text:
Why can acidification of solution lead to selective precipitation of ions such as Fe(II) and Zn(II)?
Given $K_\textrm{sp}(\ce{ZnS})= \pu{2.5E-22}$ and $K_\textrm{sp}(\ce{FeS})= \pu{3.7E-19}$, why would $\ce{Fe^{2+}}$ ions remain in solution but not $\ce{Zn^{2+}}$ with acidification of the solution?
I have little understanding of this, so I don't have a great attempt at the solution.

Are you adding an acid to $\ce{FeS}$ and $\ce{ZnS}$? If so, what acid? That might make a difference.

FeS has 1500 times the solubility of ZnS.  Given the slow addition of a small molar fraction of acid that solubilizes both Zn(II) and Fe(II), that fraction will retain at equilibrium.  It gets complicated if soluble polysufides exist. Iron gives pyrite and marcasite ($\ce{FeS2}$) via mackinawite ($\ce{FeS}$) and greigite ($\ce{Fe3S4}$), but they are insoluble.

When you acidify a solution, you increase the concentration of hydronium ions and while doing so decrease the concentration of hydroxide ions.
This decrease in the concentration of hydroxide ions cause the equilibrium of the hydrolysis(both complete and partial) of the sulphide ion to shift forward.
$$\ce{S^2- + 2H2O <=> H2S + 2OH-}$$
This forward shift in equilibrium decreases the concentration of sulphide ions in the solution and thus helps in increasing the solubility of the sulphide salts.
The solubility is increased because all the sulphide added to the solution doesn't stay in the form of sulphide ions, a major portion of it is converted to $\ce{HS-}$ and $\ce{H2S}$. This drives the solubility equilibrium of the salt forward.
$$\ce{MS <=> M^2+ + S^2-}$$
This method saves $\ce{FeS}$ from precipitating in acidic conditions but for $\ce{ZnS}$ this is not sufficient. $\ce{ZnS}$ is just too insoluble in water to salvage it by acidification, a fact which is also evident from it's very low $K_{\mathrm{sp}}$. Chemists have also used this fact to seperate mixtures of zinc and ferrous sulphide.
To precipitate ferrous sulphide, you will need to add an alkali to the solution which will alter the aforementioned equilibriums in just the reverse way.

Let's admit that some $\ce{H2S}$ is added into a solution containing both ions $\ce{Zn^2+}$ and $\ce{Fe^2+}$ with a concentration $\pu{0.1 M}$. In my table, the solubility products of $\ce{FeS}$ is $\pu{6.3 \times 10^{-18}}$, which is $10$ times higher than your value : $\pu{6.3 \times 10^{-19}}$.  With your value for $\ce{ZnS}$ and mine for $\ce{FeS}$, the maximum amount of $\ce{S^{2-}}$ for the ions remaining in solution is for $\ce{Fe^{2+}, \ce{[S^{2-}]_{Fe}} = 6.3 10^{-17} M}$, and for $\ce{Zn^{2+},  \ce{[S^{2-}]}_{Zn} = 10^{-21} M}$.
Now the amount of $\ce{S^{2-}}$ ions in a solution of $\ce{H2S}$ is dependent of the two $\mathrm p K_\mathrm a$ values of $\ce{H2S}$, which are $7$ and $13$. Combining these data gives the following maximum value for $\ce{S^{2-}}$ versus $\ce{[H+]}$ :  $$\ce{[S^{2-}] = 10^{-20} / [H+]^2 }$$ In an acidic solution of $\ce{H2S}$ where $\ce{[H+] = 0.1 M}$, this formula  gives $\ce{[S^{2-}] = 10^{-18} M}$, and this is just between the two limits $\ce{[S^{2-}]_{Fe}}$ and $\ce{[S^{2-}]_{Zn}}$ described above. So, it is enough for dissolving $\ce{FeS}$ and precipitating $\ce{ZnS}$.

Reference:
(1)    Wallace, H. G.; Stark, J. G.; McGlashan, M. L. Chemistry Data Book; Hodder Murray: London, England, 1989.


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State of aggregation of "ate" complexes
I understand that ate complexes can result from the addition of carbanions to neutral aluminum alkyls.
I was wondering what the balanced equation for this reaction is/what the structure would be in solid state - like does it dimerize?
$$\ce{2 LiCH2CH3 + Al2(CH2CH3)6 -> ?}$$

The crystal structure of lithium tetraethylaluminate has been published (DOI)
and as far as I can tell from the abstract, there is no indication for a dimeric anion here. It might be save to assume that the balanced equation therefore can be written as:
$\ce{2 LiCH2CH3 + Al2(CH2CH3)6 -> 2 Al(CH2CH3)4^{\delta-}\,Li^{\delta+}}$

That is correct, but in solid state it forms chains of alternating Lithium and Aluminium, both in tetraedral coordination. (Separated by 7Å, hold together by van-der-Waals-Forces.) However, the paper refers to the compound as $\ce{LiAl(CH2CH3)4}$, but also suggest ionic character. They assume some electron density of $\ce{Al\bond{-}C}$ is shifted towards the Lithium forming a more centre bond with some covalent character. (I am just adding that because the question points to the sold state structure.)

@Martin The -/+ added for clarity (and as separators) might indeed lead to wrong conclusions. I'll fix that.

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Why does Bismuth only create a +3 ion instead of a +5 ion?
It has 5 valence electrons, but only loses 3 of them to make a +3 ion. Why does this occur?
I believe it may have something to do with how losing 3 electrons leaves you with the p sublevel full, but I do not understand why Bismuth is okay with that.

losing 3 electrons leaves you with the p sublevel full is incorrect.  Removing 3 electrons from Bi results in an electron configuration of [Xe]6s2 4f14 5d10 or an empty 6p subshell.

Isn't it just that +5 is a lot of (positive) charge for a single cation? Which other (transition) metals do form a $\ce{M^{5+}}$ cation? Just out of my mind, I can't think of any.
Typically, these high oxidation states exist in the form of oxo-anions ($\ce{CrO4-}$, $\ce{MnO4-}$, $\ce{VO4^{3-}}$, etc.) and bismuth is no exception here: $\ce{Bi(V)}$ exists as bismutate ($\ce{BiO3-})$.

I guess you meant $\ce{Bi^{(V)}}$? Also relativistic effects may play an important role lowering the energy of the 6s Orbital.

@Martin Yes, fixed. And yes, possibly. Apparently, I need more coffee :D

See also, http://chemistry.stackexchange.com/questions/8717/what-is-inert-pair-effect/8774#8774 and http://chemistry.stackexchange.com/questions/2795/why-does-radium-have-a-higher-first-ionisation-energy-than-barium for more on the relativist effects.

I'm not sure why transition metals need to be invoked to answer this question, but a look at the Latimer diagrams of V, Cr, and Mn group transition metals (for example at webelements we see that there are a number of metals with accessible (V) oxidation states.

Bismuth can also form oxocations like $\ce{BiO+}$ which is also a +3 ion.

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Why is the 8-electron rule more important than the 2- or 18-electron rule
Why is the fulfilled electronic configuration of only $p$ orbital  is stable. I mean why $II-B $ group with fulfilled $d$ orbital,$II-A$ group with fulfilled $s$ orbital...are not stable. Why makes $p$ orbital so special?

Both the rules are differently used. Octet rule is for Lewis and only Lewis structures for the main group elements, specially the lighter ones like Carbon, nitrogen etc. The Eighteen electron rule is used in inorganic chemistry and organometallic chemistry of transition metals only. They rarely crash into each other...

By the time the $p$ orbital is filled it is still the valence shell, you can see this by filling order e.g. $3s^2$ $3p^6$ $4s^2$ $3d^{10}$ ; You can see there isn't really an '18 electron rule' because there can never be 18 electrons in the valence shell and since chemistry is under most circumstances about the interaction of valence electrons then you can see that the $d$ shell electrons rarely have much to do with reactions because they are always 'one level down' from the valence shell.
Your question is a little unclear so I can't really say much more but I think you seem to be trying to ask why Neon or an analogous noble gas is so stable whilst Magnesium with a full $s$ orbital is not. Think about the difference in energy levels between going from one shell to another and you'll realise that an $\ce{Mg^2+}$ ion and Neon are very similar in terms of their stability because they are both in a low energy state.


text:
How will the pressure of a gas in manometer change if it goes to a place with higher pressure?
Imagine a manometer where air pressure is $\pu{0.8 atm}$ and the gas pressure is more. We move it to sea level which the air pressure is $\pu{1 atm}$. How will the gas pressure change? (the temperature is constant)

I'm confused about this gas pressure. If the air pressure is 0.8 atm, how is the gas pressure more?

You have a rubber balloon partially inflated to a given diameter... A manometer's readout is a one-dimensional balloon.

Assuming an ideal gas with T constant, $$ PV = nRT $$ the entire right-hand term is constant. Therefore$$ P \propto \frac{1}{V} $$ which is Boyle's law. Since the atmospheric pressure goes up, the liquid portion of the manometer will go down on the side open to the atmosphere, and it will go up on the side open to the gas. Since the volume of the gas goes down, its pressure must increase by Boyle's law.


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Good Introduction to chemistry textbook
I want to learn basic chemistry, so which textbook do you think I should pick? It should be something both concise and clear. Thank you.

Hi user5204, welcome to Chemistry S.E. unfortunately this question is not appropriate for this site,( see here for our guidelines) is too much subjective! Keep following us!

Anyway I got an answer that's it!

The problem is how can you say that is the right answer who can assure you that is a good introduction? I know that can be annoying but these are the rules of this site.

You're right, please can you give me your opinion?

I like Kotz.

Okay thanks, I will buy it even if it's expensive.

I'd recommend "Why chemical reactions happen" by Peter Wothers and James Keeler (both lecturer-type people at Cambridge). I've read it, and it's been educational. On Amazon, it's £40, but I picked it up for £5 - probably one of the best books I've brought!!

I would recommend Zumdahl Chemistry. It discusses most of the basic topics, and contains lot of exercises.

Zumdahl is great

You need the knowledge and a working mastery of it.  Choosing a good book requires already knowing what is useful and not useful to you.  Prime the pump,
https://www.khanacademy.org/science/chemistry


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Simplest way to balance any chemical equation
In your opinion, what would be the simplest way to balance any size of the chemical equation?
For example here is an equation:  $$\ce{C12H26 + O2 -> CO2 + H2O}$$
I saw some ways by putting letters in front of each molecule, but I just don't get too much how to do this with this method! My teacher is always speaking about another method, but it simply doesn't feel logic/mathematics is enough for me.

This is a very broad question because chemical equations can be very long and complex (unlike the example you've provided).  Given this concept, the simplest way to balance ANY chemical equation would be to use a computer program that is capable of doing so for any arbitrary equation (though I'm sure this isn't the type of answer you were looking for).

@LordStryker Yes, sorry, I recognize that it's too broad ! Well, my question was before all about some pretty simple equations like the one I showed in my question but I would also like to create an algorithm to do that so in the future I will need a way to do that with any equation (that's why I was wondering if there was a pretty easy way to do that...).
But about my example, how would you have done that ?

Well, its easiest just to 'go big' then 'reduce'.  So, start by putting a 12 in front of C (on the r.h.s. of the equation).  Now scale up O on the l.h.s. to match that on the r.h.s. Continue 'scaling-up' then when everything is balanced, 'reduce' your coefficients.

You can do this in the 'guided trial-and-error' method that LordStryker showed which is probably quickest for simple reactions, or approach it in a purely mathematical fashion which is the method I will explain. This method works well for arbitrarily difficult reactions.
Your chemical equation contains 3 atomic species: $\ce{C}$, $\ce{H}$ and $\ce{O}$. This means that you need 3 equations to balance. First I write the chemical reaction as follows:
$$\ce{a C12H26 + b O2 -> c CO2 + d H2O}$$
Now I will write the 3 equalities for the 3 atomic species that we have:
\begin{align}
12a &= c \tag{for C}\\
26a &= 2d \tag{for H}\\
2b  &= 2c + d \tag{for O}
\end{align}
The numerical constants in front of $a$, $b$, $c$ and $d$ come from the number of atoms that are in the molecule for which they are included.
As you may have noticed, this system of equations is ill-defined, because we have 4 unknowns ($a$ to $d$) and only 3 equations. The reason is that we can pick any multiple of the equation without it becoming incorrect,1 just against convention. This means that we can simply set one of the unknowns to any value. Let's say $a = 1$ for now and revisit this choice after solving the equations.
With $a = 1$ it immediately follows that $c = 12$ and $d = 13$. From those two it then follows that $b = 18.5$. Our balanced reaction becomes:2
$$\ce{C12H26 + 18.5 O2 -> 12 CO2 + 13 H2O}$$
In principle this is already correct, but convention is to have only integer numbers in the equation (i.e. no decimal points) therefore we need to revisit our choice of $a = 1$ and pick it such that all numbers will become integers. In this case you can easily see that this will happen for $a = 2$ which then results in
$$\ce{2C12H26 + 37 O2 -> 24 CO2 + 26 H2O}$$
If for some reason it is not easy to see which value for $a$ you would need then you can multiply by a large power of 10 to make all the decimals disappear and then check the greatest common divisor of the 4 numbers and divide by that to obtain the same equation.

Notes

Compare $10=10$ to the multiplied version $30=30$, both are correct, but they are just different by a factor $3$.
I'm not sure whether you are familiar with linear algebra, but if you are you will probably have noticed that the set of equations is a linear set so you could solve it through matrix manipulations which makes this method applicable to arbitrarily complex chemical reactions.

But still, this method would become quite tedious in, for example, complicated ionic equations. So in these cases, what do you suggest- should we go for the traditional half-reaction method, or is there a better alternative?

If you know something like a little matlab or python with numpy or similar these linear equations should be very easy to solve

Here is an example of the process I just used:
$$\ce{C12H26 + O2 -> CO2 + H2O}$$

Start with carbon (completely arbitrary choice).  Put $12$ on the right hand side:
$$\ce{C12H26 + O2 -> 12CO2 + H2O}$$
Carbons are balanced but oxygens are not.  Therefore, move on to $\ce{O}$.  Since we added $12$ to $\ce{O2}$ on the right hand side, do same for $\ce{O2}$ on left hand side.
$$\ce{C12H26 + 12O2 -> 12CO2 + H2O}$$
Oxygen is still not balanced ($24$ vs. $25$... ouch).  What else can we do? Let's work with hydrogen for now (completely arbitrary choice).  We can return to oxygen later.  Put $13$ on right hand side to balance hydrogen.

$$\ce{C12H26 + 12O2 -> 12CO2 + 13H2O}$$

Oxygen still not balanced (even though everything else is).  This is an even/odd situation.  The oxygens on the right hand side must be multiplied by an even number. So lets multiply $13$ by $2$ to do this.
$$\ce{C12H26 + 12O2 -> 12CO2 + 26H2O}$$
Now our hydrogens are not balanced.  Multiply left hand side by $2$:
$$\ce{2C12H26 + 12O2 -> 12CO2 + 26H2O}$$
Now our carbons are not balanced.  Let's fix that.
$\ce{2C12H26 + 12O2 -> 24CO2 + 26H2O}$$
Everything but oxygen is good, $24$ on left hand side and $74$ on right hand side. We can get $74$ on the left by changing our coefficient from $24$ to $37$ (since $37\times2 = 74$).
$$\ce{2C12H26 + 37O2 -> 24CO2 + 26H2O}$$
Check to see if the coefficients can be reduced (divided evenly by the same number to get corresponding whole numbers).  Since we have an odd number ($37$), we cannot.  Therefore, the equation is balanced in Step 7.

This is an example of what I've just denoted as the 'Go-big then reduce' method (completely made this term up).  I find it to be the easiest way to balance an equation.  It may not be the most efficient but it will land you on the right answer (which, in a way, is efficient).

If the answer clarified that one should avoid starting with elements present in two places on the same side of the equation, or that you should start with one that exists only in one place on either side, and therefore it is best to start with C or H, but the choice between THEM is arbitrary then I would consider it to be an excellent answer and upvote it.     I would like to upvote it but the use of the term "completely arbitrary" is imperfect and easy to fix.

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text:
Eutectic Phase Diagram and Lever Rule

A solution having composition $p$ (left side of graph) is cooled to just above the eutectic temperature (point $s$ is at $0.18$ $x_{\ce{Si}}$ and the corresponding tie line intersects the liquidus curve at $0.31$ $x_{\ce{Si}} $); calculate the composition of the solid that separates and that of the liquid that remains.

The liquid phase is obviously 31 % silicon and 69 % gold as given in the question. The solutions manual also gives the weight percentage of the solid phase. Using the lever rule,
$$ m_\text{solid}(0.18)=m_\text{liquid}(0.31-0.18)$$
$$ \frac{m_\text{solid}}{m_\text{liquid}}=\frac{0.13}{0.18} $$
$$ \frac{m_\text{solid}}{m_\text{solid}+m_\text{liquid}} = \frac{0.13}{0.13+0.18}=0.42$$
However, the solution manual states that the wt% of solid phase is 58 %, and that of the liquid phase is 42 %. Is the solutions manual wrong, or am I losing my mind?

You are correct. Another way to calculate it is: $$X_{\mathrm{L}} = \frac{s-b}{e-b} = \frac{0.18-0}{0.31-0}=0.58$$
You can also inspect the diagram visually and see that the $bs$ line is longer than the $se$ line. The lever rule dictates that the line length is proportional to the percentage of the phase opposite from the line. Thus, it is not possible that there is more solid than liquid at this point.
There is definitely a mistake in your textbook's solutions.

Using the lever rule, the amount of liquid in the 2 phase region is given by
$$\frac{18-0}{31-0}=0.58$$
Hence for the amount of solid in the same region we get
$$1-0.58=0.42$$
since the overall sum of the liquid and solid in the two phase region is 1.

There is no mistake in the solutions of your book. I hope to help any other students that may come across this post. The formula for the weight percentage of the solid phase is
$$\text{wt% of solid} = \frac{\text{wt% of solid}}{\text{wt% of solid + wt% of liquid}}.$$
Applying this formula you get: $$\frac{0.18}{0.13 + 0.18}= 0.5806$$
Thus the $\text{wt% of solid}$ is $58.06\%$.
It is a common mistake to get the values backwards since it appears to be backwards from what it should be. I urge you to look up the formula for yourself to verify as well as check out a derivation of this formula.

@AvnishKabaj Please don't use \mathrm where it is neither necessary, nor correct.

@Martin Isn't \mathrm just a font

@AvnishKabaj No. It requests that symbols will be set in roman (upright) type, which might be correct, but it might be wrong, i.e. like here. Here it has no effect at all, as \text is always set in roman type. If you want to know more, please have a look here and here.


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Delta H of reaction and finding delta h of components?
$\Delta H$ for the reaction below is −126 kJ
$$\ce{2Na2O2(s) + 2 H2O(l) ->4 NaOH(s) + O2 (g)}$$
What is the amount of heat released by the reaction of 25.0 gram of $\ce{Na2O2}$ with water?

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Loosely put, enthalpy is the change in heat of a given reaction.  We see input of two moles of sodium peroxide to give a mole of oxygen in the balanced equation.  (What are the reduced and oxidized species?)  You have a mass of sodium peroxide input, you know how many moles that is, and you know the one equivalent of reaction requires two moles of sodium peroxide.  You then know the heat released.


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Aqueous vs. Solid
I've been having a conceptual struggle with these two states for a while now, and I'd like to get it cleared up.
A compound AB dissolves in water.
$$\ce{AB(s) + H2O <-> A+(aq) + B- (aq) + H2O(l)}$$
Now, we know that in the case of salts, there can be strong and weak electrolytes based on the degree of dissociation. For example, NaCl is a strong electrolyte because when dissolved in water it dissociates completely. However, in theory, there should be some salts that do not dissociate completely and thus are weak electrolytes, for example, I think, NaCH3CHOO, despite its high solubility.
$$\ce{NaCH3COO(s) + H2O(l) <-> NaCH3COO(aq) + Na+(aq) + CH3COO- (aq)}$$
Here's an example that confuses me. Assuming that CaO dissociates completely in water:
$$\ce{CaO(s) + H2O(l) <-> Ca^{2+}(aq) +O^{2-} (aq) + H2O(l)}$$
O(2-)(aq) + H2O(l) --> (irreversibly, I suppose because of the general formula for the combination of anhydride metal oxides with water) 2OH-(aq)
Which then finally yields:
$$\ce{CaO(s) + H2O(l) <-> Ca^{2+}(aq) + 2OH- (aq)}$$
Then, here's the tricky part:
$$\ce{Ca^{2+}(aq) + 2OH- (aq) <-> Ca(OH)2(s)}$$
Now, I don't know whether this is true. Calcium hydroxide, according to solubility rules, is insoluble in water, but as we know, all compounds are at least a little bit soluble. The KSP may be very low, but it dissolves a bit. My confusion is in how we know if a precipitate forms or not without being given any molar concentration of Ca2+ or the KSP of Ca(OH)2. The latter is necessary to calculate QSP, and this is necessary to determine whether a precipitate actually forms. Without being given any other information in a school setting, is it correct to assume that all salts are strong electrolytes and that all low-KSP compounds form a precipitate in solution?
Thanks!

$\ce{CaO}$ (lime) begins by reacting with water to $\ce{Ca(OH)2}$ (slaked lime) with a very large exotherm
http://www.ktf-split.hr/periodni/en/abc/kpt.html
  $\ce{Ca(OH)2, K_{sp}} = 5.02×10^{-6}$
If you are not concerned about pH (via ion hydrolysis) and pH buffering, soluble salts are strongly dissociated.  As you say, depending on concentration, the answer can go either way.  If you are reacting it further (lime water as a $\ce{CO2}$ detector), precipitation of much less soluble $\ce{CaCO3}$ makes the intermediate step unimportant (perhaps aside from kinetics). Add more $\ce{CO2}$ to obtain soluble $\ce{Ca(HCO3)2}$.

Up to my knowledge the reaction mechanism of CaO and $H_{2}O$ is not well understood yet, however seems that for low density CaO,  $H_{2}O$ is absorbed over the CaO. The mechanism should be something like this:
$$\ce{CaO + 2H2O -> CaO\cdot2H2O ->Ca(OH)_2 }$$
Your reaction is wrong. Reference here.
The reaction is very favorite energetically so you don't have the equilibrium you mentioned: when $Ca(OH)_{2}$ is formed you will not have CaO back until you provide sufficient heat and temperature.
The second reaction is right:
$$\ce{ Ca(OH)2(s) <=> Ca^{2+}(aq) + 2OH- (aq)}$$
The solubility of $Ca(OH)_{2}$ is 0.173 g/100 mL (20 °C). This cause the alkalinity of the solution.

How we know if a precipitate forms or not without being given any
  molar concentration of Ca2+ or the KSP of Ca(OH)2?

You can't know it: if you have less than 0.173 g/100 mL of calcium hydroxide you will not have any precipitate.

Without being given any other information in a school setting, is it
  correct to assume that all salts are strong electrolytes?

No, some salts are absolutely not good electrolytes $CaCO_{3}$ is an example. You could say that most of the halogen salts are strong electrolytes.

Is it correct to assume that all low-KSP compounds form a precipitate in solution?

No, because if the solution is not saturated you will not have a precipitate, however is correct say that probably a low-KSP compound will form a precipitate in solution.

I think what's confusing you is how you know if an insoluble precipitate will form and the simple answer would be if an insoluble precipitate can form them it will form, and this is almost a direct consequence of le Chatelier's principle.
Consider the reaction,
$\ce{MgSO4(aq) + Na2CO3(aq) -> MgCO3(s) + Na2SO4(aq)}$
What's actually happening is:
$\ce{MgSO4(s) -> Mg^2+(aq) + SO4^{2-}(aq)}$
$\ce{Na2CO3(s) -> 2Na+(aq) + CO3^{2-}(aq)}$
$\ce{Mg^2+ (aq) + CO3^{2-}(aq) -> MgCO3(s)}$
The Sodium and Sulphate ions in solution are spectators and do not 'react', they don't really do anything at all but you have a solution equivalent to that which you would have formed from the dissolution of Sodium Sulphate in water.
Sorry for the poor format, I haven't figured out how to the proper chemical font yet and also you don't have to include water in your dissolution reactions, the involvement of water is implicit since your state symbols are changing


text:
Oxidation numbers for esters?
How do I find the oxidation numbers for ester?
My main issue is about finding the numbers for ester groups.
         /O\
     -    ||
R -- O -- C -- R'
     -

I want the oxidation numbers of those two O atoms and C atom.

$\ce{CO2}$ and derivatives (e.g., carbonates and their esters, urea, guanidine, $\ce{CCl4}$) are formally C(+4), fully oxidized.  Carboxylates are then C(+3) (but not a chloroformate ester, of course).

I added a schema as much as i can can you check it please

That is an ester, yes.  The paired electrons on the oxygen are usually seen as rabbit ears on the sp^2 hybridized oxygen, normal to the pi orbitals of the double bond.

Rabbit ears is a rather bad description though, @UncleAl

By finding oxidation numbers a systematic way is to go from the element with the highest electronegativity to the lowest. Also always consider neighbouring atoms.
I the presented case of an arbitrary ester $\ce{R\bond{-}O\bond{-}(C\bond{=}O)\bond{-}R'}$, this would be oxygen. Also the $\ce{R}$ moieties are of alkyl or aryl type, so they are bonded to a carbon.
Now assign $-2$ to all Oxygens, assuming they will form a full valence shell (which is a requirement for oxidation states). Now consider that electrons between two carbons will be shared equally. Therefore you can ignore $\ce{R'}$, meaning you assign the total group a value of $0$. There is one Oxygen bonded to the carbon and to $\ce{R}$, it's effect of the oxidation number will be divided in both directions. Therefore the carboxyl carbon will "feel" only $-1$. Add up all the oxidation numbers around the carboxyl carbon ($-3$). Since it is a neutral compound the oxidation state of the carbon has to mirror its surroundings - the value is therefore $+3$.
Always keep in mind, that oxidation states are formal numbers only. They do not represent the actual charge of an atom. They will still give you an idea about the polarity of the molecule.


text:
ESI & MALDI Mass Spectrometry: particle mass vs. measurement error
For MALDI or ESI mass spectrometry of large macromolecules (e.g. in the $\approx 100$ kDa to $\approx 10 - 100$ MDa mass range), I've always how measurement error scales with particle mass.  Not having any hands on experience with mass spetrometry, and calling anything build in the last 5 - 10 years "modern", are there any published tables of general rules of thumb for how measurement error scales with particle mass (the two must surely be positively correlated)?  If we're interested in measuring the mass of large intact nucleic acids or proteins, at what single-molecule particle masses does it make sense to use ESI or MALDI?
I'd guess that at higher mass ranges (e.g. in the MDa range) ESI might become preferable since it can produce multiply charged species, allowing for smaller $\frac{m}{z}$ ratios?

The answer to your question is not an easy one, as multiple parameters need to be taken into account to give an answer. At the high masses that you quote, most of the litterature trends come to their limits in terms of extrapolation, as the major mass ranges in mass spectrometry covers an $m/z$ range 10 - 4 000.
As you indicate in your question, one should separate the question into two parts: how does mass accuracy change with high $m/z$ values and how does mass accuracy change with high mass values, considering a high number of charges, as can be achieved in ESI.
Dependence with ion charges
I will first tackle the second part of the question, because this will be needed to answer the first part. If one considers a mass analyzer in which measurement accuracy ($\Delta m/z$) is directly proportional to the measured $m/z$ result, we have:
$$\frac{\Delta m/z}{m/z} = Constant = \frac{\Delta m}{m}$$
Therefore for such a mass analyzer there is no direct advantage in using multiply charged ions (as produced in ESI) compared to singly charged ions as produced in MALDI. From this point, it comes that on this criterium alone, one should consider if measurement accuracy increases faster or slower than a proportional increase to decide whether multiply charged ions come as an advantage or a disadvantage.
Dependence with mass analyzers
Let us now consider a series of classic mass analyzers and consider how they would perform in terms of high mass measurement.
Quadrupole filters and quadrupole ion traps
I group these in the same subtopic, as when one is concerned about high masses they have fairly similar properties. Commercial instruments are usually limited to $m/z$ 4,000 (or 32,000 for the specially devised systems). So ESI is a requirement for high mass species.
Quadrupole mass filters are usually operated in the constant mass width mode ($\Delta m/z = C$, with $C$ on the order of 0.5 to 1.3 amu), which means that the mass error will be roughly only dependent on $z$ and that there is an advantage to work on the high end of the $m/z$ range, as the overall error will be lower than on the low end. One point worth notice though: if you consider a 2 MDa system at $m/z$ 1,000, it would mean $z = 2000$. Thus spacing between neighbouring charge states is of only 0.5, lower than quadrupole resolution, which means that a deconvolution of the charge state distribution will not be doable and that is will be globally not possible to determine the mass of the molecule with this technique.
Time-of-flight analyzers
These are probably the mostly used analyzers for high mass compounds, both in MALDI and ESI modes. In an ideal TOF, the mass accuracy would depend only on the digitizer (and thus scale as $1/\sqrt{m/z}$), but it also depends on the applied potentials and other factors leading to a mass accuracy that scales as $\sqrt{m/z}$. This still means that considering the above points, mass accuracy will be better achieved at low $z$ values. The mass range limit for TOF analyzers is in theory infinite, but it is in fact limited by the detector (see below).
Orbitrap mass analyzers
Technically an Orbitrap mass analyzer could benefit from working at high $m$, low $z$ conditions, as mass accuracy in an Orbitrap scales as $\sqrt(m/z)$. The instrument is nevertheless limited by the possibility to inject ions within the trap, which is currently limited to 4,000 (or maybe 8,000, I did not check the most recent developments in the field). Thus ESI is required for such high mass systems, but the highest mass accuracies would be achieved on the high end of the achievable mass range.
FT-ICR mass analyzers
In FT-ICR mass analyzers, mass accuracy scales as $m/z$. Thus there is no effect of the charge on mass accuracy. Technically, FT-ICR instruments always include a quadrupole (trap or filter) between the ion source and the FT-ICR mass analyzer and the $m/z$ range is thus limited by the range of this filter (4,000 - 8,000).
Other issues of high mass measurements
One of the difficulties that arises when using high mass ions is also ion detection: devices using a secondary electron emission (such as MCP's or conversion dynode) have efficiencies that depend on the linear velocity of the impacting ion. Therefore the standard detectors are not well adapted to high $m/z$ ions. Some detectors have been developped to circumvent this limitation, but this remains an issue for masses in the 2 - 10 MDa range that you quote in your question.
One other issue is ion desolvation: as size increases, adducts (with salt, solvent or matrix) tend to increase also, leading to changes in the total mass of the ions and a broadening of the peaks. Mass accuracy will also be reduced because there is no direct rule for removing the masses of these adducts from the measured mass to lead to the mass of the molecule.


text:
Mistake in finding bond angles in VSEPR theory

Find the bond angles as indicated, in the compounds given below, according to VSEPR theory-

I got: $109.5^\circ$(6),$109.5^\circ$(7),$109.5^\circ$(8) for the compound (d). But the given answer says it is wrong. But I don't know why. Everything else I got them right.

Looking at compound D, one of the first things we notice is the carbonyl group. The carbon atom in a carbonyl is $\ce{sp^2}$ hybridized, so angle 6 involves an $\ce{sp^2}$ hybridized carbon. In such cases the $\ce{H-C-O}$ bond angle is ~ 120 degrees.
Just to elaborate a bit more:

If you have four objects arranged in space about a central carbon atom (the $\ce{sp^3}$ hybridization case), the four bonds from carbon are going to arrange themselves spatially so that the electron - electron repulsions between the four bonds are minimized.  This is achieved by arranging the bonds such that they point towards the vertices of a tetrahedron.  In a case like $\ce{CH_4}$, where the 4 atoms bound to carbon are identical, the C-H bonds adopt the perfect tetrahedral angle of 109.5 degrees. In the case of $\ce{CH_2Cl_2}$ where the four bonds are not equivalent, the bond angles will differ somewhat from 109.5 degrees (can you guess which angle will be greater then 109.5 degrees and which angle smaller) - but still, the tetrahedral angle is a reasonable first approximation.
In the case of an $\ce{sp^2}$ hybridized carbon, the 3 sigma bonds will arrange themselves in a plane with an approximate angle of 120 degrees between the bonds.  This arrangement again minimizes the electron - electron repulsion between the 3 sigma bonds in this geometry.
Finally, what geometry will minimize electron - electron repulsion between the two sigma bonds in the $\ce{sp}$ hybridized case?

(mouse over for answer)

 Yes, the two bonds will be in a plane 180 degrees from one another, just like the $\ce{H-C-C}$ bond angle in acetylene.

Take home message:

$\ce{sp^3}$ hybridized carbon has angles ~109 degrees
$\ce{sp^2}$ ~ 120 degrees
and $\ce{sp}$ ~180 degrees.

Thank you soooooo much!!!! That's true, but how did you know?

I think in my first chemistry class I learned that sp3 hybridized carbon has angles ~109 degrees, sp2 ~ 120 degrees and sp ~180 degrees.  Seeing the C=O double bond in molecule D indicated sp2 hybridization.  Glad the answer helped, please mark it as accepted.

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


text:
Particle suspension and re-suspension of nano-particles
I am doing a senior project that synthesizes PbS quantum dots and our particles keep crashing out of solution. I was wondering if anyone can give me some advice or information on stabilizing the colloidal solution. The solvent is DI water. I am interested in the chemistry behind stabilization. like what needs to be added to lower the surface energies of particles? why it is added? and how it stabilizes the colloids?

Try mercaptoacetate surface treatment to put a negative charge on their surfaces for repulsion around pH 8 or 9 basic solution.  Uncoordinated mercaptoacetate is oxidatively dimerized to the disulfide by air (oxygen).  The drug S-mercaptocysteine as its salt (basic pH) should be more aggressive toward binding.  3-Mercapto-1-propanesulfonic acid will bind at lower pH.
The dose must be low enough that you do not disaggregate your particles, especially over time.  You can permeable membrane desalt an excess.


text:
What is the convention of numbering carbon atoms in organic molecules?
I am in high school and haven't had a formal chemistry course, however I am fascinated with biology and picked up a biology book more advanced than the one the school is using to read on my own. The book is: Biology by Peter Raven (2007) ISBN: 0073227390
I am confused about something in the organic chemistry section.
I don't understand the convention that is used to number carbon atoms in for example, glucose.
In case you don't know what I'm talking about, here is an example: a structural diagram of an a-glucose molecule courtesy of Wikipedia.

The numbers next to the carbon atoms, I don't understand how they are ordered. Can someone please explain? I know this is an elementary question for most of you, but the help is appreciated.

Um, you know, IUPAC (The International Union of Pure and Applied Chemistry) writes a 1,500-page reference book to answer this question.

Glucose occurs as the open chain form aldehyde that determines atom numbering, and as the epimeric cyclic hemiacetal(s) you provided,

The "ether" is a hemiacetal.  C-1's -OH can be up or down.  Either way, it opens and reversibly dehydrates to the aldehyde R-C(=O)H.  Number one atom in a chain is the end that bears the highest total mass.  Double bonds count double their other end atom.  The cyclic form's C=1 also has the largest summed mass attached, two oxygens plus a carbon and a hydrogen.

So they essentially are numbered from greatest mass to least?

And why do double covalent bonds cause the other end atom to count twice as much?

@JasonMills For a double covalent bond, you have 2 bonds, so you count the atom at the end twice. It as if you are counting each bond separately.

Mass priority traces back to Cahn-Ingold-Prelong (CIP) notation rules for assigning chirality and olefin configuration, etc.  CIP is self-consistent nomenclature not geometry.  L-amino acids alanine and cysteine have opposite CIP chiral labels.

How does this rule of mass work for purine (https://en.wikipedia.org/wiki/Purine)? The number 1 atom (nitrogen) has bonded mass of 36 (3 x carbon) whereas the carbon between 2 nitrogen atoms has bonded mass of 43 (3 x nitrogen + hydrogen).


text:
Law of reciprocal proportion
Is it necessary that the valency of an element must have the same valency?
For eg. If 2 elements A and B combine with the same mass of C, is it necessary that they will have the same valency as they would have when combining AB?

The following are ladders of oxidation states for two elements. Simple examination shows there is no hard and fast rule for valence and stoichiometry overall. Valence need not be integral, either, for example, the Creutz-Taube ion.
Fe(-2), $\ce{Na2[Fe(CO)4]}$, Collman's reagent
Fe(0), $\ce{Fe(CO)5}$
Fe(0), $\ce{Fe3(CO)12}$
Fe(+2), $\ce{FeCl2}$
Fe(+3), $\ce{FeCl3}$
Fe(+4), $\ce{K4[FeO4]}$
Fe(+5), $\ce{K3[FeO4]}$
Fe(+6), $\ce{K2[FeO4]}$
V(-2), $\ce{[Na(diglyme)2][V(CO)6]}$
V(-1), $\ce{(Cp)V(CO)4}$
V(0), $\ce{V(CO)6}$

Chemistry seeks to establish rules.  Then the fun starts - breaking the rules.

V(CO)6 is a coordinated compound, you cannot use the oxidant number to describe such compounds.


text:
Quotable Literature for the Enthalpy of a Proton
A proton obviously has no electronic energy, no vibrational and no rotational degrees of freedom. Therefore I think it is fair to assume, that $$H(\ce{H+})=\frac32\cdot \mathcal{R}\cdot T$$ holds for the translational degrees of freedom. I am looking for some quotable literature where they state that, or if that has not been stated anywhere, how to derive this.
I am doing quantum chemical calculations, i.e. gas phase. I define the proton affinities via
$$\text{PA}(\ce{M})=-\left(H(\ce{H+M})-\bigr[H(\ce{M})+H(\ce{H+})\bigr]\right)$$
for the reaction
 $$\ce{H+ + M -> H+M}.$$
This is respecting the (quite common) sign convention to show the energy released upon binding a proton.
I need a reliable source or at least a best-practice example(s) for $H(\ce{H+})$.

A quite complete procedure in the calculation of proton affinities from ab initio calculations is referenced in an article by H. Audier et al., Int. J. Mass Spectrom. 1996. It goes much further in the procedure than only on the calculation of the enthalpy of a proton (including corrections due to the basis set superposition error and so on), but confirms that you are right in assuming that the enthalpy of a proton is $3/2 RT$.

Unfortunately they also only state my suspicion as a fact (no proof) - but at least now I know that I am not the only one thinking so... (sorry for the delay in accepting the answer).


text:
What is the unit of pH?
I'm making some graphs and I have to label the axes. I want to be extra careful and put the units in even though the meaning of $\text{pH}$ is well known. But I have a problem (though a simple one): $\text{pH}$ is a minus logarithm (base 10) of concentration of hydrogen ions (or rather their activity). What is the unit then, is it $[-\log(\text{mol}/\text{L})]$? What should I write, could you help me?

That formulation of the definition is nonsensical -- you can't take the logarithm of a concentration since it has a non trivial unit. You first need to divide the concentration by a "standard concentration" such as $1 \mathrm{mol}/l$. Had some trouble with my Chemistry teacher who taught that sloppy definition.

Some useful information about the units of transcendental functions can be found here. The paragraphs around equation 11 are relevant to your question, although they sadly ignore @phillipp's comments about activity.

Why is "pH" not a sufficient label?

@SethBattin pH as a label for the axis of a graph is fine. (Well, assuming pH is being plotted....)

Relevant question on physics.SE

@CodesInChaos: At least that was “only” a teacher. I once failed to explain to a full professor why I had problems with taking the logarithm of something other than a scalar.

The real definition of the $\text{pH}$ is not in terms of concentration but in terms of the activity of a proton,
\begin{equation}
  \text{pH} = - \log a_{\ce{H+}} \ ,
\end{equation}
and the activity is a dimensionless quantity. You can think of the activity as a generalization of the mole fraction that takes into account deviations from the ideal behaviour in real solutions.
By introducing the (dimensionless) activity coefficient $\gamma_{\ce{H+}}$, which represents the effect of the deviations from the ideal behaviour on the concentration, you can link the activity to the concentration via
\begin{equation}
  a_{\ce{H+}} = \frac{\gamma_{\ce{H+}} c_{\ce{H+}}}{c^0} \ ,
\end{equation}
where $c^0$ is the standard concentration of $1 \, \text{mol}/\text{L}$.
If you ignore the non-ideal contributions you can approximately express the $\text{pH}$ in terms of the normalized proton concentration
\begin{equation}
  \text{pH} \approx - \log \frac{c_{\ce{H+}}}{c^0} \ .
\end{equation}
In general, there can be no logarithm of a quantity bearing a unit. If however you encounter such a case it is usually due to sloppy notation: either the argument of the logarithm is implicitly understood to be normalized and thus becomes unitless or the units in the logarithm's argument originate from using the mathematical properties of logarithms to divide the logarithm of a product which is by itself unitless into a sum of logarithms: $\log(a \cdot b) = \log(a) + \log(b)$.

Unless you have very good reason to do otherwise, treat pH as dimensionless.


text:
Activity Coefficients from Electrode Potentials
Lets say we want the activity coefficient of $\ce{Cu^{2+}}$ ions in solution. Would we simply setup one half cell of a $\ce{Cu}$ electrode in a known concentration of $\ce{CuSO4}$ and then connect this to a Standard Hydrogen Electrode.
Using the standard electrode potential of the reaction $\ce{Cu^{2+} + 2e-  ->  Cu}$  and the Nernst equation could we determine the activity coefficient of $\ce{Cu^{2+}}$ ions ?
Is this all I would need to do ?

You would not be able to easily separate the activity coefficients of the anion and cation using this method.

You will probably have to do a lot of concentration dependent cross measurements involving a huge set of different ions not only copper. You should also rather use Ag/AgCl electrode.

Sort of 'yes', sort of 'no'.  Consider the following points:

Activity coefficients are notorious for changing with $\mathrm{p}\ce{H}$, ionic strength, and concentration (among other things).  You might measure the the activity coefficient of $\ce{Cu^{2+}}$ in a $0.7~\mathrm{M}$ $\ce{CuSO4}$ solution and get one value, but try to measure the same thing in a low $\mathrm{p}\ce{H}$ and $1~\mathrm{\mu M}$ $\ce{CuCl2}$ solution and you may get a completely different value.  In practice, people usually end up measuring activity coefficients across a range of conditions that are relevant to their specific use-case.
The Standard Hydrogen Electrode is a theoretical device; and actually constructing something similar is difficult.  In practice, you would just want to use a commercially available double-junction silver-silver chloride or saturated calomel electrode. The potentials of those two are well-known and fairly stable.

See Chapter 2 of "Electrochemical Methods" 2nd Ed, by Bard and Faulkner for more information.


text:
What are the physico-chemical processes at play in ironing of clothes?
The process of ironing of clothes removes (to a large extend) the wrinkles that where introduced in the clothes while drying. Obviously this has to be some physico-chemical process that is initiated/sustained by the heat of the iron and (perhaps) the pressure applied with the iron.
My question is: what is the actual process that is going on when ironing? Let's assume we have a cotton shirt, do the polymers only physically rearrange or do they actual undergo a chemical change? If it is the latter, does ironing your clothes often cause more rapid degradation?

Ironing of clothes is a physical change (if you believe that physical changes and chemical changes are not part of a single continuum of change).
Most polymers (including the cellulose fibers that comprise cotton) have a glass transition. The glass transition is a reversible phase change from a rigid fixed solid amorphous "glass" to a solid phase where there is some mobility of the polymer chains, but not enough to be a liquid (this phase is sometimes called a "rubber," "plastic," or "elastomer" phase). The glass transition is sometimes used to refer to the phase change between an order crystalline (or semicrystalline) solid to a disordered amorphous solid.
Above the glass transition temperature, the solid material has some mobility - it can be shaped, molded, flattened, whatever. When it is cooled down again, it becomes rigid and holds its shape.
Ironing raises the temperature of the fabric to above the glass transition. The action of ironing presses out the wrinkles in the now more flexible fabric. When the fabric cools, it retains its new wrinkle-free shape.
Water is used during ironing for one important purpose. I tried to find the glass transition temperature of cellulose, but am unable to do so. Presumably it is close to the temperature at which cellulose begins to burn. Water prevents the burning, but not just in the way you would expect. Water does absorb heat from the iron in order to keep the fabric from getting too hot. However, if this were the only thing water was doing, the fabric would not reach the glass transition and ironing would not work. Water also acts as a plasticizer, lowering the glass transition temperature. Most fabrics are hydrophilic (and cellulose as a carbohydrate is especially so). The water binds to the chains, swells the fibers, and changes the physical and chemical properties of the material.

Do you think water also serves to transfer energy to fibers that aren't on the surface?  I imagine that (hot) water vapor diffuses through the pores of the material allowing heat to be transferred more efficiently to the bulk material.

The glass transition temperature of dry cellulose is about 225 degrees celsius, that of wet cellulose significiantly lower.

First iron gives water to polymer (most likely cotton). The water molecules will break the H-bonding between cotton fibers and will do H-binding with polymer themselfs.(not uniform H-bonding with next fiber causes the wrinkles in the first place)After you break these bonds you iron it, meaning apply pressure to fiber so they align uniformly.As you do this the hotness of the ironing heats up the fiber thus the water molecules avaporates and leaves the fiber.Now the fibers will do H-bonding with each other again. But this time it will be uniform. Thats it.

Uh, do you have any references for any of these claims?


text:
How do I find Rf values & length size of paper in chromatography?

A student is designing an experiment using paper chromatography. From previous results, she knows that the Rf values for two dyes that she wishes to investigate are 0.72 and 0.80.
It is proposed that the dyes be run on the same piece of paper and, for various reasons, that the sports should have a separation of a at least 1cm at the end of the procedure.
What is the minimum height of the chromatography paper that she should use in this experiment?

Can someone correct my working out. I don't know what the correct answer is.

distance travelled by dye x = $x$
distance travelled by dye y = $y$
distance of solvent front = $p$
$x/p=0.72$ therefore $x = p(0.72)$
$y/p=0.8$, therefore $y = p(0.8)$
$y-x=1$, therefore $y=1+x$
new equation: $1+x=p(0.8)$ where $x=0(0.72)$, if we solve for $p$ we get $12.5$

so the piece of paper should be at-least 12.5 cm long.

$$\Delta R_f = 0.08$$
Let the separation distance be $\Delta r$ = 1 cm and let $l$ be the length of the paper in cm.
$$\Delta R_f = \frac{\Delta r\, \mathrm{cm}}{l\, \mathrm{cm}} $$
Solve for $l$, obtain $l$ = 12.5 cm.

why is the change in rf over 1


text:
If metals want to "lose" electrons, then why will copper ions take electrons from a zinc electrode?
According to my book, metals want to "lose" electrons to achieve noble gas configuration. But at the same time, copper ions in a galvanic element are going to gain electrons from a zinc electrode. Why is it so? Is it because one of them HAS to gain/lose electrons, and zinc wins out in terms of achieving noble gas configuration?

Typically, when one uses terms such as "likes to lose" or "likes to gain", we should think of that phrase as "likes to lose relative to...".  In your case, when we think about electron configurations, it might be more helpful to think "metals like to lose rather than gain electrons to achieve a noble-gas-like configuration".  When comparing two metals, both of which tend to lose electrons, we need to look at different concepts to answer the question.
We can explain the transfer of electrons by looking at the the metal activity series such as the one shown here.  The way one reads this table is that a Galvanic cell created with two metals will have the metal higher on the list as its anode (where oxidation occurs) and the other metal will be the cathode.  Thus, we would expect Zn to be oxidized to Zn(II) and Cu(II) to be reduced to Cu.  These tables are based on standard reduction potentials:
$$\ce{Cu^{2+} + 2 e^- -> Cu(s)}\ \tag{E = 0.34 V}$$
$$\ce{Zn^{2+} + 2 e^- -> Zn(s)}\ \tag{E = -0.76  V}$$
By flipping the 2nd reaction (thus changing the sign of E) one obtains a cell potential of about 1.1 V; a positive number meaning that this reaction will happen spontaneously:
$$\ce{Cu^{2+} + Zn(s) -> Cu(s) + Zn^{2+}}\tag{E = 1.1V}$$
The standard reduction potentials tell us that the electrons would rather be on copper than on zinc.

Reactive systems seek lowest free energy equilibrium.  In aqueous solution, Zn + Cu(II) is higher in energy than Zn(II) + Cu(O) re the electromotive series.
This can be manipulated.  In hard Lewis base water, soft Lewis acid Cu(I) is unstable to disproportionation to give hard Lewis acid Cu(II) and Cu(0).  In soft Lewis base acetonitrile, Cu(II) and Cu(0) are unstable to conproportionation to Cu(I).  That last is quite a reaction.  Anhydrous CuCl2 and copper powder (slight excess) are refluxed with stirring in MeCN. The solution turns a terrible opaque inhomogeneous khaki color, then flashes colorless and transparent.
It would be interesting to try it with Fe(II) and Fe(O).

Metals want to lose electrons because the process requires very less energy. Some metals require more energy than others, e.g. consider Cu and Zn. Zn will require lesser amount of energy to lose electrons than Cu. This is easily seen by the greater Oxidation potential values of Zinc.
When a metal electrode is dipped in an electrolyte containing its own ion, then it may lose or gain electrons depending on whether it has a negative or a positive oxidation potential, again involving energy $(W=qV)$. Thus, Zn having a negative potential loses electrons and Cu having a positive potential gains electrons. Thus, a connected Zn and Cu system will allow passage of electrons from Zn to Cu.


text:
Which compound can exist as stereoisomers?
I'm having trouble in understanding how to identify chemical compounds as stereoisomers.
In the question below, D is the correct answer according to the markscheme. I've drawn out the chemical structure but am unable to see what possible stereoisomers it can form.
I also tried visualising the possible isomers on Wolfram Alpha:
http://www.wolframalpha.com/input/?i=CH3CH2CHOHCH3
There are two isomers which stand out to me, which I guess are the stereoisomers?

(S)-(+)-2-butanol
(R)-2-butanol

However, to me it looks like they are just rotated versions of each other.

ergh, dude, read these articles http://en.wikipedia.org/wiki/Chirality_(chemistry) http://www.chemguide.co.uk/basicorg/isomerism/optical.html#top

"D" is the correct answer. 2-Butanol exists as two stereoisomers, R- and S-2-butanol, just as you described.  The "2" carbon in 2-butanol is a chiral carbon atom, it has 4 different groups attached to it (ethyl, methyl, hydroxyl and a hydrogen atom).  The R and S forms of 2-butanol cannot be rotated one into the other, they can only be interconverted by reflecting one in a mirror.  They are non-superimposable mirror images like your right and left hand. Build a model of 2-butanol and then build a second model that is the mirror image of the first - you'll find that you cannot intercovert them by rotation and one is not superimposable upon the other.

Ah, so actually just by elimination A, B and C do not have any chiral carbons with 4 different groups attached, leaving only D a possible option.

Yes, that is correct.

http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
Everything you ever wanted to know about stereoisomerism.
Quick answer:  Any reasonably tetrahedral atom bearing four different groups (including isotopic substitution and unbonded orbitals) is a chiral center.  However, pairs of enantiomeric chiral centers in a molecule can cancel (meso-compounds.  Tartaric acid has three stereoisomers).
Technical answer:  Any center (need not contain an atom - helicenes) that lacks all ${S_n}$ symmetries is a chiral center.  Any molecule that lacks all ${S_n}$ symmetries is chiral.  Consider methyl phenyl sulfoxide and that sulfur's lone pair.
Rigorous answer (Michel Petitjohn, J. Math. Phys. 40, 4587 (1999) et al.):  Any collection of anonymous unit mass countable points that sum to finite principle moments of inertia, in any number of dimensions greater than zero, that cannot be superposed upon its mirror image (one coordinate with all reversed signs) strictly by translations and rotations, is chiral.  An infinite number ($\ce{Aleph_0}$) of points in a hyperbolic tiling qualifies.  Petitjohn's QCM software quantitatively calculates geometric chirality.
That last one is important.  It is a recipe for creating an acyclic, undistorted tetrahedral carbon atom bearing four rigorously identical, freely rotating subtsituents that is a chiral center.  Planar $\ce{sp^2}$ carbons can be chiral.  It demonstrates that no organic nomenclature for chirality can be complete.  It demonstrates that at least one physics founding postulate describing vacuum symmetries can be testably defective, measurably empirically falsified by chemical constructs.
The philosophical 900 lb gorillas,
Symmetry Through the Eyes of a Chemist Magdolna Hargittai, István Hargittai, 2009
Fearful Symmetry: Is God a Geometer?  Ian Stewart, Martin Golubitsky, 2011.

To exhibit stereoisomerism, a compound must have a chiral centre i.e - a carbon atom with four different groups attached.
D is the only compound that meets this requirement.

http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/symmetry/index2_helicenes.png  Where is the "chiral center?"  http://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure0/096/a_____595519.eps/_jcr_content/renditions/medium.png  Where is the "chiral center?"


text:
Simple Kinetics Clarification
I just wanted to check my understanding of this. Given the reaction,
$$ (1) \; \ce{A->B} $$
the corresponding rate law is $ v=k[A] $
Also consider
$$ (2) \; \ce{A <=> B} $$
with forward rate law: $ v_1=k_1[A] $ and reverse: $ v_2=k_2[B] $. This makes the corresponding rate equation $ v=v_1-v_2=k_1[A]-k_2[B] $.
My question is in the difference between the two. I've often seen reactions expressed in both ways. In $ (1) $, is the rate law assuming the reverse reaction not to occur (or at least to be completely negligible? That's why I drew it with $ \ce{->} $ instead of $ \ce{<=>} $. Am I correct in my thinking? In actuality, is it always more correct to use the rate law of $ (2) $? The rate law for $ (1) $ could also correspond to initial reaction velocity, right? This is because there is no $ B $ present at the start of the reaction.
If this is correct, then the integrated rate equations will become substantially more complicated using rate laws of form $ (2) $. I can't use simple equations like $ [A]_t=[A]_0 e^{-kt} $, as this wouldn't take into account the reverse reaction.

Not all reactions can or should be represented with a double arrow.  The general-audience example is that you "can't unbake a cake".  You are right that there are instances that we assume that k2 is negligible (relative to k1) and therefore a reaction is treated as if it is reversible.
Here's a sample plot of how the concentrations of A for the irreversible and reversible reactions compare for k1 = k2 = 1.

No surprises here; the reversible reaction results in an equilibrium concentration of A. We can compare the concentration of A in each of the reactions as a function of kr:

Here I've plotted Log[k2] on the x-axis and the ratio of A in the irreversible reaction to the reversible reaction.  The different lines are at various times (in seconds).  I've kept k1 = 1 and the initial concentration 1 as well.  The two equations predict the same concentration of A in the limit of very small k2, and by very small I suspect this means at least a factor of 10^4 or more (not shown on the graph).
While you are correct that the integrated rate laws get much more complicated for reversible reactions, the one you are considering is still solvable in closed form.

In $(1)$ the reaction is irreversible. For example, potassium chlorate decomposes on heating to form potassium chloride and oxygen:
$$\ce{2KClO3 -> 2KCl + 3O2}$$
In this reaction, the products cannot combine to give back potassium chlorate (I think the reason would be because of reaction not being performed in closed container). In case of irreversible reactions, the reactions occur only in forward direction. In actuality, you can use rate equation of $(2)$ for the above reaction, as rate backward reaction is zero, you will end up with rate equation of $(1)$.
Velocity of reaction $(1)$ doesn't corresponds to velocity of reaction at time $t=0$. It is the general equation for velocity of the reaction.
Consider a first order reaction $$\ce{A -> products}$$
Velocity of the reaction is given by $\frac{\mathrm dx}{\mathrm dt}=k[\ce{A}]$. Here, $[\ce{A}]$ corresponds to concentration of reactant $\ce{A}$ at a particular time and it doesn't correspond to concentration of $\ce{A}$ at $t=0$.
To understand it consider the following explanation. let the initial concentration of the reactant be $a\ \mathrm{mol\ dm^{-3}}$. Let $x\ \mathrm{mol\ dm^{-3}}$ decompose in $t$ seconds. The remaining concentration of the reactant is $a-x\ \mathrm{mol\ dm^{-3}}$.
${\therefore}$ Velocity of the reaction after a time 't' is given by $\frac{\mathrm dx}{\mathrm dt}=k[a-x]$.
Rate equations for reversible reactions will be different as given in this page (see the first order reaction part).


text:
How can I specify pyridine as a substituent?
You know how when you carry out a reaction with many compounds, when drawing the reactions you just draw a line at a right angle to the bond with a 'R' at the end (see image)?

Then you can specify what group R is in a table (eg: $\ce{OMe}$). However, if instead of adding a substituent to a benzene ring, you have pyridine? How can you specify it?

The issue, I think, is to specify how the pyridine is connected. This means you have to identify which carbon on the pyridine ring is connected. This picture shows the numbering:

So, using your example the structure and names of the various possibilities would be (using connections on the 3-position of the benzonitrile ring as an example):

I'm sure you can generalise from this example.

There is another configuration where the nitrogen is connected to the first compound and a hydrogen is added to the para position in the pyridine ring.

You can draw out the substructure within the table, so it'll be a mix of text and drawn substituents.


text:
How much hydrolysis will likely occur of the following bonds at pH 1-4 in the time period of 1-3 hours?
The bonds are: (I have given pictures of what I mean in case there's any ambiguity that I leave open in these questions. Also so that if you want to number the R groups you've got a head-start)

Secondary amides, I suspect I might know the answer, extensive. Example is the one in folic acid.

Tertiary amides. This is one that's a complete mystery to me. Example is the one in leucovorin.

Sulfonamides. Hydrolysis in this case, should lead a primary amine and a sulfonic acid. Or such is the hope.

Carboxylic esters. To give a primary alcohol and a carboxylic acid. Basically the reverse of the Fisher esterification.

Sulfonyl ester.

Secondary/tertiary amines

Ether

Sulfyl ether

Now if you're wondering how exactly is this question relevant? As it might sound pretty random for me to just ask this question but it's incredibly relevant to drugs when it comes to their tendency to be hydrolysed in the stomach.
The simple fact is that I'm likely to come up with other questions so if it's possible for you to give me a guide for figuring out this stuff, or estimating it, please do share! If you do this and I can follow it, I will instantly accept your answer as the answer! Providing me with a decent ref will also win you points.

This source contains a decent amount of information on the hydrolysis kinetics of some types of functional groups, though not all those mentioned in your question.

It has been shown that the stomach empties between 30 min and 1 hour after a meal (with some latitude). It would be better to restrict your time to 1 hour. Also the pH of the stomach is between 1 and 2 for most people.

@LDC3 My lecturers tend to teach me 1-3 hours. Plus I know I personally whenever I take my meds on a incompletely empty stomach I get severe indigestion, 1 hour doesn't do it for me, more like 2 hours at least so that gives me some evidence to the contrary to your theory. Like 1 hour with some latitude to me is +/- 30 mins so yeah I don't know where you're getting your info. Please do show me.

@NicolauSakerNeto thanks. It really does help. Still (as I see you're not putting this forward as an answer) I'd like a nice concise (like 3 paragraphs max) method for estimating the percentage of a functional group that gets hydrolysed in a set period of time.

@LDC3 read http://www.vivo.colostate.edu/hbooks/pathphys/digestion/basics/transit.html it says 4-5 hours for complete stomach emptying so it's saying my estimate is a bit low. Plus the pH depends on whether you just ate or not.


text:
Proving RTln($a_i$) equal to non PV work done by system,where $a_i$ is activity
We know that chemical potential is defined as ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$. Here $a_i$ is the activity of $i^\text{th}$ component of solution. In case of gases instead of $a_i$ it is $f_i$ (fugacity). Activity is the activity coefficient multiplied by concentration or in case of fugacity it is the product of the fugacity coefficient multiplied by partial pressure of the gas.
But now this is not the only definition of Chemical potential. Chemical potential also means the partial molar Gibbs free energy when pressure and temperature of the system are constant. We know that change in Gibbs free energy of a system at constant temperature and pressure also represents Non-PV work done by system.
How can the term $RTln(a_i)$ can be correlated to Non-PV work done by system in order  for the two definitions to be equivalent?
NEW QUESTION AFTER READING YOUR ANSWER(S)
Philipp, Ok after reading all your answers I got now modified question that still remained.First of all we know K and Kc are related but kc can be modified by adding activity coefficients and standard concentrations to give equilibrium constant that should be equal to K.This requires the activity coefficients (That we used with Kc) to always follow relation with fugacity coefficients and RT terms which is not proved in answers there and that is what actually my question is asking.Sorry but I am reluctant here to write maths but I just say refer to the second equation from last in your This Answer
Actually earlier my question has title ways of expressing concentration in activity term

No one even trying to give answers,I am disappointed.Just try to join the discussion

I'm not sure if I understand you correctly. Are you asking why I use fugacity coefficients rather than activity coefficients?

@Philipp No I understand that using fugacity for gases we can get it's relation with non-pv work but not for solutions where we use concentrations and there is no partial pressure to use.

Ah, ok. There is actually the concept of a partial pressure in solutions: the vapor pressure. In equilibrium the vapor pressure of the component in solution is equal to the partial pressure of the same component in the gas phase directly above the solution. And there are Raoult's Law and Henry's Law linking the pressure to the mole fraction/concentration in the solution. But there is some uncertainty in my mind concerning the relation between fugacity coefficients and activity coefficients...

... Maybe this Wikipedia article about activity is of interest to you.

I read it in past.Probably my question is difficult to state or way too fundamental.

So, my comment didn't clarify the matter?

I am sorry but no

I think there must be some experimental parameter so we may not prove it.

let us continue this discussion in chat

We know that chemical potential is defined as
  ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$.

Be a bit careful with your wording. This equation is not part of the definition of the chemical potential but rather it is derived from it.
The derivation of this equation can be found in this answer of mine.
The definition of the chemical potential is
\begin{equation}
 \mu_{i} = \left( \frac{\partial G}{\partial n_{i}} \right)_{p,T} \ ,
\end{equation}
where $n_{i}$ is the amount of the $i^{\text{th}}$ component in the system.

We know that Gibbs free energy of a system at constant temperature and
  pressure also represents Non-PV work done by system.

Again this is not quite correct: It is the change in Gibbs free energy of a system during some process, say a chemical reaction or a phase transition, at constant temperature and pressure that can be identified with non-PV work done by the system.
Now considering a general chemical reaction
\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C <=> \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}
the differential change in Gibbs free energy is given by
\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \mu_{i} \text{d} n_{i}
\end{equation}
or introducing the extend of reaction, $\text{d} \xi = \frac{\text{d} n_{i}}{\nu_{i}}$
\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \nu_{i} \mu_{i} \text{d} \xi \ .
\end{equation}
From this it follows that the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$) at constant pressure and temperature is given by
\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ .
\end{equation}
Here you can see that the non-PV work done by the system originates from the differences in the chemical potentials of the reactants.
Substituting the equation for the chemical potential
\begin{equation}
 \mu_{i} = \mu_{i}^{0} + RT \ln a_{i}
\end{equation}
the equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, leads to
\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}
where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced.
This is the relation between the non-PV work, represented by $\Delta G$, and the activity terms. Admittedly, it is not a very graphic explanation but I fear that it simply is a pretty mathematical matter.


text:
How much heat (in J) is required to produce 2.0L of N2(g) at 77.36 K and 1.0 atm?
The only thing that is given is:
The enthalpy of vaporization for $N_2(l)$ is 5.56 kJ/mol.
There aren't any examples similar to this problem in the textbook. I only know how to get energy using:
$q=mcΔT$ and $w=PΔV$
No clue what to do with the enthalpy of vaporization value.
UPDATE: Solved it by first getting number of moles ($n$) via $PV=nRT$ then multiplying moles by 5.56 kJ/mol. Moles cancel out, leaving only kJ, 1.75kJ to be precise.

Since nitrogen's boiling point is 77.36 K , it appears that the question is asking how much heat is required to boil nitrogen to produce 2.0 L at 1 atm.

Sorry but it doesn't help much

Using the ideal gas law, how many moles of nitrogen are in 2.0 L at 1 atm and 77.36 K? Then find the amount of heat needed to convert that amount from liquid to gas.

Once i get the # of moles, which formula do i need to apply to get heat?

@TwilightPonyInc. You should have used the equation. A favorite problem involves heating ice to the freezing point, melting the ice, and then heating the water to a defined temperature. When you melt the ice, that is the equation you want.


text:
What is the relationship between migration and the potential of double layer theory?
Double layer theories like that of Gouy-Chapman give the electrostatic potential $\ \phi\ $ in solution. For a symmetrical electrolyte and small potential drops $\ \phi\left(x=0\right)-\phi\left(x\rightarrow\infty\right)=\phi_0-0\ \text V=\phi_0$, the theory gives
$$
\phi=\phi_0\exp\left(-\kappa x\right)
$$
where $x$ is measured from the electrode/solution interface into the bulk solution.
As $\kappa$ is very large $\left(\sim10^{-9}\frac{1}{\text m}\right)$, the electric field
$$
E=-\frac{\text{d}\phi}{\text{d}x}=-\kappa\phi_0\exp\left(-\kappa x\right)
$$
falls to a constant value (0  V in this example) very quickly.
In the equations for the transport of ions in diluted solutions, the migration flux of species $j$ is
$$
J_{\text {j,migr}}=-z_ju_jFc_j\frac{\text{d}\phi}{\text{d}x}
$$
As the flux is proportional to $E$ it must be zero everywhere in solution according to Gouy-Chapman. But in every book about electrochemistry migration is said to contribute to the flux $J_j$ of species $j$ except when $c_j$ is much more smaller than the concentrations of other species. But why should there be migration in a macroscopic cell at all? In double layer theories ions gather near the electrode surface and compensate the charge on the metal, so there should be no electric field in solution. Isn't this what the equation for $\phi$ above says?
Or are there two different types of "electrostatic potentials" in solution: One associated with the double layer theories and one associated with transport?
$$
\phi_{\text{Gouy-Chapman}}\neq\phi_{\text{Migration}}
$$
If this is the case: How are both potentials related to each other (and why do books always use the same symbol for both)?


text:
Bond angle of nitrous acid
Why are there two possible bond angles for the nitrous acid molecule?
After perusing Wikipedia I found that there are both cis and trans forms of the nitrous acid molecule. I also note that the cis form is sterically hindered. Could this be a reason for the two different bond angles - there are two different geometric isomers and in one case there are larger van der Waals repulsions, which in turn expands the bond angle?

Which bond angles are you referring to?

Oops. I am referring to the O-N-O bond angle.

This ism the microwave structure of nitrous acid, 1971.  Your mileage may vary condensed phase.
DOI:10.1039/TF9716702179 (1971)
The structure of cis-nitrous acid is found to be: O—H = 0.982, N—O(H)= 1.392, N—O = 1.185 Å, ∠ NOH = 104.0 and ∠ ONO = 113.6°, and for trans-nitrous acid: O—H = 0.958, N—O(H)= 1.432, N—O = 1.170 Å, ∠ NOH = 102.1 and ∠ ONO = 110.7°.
http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf
  2007 pape with more recent citations.

Also, would you say that hydrogen bonding has to do with the two different structures? I.e. hydrogen bonding between the H and the O in HONO?


text:
Calculating the partition function based on molar heat capacity
Given the molar heat capacity of a partition function as a function of temperature, how would one determine the partition function?
Say the molar heat capacity = $\alpha T^2$

All you have to do is to use some of the well known results from Statistical Thermodynamics:
The molar heat capacity at constant volume is
\begin{equation}
  \bar{c}_V = \left( \frac{\partial \bar{U}}{ \partial T} \right)_V
\end{equation}
where $\bar{U}$ is the inner energy, $T$ is the temperature, and $V$ is the volume.
The molar inner energy in terms of the partition function $Z$ and the temperature is given by:
\begin{equation}
  \bar{U} = \frac{k_{\text{B}} T^{2}}{n} \frac{\partial \ln Z}{ \partial T} \ .
\end{equation}
where $n$ is the amount of substance.
Now, you simply have to plug the equation for $\bar{U}$ into the equation for $\bar{c}_V$ and after using the chain rule of differentiation you get the following differential equation:
\begin{equation}
  \bar{c}_V = 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} + \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} \ .
\end{equation}
Finally, you just have to use your formula for the molar heat capacity, $\bar{c}_V = \alpha T$, and solve the resulting differential equation
\begin{align}
  \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} - \alpha T^2 &= 0  \\
 \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{1}{T} \frac{\partial \ln Z}{ \partial T} - \frac{n \alpha}{k_{\text{B}}} &= 0
\end{align}
for $\ln Z$.
If solving differential equations by hand is not your thing, you can simply charge Wolfram Alpha with the task and get:
\begin{align}
  \ln Z &= \frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \\
  Z &= \exp\left(\frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \right) \ ,
\end{align}
where $C_1$ and $C_2$ are constants of integration which can be fixed by enforcing boundary conditions for the temperature.

The heat capacity is generally not proportional to $T^2$ unless over a very small range of temperatures because it becomes constant at high temperatures.
With experimental data where the heat capacity is some function of temperature $C_V=f(T) $ it is easiest to integrate directly using the definitions from statistical mechanics. Thus as $C_V=(\mathrm dU/\mathrm dT)_V$ integrating from $0$ to $T$ produces
$$U(T)=\int _0^T C_V(T)\,\mathrm dT$$
where $U(T)$ is the internal energy as a function of $T$.  If $C_V=\alpha T^2$ then the integration is straightforwards. If the function is more complex or you want to use actual data then numerical integration may be needed.
Then using
 $$U(T) = U(0) +RT^2\,\frac{\mathrm d\ln(Z)}{\mathrm dT}$$
The log of the partition function $Z$ is obtained as
$$\ln(Z)=\int_0^T \frac{U(T)}{RT^2}\,\mathrm dT$$
In your particular case $U(T)=\alpha T^3/3$ and $\ln(Z)=\alpha T^2/6$


text:
Is the end of a cigarette on fire when lit?
Is the end of a cigarette on fire when it is lit? Or is it smoldering? Is there a chemical difference between the two?

A lighted cigarette is burning, technically smoldering (no flame).  The stuff inside the paper bears only a passing resemblance to tobacco,
http://en.wikipedia.org/wiki/Cigarette#Construction


text:
Is this endo or exothermic?
$$\ce{4Fe + 3O2  -> 2Fe2O3(s) + 1625~$\mathrm{kJ}$}$$
This should be endothermic because the energy is in the product. However, if the equation is viewed from the perspective of the reactant, it is endothermic.
Is this endo- or exothermic?

The reaction is exothermic left to right and endothermic right to left. Yes, as you noted, it's wise to note which way you are looking at the reaction, but generally, we look at reactions from left to right (because the arrow generally points left to right). For example, the following reaction is endothermic as written, i.e. very hot water (steam) is very corrosive and adding heat to water encourages further ionization of water.
$$\ce{2H2O ⟶ H3O+ + HO-}$$

The reaction is exothermic:

A system that releases heat to the surroundings is an exothermic reaction.

Furthermore, forming a chemical bond releases energy and therefore is an exothermic process, such as when Iron and Oxygen bond to form Iron(II) Oxide. Sometimes it's not clear whether more bonds were formed or broken during a chemical reaction so we have to add up the energies to see, unless they are already given to us like in this example.
Lastly, always consider the equation in the direction the arrow is pointing. The standard is to point from left to right, but sometimes it can point both ways simultaneously. In this case, the question should state which direction of the chemical reaction it wants you to consider.

In the equation given, the energy is not actually in the product, but is written as a product, meaning that 1625 kJ of heat is evolved along with 2 moles of iron(II) oxide.

An exothermic reaction! As you can see that the products of the reaction involves two chemical species and energy... that is after the reaction you got energy as well as two compounds hence the reaction is an exothermic reaction...


text:
How do I find the final temperature of a calorimeter using specific heat capacities?

What is the final temperature of water and iron if a $\pu{30 g}$ piece of iron at $\pu{144 °C}$ was dropped into a calorimeter with $\pu{40 g}$ of water at $\pu{20 °C}$?
Specific heat of water is $\pu{4.184 J g-1 ^\circ C-1}$, and of iron is $\pu{0.449 J g-1 ^\circ C-1}$

Here is my work:
\begin{align}
Q &= mc\,\Delta T\\
Q_1 &= (\pu{30 g})(\pu{0.449 J g-1 ^\circ C-1})(x - \pu{144 ^\circ C}) \tag{Iron}\\
Q_2 &= (\pu{40 g})(\pu{4.184 J g-1 ^\circ C-1})(x - \pu{20 ^\circ C}) \tag{Water}\\
\text{Since,} \quad Q_1 &= -Q_2\\
13.47 (x-144) &= - (167.36)(x-20)\ \pu{J}\\
13.47x - 1939.68 &= -167.36x + 3347.20\\
180.83x &= \pu{5286.88 ^\circ C} \\
x &=  \pu{0.03420 ^\circ C}
\end{align}

This is giving me an answer that isn't correct according to my book. What did I do wrong, and how can I fix it?

$\frac {5286.88}{180.83} \neq 0.03420$

Use Kelvin instead of Centigrade/ Celsius! It would not change in this calculation as they are on the same scale and you are using differences. Also try using units throughout the whole process, this will give you a hint if you transformed your equations correctly. Apart LDC3's comment, I cannot see anything wrong.

All you did is essentially right, your only mistake is in the last step, as LDC3 already pointed out in the comments. However, I am encouraging you to use units all the way and when dealing with thermodynamics use Kelvin instead of Celsius.
\begin{align}
Q &= mc\Delta T\\
\end{align}
Now you can form the equations for each of the problem, while substituting $\Delta T$ with a temperature range, being $x$ the final temperature the whole system will end up on. Also note, that the iron will be cooled down, while the water will be heated. (I am using a different approach than you.
\begin{align}
Q_\mathrm{loss} &= m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x]\\
Q_\mathrm{gain} &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\
\end{align}
The transferred heat has to equal $$Q_\mathrm{gain} = Q_\mathrm{loss}$$
With this you can solve for $x$.
\begin{align}
m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x] &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\
%m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe})-m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x-m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})\\
% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x+m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x\\
% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= [m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})]\cdot{}x\\
x &=\frac{m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})}{ m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})}\\
%
x &=\frac{30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}\cdot{}417~\mathrm{K} +40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}\cdot{}293~\mathrm{K}}{ 40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}+30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}}\\
x &= \frac{5616.99~\mathrm{J}+49036.48~\mathrm{J}}{167.36~\mathrm{\frac{J}{K}}+13.47~\mathrm{\frac{J}{K}}}\\
x &= \frac{54653.47}{180.83}~\mathrm{K} =302.24~\mathrm{K}\\
x &\approx 29~\mathrm{^\circ{}C}
\end{align}


text:
When was oil discovered to be of biological origin?
Is there an article, book, some account of the discovery that crude oil came from decayed biological material?

If you look at a gas or liquid chromatogram of crude oil, you see a remarkable preponderance of (unbranched) even-numbered carbon compounds (re acetyl CoA homologation).  You also see traces of polycyclics that are net chiral - terpenoids, terpenes, pristane, phytane, cholestane, chlorins, and porphyrins.  Fischer-Tropsch gives a very different distribution and no net chirality.
Mikhailo V. Lomonosov, "Slovo o reshdenii metallov ot tryaseniya zemli"
Proceedings of the Imperial Academy of Sciences, St. Petersburg, 1757.
(Mikhail Tsvet was born in 1872)


text:
Can energy savings be estimated from activation energy differences?
Imagine two companies X and Y. Both companies use very similar technical equipment to carry out a biotechnological process where a chemical reaction is catalyzed by an enzyme. Company X uses an enzyme with a reaction barrier of, say, 15 kcal/mol, while company Y uses a slightly different enzyme to catalyze the same reaction but this enzyme has an activation energy of only 12 kcal/mol.
The reaction energies in the two reactors are, obviously, the same since both companies do $\ce{A} \rightarrow \ce{B}$. Still, as I see it, company X has to input 3 kcal per mole of product more into activating the enzyme.
Does company Y therefore require 3 kcal less energy to produce 1 mole of product?

The statement of this questions mixes a number of issues, so there is, in my opinion, no definite answer to this question.
First of all, one must take care that activation energies will have an influence on the kinetics (rate constant) of the reaction that is catalyzed. In a very general point of view, the overall energetic requirement for a chemical reaction does not directly depend on the activation energy (for instance the equilibrium constant is not changed).
Thus if two separate processes are considered, a different activation energy will mostly mean a reaction rate that will be slower, so overall a process that might take longer but not necessarily need a higher energy input.
Now if one would add to this problem a constraint that would be that the reaction rates of bith processes should be similar, the main method to achieve such a result would be to change the temperature at which the process takes place. This is related to the activation energy through the Arrhenius equation:
$$k = A e^{-\frac{E_{a}}{RT}}$$
If one assumes that the pre-exponential factor is the same for both processes (in effect meaning that the transition states in the catalyzed reaction have similar constraints) then one can reach some conclusions on both processes.
For instance, the reaction rates at constant temperature can be calculated. At 298 K, 3 kcal/mol difference would mean a ratio $k_{1}/k_{2}$ of about 150, so the process with lower activation energy would proceed 150 times faster than the other one. An other option would then be to increase the temperature of the slower process. In such a case, the temperature would depend on the ratio of the activation energies (and not the difference) as :
$$T_{2} = \frac{E_{a2}}{E_{a1}} T_{1}$$
If one takes the values of the qustion and considers a process that would take place at 298 K, one would need to heat at 373 K to reach the same rate constant. From this, one could consider the additional energy input required to heat the reaction medium up to this temperature.


text:
Why do the melting and boiling points of the noble gases increase when the atomic number increases?
What causes the melting and boiling points of noble gases to rise when the atomic number increases? What role do the valence electrons play in this?

Becuase London forces increase with an increase in mass and surface area.

The melting and boiling points of noble gases are very low in comparison to those of other substances of comparable atomic and molecular masses. This indicates that only weak van der Waals forces or weak London dispersion forces are present between the atoms of the noble gases in the liquid or the solid state.
The van der Waals force increases with the increase in the size of the atom, and therefore, in general, the boiling and melting points increase from $\ce{He}$ to $\ce{Rn}$.
Helium boils at $-269\ \mathrm{^\circ C}$. Argon has larger mass than helium and have larger dispersion forces. Because of larger size the outer electrons are less tightly held in the larger atoms so that instantaneous dipoles are more easily induced resulting in greater interaction between argon atoms. Therefore, its boiling point ($-186\ \mathrm{^\circ C}$) is more than that of $\ce{He}$.
Similarly, because of increased dispersion forces, the boiling and melting points of monoatomic noble gases increase from helium to radon.

For more data of melting and boiling points of noble gas compounds, read this page

Other answers have mentioned that dispersion forces are the key to answering the question but not how they increase from helium to radon (or let’s take xenon because that’s not radioactive so I feel safer breathing it).
The larger the mass of a nucleus the more protons are in there, and the more protons in a nucleus the more electrons are around the outside. Traditionally, one thought of electrons orbiting the nucleus rather like satellites orbiting a planet on more or less fixed orbits at certain heights. But that picture is wrong. It is much better to consider electrons as waves that completely surround the nucleus. If one were to translate these waves into particles (because of the wave-particle dualism at quantum levels), all one would get would be probabilities of finding specific electron $e$ at specific location $x$.
For a neutral atom that is surrounded by nothing, these probabilities depend only on the wave function, and are inherently centrosymmetric or anticentrosymmetric, leading to a net charge distribution of zero. In a sample of xenon gas, however, other atoms approach the atom we are observing. Consider an atom approaching our xenon atom at from ‘above’ (i.e. perpendicular to our viewpoint). While the atom as a whole is neutral, the could of electrons surrounding the nucleus is negatively charged. This new negative charge changes the potential energy the electrons in our observed atom are perceiving: While we originally had a centrosymmetric potential distribution (decreasing positive charge intensity from the nucleus) we now have a second negative source at 12 o’clock. Therefore, the probability distributions will shift ever so slightly and it will be slightly more likely to detect our electron $e$ at position $y$ below the nucleus rather than at position $z$ above the nucleus. Note that this simplification relies on the freezing of time at a certain moment when the other atom is approaching and a certain controlled environment that I have invented for example purposes.
With the electrons now more likely to be at $y$ rather than $z$, we can say that we have created a spontaneous or an induced dipole.  A mild positive charge is now pointing towards the other atom and ever so slightly attracting it. This will, if you advance the time flow by one spec, create another induced dipole in the originally approaching atom plus further in every other atom that is close around. We cannot freeze this picture though. Every infinitesimal change in position, movement direction or rotation will change the entire picture, meaning that our induced dipole is extremely short-lived.
It is only the combination of all these induced dipoles and their slight attraction that attracks the different atoms together. Since they rely on electron distribution changing, the more electrons we have the stronger these induced dipoles can be and the more force can be excersized between atoms. Since the number of electrons loosely correlates with mass (and strictly correlates with nuclear charge), larger atoms are said to display stronger van der Waals forces than smaller ones.

Valence electrons have not much to do with this, as their outer shell is closed.
As the other answer mentioned, dispersion forces are the ones responsible for any interaction between these atoms.
The size dependence therefore is directly coming from the size dependence of dispersion forces:
In a very simplistic way, a random charge fluctuation can polarize the otherwise perfectly apolar atoms. This induced dipole moment is then responsible for the dispersion interactions.  The polarizibility of an atom increases (easier to polarize) if the atomic number increases, therefore the interactions in nobel gases will reflect this behavior.

As mentioned in other answers the dispersion force is responsible for noble gases forming liquids. The calculation of the boiling points is now outlined after some general comments about the dispersion force.
The dispersion force (also called London, charge-fluctuation, induced-dipole-induced-dipole force) is universal, just like gravity, as it acts between all atoms and molecules.  The dipole forces can be long-range, >10 nm down to approx 0.2 nm depending on circumstances, and can be attractive or repulsive.
Although the dispersion force is quantum mechanical in origin it can be understood as follows: for a non-polar atom such as argon the time average dipole is zero, yet at any instance there is a finite dipole given by the instantaneous positions of the electrons relative to the nucleus. This instantaneous dipole generates an electric field that can polarise another nearly atom and so induce a dipole in it. The resulting interaction between these two dipoles gives rise to an instantaneous attractive force between the two atoms, whose time average in not zero.
The dispersion energy was derived by London in 1930 using quantum mechanical perturbation theory. The result is
$$U(r)=-\frac{3}{2}\frac{\alpha_0^2I}{(4\pi\epsilon _0)^2r^6}=-\frac{C_{\mathrm{disp}}}{r^6}$$
where $\alpha_0$ is the electronic polarisability, $I$ the first ionisation energy, $\epsilon_0$ the permittivity of free space and $r$ the separation of the atoms. The electronic polarisability $\alpha_0$ arises from the displacement of an atom's electrons relative to the nucleus and it is the constant of proportionality between the induced dipole and the electric field $E$, viz., $\mu_{\mathrm{ind}} = \alpha_0 E$. The polarisability has units of $\pu{J-1 C2 m2}$, which means that in SI units $\alpha_0/(4\pi\epsilon_0)$ has units of $\pu{m3}$ and this polarisability is in effect a measure of electronic volume, or put another way $\alpha_0 = 4\pi\epsilon_0r_0^3$ where experimentally it is found that $r_0$ is approximately the atomic radii. The ionisation energy $I$ arises because to estimate $r_0$ a simple model of an atom is used to calculate the orbital energy and hence radius and in doing so the energy is equated to the ionisation energy since this can be measured.
As can be seen from the formula the energy depends on the product of the square of the polarisability, i.e. volume of molecule or atom and its ionisation energy, and also on the reciprocal of the sixth power of the separation of the molecules/atoms. In a liquid of noble gases this separation may be taken to be the atomic radius, $r_0$. Thus the dependence is much more complex than just size, see table of values below. The increase in polarisability as the atomic number increases, is offset somewhat by the reduction in ionisation energy and increase in atomic radius.
If experimental values are put into the London equation then the attractive energy can be calculated. In addition the boiling point can be estimated by equating the London energy with the average thermal energy as $U(r_0)=3k_\mathrm{B}T/2$ where $k_\mathrm B$ is the Boltzmann constant and $T$ the temperature. The relevant parameters are given in the table below, with values in parentheses being experimental values:[1]
$$\begin{array}{c|c|c|c|c|c}
\text{Noble gas} & (\alpha_0/4\pi\epsilon_0)~/~\pu{10^{-30}m^3} & I~/~\pu{eV} & r_0~/~\pu{nm} & C_\mathrm{disp}~/~\pu{10^{-70} J m6} & T_\mathrm{b}~/~\pu{K} \\
\hline
\ce{Ne} & 0.39 & 21.6 & 0.308 & 3.9~(3.8) & 22~(27) \\
\ce{Ar} & 1.63 & 15.8 & 0.376 & 50~(45) & 85~(87) \\
\ce{Xe} & 4.01 & 12.1 & 0.432 & 233~(225) & 173~(165)
\end{array}$$
The fit to data is very good, possibly this is fortuitous, but these are spherical atoms showing only dispersion forces and a good correlation to experiment is expected. However, there short range repulsive forces that are ignored as well as higher order attractive forces. Nevertheless it does demonstrate that dispersion forces can account for the trend in boiling quite successfully.

Israelachvili, J. N. Intermolecular and Surface Forces, 3rd ed.; Academic Press: Burlington, MA, 2011; p 110.