text:
Do I need a license or certificate to practice chemistry or what are the advantages if did have one?
What can I do as a non-professional chemist that is legal?
If a license is not required, then what are the advantages to having one, such as ordering compounds, getting a job, etc.?

This will depend entirely on what country you are living in.

Note that being a chemist doesn't require any kind of state license or the like in Germany, neither.
But buying dangerous* substances and running a chemical lab requires that you** have certain licenses and implement safety standards.
Some of these exams a chemist has to pass during studies (e.g. on toxicology and chemical substances legislation) so having a Master/Diplom in chemistry automatically means that you have those licenses.
Having this license (Chemikaliensachkunde) and being reliable (legal meaning) and of age are legal requirements you must meet in order to be allowed to buy dangerous substances. In practice however, the suppliers put another very practical additional need: they usually do mail orders only B2B and require you to have VAT/USt number.
So if I as a small-scale freelancer who fulfills all the legal requirements want to buy from say, Sigma, I don't get anything because I don't have the tax number (for my size of side-business it would be too much hassle to go through this, so I use the small business exemtion law and for the supplier it would be too much hassle to make sure the buyer fulfills the legal requirements if it isn't at least a certain business size).
As a private person, you can buy chemicals:

at the pharmacy
at specialized stores, e.g.

photography chemicals at a good photography stores
carbide at a speleology outfitter's
gases (possibly including dry ice) at a gas merchant's, etc.

you basically cannot buy dangerous substances by mail/online order (too much burocratic hassle: the seller would need to make sure of your identity and Chemikaliensachkunde etc.)
I've never tried to buy privately anything that requires the Chemikaliensachkunde e.g. at a pharmacy so I cannot tell how difficult that is in practice.

* As a rule of thumb, there are restrictions for substances (or mixtures) that are extremely flammable, oxidizing, toxic or very toxic, (suspected of being) carcinogenic, mutagenic or teratogenic (plus a bunch of listed things)
** or some employee of yours

However, you can of course work as a chemist without (much or even any) need to handle dangerous substances. For example, I'm analytical chemist / spectroscopist / chemometrician. About 90 % of my work is in front of a computer analysing measurement data, developing improved ways of doing so and writing reports/papers. At least half of the rest is in front of a computer programming instruments. I've handled exactly one dangerous substance according to chemicals legislation at two days (actually a sample of few μg I got for analysis) during the last year. The everyday stuff I handle (methanol, ethanol, isopropanol, disinfectant solution and cyclo-hexane and paracetamol=acetaminophen as calibration standards) are not legally restricted dangerous chemicals.
I do handle biohazard material far more often - but that is not what the Chemikaliensachkunde is good for. So for my work, the more relevant safety issues are biological/human material safety and laser safety... It would be quite easy not to have any chemical lab exposure (is seldom enough as it is... - but hey, what did I become chemist for?)
Also I have some colleagues doing the same type of work who are not chemists but e.g. physicists and thus do not have the Chemikaliensachkunde.

For getting a job it is obviously of advantage if you can show some kind of education/training that makes you fit for the job. For jobs in chemical industry/academia that is obviously a chemistry degree or training as chemical lab technician, but also related professions such as physics/biology/pharmacy/toxicology depending on what the job in question is.

Over here, that is Germany, you will hardly find a job in chemical industry or academia without an academic degree (B.Sc., M.Sc, Ph.D) or a formal technical education (typically three years).
Most suppliers nowadys only deliver chemicals to universities, research facilities or other chemical companies.

and to schools with appropriately trained personel

Is there a chance to participate in an open lab science project? I have no idea whether these exist close to Sedona. Is there any university closer than Tempe?

There's Northern Arizona University in Flag, but I assume you have to pay tuition to use their labs.

Many suppliers anyways are B2B and don't deliver if the reciepient doesn't have a VAT (USt) number... (says the professional chemist with a freelancing side business that is small and thus without such a number)

Chemists are not licensed in the US.  A non-degreed "chemist" is at best a technician.  Homeland Severity bars purchase of chemicals (anything requiring a Material Safety Data Sheet) except for academia, corporations, and tis own pseudopodia.  OSHA (Occupational and Safety Health Administration) can declare a "hazard."  That can be a $10,000/incident-day fine, plus confiscation and closure.  All this is done above the law.  No prior police or court intervention is necessary.
Then the real terror begins - the War on Drugs.

How is this 100% accurate - since you can obviously buy Red Devil Lye for making soap and Muriatic acid for doing tile work, both of which obviously have a Mat Safety Datasheet somewhere?

A consumer item requires no MSDS. The US is astoundingly corrupt by convenience and payoff.  Note that BHT food preservative is two steps away from hypertoxic SF6849.  Potato chips.

"is two steps away from" - this has no chemical content whatsoever. By the same token, you could also say vinegar is a few steps away from 1080 rodenticide.

While there are many great answers here, just to be sure let's point out some implicit restrictions. While in general being a Chemist may not need a licence, you are still responsible for what you may obtain as the result of the experiment:

you are still responsible for your own and others safety, and property, in case of an accident,
you still cannot prepare drugs and other illegal substances, and you can be punished even if you didn't use or sell it.

So e.g. while obtaining NaOH is legal, you still can badly hurt yourself or others if you make soap based on Youtube videos, and you will be responsible for it. Similarly, preparing explosives or making fireworks can land you in jail depending on the country.


text:
Why is ozone O₃ diamagnetic?
Please explain how to derive the magnetism  of polyatomic molecules? If we derive it using an addition of an oxygen atom to molecular oxygen ($\ce{O2}$) it will come to be paramagnetic like $\ce{O3}$?

Ozone's ground state is a singlet.  All electrons therein are paired,

However,
http://link.springer.com/article/10.1134%2FS0018143906040060#page-1
http://link.springer.com/article/10.1134%2F1.1870062#page-1

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
reaction of diketone with ammonium carbonate at 100-115 degree celsius
Please explain what reaction is this and what are the products of reagent $\ce{(NH4)2CO3}$
at 100-115 C with ketone.

Which diketone? 1,2? 1,3? 1,4? 1,n? Saturated, unsaturated? Cyclic, acyclic?

sir straight chain 1,4 acyclic and i dont have anything like paal knorr.thanks anyway

1,4-diketones are likely to give pyrroles in a Paal-Knorr reaction.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Biochemistry for a mathematics student?
I am a second year mathematics student. I am mostly interested in statistics. Right now my curriculum doesn't offer me much choice (all second year courses are mandatory) but next semester I will be able to select the courses I want. Among the ones that are offered is a course called Biochemistry. It is actually a second year class for those how study chemistry and chemical engineering but I guess there must be applications of statistics in this field since it is also offered to third year math students.
My problem is that I am a math student: I have no knowledge or whatsoever in chemistry or biology. Granted that I had some in high school but back then I just learned everything by heart in order to pass.
My questions are the following:

With no knowledge of chemistry, is it realistic to undertake a biochemistry class?
The course description lists the following prerequisites: "Basic knowledge in general and organic chemistry". How important are these subjects in biochemistry? What are some good references (books) for self-learning these subjects?
Is Stryer's Biochemistry a good book for self-learning?
Are biochemistry and biostatistics related at all?
Are there many applications of statistics in biochemistry?

I have already searched for some information on the internet but I'd like to have some specific answers.
If it helps here is the information page of this class.
Finally I hope this kind of question is accepted here, if not please point me to a better place. Thanks in advance!

EPFL? Home of Michael Grätzel! Nice! Anyway, I can't give you a real advice, I'm "just an organic chemist". Does the faculty have a "Fachschaft" with students whom you could ask?

I will definitely try to find someone that took this class but I fear that there are not many math students that took it (it is optional after all).

Talk to the prof or to somebody who has taken it. I cannot imagine the value of memorizing amino acid structures, buffers' pH and capacity, or where the acetate carbon atoms are in cholesterol.  However, genomics is heavily statistical.  You can be of considerable value there.

It appears that you will need at least 2 semesters of O chem. I recommend Loudon for O chem. Take a look at it and if it makes sense to you, you should be ok.

It also looks like an intro class. I think that you may be ok without knowing all the o chem. However, you will have to put a significant amount of effort in filling in the gaps. It shouldnt be as bad as in mathematics but nevertheless be cautious.

I'd also suggest that you read up beforehand on pH and the related calculations and get a glimpse of physical chemistry (thermodynamics: reaction enthalpy, free energy, equilibria; kinetics: order of reactions, reaction speed, Michaelis-Menten kinetics) Maths behind that is easy, but you should be familiar with notations and useful as well as widely used approximations (I emphasize the widely used: as a maths student you may solve directly without approximation - but nevertheless approximations are employed in practice and if you don't know that it can be very confusing)

As chemist, I like the Stryer. As chemometrician I can tell you that there is a considerable need for statisticians in (bio)chemistry. Including but not limited to genomics.

With no knowledge of chemistry, is it realistic to undertake a biochemistry class?

I would say that taking a biochemistry course without any background in chemistry or biology would be very difficult, but it would be possible.

The course description lists the following prerequisites: "Basic knowledge in general and organic chemistry". How important are these subjects in biochemistry?

The info page link that you gave would not work for me, so I can't tell how in-depth this course would be, but I know that the undergraduate biochemistry courses I have experience with expect you to be familiar with (at a minimum):
Chemistry

Bonding theory
Molecular structure
Nomenclature of organic molecules
Polarity and its relation to hydrophobicity/hydrophilicity
Basic organic reaction chemistry (particularly Lewis acid/base and functional group chemistry)
Basic reaction kinetics
Basic thermodynamics

Biology

Macromolecules (proteins, carbohydrates, fats)
Cell structures
Basic anatomy/physiology

All of these are fundamental to just about every concept in biochemistry, and they will probably expect you to already be familiar with if not good at them.

What are some good references (books) for self-learning these subjects? Is Stryer's Biochemistry a good book for self-learning?

That I can't answer. I can tell you that Voet & Voet's "Biochemistry" is a good biochem book, but I am not sure if it would be suitable for self-learning. I don't have experience with Stryer's book. Voet & Voet also have a "Fundamentals of Biochemistry" book that might be better - but again, I don't have any experience with that one and can't compare to Stryer. In my personal experience, I have found that reading as many different books as possible is the best way to self-study, but your results may vary.
I can give you some suggestions for general chem and organic books:
General Chemistry

Brown, Lemay, and Bursten's "Chemistry: The Central Science"
Tro's "Chemistry: A Molecular Approach"

These are both very easy to read and clearly explain all of the fundamental concepts, with lots of example problems worked out in detail.
Organic Chemistry

Wade's "Organic Chemistry"

This is the only organic chem book I have much experience with, but I remember it being very easy to understand.
For biology textbooks, I'm afraid I can't be much help.

Are biochemistry and biostatistics related at all?

Only in the sense that biostatistics is statistical methods applied to problems found in biology, and many problems found in biology are related to biochemistry. You would certainly want to learn some basic biochemistry if you were going to work on bioinformatics problems, for example.

Are there many applications of statistics in biochemistry?

There are - particularly in areas like design of experiments, genomics, proteomics, and drug screening, and other various assays.


text:
Clarifications on the calculation of the $\Delta\mathrm{G}$ of a reaction
I'm trying to calculate the $\Delta\mathrm{G}$ of the following reaction. Are my calculations correct? I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the $\Delta\mathrm{G}$ is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?
CH3OH + H2O --> HCOOH + 4H        Delta H = 99.49 kJ/mol

% Standard enthalpy of formation of every compound

% -238.7 kJ/mol CH3OH
% -285.830 kJ/mol H2O
% -425 kJ/mol HCOOH
%    0 kJ/mol H2
% -393.51 kJ/mol CO2

% Standard entropy of formation of every compound

% 126.8 J/mol*K CH3OH
%  69.91 J/mol*K H2O
% 129 J/mol*K HCOOH
% 130.6 J/mol*K H2
% 213.6 J/mol*K CO2

delta_H_1 = -425 - (-238.7 - 285.830);
delta_S_1 = 129 + 130.6 - (126.8 + 69.91);
T = 273.15 + 20;

delta_G_1 = delta_H_1 - T * delta_S_1;

Correct version
delta_H_1 = -425 - (-238.7 - 285.830);
delta_S_1 = 129 + 4*130.6 - (126.8 + 69.91);
T = 273.15 + 25;

delta_G_1 = delta_H_1 - T * delta_S_1;

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
Gibbs Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibbs Free energy describes how much "free energy" is available, or how much work is possible.
$\Delta G^\circ$ can either be negative or positive even if $\Delta H^\circ$ is $+$. Therefore,
we have to look at all three terms, $\Delta H^\circ$, $T$, and $\Delta S^\circ$
The Gibbs Free Energy, enthalpy, and entropy are state functions. Although I won't derive it here, we can find $\Delta G$, or $\Delta H$, or $\Delta S$, by using Hess's Law.
The information provided above gives us the $\Delta H_{rxn}$ which is $99.49 \text{kJ/mol}$ and thus endothermic.
We calculate entropy from the a list of experimental tabulated values:
$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$
$$\Delta S^\circ_{rxn} = (129.0 \text{J}\cdot\text{mol}^{-1}\text{K} + 4*130.6 \text{J}\cdot\text{mol}^{-1}\text{K}) - (126.8 \text{J}\cdot \text{mol}^{-1}\text{K} + 69.91 \text{J}\cdot\text{mol}^{-1}\text{K})$$
$$\Delta S^\circ_{rxn} = 454.69 \text{J}\cdot\text{mol}^{-1}\text{K} = 0.45469 \text{kJ}\cdot\text{mol}^{-1}\text{K} $$
While solving for entropy, you forgot to multiply the entropy of hydrogen gas with the number of moles. Also, you should always include the states of your reactants and products, as thermodynamic data for compounds differs for different states.
Another thing that is incorrect is standard temperature for thermodynamic reactions is at $25^\circ C$ not $20^\circ C$ and thus your temperature should be $298.15 K$

I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the ΔG is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?

This is where a lot of people make mistakes. It is true, that in general, endothermic reactions with a positive enthalpy usually have a positive $\Delta G$ and are thus not spontaneous.
However, if we look at our equation:
$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
We have to consider our other terms and not only enthalpy, namely that of temperature and entropy.
A reaction is spontaneous if $ \Delta G < 0 $ such that:
If $ \Delta H < 0$ and (-) and $ \Delta S > 0$ (+), the reaction is spontaneous at any temperature.
If $ \Delta H > 0$ and (+) and $ \Delta S < 0$ (-), the reaction is NOT spontaneous at any temperature.
If $ \Delta H < 0$ and (-) and $ \Delta S < 0$ (-), the reaction is spontaneous at low enough temperatures.
If $ \Delta H > 0$ and (+) and $ \Delta S > 0$ (+), the reaction is spontaneous at high enough temperatures.

So I can't plot this delta_G_1 = delta_H_1 - T * delta_S_1 as a function of T because the experimental tabulated values are true just for the temperature of 25 °C right? I've added also the corrected version of the calculations, is it correct?

@FormlessCloud Exactly, because those values indexed inside the appendix are derived from experiments. If temperature were to change during those heats of formation, we would have to run an experiment to determine them. However, we can run reactions at different temperature. Those look right. I would sure that they are for the correct state. A good example is water in liquid and gas phase.

But if the experimental tabulated values are going to change at different temperatures than 25 °C how we can predict the spontaneity of the reacton in such a way? I mean this way, by temperature: (-) and (+), the reaction is spontaneous at any temperature.  (+) (-), the reaction is NOT spontaneous at any temperature.  (-) (-)the reaction is spontaneous at low enough temperatures.  (+) (+) the reaction is spontaneous at high enough temperatures.

@FormlessCloud Hi, I was referring to the calculation of the standard thermodynamic data. Entropy and enthalpy are experimentally determined at $25^\circ C$. If we run the entire reaction, we would just change the T term in the equation above. Sorry for the confusion.

Hi, don't worry : ) However ok, but if are not allowed to change the temperature because the other constants of the equation are no longer valid at different temperature than 25 °C, how this can be possible? From what I understood then the $\Delta G$ can just tell us that the raction is spontaneous or not at the temperature of 25 °C. If we say that since $\Delta H$ and $\Delta S$ are both positive (+) the reaction is spontaneous at high enough temperatures maybe at that "high enough temperatures" $\Delta H$ and $\Delta S$ can change their signs, right?


text:
Gold's atomic structure and bonding in relation to the separation techniques used
How is Gold's atomic structure and bonding related to the separation techniques used to distinguish it from other substances?
I'm just confused on what I'm actually searching for, I have read so many blogs on Gold extraction, refinement, occurences and more topics, and either I'm skipping the important parts or the answer's just not there. What can I search to find what I'm looking for? The separation technique I chose is flotation, can someone lead me in the correct path and/or give me a few hints on what to start of with? Thanks in advance, much appreciated!

There are differences in gold extraction for different mines - depending on the chemical nature of the raw ore, the impurities present, the availability of resources etc.
A good example of gold extraction comes from the Super Pit mine in Kalgoorlie, Western Australia; their gold extraction process is outlined in their website Mineral Processing, which is outlined below:
Crushing and grinding: this is a physical process where the mined raw ore is crushed, ground and water added to make a slurry.
Flotation: chemicals are added to the slurry to remove everything other than gold and pyrite.  Water is also removed at this stage.  The slurry is agitated forming air bubbles, which then rise through the slurry, carrying mineral grains.  From Introduction to Mineral Processing:

The ability of a mineral to float depends upon its surface properties. Chemical modification of these properties enables the mineral particles to attach to an air bubble in the flotation cell. The air bubble and mineral particle rise through the pulp to the surface of the froth or foam that is present on the flotation cell.

Roasting: water is re-added and the concentrated slurry is roasted to remove $\ce{SO2}$
CIL: or carbon in leach, chemicals are added to dissolve the gold and have close to pure gold precipitate on the carbon.
Elution: A mixture of caustic soda and cyanide is used to liberate the gold from the carbon.
For your specific request for resources about flotation, the example process below is for a sulphide copper ore containing iron bearing pyrite:

(Image Source)
Suggested additional resources about flotation:
Froth Flotation – Fundamental Principles
A great resource is the book Handbook of Flotation Reagents: Chemistry, Theory and Practice (chapters are behind a paywall).

Thank you so much! You have no idea how much you have helped me! Would $SaOH^- + CN^-$ be the only chemical reaction encountered in the separation process? And in which stage would I be able to establish the relation between gold's atomic structure and bonding, I am guessing that the "Elution" process plays a big part, does the sodium hydroxide and the cyanide affect gold in any chemical way?

This may be of interest for that stage http://rsteyn.files.wordpress.com/2010/07/elution-and-carbon-reactivation-basics.pdf  One clarification from your comment, the formula for sodium hydroxide is $NaOH$

Sorry! $NaOH$, I got it a bit mixed up with the cyanide formula. Does that mean that $SaOH + CN^-$ is the only occuring chemical reaction, or will I find out more after I thorougly read what's behind the link you provided? Is there a specific terminology for the relationship between atomic structure and bonding to the separation process (floation) used? Thank you again!

@SamirChahine These may help answer your question saimm.co.za/Journal/v077n12p254.pdf and http://www.eng.nus.edu.sg/m3tc/M3TC_Technical_Reports/Gold%20Extraction%20and%20Recovery%20Processes.pdf


text:
Molecular formula for 4-methyl-1-pentanol?
According to Wikipedia 4-methyl-1-pentanol is $\ce{C_6H_{14}O}$.
But here is what I got, which is slightly different.
Methyl = Think of methyl group, which is $\ce{CH_3}$.
Pentanol = Penta means we have 5 carbons, -ane ending means we have $2n+2$ hydrogens, i.e. 12 hydrogens and -ol ending means we have a hydroxyl group $\ce{-OH}$.
So the molecular formula that I arrived at is: $\ce{CH_3C_5H_{12}O}$ = $\ce{C_6H_{15}O}$.
How did I end up with an extra hydrogen compared to Wikipedia answer?

Actually, the formula you arrived at is $\ce{CH3C5H12OH}$ = $\ce{C6H16O}$.  In any case, in your "2n+2" step, you need to reduce that number of hydrogens by 2.  One is eliminated where the methyl group is attached, and a second is eliminated where the $\ce{-OH}$ group is attached.  So removing 2 hydrogens from $\ce{C6H16O}$, gives you the desired $\ce{C6H14O}$.

You’re applying the $2n +2$ method wrongly. You need to stupidly count all the carbons in a fully saturated molecule and take that number as $n$.
In your case this is:

5 because pentanol
another 1 because methyl.

That gives you six carbons in total. $n = 6 \Rightarrow 2n + 2 = 14$
Alcoholic oxygens do not introduce unsaturation, so you can safely ignore the oxygen here.


text:
Boiling-point elevation paradox
This is the boiling-point elevation equation:
$$
\Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}.
$$
If we do the limit of the above, for $b\rightarrow\infty$ we find that the boiling-point elevation is infinite. How this could be explained under a physical view?
$$
\lim_{b~\rightarrow~\infty}ib\mathrm{K}_{eb}=\infty,~\text{if}~i\wedge b > 0.
$$
For example, if we use a solute molality that is so big that the solution would be something like the pure solute, the final boiling-point of the solution, in the case of water as solvent for example:
$$
\mathrm{T}_{b}=ib\mathrm{K}_{eb} + 100\mathrm{°C}.
$$
wouldn't be like the boiling-point of the pure solute? I mean, something like this:
$$
\lim_{b~\rightarrow~\infty}\mathrm{T}_{solution}=\mathrm{T}_{solute}.
$$
Now, please forget what I've said above about $b\rightarrow\infty$. And take a look at this:
$$
T(b) = 0.86b + 64.7.
$$

As you can see is, the boiling-point of a solution of $\mathrm{CH_3OH}$ with $\mathrm{H_2O}$ as solute would be $322.7\mathrm{°C}$ for $b = 300$. This it's weird to me  'cause that temperature would be about three times higher then the boiling-point of the pure water!
So the following:
$$
\mathrm{T}_{b}=ib\mathrm{K}_{eb} + \mathrm{T_{solvent}}.
$$
holds true if
$$
b = \frac{x~\mathrm{g}_{solute}}{1000~\mathrm{g}_{solvent}} \le 1
$$
right? If yes why (in the above case):
$$
0.86\frac{1000~\mathrm{g}}{18.01528~\mathrm{g}/\mathrm{mol}}\frac{1~\mathrm{mol}^2}{1~\mathrm{kg}} + 64.7~\mathrm{°C} = 112.4~\mathrm{°C}
$$
?

After your edit... To use that equation you're imposing the constraint that the solute won't boil. You're looking at the boiling point of the solution, under the assumption that the methanol is the only component that will boil.

Could you please add units to $b$ and your $T(b)$ equation? If $b = 300$ means something like $b=300 , \text{mol}{\ce{H2O}}/\text{kg}{\ce{CH3OH}}$ then $300$ is not a realistic value as per the arguments I provided in my answer. If you considered realistic molalities ($< 50 , \text{mol}{\ce{H2O}}/\text{kg}{\ce{CH3OH}}$) you would be well in the reasonable temperature regime below 100 °C in your plot.

@Brian take a look at my updates

You have inserted the wrong values for the molality in your equation. The molality is $b = n(\text{solute})/m(\text{solvent})$. So, say, you want to know the boiling point of a methanol/water mixture where you have $1 , \text{mol}$ of water in $1 , \text{kg}$ of methanol. Then you'd get $T(b) = 0.86 \frac{\text{K} , \text{kg}}{\text{mol}} \cdot \frac{1 , \text{mol}}{1 , \text{kg}} + 64.7~\mathrm{°C} = 65.56~\mathrm{°C}$.

It's not so simple as saying it holds true if b <= 1. Rather it holds true if the solute is non-volatile and perfectly miscible. It will hold true for water with a soluble amount of NaCl for example.

@Philipp Unit corrected but the problem still there

@FormlessCloud It is not about the unit as such. You use the wrong formula for $b$. See the equation in my comment. In the numerator of $b$ you have $n(\text{solute})$ and not $m(\text{solute})$. So, your conclusion that $b = \frac{x~\mathrm{g}{solute}}{1000~\mathrm{g}{solvent}} \le 1$ is not correct.

@Philipp the correct formula for $b$ is: $$b = \frac{\mathrm{mol}{solute}}{\mathrm{kg}{solvent}}$$ since $\mathrm{mol}$ can be considered as the multiple of a given quantity of grams, I think that what I said it's correct.

@FormlessCloud To communicate over the comment section is a bit inconvenient. Would you come into the chat room, so that I can explain to you what I meant?

@FormlessCloud Your original question was about why the equation behaves the way it does - why boiling point goes infinite rather than to the boiling point of the solute, and that question has been answered. You should make a new question if you're wondering about something else :-) The reason the formula acts up is that you're using it with a volatile solute. It only works as long as your solute to a reasonable degree does not have any vapour pressure at the relevant temperatures, and is perfectly miscible. :-) It would work with water soluble salts in water for example.

""CH3OH with H2O "" Boiling point elevation is about nonvolatile substance dissolved in some solvent, nothing else. Study simple basics first.

In reality, when you're saying that the molality is infinite, you are indeed basically asking for solute with infinitesimal amounts of solvent, and it will be true that $T_\text{solution} \approx T_\text{solute}$.
This equation describes the boiling point of the solvent under the effects of the non-volatile solute. Ie. you're making the assumption that the solute does not enter the gas phase at the temperatures you're considering.
A substance boils when the chemical potential of the liquid phase equals the chemical potential of the gas phase: $\mu_\text{liquid} = \mu_\text{gas}$, or when the vapor pressure equals the surrounding pressure.
So to elevate the boiling point, we're making the claim that for a given temperature, the chemical potential of the liquid phase is lowered (it's made more stable), or equivalently, that the vapor pressure is lowered. The effect is entropy based.
When particles are interspersed between the solvent molecules, the solvent molecules become less ordered, and are thus at a state of higher entropy than the pure solvent. This translates into a lower chemical potential. The gas phase will have only the solvent molecules, because we made the assumption that the solute is non-volatile. Therefore, the gas-phase is unaffected by this effect, and the chemical potential of the gas phase remains the same.
It's easy to visualise from the vapour pressure perspective. Imagine the liquid-gas interface. At a given temperature, a certain portion of the liquid solvent molecules at the liquid-gas-interface will enter the gas phase. The gaseous solvent molecules at the liquid-gas-interface will likewise enter the liquid phase when they come in contact with it.
The amount of solvent molecules going from gas->liquid depends on the atmospheric pressure. The amount of solvent molecules going from liquid->gas depends on the kinetic energy of the molecules (the temperature). A molecule may have enough energy to break free of the solvent-solvent interactions, or it may not. The higher the temperature, the more likely.
Now imagine that we add solute molecules. There will be less liquid solvent-molecules present at the liquid-gas interface, and thus a lower rate of liquid -> gas (meaning a lower vapour pressure), simply because the probability of a liquid molecule hitting the phase interface is lower...  So the temperature has to be higher to compensate, if we want to counter the unchanged rate solvent molecules going from gas -> liquid.
Why boiling point goes infinite: At the limit where molality goes towards infinity, there will be an infinitesimal amount of liquid solvent molecules at the phase interface... And if we want that infinitesimal amount of molecules to sustain a vapour pressure that equals the atmospheric pressure, we need a temperature that approaches infinity as well.

I've heard the explanation about solute molecules "pushing" the liquid molecules out of the liquid-gas phase boundary, but something doesn't feel quite right to me. Would this explanation still be valid if I added a surfactant? Even at low concentrations, surfactants quickly cover a large part of the liquid-gas phase boundary. Does that mean surfactant solutions have a higher effect of the boiling point of a liquid compared to a normal solute, on a per mole basis?

That's an interesting thought. I don't think I can come up with anything right now. THe surfactant will form its own extremely thin liquid phase, that the liquid solvent will interact with. And I suppose it impedes both entry from liquid into the gaseous solvent phase but from the gas into the liquid phase. Does that make any sense?

I'm not sure. Honestly I feel like the whole idea of blocking the surface is sort of an over-explanation. The correct answer would simply be the change in entropy of the solution (as you mention), and the idea of covering the surface would be trying to go too far in order to provide further physical insight. I'm searching for some articles on water boiling point elevation by surfactants, but I haven't found anything enlightening yet.

The flaw in your reasoning is that the molality $b = n(\text{solute})/m(\text{solvent})$ cannot be infinite even if you use a solute molality that is so big that the solution would be something like the pure solute. Take water as an example:
How concentrated is water? Very concentrated you may say - but the concentration is limited. We know that one mole of pure water has a mass of 18 g and occupies 18 cm³. So, in one dm³, there are 1000/18 = 55.56 mol which gives you a water concentration of 55.56 mol/dm³. You cannot get more concentrated water than this (unless you did something drastic like taking it into a black hole!). For the molality it is quite similar: If you dissolve something in water the solute concentration cannot exceed the concentration of water, so $n(\text{solute})$ (and therefore $b$) has an upper bound.

What if I took 1 kg of my solvent and decided to add as many moles of solute as I like. Then I can make b as large as I like, since I would have as many moles of solute in my kg of solvent as I like. :-)

@Brian That is the nice part: At some point before reaching the 55.56 mol limit the solute would either stop dissolving and precipitate or there will be some kind of phase inverion such that water will then be dissolved in the solute. In that case the maximum concentration the pure solute could attain (analogous to the little calculation I did for water) would be the upper bound for $n(\text{solute})$.

@Brian Of course, my answer is more some kind of rationale to show that the premise of an infinite molality is impossible. Your answer goes much more to the heart of the problem (+1 for that). I think it's nice that you took the time to write out that point of view. That way FormlessCloud gets one answer that really explains what boiling point elevation is about and one that shows him a quick way to rationalize why the limit $b \to \infty$ is not possible.

We can readily agree that having a non-volatile, perfectly miscible solute can probably not be, in reality :-) and as you say that puts an upper bound on b


text:
Can oxygen burn in hydrogen?
Most combustion happens involves the fuels burning in an oxidant. But there seems to be no reason that it cannot happen in the opposite way. Are there any experiments on oxygen burning in hydrogen? Or some oxidant in solid state (like $\ce{KMnO4,KClO3,}$...) burning in a reductive gas?

Could you be more clear on what 'burning in' something means? The chemical reactions require both reactants available, so they need to mix - in what sense is Hydrogen mixing and reacting with Oxygen any different that Oxygen mixing and reacting with Hydrogen?
In some sense, we do tend to control these reactions via the fuel, not the oxidant (oxygen free environments are difficult to reproduce on Earth's surface). However there may be some examples of controlling the mixture via restricting the oxidant in a fuel-rich environment... even though I can't even get through the sentence without that sounding wasteful.
Nitrous Oxide
Would you consider the example of Nitrous Oxide injection in an internal combustion engine an example of this? Pre-nitrous, there is compressed atmospheric oxygen and hydrocarbons at a set ratio, and to further increase the burn a nitrous system will inject either NO alone (dry system) or NO and more hydrocarbon fuel (wet system) to add more oxidants that would be available through the 21% Oxygen in the atmosphere.

The Chemical: https://en.wikipedia.org/wiki/Nitrous_oxide
Internal Combustion application: https://en.wikipedia.org/wiki/Nitrous

Candles
How about a candle? There's an abundant fuel source (wax), and we frequently control the shape and characteristics of the burn with shaped glass, not to mention put it out via starving it of oxidant.
In either case, if you were to, say take a lump of solid oxygen (nevermind the practical difficulties of dealing with temperatures below 54 Kelvin) and maintained a burn with Hydrogen, what you'd be likely to find would be combustion only happening at the edges, where the oxygen was evaporating and mixing with the reduction agent, similar to lighting a can of gasoline on fire (only the vapors are flammable).

Thanks for reply. For the "burn in" thing. I think that hydrogen comes out from a tube into oxygen or air and have stable flame is a proper scene to characterize the asymmetry in the reaction. So more exactly, if we put a flow of oxygen into hydrogen and ignite it with things like electric discharge, will it have stable flame like the previous case?

I believe so, yes - however, you will find that getting a room/chamber to be devoid of Oxygen and filled with Hydrogen to be difficult to achieve and maintain (on Earth).

Are there any experiments on oxygen burning in hydrogen?

(Assuming that burning in "something" describes that "something" is more/at higher concentration than the stuff that is said to burn)
Sure oxygen burns in hydrogen: the upper flammable limit of hydrogen in oxygen is 94 % hydrogen and 6 % oxygen.
Which also tells you one case where such experiments are conducted (measuring flammability for safety sheets).
(For other fuels, the upper flammable limit is much lower, actually fuel concentration < oxygen concentration, so one would rather say the fuel burns in oxygen.)

You can do a nice experiment burning hydrogen in oxygen, well in air anyway. I have done this many times with a large instant (powdered) coffee tin.
Punch a small hole in the lid and in the base a hole large enough to put in a rubber tube connected to a hydrogen cylinder. Put the tin on a stand with the lid on top, pushed into place to seal the lid but not too securely. Allow the hydrogen into the tin till you judge it has filled it. A couple of minutes.
Light the hydrogen /air mixture at the small hole in the lid.
Turn off the hydrogen supply and wait. Your audience can see the flame burning. When the explosion limit is reached the flame dies and bang; the lid flies off. As it is so light it does no damage.
This illustrates the fact that (a) oxygen/hydrogen mixtures are stable, although the reaction is exothermic the activation barrier is so large that no reaction occurs until some extra energy is supplied (b)That hydrogen burns in air stably at some mixture ratios and (c) that it can explode when the conditions are correct. Its also great fun to do :)


text:
Conversion of potassium dichromate solution to a less toxic solution
I have a solution of potassium dichromate ($\ce{K2Cr2O7}$) and I want to convert it to a less toxic solution. What would be the best way to do this?
I was thinking of using the dissolved potassium dichromate as an oxidizer in the oxidation of alcohols to organic acids, to eventually produce the chromium(+III) ion (what is less toxic)
Or are there any other, easier methods?

Or are there any other, easier methods?

The cheapest and safest one is probably to use ethyl alcohol and some acid (acetic acid should work).

Ethyl and any other alkohol is a bad idea. Reaction of chromate with alcohols is slow. Try FeII as Kwak recommends.

That looks like one way to reduce the dichromate ion which is known to be toxic. MSDS sheets indicate that potassium dichromate has category 1B carcinogenicity.
Another way I believe is to just start from an aldehyde and convert it into a carboxylic acid:
$$\ce{3RCHO + 2Cr_2O_7^- + BH^+ -> 3RCOOH + 2Cr^{3+} + 4H_2O} $$
Alternatively, one can put dichromate in acidic solution with iron:
$$\ce{Cr2O7^{2-} + 6Fe^{2+} + 14 H^+ -> 2 Cr^{3+} + 6 Fe^{3+} + 7H_2O }$$
Dichromate oxidizes $ \ce{Fe^{2+}}$ ions and produces $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ ions.

thanks! I'll use the iron oxidation method, however : would an acid such as acetic acid be sufficient for this redox reaction?

@LievenB Hi, I believe so. As long as you have enough acetic acid to form a ratio of moles of 14:6:1, that would be sufficient for this redox reaction.


text:
Redox reaction with hydrogen peroxide with potassium iodide
I think I've managed to write down the correct reaction:
$$\ce{H2O2 + KI + H2SO4 -> K2SO4 + I2 + H2O}$$
Now I need to balance this using the ion-electron method in an acidic environment.
Normally this is no problem, but I'm a bit confused on how to work with the reduction of $\ce{O}$ from $\ce{-I}$ to $\ce{-II}$ and I'm not sure how to write down the total ionic equation. I've tried the following:
$\ce{2H^{+1} + O_2^{-2} + K^{+1} + I^{-1} + 2H^{+1} + SO_4^{-2} -> 2K^{+1}  +SO_4^{-2} + 2I^{} + 2H^{+1} + O^{-2}}$
But I'm not sure if that is correct, and if it is correct, I'm not sure what to do from here.
After this reaction and after diluting (with water?), I also need to know what will happen if I add a few drops of a starch-solution (I will actually have to perform this experiment so if it's merely something visible and not necessarily related to the reaction, then this part is less important.)
The final question is "what property of $\ce{H2O2}$ is demonstrated here?" and I suppose the answer to that is it's function as an oxidizer?

Can you provide more information as to which starch indicator that you will be using?

That's the problem, my exercise just says to "dilute and add a few drops of a starch-solution" and doesn't explicitly refer to it after that.

Hi, I now remember. If we look at the product side, we can see that we have made elemental iodine. We can't exactly see the iodine in solution, however, if we add any starch indicator, it will form a complex with iodine, and become very dark/blue solution, signifying that iodine was made in the reaction.

"We can't exactly see the iodine in solution" is not true. You can see molecular iodine pretty easily. http://www.chemistryexplained.com/photos/iodine-3326.jpg Starch just makes the initial colour change even more obvious.

With all redox reactions, we should first look at all oxidation states of our reactants and products.
$$\begin{array}{ccc}
\text{Element} & \text{Oxidation state in reactants} & \text{Oxidation state in products} \\
\hline
\ce{H} & +1 & +1 \\
\ce{O} & -1\text{ in }\ce{H2O2}; -2\text{ in }\ce{H2SO4} & -2 \\
\ce{K} & +1 & +1 \\
\ce{S} & +6 & +6 \\
\ce{I} & -1 & 0
\end{array}$$
Now, we have to look at what is being oxidized and reduced:
$$\ce{H_2O_2 + KI + H_2SO_4 -> K_2SO_4 + I_2 + H_2O}$$
Hydrogen has the same oxidation state of 1+.
Potassium has the same oxidation state of 1+.
Iodine is oxidized from -1 to 0.
Oxygen is reduced from -1 to -2.
For redox reactions, we have to write out our half-reactions.
$$\ce{H2O2 -> H_O}$$
$$\ce{KI + H2SO4 -> K2SO4 + I2}$$
For our two half-reactions, we first have to make sure everything is balanced other than hydrogen and oxygen.
For our second half-reaction, we need two moles of potassium iodide in order to balance out the potassium and iodines.
$$\ce{H2O2 -> H2O}$$
$$\ce{2KI + H2SO_4 -> K2SO_4 + I2}$$
In order to balance oxygens, we add molecules of water.
$$\ce{H2O2 -> 2H2O}$$
$$\ce{2KI + H2SO4 -> K2SO4 + I2}$$
Since we are in acidic solution, we add protons to balance hydrogen.
$$\ce{H2O2 +2H+ -> H2O}$$
$$\ce{2KI + H2SO4 -> K2SO4 + I2 + 2H+}$$
We add electrons to balance charge:
$$\ce{H2O2 +2H+ -> 2H2O + 2e-}$$
$$\ce{2KI + H2SO4 + 2e- -> K2SO4 + I2 + 2H+}$$
Now we add 'em(the half-reactions) both up and cancel out necessary terms:
$$\ce{2KI + H2SO4 + H2O2 -> K2SO4 + I2 + 2H2O}$$
This differs from your original equation since I think you forgot to balance it.
Hydrogen peroxide here is reduced to water in acidic solution.


text:
Concentrations of mixed solutions
If I mix a $\pu{2 M}$ solution with a different $\pu{1 M}$ solution, do the individual solutions reduce in molarity to $\pu{1.33 M}$ and $\ce{0.33 M}$ respectively?

Depends on the mixing ratio. Are you mixing them 50:50 or in a different ratio?

@Philipp, good question. I am unsure. The machine dispenses them for me. I will check and edit the question accordingly.

This will all depend on what you are mixing. If you are mixing 2 M NaOH with 1 M NaOH, you will have a new concentration somewhere between 1 and 2 M depending on the volumes.  If you mix 2 M NaOH with 1 M NaCl, you have a different scenario. If you are mixing 2 M NaOH with 1 M HCl, then you have a reaction, which will be a third type of scenario.

You need to know how much of each solution you add and the final total volume. The molarity of each one would be
$$\frac{C_1 \cdot V_1}{V_1 + V_2} \quad \text{and} \quad \frac{C_2 \cdot V_2}{V_1 + V_2}$$
where $C_1 = \pu{2 M}$, $C_2 = \pu{1 M}$, $V_1$, $V_2$ - volumes of $\pu{2 M}$ and $\pu{1 M}$ solutions, respectively.


text:
Why isn't nitrous acid a product of nitrite and water?

Question
  Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equations
  $$\ce{NO2- + H2O -> ?}$$

I thought it would be $\ce{NO2- + H2O -> HNO2 + OH-}$, but my textbook says that the correct products are $\ce{HNO3}$ and $\ce{OH-}$. What am I doing wrong here? The answer that they give doesn't even seem to balance.

It's probably a typo.  Nitrate is used very often and nitrite much less so.  It's an easy mistake to make and a difficult one to catch.  That's my best guess unless they give an answer of the type that @Matej suggests, though that seems to require quite a bit of insight into the chemistry for an AP Chem text.

Can you recheck and clarify if the question gives you $\ce{NO2}$ or $\ce{NO2-}$? The sign makes a big difference. If you had $\ce{NO2}$ gas, then you will get both $\ce{HNO3}$ and $\ce{HNO2}$ ($\ce{HNO2}$ further undergoes disproportionation to give $\ce{HNO3}$ and $\ce{NO}$).

It should form nitrous acid... If it was $NO_3^{-}$, nitrate ion, then it would presumably form nitric acid $HNO_3$.
Nitrous acid reacts with water to form nitrite and hydronium:
$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$
Water is an amphoteric species. Assuming we just have water the nitrite ion, the nitrite ion accepts the hydrogen forming nitrous acid:
$$\ce{NO_2^- + H_2O -> HNO_2 + OH^-} $$
This reaction is interesting as well since nitrous acid will also dissociate again into nitrite ions and water as well:
$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$

I checked several times, the book says that the products are $\ce{HNO3}+\ce{OH-}$

Chemistry, the Central Science, 11th AP Edition. Question is on page 711

It can be, since the $\ce{HNO2}$ is very unstable and decompose very easily (disproportioning):
$$\ce{3HNO2 \rightarrow HNO3 + 2NO +H2O}$$


text:
Maximum pressure that can be produced by a combustion reaction
Multiple-chamber artillery can launch a projectile with energies much higher than single-chamber guns.  I'm interested in setting a fundamental upper bound on the energy that a chamber (of a known volume) can impart to a projectile.  This would, for example, set a limit on the needed volume for a sub-orbital gun (given a projectile mass).  Since the pressure drop-off (like with single-chamber) isn't a constraint, the kinetic energy would be theoretically limited by pressure times volume.  But you can't possibly achieve infinite pressure with any chemical reaction.
For that limit, some sources claim that pressures of 60,000 psi can be achieved.  Coming from a physics perspective, I understand that a reaction should have a defined energy associated with an individual molecular transition.  We've had related questions on physics stack exchange, but my thinking is that chemical equilibrium may be be a necessary tool.
If we ignore reaction kinematics, I believe that the maximum pressure that a reaction can achieve is limited by two things:

the density of the reactants before reaction and
the energy of the reaction

Given this, we could simply model the reaction with ideal gas assumptions, resulting in simple algebra.  But I think this is wrong.  Higher pressure would push the reaction to incomplete combustion.  From a chemistry perspective, is there any theoretical maximum pressure that we could restrict all combustion reactions to?

Maximising the pressure probably isn't a good idea to maximise the energy imparted to a projectile. This isn't a chemical constraint but a physical one: if the pressure is too high the gun barrel bursts. Real projectile weapons rely on explosives that deflagrate (basically the pressure wave travels through the explosive slower than sound) as this allows sustained pressure on the projectile over time rather than a big bang.

According to New Energetic Materials, table 1.2, a list of explosives ranges from 19-48 GPa in pressure achieved.  This is two orders of magnitude greater than the 60,000 psi figure.
As far as maximum theoretical pressure, the question isn't really clear whether the maximum absolute pressure at which combustion can occur or the maximum change in pressure due to combustion is the value of interest.  The answer could be related to the maximum pressure at which chemical bonds exist.


text:
Why does gold react with cyanide and sodium hydroxide
Why does gold react with cyanide and sodium hydroxide and what is the chemical reaction's impact on gold's atomic structure and bonding?
What I think is since in the 'flotation' process, when the water, chemicals and other substances are added, gold isn't affected and neither is the pyrite, but when the cyanide and caustic acid is mixed with the slurry, it separated both substances, then the oxygen or 'air' is added into the mix, so then the gold rises to the top and frothing over the slurry. I'm not sure if i'm correct, but I don't understand how it's related to the atomic structure and bonding of gold? This process is from (Super Pit - Kalgoorlie exploration mine)

Can someone please explain how the separation of gold from the mixed compounds is related to its atomic structure and bonding? Is it to do with the 'Elution' process? Thank you in advance!

possible duplicate of Gold's atomic structure and bonding in relation to the separation techniques used

I am uncertain if this question is a duplicate. This question wants to know the chemistry behind the process.

I made sure that it not resemble my previous question, I request a much more intricate answer

Flotation and treatment of gold with $\ce{NaCN}$ in the presence of air are two distictive steps!
Flotation yields fine particles of gold (and pyrite) at the surface of a flotation tank, attached to foamy bubbles while the rest of the minerals sink to the bottom of the tank.
Flotation does not rely on differences in specific density, but on hydrophobicity!
In the cyanide treatment, the following process - similar to what Uncle Al described for the seawater case in his answer - occurs:
Solid gold is oxidized, the cations are bound by cyanide anions and form a water-soluble complex:
$$\ce{4Au + 8NaCN + O2 + H2O -> 4Na[Au(CN)2] + 4NaOH}$$
In order to reobtain solid gold, the $\ce{Au(I)}$ in the cyano complex $\ce{[Au(CN)2]^-}$ has to be reduced to $\ce{Au(0)}$ again. This can be done either by adding zinc dust, according to
$$\ce{2Na[Au(CN)2] + Zn -> Na2[Zn(CN)4] + 2Au}$$
or electrochemically.

Look at half-cell redox potentials.  Gold should be chemically inert in water.  Now, add a coordinating ligand and a mild oxidizing agent.  Gold complexes are deep thermodynamic holes. Saltwater and air will eat gold, giving chloroaurate complexes.  Cis-platin was accidentally discovered when "inert" platinum anodes in cell culture met saline and metabolic ammonia.
Gold often looks like a halogen (relativistic diddling of s- and d-orbital energies given probability near the nucleus).  Touching gold with aluminum can over time gives "purple plague" (aluminum-gold intermetallics at microchip circuit connections).  Gold and beryllium give brittle crumbly AuBe.  Gold is only inert under select circumstances.

Samir asked about "flotation", which means he does not know any basics.

You don't know about gold beneficiation that involves pulverizing, chemical conditioning agents followed by intense agitation and air sparging of the agitated ore slurry to produce a mineral-rich foam.  It works for pyritic and other sulfide gold in which the mineral matrix is grabbed.  Cyanidation and cyanide heap leaching are entirely different.

I don't quite understand this answer, predominantly due to my lack of knowledge in this field. What I really want to know is how to relate the separation process (Flotation), to gold's atomic structure and bonding. Thank you!

Samir, flotation does not work on gold as such. It works on minerals that occlude gold.  Do your chair parade - look it all up in Google, engineering and chemistry.

@UncleAl "chair parade"? I am going to borrow that! :)


text:
What is the "resonance hybrid" of benzene?
In my book, to explain the structure of benzene, it referred to the idea of resonance:

The idea of resonance is that the actual molecule is a definite structure that is a hybrid of two or more theoretical structures.

I partially understood this sentence, specifically how a molecule could have multiple structures, but am confused about what is meant by hybrid. Could someone please elaborate?

Strongly related: What is resonance, and are resonance structures real?

It means that what we usually consider to be the 'structure' of benzene is not the actual structure.
Although there are three π bonds in the structure of benzene, there is actually resonance, or electron delocalization. This means that the electrons are not localised in π bonds between two specific carbons, but distributed throughout the ring. This is one of the most important characteristics of an aromatic system.

The "localised" structures 1a and 1b above are resonance forms of benzene: neither 1a nor 1b are a true representation of benzene on their own; they are hypothetical, or "theoretical", structures.
The resonance hybrid is a weighted average of the resonance forms 1a and 1b. In this case, both forms contribute equally to the hybrid. So, each C–C bond has a bond order of (1+2)/2 = 1.5, which we can represent using dotted lines in 2 (each dotted line represents a fraction of a bond, so one solid and one dotted line means a bond order between 1 and 2).
More symbolically, we can also use the depiction with an inscribed circle 3. This means that the six π-electrons are shared between all six of the carbon atoms.

Couldn't we have a resonance structure with a positive and a negative charge in two adjacent carbon atoms? Isn't this also a valid resonance structure for the benzene?

The core feature of the benzene molecule that is 'caused' by resonance is the bond length.
You know from your previous studied that double bonds are shorted than single bonds.  In the 'resonating' Lewis structures for benzene, you would thus observe two kinds of bonds, three single bonds and three double bonds.  However spectroscopic studies show only one kind of carbon-carbon bond in benzene.  Hence 'resonance' implies some 'averaging' of structure between the two Lewis structures.  You can't really draw a Lewis structure for the true representation of benzene, which is why the circle motif was developed.

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


text:
Is there a general way to predict products of a chemical reaction
Is there a general way to predict products of a chemical reaction, given only the atomic masses of masses of all elements involved? for example:

$\ce{CuFeS2 + O2->}$
$\ce{K2CrO4 + H2SO4->}$
$\ce{FeOCr2O3 + K2CO3 + O2->}$
$\ce{K2Cr2O7 + H2SO4->}$

If there's no method to do so, are we supposed to memorize all chemical reactions?
PS: In case you are writing the mechanism of chemical reactions, please bear in mind that products are not known beforehand.

Buy a textbook on inorganic chemistry is my advice. It'll give you the basic knowledge about periodicity in the periodic table to know intuitively how a certain element is likely to react. The subject is too vast to answer here.

As @Brian has stated, this is a very broad request - there are a lot of factors to consider, depending on many factors in the reactants, conditions, temperatures, catalysts etc.  This resource could be a good start.  Take the time to learn this kind of things - it is an enjoyable process!

@ Brian, I would be grateful if you name the book

@ Brian: Is there any general method to predict the products at advanced level?

The underlying physical laws necessary for the mathematical theory of a large part of physics and the whole of chemistry are thus completely known, and the difficulty is only that the exact application of these laws leads to equations much too complicated to be soluble. It therefore becomes desirable that approximate practical methods of applying quantum mechanics should be developed, which can lead to an explanation of the main features of complex atomic systems without too much computation. -- P. A. M. Dirac

The question based on the incorrect assumption that reactions can be fully predicted based on only elements involved or must be all memorized - excluding any other options

Often times it is possible due to previous data. Trends appear and you can make deductions from such trends. Reactions may not happen at all. There may be deviations from the trends as well. The deviations may be for unknown reasons or there may be unproven theories that are included in textbooks. Inorganic chemistry reactions are not as clear cut as organic chemistry reactions. You don't have to memorize all the reactions. Know the trends, conduct the experiment, and if your prediction fails, try to think hard about the conditions in which the experiment took place. Sometimes the explanation may lay in the container that you used for the reaction, or it may be stoichiometry.
As in all sciences, you need experiments.

Is there a general way to predict products of a chemical reaction, given only the atomic masses of masses of all elements involved?

Unforturnately, no.

If there's no method to do so, are we supposed to memorize all chemical reactions?

There is some amount of reactions one have to memorize individually. However, there are classes of reactions that are quite broad and have uniform rules to check if some reactions happens. Two most common of such classes are redox reactions in  solutions, guarded by electrode potentials, and ionic reactions in solutions, guarded by compound solubility expressed as solubility product.
In your example

$\ce{CuFeS2 + O2 ->}$

Is an example of oxidation of sulfides. Most sulfides can be oxidized at elevated temperature by oxygen into stable oxides, in this case (if there is excess oxygen): $$\ce{CuFeS2 + O2 -> CuO + Fe2O3 + SO2}$$

$\ce{K2CrO4 + H2SO4 ->}$
$\ce{K2Cr2O7 + H2SO4 ->}$

Both reactions are ionic and guarded by relative acidity of chromic and sulfuric acids and led to formation of chromic anhydride.
The remaining reaction is unlikely to proceed at significant rate, as the oxide considered is kinetically inert. However, if it would react, the products would be potassium chromate (VI) and later ferrate (VI).

You can predict certain types of chemical reactions using formats.

Combination reaction: $\ce{A + B -> AB}$, for example $\ce{C + O2 -> CO2}$
Displacement reaction: $\ce{AB + C -> AC + B}$, for example $\ce{H2SO4 + Zn -> ZnSO4 + H2 }$
Double Displacement reaction: $\ce{AB + CD -> AD + BC}$ , for example $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$
Decomposition reaction: $\ce{ABC ->[\Delta T] A + B + C}$ [using heat], for example $\ce{CaCO3 ->[\Delta T] CaO + CO2}$ [using heat]

For redox reactions, I am still working

You started so well, it would be a shame if you do not finish your answer.


text:
Reasons for dry conditions in substituition reactions with benzene

All substitution reactions of benzene must be carried out in dry conditions with a catalyst that produces a powerful electrophile.

This was a statement from my book. My question is, why must it be carried out in dry conditions?
What will occur if moisture is present?

Are you sure that is what your textbook says? I am aware that there are solvent effects which make a reaction favorable kinetically or thermodynamically. But these effects vary depending on the intermediate or transition state complex. Also, I don't think that "substitution reaction" is specific enough. What textbook are you using?

Benzene and other aromatic hydrocarbons are immisible with water. So, probably there is no problem with benzene if we there is moisture.
In most of the substitution reactions which benzene undergo, requires Lewis acid catalyst, like ferric halide or aluminium halide. Most commonly used catalysts are Anhyd. $\ce{AlCl3}$ and $\ce{FeBr3}$, which become inactivated if they react with water. So, this requires dry conditions.

But isn't that only for halogenation?

@user97554. Benzene undergoes substitution reactions 1.Halogenation, 2.Sulphonation, 3.Nitration, 4.Alkylation, 5.Acylation. Almost all of them require lewis acid catalysts ($AlCl_3$ or $FeBr_3$). It is not only for halogenation.

With electrophilic aromatic substitution (EAS) you use a strong lewis acid to create an electrophile.
$$\ce{AlCl3 + R-X \rightarrow AlCl3X- + R+} $$
$$\mathrm{R = Cl,\ Br,\ I,\ alky,\ aryl,\ RC(O),\ etc...}$$
$$\ce{X = Cl,\ Br,\ I, R-C(O)O,\ etc}...$$
These lewis acids (not the benzene) are incredibly sensitive to water which will irreversably react with the acid to form a hydrate.
$$\ce{ACl3 + X H2O \rightarrow ACl3.X H2O}$$
$$\mathrm{For}\ \ce{AlCl3}: \ \mathrm{1\leq X \leq 9}$$
This hydrate is not as strong of a lewis acid as the anhydrous acid and no longer able to form electrophiles. That is why you must use anhydrous conditions with substitution reactions.

An electrophile goes after electron density, as in the two lone pairs in $\ce{H2O}$ plus poalrization.  You then require conditions wherein water is inert (hypohalite halogenations, maybe samarium triflate), micelles exclude water, or the substituted benzene is so activated (dimethylaniline) that it doesn't care.

I see a -1 here.  Somebody has never done industrial chemistry wherein "good enough" is rarely challenged by "better, but more expensive."

Actually, I upvoted your answer ;) But preparative organic (photo)chemistry in 0.5M SDS is dreadful. Been there, done that back in 1993. Lots of sat. $\ce{CaCl2}$ was used to precipitate the laurylsulfate!

Electrophiles
are reactive cations, they will react with water to form
eg an alcohol (or some weak acid etc)  and protons.
No reaction with benzene.

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


text:
Boyle's Law of pressure and volume
Boyle's Law states that pressure is inversely proportional to volume.
However, let's take an example of a balloon: as we fill air in it, its pressure increases, but its volume also increases. Can anyone explain?

(source: ucdavis.edu)

Balloon is not an ideal system to study pressure-volume relationship. Because, on expansion, the elastic skin also expands and there needs to be an additional pressure build-up on the inner side to counter that force, very similar to the extra pressure in a spherical bubble. (where the surface tension acts exactly similar to the elastic balloon skin)
But in your arguments, you are forgetting an important point. Blowing a balloon also involves increasing the amount of air in it. Boyle's law holds for only a fixed volume of gas. If you change the number of moles, Boyle's law no longer stays valid.

Let's start with the general gas equation:
$$P_1V_1 = nRT_1$$
Now to get to Boyle's law, we assume that both $T$ (temperature) and $n$ (the number of moles of the substance involved in our experiment) are both constant.  Now we can rewrite the above equation as
$$P_1V_1 = \mathrm{constant}$$
Of course if we change $P$ or $V$ in our system their product must still produce the same constant, or
$$P_1V_1 = P_2V_2$$
So we can't blow additional air into our balloon (adding more moles of material and violating one of our initial assumptions) and still expect Boyle's law to hold.  In a fixed system like the one we just described where Boyle's law holds, if we reduce the volume of our balloon $(V_2)$ by $1/2$ then the pressure $P_2$ must increase by a factor of $2$ in order to maintain the value of the constant.

As Satwik explained, a balloon is not really a great container for ideal gases because of its elastic nature. It would be better to imagine a piston like in the picture below.

As this picture shows when the piston goes down (towards the base of the container), the volume decreases. At the same time, since the number of moles of gas stays the same (a constraint that Sathwik and Ron explained well), the pressure goes up since the ideal gas particles are now more likely to interact (collide) with the side of the container. Thus we can see that as the volume of a system decreases, the pressure increases (they are inversely proportional).
Image adapted from: https://opentextbc.ca/chemistry/chapter/9-5-the-kinetic-molecular-theory/


text:
What are the mechanisms for the reduction of organic compounds employing red phosphorus and iodine?
I was recently studying oxidation and reduction reactions.
I found red phosphorus and iodine reduce many different organic compounds.
Out of interest I would like to know the mechanism for these reactions. A concerted path is preferred.

I found Red P and Iodine reduce many different organic compounds.

Actually, it is typically used for the deoxygenation in benzylic positions, using benzylic alcohols or benzoins as starting materials.
Aliphatic alcohols yield iodoalkanes, propargylic alcohols often undergo elimination under these conditions whereas allylic alcohols are typically converted to complex mixtures without synthetic value.
The deoxygenation is often carried out with catalytic amounts of iodine but an excess of red phosphorous. In principle, catalytic amounts of $\ce{NaI}$ plus one equivalent of $\ce{H3PO3}$ in water would work as well.
When using iodine and red phosphorous, hydrogen iodide is generated in situ via the formation and subsequent hydrolysis of $\ce{PI3}$.
The formal replacement by a proton is not fully understood and is disputed whether the formation of a benzylic iodide is really involved.
In princpiple, elimination from water from the protonated starting material would yield a stabilized benzylic cation. This might get reduced by a iodide to a benzylic radical and, in a second step to a very shortlived benzylic anion which is instantaneously protonated.
Both reduction steps yield iodine radicals from iodide and in summary regenerate iodine.

so the mechanism is not known correctly today right?

At least, I'm not aware of a canonical theory. People interested in the use of a particular reagent often do not care the exact mechanism and all intermediates involved. This has practical reasons: If you want to exploit the reaction, you want complete conversion, high selectivities, high yields. All this can be derived from the analysis of the stable products (and side products, if any) with standard techniques. In order to figure out the mechanism, you might need to use time-resolved spectroscopy to monitor and maybe identify the intermediates of the reaction. This is completely different.

Would the reduction work if a carboxylic acid with a benzene ring was used ? Eg. Phenylethanoic acid.


text:
Products of the o-Toluidine oxidation with glucose oxidase and peroxidase
Could anybody describe which product is formed in the following reaction (o-Toluidine oxidation)?
$$\ce{H2O2 + o-Toluidine_{uncolored} - Peroxidase -> H2O + 1/2 O2 + o-Toluidine\; Oxidate_{colored}} $$
Note: Hydrogen peroxide is one of the products of glucose oxidation to glucolactone by glucose oxydase.

is this homework?

@Amaterasu Not really. If I remember correctly, this was/is a method for the  photometric determination of glucose in blood.

@KlausWarzecha ah, good to know!

The product most likely is 2-methylnitrobenzene (3), which means that the amino group ($\ce{-NH2}$) in o-toluidine (1) is oxidized to a nitro group ($\ce{-NO2}$). The oxidation proceeds via 2-methylnitrosobenzene (2).

The formation of 2-methylnitrobenzene can be detected photometrically at $\lambda$ = 400 nm, where the aniline does not absorb.


text:
How much pressure is needed to make dry ice?
By compressing $\ce{CO2 (g)}$ the gas gains a lot of heat. When the hot compressed carbon dioxide is left to cool, it attains the temperature of the surrounding material. When it is released into the pressure outside (about 100kPa), it will get so cold (actual temperature depending on how much the carbon dioxide was compressed), that it will freeze (below 78.5 degrees celcius).
How much pressure is needed to make dry ice at 293K (20 degrees celcius)?

"" If that is left to cool, it will become so cold"" if it is left to cool, it will attain the temperature of the surrounding material. "so cold that" is not physical.

@Georg When I re-read this I see that it is a bit unclear written. I updated the question. Now it should be easier to understand what I meant. Thanks for your clarification.

Try to calculate the needed pressure from adiabatic expansion. Compare it with pressure needed to liquify the gas.If it is bigger than the liquifying pressure then the original idea gas->solid will not work. It would be gas->liquid->solid + gas instead. Note that the computation would not be accurate, as gamma will significantly change with temperature.

Isn't that a bit different?
At 293 K, you need about 56 bar to liquefy carbon dioxide. If you then release the pressure, a part of the liquid evaporates again while cooling down the rest until it solidifies at around -80 °C.

Does the carbon dioxide have to be liquid to turn into dry ice? You say that it is the evaporation that cools some of the gas down to -80 degrees. Is it the evaporation or the drop in pressure that makes it cold?


text:
Activated Complex Products/Reactants
We know that an activation complex is the temporary unstable compound formed when activation energy is reached. But what dictates whether the products are indeed going to be products or if they become reactants?
Does the activation complex form when orientation is both correct and incorrect, and only forms products when the orientation is correct? Or is the decomposition of the complex purely a chance matter?

As far as I know, it is a matter of chance in which direction the decomposition of the activated complex goes since it symbolizes a maximum of the free energy and so a structure alteration in any direction leads to a lowering of the free energy. Maybe this answer of mine might be interesting for you.

Transition state theory or activated complex theory is a very advanced topic and deals with chemical reaction dynamics. It goes beyond the scope of my knowledge, however, I will attempt my best to explain with what I know.
As stated, the activated complex is a temporary unstable product. As two particles with sufficient kinetic energy collide; along the reaction path, they distort the bonds and rearrange into new atoms. The location of the transition state for the activated complex is at the highest point; or peak, of an potential energy diagram.
As stated in my general chem textbook Principles of Modern Chemistry by Oxtoby et. al, p. 685:

The activated complex is assumed to exist as if it were in equilibrium
  with the reactants and the theory focuses on calculating the rate at which the
  activated complex passes through the transition state to form products.

Statistical thermodynamics is required to calculate all motions of particles, including both rotational and vibrational energy.
In the transition state, change from products is mediated when bonds between particles or molecules become out of phase (or antisymmetrically stretched) by vibrational energies. The frequency of the molecules becoming out of phase with one another decreases as the reaction proceeds forward and the new product is transformed. The energy transferred from vibrational energy to achieve the activated complex is turned into kinetic energy.
The Eyring formula makes it possible to calculate reaction rates using transition state theory:
$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}K^†$$
The kappa term is referred to as the transmission coefficent and refers to the probability that the particles will create the transition state. $K_\mathrm{B}T/h$ measures the rate at which the activated complex dissociates to form into it's products.
This equation is unwieldy to us; as you stated previously, it is extremely hard to determine equilibrium constants for the activated complex since the instability achieved at the height of the potential energy for the reaction exists for only a brief period of time.
Instead, we can relate the Eyring formula to thermodynamics and substitute $\exp (-\Delta G^†/RT)$ for $K^†$.
$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta G^†/RT)$$
$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta H^†/RT)\exp(\Delta S^†/R) $$
The equation that Arrhenius derived is:
$$k_r = A\exp (-E_a/RT)$$
Activation energy is expressed as activation enthalpy and activation entropy. The pre-exponential factor $A$ is substituted with $K_\mathrm{B}T/h$. This equation and the use of enthalpy and entropy as driving forces is used extensively in enzyme catalyzed reactions.

Collision with another molecule is purely chance matter but the orientation factor will be determined by the potential energy surfaces. If there is not a proper orientation then the it will be a surface with high energy surface but if there is a proper orientation then it will pass through the surface with lowest energy and that will dictate whether products are formed or the not

My answer may very well be reciprocated as a comment. Because I can't do that myself, I'll relay this as an answer.
The formation of the relevant activated complexes occurs only when the orientations are correct. This, though is not the case with its decomposition. According to the Hammond Postulate, the activated complexes take structures closer to the species (reactants or products) that have energy values closer to it. This way one can explain that in exothermic reactions the formation of transition state and the activated complexes is easier than in endothermic reactions. This allows for faster forward movement in reactions. The, activated complexes, thus, will now tend to move towards the state that bring it more stability.  So yes, it is indeed a mater of pure chance on which side the activated complexes shift, but it is not a matter of fair chance. As such, addition or removal of heat (accordingly if it is endothermic or exothermic) helps to quicken the formation of the activated complex and pushing it towards the more stable species.
Example wise, consider the free radical bromination and the free radical chlorination of ethane. Here, the first is endothermic while the latter is exothermic, both product formation and reactant formation is easier in chlorination. But product formation is what happens in the greater proportion.


text:
gas chamber for calibrating gas sensors
I have $\ce{CO2}$, $\ce{CO}$, $\ce{O2}$ sensors and need to calibrate them. Can anybody help me to produce the $\ce{CO2}$, $\ce{CO}$, $\ce{O2}$ in laboratory for calibrating the sensors (the gas should be produced in known in $\mathrm{ppm}$)

To be accurate you will need standard gas cylinder and pneumatic connections (pressure regulator, needle valves or better mass flow controllers, pipes). You will need a measuring chamber with an overflow. As a rule of thumb, if the chamber geometry is not exotic (lets say a cube with no dramatic dead volume), you need to flush it, at least  10 times, with each concentration you want to measure.
You can do a 2 points calibration for O2 by measuring atmosphere where O2 concentration can be assumed constant (span) and then flush with pure nitrogen (zero). For CO2 and CO you will need standard cylinders.
Anyway a 2 points calibration is a little bit risky because you will not be able to assess linearity and then you will have no idea in which extent your sensor is linear (dynamic range). Properly calibrating sensor requires multiple points calibration and therefore a dilution device and standards.
Caution: When dealing with gas you must take care of exhaust and prevents risk such as intoxication (especially CO and in a fewer extent CO2) or asphyxiation (due to a lack of oxygen) and explosion (CO is explosive).
You should find those materials, resources and skills in a University (Chemistry or Physics Department) or an Air Quality Laboratory.
Note: If your sensors are electrochemical ones, you need also to assess their correlations/deviations with temperature and humidity.


text:
Does benzene have isomers and resonance structures?
Does benzene have both isomers and resonance structures?
I know that it has resonance structures because you can manipulate the position of the double bonds but what about isomers?

Well, it depends on what kinds of isomers you have in mind. It certainly has constitutional isomers: For example, the sum formula $\ce{C6H6}$ does describe benzene as well as hex-1,5-en-3-in $\ce{H2C=CH-C#C-CH=CH2}$.
There are also a lot ring structures that are isomers of benzene:

As far as I know there aren't any configurational isomers of benzene since a six-membered ring doesn't allow for trans double bonds (this would be sterically impossible).

On a sidenote, Bicyclo[2.2.0]hexa-2,5-diene is also known as Dewar benzene, prismane as Ladenburg benzene. The smallest stable (at RT) ring with a trans double bond is indeed trans-cyclooctene.

As you note, benzene has resonance structures, but it does not have isomers.  All of the bond lengths in benzene are identical.  That is why 1,2-dideuteriobenzene or  1,2-dimethylbenzene (o-xylene) both exist as single compounds.  Here's an interesting way to look at it.  Look at the energy diagram below on the left.  If benzene had 2 isomers, that would be them as the two stable wells at the bottom of the energy diagram, the "resonance delocalized" version of benzene is shown as the transition state involved in the interconversion of these two "isomers".  Now look at the energy diagram below on the right.  This is the real case.  It turns out that the "transition state" is more stable than either of the 2 "isomers"!  It could be said that there is a negative activation energy for the formation of delocalized benzene from either of the 2 localized "isomers".  The negative activation energy would correspond to the resonance stabilization of benzene compared to either of the localized "isomers".  Hence, benzene exists as a fully delocalized structure with identical bond lengths at the bottom of the potential well.


text:
Half cells: difference between E(cell) = E(cathode}-E(anode) & E(cell) = E(oxidation) + E(reduction)
When do I use each of the above? If I use one or the other they give me difference results.
For example, I'm trying to put the Volts into an equation to find standard free energy change.
$$\Delta  G^\circ= -nFE$$
If I use the cathode minus anode I get a different answer to the oxide + reduct.

Hi, could you write out the problem and the calculations that you have done so far so we can get a better context of where you may have made an error?

Ah, I see what you did wrong.
The first equation is correct.
$$ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} $$
The second equation should be:
$$ E^\circ_\text{cell} = E^\circ_\text{reduction} - E^\circ_\text{oxidation} $$
Alternatively, you can write it as:
$$ E^\circ_\text{cell} = - E^\circ_\text{oxidation} + E^\circ_\text{reduction} $$
The cathode is actually the reduction step, and the anode is the oxidation step.
A useful mnemonic I learned from my chemistry professor is a red cat and an ox. Interestingly, cathodes in electrochemical cells are red and are positive. So I have found many uses for this along the way.

First, remember the following rules:

cell type
example
Where does oxidation occur?
Where does reduction occur?

galvanic cell
battery
minus pole, anode
plus pole, cathode

electrolysis cell
water electrolysis
plus pole, anode
minus pole, cathode

Gibbs free energy
Let's start with thermodynamics: The change in Gibbs free energy $\Delta G$ of a chemical reaction tells you if the reaction can occur without requiring external energy:

If $\Delta G < 0$, the chemical reaction is exergonic, meaning it can run without requiring external energy (it releases energy to the system around it)
If $\Delta G > 0$, the chemical reaction is endergonic, which means it requires external energy to take place.

Standard Reduction Potential
For electrochemical reactions, every redox pair has its own standard reduction potential $E^{\ominus}$. This is always measured as reduction reaction against the standard hydrogen electrode which has a potential of $0\rm\ V$ by definition:
$\rm 2\ H^{+}(aq) + 2 e^- \rightarrow H_2(g) \quad\quad E^{\ominus}_{\rm 2H^{+}//H_2} = 0.0\ V$
$\rm Zn^{2+}(aq) + 2 e^- \rightarrow Zn(s) \quad\quad E^{\ominus}_{\rm Zn^{2+}//Zn} = - 0.76\ V$
$\rm Cu^{2+}(aq) + 2 e^- \rightarrow Cu(s) \quad\quad E^{\ominus}_{\rm Cu^{2+}//Cu} = + 0.34\ V$
Each $E^{\ominus}$ value has a sign. You must always respect that sign in calculations! Also note that this value is always associated to the reduction reaction.
The change in Gibbs free energy for a reaction under standard conditions (indicated by $\ominus$, or $0$) is:
$\Delta G^{\ominus} = - zF\Delta E^{\ominus}$
With this you can already understand why some metals are more unreactive than others. Why is copper less reactive than zinc? Let's calculate the change in Gibbs free energy for the reduction reactions of both metal ions to the metal:
$\Delta G^{\ominus}_{\rm Zn^{2+}//Zn} = - 2\cdot 96485 \frac{\rm C}{\rm mol} \cdot (- 0.76\ {\rm V}) \approx +147 \frac{\rm kJ}{\rm mol} > 0\rm\ (endergonic)$
$\Delta G^{\ominus}_{\rm Cu^{2+}//Cu} = - 2\cdot 96485 \frac{\rm C}{\rm mol} \cdot (+ 0.34\ {\rm V}) \approx -65.6 \frac{\rm kJ}{\rm mol} < 0\rm\ (exergonic)$
Note that minus multiplied with minus makes a plus and that coulomb ($\rm A \cdot s$) multiplied with the unit volt $\rm V$ makes energy in the unit joule ($\rm J$).
You now can see that the reduction of copper(II) ions to copper metal in aqueous solution can occur spontaneously, because is it an exergonic reaction, while the reduction of zinc(II) ions to zinc metal cannot, because this reduction is an endergonic reaction.
What about electrochemical reactions?
Which of the following electrochemical reactions is exergonic?
$\rm Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s) \rm\ \quad(reaction\ 1)$
$\rm Cu(s) + Zn^{2+}(aq) \rightarrow Cu^{2+}(aq) + Zn(s) \rm\ \quad(reaction\ 2)$
As you stated, the potential difference can be calculated as follows:
$\Delta E^{\ominus} = E^{\ominus}_{\rm reduction} - E^{\ominus}_{\rm oxidation}$
So lets start with the upper reaction. In this case, copper is being reduced and zinc is being oxidized, so:
$\Delta E^{\ominus}_{\rm reaction\ 1} = E^{\ominus}_{\rm reduction} - E^{\ominus}_{\rm oxidation} = E^{\ominus}_{\rm Cu^{2+}//Cu} - E^{\ominus}_{\rm Zn^{2+}//Zn} = + 0.34 {\rm V} - (- 0.76 {\rm V}) = + 1.10 {\rm\ V}$
In reaction 2, zinc gets reduced and copper gets oxidized, so:
$\Delta E^{\ominus}_{\rm reaction\ 2} = E^{\ominus}_{\rm reduction} - E^{\ominus}_{\rm oxidation} = E^{\ominus}_{\rm Zn^{2+}//Zn} - E^{\ominus}_{\rm Cu^{2+}//Cu} =  - 0.76 {\rm V} - (+ 0.34 {\rm V}) = - 1.10 {\rm\ V}$
What does this means for the Gibbs free energy of those reactions?
$\Delta G^{\ominus}_{\rm reaction\ 1} = - 2\cdot 96485 \frac{\rm C}{\rm mol} \cdot (+ 1.10\ {\rm V}) \approx -212 \frac{\rm kJ}{\rm mol} < 0\rm\ (exergonic)$
$\Delta G^{\ominus}_{\rm reaction\ 2} = - 2\cdot 96485 \frac{\rm C}{\rm mol} \cdot (- 1.10 {\rm V}) \approx +212 \frac{\rm kJ}{\rm mol} > 0\rm\ (endergonic)$
From this you can see that reaction 2 is not taking place because it is endergonic, but reaction 1 can take place. This is in agreement with your chemical intuition, because copper is a more noble metal than zinc.
To summarize, I always like to put brackets around the values to make it clear to students that the sign of each value must be respected:
$\Delta E^{\ominus} = (E^{\ominus}_{\rm reduction}) - (E^{\ominus}_{\rm oxidation}) = (E^{\ominus}_{\rm cathode}) - (E^{\ominus}_{\rm anode})$
The half reaction with the more positive $E^{\ominus}$ value will be the reduction, because the reduction reaction for that half reaction is more exergonic.

Some people may use Ecell=Ecathode+Eanode or Ecell=Ereduction+Eoxidation if they keep each redox reaction written in the standard form, i.e. as a reduction, even though the anodic/oxidation reaction is not physically a reduction.
For example, i'f I'm working with a microbial fuel cell oxidizing acetate on the anode over a biocatalyst(E*=-0.29 for CO2 reduction to acetate) and producing water on the cathode over a Pt catalyst (E*=+1.229 for O2 reduction to water), my Ecell is 1.229+(-0.29)=0.939 V.

If the formulae seem daunting, you can just do it intuitively...
Let's consider
$\ce{A+ + B -> A + B+}$
To find the net voltage of the cell, we just add the voltages of the half reactions:
$E_{\text{cell}} = E_{A^+/A} + E_{B/B^+}$
($X^+/X$ can be thought of as $X^+ \rightarrow X$ here)
But we are given the reduction potentials $E_{A^+/A}$ and $E_{B^+/B}$. So just take negative of $E_{B^+/B}$:
$\begin{align}
E_{\text{cell}} &= E_{A^+/A} - E_{B^+/B}\\
&= E_{\text{cat}} - E_{\text{an}}
\end{align}$


text:
Why can we not use water instead of blank control solution in enzyme catalysis with Peroxidase experiment?
I have to do a lab about enzyme catalysis with peroxidase in my chemistry class. We are to use the blank control solution in the experiment. My question is, why can we not use water instead? I did some research about this and someone says something like "It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy." Is this the exact reason why we do not use water in this experiment?

It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy.

For water, the absorption coefficients $\alpha\ [\mathrm{m^{-1}}]$ at $\lambda$ = 680, 580, 535, 480 and 380 nm are 0.47, 0.09, 0.045, 0.013 and 0.011.
(Source: R. M. Pope, E. S. Fry, Applied Optics, 1997, 36, 8710-8723)
That is not a lot.
On the other hand, you never know about impurities in your buffers or other components (apart from your probe).
Therefore, you usually do not measure against "just the solvent", but against the solution of "everything except the probe" to record the complete background.


text:
Problems with precipitation of transition metal solution
I am trying to make a solution of "trace elements"--primarily transition metals-- for use in biological media; bacteria require elements like Cobalt, Molybdenum, iron, etc. in trace amounts. I am not a chemist, but my understanding is that these elements tend to form hydroxides in physiological pH ranges (slightly above 7). Therefore the authors of the protocol I am using employ EDTA as a complexing agent in 1:1 stoichiometric proportion with the metals in solution. The problem is that EDTA solubility (and complexing capability) and transition metal stability occur at opposite sides of the pH scale, and finding a balance to make both types of reagents happy is apparently really tricky; I've tried this 4 different times with no success. I am able to get all materials into solution, but any disturbance of my flask, (e.g. further stirring or shaking) inevitably causes precipitation.
Here is my recipe and protocol. I'll describe my observations after that. My question is:

Any thought what the precipitate is? How can I avoid it?

I have scant chemistry experience, so any insight you can provide would be useful.
Recipe

190 mM $\ce{EDTA}$ $\ce{Na4.2H2O}$ (equimolar to subsequent di- and trivalent cations)
29 mM $\ce{FeCl3.6H2O}$
80 mM $\ce{CaCO3}$
6 mM $\ce{MnCl2.4H2O}$
0.5 mM $\ce{CuSO4.5H2O}$
0.5 mM $\ce{CoCl2.6H2O}$
5 mM $\ce{ZnO}$
2 mM $\ce{H3BO3}$
66 mM $\ce{MgCl2.6H2O}$
5 mM $\ce{NaMoO4.2H2O}$

Protocol (making 1 L)

Dissolve EDTA in $\ce{750 mL}$ of water. Bring pH down to about 5.8 with $\ce{HCl}$.
Add $\ce{FeCl3}$ and watch pH; don't let it drop below 4.8.
Add $\ce{CaCO3}$. Allow to dissolve (i.e. $\ce{CO2}$ to evaporate) by maintaining pH at 4.8 with $\ce{HCl}$ and occasionally shaking vigorously (not sure why the recipe doesn't just call for something more soluble like $\ce{CaCl2}$.)
Add remaining elements in order given in the recipe, watching the pH. It should drop to about 4.2 without any addition of acid or base.
At this point solution appears clear. Bring volume up to $\ce{1L}$.
Autoclave solution (heat under pressure at $123^{\circ}\ce{C}$ for 15 minutes) ---this is required for biological sterilization, and seems to improve solubility for a time.

Observations
Solution remains transparent throughout preparation, stirring constantly throughout at room temperature. Color is orangish-green, due to iron and copper. Once all ingredients have been added, additional stirring before autoclaving leads to gradual (over 15 minutes) accumulation of a white powdery precipitate. I thought this was EDTA coming out, so I added $\ce{NaOH}$, but this only caused the iron to precipitate, forming a brown precipitate. Letting the solution sit also seems to lead to white precipitation, although less rapidly. When I autoclaved the solution, it remained transparent (color had turned to deep rust-orange) even after sitting overnight, but as soon as I disturbed it by transferring from an Ehrlenmeyer flask to a glass bottle it precipitated again. The pH of the precipitated solution seems stable at 4.2, what it was after all ingredients had been added. I have also tried this preparation adding EDTA after the boric acid and keeping pH below 6 with $\ce{HCl}$, but this caused even more rapid precipitation. I have also tried EDTA (fully protonated) after the boric acid and adding NaOH to deprotonate, but the results were even worse.

Here's an idea.  Add the cations as their acetates or nitrates.  Control pH with acetate buffer. Fe(III) is a bad player.  "Ferric acetate," $\ce{[Fe3O(OAc)6(H2O)3]OAc}$, is water-soluble. You might sparge the solution with argon before sterilizing (sealed vessel) to prevent oxidation, and/or add ascorbate (or erythorbate, often biologically inactive) as anti-oxidant - if that is the problem.
Lactate chelates, and may give greater solubilities.  Fe(II) lactate is E585 via the European Food Safety Authority.
(Note that my singular contribution to biochemistry was telling a lab its phosphatase assay bombed because...they used phosphate buffer.  I can't say they were grateful for acquiring the knowledge.)

If solids have formed then it can sometimes be hard for them to redissolve. A classic case is plutonium(IV) colloids. Once they form they are often very hard to redissolve. I think it would be best to chose a order of addition with care to prevent any solids from forming. Also the Mo as molybdate can precipitate in acid.

Is it conceivable that the white precipitate is $\ce{Zn(OH)2}$? Usually it precipitates more jelly-like, but anyway...
It will however dissolve both on lowering the pH, forming colourless $\ce{[Zn(H2O)6]^{2+}}$ and on increasing the pH, yielding the likewise colourless $\ce{[Zn(OH)4]^{2-}}$.