the vapor pressure is larger that atmospheric pressure like in cooking? My book says that the boiling point of water increases...?
ِAnd why the temperature changes in solutions during boiling?

I think you meant cooking under pressure. When we cook under pressure water molecules just cant vapourize at boiling point because now its facing much greater pressure which water molecules have to overcome to boil. So we have to heat water more and more till it has enough energy to overcome the increased pressure, giving us the utility to cook with water above 100 degrees Celsius.

Yea that is what I thought.
So as you said the pressure was created by vapor pressure then why it helps to boil the liquid instead of decreasing the number of the molecules that changes to gas.
Boiling is a point where vapor and atmosphere pressure are equal. Shouldn't that just increase the pressure on the molecules on the surface preventing it to change to gas
That was my initial question.

The vapour pressure isn't preventing other molecules to change to vapours. Vapours and liquid are in equilibrium with each other. That means gas particles are constantly being converted to liquid and vice versa. When we increase temperature, the vapour pressure itself increases because more liquid has turned to vapours, this in no way preventing other liquid molecules to vapourize. You can say they are on one team.

Let us continue this discussion in chat.


text:
weird patterns in solutions
When you mix a solid into a solvent you can see weird patterns right after the mixing.
A common example is mixing sugar in water.
What are those wave-like patterns and why do they happen?

This I think might be a better question for physics.SE, providing that they want it.

"Schlieren" or "schlieren lines." Definitely physics.

Those rippling transparent patterns are caused by the variations in refractive index, which in turn are caused by the variations in concentration of the solute. As soon as the dissolution and mixing is over, and the solution is uniform again, the patterns are gone. I don't know the English word for it, or whether there is one; "Schlieren" seems to be much more general.
It's a physical phenomenon indeed, but chemists might be more used to seeing it.
You may observe a remotely similar pattern around a heated object in the air (see "heat haze"). In that case, it is caused by the variations of temperature and not concentration.

I think English uses the same German(?) term, 'Schlieren'. It's the only term I can recall having been used for it.

Schlieren is very general, but it happens for the same reason as any other schlieren.


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Steric factor v's electronic factor?

When water approaches both of them, no reaction is seen in the first compound, but the second compound react readily with water.
In my notes, it is written that CH3 causes steric effect and also +I effect which decreases the positive charge on the adjacent (sp2-hybridised) carbon. Hence, the first compound doesn't react with water.
But in CF3, I know it causes lots of -I, which increases the positive charge on the adjacent carbon, but won't it also cause steric effect too? The lone pair on Fluorine and approaching oxygen of H2O shall hinder each other? (But it does react.)

Water definitely can react with acetone. A solution of acetone in water is in equilibrium with 2,2-propanediol

Anyway, acetone in water favors mostly acetone. Induction leads to a very positive carbonyl carbon in hexafluoroacetone. Fluorine has a smaller atomic radius than hydrogen. Lower sterics is a slight factor, maybe, but the big thing here is induction.


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Substitution reaction
If the reaction takes place in presence of sunlight ,
CH3CH2CH3 + Cl2 gives CH3CH2CH2Cl &CH3CHClCH3
In this second compound is major product with 2° C atom attached chlorine BUT ,
In this reaction ,
CH3CH(CH3)CH3 + Cl2 gives CH3CH(CH3)CH2Cl & CH3C(CH3)(Cl)CH3
In this first compound is major product .
Why ? I cannot understand why major product changes even though reaction mechanism is same.


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Protonation of tetrahydrofurfuryl alcohol

I think the ethereal oxygen in THFA gets protonated. Due to greater +I effect, the electron density on it is higher, hence it is more likely to behave as a Lewis base. Consequently, the product formed is pent-2-en-1,5-diol:

Apparently, this is wrong. What exactly happens here and why my answer is wrong?

Cleavage of ethers is rather difficult http://chemistry.stackexchange.com/questions/14860/cleavage-of-ethers

Also alcohols are stronger bases than ethers http://chemistry.stackexchange.com/questions/46514/why-is-dimethyloxonium-cation-stronger-acid-than-hydronium/46517?s=2|0.0667#46517

but in this case, there is greater +I effect on the ethereal o than on the alcoholic O. Would that not make the electron density on the ethereal oxygen greater?

The alcohol group of your educt (tetrahydrofurfuryl alcohol) is protonated, and after subsequent loss of water, the resulting primary carbocation can rearrange to a more stable 6-membered ring cation, which is stabilized by resonance. Initial protonation at the ether oxygen is therefore disfavored, as it will not result in the formation of a similarly stable carbocation (with positive charge adjacent to oxygen). After final deprotonation, 2,3-dihydropyran is obtained.

The arrow in the third step should be from the C-C bond, not the C-O bond.

@Lemoine Thanks for catching that, I've corrected the mechanism.


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change of pH in urine
I'm looking to lower the pH of a urine solution from 12 to 7. I'm planning to do it in 2 ways. First is to use a strong acid such as HCl. Secondly, using vinegar, a weak acid. However, I am having troubles calculating the pH in both methods as it other chemicals in urine such as ammonia may react to it. Since it's a theoretical experiment, can I just assume that the acid would only react to water? But how will I do it either way?

Urine certainly does contain other chemicals (and the content varies wildly, mind you), which certainly would react with acid. There is no going around that.

Not knowing what concentrations of the other chemicals that are present, it is impossible to do a theoretical calculation. The only way to do this would be experimentally.


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Maximum amount of P4
$\ce{2Ca3(PO4)2 +6SiO2 +10C->6CaSiO3 +P4 +10CO}$
What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phosphorite sample is 75% $\ce{2Ca3(PO4)2}$ by mass

The wording is confusing. I do not know what to do. Would you start by finding the limiting reactant?

First you find the number of moles of $\ce{Ca3(PO4)2}$ reacting. You have 75% of 1 kg, i.e. 750 grams, of calcium phosphate. The rest of reagents are assumed to be in excess as we have to find the maximum amount of $\ce{P4}$ obtained, hence reaction should not have other limiting reagents.


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What is the difference between Alkanes and Polymers?
As far as I understand, Polymers are long molecules made of smaller monomers which are Alkenes. But polymers don't have double bonds, so they are saturated. What is the difference between Alkanes and Polymers structures.

The first one is an Alkane, but if the chain was much longer, it would be structurally the same as the Polymer (Second image).

Polymers aren't necessarily made from alkenes ! What is more, you can have all kinds of functions on polymers.

Some polymers are by definition alkanes.  The difference is basically size.  We wouldn't normally consider something a polymer without a molecular weight of at least 1000 g/mol.  However this is an arbitrary distinction.

Some points to note: You said polymers are saturated. This generalization is not true in many cases, e.g. polystyrene, nylon, and much more. Alkanes are always saturated.
Alkanes can have branchings. The ones you have in mind are called normal alkanes, in which no carbon has higher order than two. An alkane can have any kind of branching, as long as it is saturated.
Now, maybe the most important one: polymers are made of so-called monomers. All polymers can be constructed from exactly one monomer (even co-polymers). An alkane which is sort-of tree-like randomly branched, will not have such an unit.
A normal alkane has some similarities with polymers, tho. As others mentioned it, the size is the deciding factor there. A normal alkane with a carbon count of 10000 is for sure a polymer.

"An alkane which is sort-of tree-like randomly branched, will not have such an unit." Please note that many polymers made by free radical polymerization are branched excessively. Branching does not contradict that an ensemble of the corresponding macromolecules can be considered a polymer.

There are two types of common polymers that are basically just alkanes with very long chains: polyethylene and polypropylene. But there are a large number of other polymers that are made of much more complex parts.
The essential differences between a polymer like polypropylene or polyethylene and an alkane are the size of the molecule and the purity of the molecules. Polymers are big, really big, often containing from thousands to hundreds of thousands of carbons. And, they are often made of mixtures rather than a pure alkane of a fixed number of carbons. These properties arise from the mechanisms of their creation which usually involve long chains of reaction where small molecules (monomers such as ethylene or propylene) are joined together often via a radical reaction to create long chains. These reactions can't be controlled precisely enough to give a single product so a mix of products with a distribution of chain lengths and "branchiness" usually results. To some extent the branchiness can be controlled and this yields different grades of polymer (some clever catalysts can give much more control of chain length).
The properties of polymers differ from those of alkenes mainly because of the size of the molecules. Though you can get some idea of the types of property to expect by following the properties of progressively larger alkanes. Small alkanes are gases, middle-sized alkanes are volatile liquids, larger alkanes (dozens of carbons) are often waxy solids. Polyethylene is just like an extreme version of the waxy solid (depending a little on how it is processed).

For answering the question, it makes sense to look at the definition of the term polymer as recommended by the International Union of Pure and Applied Chemistry (IUPAC) that they give in their Gold Book. It says a polymer is...

A substance composed of macromolecules.

So the word polymer does not refer to individual molecules although it is colloquially used like that. Then, a macromolecule is defined as:

A molecule of high relative molecular mass, the structure of which essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass.

The concept of the repeating units is already described in the other answers. Here, I would like to focus more on the question what is meant by a high molecular mass. This is covered by the first note on the IUPAC macromolecule definition:

In many cases, especially for synthetic polymers, a molecule can be regarded as having a high relative molecular mass
if the addition or removal of one or a few of the units has a negligible effect on the molecular properties. [...]

So focussing now on your example, we can have a look at the boiling point of some alkanes:

Alkane
boiling point [°C]

n-butane
0

n-pentane
36

n-hexane
69

n-heptane
98

n-octane
126

As you can see, the physical property of the boiling point changes considerably with the number of carbons per molecule and with the molecular mass, so these low molecular mass alkanes are no macromolecules, and an ensemble of these molecules will not be a polymer.
Now, for the example of polyethylene I found a nice paper, unfortunately behind a paywall.[1] It shows a graph of melting temperatures of n-paraffins and polyethylene as a function of number of carbons per molecule. The melting temperature inceases at low carbon counts, but levels off around approx. 1,000. This would correspond to a polyethylene macromolecule with 500 repeating units with a molecular mass of around 14,000 g/mol which could be considered the lower molecular mass limit for polyethylene. This limit is far from being a strict limit, there is always some ambiguity, but for all practical purposes this is sufficient.
When I was a student, we were taught that as a rule of thumb, macomolecules will have molecular masses above around 20,000 g/mol, so the estimation above fits in quite well.
Molecules with lower molecular masses which still are composed of repeating units would be called oligomer molecules, and ensembles of these molecules would be called oligomer. So an alkane like dodecane could be considered as a low molecular mass oligoethylene.

[1] J. Phys. Chem. 1965, 69, 2, 417–428


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Understanding basicity of heterocyclic compounds
I want to know the reason why,
Pyrrolidine

is more basic than,
Piperidine

Mainly I want  to know what affect  does  the ring structure have on basicity  and how does 5 membered rings and 6 membered  rings containing a electron rich atom,  affect the basicity?


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pKa and pH behaviour of a drug lignocain
Lignocain is a weak base with pKa of 7.9. if injected into an acidic ( low pH medium or tissue ) it does not act adequately. Reason probably is that it becomes more dissociated and ionized drugs donot penetrate tissues adequately.
But according to relationship between pH and pKa in this forum it is given to understand that substances with high pKa in low pH solution get protonated or undissociated ? why the difference in interpretation?

If I understand your question correctly, this should answer it:
An "uncharged" base with a high pKa is protonated to form an ionic species:
$$\ce{B + HX <=> HB^+ + X^-}$$
$$Ex:\ \ce{NH3 + HCl <=>> NH4^+ + Cl^-}$$
A negatively-charged base with a high pKa is protonated to form an uncharged species:
$$\ce{B^- + HX <=> HB + X^-}$$
$$Ex:\ \ce{OH^- + HCl <=>> H2O + Cl^-}$$

Thanks for explaining . Actually it does not clarify what I asked. Pharmacokinetics says that substances( read drug) which are weak bases having higher pKa in low pH get associated or undissociated or unionized, but clinical book says the substance ( read drug) which are weak bases , if put  in low pH  get  ionized  or dissociated.

Thanks for answering . Does this mean that the drug ( a Local anaesthetic) , usually a weak base is existing as negatively charged when manufactured). Are we injecting a charged drug or uncharged drug. I thought that drugs do not have charges on them when they are in bottles.

@kamalfotedar You might want to reread what you wrote in your question and make it more obvious what you're asking.

So, specific to lidocaine, when it is in a solution with a pH lower than 7.9, it will be protonated to form a positively-charged ammonium ion.


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Is the pressure of a gas proportional with the concentration of a gas?
Is the pressure of a gas proportional with the concentration of a gas?
Because you can also use the pressure instead of the concentration in the  equilibrium condition.

What do you call the "concentration" of a gas ?

@Hippalectryon mol/l?

I'm feeling generous: use the ideal gas law.

Assuming it follows $PV=nRT$, then if the temperature is fixed then indeed it's proportional to $n/V$

Assuming your gas follows the ideal gas law, you have $PV=nRT$. Setting $c=\frac nV$ yields $P(c,T)=cRT$.
If the temperature is fixed, then $P(c)=(RT)c$ indeed tells us that the pressure of a gas is proportional to its concentration.


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Why are amines more soluble than ethers in water?
For example, diethyl ether ($\ce{C2H5OC2H5}$) has a limited solubility of $6.05\rm~\frac{g}{100~mL}$ water at $25\rm~^\circ C$. However, diethylamine ($\ce{(C2H5)2NH}$) is a lot more soluble in water. Why are ethers less soluble than amines? Both can form 2 hydrogen bonds with water (ethers accept two H-bonds, and amines donate an H-bond and accept an H-bond). Oxygen is more electronegative than nitrogen, so shouldn't ethers be more soluble?

Consider chemical interactions between water and each of these, not just physical interactions.

Diethyl Ether can accept hydrogen bonds readily, and you are correct in pointing out it would be better than diethylamine at doing such.  However, diethylamine can also donate a hydrogen bond, giving it two modes of solvation, possibly both simultaneously.
Also, there is some minor acid-base reactions that may occur forming diethylammonium hydroxide, which being an ion, would certainly be more soluble than diethyl ether.

I'm guessing having an H-bond donor is probably more than the acid-base stuff. Actually, I'm not sure. Constant equilibrium between all the amine molecules with water is probably fairly important.

Because they both can form a total of 2 H-bonds with water, would having both donating and accepting capabilities be better than just having accepting capabilities? Or is the reason for more solubility mainly that amines are bases?

I would expect the hydrogen bonding is more important, but that is a guess.

So why is having one donating and one accepting better than having two accepting?

A molecule won't accept two hydrogen bonds.  Once the electron density is donated, the molecule is somewhat electron deficient and which won't donate more.  OTOH if a molecule is donating a hydrogen bond it is more able to also accept one.

Wait, but I thought that water can donate two H-bonds and accept two H-bonds, no? It says so in this Wikipedia page https://en.wikipedia.org/wiki/Hydrogen_bond#Hydrogen_bonds_in_water

Water can donate and accept at the same time.  Ether cannot.


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Why does selenophene not undergo aromatic substitution?
Other derivatives like thiophene, pyrrole give electrophilic substitution but why not selenophene? Is it because selenophene does not have enough electron donating effect and not having its electrons in conjugation, making the ring anti-aromatic?

The Se atom has such diffuse valence orbitals compared with carbon that there is not good pi overlap, so we don't really have a conjugated cycle.  By definition when there isn't a conjugated cycle, it's "not aromatic", not anti-aromatic or non-aromatic as they both require cyclic conjugation. Words involving 'aromatic' are tricky.

We can, however, make aromatic selenium compounds if the selenium is placed outside the ring, see for instance http://www.google.je/patents/US4755589


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Stoichiometry Assignment
$\ce{2Ca3(PO4)2 +6SiO2 +10C->6CaSiO3 +P4 +10CO}$
What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phosphorite sample is 75% $\ce{2Ca3(PO4)2}$ by mass

I am confused. How do I find a maximum amount of a reaction

Phosphorite is a mineral containing calcium phosphate. This particular 1kg sample has 75% percent of it by weight i.e. 750 grams. All others reagents are assumed to be in excess as you have to find the maximum p4 obtained which would be at a point where all of our mineral is used.


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Reactivity of Alkanes, Alkenes, and Alkynes
In my textbook, it says that alkanes are generally the least reactive of the three, alkenes are slightly more reactive, and alkynes are even more reactive. However, alkanes have single bonds, alkenes have double bonds, and alkynes have triple bonds. How can the molecule with triple bonds be the most reactive if triple bonds require the most energy to break?

When you break a triple bond you have to break all three bonds. To convert an alkyne to an alkene you just have to break one bond.

You don't always break both pi bonds and a sigma bond in alkyne reactions. The "weaker" pi bonds can react, but the sigma bond often stays intact.

Tables of bond strengths give you energies for all three bonds together, so they look stronger.  The other comments have correctly indicated, the bonds react individually.  So the first bond of an alkene to react is more reactive than an alkane, etc.

Maybe it will help if you look at reactivity in terms of ability to attract electrons instead of bond enthalpies. Alkanes are $sp^3$ hybridized, and hence have $25$% $s$ orbital character and $75$% p character. $s$ orbitals are closer to the nucleus and thus have a 'contracting' effect on the hybrid orbital. Greater the $s$ character, 'smaller' the hybrid orbital. This means that electrons are more closely packed. Alkanes have only 25% $s$ character, hence the hybrid orbitals are comparatively larger, and the effective nuclear charge on outermost electrons is less. Alkenes ($sp^2$) and alkynes ($sp$) have 33.3% and 50% $s$ character respectively. Thus their hybrid orbitals are SMALLER; i.e. effective nuclear charge is more. This means that it is easier to accept electrons, as now the effective strength of nucleus is more. Hence, alkynes can easily accept electrons, followed by alkenes and then alkanes. Thus, alkynes are most reactive, followed by alkenes and alkanes. Even though increase in $s$ character increases bond enthalpy, but it also increases the electron withdrawing capacity; and it is the latter which predominates.
(In fact, in general, greater the $s$ character, more the electronegativity and more the acidic nature).

But this is just acidity, and thus only applies to terminal alkynes. There is much more to alkyne reactivity than just acidity.

The three bonds consist of two pi and one sigma bond.. Pi are very easy to break and sigma is difficult.. So alkynes can be easily broken as they have a more number of pi bonds while alkanes have only one sigma bind which is difficult to break.

Your text book is wrong alkenes are most reactive and alkynes are most acidic as they have the most s character due to sp-bonds .. The fact that alkenes are most reactive is due to a single pie bond but alkynes have two π-bonds which contribute in the electron delocalising, which would reduce the energy of the π-system

The two $\pi$ systems in alkynes are orthogonal so their energy is not reduced by interaction with each other. Additionally, it is meaningless to say that alkenes are more reactive than alkynes without specifying reactivity towards what. There are reactions that alkenes won't do which alkynes will and vice versa.


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Does diamagnetic anisotropy only apply to a proton bonded to triple bonded carbon?
Lecture tells me that double bond, with the sp^2 bonding will have more s-character, which will hold the electons closer to nucleus, which means a more deshielded proton which means downfield chem shift.
But it doesnt follow the same idea with triple bond, with sp bonded carbon, and this is where diamagnetic anisotropy becomes a factor.... But is D.A. a factor with double bond also and why? Thats what I dont understand, because both double and triple will have pi bonds.. thanks for any help

Diamagnetic anisotropy depends on having some sort of ring of double or triple bonds which can setup a current moving through the molecule.  The moving charges generate an additional magnetic field.

thanks for answer maxW. I understand that the moving charges generates additional field, but  why do we assume the additional field will add to external field? If we change the orientation of the ring then wouldnt it oppose the external field? Or if the direction that the pi electrons circulate is reversed, would that also change the additional field?

sigma value (ppm) for triple bonded acetelyne is about 2 - 2.5ppm, yet sigma for double bond is about 4.5-6.5 .. yet dont both double bonded and triple bonded molecules have circulating pi electrons? if so, then why would H attached to sp Carbon be more shielded than an H attached to a sp2 carbon? thanks for any more help

I understand now that its the external field that causes the pi electrns to circulate, so this is just right hand rule if it opposes the field. But I still dont understand why H attached to sp triple bonded carbon will have a lower sigma value than H attached to sp2 double bonded carbon?

You're going to want to make different questions or include those questions in your original post somehow.

For diamagnetic anisotropy to occur the current has to a "DC" current in a ring, not an oscillating "AC" current in a sequence of linear double bond.

In principle, diamagnetic anisotropy can be invoked for any bond or any functional group; however, there's only a few situations where it's commonly used to understand patterns of shielding/deshielding.  A decent discussion of this can be found in Silverstein's textbook (Spectrometric Identification of Organic Compounds), but I seem to recall that a couple of his figures are drawn in a confusing manner...
In short, this picture relies on the fact that pairs of electrons are diamagnetic, i.e., when immersed in a magnetic field, they get magnetized against the direction of the magnetic field:
$M=\chi B_0$
(M is the induced magnetization, B0 is the applied magnetic field, and $\chi$ is the magnetic susc., which is negative for diamagnetic objects).  Thus, any electron density in a stable molecule (ignoring radicals) will become magnetized against the magnetic field (this is why we speak in terms of shielding).
We can picture molecular orbitals or valence bond orbitals of interest to us simply as a cloud of electron density that becomes magnetized in a magnetic field.  Anisotropy means that effect depends on an angle: we're familiar with the picture of a magnetic dipole having magnetic field lines that curve around from the N end back to the S end – so something below or above the electron cloud will be in the (so called) shielding cone whereas things off to the side will be in the deshielding plane.  Note, this is true for anything that is diamagnetic - there's no need for mystical ring currents or anything like that; any electron cloud placed in a magnetic field will be magnetized and, because it's diamagnetic, will reduce the local magnetic field above and below the electron cloud, but increase it off to the sides.
However, the molecules are tumbling rapidly in solution, so most of this effect simply averages away to zero...  We need special situations where a proton of interest is held in a fixed position relative to the electron cloud of interest to get an noticable effect.  Common example:

Flat conjugated systems like benzene, but also larger annulenes and related molecules.  Most chemists are familiar with the argument that hydrogens outside the ring are deshielded by the "ring current" – while some may debate the existence of the ring current, the pi system is, by its nature, diamagnetic, so it's not surprising that any protons held in the outer edge are deshielded while methylenes held in a position above or below the ring are significantly shielded.  There are beautiful examples of annulenes with protons inside the large aromatic ring that have negative chemical shifts due to the large shielding anisotropy of the pi system.
Triple bonds have two pi bonds that are 90º to each other.  It's a little harder to "see", but you can look at various possible orientations in a magnetic field and see that you can rationalize some shielding of the carbons in the triple bond.
Cyclohexane derivatives.  While not as dramatic as the benzene example (and confusingly illustrated in the Silverstein book), you can picture the sigma skeleton of the cyclohexane as a large cloud of electron density, so it should have a shielding cone above/below the ring and a deshielding plane off to the sides of the ring.  This turns out to be true!  For a system that has a relatively fixed chair conformation and a diastereotopic methylene, you'll always find the equatorial hydrogen deshielded relative to the axial hydrogen.

I'm sure there are more examples, but those are the only examples that I have seen work consistently enough to be useful.

In short, any bond arrangement that is non-spherical will have an element of anistropy associcated with it. The key, as S.Burt quite rightly points out above, is that in solution state NMR, this anisotropy usually averages out to zero through rapid tumbling. However, when the orientation of the nucleus of interest with respect to the electron cloud of the bond orbitals is fixed, this anisotropy will prevail, and generate an influence on chemical shift.
Many common functional groups have proposed anisotropy cones; especially aromatic systems, alkenes and alkynes. Even C-C single bonds have an element of anisotropy. As usual, I would refer you to the excellent online resource of Hans Reich for more detailed examples. (https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-02-delta.htm#05-hmr-02-delta-anisotropy)
However, this provides nothing new to the answer above. What is an interesting addition comes from the 2104 review article by Marija Baranac-Stojanovic (RSC Adv., 2014,4, 308-321. DOI: 10.1039/C3RA45512B) 'New Insight into the anisotropic effects in solution-state NMR spectroscopy'.
This review examines a number of computational methods which brings into question a number of traditional models of anisotropy. Specifically, it suggests, for example, that the ring current of benzene is substantially overstated, with the influence of the σ electrons making the dominant contribution to the local shielding. Also, it is suggested that the overall anistropy cone for the single C-C bond should be reversed, which, by traditional models, would result in Hax being deshielded relative to Heq. However, when total contributions of all local bonds are considered, and not just the C-C bonds, experimental and theoretical chemical shift values align. There are a number of good examples also of the role of sterics in influencing chemical shift over the customary justification of anisotropy. This review makes for a good read which may cause some of us to reconsider the strength with which we promote anisotropy as a factor in chemical shift at the undergraduate level.


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Why are molecular reactions endo- or exo-thermic?
It seems strange to me that chemical reactions should be exothermic, meaning the molecules move faster after the reaction.
Normally, in physics when two moving objects collide and stick together, the resultant velocity of the combined object is less than that of the constituent objects, due to conservation of momentum.
Yet, when some molecules combine they move faster. An extreme example of this are fast reactions like explosives in which the chemicals combine and move off at high speeds, sometimes greater than the speed of sound (a "detonation").
Why would this be? Is there a theory that positively can predict this behavior? In other words, if I were to supply you the names of two chemical species of which you had no prior knowledge, could you predict ahead of time whether their reaction would be exothermic or not, and by how much? Or, can this only be determined experimentally?

Please refer to the spontaneous crystallization of a super saturated solution of sodium acetate.  This is a single example which invalidates your statement, so clearly we need to adjust your view on exothermic.  Although I'm not quite sure how.  Your view of what makes something exothermic is a likely consequence of a reaction being exothermic, not a rigorous requirement.

The short answer is that the energy change is the result of making and breaking chemical bonds not the collisions between molecules.

You are thinking it in a pseudo classical way. In those terms, imagine bonds like some kind of springs with some potential energy. During the reaction there can be rupture or formation of springs, changing the potential energy. In consequence, due to energy conservation kinetic energy must change.
A pictorial argument, that may help comparing your view point about collisions and the chemical reaction case, that is not always true, can be go like this: If we think of two balls bonded by an ordinary spring, in the "relaxed" position (that occurs somewhere between max and min elongation) we have the minimum potential energy. This is also true for a diatomic molecule (look some graph of potential energy curve for a diatomic molecule). Imagine that two of this balls bonded by the spring collide in the region of the (ideal, non mass, non perturbed by the collision etc.) spring and cut them out:

If you make this collision the speed of each ball will remain the same due to the spring is mass less. If molecules were like the balls no change in temperature would occurs, because kinetic energy does not change.
The key point is that molecules does not behave like that, spring never cut so: potential energy always is present. So if they where molecules, as products the separation between balls is larger and so the potential energy is greater and the kinetic energy turns to be smaller.
Now, think in the backward reaction, this reaction would be an exothermic reaction.
I am not sure if that is the kind of answer you need. If not or if something is unclear leave a comment.
Edit
I forgot answer that last questions. Yes, the temperature change can be calculated. Normally what is calculated is the enthalpy change, but they can be related trough heat capacity. Of course there are limitations in precision and size of the molecule, but in essence the answer is yes, it can be calculated following the quantum mechanics principles.

First, your question. Needless to say, you are right in your physics, but let me just restate. If two "objects" collide and combine, the law of momentum conservation does indeed dictate that the velocity of the conglomerate will be lower than the velocities of each of the starting objects.
However, what objects are we speaking of? The basic principles of mechanics, like above, all assume that the "object" is simply some ideal perfectly spherical substance that exhibits no properties that require further consideration. Molecules and atoms, however, do not behave like macroscopic particles such as beach balls or lumps of clay.
In classical physics (and physical chemistry), there is kinetic energy and potential energy (gravitational, electrical, elastic...). In chemistry, there is at least one more consideration--chemical potential energy. Take the collision of two atoms to make a molecule, for instance. What holds the two atoms together to make that molecule? It is because a bond formed between the two atoms when they collided in the right manner. They end up sharing an electron, thus releasing chemical potential energy as they formed a (covalent) bond. Chemical potential energy changes when bonds break or form. Formation of stronger bonds is exothermic, and destruction of stronger bonds (to form weaker bonds) is endothermic. That molecule that formed has less chemical potential energy than the sum of the two atoms before. This is an exothermic reaction, because energy was released to form the bond (otherwise, why would it the molecule be more stable, given that entropy is already against its formation?). As a summary so far, considering only postulates of physics when dealing with atoms and molecule fails the same way that classical physics could not explain what became the domain of quantum physics.
OK, so that was a exothermic reaction. Explosive reactions are too, such as the decomposition of nitroglycerine. When this happens,
$\ce{4 C_3H_5N_3O_9 -> 6 N_2 + 12 CO_2 + 10H_2O + O_2}$, a lot of chemical potential energy as released because the products have stronger bonds.
The reaction is exothermic. What does that mean? Some energy has been introduced as thermal energy. Suppose that you had 20 atoms in a container. Two of them combine to make a molecule, releasing energy. That exothermic reaction means that chemical potential energy has decreased. By the conservation of energy, thermal energy (kinetic energy of individual particles) will increase. True, for a very short time after the formation of the molecule, that newly formed molecule will hold that extra energy. However, the ensuing collisions with the other particles in the container means that the liberated energy is shared with all of its buddies, and the kinetic energy of all of the particles increases, including the newly formed molecule.
To summarize, you need to aknowledge chemical potential energy. Perhaps also reread enthalpy with that in perspective.

Rather than try to answer your points directly I try to explain why reactions can be exothermic or endothermic. Chemical reactions depend on the different amount of chemical potential (energy) molecules have bound up in their chemical bonds.   Suppose, as a thought experiment, that we could take the atoms that will form a molecule and separate then by some huge distance, and we will call the energy they have, zero, (we can assume that they are stationary). Next we bring them together to form the molecule, in doing so the electrons in the atoms interact and release energy as chemical bonds are formed. Finally the molecule will have some total negative energy $\Delta H_1$, this is called the heat of formation and the subscript indicates its our first molecule.  (In chemistry negative energy denotes stability). We can repeat this process with all sorts of atoms to form any types of molecules we want, each time we obtain a heat of formation, $\Delta H_2$,$\Delta H_3$,$\Delta H_4$  etc, but these heats are not all the same, some large, some small. The reason is that not all types of chemical bonds have the energy,(C-O is different from C-C etc.) and not all types of molecules have the same number of bonds thus not all types of molecules have the same heat of formation.
The important point is that different types of molecules have different amounts of internal energy bound up in their bonds. This is released when a reaction occurs and redistributed in making the chemical bonds in product molecules. Sometimes more energy is available from the reactants than is needed to form products and so this is released as heat into the solution or gas depending on how the reaction is carried out. Sometimes its the other way round. Finally, the rates of reactions (how quickly they react) depend on the size of the potential energy barrier that exists between starting (reactants) and ending(product) molecules. The barrier is called the activation energy. If there were no barrier, all molecules would react immediately on contact and we, or any other living thing, would not exist).


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Precaution steps on qualitative analysis (inorganic salts)
I will carry out the experiment on determine the cation and anions of inorganic substances (magnesium thiosulphate, aluminium ion and ammonium ion) on Wednesday. I am looking for the precaution steps for this experiment. Anybody can answer my question?

What are precaution steps? You should be careful not to ingest toxic compounds and people call that precautions. You should also be careful not to mess up the experiment which people still call precautions. Please [edit] and be more clear about what exactly you need to know.


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Difference between oxidation and formal charge
What is the difference between oxidation number and formal charge? I am confused about the two.

Those are both assigned values referring to the definitions, but think of it this way: formal charge assigns charge according to the electrons around the nucleus, assuming that electrons in all chemical bonds are shared equally between atoms. Well, the oxidation number doesn't go for the same assumption, making that $\ce{CO2}$ has for the carbon atom a formal charge of $0$ and an oxidation number of $+4$

This is best explained in an example. Let’s take methyl isocyanide:
$$\ce{\overset{-}{C}#\overset{+}{N}-CH3}$$
This structural formula already includes the formal charges. How do we arrive at these? First, we draw a Lewis structure with the correct, required connectivity.
In this case, we have to know that a $\ce{CN}$ fragment is connected to the methyl group via the nitrogen. We can also find out that we need to form a total of seven bonds, three of which must be $\ce{C-H}$ bonds in the methyl group. Meaning that from the intermediate of $\ce{C-N-CH3}$ we need to add two more bonds but the carbon on the right-hand side already has a completed octet so these can only be between nitrogen and the left-hand carbon.
We also know that we should have a single lone pair; the only atom that can accomodate this is the left-hand carbon. All atoms now have a completed octet, our new intermediate solution is $\ce{C#N-CH3}$.
To determine formal charges, each bond is split homolytically, meaning that each bonding electron pair is split apart giving each atom one electron. Lone pairs are left alone, since they are, well, lone. The total we get is compared to the number of electrons an atom should have according to its position in the periodic table. This gives us:
$$\begin{array}{lcccc}\hline \text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{charge}\\ \hline
\ce{C} & 5 & 4 & 1 & -1 \\
\ce{N} & 4 & 5 & -1 & +1 \\
\ce{C(H3)} & 4 & 4 & 0 & \pm0 \\ \hline\end{array}$$
This concludes the determination of formal charges; the left-hand carbon gets a single formal negative charge while the nitrogen gets a single formal positive charge.

For oxidation states, a different method is employed. Here, all bonds are split heterolytically in such a manner that the more electronegative partner gets both electrons. Nitrogen is more electronegative than carbon which in turn is more electronegative than hydrogen. The resulting number is again compared to the number an atom should have; it will likely be different than before. This method will give us the following:
$$\begin{array}{lcccc}\hline
\text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{oxidation state} \\ \hline
\ce{C} & 2 & 4 & -2 & \mathrm{+II} \\
\ce{N} & 8 & 5 & 3 & \mathrm{-III} \\
\ce{C(H3)} & 6 & 4 & 2 & \mathrm{-II} \\ \hline \end{array}$$
If we add the three $\mathrm{+I}$ of the three hydrogens, we arrive at a neutral $\pm0$ again.

It is important to note that both resuts are quite different and the only thing in common is that they both add up to zero for a neutral molecule or the ionic charge of a molecular ion (unless you did it wrong). The formal charge rarely has any non-formal meaning except in ammonium derivatives where it denotes the actual centre of positive charge. The oxidation states corresponds to the actually observed electron density in an atom somewhat.
Both however are formal concepts and you need to know how to cope with them to use them to your advantage.


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Calculating current for electrolysis process
Is there a way to calculate the current during electrolysis of a dilute NaCl solution.
I need to oxidise the chloride ions to create hypochlorite ($\ce{OCl-}$), so I was thinking of using 2 inert electrodes (Graphite) with an applied voltage slightly greater than 2.19 V.
I want to calculate the current because I want to see how long I will have to run the electrolysis for to produce 0.5mg/L of ($\ce{OCl-}$), and how long the batteries will last. I was thinking of estimating the resistivity of the liquid but I'm not sure.


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Doubts on Diffusion/ Effusion of gases at different pressures
Let's start with a question:

"One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of Xenon with fluorine at 1.6 atm  takes 57 s to diffuse through the same hole. Calculate the molecular formula of the compound."

It is a question where pressures of the gases are not similar. Could be solved quite easily, if we calculate the rates of diffusion ($r_{N_2}$ and $r_{un}$) of each gas separately as no. of moles of the respective gas diffused per unit time, find the ratio of the rates and then equate this ratio, with another ratio found using pressures and molar masses of the given gases, i.e, with
$$\frac{r_{N_2}}{r_{un}}= \frac{p_1}{p_2}\sqrt{\frac{M_{un}}{M_{N_2}}}$$, ($M_{un}$ and $M_{N_2}$ are molar masses of the unknown gas and nitrogen gas respectively) and finally find the no. of fluorine atoms attached with xenon from the value of molar mass of the unknown gas.
It's fine that way. But we also know that, volumes of a gas diffused per unit time gives us rate of diffusion too. And in the question above, the volumes (in case they were given instead of moles) would have been different because of different pressures. Consequently, the first ratio (the ratio involving volume and time) would have been different compared to the ratio involving moles and time. And that would have affected the final value of molar mass and thus messed up our solution.
So my question is, what would be the actual cause of such a situation? And how to fare in those situations?

If you write the rate of diffusion as number of moles divided by time then multiply both sides by p2 and divide by p1. Now on the left side of equation, multiply divide by RT. Now use gas laws.

You finally get that volume rate of diffusion is not proportional to pressure. I still don't know why :/

@jatin- Could you write the steps for me? i am getting an equation with squares on the volume variables...

Just write the no. of moles you've written in the equation as P1*V1/RT.

Yeah, P*V is more important than just V. I get it..

"But we also know that, volumes of a gas diffused per unit time gives us rate of diffusion too. And in the question above, the volumes (in case they were given instead of moles) would have been different because of different pressures."

If you wanted to measure the effused gas in terms of volume, you had to somehow calculate this volume from the number of moles effused. This volume has nothing to do with the volume or the pressure inside the vessel from which the gas effused.
[edit]

From the number of particles that escaped to the right, you could calculate a volume, given an arbitrary pressure (e.g. standard pressure). With this you could express the rate of diffusion in units of volume per time unit.

Are you saying no. of moles of a gas effused or diffused could be treated as volumes themselves and in any case, equal moles of two gases correspond to equal volumes irrespective of pressures? Please elaborate and state reasons.

After your answer I have begun to doubt my conceptual depth on 'pressures'.

The rate of effusion is the number of particles that escape through the orifice per time unit. If you wanted to measure the escaped quantity as a volume you had to calculate this volume from the number of particles escaped. Hereby the pressure inside the vessel from which the gas effuses through the hole is considered to be constant, and the volume of the vessel is considered large enough to keep pressure constant.

So an answer to how i would fare in such situations would be that i would multiply the volumes with their respective pressures because that would give us P*V which remains same at constant temperatures and is only dependent on no. of moles?

Isn't that what you mean by "calculate volume from the number of moles"?

Added an image. Hope this helps to clarify what I meant.


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Hydrogen bonding in complex compounds?
Does the water molecules in coordinated water (where water acts as ligand) form hydrogen bonding with other complex molecules? Will the water molecules inside the complex form intramolecular hydrogen bonds?
Consider the examples coordinated water and hydrogen bonded water molecules like $\ce{[Cr(H2O)6]^3+}$ and $\ce{CuSO4.5H2O}$.

In the generality the question cannot be answered with a definititve statement. You have to look at the cases. For example in the solid state structure  of CuSO$_4\cdot$5 H$_2$O, which is actually more like [CuSO$_4$(H$_2$O)$_4$]$\cdot$H$_2$O, the single crystal water (non-coordinating water) in deed forms hydrogen bonds to oxygen molecules of other coordinating water molecules. In general one could maybe predict, that coordinating water molecules for steric reasons are less likely forming hydrogen bonds to other coordinating water molecules: The oxygen points to the metal and the protons will point to outside so the likelihood of protons from coordinating water molecules  meeting oxygens of coordinating water molecules is small.


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Is adsorption exothermic, and if so, why?
Why is adsorption exothermic?
The explanation given in my textbook is:

For a process to be spontaneous, the thermodynamic requirement is that, at constant temperature and pressure, $\Delta G < 0$.
As the entropy change of adsorption is negative ($\Delta S < 0$), the enthalpy change must therefore be negative ($\Delta H < 0$).
If $\Delta H$ were to be positive, then $\Delta G = \Delta H - T\Delta S$ would always be positive.

It basically says adsorption is exothermic because $\Delta G$ has to be negative. Isn't there any other reason?

This is just an educated guess, but the adsorption of atoms on a surface is akin to creating weak bonds; the creation of bonds is exothermic (since it stabilizes the system), hence adsorption is exothermic too.

There are two common arguments presented as for why $\Delta H < 0$:
Argument (1): Well, indeed I see nothing wrong with the argument presented by the textbook. If adsorption takes place spontaneously, then one can conclude that the change in Gibbs free energy of the process is indeed negative. Since, the entropy change associated with process is necessarily negative (if we assume the entropy of the adsorbent is necessarily greater in the gaseous or liquid state than it is in the adsorbed state), we need a sufficiently large negative value for the change in enthalpy to ensure spontaneity.
Argument (2): Now, adsorption as a phenomenon is associated with "surface energy" (not, unlike surface tension). The surface of the adsorbent molecule attracts and "attaches" adsorbates either via weak van der waals forces (physisorption) or stronger chemical interactions (chemisorption)--in either cases, the surface energy of the system is minimised due to the formation of these new attractions. Thus, one would say adsorption would be exothermic in nature.
The key difference in physisorption and chemisorption is that the electronic structures of atoms/molecules of the adsorbent and adsorbate remain largely unperturbed in the first case (i.e no chemical reaction takes place so to speak).
Anyway, these seem to be the oft cited reasons. Both these arguments are generally sound and hold true under most, but not all conditions. These arguments are often introduced in most introductory level courses, but they do have some holes.
Food for thought - what if a positive entropy change occurs? Or, the chemical interactions that take place during chemisorption are endothermic in nature? So is adsorption always exothermic, no exceptions? Well, short answer: No.
The following discussion is inspired (largely) by ref 1.
If we just look physisorption exclusively, since this is essentially a condensation process analogous to liquefaction on the absorbent.
This is so because the number of degrees of freedom of the adsorbed species is less than the number it possessed prior to adsorption, and because the entropy of the absorbent is unaltered, since there is no chemical interaction. So argument (1) can be applied.
The same need not hold true for chemisorption, since a chemical reaction is indeed taking place. In argument (2), the assumption was that whatever chemical reaction that is taking place at the surface is exothermic, however this need not be the case. If an endothermic reaction takes place then for the process to remain spontaneous the entropy change should not only be positive, but $T\Delta S$ must exceed, numerically, $\Delta H$ so that the net Gibbs free energy change as a whole is negative
Recall: $\Delta G= \Delta H-T\Delta S $
This can, indeed be realised. For instructional purposes consider the following hypothetical system: A molecule, $A_2$, is dissociatively chemisorbed on the surface of a solid, M; consider strength
of the M-A bond equal to half that of the A-A bond. This would give us a net enthalpy change of zero. If the adsorbed atoms, A, have complete two-dimensional mobility, it would then follow that a positive entropy
change, would result, corresponding to a net gain of one degree of freedom, and the associated free energy change is a) negative and numerically equal to $T\Delta S $.
Similarly, one can envisage a system where the net positive entropy change is caused not because of increased mobility of the adsorbate molecules; the entropy of the adsorbate may very well decrease, but an associated large positive entropy change of the adsorbent (consequence of the chemical reaction taking place at the surface) can make up for it (possible, again).
Again, one can expect endothermic adsorption in such a scenario.
I'll quote the conclusion from the paper:

Summarising, it may be said that in all adsorptions entropy changes due to surface-structural changes in the adsorbent itself must be considered along with the entropy changes of the adsorbate. In physical adsorption, the contribution of the first of these two entropy changes to the total entropy change is always likely to be insignificant compared to the contribution of the second (though further investigations alone can establish this unambiguously). In chemisorption, however, there is ample evidence already available to show that the entropy change due to surface structural changes in the adsorbent contributes significantly to the total entropy of adsorption. Whenever the positive entropy change due to adsorbent surface changes exceeds numerically the negative entropy change due to loss of freedom of the adsorbate, the net entropy of adsorption is positive. Failure to recognise this has led to serious error when using equation (1) to arrive at the sign of the heat of chemisorption. This, in essence, is why endothermic chemisorption has been discounted until very recently, though its existence should be considered as normal as the existence of endothermic solution.

You can look up specific examples, which are themselves referenced within ref 1. One such example is in ref 2, which provides some detailed potential energy diagrams illustrating endothermic chemisorption.

References
(1) Thomas, J. M. The existence of endothermic adsorption. J. Chem. Educ. 1961, 38 (3), 138. DOI: 10.1021/ed038p138.
(2) Dowden, D. A.; Mackenzie, N.; Trapnell, B. M. W. 9 Hydrogen-Deuterium Exchange on the Oxides of Transition Metals. Adv. Catal. 1957, Vol. 9, 65–69. DOI: 10.1016/S0360-0564(08)60154-6.

"If the adsorbed atoms, A, have complete two-dimensional mobility" I can't envision what you mean here by two-dimensional mobility. Do you mean stretching?

If adsorption happends to minimize surface energy then why multilayer adsorption happens?


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Is it possible to create a ketone using ozonolysis, when you have an alkene with no alkyl groups?
Can Butan-2-one ($\ce{CH3-CH2-CO-CH3}$) be created from Butene ($\ce{C4H8}$) using ozonolysis?

As far as I know, the ozonolysis of butene will simply yield ethanal (or ethanoïc acic, depending whether you use $\ce{Zn}$ or $\ce{H_2O_2}$)

1-Butene or 2-butene? In both cases, the anwser is most likely that you cannot. But you could catalytically oxidize 1-butene to butan-2-one (DOI).

I don't think that's possible.
The general reactions scheme for ozonolysis is as follows:
Consider the case of butene:

Herein $R_1$ and  $R_3$ can be identified as methyl groups, and similarly $R_2$ and $R_4$ are plain old hydrogens. Following the reaction scheme, one would naturally expect Ethanal as a product (if you use $\ce{Zn}$, ethanoic acid may be obtained if you use something else).
I'll drop by later and write out a mechanism perhaps.


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Is the voltmeter in a galvanic cell in parallel?
Often, a galvanic cell is illustrated as having two electrodes, two electrolytes, a salt bridge, and wires linking the electrodes to a voltmeter. However, isn’t a voltmeter connected in a circuit in parallel? This set up makes it seem as if it is being set up in series. How would it work, then?
I would much appreciate it if a circuit diagram (much like the ones you get in physics) could be included in the explanation.

Short answer, to the question title:
Yes, always.
Long answer:

Often, a galvanic cell is illustrated as having two electrodes, two electrolytes, a salt bridge, and wires bridging the electrodes to a voltmeter.

The piece missing from your list of components, in terms of a galvanic cell being put to practical use, is the load.  You're correct, if a load were attached to the cell, it would be wired in parallel with the voltmeter.
A quick search turns up the following schematic along the lines of your description, from the UC Davis ChemWiki:

From a brief scan, the construction of the half-cells and salt bridge look okay to me. I have not double-checked the half-cell potentials for accuracy.
If I read the schematic correctly, there is no load attached, and so the only thing completing the circuit with the galvanic cell is the voltmeter.  In this case, the appropriate cell voltage would be read on the voltmeter. However, the current through the system would be negligibly small, due to the extremely high resistance of the voltmeter. The $\leftarrow e^-$ notation here indicates only the direction that current would flow through a lower-resistive load that were connected (again, yes, in parallel) with the voltmeter.


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Empirical formula of Glauber's salt
I'm not too sure on how to get the empirical formula for Glauber's salt because water ($\ce{H2O}$) is included.
The question is:

Glauber's salt contains $14.3\%$ sodium, $10\%$ sulphur, $10.9\%$ oxygen (not in water) and $55.8\%$ water. Find this substance's empirical formula.

This is what I did. I got $\ce{Na4SO.H2O}$ as my answer.

$\ce{Na}$'s molar mass is $23$ g/mol not $11$ g/mol

Oh yeah totally forgot about that! Thanks for noticing my silly mistake. I'll change it.

Also, why do you have $n(\ce{H_2O})=1$ ? It should be the same as all the others, $n=\frac mM=\frac{55.8}{2+16}=3.1$. That gives you $3.1/.3\approx 10$ molecules of $\ce{H_2O}$

I was thinking that it was just one mole.

Your percentages add up to 91. You are missing something.

I'm not too sure on that. My teacher came up with the question

Your approach seems correct, however you seemed to have made a few computation errors. First of Sodium's molar mass is $ 23 \ce{g/mol}$ and not 11.
So assuming a total mass of 100g, we can re-interpret the given percentages as follows:
mass of sodium = 14.3 g
mass of sulphur = 10 g
mass of oxygen = 19.9 g
(note: the percentage you provided seems incorrect, the percentages add up to 91, not a 100. I've changed this data point to get the correct result.)
mass of water = 55.8 g
Dividing these by their respective molar masses to get no. of moles:
no. of moles of sodium  = 0.62
no. of moles of sulphur = 0.31
no. of moles of oxygen  = 1.2
no. of moles of water   = 3.1
The ratio we get is 2:1:4:10. Thus, the empirical formula is $\ce{Na_2SO_4.10H_2O}$

Thanks for the assist! My teacher must have made a mistake when she was coming up with the percentages. I'll tell her about it when I get to school tomorrow. I will also correct the molar mass for sodium. Thanks again! :)


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Which will be the most acidic hydrogen in the following organic compounds?

Which will be the most acidic hydrogen in both cases? Please explain.

According to me, in the first compound 2 should be most acidic as in both 1 and 2, resonance occurs but 2’s carbon is closer to the oxygen, which can stabilize the negative charge on carbon.
In the second compound, 1 should be the most acidic as there is $+I$ at 2 by carbon 3, which destabilizes the negative charge on it after removing hydrogen from carbon 2.
Am I correct?

Resonance effect is a more dominating effect than inductive effect. Here I think in both the structures hydrogen attached to carbon 1 will be more acidic. The reason is the negative charge formed on the carbon after removal of hydrogen is resonance stabilised to a greater extent.
In the first structure though after removal of hydrogen from the second carbon, the negative charge formed will be resonance stabilised, but its delocalisation is less compared to the delocalisation of the negative charged which would have been formed from the removal of hydrogen from the first carbon. So, if the negative charge forms from the first carbon ( from the removal of hydrogen), then it is more stabilised. Hence the hydrogen attached to the first carbon is more acidic.
In the second structure also the hydrogen attached to the first carbon is more acidic. This is because if negative charge forms on the second carbon, there is no stabilisation effect. But if the negative charge forms on the first carbon, then again by resonance it will be stabilised. So in the second structure also the hydrogen attached to the first carbon is more acidic.

Somewhat surprisingly, given both the commonness of the compounds in question and the otherwise comprehensiveness of published $\mathrm{p}K_\mathrm{a}$ data, I was unable to find any published acidity data for any of the four protons. Most notably, even looking up the structures in SciFinder, ChemSpider and the NIST Chemistry Webbook returned no acidity results. I must conclude that vinylogous ketones (possibly even aldehydes) have not yet been a research topic and would assume a low-hanging physical chemistry paper to be up for grabs.
Therefore, we must use rationalisation and chemical intuition to determine which factors dominate in which conjugate base of molecule.
To state the exceedingly obvious first: both protons marked 1 should have practically the same acidity since the distal methyl group will have practically no steric and absolutely no electronic effects that we must account for. That doesn’t really help us, though, since we are supposed to compare 1 and 2 within each molecule and not across them.
Upon deprotonation of the left-hand structure, we will generate one of the following enolates:
$$\begin{align}\ce{H2C=CH-CH=C(-O^-)-CH3} && \ce{H3C-CH=CH-C(-O^-)=CH2}\end{align}$$
Both of the two feature a five-atom π system with one internal and one terminal double bond; but one π system is branched while the other is linear. Also, the internal double bond is triple-substituted in the left-hand anion while it is double-substituted in the right-hand one. These two features lead me to assume that proton 1 in your first compound is more acidic than proton 2.
The deprotonation of the right-hand structure will generate one of the following enolates:
$$\begin{align}\ce{H2C=CH-CH=C(-O^-)-CH2-CH3} && \ce{H3C-CH=CH-C(-O^-)=CH-CH3}\end{align}$$
(I have immediately disregarded proton 3 as it is purely aliphatic and unstabilised; the race is only between protons 1 and 2.) As I mentioned above, the left-hand enolate is practically identical to the one above but the right-hand enolate is now also trisubstituted. Therefore, when purely comparing double bonds the left-hand structure (terminal & trisubstituted) is now less favourable than the right-hand structure (internal & trisubstituted) — this would favour proton 2. On the other hand, the π system after proton 2’s removal is branched but that upon proton 1’s removal is linear. I would assume that these two effects work against each other — i.e. that a linear π is better than a branched one.
However, when comparing the two effects with each other I think the higher substituted double bonds should win the contest over linear vs. branched. Therefore, I assume that in your second compound proton 2 is more acidic.
As I mentioned above, I unfortunately do not have experimental data to back my claim.

To just quickly comment on your suggestions:

As I stated implicitly, neither of the anions will actually be carbon-centred; the predominant mesomeric structure will always be the oxygen-centred anion. Thus, which carbon is closer to oxygen is a moot discussion as they do not carry any charge.
Methyl groups do indeed destabilise anions by $+I$ effects. However, they also serve to stabilise double bonds by the same $+I$ effect. Since we are not dealing with a carbon-centred anion, the destabilising effect is practically non-existant.


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Most acidic compound amongs the following unsaturated ketones?

In the given picture, I know that the most acidic will be (a), because of potential of ring expansion. I tried a few times but wasn't able to get a satisfactory six membered ring. I think (a) is the most acidic compound among the 4 due to ring expansion only. Or if I am wrong, please correct me with a detailed solution.
The answer given is (a) but explanation is needed

What exactly do you call "ring opening"?

Five member ring to six member ring, like that.

I would not expect that to happen.

Definitely no ring expansion going on.

My bad. I was always removing the H from CH3. Yup no ring expansion. Realise this from Aditya's answer.

And that term should be ring expansion, not ring opening :)

Its not because of any ring expansion but because of formation of an aromatic compound.
Lets Start with (C) and (D). Both compounds after formation of carbanion are stabilized by extended conjugation:

But (A) forms an even more stable carbanion because of aromatic nature of:

Cyclopentadiene has a $p_{ka}$ of 16. Its high acidic strength is because of the above reason.
You can analyse (B) yourself. But it wont form a carbanion as stable as (A).

Maybe the ring does not open. Maybe (a) is a stronger acid because removing the proton from the ring carbon that's attached to the acetyl group produces a strongly stabilized structure.
What type of strongly stabilized structure may involve a five-membered ring with a negative charge (left behind when the proton is removed)?

Nice shot people.  I wanted not to give away the answer and this is what happens.

Haha. Keep trying.

I think you were saying right, but weren't sure if you were correct. Yes, removing Hydrogen from the ring will make it Aromatic. Otherwise your answer had been acceptable :)


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How is it possible to safely use carbon monoxide as a fuel?
Is there anything that will allow carbon monoxide to be used as a fuel? It is quite toxic, but has attractive properties otherwise (only produces CO2 with no water, also no smoke or soot). Aside from a permanent spark, or a pilot light, I can't see any way of doing this.
Is there some catalyst that will help the reaction? Apparently, platinum can be used as catalyst between hydrogen and air.

A fuel for what?  If you wanted to heat with it, then its conceivable that a pilot light would do the job.  If you're wanting to use it in an IC engine, that has been done via wood gasification for some time, though that yields a blend of CO and H2.  Any leaks in the system could prove fatal though, so I wouldn't count on it being the next big thing.

A CO driven airplane will not leave contrails because of the absence of hydrogen forming water vapour in the exhaust. That might be interesting for reconnaissance aircraft flying high.

Carbon monoxide has been used as a fuel but not usually by itself.
The reason why we don't use it much now is that there are better, safer alternatives.
Before the widespread use of natural gas as a fuel (which consists mostly of methane) many cities piped coal gas (sometimes called town gas) to consumers as a fuel. Coal gas is made (obviously) from coal by a process that combines incomplete oxidation and the water gas shift reaction (see the details of how it evolved). The result is a gas containing both CO and H2 with some other ingredients in smaller quantities. Before electricity this gas was the dominant source of street lighting in cities and was commonly supplied for heating and cooking for much longer than that (the UK only phased it out in the 1970s and 1980s when cheap supplies of natural gas were discovered in the North Sea.)
The disadvantage of coal gas is that it isn't as good a fuel as natural gas (having a much lower calorific content) and that it is poisonous because of the CO content (which is where the trope of suicide by putting your head in the oven originated). But it is clearly possible to use a gas containing a lot of CO as a successful fuel and you don't need expensive catalysts to achieve this as it burns well in a suitable device.

During WWII many cars were powered by a gasogene engine, which transformed wood or charcoal into a synthetic gas composed by $\ce{CO}$, $\ce{N2}$, $\ce{H2}$ and $\ce{CO2}$. Thus the use of $\ce{CO}$ (albeit not as a pure gas) as a fuel is not something new (the former use of piped coal gas for house heating and cooking has already been mentioned above). I am not sure however about how safe those gasogene engines were.


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heat capacity at constant volume and internal energy
I read that heat supplied at constant volume to a system containing ideal gas changes the internal energy of system i.e $q=n C_v \Delta T= \Delta U$
But what happens if I heat a system and change its temperature from $T_1$ to $T_2$ with system not being at constant volume,can I still say that $\Delta U = n C_v \Delta T$?
Is there any proof regarding this?

Yes, you can apply the formula $\Delta U=nC_V\Delta T$ for an ideal gas in a container whether or not the process is isochoric.
This is because the change in internal energy for an ideal gas undergoing some mechanical process is mainly affected by the change in kinetic energy of the gas. It does not depend on the change of  volume of the gas. So as long as the temperature difference is same the change in internal energy will be same. Heat capacity is the amount of heat required to raise the temperature of a substance by 1 unit so $\Delta U$ is directly proportional to heat capacity and $\Delta T$ along with the number of moles of gas.
When the process is carried with constant volume only then $\Delta W$ is $0$ so in that case $\Delta Q = \Delta U$ whereas $\Delta U =nC_V\Delta T$ holds good always for an ideal gas under some mechanical process.