text:
Electrolytic cell chemical composition during/after charging?
I have a pot with some (bottled) water and added (sea) salt. It has a pencil (graphite) and tin foil (aluminum). The pencil is connected to the positive outlet of a solar panel and the tinfoil is connected to the negative outlet of the solar panel (effectively 'charging' the electrolytic cell).
To ensure I don't come to harm, I would like to know:
A) What chemical(s) gas(es) are produced at the pencil?
B) What chemical(s) gas(es) are produced at the tin foil?
C) What will the chemical composition of the water generally be (assuming I've been charging it a while)? (Acidic, toxic, etc?).

Voltage determines the kind of reactions, amperage determines the amounts.  The positive graphite electrode is the oxidizing anode.  It will evolve oxygen and chlorine and be eroded to carbon dioxide, carbon monoxide, and chloro-organics.  The negative aluminum electrode is the reducing cathode.  It will reduce Na+ to elemental sodium that immediately reacts with water to make lye and hydrogen.  The lye will erode the aluminum into more hydrogen and aluminum hydroxide gel, plus some black slime for about 1% alloy-reinforcing precipitated intermetallics in the aluminum foil (usually 8111 alloy).  Lye plus chlorine gives hypochlorite bleach that will erode both electrodes, and do a little disproportionation into chloride plus chlorite, then that perhaps into chloride and chlorate.
Sodium carbonate (washing soda) is a more benign electrolyte, but it is an alkali hazard to eyes and skin.  A glass is a more inert container.  Do not short the electrodes, do it outdoors and stay upwind.  Avoid electrolyte or product contact with your eyes and skin.

Do you know how I would be able to easily test for the presence of lye (assuming I have no scientific equipment to hand)?


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'Hard' soap vs. 'Soft' soap: Why do they work this way?
In school, we did an experiment using fat and $\ce{KOH}$ to create 'soft' soap.
Afterwards, we added $\ce{NaOH}$ to change it into 'hard' soap.
I read on multiple places on the internet that the difference between using Potassium and Sodium is what determines if it becomes a 'hard' soap or 'soft' soap.
Am I correct in the notion that the $\ce{K+}$, resp. $\ce{Na+}$ ions connect to the $\ce{O-}$ ends of the Fatty Acid?
Is it possible to skip the first step and create 'hard' soap by using $\ce{NaOH}$ right away?
Why does 'Soft' soap dissolve so much easier than 'Hard' soap? Or in other words: Why is it much easier for the $\ce{K+}$ to fall off than for the $\ce{Na+}$ ?

Basically, yes, by common terms, soft soap has $\ce{K+}$ and hard soap has $\ce{Na+}$.
They are both alkali metals, but $\ce{K+}$ is below $\ce{Na+}$ on the periodic table so maybe we should consider periodicity trends. As you go down a table the atoms become more reactive, so $\ce{K+}$ is more likely to react with polar parts of water than $\ce{Na+}$. Here are solubilities for both (though there are other values that would tell this story, perhaps better than solubility of the product):
[Solubility for $\ce{KOH}$] in $\ce{H2O}$ at 25C: 121 g / 100 mL
[Solubility for $\ce{NaOH}$] in $\ce{H2O}$ at 25C: 100 g / 100 mL
Regarding your question of going straight into hard soap, based on this link it seems doable; he just used $\ce{NaOH}$ (lye) and oils with no mention of $\ce{KOH}$ (potash?) to speak of.
And yes, the positive ions connect to the $\ce{O-}$ of the fatty acid as seen in this diagram.

(img src: http://jan.ucc.nau.edu/~doetqp-p/courses/env440/env440_2/lectures/lec19/lec19.html)

This is not an answer to the question, and the part about the reactivity is also complete nonsense here.

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Can a organic compounds such as hydrocarbons contain an ionic bond?
Can organic compounds like hydrocarbons have types of bonds other than covalent bonds? Can they also possess ionic bonds?

Yes, compound made of very stable carbocation and carboanion has been synthesized

Azulene has a 1.05 debye dipole moment,

A hydrocarbon as such bonded as an isolated cation-anion pair with no heteroatoms could happen in principle given sufficient steric hindrance.   Penta-tert-butylcyclopentadienide or pentakis(2,4,6-tri-tert-butylphenyl)cyclopentadienide with like fully substituted cyclpropenium or tropylium might work.
Similar substitution on fulvalenes' four inner carbons would dihedral angle rotate the rings by steric hindrance.  This prevents the double bond from forming, giving sort of a sigma covalent plus pi-dative bond.

Very strained theoretical hydrocarbons such as tetrahedrane, pagodane derivatives and higher-order prismanes show considerable acidities and basicities, so I imagine it could be possible for a very special pair of hydrocarbons to directly react in an acid-base reaction and produce a hydrocarbon salt?

Enthalpy of reaction for undesired reaction must be small (e.g., direct electron transfer), steric barriers to undesired bimolecular reaction must be high. Both cyclopentadienide and tropylium have pKa,pKb near water and are very soft, delocalized ions. I doubt (homoaromatic, etc) sigma systems can do better.  If you have one....

"Ionic bond" does not necessarily mean the same thing as "bond that is not covalent". For example, I would have a difficult time classifying a hydrogen bond as being either ionic or covalent in nature.
Nevertheless, four different hydrocarbon salts (i.e. salts containing only carbon and hydrogen) have been reported by Okamoto et al. (J. Org. Chem. 1990, 55 (3), 996–1002). The four salts involve carbanion 1 being paired with carbocations 2 through 5:

As you can see, a great deal of aromaticity/conjugation is "employed" in order to stabilise the charges. I don't think the cyclopropyl substituents are a coincidence either, since positive charges $\alpha$ to a cyclopropane ring are stabilised by $\sigma_{\ce{C-C}} \rightarrow p$ donation (see this question).
According to the paper, Kuhn's anion, 1, has a $\mathrm{p}K_\mathrm{b}$ of 8.1. The stabilities of the carbocations are measured differently, using a $\mathrm{p}K_\mathrm{R^+}$ scale:
$$\mathrm{p}K_\mathrm{R^+} = \log \left(\frac{[\ce{R+}]}{[\ce{ROH}]}\right) + H_\mathrm{R}$$
for the equilibrium
$$\ce{ROH + H+ <=> R+ + H2O}$$
where $H_\mathrm{R}$ is an acidity function defined for the medium (for more details, see Carey & Sundberg, Advanced Organic Chemistry: Part A, 5th ed., p 426).
One would expect stable salts to form between Kuhn's anion and carbocations with $\mathrm{p}K_\mathrm{R^+}$ values near 9 to 10. This is indeed so:
$$\begin{array}{cc}
\text{Carbocation} & \mathrm{p}K_\mathrm{R^+} \\
\hline
\mathbf{2} & >10 \\
\mathbf{3} & ~10 \\
\mathbf{4} & 8.9 \\
\mathbf{5} & 8.7
\end{array}$$
The fact that the compounds existed as ionic salts was proven by IR spectra, UV-VIS spectra, and conductivity measurements.
If you want to expand your question to include "salts made of organic compounds", then another reference is J. Am. Chem. Soc. 1963, 85 (9), 1354–1355.

This is brilliant, great find!

Yes, of course. Look up for Grignard reagents, it is the most obvious. But there is other type of bounds. Cyan-hydride ligand is another bound type. Generally speaking this kind of not-exactly-covalent-and-close-to-ionic bound occurs with metals.

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How do we calculate N factor and why is it useful in calulations?
I don't understand the concept of N factor at all. Why is it important?
I know that it is something dependent upon the oxidation number, for example N factor of Na would be 1 since it has oxidation number of +1.
For acid and bases we look at the replaceable $\ce{H+}$ or $\ce{OH-}$ present in them.
Apart from this, I didn't really understand how I can use N factor for calculation or how to find the N factor for whole chemical reactions. Please correct me where I'm wrong.

There are some different applications.
For one, it tells you how many times an acid can lose a proton. For instance, you can have a diprotic ("with $\ce{2 H+}$") molecule, with $n=2$. For instance, $\ce{H2SO4}$, deprotonates to $\ce{HSO4-}$, which can then go to $\ce{SO4^{2-}}$.
Why is that important? Take an acidic reaction with $\ce{H2SO4}$. If the extent of reaction is strongly forward, you would treat $\ce{H2SO4}$ as if it had twice its concentration to take into account that it has two protons it can lose. If you only accounted for the first proton you would be underestimating this effect.
It's also known as equivalent concentration, which might make it sound less foreign to you.
In redox reactions, it shows how many electrons an agent can accept/donate, and in precipitation reactions it measures the # of ions that'll precipitate out of a solution. For all these situations the n-factor or equivalent concentration corrects for multiplicity, since now we know not all molecules are made equal. Some acids can deprotonate twice, some reagents will lose 4 electrons in redox while others may lose just 2. To consider these differences we use this concept.
Sources: Wikipedia

n factor is atomic weight or molecular weight by equivalent weight. n factor for acids is its basicity, ie, number of replaceable hydrogen ion per molecule. For exampleg $\ce{HCl}$ has n factor 1 n $\ce{H2SO4}$ has 2 $\ce{H3PO3}$ has n factor 2 because when $\ce{H3PO3}$ dissociates to form H+ and HPO3^-2. Same ways for bases its n factor is its acidity or no of replaceable OH ions per molecule. Eg: for $\ce{NaOH}$ it is 1 n for $\ce{Ca(OH)2}$ its 2.

N factor is very important for numerical problem in chemistry. N-factor of an acid is its basicity. It is a product of molarity\normality. N-factor helps in determining the gram equivalent of compound. Total H+ ion in a compound shown in reaction is its n-factor. Eg:- $\ce{H2SO4}$ has two hydrogen if one hydrogen is utilised in a chemical reaction then n-factor is 1 if both hydrogen is utilised then 2 is n-factor. Similarly in bases too. In $\ce{NaOH}$ acidity of this compound is. This means n-factor is 1. let a reaction be :- NaOH reacts with $\ce{H2SO4}$ to form $\ce{Na2SO4}$ with water. Here both hydrogen of acid is used so n-factor is 2 for acid and always n-factor of $\ce{NaOH}$ is 1.

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Electrochemistry lye quality
With the following setup: plastic container, water (impure), pencil (anode), tin foil (aluminum, cathode) and solar panels supplying electrical charge (roughly 36v at about 400-800ma), with the intention of producing lye via electrochemistry:
1) Would it be possible to produce the right kind of lye to a level sufficient to produce handsoap with this current setup?
2) How would you test the lye quality to be sure?
If not possible to make the right quality lye, is there a way to modify the setup so it's able to produce suitable quality lye for soapmaking?

You can purchase Red Level lye in a supermarket.

A common way outside the lab to test if the strength of the lye before soap-making is to put an uncooked egg in the lye; if around 1/4 inch sticks out floating at the top, it's fine

@UncleAl I want to see if I can produce reasonable quality lye for free.

@UncleAl Also if you mean "red devil" lye, it's apparently been discontinued: http://candleandsoap.about.com/b/2005/09/19/its-not-a-lye.htm

http://www.essentialdepot.com/servlet/the-Sodium-dsh-Hydroxide-dsh-Lye-dsh-Food-dsh-Grade/Categories
http://www.essentialdepot.com/servlet/the-Sodium-dsh-Hydroxide-dsh-Lye-dsh-Tech-dsh-Grade/Categories

Knowledge would be a better reward than convenience for me.

I'm quite curious if you had any success @user4838. But in my opinion, power from power network would be better than power from solar panels.


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How many lumens from 1 gram of flash powder
Does somebody know how many lumens we get from lighting 1 gram of flash powder in pure oxygen (or normal) environment? I'm thinking about $\ce{KClO_4} $ (60%) + $\ce{Al}$ (40%)  flash powder and quite fast flaming due to lighting it up with electricity.


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How to speed up flash powder flaming reaction
Do you have an idea how can be flaming of flash powder (probably $\ce{KClO_4} $ (60%) + $\ce{Al}$ (40%) ) can be speeded up? Of course we consider the pure oxygen-in-air conditions. So how? With some kind of catalyst?
Thanks.

First, we postulate safety is not an issue. If you try any of this you will probably die.  If you need a short duration event-triggered flash, explode a wire.  A Pockels cell plus a polarizer is a nanoseconds shutter.
Ball mill potassium perchlorate all by itself.  Mix well with micronized aluminum.  Compact.  (Impact, friction, or a static discharge may kill you.)
Aluminum has a high boiling point, 2519 C.  Magnesium is 1091 C and burns under nitrogen.  As above, with micronized Mg, adjusting stoichiometry for the new equivalent weight.  Same warning.
Use a more potent oxophile, like titanium, adjusting stoichiometry for the new equivalent weight.  Micronized titanium can flash in air, nitrogen, or carbon dioxide all by itself.  Same warning.
DO NOT TRY CHLORATE, as opposed to perchlorate.  There is almost no activation energy barrier to reaction.


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Why are bonds ionic when the electronegativity difference between bonded atoms is greater than 1.7?
I'm learning about how to recognise whether a bond is ionic or covalent, based on the difference in electronegativity between the two bonding partners, $\Delta \chi$.
What I have now is a formula:

If $\Delta \chi = 0$, then the bond is nonpolar
If $0 < \Delta\chi \leq 1.7$, then the bond is polar covalent
If $\Delta \chi > 1.7$, then the bond is ionic

But I don't know how scientists determined that formula, the history of it and which experiment indicates that formula.

I dont really think it has much meaning as you think, this value is given for intro chem courses so you don't have to go very deep into how bonds actually work. Real chemical behavior isn't so binary. (Maybe 1.7 has SOME significance though)

Ionic and covalent lie on a continuous spectrum, and no "ionic" compound will have 100% ionic character.

The first thing to consider is the difference between covalent and ionic bonding, from the UCDavis ChemWiki site Ionic and Covalent Bonds,

In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms that have the same electronegativity share electrons in covalent bonds since donating or receiving electrons is unfavorable.

The electron donor has a low electronegativity and the electron acceptor has a higher elelctronegativity - so there is a difference in electronegativity $\Delta{EN}$, effectively creating a positive and negative end, an example is below:

Image source: Japan Synchrotron Radiation Research Institute (JASRI)
When the differences in electronegativities of various compounds are graphed against % ionic character, as shown below:

Image source: University of Florida Chemical Bonding page
Values of $\Delta{EN}$ greater than 1.7 correspond to an ionic character of greater than 50%, from the University of Florida website:

What determines how the electrons are shared is the relative electronegativity (electron greed) of the bonding atoms. The degree of polarity or degree of ionic bonding of any given bond can vary continuosly zero to nearly 100%. We normally say that bonds between atoms with electronegativity difference ($\Delta{EN}$) greater than 1.7 are ionic, although this really means only more than about half ionic in character.

Another resource is from the University of Washington Lecture 23: Ionic to Covalent Bonds.


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Reaction between ammonia and iron
I am not a Chemist or did not learn chemistry as a student. I have faced this phenomena from a natural visualization. So this question.
Experiment: What will be happen if we add a solution of ammonia and water over bad quality iron? When it shall be dried I shall pour again. Several day It shall be done regularly in open nature. Some reaction will take place. Rust will be created on iron.
Here is the main question:
Common people will not see anything significant except rust. But As a chemist how do you understand after three to four days after completion of the experiment that I pour ammonia solution on it? Assume that this three or four days it was in open nature. Something will be produced after reaction except rust. How much of them is water soluble?

try this http://www.corrosionist.com/Iron_Corrosion.htm. Quite simple explanation.

In fact metallic Iron doesn't react directly with ammonia. However ammonia can influence the corrosion process because tend to increase the pH. From the Wikipedia Pourbaix diagram (below) you can see which are the products of a corrosion at an higher pH.Of course the pH of you solution depends on the ammonia concentration of your solution (a 1.0 M ammonia aqueous solution has a pH of about 11).

Corrosion is in fact inhibited at higher pH, and the products of the corrosion are not specific of the ammonia reaction, so (I think) is hard do identify the presence of ammonia from the corrosion process.
Furthermore ammonia is in fact a gas and many of the compounds it forms tend to became ammonia (e.g. ammonium carbonate) and then flow away. However I think that there are many chances that some ammonium salts are produce by the reaction with other species, (e.g. acids) present in the surface of the metal or in the environment.
In this case you can look for ammonium ions: $\ce{NH_4^{+}}$ you can identify them with a Ion chromatography or with Nessler's reagent (keep in mind that Nessler's reagent contain mercurium $\ce{Hg}$ so is toxic and not very environmental friendly) that detects even ammonia trace.

Thank you for your answer. So it is not a easy task to identify the ammonia. May I come to the conclusion?

@Dutta ammonia in a bottle is easy to identify, identify ammonia as a product of a determinate process in an open environment no so easy! You will not have specific products from reaction with iron however if has been used a lot of ammonia some secondary products with other chemical species could be present. If you are going to analyze the soil keep in mind that ammonium ion is a common ion in the soil but maybe confronting adjacent area you could find an increase in the concentration. In this case many other factor like biological activity should be taken in account. Good luck!

@Dutta you are welcome! Maybe the simple thing, if there are no restriction in your country, is to use Nessler's reagent I think you can find it on ebay  however you should previously perform with this some test at different time interval with pieces of metal you have treated with ammonia to understand if you can have good results.


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The Crisscross method for finding the chemical formula
I am reading this wikipedia article that I don't understand. What I don't understand is:

suppose we have two elements $X$ and $Y$ having oxidation numbers $x$ and $y$ respectively. Can we prove that the compound formed will be $X_yY_x$.

I tried to understand this method as: I think this method comes from the fact of charge neutrality. Suppose the compound to be formed has formula $X_{y'}Y_{x'}$. The charge on $y'$ number of $X$ elements is $xy'$. The charge on $x'$ number of $Y$ elements is $x'y$. Now by charge neutrality we should have : $$|x|y'=x'|y|$$
 this implies $\dfrac{|x|}{|y|}=\dfrac{x'}{y'}$.

But I could not further understand why the we divide $x$ and $y$ with their greatest common factor to find the required solution.

I have also found some contradiction examples to this Crisscross rule. Example is : Solid-state-semiconductor $GaAs$, in this Gallium has valancy $3$ and $As$ has valancy $5$.
 The main points on which I want answers are:

What is the logic behind this method, how to prove this method?
Who, how and when developed this method?
Is this method only applicable to ionic compounds?
Why compounds like $GaAs$ do not follow this rule, how charge neutrality is preserved in these compounds. I think these don't follow this rule because these are not ionic compounds, if yes then how the chemical formula of compounds like solid state $GaAs$ is known(or decided) and how it is justified, how one atom of $Ga$ balances one atom of $As$ ,their valencies are different, how they are bonded(covalently I guess) together?

I think you mean the oxidation number for N, As, and P are +5, not -5... https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements
I've never heard of this method, but it's just simple math. You know that the net charge of a neutral ionic molecule is 0: For example, say you have Al(+3) and O (-2). You end up with 2 Al and 3 O because +32 = -23 = 6; 6 is the least common multiple of 2 and 3. You can do this with any binary ionic molecule.

@rch We could also have $Al_4O_6$ why always take the least common multiple? Also in case of $GaAs$ shouldn't we have $Ga_5As_3$.

Oh I see what you mean. Yeah, this sure does sound like something a high school teacher thought of to make this easier to teach.

What is the logic behind this method, how to prove this method?
The logic of the method is in fact trivial. Suppose you have to write a compound knowing his oxidation numbers :
$$\ce{xA^{a} + yB^{b} -> A_{x}B_{y}}$$
Where $a$ and $b$ are oxidation numbers  that have opposite sings,  $A$, $B$ chemical species and $x$, $y$ stoichiometric coefficient $ \in \mathbb{N}$.
We can start from this assumption: if a compound is neutral the algebraic sum of the oxidation numbers of his atoms must be zero ( point .4 of current IUPAC definition of the oxidation state) so we have to find when:
$$ ax+by=0 \space [1] $$
But we assumed that $a$ and $b$ have opposite sings so we can write:
$$ ax - by=0$$
This is a single equation with two variables so you can't solve it easily because it can have several solutions. However when $\textbf{x=b}$ and $\textbf{y=a}$ the equation is always solved, because:
$$ ab - ab=0$$
And so you can always write a compound formula that satisfy equation 1 using this formula:
$$\ce{xA^{a} + yB^{b} -> A_{b}B_{a}}$$
This works for $\ce{Al^{+3} , O^{-2}}$:
$$\ce{xAl^{+3} + yO^{-2} -> Al_{2}O_{3}}$$
 but what's about $\ce{Ti^{+4} , O^{-2}}$?
This turns out $\ce{Ti2O4}$ and not $\ce{TiO2}$ we need another step, we need to divide the subscripts by their greatest common factor gcf(a,b) because we have found another solution of the equation 1 with higher coefficient. In this case the gcf(2,4) is 2 and dividing the stoichiometric coefficient by two we get $\ce{TiO2}$.
Who, how and when developed this method?
I've found it the first time in Chemistry and Human Affairs by William Evans Price, George Howard Bruce, World Book Company, 1949. I don't know who is the inventor.
I think you can easily deduce the crisscross method simply looking at many compounds and their oxidation number! Is not so hard to imagine!
Is this method only applicable to ionic compounds?
No, it works for every compounds that can be describe with oxidation numbers. Oxidation numbers are the real charge in ionic compounds and a formal (or imaginary) charge in compounds with covalent bonds, look at $TiO_2$ example.
Why compounds like $GaAs$ do not follow this rule?
Is not right if we assume that we are dealing with $Ga(+3)$ and $As(-3)$ the equation is this:
$$\ce{xGa^{+3} + yAs^{-3} -> Ga_{3}As_{3}}$$
if we divide the subscripts by the gcf(3,3)  that is 3, we get $GaAs$. So this method will be always valid until you can describe the compound effectively with oxidation numbers, unfortunately not all the oxidation numbers are so easy to determine.

In fact I don't find the criss cross method useful I think is better to calculate the least common multiple of $a$ and $b$ and then divide it by $a$ for $x$ and $b$ for $y$.
I think this photo, from J. Chem. Educ., 2012, 89 (11), pp 1436–1438, summarize the best way to do it:

Edit:
The general solution of |a|x=|b|y  s.t.  x,y∈ℕ is x=n|a| and y=n|b| where n∈ℕ, that is we can use x=2|a|,3|a|...n|a| and similarly y=2|b|,3|b|...n|b| then why use x=|a| and y=|b| only?
(rearranged after the advices in the comment of mannaia)
In fact you can write $Na_{2}Cl_{2}$ without breaking any charge neutrality or valence rule. NaCl is a crystal with a lattice and you  can't consider $NaCl$ like a stand alone unit, what matter is the charge balance, the proportion between positive and negative ions. Maybe if you are dealing with a crystal made of 23456 atoms of $Na^{+}$ and 23456 atoms of $Cl^{-}$ you could  write $Na_{23456}Cl_{23456}$ this is in fact a solution of the equation and is not theoretically wrong however if you are talking about a generic crystal of NaCl you should write the ratio between $Na^{+}$ and $Cl^{-}$ ions it has no sense to choose randomly one of the solutions because has stated mannaia: "there's no need to choose a larger unit, whereas with a smaller one you can fully explain how the crystal is formed". Mathematically speaking we write the ratio to Lowest Terms.
Why always take the "gcf"?
In fact we don't take the "gcf" we divide for the "gcf" because this is the best way to get the ratio in the lowest terms.
Furthermore as explained in the $TiO_{2}$ example.
$$
A=\{(b,a),(2b,2a),(3b,3a),\ldots\}.
$$
Is a set of solutions of the equation but not all the solutions. If you have a=6 b=-4 you have the solution (2,3) too, here is the plot of the solutions in $\mathbb{N}^{2}$.

For compounds with covalent bond is different, because you are dealing with different stand alone units called molecule. For example if you have $C^{-4}$ and $H^{+1}$ the solutions not divide by gcf of the equation have no sense. For example $C_{2}H_{8}$ doesn't exist due the valence of the atoms involved.

"However when x=b and y=x the equation is always solved, because:": Did you mean y=b?
The correct equation is infact $|a|x=|b|y$. The general solution of  $|a|x=|b|y\ \  s.t.\ \  x,y \in \mathbb{N}$ is $x=n|a|$ and $y=|n|b|$ where $n \in \mathbb{N}$, that is we can use $x=2|a|,3|a|...n|a|$ and similarily $y=2|b|,3|b|...n|b|$ then why use $x=|a|$ and $y=|b|$ only? Why always take the "gcf"?
Note: In solid state $GaAs$, $As$ does not have -3 oxidation state. So the last part of your answer is technically incorrect.

@Anupam 1. yes i've correct the typo. 2. Yes! I've simplify it to reduce the length of my answer but if you want I will add it when I've more time to improve it 3. "I've said if we assume" in fact I don't like very much talk of oxidation number or charge for this compound however normally is arsenide has charge -3 see here and here.

@GM I wouldn't say that it's just a mathematical convention...where is the chemistry behind ? From the chemistry point of view, $\ce{Na2Cl2}$ is definitely not correct: because there's no need to choose a larger unit, whereas with a smaller one you can fully explain how the crystal is formed (3D repetition of the least unit, according to rigorous symmetry rules).

@mannaia I agree with you what I mean is that  NaCl is not it self a stand alone unit. And Na2Cl2 do not brake any stoichiometric or valence rule is only not the right way to describe it. However I will edit the answer to make it more clear thank for the comment.

Does this also work with compounds other than only ionic compounds? Yes - take a look at this : in water hydrogen has a ($\ce{+I}$) partially positive oxidation state and oxygen has a ($\ce{-II}$) partially negative oxidation state. So if you work out the rule then you get : $\ce{H2O1}$ (and you can leave the 1 away so you get $\ce{H2O}$).
Other examples include :
$\ce{HCl}$ :  ox.state (H) = $\ce{+I}$ and ox.state (Cl) = $\ce{-I}$   ... so... : H 1 Cl 1 ->remove ('1')-> HCl
This rule even works on certain, more complicated molecules :
$\ce{H2SO4}$ : ox.state(H) = $\ce{+I}$ and ox.state(SO4) = $\ce{-II}$ ..... so... : H 2 SO4 1 ->remove('1')-> H2SO4
$\ce{KMnO4}$ : ox.state(K) = $\ce{+I}$ and ox.state(MnO4) = $\ce{-I}$ .....so.... : K 1 MnO4 1 -> remove ('1') -> KMnO4
(of course all the charges mentioned above are partial(+/-) charges, because they are covalently bonded molecules)
And why the greatest common factor is taken... :
Take a look at calcium oxide. Here you have one calcium ($\ce{+II}$) ion and one oxygen ($\ce{-II}$) ion. When the rule is applied you get $\ce{Ca2O2}$. Now, you can divide both indexes by 2 to get $\ce{CaO}$. But why? Look at $\ce{CaO}$ : Here the one calcium atom has lost 2 electrons to achieve a stable noble gas configuration. And oxygen has gained 2 electrons to achieve a stable noble gas configuration. Both atoms are 'happy' and there is no need for another calcium atom or oxygen atom. So $\ce{CaO}$ is enough.
This is the bonding in $\ce{GaAs}$, I found this online: (source)

So the crisscross method wouldn't work with semiconducting materials. However it does (as mentioned above) work for certain normal molecules such as H2O, HCl,... these semiconducting materials just are exceptions to the rule...

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I think the contradiction you mention in your question is only apparent and is due to a misunderstanding around the word valency, which is by itself a rather ambiguous term (valency vs oxidation state).
According to Wikipedia, valency yields the number of bonds an element tends to have. For instance, oxygen tends to form $2$ chemical bonds: its valency is then $2$.
Valency is different from valence electrons: oxygen has $6$ valence electrons: a valence electron is an electron that can participate in the formation of a chemical bond.
If we consider arsenic, its valency is $3$ and not $5$, as you write in your question. $5$ is the number of valence electrons. Of course, this value of $3$ is based on the octet rule, that's just a rule (that often works) not a law (that should always work).
Considering $\ce{GaAs}$, $4$ bonds are actually formed between $\ce{As}$ and $\ce{Ga}$, as discussed in the text given by LievenB in his answer. One should more precisely say that each $\ce{As}$ atom is bonded to $4$ $\ce{Ga}$ atoms and each $\ce{Ga}$ atom is bonded to $4$ $\ce{As}$ atoms, in a diamond-like blende structure.  This is discussed in the book Fundamentals of Solid State Engineering of M. Razieghi, here below:

The text also says that $\ce{As}$ attracts more strongly electrons. It adds also that for the fourth bond, $\ce{As}$ contributes two electrons, while $\ce{Ga}$ contributes none.
Now, knowing that each $\ce{As}$ is bonded to $4$ $\ce{Ga}$ atoms. Let's consider a given $\ce{As}$ atom.

That $\ce{As}$ atom and one $\ce{Ga}$ will share $2$ electrons to form a covalent bond: these $2$ electrons are closer to $\ce{As}$: oxidation number will become $-1$.
The same $\ce{As}$ atom and a second $\ce{Ga}$ will share $2$ electrons to form a second covalent bond: these $2$ electrons are again closer to $\ce{As}$: oxidation number will become $-2$.
Again the same $\ce{As}$ atom and a third $\ce{Ga}$ will share $2$ electrons to form a third covalent bond: these $2$ electrons are again closer to $\ce{As}$: oxidation number will become $-3$.
The last bond, the fourth, will built up with the $2$ remaining valence electrons of our $\ce{As}$. Like the others, those $2$ electrons will stay closer to the $\ce{As}$: as a consequence, no change in  the oxidation will further occur, since those $2$ electrons are both coming from $\ce{As}$ and not from $\ce{Ga}$.

Such a reasoning would suggest that $\ce{As}$ has an oxidation number of $-3$ and $\ce{Ga}$ of $+3$ (again Wikipedia).
Now, coming back to the so-called CrissCross method: if you apply the definition you quote in your question:

suppose we have two elements $X$ and $Y$ having oxidation numbers x and y respectively. Can we prove that the compound formed will be $X_{y}Y_{x}$.

and you use oxydation number $+3$ for $\ce{Ga}$ and $-3$ for $\ce{As}$, the logic formula $\ce{GaAs}$ is finally obtained.
For other arsenides, this method will not be easy to apply, since the oxidation number may not be clear for the elements involved, especially if one has no idea about the structure by which atoms are organized.

Comments are not for extended discussion; this conversation has been moved to chat.


text:
What are the differences between the various household surfactants?
In a typical household, the following surfactants are found, among others:

Soap / handwash
Shower gel
Shampoo
Dishwashing agent
Washing powder / liquid

From my understanding, the role of all these is to lower the surface tension between water and fatty acids, enabling one to "wash" things.
Are they all the same then, and is the only difference between them marketing? Can I wash my hair with dishwashing agent, or pour shower gel into my washing machine?

Can I wash my hair with dishwashing agent [...]!

If it's an agent for dishwashing machines: Please DON'T!
Cleaning in dishwashing machines is performed in alkaline medium.
The tabs usually contain (among others):

alkaline compounds, such as sodium carbonate and sodium hydroxide
bleaching agents, such as sodium perborate and sodium percarbonate
pentasodium triphosphate (to keep calcium and magnesium in solution)

The solution of these tabs is not meant to get in contact with skin and eyes.

Exactly, as someone who has seen the result of a particular nasty interaction between a strong detergent and someone's eyes (was a first responder to it), I can definitely echo this answer (I will upvote when I can - hit the vote cap)


text:
What is LE (plant contents)?
I am reading the contents of the oatstraw:

.. high in vitamins A, C, B complex, LE, G..

I am unable to find on the Internet what LE refers to.

What brand is this?  It could be a way the company notates a particular chemical.

No idea, I just have access to the list of ingredients and unfortunately no more information. I have started thinking if they have missed the comma "," between L and E. Is there a vitamin L? Never heard of L.

"Vitamin L": http://micro.magnet.fsu.edu/vitamins/pages/anthranilic.html

No problem; that being said I have no clue what LE is supposed to be... Seems a bit odd to have a missing both a comma and a space in a nutrient facts sheet.

My answer will unfortunately have almost nothing to do with chemistry.
The quoted statement is BS, there's no vitamin LE.
This claim appears to appear on several web pages, the common source seems to be a post from an nutrition/herbal oriented mailing list from 1998:[1]

…
OATSTRAW
Oatstraw as any livestock breeder will confirm, builds the strongest possible
body, with the greatest resilience. Used extensively in European cultures
throughout time for health and beauty, our country has let this valuable herb
slide into disuse.
Contains: calcium. high in Vitamins A, C, B complex, LE, G, phosphorus,
potassium, proteins.
Oatstraw contains Avenin, an amorphous alkaloid which is highly nutritive to
cells, improving normal cellular reproduction.
…

(also note e.g. that avenin is not an alkaloid but protein)
I have an evolution theory how the author could have made this typo, when (hopefully) compiling a literature:

… vitamins A, B complex, C, E, …
… vitamins A, B complex, C E, …
… vitamins A, B complex, CE, …
… vitamins A, B complex, LE, …
… vitamins A, C, B complex, LE, …

(Some intermediate forms might be purely hypothetical. An intermediate form “… vitamins A, B complex, C E, …” can be seen alive at the page mentioning oatstraw, at a different product though[2])
References:

http://www.geocities.ws/chdigest/herbs/pages/apr1998
https://www.healthylife-solutions.com/blog/2017/5/15/womens-herbal-tea-recipe

That's indeed quite plausible and may serve a great illustration to why choosing a good typeface is important (also, congrats on 4000 rep. :) ).


text:
Conversion of units for concentration: mol/l to ppm
The concentration of an aqueous solution of $\ce{Mg(SO4)2}$ is  $0.001\, \textrm{mol} \cdot \textrm{dm}^{-3}$.
Find out the concentration of solution and the concentation of $\ce{Mg^{2+}}$ ion in ppm.
How to do this? How to find moles in ppm?
If anyone could clearly describe, it would be great.

Convert mol-dm^2 to grams/volume.  A decimeter is 10 cm.  A dm^3 is then a liter.  Water has density ~1 g/cm^3.  Multiply the g(Mg)/g(water) by a million.


text:
Symmetry of an in-phase and an out-of-phase wavefunction "approaching" each other in an MO
Hiho.
I have a question concerning this image (Chemistry, Catherine E. Housecroft et al., Pearson Education, 2010).

In the text, it states the following:

"Each of the new molecular wavefunctions, $\Psi$(in-phase) and $\Psi$(out-of-phase), has an energy associated with it wich is dependent of the distance apart of the two hydrogen nuclei. As Figure 4.13 [provided image] shows, the energy of $\Psi$(in-phase) is always lower than $\Psi$(out-of-phase)."

The in-phase-function has a stabilized minimum, but the out-of-phase-function seems stable only at an infinite distance. For my intuition, the two graphs should be symmetric, since the two wavefunctions are alike and approach each other equally. Why do they behave like this?
Why are those functions not symmetric?
A following question would be, when I combine those two functions, I'll get a minimum at the x-value I marked with "minimum". Would this be the distance at which a nucleus in a stable diatomic molecule would be found like in this figure?

===
Thank you for your help and have a nice day :)

Yes.  As per the axis label on the chart, 'x' is the optimal inter-nuclear distance for a wave function in-phase.  Note that when considering a diatomic, the wave function is represented by a linear combination of atomic orbitals to form molecular orbitals, not just two 1s orbitals.

OK, my second question is answered. But then again, how does it come, the two functions are not symmetric?
PS: Yes, thanks for the hint; in this particular example, it's H2.

Considering this diatomic molecule, the total wavefunction $\Psi$ is formed via a linear combination of the atomic orbitals $\psi_1$ and $\psi_2$ (which is an approximation, but not a too bad one).
$$ \Psi^+ = c_1 \psi_1 + c_2 \psi_2 $$
$$ \Psi^- = c_1 \psi_1 - c_2 \psi_2 $$
The (electronic) energy of the system can be described using the following (stationary) Schrödinger equation:
$$ H_\text{el} \Psi_\text{el} = E_\text{el} \Psi_\text{el}$$
So somehow this all factors into each other and in the end gives you the curves (for the potential energy of the system vs. internuclear distance) that you posted. Where is the symmetry? Well, it is still there, with respect to the center of one atom. If you imagine rotating the functions around the $y$ axis, you will get a symmetric image. And you're allowed to do this because space expands in all three directions. So in theory (if you follow through and expand the function to everywhere) you get a 3D image of the energy at every point around one atom.

But: The wavefunction in itself is not all that interesting to us, because there is no "real" physical information that can be directly extracted from it. But fret not, we can at least get out the electron probability density by complex-squaring it:
$$ \rho = |\Psi|^2 = \Psi \Psi^* $$
And this is where those nice small bubbles come into play that we see in your second figure. And we have again found the symmetry that you seek, because both linear combinations have $D_{\infty h}$ symmetry (I think... I'm not all too good with point groups and the such).


text:
What is the range of sizes for molecular systems?
Molecular systems - are in a simple count systems made up of molecules. But almost every system, surrounding us are molecular than. Considering the molecular systems in terms of chemistry compels us, firstly, impose dimensional constraints on the molecular system (which?), molecular system could not be macroscopic (How big?), otherwise you'll have to talk about it as a solid body; secondly, energies for such systems are limited (<1 eV); in third place we will be interested in properties of such system as whole, but not as sum of its parts properties.
So the main question is raised: What is the range of sizes for molecular systems?

Well apportioned, mixed, and cured epoxy glue is a single covalent molecule.  Single crystal white diamond is a single covalently-bonded molecule.  On the small end, a dihydrogen molecule, is 0.7414 Å (ortho or para, your choice to trim the decimal).
The largest human chromosome is ~220 million base pairs, or about 1.5x10^11 g/mol.

There is a confusion here: "molecular system" is just a figure of speech, and there is no hard line between molecular systems and e.g. macroscopic solid bodies. It is not strictly about size of bonding type. If you use the definition of molecules as entities where the atoms are connected with covalent bonds then you will end up indeed with molecules like a chromosome, or a piece of silicon single crystal.
In general, molecular systems are just a way to speak about systems which are smaller then lets say 5-10 nm, orbital picture is good to describe physical / chemical properties. The things that chemist are comfortable with. Polymers, DNA, proteins are treated generally as large molecules, because generally we look at smaller units where the molecular approach is useful. No one cares about the bangap of PE.
Solids are generally systems where the band picture is a much more useful description, and generally needs to be 100s of nanometers. A piece of silicon can be discussed a a molecule, yet, people are generally much more interested on questions like band structure, dopant effect etc, where the solid state approach is much more useful.
In between: we call them nanostructures, supramolecular systems, colloids, depending on who speaks about them and how to approach them. Again, it is not necessarily about the exact size, but what kind of theory you try to apply. There are gold nanoparticles of 1-3 nm size, which is much smaller then most of the enzymes and proteins. There are even nanoparticles with fixed number of atoms, i.e. well defined stoichiometry! Yet, I heard no one calling a gold nanoparticle molecular system, just because we prefer to look at them as a small piece of metal instead of a large molecule. If you prepare a silicon particle that size, most probalbly people will categorise it as nanosystem, and not as a large molecule, in spite the fact that the atoms are connected with covalent bonds. On the other hand, we prefer to call protein complexes to supramolecular systems, because the secondary interactions between building blocks is the defining feature we care about.
Half-seriously: if you don't know if is is a molecular system, ask yourself, who would publish about this? If it is in the field of physics, than it is solid mater, maybe nano-system. If it sounds more like chemistry, then is it a molecular system, a colloid system (if it is something old and classic), or a supramolecular system.


text:
What are some good examples of rate equations for a math class?
I'm a mathematician who's currently teaching a course on differential equations.  Though I don't know much about chemistry, I like to include examples from chemistry in my course, and I prefer for the details to be accurate.  Here is a typical exam problem:

When a container of gaseous nitrogen dioxide is heated above 150 °C, the gas begins to decompose into oxygen and nitric oxide:
  $$\ce{2NO2 -> O2 + 2NO}$$
  The rate of this reaction is determined by the equation
  $$\frac{\mathrm d\left[\ce{NO2}\right]}{\mathrm dt} = -k\left[\ce{NO}_2\right]^2$$
  where $k$ is a constant.
(a) Find the general solution to the above equation.
(b) A large container holds 50.0 moles of $\ce{NO2}$ at a constant temperature of 600 °C.  After one hour, only 34.3 moles remain.  How much $\ce{NO2}$ will there be after another hour?

So my questions are:

Is the science in this problem reasonably accurate?  Is there anything you would change?  (I looked up a suitable value of the rate constant $k$ to make sure that the time in part b was reasonable.)
What are some other examples of reactions that are governed by simple rate laws?  Ideally, I'd like to have several examples each of reactions governed by the equations
$$
\frac{\mathrm dy}{\mathrm dt} = -ky,\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^2,\qquad\text{and}\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^3.
$$
(Is $y^3$ really possible?  Are non-integer powers of $y$ possible?)

Check out this reference I found to answer another question: http://science.widener.edu/~svanbram/chem146/ch15/lm_c15_graphs.pdf It's probably well below the level that you are looking for, but it will give you an idea of how the kinetics can become more and more complicated as the number of reactants increases.

This is that question, if you have a better demo or find an egregious error in mine, please leave another answer :)

Since it was somehow forgotten in the joy over the nice question: Welcome to Chemistry.SE! (And yes, the idea to use real life examples is nice!)

Take a look to the wonderful iodine clock reactions I think you could find interesting this paper for the ultimate example!

Here's a somewhat relevant discussion

How thoughtful of you to include chemistry in your differential equations course! We appreciate your effort, especially going the extra mile to make it realistic.
I hope you've seen the rate equation page on Wikipedia as it contains a good deal of the mathematics in several examples of reactions. You should also be interested in the reaction order page (you can find examples here of reactions with fraction or negative order with respect to some reagents).
Let me do a quick recap on basic chemical kinetics theory. Consider the general reaction equation:
$$\ce{a\ A + b\ B + c\ C + d\ D + ... \longrightarrow w\ W + x\ X + y\ Y + z\ Z + ...}$$
, where uppercase letters indicate different molecules in a gas or solution, and lowercase letters indicate the stoichiometric coefficients (negative for the reactants, positive for the products). Naively, we can assume the reaction happens because, while the molecules are jostling about, at some point all of the reactants ($a$ molecules of $\ce{A}$, $b$ molecules of $\ce{B}$, $c$ molecules of $\ce{C}$, etc.) will bump into each other all at the same time and undergo reaction. If you think of collisions as independent events with a probability of happening proportional to the concentration of each species, then it isn't too hard to understand that the frequency with which reactions happens is proportional to $[A]^a[B]^b[C]^c[D]^d...$, and so the rate equation for the reaction would be:
$$r=k[A]^a[B]^b[C]^c[D]^d...$$
, where r is the reaction rate and k is a proportionality constant. The consumption or production rate for any species $\Gamma$ with stoichiometric coefficient $\gamma$ is trivially related to the reaction rate by:
$$\frac{d\Gamma}{dt}=\gamma r$$
However, experimentally things aren't so simple (fortunately for nature, unfortunately for our minds). A reaction that requires the simultaneous collision of many molecules would be highly unlikely, meaning that it would happen very, very slowly. Even so, experimentally we observe many reactions that involve a large amount of molecules. For example, the combustion of cyclohexane in air is formally given by the equation:
$$\ce{C6H12 + 9 O2 → 6CO2 + 6 H2O}$$
If the reaction really required ten molecules to bump into eachother at the same time with sufficient energy and in the right geometry, then this reaction probably could not happen in air, as oxygen would be too rarefied to compensate the extremely low proportionality constant k for a single-step reaction. In actuality, a gaseous mix of cyclohexane and air can react so fast as to cause an explosion, converting all the reactants into products in a miniscule fraction of a second. Clearly something is wrong.
It turns out the assumption that reactions occur in one step is incorrect in general. Usually, there is more than one step involved, and indeed there generally is more than one path from reactants to products. The speed at which a reaction occurs is really described by considering all paths at once, with all their steps, and adding their contributions (like a start and end point in a complex tree diagram). In general this is pretty difficult and unnecessarily complex. It is often a reasonable approximation to model the reaction by selecting only the fastest route from reactants to products, and the reaction rate is bottlenecked by the slowest step of the fastest route, which becomes responsible for determining the rate equation.
Now, since the reaction rate depends on selecting the fastest route, you shouldn't expect any of the steps in it to contain some very huge bottleneck, such as requiring six molecules to react simultaneously; there would likely be a slightly different route, perhaps with more steps, but in which each step doesn't involve so many molecules interacting at once, and hence being a faster route. Because of this, it turns out that most of the fastest reaction pathways involve slowest steps that very rarely depend on more than two molecules at once. Most reactions which can be modeled simply like this therefore have rate equations of the type $r=k[A]$, $r=k[A]^2$, $r=k[A][B]$ or even $r=k$. Few reactions rates are of the third order, of the type  $r=k[A]^3$,  $r=k[A]^2[B]$,  $r=k[A][B][C]$ or similar. I don't know of any reaction that goes as $r=k[A]^3$ specifically. Fourth order reaction rates are so rare that they are the focus of research when found. I don't expect anyone to know a fifth order reaction.
At the end of the day, kinetic theory is hard enough that reaction kinetics end up being determined simply by parameter fitting from experiments. Even if the underlying reaction mechanism is not completely understood, if a fit of the type $r=k[A]^{3/2}[B]^{-1}$ happens to be good, then so be it. Reactions which have been studied in depth are significantly more complex, such as the misleading simplicity of the reaction between hydrogen and bromine.
Edit: I don't know why I didn't search for a fifth order reaction rate! Turns out some exist, such as this one. The reaction rate is of the type $r=k[A][B]^4$, so you could have an example where $\frac{d[B]}{dt}=k[B]^4$.

Thank you -- this is very helpful for my overall understanding.  In particular, I had been wondering why the order of a reaction is only sometimes equal to the number of reactants.

The Wikipedia pages that you mentioned are also quite helpful.  The "rate equation" page lists three reactions (decompositions of hydrogen peroxide, sulfuryl chloride, and dinitrogen pentoxide) that are first order, and which I ought to be able to use.  However, the only second-order reaction listed is the decomposition of nitrogen dioxide.  Do you happen to know any other simple examples of second-order reactions?

By the way, the link you gave on the hydrogen bromine reaction (with a 3/2 exponent for [$\mathrm{Br}_2$] at low $\mathrm{Br}_2$ concentrations) is quite helpful.  I will certainly be using this example!

I imagine you want a reaction that simply has at least one reagent with a second-order dependence (so $r=k[A]^2$ would work just as well as $r=k[A]^2[B]$)? I'm actually having a bit of a hard time finding a simple example of the former, even with @jlandercy's excellent source. Perhaps someone else has an example they could share?

It's true that the former would be better, but the latter ($r=k[A]^2[B]$) would be fine as long as I could describe in words a realistic situation for which $[B]$ is much larger than $[A]$, and hence essentially constant.

I know this was quite a while ago, but an example of $r = k[A]^2$ would be some sort of dimerization reaction such as $BH_3$ dimerizing to form $B_2H_6$. :) https://classesv2.yale.edu/wiki/site/chem124_f08/bh3_dimerization.html

I have no example in mind but fractional and high partial order (exponent) are possible. When this kind of relationship appears in chemistry they often result from complex reaction schema where more than one steps are required.
Anyway have look to Kinetic NIST Database, you will find many reaction kinetic constants and models.

Thanks.  The NIST database seems to be much more useful than my old method (Google Books) for actually looking up the reaction constants.


text:
Williamson ether synthesis
Williamson ether synthesis is conducted using vanillin, $\ce{KOH}$, $\ce{EtOH}$, water and benzyl chloride.

Why is water needed for the synthesis?
Why must the solution be refluxed for an hour to gain the desired product?
Why is water added after reflux to create a cloudy solution?

1.

Consider the phases for potassium hydroxide and benzyl chloride at room temperature.
Why might you not want to use ethanol as the solvent to form the benzyl alkoxide of vanillin in?
(As an aside, think about why you might not want to use water over, say, acetonitrile as a solvent for this reaction.)

2.

What governs the rate of the Williamson ether synthesis?
Why would you want to heat a chemical system you want to have a reaction in?

3.

What solvents would the product ether dissolve in?
Which by-product(s) would dissolve in water but not ethanol?
How would you remove the product ether from the solvent mixture?

I'm trying not to give away the answers directly but these bullet points should help you work out each point.


text:
Phase transition from gas to solid
Question 13 from the ACS chemistry olympiad exam:

The phase transition from gas to solid is called
  \begin{align}
&\text{(A) condensation.} & \text{(B) evaporation.} \\
&\text{(C) polymerization.} & \text{(D) sublimation.}
\end{align}

I am pretty sure the answer is deposition, which is not listed, but yet the answer is (A) condensation [scroll to the end of the question paper to see the answer key].
Did the chemistry olympiad make a mistake, or is there something obvious that I am missing?

I guess the diagram of phase changes you have in mind looks like this:

Indeed this says deposition is the right name for a gas-to-solid phase transition and I believe that this is the only correct term.
The only thing I can argue why it is obvious here that condensation is the right answer (at least the closest) is that this is the only answer that gives a transition from fluid to solid. B and D are the other way around, and C is not really a phase change (although it could result in one).

Good point concerning the condensation as the only "real" phase change in a classical sense.


text:
Changes in which factors affect both the rate and the rate constant of a first order reaction?

Changes in which factors affect both the rate and the rate
  constant of a first order reaction?
I. temperature
II. concentration
(A) I only (B) II only
  (C) Both I and II (D) Neither I nor II

I would think that the answer would be (C), both temperature and concentration.
Concentration would obviously affect it (from the rate expression), and the temperature would also affect it because the Arrhenius equation for the rate constant is $$
k = A e^{-E_a/(R T)}
$$
and thus temperature would have an influence on the rate constant, and since the rate constant is part of the rate equation, it would also affect the rate of the reaction itself.
However, the given answer is

A

I don't understand why this is.

If you see the rate equation for the Arrhenius equation:
$$
k = A e^{-E_a/(R T)}
$$
you can see that the rate constant increases for an increase in temperature (and as activation energy increases, the rate constant decreases.
Now, since the rate constant increases, this implies that the rate of reaction increases with temperature.
Regarding concentration: you are correct in that concentration affects the rate of reaction. However, a rate constant does not change according to concentration. You can observe this in the Arrhenius equation, where A is independent of the concentration of the substrate. The fact that the rate increases does not imply the rate constant changes.
Possibly more clearly: it is implied that increasing the rate constant will increase the rate of reaction. However, increasing the rate does not increase the rate constant.
An increase in temperature increases the rate constant and hence the rate.
An increase in concentration increases the rate but not the rate constant.

the rate of a reaction depends on the rate constant and concentration R=k[A][B]..
the rate constant according to Arrhenius is as u stated. from ur equation u can see that concentration doesnt affect k but affects R. Temperature affects k and k affects R so, temperature affects both while concentration affects only Rate of reactions.

It is necessary to distinguish between rate and rate constant (also called rate coefficient). The rate of a second order reaction A+B$\rightarrow$C is expressed as   $$d[A]/dt = -k_2[A][B]$$ where $k_2$ is the rate constant. (The negative sign is because A is consumed in the reaction). In the case of a first order reaction, e.g. a cis-trans isomerisation, the rate is $$d[A]/dt=-k_1[A]$$  and the rate constant $k_1$. As you can see from the Arrhenius equation the rate constant ($k_1, k_2$) depends on temperature and the rate depends on concentration of both A and B and the rate constant. So temperature affects rate and rate constant, and concentration only affects rate. So the answer is (A)

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


text:
What are the products of the thermal decomposition of potassium cyanate with carbon at elevated temperature?

What are the products of the thermal decomposition of potassium cyanate ($\ce{KOCN}$) with carbon ($\ce{C}$) at elevated temps ($\ce{300^\circ C - 1300^\circ C}$)?

I guess some of the nitrogen is lost and forms something. I'm trying to find out the full equation, but a google search brings up nothing concerning reduction of potassium cyanate with carbon.

[...] some of the nitrogen is lost and forms something [...]

Carbon is a good and cheap reductant, often used in the processing of oxidic ores - except when metal carbides are formed. Carbon dioxide a reasonable product in these reactions.
$\ce{2KOCN + C ->[\Delta] 2 KCN + CO2}$
Let's agree that potassium cyanate is reduced to potassium cyanide, while carbon dioxide is released.

wouldn't it be CO and not C02?

I found this: thermal decomposition of KCNO always causes some loss of nitrogen and leads to products that are somewhere from 1/4 to 1/3 K2CO3, the balance being KCN. So could the products be some K2C03 and also carbon monoxide?

You did not ask for the decomposition of $\ce{KOCN}$ in the first, but for its reaction with carbon at elevated temperature.


text:
How well known is the formula of carbonic acid?
My chemistry teacher in high school didn't know the formula for carbonic acid, which struck me as something someone with a degree in chemistry (as they allegedly do) should know instantly, or at least after some time thinking. However, to give them the benefit of the doubt - is carbonic acid a relativly common chemical, or alternatively is there a well known way to derive its formula from the name (like how you can work out the formula for a molecule called something somethingelseide for instance)? Should someone with a degree in chemistry know it off the top of their head?

There are rules for naming acids, one of which is that any anion ending in -ate changes to an -ic when it becomes an acid. So for example, carbonic acid would have the anion as carbonate. Carbonate should be relatively well known as CO3, and since it has a 2- charge there must be 2 hydrogens to make it neutral, thus making the formula H2CO3.

Just a side note - not all chemistry teachers have chemistry degrees.

@Helios This teacher has told us that she has.

Carbonic acid has the formula $\ce{H2CO3}$ - you can't always immediately derive the formula from the (trivial) name - you have to learn it. And - yes - carbonic acid is a relatively common chemical : you can find it in all types of fizzy drinks (fanta, cola,sprite,....)
However, you can derive the formula from the other name (systematic) of carbonic acid, dihydrogen carbonate =  $\ce{H2CO3}$.

Actually, you can indeed derive the formula from the name. See my comment to the the original post.

@user2612743, that's correct:) - in the case of carbonic acid that is a good solution

If there is open water and air, there is carbonic acid; carbonated beveradges including beer and champagne; and you are filled with it from aerobic respiration.  pH vs. pKa determines ionization.  (The leading lowercase "p" means "negative logarithm of.")
Naming simple acids:  The root is the element- bor(on), car(bon), nitr(ogen), sulfur, chlor(ine) phosphor(us).  Suffix shows oxidation state, ous/ite (low) and ic/ate high.  Prefix intensifies suffix.
Hydrochloric, Cl(-1) ("hydro" in context means "no oxygen") chlorides
Hypochlorous, Cl(+1) hypochlorites
chlorous, Cl(+3) chlorites
chloric, Cl(+3) chlorates
perchloric, Cl(+5) perchlorates
Simple names become not simple when complex structures form from concatination (peroxysulfuric acid,polyphosphoric acid), element substitution (thiosulfuric acid), substituents (methanesulfonic acid, chlorosulfonic acid), and general diddling around (adenosine triphosphate).
http://en.wikipedia.org/wiki/Oxyanion


text:
Rate and order confusion
I just learned about rate and order... I just want to confirm if these three ideas are correct:

If the first elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be EQUAL to the order the concentration of the reactants in the rate equation is raised to?
e.g.
$$\begin{align}
\ce{2A + B &-> C} &(\text{slow})\\
\ce{C + B &-> D} &(\text{fast})
\end{align}$$

Will the order with respect to $[\ce{A}]$ be 2 and $[\ce{B}]$ be 1 if this proposed mechanism is correct?

If the balanced equation is the ONLY elementary step in the reaction mechanism, the stoichiometric coefficients of the reactants will be the same as the order the concentration of the reactants are raised in the rate equation
If the second elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be different to the order the concentration of the reactants in the rate equation is raised to?

Could this answer be of any use to you?

@Nicolau Saker Neto: This answer really helped. However, I am still unsure if the  stoichiometric coefficients of the reactants of the first elementary step equal to the order in the rate equation if the first step is the slowest step

Although Nicolau Saker Neto's answer provides the essential, I wish here to propose two emblematic examples for the situations cited in your question:
First case: two steps process, the first being the slowest.
Reaction between $\ce{H2O2}$ and $\ce{HBr}$:
\begin{align}
\ce{H2O2 + H+ +Br- &-> HOBr + H2O}  && \text{slow}\\
\ce{HOBr + H+ +Br- &-> Br2 + H2O}   && \text{fast}\\
\end{align}
The rate of this reaction, determined by the slow step, equals:
$$
\frac{d[\ce{Br2}]}{dt}=k_{1}\cdot[\ce{H2O2}]\cdot[\ce{H+}]\cdot[\ce{Br-}]
$$
This is therefore a third order reaction.  Stoichiometric coefficients of the slow elementary step yield the specific order in each of the three reacting species.
Second case: two steps process, the first being the fastest.
Reaction between $\ce{NO}$ and $\ce{O2}$:
The initial equilibrium is rapidly established:
\begin{align}
\ce{2NO &<=> N2O2}      && \text{fast}\\
\ce{N2O2 + O2 &-> 2NO2} && \text{slow}\\
\end{align}
In this case, the rate is:
$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot[\ce{N2O2}]\cdot[\ce{O2}]
$$
This is therefore a second order reaction, with specific orders coinciding, as usual, with the stoichiometric coefficients in the slow step. Using the equilibrium constant $K$ of the first step, one can finally express the rate of the global equation as a function of $[\ce{NO}]$ and $[\ce{O2}]$, say:
$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot K \cdot[\ce{NO}]^{2}\cdot[\ce{O2}]
$$

Summarizing, as you correctly say, the stoichiometric coefficient of a given reactant, in the rate determining elementary step ("slow" step), directly yields the order by which the concentration of that reactant is raised in the rate equation.

Just to notice that the global reaction $\ce{2NO + O2 = 2NO_2}$ is known to be a third order reaction (very rare). Another specificity, its kinetic constant decrease with temperature (this is why it is modelled by a negative activation energy). See NIST Kinetic Database for detailed data and article references.


text:
Why is phenolphthalein an appropriate indicator for titration of a strong acid with a strong base?
How does the pH at the equivalence point, as well as the pH range over which the colour of phenolphthalein changes, make it a suitable indicator for titrations between strong acids and strong bases?

If you look at the titration curve, which plots the volume of base added vs pH (source):

you can see that the equivalence point occurs at pH = 7.
Phenolphthalein is fuchsia in pH's roughly between 8.2 and 12, and is colorless below pH 8.2. When the number of moles of added base is equal to the number of moles of added acid (or vice versa; example valid for strong monoprotic acids/bases assuming 100% dissolution), the pH is equal to 7.
You might say, if the pH needed is 7, and phenonphthalein changes at pH's around 8.2, how can you use this indicator?
Well, again looking at the curve, from pH = 11 to about pH = 4, pH changes very rapidly with from an infinitesimally small change in the volume of base added. Since one drop of added titrant will cause this large change, even though the change in color of phenolphthalein does not occur right on the equivalence point, it is within approximately one drop. This kind of uncertainty is "acceptable uncertainty" in using titration to volumetrically determine concentrations.
To clarify what I mean by "acceptable uncertainty", you should realise that each of your measurements has some kind of uncertainty to them:

When you weighed out the primary standard to titrate against, was the balance perfect?
Was your solution made up precisely to the graduation in the volumetric flask?
Did you pipette the exact volume of the aliquot or were you off by a drop or two?
Are you able to state the volume added from the burette to arbitrary precision or is there some uncertainty beyond the two decimal places given by the graduated lines?

In the scheme of things, ~1-2 drops will not be a significant factor in getting an accurate result, but you should most definitely acknowledge that there is uncertainty in your answer.

This is true for a strong acid and a strong base that have no buffering capacity.  It will, of course, not be nearly as clean when one more drop of titrant is pH-buffered.

@UncleAl What do you mean by strong acids/bases with buffering capacity? Do you mean acids like H2SO4?


text:
Entropy of solid, liquid, and gas at triple point of water

At the triple point of water how do the entropies of solid,
  liquid, and gas compare?

I think that either they will be equal or it will be that gas > liquid > solid.
I don't know if entropy is influenced by the fact that it will be at the triple point. Can someone please elaborate?

Triple point defines a situation of simultaneous equilibrium between the solid, the liquid and the gas phases.
For such an equilibrium, you simply write:
$$
\Delta S= \frac{\Delta H}{T}
$$
that rises from the fact that $\Delta G=0$.
Considering water molar ($m$) enthalpies for each phase transition at $273\,K$:
$$
\Delta H_{melting,m}=6.01\,kJ\,mol^{-1}
$$
$$
\Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1}
$$
$$
\Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1}
$$
one realises that

the entropy of the gas phase is higher than the entropy of the liquid phase.
the entropy of the liquid phase is higher than the entropy of the solid phase

In this respect the triple point has no peculiar behaviour compared to other points where a two-phase equilibrium is established.


text:
Decreasing rate of reaction between I2(s) and H2(g)?

Which change will decrease the rate of the reaction  between
  $\ce{I2(s)}$ and $\ce{H2(g)}$?
(A) Increasing the partial pressure of $\ce{H2(g)}$
(B) Adding the $\ce{I2(s)}$ as one piece rather than as several small ones
(C) Heating the reaction mixture
(D) Adding a catalyst for the reaction

I can eliminate D as a catalyst would speed up the reaction.
I am not sure about the others though. I could look up the enthalpy for this reaction and eliminate C, but this is from the Chemistry Olympiad and I can't look up enthalpies during the test. Is there any other way to determine that this reaction is endothermic?
Once you eliminate C, you are left with A and B. How would you go about finding the right answer?

HINT 1: If you increase pressure in a fixed volume, what happens to temperature? HINT 2: What does surface area do to reaction rate?

By guy lussac's law increasing pressure will increase temperature. How would you determine if the equation is endothermic or exothermic though? Greater surface area results in faster reaction speeds. However, doesn't the larger piece have a greater surface area than the smaller pieces?

Regarding the first hint, after increasing the partial pressure, even if the temperature is allowed to decrease to its previous value, an effect on the reaction rate is still observed. Why could this be?

"Greater surface area results in faster reaction speeds. However, doesn't the larger piece have a greater surface area than the smaller pieces?" Take a 1" cube and measure its surface area. Now cut the cube into 8 1/2" cubes and measure their total surface area. Note that the latter includes all the prior surface area PLUS new surface area from the cuts. We conclude?


text:
Are all pure alkanes, alkenes and alkynes tasteless, odorless and colorless?
Non-pure alkanes have a taste.
Are all pure hydrocarbons tasteless, colorless and odorless? Are there hydrocarbons that do have a taste, color and/or odor?
If so, what makes these ones different, even though they are only made of carbon and hydrogen?

There are numerous hydrocarbons which are neither odorless nor tasteless. Some are moderately complex and exhibit chirality, such as pinene and limonene. However, even simpler achiral ones can, such as benzene and its various derivatives (the name "aromatic hydrocarbon" is not coincidental).

I think this is a good question maybe you could write the body better because it seems is going to be closed "for length and content".

Why would a mixture of two odourless alkanes have an odour?
As already pointed out by Greg E and Uncle Al, numerous hydrocarbons without any (heteroatom-based) functional groups do show characteristic smells.
Naphthalene is always remembered for the typical smell of grandma's mothballs.
More pleasant smells are found in the group of monoterpenes, i.e. in those molecules with the sum formula $\ce{C10H16}$ that are derived from the combination of two isoprene units.

These volatile compounds were/are used as fragrances for perfumes. Typical examples are myrcene (1), $\alpha$-terpinene (2), limonene (3), $\alpha$-pinene (4), $\beta$-pinene (5), $\alpha$-thujene (6), sabinene (7), and camphene (8).

As far as the colour of hydrocarbons is concerned:

A sufficiently conjugated system is far from being colourless, take lycopene (9), the deep red coloured dye in tomatoes as an example.

Adamantane smells of camphor.  n-Nonane is in the scent of some roses, see here.

This page says that it's odourless I think that the non-pure one smells of camphor and the pure one is odourless ..
http://chemyq.com/En/xz/xz1/4626inphf.htm

@user37421 "This page" is a sloppy translation from Chinese to English with a garbled CAS no. In contrast  to the article in a peer-reviewed journal, that Uncle Al linked in his answer, it is very unreliable.


text:
Which factors best account for the increase in boiling point?

The boiling points of $\ce{CH3COCH3}$, $\ce{CH3COC2H5}$, and  $\ce{CH3COC3H7}$ are $56˚C$,
  $80 ˚C$, and $102 ˚C$, respectively.  What factor(s) best account(s) for this?

I believe london dispersion forces is one of them. Is dipole-dipole interactions another one? It seems to me that there are dipole-dipole interactions since none of the structures of symmetrical, so can they account for this increase in boiling point?

You're definitely right on London. Between the three molecules, the only thing that's changing is the carbon chain on the right hand side of the Oxygen. That plus the corresponding increase in boiling temperatures together form a good indicator that London forces are at play here.
Dipole-dipole (or Hydrogen bonding for that matter) are non-significant in this case, because you're not changing the electronegativity of anything here, only adding more of the same repeating atoms.
Source: This was on the ACS 2011 local exam, question 55.
Further information here


text:
Why do only unsaturated hydrocarbons undergo addition reactions?
Why do only unsaturated hydrocarbons undergo addition reactions? For example, ethene or ethyne will undergo an addition reaction with chlorine, whereas ethane will not.
Furthermore, why does benzene not undergo addition reactions, even though it is not saturated?

To what would you add given a saturated hydrocarbon? You must substitute, abstract, or displace.
Benzene is aromatic — different pattern of reactivity.  It photochlorinates.  It UV reacts with maleic anhydride, then adds another mole thermally.  Palladium catalysis will swap appropriate substituents.


text:
How easy it is for MoS2 to become oxidized
I made I high pressurized $\ce{MoS2}$ pellet, using a 6 ton pressurizer. I used very thin powder to create the pellet.
I am using this pellet on beam, with some nuclear techniques. In my spectra I can see oxygen.
How easy it is for $\ce{MoS2}$ to become oxidized?

I think you have to use XPS to determine if the Oxygen is from adsorption or oxidation.

It seems that the problem has been examined even 50 years ago, probably in order to determine if/how oxidation would alter the the properties of $\ce{MoS2}$ as a lubricant.
Here is an old NASA document from 1949, and in Surface Oxidation of Molybdenum Disulfide from 1955, the authors state in the abstract, that the oxidation is limited to the outmost surface layer, which then passivates and protects the rest of the batch.

Just as for almost every compound of d-elements, $MoS_2$ tends to adsorb molecules from gases on its surface. This is especially true for fine powders, that have high surface-per-gram. MoS2 is especially troublesome, because in strongly reductive atmosphere it is reduced to metallic molybdenum. So, ideally, to get a pure MoS2 you have to

threat powder with $H_2S$ at about 300-400 Celsium before use
press powder into pellet under argon
not allow it to touch air for any significant amount of time (more that several seconds)

In general a very brief exposure to air may significantly oxidize $MoS_2$ catalysts, so you can safely assume that if your $MoS_2$ touched air, it contains oxygen, especially if it is in fine powder.


text:
Why are NMR typically taken for a power of 2 scans?
Why do we typically take $n^2$ scans, even in the basic NMR experiment? I've heard various explanations, including that it helps the fast Fourier transform, and would like to know the real reason.

If you mean why do we process raw NMR data using $n^2$ data points, then @Klaus's answer is correct. Although, it is actually usually possible to record an arbitrary number of time domain data points; it is just the processing algorithm is limited to how many data points it can handle. You will either lose the extra time domain data (if using less frequency domain data points) or add zero filling (if using more frequency domain points).
If you mean the number of scans as the number of transients added together to give the final raw FID, then this comes about from the requirement for phase cycling a pulse sequence in order to remove unwanted NMR signals that may arise from

imperfections in hardware phase (giving rise to quad images)
coherent noise
generation of undesired magnetization arising in multi-pulse experiments

Here, though, the number of scans is not limited to $n^2$. Historically, the number of scans for a 1D spectrum would typically have been confined to a multiple of 8, as a phase cycle requirement. As you start to need to run greater numbers of scans due to sensitivity issues, then you would need to consider the accumulation of noise as well as signal, as @ron has alluded to (doubling the number of scans only give a $\sqrt{2}$ improvement in S/N; you need to run 4 times as many scans to double S/N). It is through this combined relationship that the number of scans, then, adheres to a $n^2$ increase. If 8 scans (required by phase cycling) didn't give sufficient S/N, then we need to double or quadruple the number of scans to see a real improvement in S/N. And so of for ns=16 or 32 or 64.
For more complicated experiments, phase cycling may have required multiples of 16 or more. I remember running a dqf-COSY experiment which had a phase cycle of 24, so the number of scans had to be a multiple of 24. Yes, these experiments took many, many hours to run.
However, (there's always a however) in modern spectrometers (anything newer than about 20 years old), the phase cycling restrictions are largely removed through improvements in hardware design and implementation of gradient pulses. So now we see for most basic 1D experiments, such as a 1H, there is a minimum number of scans of 1. We run many high quality, high resolution, quantitative analyses of samples using long acquisition, single scan experiments.
For multi-pulse experiments, there is still a requirement to phase cycle, but even for many 2D and 3D experiments, this is limited to multiples of 1, 2 or 4. Very few experiments these days require large numbers of scans to account for phase cycling.

So if I do say, 10 scans (n = 10 in brukeres) in a simple 1D, 1H experiment, then process it, would I be throwing away data points to get it to be 2^n?

@Canageek - No - you are confusing two separate and independent parameters: Number of data points in the FID (on Bruker systems called TD), and the number of scans (on Bruker systems called ns). Each scan is collected over TD data points, and this is repeated ns  times. Each scan (consisting of TD points) is added together to give the final FID. S/N increases as you increase the number of scans. Processing the final FID (of TD data points) is done by the FFT algorithm, and has limited allowed solutions, consisting of SI data points, and which can only handle $2^n$ data points.

The number of data points in the FID (TD) defines the resolution of the final processed (frequency domain) spectrum. Over a given sweep width, the more data points will result in higher resolution.

The number of scans defines the number of times you will repeat the acquisition and add all the individual FIDs together to give a final raw (time domain) FID. The final FID still contains TD data points. S/N improves at a rate of $/sqrt 2$ for every doubling of the number of scans, because noise also accumulates over time. For most 1D 1H experiments, the FID consists typically of, 16k, 32k or 64k points. In your example an experiment of 10 scans results in 10 individual FIDs (each of, say, 32k points) added together to give a final FID of 32k points.

I should add that 16k=16384 (2^14), 32k=32768 (2^15) and 64k=65536(2^16). These are the common abbreviations used in NMR

I'm aware of that long. However, the spectrum is the sum of all the data points, right? So if I have 1k data points in a spectra, then do 3 scans I have 3k scans, which is not a power of two.

Not quite right. The FID is a measurement of signal intensity as a function of time. The intensity of each data point in a single transient FID is added to that same data point number over all scans to give the sum FID. This is how S/N builds up. So the intensity of the 1st data point in each scanned FID is added together to give the sum intensity of the 1st data point in the resultant FID. The same for the 2nd, 3rd and so on data points. If you define your TD as 16k, then each scan will acquire 16k data points, and add together to give a final FID of 16k data points.

I see, so there is no reason not to use say, 5, 34 or 300 scans in the standard 1H experiment that comes with a Bruker NMR? (Other experiments, like an STD NMR of course need a multiple of their pulse sequence).

Exactly. Simple 1D experiments such as 1H or 13C{1H} will have a phase cycle of 1, meaning you can do exactly as you say. Many other experiments still have phase cycling requirements which require the number of scans to be multiples of $x$ where $x$ is the phase cycle; this may be as little as 2, but for complicated experiments, and those not using gradient pulses, will still be multiples of 8 or 16, for example. For Bruker pulse programs, the number of required scans is listed in the pulse program (use edpul or edcpul to view), usually right near the bottom; it will be listed as $ns = x * n$.

The signal to noise ratio (S/N) is proportional to the square root of the number of scans.  Hence the more scans, the better the S/N ration

The processing of the FID uses the Cooley-Tukey FFT algorithm, which can only process $2^n$ data points.

This is the reason why we typically process raw NMR data with n^2 data points, not why we record n^2 scans


text:
Nicotine concentration in e-liquids
E-liquids are blends of nicotine, propylene glycol, vegetable glycerin, and aroma.
I need to develop a method for determining the concentration of nicotine in a vial.
(to verify if the information listed on the bottle are correct.)
Does anyone have any idea how to proceed?

Which analytical equipment is available?

Gas chromatography would be an obvious choice.

I tested the gas chromatography but the curve I get is polluted by aromas. currently there a way to separate ?

Maybe my first idea was just a bad one. In this article on the determination of nicotine traces in "nicotine-free" e-liquids, HPLC was used.

Nicotine concentration is a serious concern, as the material is a serious poison (dose matters).  Liquid Chromatography/Tandem Mass Spectrometry is a little pricey.
http://oxfordchemserve.com/e-cigarette-refill-nicotine-quantitative-test-kit/
  Specific test kit "designed with the non-chemist in mind"
http://oxfordchemserve.com/wp-content/uploads/products_img/Oxford_Nicotine_Test.pdf
   Explanation - it is a pH titration of nicotine base.
http://www.eliquidtest.com/
   Another titration kit.  Google for the product spectrum.
Saliva or urine nicotine test kits detect nicotine persistent metabolite cotinine and are therefore irrelevant.
http://www.medpagetoday.com/Pulmonology/Smoking/44238
Warning about E-liquid refills.
http://www.veppocig.com/product_images/uploaded_images/eliquid-nicotine-strength-guide-2.png
Claimed dosage guide.  Caveat emptor.
http://www.learn.eversmoke.com/nicotine-strength.html
A second dosage opinion.
Glycerin can pyrolyze to acrolein.  This is not a good thing to inhale.  NEVER inhale butter flavor.  Butanedione destroys lungs: "popcorn lung," bronchiolitis obliterans, obliterative bronchiolitis, constrictive bronchiolitis are life-threatenting, cumulative, and irreversible.

Thanks you for these links. Do you know the accurency of the first test?

Ask the manufacturer.  It also depends on the skill of the user at all stages.

If you need a quantitative method witch can determine in the same time the quantity of all component, gas chromatography is good way. Only mass spectrometry detection can resolve aroma problems on your curves.
The most complicate aspect is chromatographic conditions : column, temperature...
I don't know the best column for these compounds (propylene glycol, glycerin and nicotine).
Maybe someone can help you to determine how to start your method on GC.
;)


text:
Endothermic and exothermic reactions
Today, in class, I learned about the two reactions.
I know connecting bond releases energy, and breaking it requires and aborb/ consumes energy.
But on my reference table I, 2C+H2 yielding C2H2 realeasing 227.4 kJ of energy. My point is why it releases energy? Also, where the activation energy comes from?
The paradox exists in that lower energy level means more stability, but nature also likes to form bonds thus to increase stability. However,  sometimes forming bonds increases the energy level.

You should remember that stability depends on lower energy and greater entropy. It is to be noted that both are independent factors.

You got it all wrong. Acetylene formation is endothermic; it consumes energy rather than releases.

Your idea is in principle correct. The reaction you mentioned works well, you break one H-H bond and create 1 C-C bond (triple bond, in fact) and two C-H bonds. The energy is released. And in most crude approximation it also says where the activation energy barrier comes from - you have to break a H-H bond.
As for the paradox - that is the principle of chemistry, to make bonds we wish for. If it means increasing the energy level, thus being less favored by nature, you have to pay for it. By sacrificing other bonds. Those are the highly reactive chemicals humbly stated above reaction arrow, which quite often do all the magic.
As the most revealing example in this I consider the Nucleophilic Acyl Substitution reaction, where you have a series of derivatives of increased stability. You can always and simply make the more stable out of the less stable, having wide choice of reactions. And there are just a few to climb up, using very nasty chemicals.

The energy released that you mention is the free energy between reactants and product. This is a thermodynamic quantity and in this case only tell us what the difference in free energy is between reactants and products. It does not allow us to predict the activation barrier or how fast a reaction will be.  As molecules do not react instantly on meeting one another this means that there is an energy barrier between reactants and products. The height of this (potential energy) barrier is the activation energy. The higher the barrier the slower the reaction. Its value cannot generally be predicted except for very simple reactions such as atom + diatom reactions, e.g. H+HCl. Reaction occurs because occasionally a reactant will randomly accumulate enough energy from the surrounding solvent (or gas) to surmount the barrier and so react. Most collisions between reactants do not lead to reaction as there is often not enough energy. Thus only occasionally when molecules collide does reaction occur.


text:
For which pair of species is the difference in radii the greatest?

For which pair of species is the difference in radii the
  greatest?
(A) $\ce{Li}$ and $\ce{F}$
(B) $\ce{Li+}$ and $\ce{F^-}$
(C) $\ce{Li+}$ and $\ce{O^2-}$
(D) $\ce{O^2-}$ and $\ce{F^-}$

I am pretty sure it is not (D) as $\ce{O}$ and $\ce{F}$ are quite close to each other on the periodic table and should have atomic radii pretty much the same.
I also know that gaining an electron increases the atomic radius whereas losing one decreases atomic radius. Also, as you move to the right of the periodic table atomic radius decreases.
I think that the answer could be (A) because $\ce{Li}$ is larger than $\ce{F}$, and the other choices result in $\ce{Li}$ becoming smaller and $\ce{F}$ becoming larger which decreases the difference in radii, but this is not the correct answer.
The answer given was C. How can this be explained?

Your argument that $r_\ce{Li} > r_\ce{F}$ is valid only for the neutral atoms.Consider the lithium ion. Its electronic configuration is exactly similar to helium but the nuclear charge is greater by one unit, and hence is even smaller than the helium atom. On gaining an electron, fluorine now has the electronic configuration of neon with lesser nuclear charge and hence is larger than a neon atom. Similar, oxide ion, with even lesser nuclear charge is the largest. The decrease in radius of lithium on losing one electron (reduction by one entire shell), more than compensates for the difference in the radii of neutral lithium and fluorine. Finally, the order appears to be:-
$$\ce{Li+ < He < F < Li < F- < O^{2-} }$$


text:
Are there any trends for electrical conductivity?

Which element has the greatest electrical conductivity?
(A) $\ce{As}$
(B) $\ce{Ge}$
(C) $\ce{P}$
(D) $\ce{Sn}$

I am unsure of how to approach this problem. Is there some sort of trend for electrical conductivity or some other method to determine which of these answers the question?
I am pretty sure it is not $\ce{Ge}$ as that is used for semi-conductors, and am pretty sure it is not $\ce{P}$ because that is a non-metal but I don't know how to narrow down between the others.

Phosphorus allotropes are white, red, and black.  Arsenic allotropes are gray, yellow, and black.  Germanium allotropes are alpha- and beta-.  Tin allotropes are white and gray (Napoleon learned about that outside Moscow).  Have some fun with the question, doi:10.1063/1.1729699

I am unsure of how to approach this problem.

No, you are not! You have already reduced the number of candidates by good reasoning!
At this point, one

could have a look at conductivity tables or remember in which period the metal blocks start for every main group or
remember typical electricity-related uses of $\ce{As}$ and $\ce{Sn}$

Did you ever form a conducting connection between electronic components (or just some wires) by soldering? $\Leftarrow$
(Do you recall that $\ce{GaAs}$ is used as a semiconductor material in thin film solar cells?)


text:
Is this a valid structure for the nitrate ion?
Why is the below structure not considered a valid structure for the nitrate ion? Is it because adding a double bond reduces the formal charges present on the nitrogen? Does the above form exist anywhere?

Nitrate actually exists as a superposition of these three resonance forms:

(Source: Wikipedia)
Essentially, a lone pair from an oxygen with a negative formal change forms a double bond to nitrogen. However, as nitrogen is a period II element, and hence must obey the octet rule, it can not have 5 bonds to other atoms. The already-present double bond then turns into a single bond and the pair of electrons moves onto the oxygen. This "movement" can happen with all three of the oxygen atoms and all three of the bonds. This is called "resonance".
Due to resonance, the energy of the system is minimised making it more stable.
Your Lewis structure is not far off. The "true" structure of nitrate is:

(Source: Wikipedia)
Approximately, each oxygen atom contributes one third of a bond with the nitrogen, lowering the formal charge on each atom from what you have written, which does not exist in nature.
The true structure is sort of an average of the properties of each possibility for where the electrons could be located, including the three resonance forms above, your structure, and any other configuration. However, these other forms are of minuscule probability to be formed (perhaps read up on wavefunctions and electron distributions if you're interested), and so the main contributors to the "real" structure are the three resonance forms.
Note that none of these structures ever exist in isolation for more than the most infinitesimal of time periods.

Technically, the asker's suggested canonical structure (and indeed many others) also plays some role in describing the nature of the nitrate ion, but its importance is significantly smaller than the three main forms with one double bond.

Edited to address that. Hopefully this is more accurate and still clear now.

None of the resonance structures exist, not even in the most infinitesimal time frame. That's why it is a superposition. The OP suggested a valid Lewis structure, which I think has an important (non-negligible) contribution to the overall electronic structure. Probability should not be taken as an indicator that this structure exists, the probability expresses the contribution of this hypothetical state to overall state. This Q&A is related and explains the structure in the molecular orbital picture.

I'm a bit confused about resonance structures. Does the electron "circulate" between the oxygen atoms so that the ion constantly switches between the resonance structures, or do all three resonance structures exist for a single ion at the same time? (Because I don't think 1/3 of an electron can exist)

Downvoting because this answer suggests that a small, finite number of resonance structures can be drawn and that were it.

@user3932000 no. Nothing move if not in our drawings.

Your resonance structure is valid. There are many more valid resonance structures, the most common ones being the three which feature one $\ce{N=O}$ double bond and two $\ce{N-O}$ single bonds with a single formal positive charge on nitrogen and two single formal negative charges one on each of the single-bonded oxygens.
These four resonance structures are by no means the end! A multitude of further, more minor contributors can also be written. Each of these can be considered to contribute to the overall picture which cannot really be displayed accurately in the Lewis formalism. While the structure you propose contributes less than the three with a single double bond, it still contributes more than some others.

Apart from the structure with two positive charges on the nitrogen as in Martin's answer (which he linked to in his comment above), I don't get how more structures can be drawn. Could you show me an example or two?


text:
How do I perform a gravimetric analysis of Portland cement?
To determine the silica ($\ce{SiO2}$) content in Portland cement by gravimetric analysis it is digested in $\ce{HCl}$ and the silica residue is eventually measured.
The chief constituents are $\ce{Ca2SiO4}$ (or $\ce{2CaO.SiO2}$), $\ce{4CaO.Al2O3.Fe2O3}$, $\ce{Ca3(AlO3)2}$ (or $\ce{3CaO.Al2O3}$), and $\ce{CaSiO3}$ (or $\ce{CaO.SiO2}$). Gypsum is also present. What species are present in solution after the $\ce{HCl}$ has reacted with the cement? I assume $\ce{CaCl2}$ and $\ce{H2O}$ would be present, but would $\ce{AlCl3}$ also form? How would I approach this problem?

http://books.google.com/books?id=VeXQAAAAMAAJ&pg=PA2&lpg=PA2&dq=Gravimetric+Analysis+of+Portland+Cement+%22aluminum
Copy whole URL. Page 2 for clickable index, then page 18.
DOI: 10.1039/AN9709500051
Digest it in "strong" HCl (tough lab for employment, worse if you service the fume hoods).  Everything except silica comes out as soluble chlorides (hydrated complex ions).  That says something about Portland cement in acid environments and, unless modified, in seawater.