text:
Why isn't 4-hydroxybenzoic acid formed along with salicylic acid in Kolbe's reaction?
I've read that in Kolbe's reaction of phenol, i.e. reaction of phenol $(\ce {C6H5OH})$ with $\ce{NaOH}$ forming sodium phenoxide $(\ce {C6H5O^{-}Na^{+}})$, followed by reaction with $\ce{CO_2}$ and acidification; salicylic acid (2-hydroxybenzoic acid ) is formed, but when $\ce{KOH}$ is used instead of $\ce{NaOH}$, 4-hydroxybenzoic acid is formed.
I don't understand why two different products are formed in the same reaction using very similar bases. Moreover, there is electrophilic aromatic substitution on the phenoxide ion in both reactions and  $\ce{O^{-}}$ being an ortho-para directing group must form both products in both the reactions.

At temperatures above 410 K, 4-hydroxybenzoic acid is also formed if you use NaOH.

Why the temperature difference? In fact there should be steric repulsion between bulky COO- and O- hence the other product should be easy to form right?

Intramolecular hydrogen bonding maybe?!

I don't understand why two different products are formed in the same reaction using very similar bases.

If the bases make a a difference, there must be a difference between them.
If the anion is the same, the cation must make the difference.

The difference between $\ce{Na+}$ and $\ce{K+}$ lies in the ionic radii of 0.95 and 1.33 $\mathring{A}$, respectively.
It is believed that $\ce{Na+}$ serves as an anchor and guides $\ce{CO2}$ into the ortho position during the formation of the Wheland intermediate. In the latter, $\ce{Na+}$ is chelated between the oxygen atoms of carboxylate and the phenolate. Apparently, this is very effective.
This does not mean that the phenolate doesn't attack $\ce{CO2}$ via C-4 at all. However, the formation of a Wheland intermediate (= the rate-determining step) is normally (and definitely in the case of the para attack) reversible.
It is open to question if full reversibility can be claimed for the $\ce{Na+}$ directed ortho attack under modest reaction conditions.
It is however likely that the back reaction (= cleavage Wheland intermediate) will stronger come into play at elevated temperatures. Consequently, some 4-hydroxybenzoic acid will be formed then.
With the bigger $\ce{K+}$, the anchoring and chelating effect is seemingly not possible, $\ce{K+}$ doesn't fit and the para direction effect of the phenolate prevails.

That's a nice explanation.can you explain why 4-hydroxybenzoic acid isn't formed in the NaOH reaction?It should be easy, given that steric factors don't come into the way,right?  I assume that steric factors also aid in difficulty of formation of salicylic aid in the reaction with KOH

@scienceauror I added some lines to explain the possible reasons.

Okay so that means in each reaction the other product can form in difficult-to-reach conditions like high temperature, and are easily reversible; due to the reasons you specified.

In reactions, in which the first step is reversible and the following one is not, the product formed first (fastest) is not necessarily the most stable one. At low temperature, the faster one makes the race. At higher temp, the system equilibrates and the more stable product is eventually formed.

@Klaus-DieterWarzecha But using the same analogy as you told, it suggests that in reimer tiemann reaction, the para salicylaldehyde shall be major product if KOH is used. However on wikipedia, ortho product is formed with NaOH as well as KOH in reimer tiemann reaction. Can you please explain why this don't work in reimer tiemann reaction?

The question: Why isn't 4-hydroxybenzoic acid formed along with salicylic acid in Kolbe's reaction?

There was a good answer provided to OP's question elsewhere using the cation sizes. However, it didn't provide the clear answer to the main question about why isn't 4-hydroxybenzoic acid formed when $\ce{NaOH}$ is used as a base. And there is also another question within the text: Why isn't salicylic acid formed  along with 4-hydroxybenzoic acid when $\ce{KOH}$ is used as a base. Yet, OP has not provided any reference to these claims. Therefore, we can conclude that there are no scientific evidence to support them.
The Kolbe-Schmitt reaction is a carboxylation reaction of alkali metal phenoxides with carbon dioxide where hydroxybenzoic acids are formed. This reaction has been used for more than a century for industrial production of aromatic hydroxy acids such as salicylic acid (2-hydroxybenzoic acid). The accepted general mechanism of the formation of salicylic acid is presented in the bottom of following image:

The Kolbe-Schmitt reaction is a classical example of a reaction in which the nature of the reaction products is dependent on the alkali metal cation (Ref.1). For example, Marković et al. have investigated the mechanism of the Kolbe-Schmitt reaction (using phenolates of lithium, potassium, rubidium, and cesium phenoxides for the carboxylation) by means of computational analysis. It is shown that the reactions of all alkali metal phenoxides with carbon dioxide occur via very similar reaction mechanisms and the reactions can proceed in the ortho- and para-positions. The exception is lithium phenoxide, which yields only salicylic acid in the Kolbe−Schmitt reaction. It is found that the yield of the para substituted product increases with increasing the ionic radius of the alkali metal used.
Experimentally, under comparable conditions, lithium and sodium phenoxides have yielded salicylic acid as the major product (Ref.2). In contrast, potassium, rubidium, and cesium phenoxides have yielded the mixtures of salicylic acid and 4-hydroxybenzoic acid in different ratios (Ref.2 & 3).It is worth mentioning here that before the work in Ref.3 (1954), it was generally accepted that carbonation of sodium phenoxide leads to the formation of salicylic acid, while potassium phenoxide, under similar conditions, converts primarily into 4-hydroxybenzoic acid. However, Ref.3 has shown that this is not always true.
Because of these new insights the mechanism of the Kolbe−Schmitt reaction has been the focal point of both chemical and computational studies. A mechanism involving an intermediate chelate complex between carbon dioxide and metal phenoxide or 2-naphthoxide ($\ce{MOPh–CO2}$ and $\ce{MONaph–CO2}$ where $\ce{M}$ stands for a metal) is a postulation of early researchers. The presence of the intermediate $\ce{PhONa–CO2}$ complex was confirmed on the basis of the IR absorption spectra (Ref.4). Based on these results and others  indicated that direct carboxylation is not an operative mechanism of the Kolbe–Schmitt reaction.
Mean time, a standpoint of initial formation of the $\ce{MOPh–CO2}$ or $\ce{MONaph–CO2}$  complex has been supported by a number of theoretical works (Ref.5). It was shown that the $\ce{CO2}$ moiety of the $\ce{MOPh–CO2}$ complex performs an electrophilic attack at the benzene ring in the ortho- and para-positions, leading to the new intermediates and final products (similar to pathway 1 in the image; e.g., Ref.5&6):

This mechanism clearly shows that the ortho-product is possible whenever $\ce{KOH}$ is used as a base. When 2-naphthol is the substrate, Ref.4 listed following results (Reference reaction is at the top of ist image):
$$
\begin{array}{c|ccc}
\hline
\ce{MOH}^a & \text{Rxn Temp.} & \text{Rxn Time} & \text{% yield ($\bf{1}$}) & \text{% yield ($\bf{2}$}) & \text{% yield ($\bf{3}$}) & \text{% SM recovery} \\
\hline
\ce{KOH} & \pu{293 K} & \pu{1 h} & 52.0 & 0.0 & 0.0 & 47.0\\
\ce{KOH} & \pu{503 K} & \pu{10 h} & 4.0 & 72.0 & 0.0 & 24.0\\
\ce{KOH} & \pu{523 K} & \pu{10 h} & 2.0 & 87.0 & 0.0 & 10.0\\
\ce{NaOH} & \pu{503 K} & \pu{10 h} & 14.5 & 71.5 & 2.5 & 11.0\\
\ce{NaOH} & \pu{523 K} & \pu{10 h} & \text{Trace} & 84.5 & 2.0 & 13.0\\
\hline
\end{array}\\
^a: \text{Carboxylation of pottasium and sodium 2-naphthoxides with $\ce{CO2}$ at $\pu{5 MPa}$ pressure.}
$$
The above Table shows there is not clear preference for pottasium and sodium 2-naphthoxides. The recent green approach shows adjusting conditions can make different isomers (Ref.7):

References:

Zoran Marković, Svetlana Marković, Nebojša Begović, "Influence of Alkali Metal Cations upon the Kolbe−Schmitt Reaction Mechanism," J. Chem. Inf. Model. 2006, 46(5), 1957–1964 (https://doi.org/10.1021/ci0600556).
Alan S. Lindsey, Harold Jeskey, "The Kolbe-Schmitt Reaction," Chem. Rev. 1957, 57(4), 583–620 (https://doi.org/10.1021/cr50016a001).
Ogden Baine, G. F. Adamson, J. W. Barton, J. L. Fitch, D. R. Swayampati, Harold Jeskey, "A study of the Kolbe-Schmitt reaction. II. The carbonation of phenols," J. Org. Chem. 1954, 19(4), 510–514 (https://doi.org/10.1021/jo01369a006).
J. L. Hales, J. Idris Jones, A. S. Lindsey, "Mechanism of the Kolbe–Schmitt reaction. Part I. Infra-red studies," J. Chem. Soc. 1954, 3145-3151 (https://doi.org/10.1039/JR9540003145).
Svetlana Marković, Igor Đurović, Zoran Marković, "Revisiting the Kolbe–Schmitt reaction of sodium 2‑naphthoxide," Theoretical Chemistry Accounts 2015, 134, Article # 45 (9 pages) (DOI 10.1007/s00214-015-1648-0).
Zoran Marković, Svetlana Marković, Nedeljko Manojlović, Jasmina Predojević-Simović, "Mechanism of the Kolbe–Schmitt reaction. Structure of the intermediate potassium phenoxide-$\ce{CO2}$ complex," J. Chem. Inf. Model. 2007, 47(4), 1520–1525 (https://doi.org/10.1021/ci700068b).
Khakim A. Suerbaev, Mayliby K. Aldabergenov, Nurbolat Zh. Kudaibergenov, "Carboxylation of hydroxyarens with metal alkyl carbonates," Green Processing and Synthesis 2015, 4(2), 91-96 (https://doi.org/10.1515/gps-2014-0098).

Decarboxylation of 4-hydroxy-3,5-diisopropyl benzoic acid with NaOH/ethylcellosolve at 130 deg. C gives 2,6-diisopropyl phenol, propofol. [https://pubs.acs.org/doi/10.1021/op400300t]. Table I, pg. 588, in ref. 2 (vide supra) gives data on the effect of temperature of the consumption of p-hydroxybenzoic acid with potassium counter ion in the K-S-M reaction.


text:
What is a complete reaction?
For a lab assignment, I combined $\pu{10 mL}$ of $\pu{1 M}$ sodium sulfate solution with $\pu{10 mL}$ of $\pu{1 M}$ calcium chloride solution. The ionic equation I came up with is:
$$\begin{multline}
\ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + 2Cl^- (aq) ->}\\
    \ce{CaSO4(s) + 2Na^+ (aq) + 2Cl^- (aq)}
\end{multline}$$
I was then asked, in a lab question, the following:

If the reaction called for $\pu{1 M}$ chloride ion with $\pu{1 M}$ calcium ion, would you still use $\pu{10 mL}$ of each solution for a complete reaction? Explain your reasoning.

To my understanding, I initially had $\pu{2 M}$ chloride ion and $\pu{1 M}$ calcium ion. With only $\pu{1 M}$ chloride ion instead, the ionic equation would change to this, or so I'm led to believe:
$$\begin{multline}
\ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + Cl^- (aq) ->}\\
 \ce{CaSO4(s) + 2Na^+ (aq) + Cl^- (aq)}
\end{multline}$$
As a (balanced) molecular equation, to my understanding, this would be:
$$\begin{multline}
\ce{Na2SO4 (aq) + Ca (aq) + Cl (aq) ->}\\
\ce{CaSO4(s) + NaCl (aq) + Na (aq)}
\end{multline}$$
Is this a complete reaction? I have seen conflicting definitions of a complete reaction online:

answers.com says that a complete reaction means all of the reactants are reacted into products.
However, someone on chemicalforums.com says that a complete reaction means that all of the limiting reactant is used up. The University School of Milwaukee seems to back up what the user on chemicalforums.com said ("all of at least one of the available reactants is used up and converted into products").

My main question is thus: what is a complete reaction? No matter my final equation for this lab question, I still need to know what a complete reaction is so that I can identify them in general.
My intuition says that this is a complete reaction, since all of the calcium ions and the chloride ions are converted to products. Is this correct? The leftover sodium ion makes me suspicious, so I don't know how to answer.
I later spoke to my lab instructor and he clarified that the question is asking for me to come up with two arbitrary aqueous solutions, one which contains $\pu{1M}$ of chloride ion and another which contains $\pu{1M}$ of calcium ion, e.g. $\ce{KCl(aq)}$ and $\ce{Ca(NO3)2 (aq)}$.
So in this case, I would need twice the volume of potassium chloride to get the number of chloride ions I need to balance the equation and prevent explosions, i.e. no extra sodium ions leftover.

This question has an archived discussion in chat.

I suspect that you would have both these molecular equations:
\begin{align}
\ce{2KCl(aq) + Na2SO4(aq) &-> K2SO4(aq) + 2NaCl(aq)}\\
\ce{Ca(NO3)2(aq) + Na2SO4(aq) &-> 2NaNO3(aq) + CaSO4(aq)}
\end{align}
To answer your question with "What is a complete reaction?":

A complete reaction is one where all of at least one of the available reactants is used up and converted into products.

So for example $\ce{KCl}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{2 M}$ $\ce{KCl}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. However, you only have $\pu{1 M}$ $\ce{KCl}$ (so $\pu{1 M}$ chloride ions). This means that, as you mentioned, you will need the double concentration for the $\ce{KCl}$ solution to achieve a complete reaction.
So in this case the only way for all $\ce{Na2SO4}$ molecules to react is to add the double amount of $\ce{KCl}$ solution. Then this is a complete reaction : at least one reactant is used up in the reaction.
But say we have a situation in which we have a $\pu{2 M}$ $\ce{KCl}$ solution, but a $\pu{2 M}$ $\ce{Na2SO4}$ solution, then all the $\ce{KCl}$ would react, but still \pu{1 M} of $\ce{Na2SO4}$ will not have reacted.If in your lab assignment it is the meaning that you react all $\ce{Na2SO4}$, then in this case you don't have a complete reaction. Only when a two times larger volume of the $\ce{KCl}$ is added (to 2M $\ce{Na2SO4}$), then you will once again achieve a complete reaction. Furthermore you would just get a solution of solvated ions.
In the case of $\ce{Ca(NO3)2}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{1 M}$ $\ce{Ca(NO3)2}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. You have $\pu{10 ml}$ of both compounds, each $\pu{1 M}$ in concentration: so in this case there is no need to use a larger volume of aqueous $\ce{Ca(NO3)2}$.

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text:
How many Ga atoms are connected to one As atom in solid gallium arsenide?
In Boylestad and Nashelsky's Electronic Devices and Circuit Theory, the structure of GaAs is shown as follows:

Arsenic is shown making five covalent bonds to neighbouring gallium atoms.
However, I recently also saw an alternative depiction in Razeghi's Fundamentals of Solid State Engineering:

Here, arsenic is shown making four covalent bonds to neighbouring gallium.
Which source is correct? Please give appropriate references or experimental proof.

It is quite well established that stoichiometric gallium arsenide (CAS 1303-00-0) forms a zincblende analogous structure. It consists of two stacked face-centered cubic lattice systems (Ioffe Physical Technical Institute database). See also Uncle Al's Answer. In other words, the arsenic forms the lattice and the gallium fills half of the tetrahedral positions (and vice versa). Therefore every gallium has four arsenic nearest neighbours.
This does not necessarily mean that these are bonds. In this case however, the bonds are best described as covalent (having a somewhat ionic character).
The Britannica Online Encyclopedia (Academic Edition) (You might need access provided through an educational institution) states:

Besides the elemental semiconductors, such as silicon and germanium, some binary crystals are covalently bonded. Gallium has three electrons in the outer shell, while arsenic lacks three. Gallium arsenide (GaAs) could be formed as an insulator by transferring three electrons from gallium to arsenic; however, this does not occur. Instead, the bonding is more covalent, and gallium arsenide is a covalent semiconductor. The outer shells of the gallium atoms contribute three electrons, and those of the arsenic atoms contribute five, providing the eight electrons needed for four covalent bonds. The centres of the bonds are not at the midpoint between the ions but are shifted slightly toward the arsenic.
Such bonding is typical of the III–V semiconductors — i.e., those consisting of one element from the third column of the periodic table and one from the fifth column. Elements from the third column (boron, aluminum, gallium, and indium) contribute three electrons, while the fifth-column elements (nitrogen, phosphorus, arsenic, and antimony) contribute five electrons. All III–V semiconductors are covalently bonded and typically have the zincblende structure with four neighbours per atom. Most common semiconductors favour this arrangement.

This is still a fluorishing field for researchers in the field of materials design semiconductors, semiconductor lasers, computational physics/forcefield development, and more.
The book "Electronic Devices and Circuit Theory" clearly gets it very wrong. I had a look into it and noticed they also argue with the Bohr model, which might not be a very good starting point at all. However, the focus of the book is clearly not solid state chemistry, and therefore it might fit the purpose.

The two-dimensional representations you have given are inadequate and should not be taken too seriously; crystal structures are nearly always three-dimensional. Gallium arsenide has the cubic zincblende structure, shown below. Here, blue atoms can are Ga and orange atoms As (it can be the other way round, too, as these atoms are related by symmetry).

(Image source)
(As a side note, this structure is similar to that of diamond: in diamond, both types of atoms are carbon.)
Both gallium and arsenic have coordination number 4; the coordination geometry is tetrahedral about both.
In fact, 5-fold symmetry is very rare in crystals (see quasicrystals for more information).

This diagram appears to imply 4 orange atoms per unit cell and 6(1/2)+8*(1/8)=4 blue atoms per unit cell, making the GaAs ratio 1:1. That's correct, right? And so by a translation of origin Ga and As positions can be interchanged?

The somewhat standard reference (~1800 citations in Google Scholar) for bulk GaAs structure is a 1982 paper by J. S. Blakemore: J. Appl. Phys. 53, R123 (1982). The experimental/numerical data in it (via X-ray diffraction) establishing the crystal structure of bulk GaAs goes back to the 1960s though. One of the more helpful visualizations of zincblene structure (which bulk GaAs shares) can be found on YouTube.
Note however that nanowire GaAs can have wurtzite structure as well; here is another YouTube animation. Nanowires weren't practical in the 1980s, so this is more recent research; see for example S. Conesa Boj's Masters thesis at the University of Barcelona, "HRTEM assessment of Wurtzite and Zinc-Blende phases in GaAs nanowires for optoelectronic devices". (HRTEM is explained well enough on Wikipedia, in case you wonder how that works). In any case, the coordination numbers are still 4 for both Ga and As in this structure.
In both cases (crystal structures) the ratio of the Ga:As atoms in the unit cell is the same 1:1. I've actually checked the 11th (2012) edition of Boylestad & Nashelsky and they have the same old explanation. My guess is that they don't want to explain polar bonds to first-year EE undegrads, so they engage in some "lies to children" by sticking with the simplified covalent bond model.
The book by Razeghi (which gives the right ratio and explanation for bulk GaAs) does call them "polar covalent bonds"/"mixed bonds" (pp. 65-66), but from its preface this book seems aimed at senior-level undergraduates. Razeghi does mention prior to that in his book (p. 30) that GaAs has the zincblende structure, and goes into some detail on that. The flattened 2D structure on p. 66 is a bit strange in view of that prior level of detail; I'm guessing he wanted to emphasize the bond polarity (on p. 66) and he thought 2D was a simpler way to show it.

If you extend the drawing in figure 1, you will discover that there are 5 Ga ions for 3 As ions. This would be suitable for a alloy. Since you are probably interested in a semiconductor, the arsenic needs to have a formal charge of -3. This means that the ratio of Ga to As will be 1:1.

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text:
How to determine polarity in large molecules?
In case of small molecules like water things are more or less clear, but when I see a large complex molecule like Thyroid Hormone - how should I find out that this molecule is non-polar?

I've been taught about the "1:4"-rule; 1 non-polar atom "makes up" for 4 polar atoms. Water, consisting of only hydrogen and oxygen (where the difference in electronegativity is large) is rather polar. Octane, in the other hand, consists of 8 carbon and 18 hydrogen, making the ratio 4:9, hence quite non-polar. Obviously, this is just a rough estimate, although it's rather useful for a simple TLC-analysis for example in order to determine the right eluent.

For what reason you want this estimate? If it is water solubility, it is reasonably simple. If for other reason, you can try to roughly estimate the dipole moment of whole molecule (thinking little bit of 3D structure).

You can be very certain that the larger a molecule is, the more likely it will be somewhat polar, esp. if it contains (so called) hetero atoms.

Once you move up to "larger" molecules like this, the definition of a "polar" molecule becomes a bit fuzzy. Rather, you will find that molecules contain polar and non-polar groups which all contribute to the overall characteristics of the molecule. It really depends on the context that you are interested in.
If you are curious about more macroscale properties, such as boiling point or solubility, a variety of group contribution methods have been developed to provide ballpark estimates based purely on the contributions of individual components of the molecule.
While group contribution methods will primarily give you an idea of bulk properties, you may also be interested in lower level behavior such as micellization or specific protein binding which will depend on the location and arrangements of all of the polar and non-polar groups on a molecular level.

A practical tutorial with Avogadro

It is not so easy to understand from the structure because polarity in fact depends on the arrangement of the groups in tridimensional space relative to one another and not simply on the groups present (e.g. $\ce{trans-C2H2Cl2}$ is not polar because the individual bond dipoles cancel one another, $\ce{cis-C2H2Cl2}$ is polar) so you have to use a geometry optimization tool first. You can perform this optimization and make some assumptions from the model calculating the electrostatic potential with Avogadro. I've made this short tutorial.

Draw you molecule.
Check it (doing a lot of calculations with the wrong molecule is not fun...)
Start optimize geometry tool, wait until dE=0 or near 0
Stop optimize geometry tool
Go to Visualize - Properties - Molecule properties read the polar momentum in Dalton
If you want, you can create the potential surface from extension - surface
Eventually you can try to search for other conformers and repeat the process

How accurate is this method?
From my attempts unfortunately not very. If you try with $\ce{C2H2Cl2}$ you get about 1.086 D more for the cis conformer and 0.22 Dalton, more sometimes depending on the optimization, for the trans conformer after the research of other conformers.

This is a really excellent way to get a general idea of polarity in simple molecules. But as you pointed out, getting dipoles correct in molecular modeling is a tremendously challenging problem and an area of active research.

@CTKlein you are right, I would like to compare it with other software and methods! I am very grateful  for the edit! Thanks a lot!

Avogadro is a nice first estimate, but results are questionable. Anyway, the dipole moment is 'only' a number and might not say anything about polarity. For example $\ce{CO2}$ has no (permanent) dipole, but is very polar.

@Martin I was thinking about it yesterday but I've read "A polar molecule has a net dipole as a result of the opposing charges" reading wikipedia article.

@GM Unfortunately polarity, or in this case nonpolarity is not very well defined (goldbook). It usually is based on a relative scale as every molecule is somewhat polar (even $\ce{H2}$). When dealing with surfactants, one usually describes part of the molecule as nonpolar (tail, hydrophobic) and polar (head, hydrophilic).

@Martin wow gold book still works, was about a mouth that I could not access to his ambiguous definitions..however i think you are right, I hope the tutorial is still useful for understand more about this topic... Thanks for the comments!

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text:
How much water has boiled off when precipitation begins?
Here's a challenge problem I created! None of my students could get it; can you?
$25$ grams of table salt (NaCl) is dissolved in $150$ grams of water. The temperature of the solution is raised to its boiling point until sodium chloride begins to precipitate.
How much water has boiled off when precipitation begins? Also, what would be the temperature of the boiling solution when precipitation begins?

One expects 25 g of $\ce{NaCl}$ to saturate 62.5 g of water at 100 °C. However, 25 g of $\ce{NaCl}$, 58.443 g/mol, is 0.4278 mol in 62.5 ml of water at 100 °C. Colligative property boiling point elevation is +0.52 (°C kg)/mol. 0.8556 mol of particles in 62.5 ml of water is 13.69 mol/kg or 7.1 °C elevated.
$\ce{NaCl}$ solubility/temp increases from about 36 g/100 g at 0 °C to 40 g/100 g at 100 °C or 0.04 g/°C. This gives first approximation 0.2847 g more solubility. "25.28" g $\ce{NaCl}$ in 62.5 g water at 107 °C is then 25 g $\ce{NaCl}$ in 61.8 g water. 88.2 g water boiled off (with some decimal trim for Zeno’s paradox).
Repeat the problem with ammonium cyanate.


text:
Why do electrons follow from Zn to Cu but not Cu to Zn in the lemon battery experiment?
In the lemon battery experiment, as I understand, both $\ce{Zn}$ and $\ce{Cu}$ meet acid in the lemon, cause $\ce{Zn}$ and $\ce{Cu}$ release electrons to form $\ce{Zn^{2+}}$ and $\ce{Cu^{2+}}$.
So why do the electrons follow from $\ce{Zn}$ to $\ce{Cu}$ but not from $\ce{Cu}$ to $\ce{Zn}$?

In fact $\ce{Cu}$ doesn't release electrons, only $\ce{Zn}$ does. For the usual explanation see this Wikipedia article, I will try to explain it in other words to make it much easier.

One important thing that you should know is that energy comes from metal not from the acid.
Some species have the tendency to donate electrons other species the tendency to acquire them. Chemically speaking some species have more internal energy when their are in the reduced form (when they have the maximum number of electrons) others species reach their minimum internal energy when they are in the oxidized form (when they have the least number of electron). In this world every system tend to go to a state of lower energy spontaneously so if two species are not in their lowest energy form if they can exchange their electrons freely they acquire (are reduced) or give (are oxidized) electrons until they reach the lowest energy.  There are also species that do not have any particular tendency or are only little more prone to acquire electrons or to donate them. In all these cases the exchange can occur only if the electrons can move easily from one specie to another, for this reason a good electrolyte is many times needed.
An easy comparison
We could imagine chemical species as men and electrons as money.
Imagine that there are different kind of men at the antipodes there are avaricious men that needs more money and charitable men that feel the urge to give money. I think that you have no doubt which of them do what when one meet the other. The problem is that many times the transition occurs only if there is the possibility that the flow of money goes from who want to give them to who want them so you need a third agents to create the right environment and conditions for the transition. Furthermore the loss of money must be counterbalanced by something similar.
Then there are others type of men that simply don't want to give or take money we can call them noble men and then there are other men that can take money without actually feeling the need, they simply take them, I will call them hydron men.
Your case
Lemon batteries are in fact one of the more complex and debated battery you could examine. When you plug the zinc and the copper in your lemon we can say that we have three main active species: copper, zinc and acid (citric acid). Zinc is one of that metals we compared to a charitable man it tends to donate electrons:
$$\ce{Zn <=> Zn^{2+} + 2e^{-} }$$
Copper however is a noble metal, physics explain this because it has the d-bands of the electronic structure filled, so it doesn't accept zinc's electrons. Instead hydrogen from acid, can accept the electrons in this way:
$$\ce{2H+ + 2e^{-} -> H2 ^}$$
And this is in fact what actual experimental results show, see J. Chem. Educ. 2001, 78 (4), 516 for reference.

The short answer from the website Experiments in Electrochemistry (Carboni, 1998) is that

The Copper ($\ce{Cu}$) atoms attract electrons more than do the Zinc ($\ce{Zn}$) atoms.

According to the Naked Scientists website Lemon-Powered iPod - Fruit batteries, they state that

Zinc metal can dissolve to form zinc ions, releasing energy; it also loses electrons. If the zinc is connected to the copper by an electric circuit, these electrons can flow around the circuit and reduce copper ions in the lemon.

The reason why this occurs, is well described in the UC Davis Chemwiki Standard Reduction Potential: Activity Series, that as copper has a standard reduction potential of 0.34V, hence will be reduced (as described above), as zinc's standard reduction potential of -0.76V is considerably lower.
However, some copper ions dissolve into the lemon's citric acid (Naked Scientists).
The process is shown in the image below:

this continues until the electrons (according to the Carboni, 1998 article linked above):

concentrate on the copper, the electrons repel each other. When the force of repulsion between electrons and the force of attraction of electrons to the copper become equalized, the flow of electrons stops.

Some more explanation is available in the question and answer at Electronegativity of zinc vs copper in galvanic cell

I've down voted you because in my opinion is not right see this article

@GM When opinions get in the way of true science, then it is time to go from the place that allows such nonsense.

"Science is Truth through verifiable evidence" :-), I hope you did not bother for the down vote I've done it  in the name of science, people tend to agree without reasoning if there are more then two upvotes. Thanks for the motto!

@GM you downvoted based on an opinion, not on scientific information that are obviously from the links in the answer.

:-) My down vote is based on scientific evidences from the links in the answer, in fact there no evidences in the other sites that propose the $Cu^{++}$ reduction, there is only their opinion... Thanks for the patience, really I'm interested only in find the right answer otherwise I simply downvoted your answer without saying why. I've upvoted many of your answer so don't take it personal!

@GM there are several that do actually, just because you did not read the information properly.  However, you said in your first comment that it was your opinion...

I'm sorry it was not my intention to annoy you, as I told you my downvote is based on these articles and related experimental evidences I don't think I  can separate my opinion from these because  consider them reliable means in fact have an opinion. I hope the fact you delete your account is not related to this episode, when I was learning programming downvote for me were more useful then upvote if somebody explain the motivation. Best regards.

I'd be interested in any comments regarding the effect of temperature on current. Does heat simply reduce it?


text:
How much UV lamp power is needed to initiate radical polymerization?
I'm looking to purchase a UV lamp for initating some radical polymerizations. Is a hand lamp like the type for examining TLC plates (5-10W, like this) enough power to do this, or do I have to buy something a lot more expensive?
In my case, I'm looking to polymerize a fairly small sample (1mm thick, about 100 mm$^2$ of area, encased in 1mm-thick lab slides) and the chemistry will probably be MMA (initiated with AIBN). I'd like to know for my particular case, but it would be great to know if there's some rules of thumb involved (does it depend on the chemistry? does it depend on the amount? Will different power levels give different final results?)

A photopolymerization requires a photoinitiator to capture the energy.  Where does AIBN absorb versus your source?

OK, that looks good.  AIBN has a strong cage effect.  2,2′-Azobis(2-methylpropionitrile) extrudes nitrogen and the radicals combine without initiating polymerization.  The dimer, tetramethylsuccinonitrile or 2,3-dicyano-2,3-dimethylbutane, is quite toxic.  The superior initiatior is 1,1'-azobis(cyclohexanecarbonitrile), VAZO catalyst 88; J. Org. Chem. 52 2958 (1987) (e.g., Sigma-Aldrich Chemical Company).
Camphorquinone (yellow powder) can replace AIBN, shifting absorption into the visible, 380 - 510 nm.  That is a tungsten-halogen bulb or a middling blue LED, GaN or InGaN maxing at 450 - 460 nm.  The graph's yellow line is camphorquinone.

Molecular oxygen quenches free radical polymerizations.  If things stay gooey, add a little more initiator to the starting mix to burn out the oxygen.  Now, the fun part!  If you treat the clean glass slides with Rain-X, (spray, wait a few minutes, wipe, lightly buff; maybe lock in at 110 C for 15 minutes) you silanize the surface and polymer will not stick.  It's a miracle for treating vacuum polymerization ampoules.  If you want a cleaner surface and can play chemist,
http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx
Gelest  Aquaphobe CM, diluted by you to 8% in dichloromethane or toluene as a 5-minute soak in a closed glass container.  Soak, rinse with solvent, lightly buff, oven.  The active material gives off HCl, including exposure to humidity.

Sounds interesting! These hand lamps typically have two tubes with emission maxima around $\lambda$ = 254nm and 360 nm, respectively. The latter should be just right to excite AIBN and isn't cut off by Pyrex or comparable glas.
More power is obtained from a high pressure mercury lamp, such as the classical HPK 125W, but these lamps are normally used in cylindrical irradiation vessels with immersion cooling.
If you want a complete system with top- or side irradiation, replacable tubes, metal housing and ventilation have a look at the portfolio of companies like LuzChem.

It was recently pointed out to me that there's a way to do this in a principled manner. Let's assume the lamp is $4\ \mathrm W$ strong and distributes its power over 10 plates ($1000\ \mathrm{mm^2}$). The energy of a $254\ \mathrm{nm}$ photon is $7.8 \times 10^{-19}\ \mathrm J$. If the lamp is at $4\ \mathrm W$, it's putting out $4\ \mathrm{J/s}$ of energy. This corresponds to $$ \frac{4\ \mathrm J}{\mathrm s}\times\frac{\text{photon}}{7.8 \times 10^{-19}\ \mathrm J}  = 5.11 \times 10^{18}$$ photons per second.
Now, if the absorbance of the initiator is 0.0001, we have that the absorbed fraction of the photons is: $$ \% Abs = 1 - \%T = 1 -  \frac{I_1}{I_0}  = 1 - 10^{-A} = 2 \times 10^{-4} $$
Since we have $5.11 \times 10^{18}$ photons per second are incoming, roughly $1 \times 10^{15}$ of them will be absorbed, which should be more than enough to initiate polymerization.
Of course, this is a very, very rough calculation, but I thought it was cool that you could, in theory, calculate that this experiment should work (which it did!)


text:
What is the oxidizing agent for Fe in the following redox reaction?
Redox: $\ce{Fe + 2HCl  ->  FeCl2 + H2}$
Oxidation:  $\ce{Fe -> Fe^{2+} + 2e-}$
Reduction: $\ce{2H+ + 2e- -> H2}$
The $\ce{HCl}$ and $\ce{H2}$ are formed by covalent bonds. The $\ce{FeCl2}$ forms an ionic bond.
The iron actually loses 2 electrons to the two chlorines, so the oxidation equation makes sense to me. But the hydrogen is always in a covalent bond with the chlorine or another hydrogen atom, it is always sharing electrons, never losing or gaining, so the reduction equation doesn't make sense to me. When the $\ce{H}$ is bonded with the $\ce{Cl}$, it "has" two electrons because it is sharing a pair with the $\ce{Cl}$. After the reaction, when the $\ce{H}$ is bonded with another $\ce{H}$, it still "has" two electrons because it is sharing a pair with another $\ce{H}$, so no electron was gained as the reduction equation indicates.
After a little a thinking I came up with an answer, which raised even more questions:
Since the covalent bond between the $\ce{H}$ and the $\ce{Cl}$ is a polar bond and it behaves more like an ionic compound because of that, when $\ce{HCl}$ is dissolved in water, the $\ce{H}$ will lose an electron to the $\ce{Cl}$, making $\ce{H+}$ and $\ce{Cl-}$ ions, so there is no covalent bond when $\ce{HCl}$ is in water. The $\ce{H+}$ ions can gain one electron and then make a covalent bond with each other, making $\ce{H2}$ molecules.
But what about the $\ce{Cl-}$ ions in the solution? they have already formed an octet in their outer shell by getting an electron off the hydrogen when they got separated, so they are satisfied. Then why $\ce{Fe}$ atoms oxidize? who is the oxidizing agent? The $\ce{H2}$ molecules are already comfortable with their covalent bond, and the $\ce{Cl-}$ ions are also satisfied with their octet.
I'm very confused.

To answer your initial question:
The oxidizing agent is the reagent that gets reduced. Conversely: The reducing agent is the reagent that gets oxidized.
Now, addressing your trouble with the concept:
You are right that hydrochloric acid is found in the ionic form when in aqueous solution, so:
$$\ce{HCl ->[][\ce{H2O}] H+ + Cl-}$$
Now we have the reagent that gets reduced, which is the proton. Together with another proton and two electrons, they form hydrogen gas that then evolves from the solution:
$$\ce{2H+ + 2e- -> H2 ^}$$
This is basically our electron sink, because that's where they go "to die", i.e. to vanish from our solution.
But where do these electrons come from? Well, we have two potential electron donors here: The chloride anion and the iron. Which one will give up his electrons more easily?
The answer we already know: Iron.
Why? Because as you correctly stated, the chloride anion has all the shells filled and is very happy because of that filled octet state. So the energy to get out an electron from it is higher than to rip it away from an unsuspecting iron atom.1)
$$\ce{Fe^0 -> Fe+ + e- -> Fe^{2+} + 2e-}$$
The electrons then vanish (because of the formation of hydrogen, remember?) which leaves us with some iron cations and chloride anions. They will then start to mix with each other. Don't judge them, they don't know any better!
$$\ce{Fe^{2+} + 2Cl- <=> FeCl2}$$
Since ferric chloride (II) is quite soluble in water, it will most probably not precipitate.

Footnotes:

Actually, you could also compare the standard reduction potentials of the reactions $\ce{Fe -> Fe^{2+} + 2e-}$ and $\ce{Cl- -> Cl^{.} + e-}$. The one that has the lower standard reduction potential is the winner (because the energy of a system is always minimized). This is also what you would usually do, but in this example it wasn't necessary because you could get to the solution via chemical concepts.

Thank you!! It is much clearer now. I was forgetting about the fact that the Chlorine atom had already gained an electron from the Hydrogen when the Hydrocloric acid was dissolved in water, so I thought they were getting the extra electron from the Iron (to form the ionic bond in FeCl2), thus leaving the Hydrogen ions with no source to get the electrons from.


text:
Understanding Mesomeric Effect
I am getting confused between the type of mesomeric effect (+M or -M) shown by compounds containing $\ce{-COX}$ where $\ce{X}$ is an electronegative element.
Examples: $\ce{-CONHCH3}$, $\ce{-COOCH3}$, $\ce{-COOH}$.
According to me, all these compounds must show both a +M as well as a -M effect. Since the group (e.g. $\ce{-COX}$) can accept as well as donate $\pi$ electrons (as in $\ce{CH_2=CH-COOH}$).
However, the book says these compounds show only -M effect. Please explain me the reason behind it.
In the above example, isn't the group showing +M effect, since it's donating the electrons?

carbonyl group does not show +M effect because in it carbon is highly electron-deficient. In it oxygen cannot donate electron to pi-system from electron pair and donating electron from double bond will leave oxygen with electron sextet, which will not happen under any circumstances, period.

Following up on Permeakra's comment, the relative importance of resonance structures is typically determined by:
1) Maximize atoms that fulfill the octet rule.
2) Electronegative atoms have negative formal charges and avoid positive charges.
3) Minimize charge separation.
Carboxylic acid derivatives (esters, amides, etc.) have resonance structures that break #2, but since all atoms fulfill the octet rule, they are important contributors to the overall electronic picture of the compounds. Hopefully the image below is legible.

In general, resonance (mesomeric) structures that contain an electronegative atom that both breaks the octet rule AND bears a formal positive charge are not reasonable resonance structures and do not contribute to the overall electronic picture of the compound. The resonance structure of the carboxylic acid drawn in the original question falls into this category as the 'top' oxygen both breaks the octet rule and contains a positive charge.


text:
Why is the orbital angular momentum of a pi electron along the axis of two atoms' molecule one?
I'm reading quantum chemistry. The book says that the orbital angular momentum of a $\pi$ electron along the symmetry axis of a molecule made up of two atoms is $\pm 1$. I think this is a primary question, but I do not know why (I'm a student studying physics)

I currently have a preliminary understanding of this:

In a molecule inculding two atoms, potential energy axisymmetric about z' axis(the line connecting the two atoms). So, the angular momentum along z axis is quantized. That is to say $m_z$ is good quantum number. Let's consider $\pi$ orbit made up of two $p_z$. The orbital angular momentum along z axis(the symmetry axis of $p_z$) of an electron in $p_z$ is 0. So considering orbital angular momentum along z' axis of this electron, the electron is in $\frac 1 {\sqrt 2}(|+>-|->)$. So the orbital angular momentum along z' axis is either 1 or -1. So the orbital angular momentum along z' axis of a $\pi$ electron is either 1 or -1.
Is my understanding correct?

Could you elaborate on your question? Are you asking why the angular momentum l is plus or minus one? Or are you asking why the pi orbitals look like dumbells? And what do you mean by line connecting the atoms? A sigma bond? A pi bond? An exotic bond that was published recently? Are you adking why orbitals have to be oriented a certain way to form bonds? I suspect that you may be using the terms wrongly.

@user97554 I'm asking why the angular momentum l is plus or minus one. Thank you for your advice. I have improved my description. Is it more clear?

Are you familiar with Laguerre polynomials and spherical Bessel functions? Eigenvalues?

@user97554 Yes.

I am so sorry. But turns out I misunderstood my knowledge. I thought I knew it but in reality I didn't. I apologize. I think you may have better luck at Physics SE. Again I apologize.

@user97554 Do not care. I'm afraid that they are not familiar with pi electron, but I will try. Thank you for your advice!

Let the molecular axis be the $z$ axis, A and B the two atoms forming the bond. The $2p_z$ AO has atomic quantum number $m_l = 0$ and so has $L_z =0$. The MOs formed from
2pz AOs are therefore $\sigma$ MOs, bonding and antibonding.
The $p_x$ AO is a linear combination of $m =+ 1$ and $m = -1$ AOs and has $|m| =1$.
$2p_x \equiv \frac{(2p_1 + 2p_{-1}) } {\sqrt 2}$
$2p_y \equiv \frac{(2p_1 - 2p_{-1}) } {i\sqrt 2}$
Therefore the MOs made from the $2p_x$ and $2p_y$ AOs have $|m| = 1$  and are $\pi$ MOs.

+1 for correctly rendering the $m$ quantum number of the $p_x$ and $p_y$ orbitals.  Too often I see one of these orbitals specifically assigned $m=+1$ and the other rendered as $m=-1$, causing my eyes to pop out and my face to turn purple (OK, I made up that last part but I do take it badly).

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


text:
Is there a relation between the ionization energy and the electron affinity?
I am aware of the definition and equation of IE and EA. I was wondering if there is a direct relation between this two quantities with the energy of the neutral specie.

Okay, here's what I did (and I highly encourage everybody to do this along with this tutorial-style answer, just so you get the feel for prepping and looking at data from the world wild web).
Data Acquisition
First, I went and copied the data from the tables at Wikipedia: This one for the ionization energies and this one for the electron affinities.
Next I had to clean up this data somewhat. For the electron affinity data I first had to get rid of all the unnecessary data columns. This was quickly done with awk:
   awk '{print $1, $2, $4}' raw-affinity.dat > affinity.dat

Next, I filtered out all the lines containing "-" and "?" using inverse search with grep:
    grep -v "?" affinity.dat | grep -v "-" > affinity2.dat

And finally I got rid of all these pesky statistical uncertainties with either one or two numbers between parentheses with sed:
    sed -e 's/(.)//g' -e 's/(..)//g' affinity2.dat > affinity3.dat

Now the data file looks somewhat like this:
1 H  0.754195
3 Li 0.618049
5 B  0.279723
6 C  1.262118

Now for the ionization data file. Since we only want to look at the first ionization energy we can scrap all the other data. Again, we'll use awk for this:
    awk '{print $1, $2}' raw-ionization.dat > ionization1.dat

Next we'll filter out all the data we don't want in there (again with grep and inverse searches):
    cat ionization1.dat | grep -v "use" | grep -v "WEL" | grep -v "talk" > ionization2.dat

In this file we still have some nasty linebreaks... So I loaded the file in vim and substituted the line break + "CRC" with nothing and saved it as ionization3.dat:
    :%s/\nCRC//g
    :saveas ionization3.dat

I also corrected the value for At manually in vim while I was at it, because it wasn't entered like all the other data in the Wikipedia table.
The ionization data file now looks like this:
1 H  13.59844
2 He 24.58741
3 Li 5.39172
4 Be 9.3227

All right, so we have the data nicely prepared and we're ready for the next step, which is
Data Analysis Shenanigans
I did everything in python using the numpy and matplotlib packages. The code used can be found here.
I didn't really notice some large trend of something you might expect. The plot gets pretty messy somehow. The darker the scatter points, the heavier the element.

Maybe someone else notices a pattern that I'm missing right now.


text:
If two electrons move up to a higher energy level in an atom, is it considered excited?
In Germanium, if two electrons go from 1s go to 4p is it in its excited state?

http://physics.nist.gov/PhysRefData/Handbook/Tables/germaniumtable5.htm
Energy levels of a neutral germanium atom
https://www.webelements.com/germanium/orbital_properties.html
Middle, electron binding energies for germanium
It will certainly be in one of its excited states, yes, probably followed by an Auger cascade.

Label   Orbital   eV
K       1s       11103
L I     2s       1414.6
L II    2p1/2    1248.1
L III   2p3/2    1217
M I     3s       180.1
M II    3p1/2    124.9
M III   3p3/2    120.8
M IV    3d3/2    29.8
M V     3d5/2    29.2

i thought an atom can only become excited when only ONE electron moves to a higher energy level? so why does it get excited when two electrons move up?

"Excited" means "higher energy state". Everything higher than the ground state is therefore considered excited.

http://en.wikipedia.org/wiki/Rydberg_atom  "A Rydberg atom is an excited atom with one or more electrons that have a very high principal quantum number,"

@Jonie: People commonly refer to such states as "doubly excited".


text:
Calculate the wavelength of the radiation released when an electron moves from n= 5 to n=2

Calculate the wavelength of the radiation released when an electron in a hydrogen atom moves from $n = 5$ to $n = 2$.

How can I find this wavelength?

The Rydberg formula (for the hydrogen atom) is usually given as
$$\tilde{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_\mathrm f^2}-\frac{1}{n_\mathrm i^2} \right)$$
Here, $R$ is the Rydberg constant; take care of the unit ($\mathbf{m^{-1}}$  or $\mathbf{cm^{-1}}$) here! Wikipedia quotes a value of $\pu{1.097 \times 10^7 m-1}$ for a nucleus of infinite mass. In principle, the value for a nucleus of finite mass (as is the case here) is slightly different, but for the purposes of this question, we can ignore that.
$n_\mathrm f$ is the final state, and $n_\mathrm i$ is the initial state.
So, if you substitute $R = \pu{1.097 \times 10^7 m-1}$, $n_\mathrm{f} = 2$, $n_\mathrm i = 5$, you can arrive at $$\tilde\nu = \pu{2.304 \times 10^6 m-1}$$
and the wavelength is $\lambda = 1/\tilde\nu = \pu{4.341 \times 10^-7 m}$, or if you convert to nanometers, $\lambda = \pu{434.1 nm}$. Double-checking against the Wikipedia page on the Balmer series, this value is indeed correct.


text:
Alginate nanoparticles
I want to produce alginate nanoparticles from NaAlg. I choosed nebulizing NaAlg solution into $\ce{CaCl2}$ solution but 2% (m/v) NaAlg solution is quite viscous. What should I do to make it inviscid?

Dilute the feed solution tenfold, etc.  Inelegant.  Try solution engineering,
http://www1.lsbu.ac.uk/water/kosmos.html
Chaotropes and kosmotropes.
http://en.wikipedia.org/wiki/Theta_solvent
Add a little "poor" solvent like ethanol or acetone to ball the polymer, disrupting hydrogen bond crosslinking and reducing chain entanglement.
You want a novel solution not obvious to one skilled in the art.  That qualifies it for patent protection.  A large fraction of the hard answers you need already exist.  They were done by other people doing other things.  Accumulate bone piles of knowledge.  The process for creating ceramic dislocation-pinned nickel superalloys works for Plexiglass, which "dissolves" 30 wt-% copper phthalocyanine as a dye masterbatch.  So it can't be done, so what?  Do it the "other" way.

Heating most viscous solution will decrease their viscosity. Most agar gels will liquefy when heated. Also, some polysaccharide solutions don't become viscous again when they cool down. I think alginate will become viscous, but you will have to test that out.


text:
How to convert an amide to an imine?
We haven't learned this in class yet, but after looking online, and through my textbook, I can't seem to find a definite answer. I am hopelessly and utterly lost.

The problem can be solved using retrosythesis.
You look at the product and imagine how that was made.
It's like looking at Lego construction, trying to identify the pieces - or looking at a pair of jeans, trying to identify how it was stitched together.
Then you go back and do the same for the previous building blocks.
The final product apparently is an imine. I'm sure we have seen that before among one of the exercises you worked on - in that case it was an intramolecular reaction.

Imines are usually made from carbonyl compounds and primary amines under acidic conditions. This one is made from propanal and ethylamine.
Looks good here, I see $\ce{H2NCH2CH3,\ H3O+(cat)}$ among your reagents.
What about the aldehyde then? The number of carbon atoms in the aldehyde and the amide (propionamide is the starting material) are the same, just the (formal) oxidation of the carbonyl/carboxyl group are different.
Let's think about a direct reduction of the amide with $\ce{LiAlH4}$:

OK, that isn't it. Just two steps would not have been enough anyway ;-)
But we do remember that

esters can be reduced to alkanols with $\ce{LiAlH4}$
alkanols can be oxidized to aldehydes with PCC (pyridiniumchlorochromate)

All this is among the list of reagents. Nice, we know to make the propanal from a propionic ester!
How to turn propionamide into a propionic ester then?
Direct reaction with methanol under acidic conditions does not work, but after alkylation of the amide with an alkyl halide, we should be ok.
Lets's summarize our five steps then:

The reagent for the first step it says is incorrect, it says the amide must be turned into a carboxylic acid first

This is probably not the only way: acidolysis of the amide, followed by esterification is the alternative.

solved!
All it took was a little reading and understanding :)


text:
Halogen bonds in QuteMol
I've been using QuteMol for generating 3D animated molecular images and I've encountered a problem; for some reason that's a complete mystery to me when I create a pdb file for etravirine (using MarvinSketch to create a sdf file and then using OpenBabel to convert the sdf to pdb and add implicit hydrogens) and load it I found the bromine atom was all by itself with no bond connecting the two and I would like a bond in the image. Any ideas? Thanks in advance.
The following image is the result I'm getting from QuteMol.

I haven't used MarvinSketch in a while, but isn't it a 2D editor?
Generation of 3D coordinates from 2D files often means trouble. I tried that recently with a 2D BKChem drawing of a trypticene-based ligand for another answer and the result was useless.
I definitely recommend the use of Avogadro as a free 3D editor. Please have a look at their website or the article in Journal of Cheminformatics.
My typical workflow here is:

switch off auto-adding of hydrogen
draw the skeleton with the right multiplicity of bonds, but using carbon only
change the "wrong" carbons to heteroatoms
let the program add the hydogen atoms
run a force-field optimization
manually adjust some dihedral angles
run the force-field optimization again
(optional) align a bond with one of the principal axis
save as CML or MOL

I put a MOL file of etravirine created that way on pastebin. Please give QuteMol a try at it.

I generated a PDB from the molfile mentioned above and ran it through QuteMol. The image below shows that the $\ce{C-Br}$ bond does exist in the PDB file.

Admittedly, checking the PDB file for the respective CONNECT x y line would have been a (less flamboyant) proof too ;-)

MarvinSketch can do both 2D and 3D SDF files. If I had accidentally used a 2D image what would have happened if that the image I shared wouldn't have any 3 dimensionality at all, it'd all be in the same plane. So that's not the problem. I've created 3D SDF files for conversion to PDB using OpenBabel with MarvinSketch before, without a problem. Additionally I use QuteMol to create animated gif structures. Can avogadro do this? Animated GIF file is pretty important, other formats will not work for my purposes.

@BrentonHorne I wasn't aware that MarvinSketch could do that. Anyway, something apparently went wrong along the line: SDF/MOL file output broken, PDB conversion faulty, PDB reading by QuteMol erroneous... I didn't suggest to replace QuteMole with Avogadro - but using an input file generated differently will help to lacalize the problem.

Ah, so you're saying I should use Avogadro to open the sdf and then use it to create a pdb? Because whenever I open my pdb file (generated with MarvinSketch & OpenBabel) in Avogadro it shows the bromine bond just the way I want it. I think it must be a bug in QuteMol.

@BrentonHorne Actually my suggestion was to first use the SDF/MOL file that I uploaded to pastebin for you to see if that  works :D


text:
Does partial charge violate the law of quantization of charge?
Well, we know that charge is quantized. Quoting wikipedia below,

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

And we know that there is partial charge. Quoting wikipedia below,

A partial charge is a non-integer charge value when measured in elementary charge units.

Don't the above two things (and both being from the same respectable source) contradict with each other?
If yes, then which one is correct? And if both are true, then in which conditions do they follow?

Definitions are powerful words, you have just proved that definition of wikipedia to be wrong.  That definition needs to be improved based on the exceptions as you mentioned.

Related question on Physics StackExchange : Van der Waals forces & Partial charges

Partial charges are like resonance structures: they are a convenient notation which captures the most important aspects of a phenomenon, but they are not what is "really" going on.
When we say that a molecule contains partial charges, what we mean is that the electric field surrounding a bond is polarized as if some fractional number of elementary charges had been displaced from one end to the other.  The phenomenon — the polarized electric field — is real, but the mechanism that produces it does not violate charge quantization.
All of the electrons always have charge −1, and all of the protons always have charge +1.  But the electrons' positions are indeterminate.  (The protons' positions are also indeterminate, but on a much smaller scale which is usually neglected in this scenario.)
A standard technique for calculating the electric field of the molecule is to take the nuclei as fixed and then allow the valence electrons' wave functions to range over the entire molecule.  When you do this, you often discover that some bonds are polar, with electrons (formally belonging to the atoms at both ends of the bond) more likely to be found at one end than the other.
The electric field of such a bond is numerically the same as the electric field that would be produced if you could transfer a fraction of an elementary charge from one end of the bond to the other.  But that is not what has happened.  You can think of what is really going on in these terms: the electrons are moving rapidly back and forth along the bond, but they are slightly more often at one end of the bond, so the average charge density at one end is greater, but not by a full elementary charge.  (This too is inaccurate, but is a standard way to bend quantum theory to make it more comfortable to intuition.)
(N.B. this is the same as permeakra's answer, but in less technical language.)

But I still can't get why that should give result to partial charges. I see it better as forces which could only be exerted by partial charges, and the reason, as you said, would be the redistribution of elementary charges (or their center of masses) in space. Am I correct with this point of view? Or are you trying to say the same thing? Please comment back...

I'm afraid I have no idea what you mean by "I see it better as forces which could only be exerted by partial charges, and the reason, as you said, would be the redistribution of elementary charges (or their center of masses) in space." so I can't tell you if it is correct or not.

I am trying to say that there is actually nothing like partial charges. It is only that the atoms participating in the bond are exerting forces on each other, which are not complete. I am trying to say that its not the charge which is partial, but the bond or its force has some partial character induced in it due to electro-negativity.

Ah.  Yes, that is correct.  The charge on each electron remains exactly −1 elementary charge unit, but the electric field around the bond is polarized, and the partial charge notation accurately describes the polarization (to first order, anyway).  Partial charges are like resonance structures: they are not what is actually going on, but they capture enough of the important details to be convenient notation.

Just one last thing, I can't precisely understand what you wrote in the answer, but, were you trying to say something like that, because, this view of mine, arose from what I inferred from the key terms in your answer...

Yes, that is more or less what I was trying to say.  If you are just beginning to study chemistry, do not worry about the details right now.  Also, please take from this that Wikipedia is not entirely to be trusted.  It's usually more accurate than not, but pages often contain minor errors, and even more often were written by people who understand a topic very well, but don't know how to explain it to people who don't already understand.

Both are.
Any isolated object has integral charge, as any elementary particle has integral charge. However, as we know, elementary particles are not truly located in one point.
When we talk about stationary states in physics, we can assume, that particles inhabit some volume with special density function indicating how much of the particle in question is in any given unit of volume. When we consider isolated molecule, we can assume that it is in a stationary state. This means, that electrons may be viewed as electron clouds, and different parts of said cloud may be assigned to different atoms in the molecule. In case if total amount of electron density, assigned to specific atom is non-integral, we gain a partial charge.
Whenever atomic partial charges are estimated, there is a significant arbitrariness in the process, as there is no unquestionable way to define, which parts of electron density assigned to any given atom, as there is not strict definition of atom borders (atomic radii are estimated using equilibrium positions in some circumstances, but in truth electron density is non-zero even far beyond that radii). Still, the concept is useful enough.

I can't get what you are trying to say. Are you saying that partial charge does not actually exist but it is said to be so on an average position?

I'm trying to say that electrons in atom are more like clouds than dots, and such a cloud may occupy volume of two or more atoms. In that case, elementary charge of the electron is distributed between two (or more) atoms.

But I still can't get why that should give result to partial charges. I see it better as forces which could only be exerted by partial charges, and the reason, as you said, would be the redistribution of elementary charges (or their center of masses) in space. Am I correct with this point of view?

It doesn't have to. It may in some cases. You see, there is no reason why, if one electron have to occupy two atoms, exactly one half of it will be in both. For example, consider $HF$ molecule. It can be considered as having 2 core electron in core of $F$ atoms, 6 electron clouds in F exclusively, and 2 in both $H$ and $F$ atoms. However, 2 latter have more density in $F$ atom than in $H$ atom, say 2/3 of density in $F$ atom. this leads to partial charge of F equal +9 (charge of nuclei) - 2 (core) -6 (exclusive) -2 * 2/3 = -1/5, while hydrogen have +1 - 2 * 1/3 = +1/3.

Oh, yes, I was trying to understand it that way and you made it easier for me that way. I am sorry if I am overdoing things; but I still don't get the answer to my original question.

At the beginning. Both statements are correct. However, they describes situation in two different models. Say, partial charges belongs to approach, developed in molecular mechanics (see wikipedia), that, having absolutely no physical sense, works reasonably well for prediction of molecular geometry and interaction for many biological molecules. Charge quantization, on the other hand, belongs to the field of quantum mechanics, that, making absolutely no common sense, is correct, but hard to understand. So, in general we use simpler model if we can and switch to complex one if simple one fails.

It does violate that particular statement (and not quantization of charge). You just need to understand that charge is quantized mean that there exists a unit of charge which is indivisible.
We may say that charge on an object is always integral multiple of multiple of elementary charge, but it is wrong, as you have pointed out. It is just used to say that elementary charge is indivisible.
In your case, when an electrically neutral atom bonds chemically to another neutral atom that is more electronegative, its electrons are partially drawn away. This leaves the region about that atom's nucleus with a partial positive charge, and it creates a partial negative charge on the atom to which it is bonded. It is to be noted that elementary charge (electron in your case) is not divided.
Principle of quantization of charge can be stated as: Charge on any isolated object is always integral multiple of elementary charge.

It is to be noted that electron is still elementary particle. Proton and neutrons are not elementary. Quarks are now believed to be smallest indivisible unit of charge.

Your comment to the main question seemed more appealing to me! Shorter and better. Anyways, thanks for the great explanation...


text:
What is "stuffy"/"stale" air composed of?
When a room is not ventilated for a long time (days if people are present, weeks if not), humans perceive a change in air quality commonly described as "stale" or "stuffy" air.
What is going on here chemically? How is "stale" air different from normal air?
Note that the difference cannot be the mere presence of biological exhalations from the inhabitants, as the phenomenon definitely also occurs when nobody lives in the building at all (though it takes a lot longer in that case).

It may not be a complete answer but I can think of a few distinguishable features that may or may not be chemical reasons:

chemicals created by bacteria and organisms that are present in everyday households (mold, light activating some production from bacteria and micro-organisms, $NH_3, S^{2-}$ related compounds (they smell bad), etc).
Compound this with the acknowledged lack of circulation and you have funky smelling air.
high humidity: how one can "feel" "stale, muggy" air.
high carbon dioxide concentration can make the air feel heavier (from interior design sources).


text:
Does water really 'go bad' after a couple of days?
Among my friends it is a sort of 'common wisdom' that you should throw away water after a couple of days if it was taken from the tap and stored in a bottle outside the fridge, because it has 'gone bad'.
First of all, the couple of days is not very well defined, which already makes me a bit suspicious. Second, I cannot think of anything in tap water that would make the water undrinkable after a couple of days already.
Can someone clarify this issue for me? Does tap water really 'go bad' after a couple of days outside the fridge? Why?

I suspect that "going bad" has more to do with changes in taste than actual health hazards. I believe chilled water tastes different from room-temperature water due to the lower amount of dissolved gasses in the latter. If water is left in an open cup over a few days, then it starts gathering matter which floats in the air, which may impart a disagreeable taste if a sufficient amount accumulates. Tap water probably isn't a very good culture medium, as it is relatively poor in organic nutrients. Of course some microorganisms will develop in exposed water, but they shouldn't multiply too much.

Tap water outside the fridge but in a closed container should be perfectly fine for months though. After a very long time, eventually some photosynthetic organisms will make it inside and very slowly reproduce by creating organic matter out of $\ce{CO2}$ in the air and whatever residual light they can gather.

I'm wondering if part of the origin of this idea is due to reusable water bottles (you didn't say if they were sterile or not) that are often not properly cleaned between uses as 'they've only had water in them' and are contaminated with bacteria and microscopic food particles from your mouth, the bacterial colonies then produce the same compounds that cause bad breath and other unpleasant tastes and the water in the bottle starts to taste and smell bad. Refreshing the water temporarily solves this problem, though washing the bottles properly would be better.

Pseudomonas aeruginosa seems to always find a way, even in bottles of deionized water that are left out.  In such an environment, this microbe can produce fluorescent green toxins.

First of all, it depends on how the tap water was treated before it was piped to your house. In most cases, the water was chlorinated to remove microorganisms. By the time the water arrives at your house, there is very little (if any) chlorine left in the water. When you fill you container, there is likely to be some microorganisms present (either in the container or in the water).
In a nutrient rich environment, you can see colonies within 3 days. For tap water, it will probably take 2 to 3 weeks. But that doesn't mean that the small amount of growth doesn't produce bad tasting compounds (acetic acid, urea, etc.).
BTW Nicolau Saker Neto, cold water dissolves more gas than hot water. Watch when you heat water on your stove. Before it boils, you will see gas bubbles that form on the bottom and go to the surface (dissolved gases) and bubbles that disappear while rising to the surface (water vapor).

In the Netherlands, where I live, the chlorination of water completely stopped in 2005. At the moment they use ozone or UV-light for the disinfection. So it is safe to say that there is no chlorine (left) in the water ;)

@Michiel Even if primary disinfection is done without chlorine I doubt there is no residual chlorine dripped into the water before it leaves the treatment facility. This is not meant to clean the water but to keep it clean. The amount of chlorine put into the water is a function of how long the operators expect the water to be in a pipe before coming out of your tap.

@Brad It's apparently really the case but here is a critical view.

Also in Germany tap water is not chlorinated at all. Chlorination is actually forbidden by law since 25 years. A reasonably well maintained pipe system in a country with  median temperature around 10°C never fouls.

I cannot think of anything in tap water that would make the water undrinkable after a couple of days already.

Tagging the question with biochemistry probably points in the right direction. The only effect I can think about is growth of anaerobic microbes, e.g. Escherichia coli, causing diarrhea.

There's good chance to keep the water from 'going bad' and even 'turn it good' again when keeping the water in PET bottles in the sunlight, letting SODIS (solar disinfection) work for you.

Ok, seems logical that it needs to be anaerobic when the bottle is closed, but how does the E. coli. end up there in the first place?

E. coli grows in the lower intestines of most mammals, leaves the body via defecation. Sewage contamination of drinking water will most likely lead to elevated concentrations. Lack of proper water treatment will keep it there, giving it a chance to grow.

@Michiel Unless something has been recently gamma irradiated or heavily boiled you can pretty much assume a very small amount of all common bacteria is in everything

A few lazy weeks of summer cottages, camping far from the beaten track or boat excursions in the archipelago: When summer comes, we seek to be closer to nature. Sometimes that means we may waive some of everyday life's certainties, like drinking water straight from the tap.
Do this!
Fill the drinking water of good quality, such as municipal drinking water in sterilized bottles or cans. Keep the water dark and as cool as possible. That keeps drinking water fresh for at least one week.
Smell and Taste

Putting a limit on how long drinking water can be stored has more flavoring to it than health risks, says Torbjörn Lindberg, State inspector at NFA.
The risk of getting sick of water is small in Sweden. But you should of course do not drink water that tastes or smells bad, is turbid or highly colored.

Get microorganisms in drinking water
Good drinking water does not contain pathogenic microorganisms, such as viruses and bacteria. It is also a nutrient-poor environment that makes it difficult for such microorganisms to survive. They are simply ill-suited to the aquatic environment. If they are there from the beginning, it is therefore likely that they will disappear over time, and the risk of getting sick is reduced.
Drinking water can be bad
Sometimes, bacteria, micro-sponges and algae that are adapted to the aquatic environment are found in the drinking water from the start. They can also multiply over time. It is completely natural and gives no health risks. However, drinking water can start to smell or taste badly, especially if kept long and hot, for instance in the sun. It can also be colored and can form a slimy layer on the inside of the container.

Please link to the sources that you used or indicate if this comes from your own experience.

I am not an expert and its been a long time since I last studied chem and bio. The best answer will probably only come from a biologist. But out of interest, I will try to answer. Also, I suggest that you post this question on websites such as http://www.askabiologist.org.uk/ . Its likely that you will get some
good answers there as well.
Assuming that the water in the bottle is thoroughly "purified" at the water plant , there has been absolutely no contamination (pipes, faucet etc.) on the way, and the bottle and handler are clean, then it is unlikely that water will go bad. Example that suggests contamination at faucet - http://www.journalofhospitalinfection.com/article/S0195-6701(98)90295-X/abstract
I don't think that most bottles are 100% air-tight. So, the next thing I would check is if there are any water-borne pathogens that can spread by air. If such pathogens exist, then is it possible for them to enter the bottle through small "air spaces" ? That seems unlikely to me.
So, if water is microbe free and can be kept that way by a bottle, then it should be good for a "long" time.

It depends on the water. If the water is already contaminated, then after a couple of days, there are more chances of water to go bad.
But if the water is mineral, already purified, then it takes too many days.
Also the other fact about the sunlight, if the water is directly faced with sunlight, then sunlight causes formation of the algae in water.
In simple words, you can drink a water, if it is a mineral water, then after couple of days, if it is a tap water, 2-3 days normally.
This is my individual opinion, I am not a doctor. You follow what you want. :)


text:
can an alloy be used to make another alloy?
The answer is likely very obvious but I know pretty much nothing about this so:
Can I make an alloy, using an alloy? That is, could I make an alloy of steel (in itself an alloy of iron and carbon) and silver?

An alloy is purely a mixture of metals (and sometimes non-metals) which are non-chemically bonded.
So yes, you could mix steel and silver to create an alloy. There would be nothing stopping you from doing this, but I cannot say whether or not it would produce any desirable results. Mixing alloys with other materials is merely creating another alloy. Stainless steel is actually just regular steel combined with Chromium.
An alloy is usually created to increase certain properties of its constituents. For example, iron is mixed with carbon in order to create steel, which is far stronger and can be used to manufacture items such as girders. Or, in other cases, Chromium may  be added, if the steel is to be used in, say, cutlery.

Thank you very much for the quick answer and for answering without making me look a fool haha :) if I wasn't a complete muggle I would have thought of stainless steel. Thank you so much  :)

No problem. Asking a question shouldn't make you look like a fool. I gave it a quick Google (alloys of alloys) and Stainless steel was the first result. :)

I did try wikipedia list of alloys on the search for steel but Google hates me :) I wanted to add a thanks also for telling me the purpose of alloys. I was under the impression that an alloy reinforced the material, as opposed to increase a specific property. So thanks again!

Well, in steel's case adding carbon is both reinforcing the iron and increasing a property (its hardness). The two can mean the same thing in some cases.

That's true. Would I be right in saying that chromium does NOT reinforce the steel? As in, it makes it stainless but does not strengthen the bonds?

Yes, I believe so. Chromium merely reduces steel's susceptibility to corrosion.

True- Chromium main reason for addition to steel is reduction to susceptibility to oxidising corrosion- needs at least 12% to be effective but it also increases its hardenability- best to search yourself for that term.

Silver and Iron are almost immiscible- 1% in the solid and liquid phase.  Steel  which is iron plus  perhaps only 0.03% carbon and some manganese. You aren't going to able to mix much steel into silver or much silver to steel as a liquids or after it freezes. Any more than 1% of either and it will just separate.
http://www.e-journals.in/pdf/v2ns1/s153-s156.pdf

I don't suppose you could elaborate? Honestly, your answer makes no sense to me at all, though it's a fault on my end. I barely have a starting knowledge of this kind of thing.

@MacCooper the point is that the random example you chose (iron+silver) happens to be one of the few cases where making alloys doesn't work as usual.

@Peteris, which is just about typical, yeah ^^ The purpose is completely fantastical though, so I'm just happy that it's sound in principle :)


text:
Halogen Oxyacid Trend (acid strength)
Why is $\ce{HClO4}$ more acidic than $\ce{HBrO4}$? But $\ce{HCl}$ is less acidic than $\ce{HBr}$? What determines the acidity? Is it the concentration of $\ce{H+}$? If so, how do these trends develop along the halogen group?

Would you mind to give the $pK_a$ values and the solvent in which they were measured for the acids you are asking about?

In the determination of the $\mathrm{pK_a}$ values of really strong acids, water as a medium is irrelevant at a certain point.
Ivo Leito and coworkers have examined the acidity of various compounds in 1,2-dichloroethane (DCE).
In this medium the $\mathrm{pK_a}$ values for $\ce{HCl}$, $\ce{HNO3}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HBF4}$, an $\ce{HClO4}$ have been determined as -0.4, -1.7, -4.9, -7.7, -10.3 and -13.0.

In this survey, 1,1,2,3,3-pentacyanopropene (1), was found to be even more acidic with a $\mathrm{pK_a}$ of -15.0!

EDIT
It is important to note that these values are likely to be correct in their relative order. But the values are based on picric acid, for which the $\mathrm{pK_a}$ was arbitrarily set to zero. This initially escaped my attention!
Thanks to Martin for pointing out this important detail in his comment below!

Being an oddball, the rather unstable perbromic acid ($\ce{HBrO4}$) was not in the list!
It was synthesized by Evan Appelman (even on a larger scale) by bubbling fluorine gas through an alkaline solution of 1M $\ce{NaBrO3}$ and 5M $\ce{NaOH}$ until the solution becomes acid.
Amazingly, the author further states in a footnote:

The reaction is not smooth, and small explosions may take place in the vapor above the solution.

Could you clarify/ prove your first statement. From my understanding $\mathrm{pK_a}$ values always involve a solvent. For superacids they are measured relative to each other. Your cited paper uses 1,2-dichloroethane as solvent and the $\mathrm{pK_a}$ values have to be anchored to an acid with a known value. For the cited case this was not possible as exact measurements in DCE were not accessible. So your posted numbers are quite arbitrary.

@Martin Good observation! One does indeed need a reference. In the paper cited, pikric acid was arbitrarily set to 0 an I forgot to mention that. This does however not invalidate the relative order.

Assuming you are talking about acids in water, and taking into account that the acids you mention are all regarded as strong acids, I would focus on the relative stability of the corresponding conjugated bases.
$\ce{Br-}$ being larger than $\ce{Cl-}$ can better allocate the negative charge. Then, $\ce{HBr}$ ($pK_{a}$= −8.7) is stronger acid than $\ce{HCl}$ ($pK_{a}$=-6.3).
On the other hand $\ce{ClO4-}$ is more stable than $\ce{BrO4-}$. Here, another factor plays a role and should explain the difference in stability: the charge dispersal associated to resonance. Being the structure similar, we would expect the same effect: however, the resonance stabilization that characterizes such kind of ions has to be more effective for $\ce{ClO4-}$, due to a better overlap between $p$ orbitals ($2p$ of oxygen with $3p$ of chlorine, better than with $4p$ of bromine). As a consequence, $\ce{HClO4}$ ($pK_{a}$= −8) is stronger acid than $\ce{HBrO4}$ ($pK_{a}$= ? haven't found it, but here $\ce{HBrO4}$ is said to be an "almost strong acid", therefore not stronger than $\ce{H3O+}$).
These considerations tentatively explain the difference in $pK_{a}$ for such strong acids.

Perbromic acid: the Wikipedia article has been heavily edited.  You sure it still says perbromic acid is almost strong?

There are two different kinds of acids mentioned. First, binary acids such as $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, etc. The other acids mentioned were inorganic oxyacids or oxoacids.
The most important factor in oxoacid strength is the ability to donate a proton which is influenced by electronegativity and bond polarity.
Oxoacids are usually designated with X as the central atom, and thus, their structure can be written as:
$$ -\text{X}-\text{O}-\text{H} $$
X can be bonded to other things. i.e. hydrogens, more -OH groups, etc.
If X was an alkali or alkaline earth metal, and thus electropositive, the corresponding compound would be a base. This is not an acid. The reason for this is the $X-O$ bond is extremely covalent, almost to the point where the strength is equivalent to bond in ionic strength. The oxygen atom thus has a large partial negative charge $\delta^- $ and the electron density accumulates on the $-O-H$ while the hydrogen, $\ce{H^+}$, binds tightly with the oxygen, preventing deprotonation.
As the number of lone oxygens on $-X$ increases, the strength of the oxoacid increases as well.
Taken from Principles of Modern Chemistry by Oxtoby et. al 7th Edition:

If the formula of these acids is written as $XO_nOH_m$, the corresponding acid strengths fall into distinct classes according to the value of n, the number of lone oxygen atoms  Each increase of 1 in n increases the acid ionization constant $K_a$ by a
  factor of about $10^5$.

Another way to examine the strength of acids is to look at the stability of their conjugate base which can be written as $XO_{n+1}(OH)^-_{m-1}$. As more oxygens are added on the central atom, the net negative charge is distributed more evenly around the base. As bases become more stable, they are less prone to accepting a proton. Remember that conjugate bases of strong acids are weak so intuitively, this makes sense.
Put briefly, as $-X$ becomes more electronegative, electron density is pulled towards the central atom. As the electron density is pulled from the $-H$ bond to $-X-O$, the $O-H$ bond breaks easily, and a proton is produced.
If we discuss organic acids and not binary nor inorganic acids, we not only have to talk about electronegativies, but steric hinderance, and resonance structures as well. If you would like, I can explain those as well.


text:
The effect of concentration on the rate of a homogenous reaction
I have to do an experiment where I will find out what the influence of the concentration on the rate of a homogenous reaction is.
We are asked to create 2 solutions, respectively A and B:
$$\ce{KI + Na2S2O3 + Starch + water}$$
and
$$\ce{H2O2 + H2SO4 + water}$$
As long as A and B aren't added together, the reaction cannot start.
Now what I'm supposed to do is make 10 different A and B solutions and add these together and measure the time it takes for the solution to turn blue (Iodine + Starch as an indicator).

Now, my problem is that the following questions are also asked:
"Write down the different reactions that take place"
This seems dubious to me but might not be.
e.g. Does dissolving into ions also count as a reaction?
Does the Starch (for which I have no given chemical formula) react in solution A?
I've also read online that adding $\ce{H2O2}$ and $\ce{H2SO4}$ creates a "Piranha" solution ($\ce{H3O+ + HSO4- + O}$ or $\ce{H2SO5 + H2O}$, I've found both online but I don't know which is correct here.)
In conclusion I just don't know which reactions I should give (Everything that happens in A and B isolated from eachother or everything that happens when I add A and B together? Either way, I'm also not sure what the reactions would be because, for instance, I don't have any certainty on what the chemical composition of the Starch is.)
Is the reaction a multistep reaction or a single step reaction? Why? If it's a multistep reaction, write down the partial reactions
Assuming that adding B to A is "The" reaction, how do I find out if there are multiple steps or not? How do I find these steps?
Given that this question is asked already makes me think that it is in fact a multi-step reaction.

You could look up "iodine clock reactions" and get the straight skinny on the chemistries and their kinetics.  You might also look up "Theory of Experimentation" and see how confounded variables cannot be teased apart by varying them individually.  Instruct your instructor.
http://en.wikipedia.org/wiki/Iodine_clock_reaction

I've also read online that adding $\ce{H2O2}$ and $\ce{H2SO4}$ creates a "Piranha" solution [...]

This is true - in principle. In reality, Piranha solution is made from concentrated sulfuric acid and 30% hydrogen peroxide.
In the case of your experminent, you will most likely use diluted sulfuric acid and aqueous solutions of hydrogen peroxide with a much lower concentration. Consequently, the Piranha effect will not play a role in your experiment, neither is the oxidizing power of sulfuric acid relevant here.
On a side side note, treating a sample with conc. $\ce{H2SO4}$ is a method for the qualitative determination of iodide: Upon heating, one can observe the violet vapour of iodine.
In your case, you need the sulfuric acid to keep the pH value low. Under these conditions, $\ce{H2O2}$ acts as an oxidant:
$$\ce{H2O2 + 2 H+ + 2 e- -> 2 H2O}$$

Does dissolving into ions also count as a reaction?

No.

Does the Starch (for which I have no given chemical formula) react in solution A?

Starch is a mixture of two natural polymers (amylopectin and amylose), both made from D-glucose. Amylopectin doesn't play a role here, but it doesn't disturb either and it's just cheaper to let you work with the natural mixture.
It's the helical amylose that is relevant here - and it acts as a reagent.
What actually happens in the reaction is the following:
$$\ce{2 I- -> I2 + 2 e-}$$
$$\ce{I2 + I- -> I3-}$$
The linear triodide ion forms a strongly coloured inclusion complex with the amylose in the starch.


text:
Evaporation of a mixture of salts
I was wondering what happens when you boil dry a solution of several salts for example a solution of KCl and NaNO3. Since the ions dissociate I was wondering if the ions would combine randomly upon being boiled out of solution making the original salts as well as KNO3 and NaCl? Does this happen? If not why?
Also in a related question when you mix two solutions to form a precipitate in a double displacement reaction will the insoluble salt always form or are there situations where an insoluble salt can form but doesn't?
Thanks in advance.

As the solvent evaporates the least soluble salts precipitate in succession.  This is used during seawater and salt lake evaporations to isolate values beyond NaCl. Overall, you will get a mixture of combinations of cations and anions.
Supersaturation is not uncommon.  Metastable solutions usually require nucleation sites to get over their initial activation energy for solids' crystallization.  High melting points (strong lattice forces) are difficult to supersaturate.  Benzil in xylenes remarkably supersaturates, and is stable for a week or more in silanized glassware and cooled very slowly within styrofoam and rags.

Thank you for both explanations. One further question. If the salts precipitate based on order of solubility then how do you know when on has stopped precipitating and another starts? Is there  a physical change to observe?

Serial precipitations have no visible great change (barring colored solids). Such purification of lanthanoids (liquid-liquid extractions by extension) are theoretical plate separations in both directions, http://www.chem.uoa.gr/applets/AppletCraig/Appl_Craig2.html  Replaced by chromatography and ion exchange columns.  A classical theoretical plate scales centimeters.  A modern theoretical plate scales microns.  Tonnes get complicated, http://www.thermopedia.com/content/680/


text:
Why Ionic compounds dissolve into water
First, I would like to say that I understand the basic idea of disassociation and solubility. However, as I study intermolecular forces I feel like this basic Idea is being challenged. My understanding from basic chemistry is that ionic bonds are much stronger than covalent bonds, due to the actual exchange of electrons and the formation of cations and anions, causing an attractive force between the opposite charges to "hold" the compound together. When explaining solubility with IMF theory, however, I fail to understand how the IM forces present in water--dipole-dipole, H-bonds, and dispersion--are able to overcome that attraction between the cation and anion present in the ionic bond; because, I was under the impression that actual bonds are always stronger than IMF bonds. So, for instance, does it not infer that if, say, Sodium Chloride is placed into water, and the attraction of the h-bonds in water cause the NaCl to essentially break apart into ions, that the inherent attractions between cations and anions in an ionic compound are weaker than the attraction force of the h-bond water molecule?

Does this help?

It's not that ionic bonds are weaker than hydrogen bonds, but that the energy cost of breaking the ionic bonds is well compensated by the gain in solvation energy when you put NaCl in solution.  Solvation energy drives the spontaneity of this dissolution process, because of entropic considerations.

An individual IMF bond is weaker than an ionic bond. But when a salt dissolves, at least 4 IMF bonds are formed for each ionic bond. This is why some salts (e.g. CaCl2) release heat when they dissolve.

I was also confused at this same problem, but what I found out is that there are 2 possible explanations for this conundrum:

Entropy causes everything to become more disordered, and so ionic compounds will dissolve to increase entropy.

Each ion can participate in multiple ion-dipole interactions when dissolved, and these multiple ion-dipole interactions could potentially store more energy than the ion-ion interactions, thus making the dissolving of ionic compounds more thermodynamically favored. You can see proof of this through a few exceptions in solubility (such as $\mathrm {Li_3PO_4}$ not being soluble) and some types of ionic compounds never being soluble (ionic compounds with $\mathrm{OH^-}$ are almost never soluble), where some ionic compounds will not dissolve due to it not being more thermodynamically favored. In fact, this also explains why some ionic compounds (such as $\mathrm{CaS}$) decompose in water: because it is more thermodynamically favored.


text:
How to convert CO2 ppm volume to ppm weight?
I'm using a sensor that measures $\ce{CO2}$ in ppm by volume aka atmosphere.
What I'm using it for is to measure $\ce{CO2}$ concentration in water which is measured in ppm by weight.
This is what I have so far
$$\text{Ppmwco2} = \text{ppmvco2} \times 0.8317 \times 44.01 / 10000;$$
This is quicker than $\text{co2}/1000000 \times 0.8317 \times 44.01 / 1000000$. 0.8317 accounts for Henry's Law and max ppmw will be 36.6 with 10,000 ppmv.
The sensor floats on the waterline detecting any $\ce{CO2}$ bouncing out of the water. Do I even need to account for Henry's law? Is the math right?

You can't determine the concentration of CO2 in the water just from measuring the concentration in the air above the water.
The solubility of CO2 is extremely dependent upon pH. See: http://ion.chem.usu.edu/~sbialkow/Classes/3650/CO2%20Solubility/DissolvedCO2.html
There is no indication you are considering pH at all.
Solubility of CO2 also depends upon temperature.
Also, you need to consider whether or not the phases are in equilibrium.

Actually from the link you provided, your statement is false. pH is dependent on CO2 species like HCO3, not CO2. After reading further on wikipedia, Henri's law can't be used accurately because CO2 reacts with water to form HCO3 and the like.

From wikipedia:The hydration equilibrium constant at 25°C is called Kh, which in the case of carbonic acid is [H2CO3]/[CO2] ≈ 1.7×10−3 in pure water[3] and ≈ 1.2×10−3 in seawater.[4] Hence, the majority of the carbon dioxide is not converted into carbonic acid, remaining as CO2 molecules

For those looking, I think I solved it.
$$\text{ppmw} = \frac{(\text{ppmv})(\text{MW})}{((0.08205) (¦K))(\text{density of Water})}$$
where:

ppmv = air pollutant ($\ce{CO2}$) concentration, in parts per million by volume
mg/m3 = milligrams of pollutant per cubic meter of air - (ppmv)(MW) /
[(0.08205) (¦K)]
¦K = atmospheric temperature in degrees Kelvin =
273.15 + ¦C
0.08205 = universal gas law constant in (atm+liter)/(gmol+¦K)
MW = molecular weight of the air pollutant (dimensionless) - mol weight of
$\ce{CO2}$ ($\pu{44.01g/mol}$)
$\pu{1000 kg/m^3}$ (density of water)

density of Water = $\pu{1000 kg/m3}$ - use 1000 as default since you'll need to measure the density to get accurate numbers. The density does vary due to temperature but very minor.
400ppmv of CO2 = 0.720ppmw