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Storage of Urine
Not all may be favorable to this project, but I will explain what I am trying to do. I work at home, and instead of walking a moderate distance to the bathroom and loosing my focus, I've been, at times, peeing in a 3 Quart Poland Springs water bottle. If you take offense at this, please do not continue reading except to be helpful in the scientific goal. I know this subject won't suit many types of people, so just ignore it if that is your case.
I noticed first of all that urine is not at all as sterile as people say that it is. The rate of growth of bacteria is relatively slow, but as a precaution, I found the need to use additional measures to prevent the growth of bacteria. I settled on the following method: I have two bottles and I add to each bottle about enough salt as can be soluble in the urine and sometimes maybe a little more. The one bottle then fills up throughout the day and is emptied, washed, and refilled with salt. The salt helps to kill the bacteria which would be lingering in the empty bottle. The next day, the bottle stays empty and the other is used.
I would add that I discovered that the bacteria (without the salt) does not usually grow unless the bottle is left with urine for two days. After this, however, that same bottle (without the salt) would retain the bacteria and immediately grow, if used again.
This system works relatively well, so long as it is done every day. It will even withstand 2 days with only moderate growth. (If I should leave it by mistake for longer it can get ugly). Nevertheless, I am still looking to improve upon this. One reason is that, if I drink less water or relieve myself normally, the bottle does not fill in one day. I am looking for someone with knowledge of chemistry to help me find a substance that can be added to this solution which fits a number of common sense criteria. I will also add a list of the substances that I have tried or already considered.
Necessary qualities

Safe to be handled and mistakenly touched. Implied that it will not eat through the plastic bottle. I can obtain a suitable glass bottle only with difficulty.
Something that will not kill a dog or a human through trace amounts, should they drink from the (clean) toilet bowl. For example we put ammonia in toilets, flush it, no problem.
Very important - something that will not react with the urine to produce gas, thereby causing the bottle to explode or expand. Something that will not (implied) produce a distinctively offensive smell. An example of this would be chlorine.
(Obvious) That it will be suitable to flush down the toilet.
Something that is not too expensive, physically large, or heavy in the quantity that I intend to use it. An example of this would be diluted vinegar.
Something not involving much preparation, easy to obtain, etc.

Substances considered

Vinegar: A moderate amount of vinegar did not have any effect. More than this would be difficult to obtain and store. Even at 5x the strength, I think it would still require a large amount, and this is not sold in stores that I visit.
Chlorine/Bleach: I understand that this substance reacts with (things in) the urine and produces gas. It would produce a distinctive pool odor and I believe it can also cause burns.
Ammonia: Same issues as chlorine - it would produce an offensive gas that burns my eyes.
Acetic acid: similar to vinegar, except dangerous to handle. I don't even know that it would work.
L-ascorbic acid (Vitamin C): probably dangerous to handle in a concentrated form and I don't know that it would work. Basically the same as vinegar. I know that it is used as a preservative, so it came on this list.
Alcohol - might work, smell might be tolerable. I would need to know more about what concentration and amount to buy/use.
Hydrochloric Acid - no

There is a list of preservatives on Wikipedia, and one that just now caught my eye was Sodium Propionate ($20/lb on eBay). Might not be a bad idea; maybe someone knows about it. I would much prefer something that could be bought in a store. Since this urine won't ever be consumed by humans, we should have more options than common preservatives - maybe I/we could push the PH range out to one extreme to inhibit all growth. Someone more knowledgeable could tell me how/if that could work without becoming dangerous.
Ideally, I think would be something that stops the bacteria growth completely. Next best would be something that gives me an extra day or two if I need it.
Thanks in advance.

Sodium Propionate or Sodium Sorbate are both suitable for inhibiting bacteria growth and would not pose any problem (unless consumed in large quantities). The metal ion (calcium, potassium, sodium) associated with the organic acid is not very important. I'm not sure how much you would need to use, but a teaspoon should be sufficient. If you search for "food grade ...", you will be able to shop from your browser.

I'll go with that. Your link is actually calcium propionate (doesn't matter). I'll leave the answer unmarked for now in case others posts come along. Also, I noticed that Sodium Sorbate has PH requirements, but Calcium Propionate did not.

@user1122069 I know, I did a search for food grade propionate and that was one of the cheapest. I don't remember the pH range for urine, but it should be between 5 and 8. Adding sodium sorbate will make the solution more basic.

I may check out some boric acid also. I've been a day and a half with the calcium propionate and it looks like nothing has grown.

Summary
In this answer I will present to you several ways of optimizing your method, quickly summarized:

UV irradiation
Better Sanitization of the Bottles

Wait, shouldn't urine be sterile?

I noticed first of all that urine is not at all as sterile as people say that it is.

It has long been a myth that urine is sterile, but it was debunked this year in a study. There are in fact low amounts of bacteria present in urine. That being said, a high bacterial load in urine is an indication of an infection in the urinary tract and should not be left unchecked.
How do we know what is growing in there?
I don't know how you checked what kinds of microorganisms are developing in the bottled urine, it might also be more fungal instead of bacterial. This might also affect the way to counter the problem, so without knowing exactly what is growing in there, the solutions might be worthless. For now I consider bacteria as the sole contamination.
Now, let us proceed to the interesting part.
UV Irradiation
Let's start off fancy. If you continue with your workflow of intermediate urine storage in bottles, it might be worth looking into solutions involving UV irradiation as a way to reduce the amount of microorganisms in the bottle. Something as simple as a UV diode screwed into the cap might make a huge difference. Maintenance will be low, but the higher initial cost could put you off.
It is always worth considering long-term solutions.
Reduce Contamination of Bottle
I don't know how you prepare the bottles prior to peeing in them, but you might well reduce the problem by simply sanitizing the bottle properly. One way would be to boil the bottle in hot water for about 30 minutes. However, this is quite time consuming and thus defeating the purpose of the exercise.
Another method would be to rinse it well with (warm) water and diluted ammonia (which you then can flush down the toilet) and then swirling some sanitizing solution (think of the commonly available hand sanitizers) around the bottle, even leaving that in there. An advantage of this method is that hand sanitizers usually contain fungicides and viricides, so you kill several birds with one stone.
I don't know if that already is sufficient to guarantee a two-day period where bacterial growth is inhibited enough that you will not observe it, but it might be well worth checking out, as hand sanitizers are widely available and should keep the cost of maintenance low.

I use potassium metabisulfite, sodium sorbate, or sodium benzoate as preservatives in winemaking.  I have to think it would work here as well.  All three should be readily available from a homebrewing or winemaking shop, and they are fairly cheap.  They can also be purchased online fairly easily.
Amazon sells sodium benzoate for $14.00 per pound, which should be month's worth of preservative.

I work at home, and instead of walking a moderate distance to the bathroom and loosing my focus, I've been, at times, peeing in a 3 Quart Poland Springs water bottle.

So instead of walking into the bathroom... you just pee into the Poland Springs water bottle next to you? And then you decide to leave it next to you the next few days?!?!

All jokes aside, it depends on what you want to do with the sample. If it is strictly storage and you want to disinfect and neutralize bacteria while keeping odors at a minimum, probably the easiest and most efficient method would be either to raise or decrease the pH to sufficient levels such that bacteria growth is inhibited.
Urine contains a lot of compounds that are excreted from the body. Protens, sugars, nitrites, ketones, bilirubin, urobilinogen, erythrocytes, leuckocytes, etc, etc. (1)
Obviously, a concern that is immediately raised is disposal of such acidic or basic solution. Since you are working at home, you will generally not have designated or specific areas for such waste disposal. The easiest method is to dilute the urine solution with water sufficently such that the solution is effectively neutralized and is ~7.0 pH. You can check the pH of the solution by using Litmus paper or just standard pH paper which is relatively cheap to come by. Once you reach the desired pH limit, you can then pour the solution after "walking the moderate distance to the bathroom" and dispose of it in the toilet or sink, whatever floats your boat.
If you don't want to do that and instead want to hang on to the urine/acidic/or/basic/solution, you can order a carboy. (2) Needless to say, it can be quite difficult to pour into the carboy depending on the original container you have. Then you can send it off to your local hazardous chemical disposal facility. Yay. They might wonder why your carboy's are all filled with copious amounts of urine. Be prepared to answer honestly so they can in fact, properly dispose of such material since it is biological in nature.
Collection of Urine
Urine is prone to collecting bacteria and other cellular contamination. Yes, the urine that comes out of your bladder is sterile. Your urethra, I don't know about you, may in fact not sterile. Common procedures are to stick a sterile catheter first up in through your urethra and clean it, although, that is in fact highly unpleasant, and is very painful. The most easiset way to collect a more or less "clean" sample midstream clean catch specimen. First pee into one container for a few seconds. Hold and clench, and then pee into the container where you wish to collect your urine. Then hold and clench again, pee the remainder back in the first container. Ok, I'm wondering what you do about washing your hands...
Container
Your container, must, must, must, must, be sterile. Every time you collect, you must use a new container. Can I say that again? It must be sterile. Otherwise it would defeat the purpose... so yea. The next question is, how long do you wish to store your urine next to you?
The most common containers are urine collection containers. These are generally the cups you pee in when you have a physical at your doctors. These samples are usually analyzed within the first few hours of collection.
If you wish to store for a longer period of time, say 24 hrs, a urine collection container is more suitable. This is a 3L container and is amber. The amber color is necessary since analytes such as porphyrins and urobilinogen are photosensitive.
Preservation
Common 24-hour preservatives are hydrochloric acid, boric acid, acetic acid and toluene.(3) A lot of the literature is pointing towards boric acid as the preservative. UCSF medical center has a nice protocol which you can follow here. (4)
Boric acid is pretty cool since it's powdery and not so soluble in room temperature as other acids. Yea, get the tablet's since they are way easier to deal with than the powdered form.
Other methods include preservation at $4^\circ$C for first 24-hr. Then deep freeze at $-80^\circ$C for longer term storage. Those refrigerator's cost tens-of-thousands so yea... dry ice might work.

Commercial multi-surface cleaners are available in spray bottles that claim to kill "99.9 %" of bacteria. A squirt or two (or 3 or more) could do the job, especially when the urine is fresh and there are only one or two (million?) bacteria present.
The cleaners containing quaternary ammonium salts at less than 1% and seem to be safe, even if ingested in tiny amounts. These are to be distinguished from the ordinary toilet bowl cleaners, which often contain HCl to dissolve lime deposits.

I asssume the problem with the bacteria is that there is a problem with odor. Since you are going to flush away the urine anyway the problem must not be the bacteria itself but what the bacteria is causing. The odor problem is solved simply by not putting a lid on the bottle, and covering it with cloth instead. Leave it open to the air and there will be no noticeable odor. It will not be necessary to add salt. Urine already has salts in it. They will be deposited in the bottle. They can be washed out with very hot water from time to time.


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Do elements actually transmute in nuclear reactions?
Whenever a nuclear transmutation happens, do we get a new element or do we have the same element with chemical properties of a different element?
Isn't it very much against the intuition?

By definition, a nuclear transmutation involves transformation of one element into another. This means the element has undergone some sort of change in the nucleus (either by nuclear reaction or nuclear decay), hence a change in atomic weight and/or number.
Whether or not this goes against intuition depends upon what experiences one's intuition is based on.

To me, a 'transmutation' requires a change to the number of protons in the nucleus, which clearly changes the electronic structure, which in turn changes the chemistry. Nuclear reactions such as neutron absorption only change the neutron number, and do not change the chemistry (well, except for hydrogen isotope chemistry where the mass difference can cause chemistry differences, particularly biochemistry). So, I will disagree with Encyclopedia Brittanica that neutron capture is a transmutation event...

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Odd color reaction with Marquis and MDMA
The textbook colour reaction of Marquis reagent (mixture cc. $\ce{H2SO4}$ + formaldehyde, optionally with some methanol to slow down polymerization) with MDMA (3,4-methylenedioxy-methamphetamine, the party drug "ecstasy") is a gradually appearing dark purple/blue colour.
Recently three different samples were examined with marquis. The samples were taken in different countries, at different times. They looked a bit different too, one was light brown, the other two were clear.
TLC results suggested they are the same compound with $R_f$ values that of MDMA.
Reaction with Mecke reagent was also MDMA-like.
Reaction with Marquis started OK, with fizzing (a sign of a HCL salt) a light and then dark purple colour appeared.
After maybe a minute or two, the dark colour started to fade, it turned into light purple. As the colour faded even more, it turned green, and then to light yellow.
The test was repeated with two different reagents and all three samples, the results were the same.
I found no documented nor anecdotal record of marquis turning green/yellow in the presence of MDMA after a few minutes.

Is this normal? Should Marquis turn green/yellow after a few minutes in the presence of MDMA?
If this is really unusual, then what could possible cause fading & color change?

Were your samples solid, or liquid? I can't help thinking about impurities that dissolve over time, changing pH, and color. Or maybe even release some sort of bleaching agent. What happens if you dissolve the sample in formaldehyde first?

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how many electrons are in the second subshell of this element?
first of all sorry if I have some mistakes in my "chemistry words", I'm Iranian and I know the Persian terms but only some of the English ones.
So suppose we have an element (x) with the atomic number of 24, so the electron configuration would be something like this :
$_{24}x : 1s^2\ \ 2s^2 \ \ 2p^6\ \ 3s^2\ \ 3p^6\ \ 3d^{5}\ \ 4s^1$
So the question is : in the element above, how many electrons are there in the subshell of l=2 (which is the subshell of d) and how many are there in the l=0 subshell (which is the subshell of s)
P.S. I just took an exam and this exact question was in the exam, my answer was 5 electrons for 'd' and 7 electrons for 's'

You're answer is correct. The s-shell has 2 electrons in the 1s subshell, 2 electrons in the 2s subshell, 2 electrons in the 3s subshell and 1 electron in the 4s subshell. The d-shell has 5 electrons in the 3d subshell.
The element with atomic number 24 is chromium, who's electron configuration can be shown diagramatically as follows:

Note that the 3d shell is 'half-filled' before the 4s subshell is completely filled.
Chromium is one of the 'exceptions to the predicted electron configurations' when applying Madelung's rule for quantum numbers and electron configuration. The 'Aufbau' principle (represented in diagram below) can be used to determine the electron structure of relatively simple atoms, by hand, as long as one remembers the anomalies (such as chromium and copper). The 'anomaly' is typically 'explained' by stating that the a 'half-filled' or 'completely filled' d-subshell is more stable and will therefore happen before 'completely filling' the 4s subshell.

In modern quantum computational chemistry, various numerical methods (such as Dirac Hyper Hartree–Fock method and density functional theory) have been devised to iteratively solve the Schrodinger equation for more accurate calculation of electronic structures of larger atoms and molecules.


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How report metal amounts as amount of their respective metal oxides?
I have performed an element analysis, and I I have calculated the amount of each metal in my sample, given in milligrams. Let's say I weighed in 10 mg of the powder. I want to see how much of that I have accounted for by the major components in my sample (those > 1-2 %).
Since my metals were not free metals in my sample, but rather as various oxides, I need to convert back to the most common oxides. This is the idea:

If I have 1 mg $\ce{Mn}$ in my sample, how many milligrams would that be if I
  have $\ce{MnO2}$ instead?

Should I use the number of atoms? Like this:
I have 0.002648 g of $\ce{Mn}$ = 4.82×10-5 moles = 2.90×1019 atoms. I then assume I have 2.90×1019 molecules of $\ce{MnO2}$. Going backwards, this gives me 4.815×10-5 moles = 4.187 mg. Would this be the correct way of doing it?

Yes, your calculation is correct. But how do you know you have 1 mg of Mn in your sample? You have oxides, and that's what you weigh.

That was just to express the problem. The software gave me a concentration of Mn given in mg/L. I then multiplied with the dilution volume, and obtained the amount of Mn in my sample. This was 0.002648 g. I weighed in oxides, but these have been digested in acid and atomized in the argon plasma.

So your calculation is correct. Figure out how much moles of Mn is that, and then calculate how much would a similar amount (in moles) of MnO2 would weigh.

Well, not really. There are a lot of manganese oxides. Look at the mineral Hausmannite. https://en.wikipedia.org/wiki/Hausmannite So you might very well end up with a total weight which is statistically significantly different that 100%.

Just digging up an old one here...Do you know if your sample was prepared for ash analysis? In other words, Was it combusted in a furnace to give anything from a light to dark gray powder prior to digestion and analysis?


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Why isn't it commercially suitable to dissolve sulfur trioxide in water to produce sulfuric acid?

Choose the two correct options from A-F, why isn't it commercially suitable to dissolve $\ce{SO3}$ directly in water to give concentrated sulfuric acid instead of using the contact process? The major reaction of the contact process are
$$\ce{S + O2 -> SO2}$$
  $$\ce{2 SO2 + O2 <=>2 SO3}$$
  $$\ce{SO3 + H2SO4 -> H2S2O7}$$
  $$\ce{H2S2O7 + H2O -> 2H2SO4}$$

A. To reduce the hazards of spillage
B. Because the density of the product is too high
C. To minimise transport costs of large volumes
D. Because the last step is too expensive.
E. Because the last step of the process is too exothermic.
F. An aerosol of the sulfuric acid rapidly fills the container.

The solution is C and F, but I think it should be A and D. Is it because to make gas react with $\ce{H2O}$, too much pressure is needed and it will be expensive? Can someone explain why each choice is false and true?

Part of the difficultly with this question is that it is asking about what is commercially viable as opposed to what is merely possible. It is, of course, quite possible to dissolve SO$_3$ in water, but doing so creates a runaway chemical reaction. So let's go through each of the points and explain why they don't make commercial sense.
(A) The contact process produces SO$_3$, as well as several other noxious gasses anyway, so they is no reduced risk as regards spillage.
(B) The density of the product is totally irrelevant, as you get the same product in both cases.
(C) Sulphur is a solid at standard temperature and pressure (STP), so one can transport a large amount using a standard container and transport vehicle. As a gas at STP, one can transport far less (many millions of moles less) SO$_3$ at STP for the same size container, so an unfeasibly large container would be needed unless a highly pressurized or very low temperature container is used. Either method is considerably more expensive, so less commercially viable.
(D) The last step of the contact process is to react oleum (H$_2$S$_2$O$_7$) with water. This is in fact relatively safe to do if done slowly and water is extremely cheap, so the process is not expensive
(E) The reaction of oleum and water is exothermic. However, in practice water is added relatively slowly to the oleum (both of which are controllable liquids), which controls the process.
(F) As said earlier, the reaction of water with SO$_3$ gas cannot be easily controlled. It runs away with itself, and the water starts to boil. This throws up lots of small droplets (an aerosol) of conc. sulfuric acid. This is highly dangerous, can increase the pressure in the reaction chamber, and also makes it difficult to cheaply extract the product from the reaction chamber.
Therefore, on economic grounds, the answer is indeed C and F.

$\ce{SO3}$ is not a gas at STP

A is false because even if you are doing it by contact process the hazard of spillage would be the same.
B is false because the product in both processes are the same. I don't know that there is a problem with high density material.
C is true because sulfur is solid and SO3 is gas and gases can occupy a large volume when same no. of moles are taken.
D is false because if the 2nd and 3rd step have same condition requiring high pressure to force its equilibrium forward according to Le Chattelier Principle
E is false because      SO3(g) + H2O(l) ---> H2SO4(aq), ......ΔH = -227.8 kJ
                        S(s)+O2(g)-->SO2(g), ......ΔH = -296.8 kJ ( even more exothermic )
F is true as it is only other option left given that two answers have to be selected and also it is a fact which might be explainable, but I cannot explain with tools in my hand at present.


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How can I balance the redox reaction of sulfur dioxide with oxygen?
How do I balance this reaction using half-redox equations?
$$\ce{2SO2 + O2->2SO3}$$
My work is:
Oxidation: $\ce{SO2 + H2O -> SO3 + 2e- + 2H+}$
Reduction: $\ce{H2O + 2e- + O2 -> SO3 + 2H+}$
Total: $\ce{SO2 + 2H2O + O2 -> 2SO3 + 4H+}$
BUT it's false. What did I do wrong?

Your reduction formula is wrong. For one thing you have no sulfur containing species on the reactant side and the reactant side has a charge of -2 while the product side shows a charge of +2, this is a violation of charge neutrality, both sides must have the same overall charge.

WELL I think you put something erroneous in your reduction formula.
You will never get SO3 out of just O2 and water. Instead of SO3
The reduction formla will be 4e- + O2 + 4H+ --> 2 H2O
Multiply oxidation formula by 2 and add it to new reduction formula and you got your answer


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Filling a buret tip before an experiment
I have a question that asks if the buret tip is not filled before the experiment what would happen (effect on molarity for acid/base titration). The way I'm thinking about it, it's not like you're putting a pre measured solution into it. You fill it as high as you fill it and you can't put anything in without getting the tip filled up too, so I don't see where that will change anything. Am I thinking about that right?

You fill it as high as you fill it and you can't put anything in without getting the tip filled up too,

No I don't think it will work like that. Usually you have the stock or valve closed at the bottom. Consequently, the tip will be empty until you bleed the buret. Further, you will often bleed (allow the titrant to flow) a buret while thumping the tip to remove any air bubbles that may form directly under the stop-cock when your titrant flows initially through the tip. Once you have titrant in the tip and the tip is devoid of air bubbles you will then close the valve and fill the buret to the desired level.
At this point you can better determine just how much you have used.
The measure marks on the buret itself take into account what is in the tip below the stop cock. But if you do not bleed the tip initially, filling it with your titrant then your first addition will be a bit off because the tip does not fill entirely with liquid before you allow some to drip through.
Here is a good write-up on titrations with burets.

A bubble in the nozzle of a buret will produce an inaccurate volume reading if the bubble escapes during a titration. Bubbles may be large and visible as shown above left or so small as not to be seen, above center. During a titration such small bubbles begin to move in the direction of the nozzle but may remain in place even though there is a moderate flow of titrant (above right). Even when the buret valve is wide open some bubbles remain in place until you take your eyes off them. Then they sneak through the nozzle and ruin your titration. Also if you let a an air bubble stay because you think it will remain and then later it leave to the tip you will have to start over because you have no idea what

I'm assuming that the question is talking about leaving air in the tip of the buret once you've filled it. In that case you would actually be adding less solution than you would have measured because the first half mL or so would go into filling the tip.

Yea, but in the lab we didn't measure any amount ahead of time, just filled it to the markings on the buret that we wanted.
(sory for the  annonymous post as an answer, away from my comp and coudn't remember my password to log in on this one.)


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The "rules" for LCAOs in Molecular Orbital Theory
In our course on physical chemistry, which involves MOT, we have been taught that in the LCAO approach, the wave function for a molecule … say hydrogen ion ($\ce{H2+}$), can be approximated by a linear combination of two atomic orbitals.
From what I could gather from the very little I know about variational method is that this linear combination is going to be then optimised to give the minimum energy to find the ground state of the molecule in question. Is this correct?
Also, if we are simply using the linear combination as a test/guess function for a variational method, why should we restrict ourselves to only (say $+$ and $-$) two combinations?
That is, how does the rule "Number of molecular orbitals = Number of atomic orbitals constituting them" arise?
Also, why is there the restriction in form of the rule that "Only atomic orbitals of close energies and proper symmetries combine to give MOs"?

I think the answer is great. The short answer to "proper symmetries" is that orbitals of different symmetry are orthogonal, so they can't combine to give MOs.

By saying that we restrict to only $+$ and $-$ combinations, are you considering only the specific case of $H_{2}^{+}$? If so, the reason that we always have a symmetric or antisymmetric combination of the two $s$ orbitals is the mirror symmetry with respect to the plane bisecting the molecule.

OP, I'd like to suggest you take a look at the first 8-9 lectures of Chemistry 107 (UCI Open Chem). It might prove helpful.

Does LCAO form only sigma bond not pi bond?

You are right to be questioning the validity of this method, and I congratulate you for doing so. This is actually an extremely important skill, that differentiates the best students from the rest.
The variational method, like many methods from physical chemistry, is a method of approximation (a model) to what really happens. There exist a whole hierarchy of methods for computing the molecular orbitals of molecules (which are themselves models, being the stationary states of the Schrödinger equation), including at the top end post-Hartree-Fock theory and density functional theory. These methods provide quantitative information about molecular orbitals, but also require serious computer power (which is generally unavailable to undergraduate students). However, all of the basic physics can be explored and understood with simpler models that can be solved on a few sheets of paper, such as LCAO theory, hence why we teach them to undergraduates.
As regards to your question, we do indeed optimise a linear combination. We start off by assuming that the total molecular orbital wavefunction can be approximated using a linear set of atomic orbitals:
$$|\Psi\rangle = c_1|\phi_1\rangle + c_2|\phi_2\rangle + \cdots + c_n|\phi_n\rangle$$
We then need to find the coefficients. The variational principle (also known as the Rayleigh-Ritz method) states that the coefficients that give the best approximation to the wavefunction will minimize the energy, given by
$$\mathcal{E} = \frac{\langle\Psi|\hat{H}|\Psi\rangle}{\langle\Psi|\Psi\rangle}$$
Now, the second part of your question involves how to compute the two terms in this fraction. Without going into massive mathematical detail, the LCAO method can be recast into a matrix problem rather than an integral problem using the atomic orbitals as a form of basis vector. In the simplest case (Hückel theory) we assume that they are normalized, such that the denominator is always 1 or 0 (see later). The problem now is mostly how to determine the numerator.
In brief, each element of the Hamiltonian matrix is given by
$$H_{ij} = \langle\phi_i|\hat{H}|\phi_j\rangle$$
The variational principle applied to this matrix implies that the optimized energies are the eigenvalues of the Hamiltonian matrix, and the coefficients are given by the eigenvectors. Since an $n$ x $n$ matrix has precisely $n$ eigenvalues, this implies the "number of molecular orbitals" rule. Notice that the reason for the +/- combinations is by our own design (LCAO). We could have picked more complex trial functions by assumption, but this would make computation far more difficult.
It most certainly is true that in order for there to be a significant interaction, two orbitals must be close in energy. The detailed reasons are complex, but essentially it comes down to the size of the $\langle\phi_i|\hat{H}|\phi_j\rangle$. Orbitals that are far apart in energy have small values of this term hence interact weakly. This doesn't necessarily mean they can't form molecular orbitals, however, but these effects are negligible and are not important in understanding the chemistry, so are neglected in simple models (although are often included in some of the most complex modern methods).
The reason that only atomic orbitals of the same symmetry give molecular orbitals is the overlap integral $\langle\phi|\phi\rangle$. This is the total sum (integral) of the product of the AOs. This is zero for different symmetries. This must be non-zero to give an MO. Consider a $\mathrm{2p_z}$-orbital and a $\mathrm{1s}$-orbital for instance (for illustration purposes only!). The s-orbital is spherically symmetric, with all points the same sign. The $\mathrm{p_z}$-orbital has a "dumbbell" shape, with equal areas of different signs above and below the $xy$-plane. Therefore, the total sum of the product is zero. No overlap = no interaction.

Thank you very much for your answer! What I don't get is how does variational principle produce the energy of the higher states? After you answered, I searched for variation principle and the Hamiltonian matrix and I came across linear variation method, which minimises the energy wrt each coefficient and from condition for non trivial solutions we get the values of energy. How do we know that these correspond to the higher states of the molecule?

Whilst it is true that the variational principle is only rigorously true for the ground state, one can in principle use this to extract higher states based upon the orthogonality of the wavefunctions: a consequence of the Hermiticity of the Hamiltonian operator. This is possible if you know a quantum number (such as energy) that differentiates the states. For some trial wavefunctions, this involves complex projection operations, however due to the way the LCAO can be formulated using Hermitian matrices, we can get a bunch of orthogonal higher states "for free"

@DrHarps the variational principle is perfectly valid for excited states.  The issue is that there is a difference between molecular orbitals $\psi$ and the state that they describe $\Psi$.  This discussion has focused on the solution of one $\Psi$ (the ground state), but it is possible to solve for higher states $\Psi$ using techniques such as multiconfigurational HF theory. A somewhat crude description is that an excited state is the same MO set as the ground state, but with different occupations, orbital coefficients, and CI coefficients, cf. the State-Average CASSCF method.


text:
Why is the methyl radical planar?
Why is the methyl radical planar? The VSEPR theory would predict an angle between 120° and 109.5°, while it is actually 120°.

VSEPR is about "electron PAIR repulsion". I would say it does not predict anything for radicals. Or at least be very careful to not make definite statements. ;-)

Related: Shapes of methyl and silyl radicals

This question unfortunately illustrates the significant shortcomings of the VSEPR model. This model is based upon classical electrostatic arguments. The molecular geometry will seek to maximize the distances between the repulsive electrons in the valence (outer) shell, that are completely localized in their chemical bonds and their lone pairs/singly occupied molecular orbitals.
This model works well in some cases. However, when quantum mechanics comes into play, it falls pretty flat, as it totally neglects (the really important) quantum effects. A better way of finding the molecular geometry is to find the geometry that minimizes the total energy of the molecular orbitals. In this case, quantum mechanical calculations find that a set of three $\mathrm{sp^2}$-hybridized molecular orbitals arising from a $D_\mathrm{3h}$ symmetry has a lower total energy than breaking the symmetry to form the pyramidal $C_\mathrm{3v}$ geometry.
However, this does not mean that all $\ce{\overset{.}{C}X3}$ radicals follow this pattern. Put an extremely electronegative substituent on in place of $\ce{H},$ e.g. in $\ce{CF3},$ and the situation changes. The lowest energy geometry is now the pyramidal one.
Sadly, quantum chemistry doesn't always care to follow general rules.

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text:
Gaussian Calculations - What is the Basis?
What am I actually changing when I click 631G(d) or 631G(d,p), etc. when I choose a "basis" in restricted Hartree Fock calculations? What is the difference between using an alternative basis?  Many thanks.

The basis set is used to describe the molecular orbital wave functions.
I suggest reading through a number of links:

Wikipedia
The Gaussian manual - Basis Sets - a nice description of the different basis sets, if a bit technical at times.
Naming of quantum chemistry basis sets
GAMESS Crash Course - applies to Gaussian or other packages

Really nice links thank you! I'm still a little confused as to what the "diffuse function" is though.

In some cases, the electron distribution is far from the nucleus. A great example is $\ce{H-}$, where the extra electron is really diffuse. So a "diffuse function" might be a very large "s" or "p" function to describe the wave function far from the nucleus.

Diffuse functions are critical when dealing with anions (among other situations) since the extra electrons are usually.. well, diffuse.

ah thank you I see now, its just how spread out the electron density is then? It seems though that if I change the Basis set to whatever I want I will get a different answer for the energies, how do we know which to trust, let alone which programme?

@RedPen If you're using Hartree Fock, you can use the variational principal, which clarifies that the lower the energy, the closer to the "true" correct total energy. In general, you use the largest basis set you have time to use, e.g. http://chemistry.stackexchange.com/questions/7920/suggest-methods-and-basis-sets-for-a-variety-of-systems?rq=1

Thank you very much for your help, I'll read over these links over the next few days and see if anything comes up I don't quite understand :)


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Symmetry C3V For MeCl
Methyl chloride belongs to the $C_{3\mathrm{v}}$ point group. This is because it has a $C_3$ axis down the $\ce{Cl-C}$ bond.
However what I do not understand is how it has three vertical mirror planes? It has three planes down each of the $\ce{C-H}$ bonds, but don't these bisect a bond making them dihedral planes? Although I'm wrong I can't fathom why it isn't a $C_{3\mathrm{d}}$ point group. I have used a molecular viewer to visualise it but I still get the vertical mirror planes bisecting a bond...

Confusingly, C3d groups don't exist in the nomenclature I learned. The "v" subscript is used for C groups and "d" is used for D groups, but the difference between the notations (or for that matter, the exact distinction between a vertical and dihedral reflection) has never been made clear to me.

In principal, the "vertical" mirror planes go along bond axes, and the "dihedral" mirror planes go between bond axes. (Consider, for example an octahedral complex.) In practice, $C_n$ point groups don't have dihedral "d" mirror planes, and plenty of exceptions exist (e.g., ferrocene, where the mirror planes are both "d" and "v" at the same time).

You've correctly identified the $C_3$ axis that contains the $\ce{C-Cl}$ bond.  Methyl chloride also has 3 planes of symmetry that contain the $C_3$ axis and each of the $\ce{C-H}$ bonds.  Since these planes contain the $C_3$ axis they are referred to as $\sigma_{v}$ planes of symmetry.  There are no other symmetry elements in methyl chloride so it belongs to point group $C_{3v}$ (n $\sigma v$ planes plus a $C_n$ axis).
There is no $C_{nd}$ point group.  There is a $D_{nd}$ point group which contains a $C_n$ axis, n $\sigma_{v}$ planes (so far methyl chloride fits the description), but also n $C_2$ axis.  Methyl chloride does not contain any $C_2$ axes, so it does not belong to the $D_{nd}$ point group.

Thank you very much for your answer. I think I understand now! The $n$ $C_2$ axes would have to be perpendicular to the principal axes right?

Yes, that correct.  If you want to apply it to some molecules, allene is $D_{2d}$ and the chair form of cyclohexane is $D_{3d}$.

If I remember my symmetry correctly then a dihedral plane by definition is a vertical plane that bisects a perpendicular $C_2$ axis. Since we have a perpendicular $C_2$ axis then I think that would mean you label it $D_n$.
In this case there are no perpendicular $C_2$ axes so I think for that reason they are just vertical planes until such time that you they bisect a perpendicular $C_2$.
I don't know what would be the case if you had $C_n$ on the principal axis and then $m$ perpendicular $C_2$ axes, where $m\neq n$, maybe someone could shed some light? My guess would be that it wouldn't be $D$ but would qualify for the $\sigma_d$ plane.

You can't have $m$ perpendicular $C_2$ axes where $m \neq n$ unless we're talking about a highly symmetric point group, like $I_h$ or possibly $T_h$.. I'd have to think about those.

The picture shows the vertical symmetry planes along the principal $\ce{C3}$ axis in the $\ce{C_{3{\mathrm{v}}}}$  molecule $\ce{CH3CN}$ which has a $\ce{CN}$ group instead of Cl.


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Grapefruit and sugar gelatin like substance
More or less I am trying to figure out how this reaction occurred, so I can reproduce it.
On a plate I had an excess amount of sugar with a fair amount of grapefruit juice which was stuck to the plate. I filled the plate with water and left it for a little over 24 hours. When I finally came back to do the dishes, I poured the contents of the plate into the sink only to find that the bottom of the plate was covered in this slime very similar to clear gelatin. This slime was about as thick as honey, but was not sticky in any way, and was completely clear.
What could have caused this?

If the plate had mostly dried, it might have been a gel formed from the pectin in the grapefruit.  That's just a guess though.


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Why do principal energy levels in an atom get closer together as n increases?
The title says it all. Reasons that I can supply include: increased nuclear charge increasingly catches up in terms of influence to the increasing shielding and proof by contradiction in that if the space remained the same there would be negligible nuclear pull on the valence electrons and so this must not be the case. What is a comprehensive explanation for why this phenomenon is so?
Thank you!

It's not at all clear what system you're talking about in the question? I guess the hydrogen atom? (The same thing would be true for particle-in-a-box, metals, etc.)

I'm talking about any atom.

That's an artifact of Euclidean space and the geometry of the problem (as explained in another answer), and the fact that the energy of an electron is constrained. As you move an electron away from a nucleus, the Coulombic attraction decreases, and the total energy of an electron in a state further removed from the nucleus correspondingly also decreases. A bound electron infinitely far from the nucleus has zero potential energy and zero total energy (since it is bound).
That leaves the question of how to decrease the energy between zero and a finite value in an infinite number of steps. There are many ways of doing this (any potential function with the form $-kr^{-n}$ will do the trick), but ultimately you have to allow the separation between energy levels to go to zero, otherwise you can't have infinitely many energy levels.
That is not entirely satisfactory. It kicks the can down the road and requires an answer to "why are there infinitely many energy levels?" I'll let someone else answer that in detail, but it is related to the fact that for high energies (a near free electron, say) the behavior of the system approaches the classical limit. In a classical Coulombic potential the allowed levels form a continuum, a key difference between classical and quantum systems (although that is again a bit of a circular explanation).

The energy levels for an atom are similar to a "particle in a box", except that the walls have a different shape. When you analyzed the "particle in a box" situation, you used a rectangular box. An atom is spherical, so the walls of the atom open outwards at the same rate as the sphere increases in surface area. You should have noticed that the energy levels for a "particle in a box" decreased in distance at higher energy levels. The surface area of the sphere brings the energy levels even closer.

It is to do with the way the potential energy changes with distance, which is different in different cases so there is nothing 'special' about an atom. As quantum number increases (a) in a harmonic oscillator, energy levels are equally spaced, (b) in rigid rotor rotational motion they diverge, (c) in the anharmonic oscillator (e.g. Morse potential) they become closer and (d) in the particle in a box they diverge.

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text:
Sulfur trioxide structure
What resonance structure is the most abundant in sulfur trioxide: the one with one double and two single, the one with two double and one single or the one with all three double bonds? Is it actually correct to write an expanded octet when it is not necessary?

Have a look at this excellent answer. It contains some interesting information about the bonding in SO3.

So it is just one double bond. I don't get the sulfite ion explanation. There are only single bonds in it. Is it correct to write expanded octets by Lewis when they aren't necessary, or it is octet over expanded octet?

The problem is that Lewis structures very rarely give a realistic account of the actual bonding situation in a molecule. It is rather good for organic molecules but for molecules like SO3 it is not suitable. Fact is that all S$-$O-bonds in SO3 are the same and they are neither typical single nor double bonds. So maybe it would be best to use some delocalised double bonds like you would with benzene.

I know that they are delocalised, but what do you suggest me to draw, normal octet or extended octet? What is more useful from these two in organic chemistry?

It does not really matter which one you choose, because they don't carry that much meaning anyway. The most common respresentation I know is the one with three double bonds - it reflects that all bonds are equal. If you choose a Lewis formula with one or two double bonds, you must in principle include the other mesomeric structures as well. That's why I suggested using delocalised double bonds, i.e. dashed ones, they convey the more involved bonding situation better.

Using an expanded octet and providing three equivalent double bonds minimizes the formal charges on all the atoms to zero.  I personally am partial to this particular representation of SO3 in a Lewis dot structure for that very reason.

Yeah, that's true, charge separation should be avoided, but on the other side, MO theory implies a structure that corresponds mostly to the lewis structure with one double and two single bonds.

The answer depends on which teaching you follow.
Traditional teaching, dating back to Linus Pauling, would prefer three double bonds to minimise charge separation, requiring an expanded octet.
Even older traditional teaching would favour the validity of the octet rule and take charge separation as a tolerable by-product.
Both sides have their advantage and disadvantage:

The Pauling method

correctly shows that all three bonds are equal without resorting to mesomery
incorrectly assumes a bond order of $2$ rather than the actual $1.4$

The charge-separation method

predicts a bond order of $1.33$ which is much closer to $1.4$
requires mesomery of three structures to explain the three equal bond lengths.

I for my part prefer the mesomeric charge-separation description, because we have no trouble drawing benzene in the same way and because I like the octet rule. Others will disagree and prefer the three double bonds.
I wouldn’t know anybody who prefers a description with two double bonds and one single bond. It is just between the two models making it all the less correct.
Please also see this excellent answer to a related question by Martin which includes molecular orbitals for $\ce{SO3}$


text:
How to calculate the concentration of lead(II) liberated from lead(IV) oxide under acidic conditions?
What is the concentration of $\ce{Pb^2+}$ in a lake that has a pH of $6.0$ and is in equilibrium with $\ce{PbO2(s)}$ and atmospheric oxygen?
\begin{align}
\ce{PbO2(s) + 4H+ + 2e- &-> Pb^2+ + 2H2O} &
\log K &= 49.2
\end{align}
I spent so much time to solve this, but I could not, my final answer does not appear to make sense:
\begin{align}
  \mathrm{pH} &= 6 &
  \implies&& [\ce{H+}] &=10^{-6}\\
  \log K      &= 49.2 &
  \implies&& K &= 10^{49.2}\\
  K &= \frac{[\ce{Pb^2+}]}{[\ce{H+}]^4} &
  \implies&& 10^{49.2} &=\frac{[\ce{Pb^2+}]}{(10^{-6})^4} \\
  &&\implies&& [\ce{Pb^2+}] & =10^{25.2}
\end{align}
What is my mistake?

I think that the problem here is one of notation. I'm pretty sure that the logarithm intended to be used is the natural logarithm, $\log_e(K)$ also represented as $\ln(K)$, and not the base 10 one. Otherwise your method is good.
If you use this, you get a much more reasonable answer (work it out). It also makes sense that the question was worded in this way, as the equations in which one tends to use the equilibrium constant $K$ actually use its natural logarithm, for example the Gibbs's free energy and the Nernst equation.
By the way, this reaction is also part of the reduction process at the cathode of a lead-acid battery.


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Coloured compounds on reacting with oxygen
In the reaction of chlorine gas with water, Hydrogen chloride and nascent oxygen is produced.The reaction of any coloured substance with this nascent oxygen makes it colourless.
Is it true?if yes, how?

The feature that makes colored compounds colored is the one that makes the compounds easily oxidized. When the compound is oxidized, something does not exist anymore. I'm not giving you a straight answer since you will experience a nice enlightenment when you figure that out. Take a look at the structure of beta-carotene, for example. Where would the oxygen go and what would the structure look like after that?


text:
Calculating mass and volume of two different substances required for one solution
I want to produce a solution, which contains $0.02~\mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ and $~0.2~\mathrm{mol~L}^{-1}~\mathrm{NH}_3$. The following substances are given:

copper(II) sulfate pentahydrate
$100~\mathrm{mL~H}_2\mathrm{O}$
diluted ammonia solution with a concentration of $2~\mathrm{mol~L}^{-1}$.

I need to determine the required volume ($V$) of $\mathrm{NH}_3$ and the mass ($m$) of $\mathrm{CuSO}_4\cdot 5\mathrm{H}_2\mathrm{O}$.
I have no idea how to calculate $m$. However, to calculate $V$, I solved the following system of equations:
$\dfrac{2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V}$
$\dfrac{0.2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V+0.1~\mathrm{L}}$
Therefore, $V = 0.01111~\mathrm{L}$. Is this correct?
Is $m=555~\mathrm{mg}$ correct?

To calculate the molar mass of a molecule you simply add the masses of each atom in the molecule. Eg H2O weighs 18.016 g mol^-1. So Cu is 63.5 g mol^-1, S is 32, O is 16and H is 1.008. 63.5+32+(416)5(21.008+16)=249.58g mol^-1

To calculate the molar mass of a molecule you simply add the masses of each atom in the molecule. Eg H2O weighs 18.016 g mol^-1. So Cu is 63.5 g mol^-1, S is 32, O is 16and H is 1.008. 63.5+32+(416)5(21.008+16)=249.58g mol^-1. You want 0.02 mols L-1 or 0.002 mols in 100mls. So 0.002mol *249.58 g mol-1= 0.499g.

We have direct formula for calculating the weight of the substance as,
$$\text{Weight}=\frac{\text{Molarity} \times \text {molecular wt} \times \text{volume in mL}}{1000}$$
$$=\frac{0.02\times 250 \times 100}{1000}$$
$$=0.5 \text{ g}$$
So to prepare 0.02 $ \mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ ,0.5 g of copper(II) sulfate pentahydrate should be dissolved in 100 mL water.
To find the volume of ammonia,
Initial molarity = 2 M
Initial volume= ?
Final molarity = 0.2 M
Final volume = 100 mL
We have formula,
 $$ \text {Initial molarity} \times \text{Initial volume} = \text {Final molarity} \times \text{Final volume}$$
$$\text{Initial volume} = \frac{0.2 \times 100}{2}$$
$$= 10 mL$$
Hence, 10 mL of 2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$ should be added in 100 mL water to obtain 0.2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$.


text:
Synthesise the CBZ derivative of glycine from glycine and benzyl chloroformate

Question
  The benzyloxycarbonyl group (CBZ) is widely used to protect amines, especially in aminoacid chemistry. A procedure to synthesise the CBZ derivative of glycine from glycine and benzyl chloroformate is given in the description below.
Glycine ($0.1~\mathrm{mol}$) was dissolved in aqueous sodium hydroxide ($50~\mathrm{mL}$ of a $2~\mathrm{M}$ solution)
  and the solution was cooled in an ice bath. Benzyl chloroformate ($1.2~
\mathrm{eq}$) and aqueous
  sodium hydroxide ($25~\mathrm{mL}$ of a $4~\mathrm{M}$ solution) were added dropwise simultaneously to the
  glycine solution at $0°\mathrm{C}$ dropwise over thirty minutes. The mixture was stirred for an extra
  ten minutes, then warmed to room temperature. The solution was extracted twice with
  ether. The combined organic layers were acidified to $\mathrm{pH}=1$. The precipitate was filtered,
  washed with a little cold water and dried to give carbobenzoxyglycine ($18.5~\mathrm{g}$), m.p. $119-120°\mathrm{C}$.
(a) Relative to glycine, how many equivalents of sodium hydroxide are used in this reaction? Explain why this number was used.
(b) Can sodium carbonate ($\ce{Na2CO3}$) be used in place
  of sodium hydroxide? Explain why it can or cannot be used.
(c) What inorganic by-product will be formed (write its chemical name)?
(d) The procedure given includes a serious mistake, which would result in a yield of $0\%$.
  What is the mistake, and how might it be corrected?

My answers
(a) I mentioned $1~\mathrm{eq}$ of sodium hydroxide as sodium hydroxide reacts with glycine in a 1:1 ratio.
(b) I mentioned that sodium carbonate cannot be used in place of sodium hydroxide as sodium Carbonate is not basic enough to catalyst the reaction.
(c) Sodium chloride
(d) I mentioned that during the process when the combined organic layers were acidified to $\mathrm{pH}=1$, it could destroy the product by hydrolysing it hence giving a yield of $0\%$.
I'm unsure whether my answers are correct though, could anyone explain? Also, are there any resources on the web that contain practice questions of a similar nature to this?

I agree with Ron's answer except for d. Using an extra equivalent of NaOH is the right procedure since you need to "mop up" the HCl that is produced when $\ce{CbzCl}$ reacts with the amino group. What is not correct is this:
When you wash the reaction mixture with ether you get rid of any impurities and your product should stay in the aqueous phase. So you have to acidify the aqueous layer and not the ether one. pH 1 should be fine for the Cbz unless you leave it there and go away for the weekend.

Nice answer K-P, +1.  When I read the question I was thinking Cbz and $\ce{NaOH}$ in the same dropping funnel, but now I see that's not the case.

Under neutral conditions glycine exists as a zwitterion as shown below.

In this form, there is no "amine" available to react with the carboxybenzyl (or benzyloxycarbonyl, or Cbz, or Z) protecting group.
a) 1 equivalent of sodium hydroxide ("dissolved in aqueous sodium hydroxide (50 mL of a 2 M solution)") was used to convert all of the glycine zwitterion to the glycinate anion.  In the anion, we do have a free amine group that can react with Cbz.
b) The $\mathrm{p}K_\mathrm{a}$ of the ammonium group in the glycine zwitterion is around 9.8 (see here for a nice discussion of the acid-base equilibria involved with amino acids), so, as you noted, a strong base is needed to convert the ammonium ion to the amine.  Sodium carbonate is not sufficiently basic to accomplish this.
c) After adding sodium hydroxide to the glycine zwitterion, we form sodium glycinate and water.  After the Cbz is added it reacts with the free amino group to produce the Cbz protected derivative of glycine plus hydrochloric acid.

The hydrochloric acid and the sodium cation react to produce sodium chloride as the inorganic by-product.
d) the mistake involves the simultaneous addition of Cbz and additional sodium hydroxide ("Benzyl chloroformate (1.2 equiv.) and aqueous sodium hydroxide ($\pu{25 mL}$ of a $\pu{4 M}$ solution) were added dropwise simultaneously").  For one thing, enough sodium hydroxide was added earlier to convert all of the glycine zwitterion to the glycinate anion, there is no need for additional sodium hydroxide.  Second, sodium hydroxide and Cbz will react immediately to produce the mono-sodium salt of benzylformate.  There won't be enough Cbz left to react completely with the amino acid.  Once the free amine has been generated the Cbz should be added separately in a second step to react with the free amine group.
Sorry, not aware of any websites with similar practice questions.


text:
Can someone recommend me a book ( preferably written long ago) that includes all the details of high school chemistry?
The textbook I got from my teacher is shown to be too general. Part of the fun of learning chemistry, for me, comes from knowing the derivation of equations. But most of the derivations are missing out from the textbook, like the derivation of kinetic molecular equation and real gas equations.
It will be fine to recommend a college level textbook as long as it does not include calculus.

How about Linus Pauling's "General Chemistry"? Also if you want derivation of equations check out any physical chemistry book. I.e. Atkins.

But some of them I cannt afford

@Doeser Pauling's book is a Dover book and goes for about 10 bucks.

You can find Paulings book for free on the web easily. I think it's copyright has long expired.

I second Atkins' Physical Chemistry. Also, it's so ubiquitous that I'm sure your local library has or can get hold of a copy from somewhere if you can't afford to buy it.

"College Chemistry - An Introductory Textbook of General Chemistry" by Linus Pauling (2nd Ed) is available as a free download (pdf & other formats) from https://archive.org/details/CollegeChemistry.


text:
Can ketones next to a nitrogen hetero atom tautomerise?
I worry that the electronegative N will stabilise keto form so much that equilibrium will be so far towards keto it would be unreasonable to call the enolisiation process a factor

If I understand your question correctly, I think you're asking if an amide can enolize like a ketone.  The answer to that question is yes.

The enol content of most organic compounds is very low and special methods must be used to measure the enol content.  Perhaps the question becomes, "is there more or less enol content in an amide than in an ester, for example"?
We can draw resonance structures for an amide (or ester) showing that the nitrogen stabilizes the amide.  But above we also drew resonance structures showing how the nitrogen can also stabilize the enol.  We can draw similar structures for both an ester and its enol as well.  Although in the ester case we wind up putting a positive charge on oxygen in some of the resonance structures, so they might not count for as much as they do in the amide case - but both the starting ester and its enol would both be affected similarly.  So as a first approximation I might guess that the enol content in an ester and an amide would be similar.

Indeed the enol content in malonamide is similar to the enol content in diethyl malonate (see page 34 here)

With a nitrogen adjacent to a carbonyl moiety, the functional group is an amide, and the tautomer involving a $\ce{C=N}$ bond is called an imidic acid. The amide is usually vastly more stable and typically predominates by many orders of magnitude. This can be rationalized by several considerations. First, the two tautomers differ by the exchange of a $\ce{C=O}$ bond with a $\ce{C=N}$ bond and an $\ce{N-H}$ with an $\ce{O-H}$, but the $\ce{C=O}$ bond is particularly strong, making the amide enthalpically more stable. Second, if you consider the resonance structures (i.e., mesomers) available to the two respective tautomers, the amide has a contributing structure in which a negative formal charge is placed on the more electronegative oxygen atom, whereas the corresponding structure in the imidic acid has the charge placed on a less electronegative nitrogen atom. Granted, resonance is not a real physical process, but examination of these structures gives some idea of the distribution of electron density in the frontier orbitals, which too points to favorable increase in stability for the amide. You can apply similar logic to the case of tautomers involving a $\ce{C=C}$ bond, where the ultimate conclusion should be the same (and the difference in stability between tautomers even greater).

Greg, maybe I'm misunderstanding the question, but isn't the OP asking about enolization away from amide nitrogen?  In that case which tautomer has a C=N  double bond?


text:
How to interpret a luminescence intensity vs wavelength graph?

Luminescence is defined as the amount of light emitted by a atom. But what confounds me of this graph is the fact that all the peaks are of the same height. In addition, there are no peaks in between two peaks.
 why are the peaks of the same height?
Andwhy are there only six peaks, if photons can be emitted having the wave length of say, 110nm?
I am sure those are ultraviolet rays.

I got the second question. Because of the quantum theory

What are the units for the x-axis?

Wave length in bm

Nanometer not bm, sorry typo

The first question might has something to do with possibility

But I still have no idea about the answer

This is a partial spectrum of the luminescence from a hydrogen atom, the graph presents only the Lyman series of emissions.
When a hydrogen atom is excited (absorbs energy), the electron can be promoted to the n=2, 3, 4, 5...infinite level.  This excited electron can eventually emit its excess energy and return to the n=1, 2, 3, 4...etc. state.  All returns to the n=1 state (ground state) are termed the Lyman series,  all returns to the n=2 state are termed the Balmer series and so on.

For the hydrogen atom, the wavelength of these various emmissions is given by the Rydberg formula
$${1 \over \lambda} = R \left( {1 \over (n^\prime)^2} - {1 \over n^2} \right) \qquad \left( R = 1.097373 \times 10^7 \ \mathrm{m}^{-1} \right)$$
for the Lyman series this equation becomes
$${1 \over \lambda} = R \left( {1} - {1 \over n^2} \right) \qquad$$
and six lines can be seen in the ultraviolet.  After the 5th line (n=6 -> n=1), all other transitions are so closely spaced that they appear as one line (the sixth line).
\begin{array}{|c|c|}\hline
n & λ (nm) \\ \hline
2 & 122 \\ \hline
3 & 103 \\ \hline
4 & 97.3 \\ \hline
5 & 95.0 \\ \hline
6 & 93.8 \\ \hline
\infty & 91.2 \\ \hline
\end{array}
As to the relative intensities, I'm not sure why they are all the same in your graph.  The intensities do depend upon how many electrons are excited to the various excited states.  If for some reason all excited states were created in a manner such that they were equally populated, then something like your spectrum would result.  Usually the relative intensities differ and are used much like a fingerprint.  For example, the relative intensities of the Lyman series from the hydrogen atoms in one galaxy differ from those in another galaxy due to the difference in temperature of the two galaxies.  My guess is that your data was artificially created (e.g. the title in your graph says "Model").

Is Z factor there in the equation if it is not hydrogen?

Actually it is $Z^2$. Extension of the Rydberg formula to other elements is discussed here (see the section "Rydberg formula for any hydrogen-like element", also follow the link to Moseley's law).

Does luminescence intensity means how many electrons are released as a result of photon bombardment?

@Doeser  It's photons released.  When an electron in an excited state falls back to a lower energy level a photon is emitted.


text:
Decarboxylation under acid, water, and extreme heat
I am asked whether dimethyl malonate or methyl acetoacetate decarboxylate when hit with acid, water, and extreme temperatures.
I believe the answer is both because both are esters, and esters, upon being hit by such conditions, return to their respective carboxylic acids.
From there, both carboxylic acids can undergo decarboxylation because both resulting carboxylic acids are beta-keto acids. Technically, dimethyl malonate becomes malonic acid, which can still undergo decarboxylation (but malonic acid isn't of the beta-keto class; rather the dicarboxylic acid class).
Acid and heat treatment of methyl acetoacetate initially gives 3-oxobutanoic acid but this can decarboxylate under the heat as well.
Is my analysis correct?

Yes, your analysis is correct. This is a common sequence in the malonate synthesis of carboxylic acids and acetoacetate synthesis of methyl ketones. After taking advantage of the acidic, doubly alpha-proton and using the resulting enolate as a nucleophile, the ester(s) are hydrolyzed with acid. A carboxylic acid with a carbonyl in the beta-position can be thermally decarboxylated to give an enol, which typically tautomerizes to the keto form.


text:
Acid-catalyzed hydrolysis of diphenyl malonate

What are the complete, acid-catalyzed hydrolysis products (with application of heat) of diphenyl malonate?

According to me, they are phenol, carbon dioxide, and acetic acid.
The aqueous acid and heat conditions indicate to me that we are returning from the carboxylic acid derivative (ester) to the carboxylic acid. The mechanism involves protonation of the carbonyl oxygen and nucleophilic attack by water. This ejects alcohol (well, in this case, phenol).
The heat induces the malonic acid intermediate to decarboxylate and this results in the production of carbon dioxide and acetic acid.
Is my analysis correct?

If the question is "What are the products of hydrolysis of diphenyl malonate?", then my answer would be 2 equivalents of phenol and one equivalent of malonic acid.
If the question is "What are the products of diphenyl malonate when heated in an acid aqueous solution?", then my answer would be 2 equivalents of phenol, one equivalent of acetic acid, and one equivalent of carbon dioxide.
I see it sort of as a semantic question, since the hydrolysis of an ester gives an alcohol and a carboxylic acid, as you suggest. I don't consider decarboxylation to be a hydrolysis reaction, literally defined as cleavage of bonds by addition of water. Water isn't incorporated into the products of the decarboxylation step.


text:
If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products?
I'm wondering if this is possible because could there not be a situation in which the reactants were pure solids/liquids and would therefore not contribute to the value of k but still contain more total moles that the product side?

Here's a related question.

Yes, even if the equilibrium constant is much greater than 1, there can be almost all reactants and very little products.
A wood pile can just sit outside (in an oxygen containing atmosphere) all winter, but until you stick the wood in the fireplace and supply activation energy almost no reactants will be converted to product.
If all the reactants are mixed in the same phase, the same is true.  Mix methane and air in a Bunsen burner or gas stove, and nothing noticeable happens until you provide a spark or flame.

Ok. If however, the reaction is allowed to go through to equilibrium when K is greater than 1, there will always be more products than reactants?

If the sum of the exponents of the product activities (concentrations) does not equal the sum of the exponents of the reactant activities, then it is arbitrary when K is 1. In other words, it depends upon the choice of units and the definition of the "concentration constant". So, at least for this reason there could be more reactants than products at equilibrium even though K is greater than 1.

@Emu27 for example, if $K = 10 = ([B]/[A]^3) \times M^2$ and [B] = 0.01M and [A] = 0.1M then there are more moles of reactants than products.

The equilibrium constant does only contain the activities/concentrations of the species that are actually available to the reaction, e.g. in solution it would only contain species that are actually dissolved in the solvent or that are present at the interface between the solution and another phase. So, solids would not enter the equation with all there moles but only with the small percentage that is acutally available to the reaction: if some of the solid is dissolved and those dissolved species take part in the reaction then it is those (and only those) molecules that enter into the equation for the equilibrium constant; in the case of a heterogeneous reaction the solid can only enter the equation via its available surface sites. So, the bulk of a solid would not contribute to the equilibrium constant in a reaction taking part in a liquid phase that shares and interface with the solid.
Now, concerning the main question: If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products? This can certainly be the case if you already have some of the products present and keep their concentrations constant. Look at the example reaction
\begin{equation}
  \ce{A + B <=>> C + D}
\end{equation}
which shall have an equilibrium constant of $K_{c} = 10$. Let's assume you keep the products at a concentration of 1 mol/L each, so $[\ce{C}] = 1 \, \mathrm{mol}/\mathrm{L}$ and $[\ce{D}] = 1 \, \mathrm{mol}/\mathrm{L}$. The equation for the equilibrium constant is
\begin{equation}
  K_{c} = \frac{[\ce{C}] [\ce{D}]}{[\ce{A}] [\ce{B}]}
\end{equation}
Since you've already fixed the product $[\ce{C}] [\ce{D}]$ at $1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}$ the product of your reactants must fulfill
\begin{equation}
  [\ce{A}] [\ce{B}] = \frac{[\ce{C}] [\ce{D}]}{K_{c}} = 0.1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}
\end{equation}
So, if you use only very little $\ce{B}$, say $[\ce{B}] = 0.01 \, \mathrm{mol}/\mathrm{L}$, your concentration of $\ce{A}$ in the solution will be
\begin{equation}
  [\ce{A}] = \frac{[\ce{C}] [\ce{D}]}{K_{c} [\ce{B}]} = 10 \, \mathrm{mol}/\mathrm{L}
\end{equation}
Now, if you conduct this reaction in a vessel with a volume of 1 L you'll find a total of 2 mol of products and 10.01 mol of reactants in your reaction mixture, i.e. more reactants than products.

Does that mean that it is not possible?

It does mean that once equilibrium is reached the product of the concentrations of the available reactant species raised to the power of their respective stoichiometric coefficients will be smaller than the product for the product species.

@Emu27 I've extended my answer.

If we do not cheat as you put it, your answer is at odds with the one below.

@Emu27 Ups, sorry, it was quite early and I didn't really consider cases with different stoichiometric coefficients. Forget the last sentence of my answer.


text:
What is this equation for?
I have been given a formula on the topic of alloys and temperature, however am unsure specifically what this equation calculates. The equation is as follows:
$xC_{1} + (1 - x) C_{2} = C_{3}$
I am assuming that the C's will be given (and possibly represent Specific Heat?) - however exactly what is this equation to represent? What does solving for $x$ achieve?
I know that the letters used may vary, however our teacher has provided no more information due to sickness. I have provided all the information I know.

$x$ would be the fraction of C1 in C3; the fraction of C2 in C3 is 1 minus the fraction of C1. I don't think physical attributes of an alloy can be determined from the fractional amount of their parent metals.

I fear that without additional information it will be difficult to provide you with a definitive answer. The $x$ usually represents the mole fraction and in the context of a 2-component alloy this seems plausible. The $C$s might indeed be specific heat constants, i.e. $C_{3}$ would be the specific heat of the alloy and $C_{1}$ and $C_{2}$ would be the specific heats of the components making up the alloy, whereby component 1 would have a mole fraction of $x$  and component 2 would have a mole fraction of $(1-x)$. For a mixture of ideal gases such a formula would be correct but I doubt that it will work well with alloys in the real world. But if this came up in the context of a school lesson then the approximation might be ok.


text:
How to find Gibbs energy for a real gas
I'm asked to find $\Delta G$ for the isothermal compression of $\ce{N2(g)}$ (real gas) at $0\ \mathrm{^\circ C}$ if the gas expands from $P=1\space\mathrm{atm}$ to $P=1000 \space\mathrm{atm}$. The fugacity coefficients are $\phi=0.9996$ and $\phi=1.84150$ respectively.
I'm confused about two things:

My professor makes a distinction between $\phi^*$ and $\phi$, the first one being the fugacity coefficient for a pure gas and the second for a gas in a mixture. Are they the same in this case or did he make a mistake since there's not a mixture to begin with? (Not sure how Lewis—Randall rule can relate here since, again, there's no mixture to apply the relation between the two coefficients.)
I already calculated the Gibbs free energy if the gas were an ideal one, I got $\Delta G=15.688\ \mathrm{kJ\over mol}$. Since $G^E=RT\ln(\gamma)$, and I have the Gibbs energy for an ideal gas, I could find the Gibbs energy for the real gas when I get a value for $G^E$; but again, in order to find $\gamma$ I need both coefficients, the one for a mixture and the one for a pure gas.

Can anyone please let me know what am I doing wrong? Or if my approach is far from the right path? Thanks in advance.
EDIT:
I assumed that the data that was given corresponded to the values of fugacity coefficients for the pure gas, making it all simpler since $G^E=RT\ln(\phi)$, so I got $G^E_{N_2}=17.075765\ \mathrm{kJ\over mol}$ and then $\Delta G^R=1.387765\ \mathrm{kJ\over mol}$
What I don't understand is that according to those two definitions of $G^E$, I get that $\gamma=\phi$, which I know it's not right, how's it possible?


text:
What is the carbon footprint of a diamond?
I think it would be interesting and illuminating to calculate the carbon footprint (per carat) of diamond. The answer would undoubtedly be different for artificial (industrial grade) diamonds vs. natural ones. Is anyone out there familiar enough with either natural diamond mining and processing or the high-pressure process for making artificial diamonds to venture an answer to this question? Both surely involve expenditure of a considerable amount of energy.

Perhaps this question fits better on Sustainability SE?

I’d like to answer this. I found an answer on the inter-webs, at  https://www.cleanorigin.com/diamonds/#p=7


text:
Assign symmetric and antisymmetric stetches in IR spectroscopy
Typically the antisymmetric stretch in IR spectroscopy is higher than the symmetric stretch for a given functional group. For example for $\ce{NO2}$ the antisymmetric stretch falls at $\sim 1530\ \mathrm{cm^{-1}}$ compared to $1350\ \mathrm{cm^{-1}}$ for the symmetric stretch. Likewise $\ce{NH2}$ has a similar pattern ($3400\ \mathrm{cm^{-1}}$ antisymmetric and $3300\ \mathrm{cm^{-1}}$ symmetric).
However, in some cases this pattern is inverted. e.g. $\ce{[HCO]+}$ has an symmetric stretch at $3896\ \mathrm{cm^{-1}}$ and an antisymmetric stretch at $2814\ \mathrm{cm^{-1}}$. How can this be explained?


text:
Why does radioactive decay occur?
Is there a reason radioactive decay is exponential, as in after an extremely long time some of the original compound remains, or is this just an empirical model? Also, why does $\ce{^{238}U}$ have a much longer half life than $\ce{^{210}Pb}$, for example?

The simplified version of why radiation occurs is a balance between attraction of the protons and neutrons in the nucleus and the positive charge of the protons repulsing each other. Get the balance right, and you end up with a stable isotope, get it wrong, and the isotope is unstable and will decay.
As for the exponential decay: a nucleus falling apart is stochastic process, meaning it is only dictated by chance! When you take an unstable nucleus, after it's half-life there is a 50% chance that it will have decayed. Then, you can add more nuclei the chance of half of them decaying after it's half life is far, far larger than all of them being stable. The larger amount you start of with, the better this theory describes the decay. If you then measure a vast amount of atoms (a mole for example) the curve you find will describe an exponential decay.


text:
What happens if the electronegativity difference is exactly 2.1?
I'm learning about electronegativity in atoms and how their difference affects the type of bond they make. I know that no difference makes a non-polar covalent bond, a difference less than 2.1 is a polar covalent bond, and a difference above 2.1 is an ionic bond. But what kind of bond forms if the difference is exactly 2.1?

As mentioned above there is no hard line between an ionic and a covalent bond, they are just two ends of a spectrum where a bond can have a certain degree of ionic or covalent character. At the lower end of this spectrum, bonds with small electronegativity differences between the atoms have very little ionic character because the charge distribution in the bond is almost symmetrical between the two atoms. As the EN difference increases the charge distribution becomes asymmetrical and this results in an increasingly polarised bond, which therefore has an increasing degree of ionic character. Eventually at the other end of the scale you have a bond where almost all of the charge is concentrated on one atom and this is generally termed to be an ionic bond.
So in answer to your question; what happens if the EN difference is exacly 2.1; nothing happens, you have a relatively strongly polarised bond which under some definitions might be classed as ionic and under others as strongly polar covalent but in reality it has a large degree of both ionic and covalent character.

Based on the the answer to this question:

Why electronegativity difference greater than 1.7 are ionic?

Here is a very nice graph of percent ionic character as function of electronegativity difference for some common binary compounds. Here ionic character can be thought of as the degree of charge separation across the covalent bond. Compounds with 50% or more ionic character are considered "ionic". This division of classes of compounds is arbitrary.

Image source: University of Florida Chemical Bonding page


text:
Nernst equation for a fuel cell when gases are stored separately
I have run some tests with a PEM fuel cell and the voltage we have found was $1.44\ \mathrm{V}$. Given that the standard potential for a fuel cell is $1.229\ \mathrm{V}$ what is causing the difference in our measurement?
Initially we provide power to the fuel cell so it can generate oxygen and hydrogen into separate containers. These containers are submerged in water so there is water pressure keeping them contained.
All measurements were done in standard conditions, except that the pressure of each container is slightly different (hydrogen displaced twice the amount of water since there are twice the moles of atoms present).
I am not sure how to calculate this using the Nernst equation since the conditions of both the oxygen and hydrogen are slightly different (each gas is of a different concentration in its respective container since there is different water pressure and water vapour pressure present).
My equation is:
$$E_\text{cell} = 1.229\ \mathrm{V} - \frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}]}{[\ce{H2}][\ce{O2}]^{0.5}}\right)$$
For $[\ce{H2}]$ I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).
I also do the same thing for $[\ce{O2}]$
I'm not sure what to use for $[\ce{H2O}]$
The answer I am getting is $1.24\ \mathrm{V}$. What am I missing?

At first glance your approach looks decent.  Is your solution concentrated?  The Nernst equation only works well with dilute solutions if using concentrations, otherwise you have to use activities.

When you are applying power to the cell, you are most definitely applying a higher voltage. How long do you wait after you turn it of before you measure the potential? It might be that the cell didn't come to equilibrium after the electrolysis and the voltage you are measuring is just leftover from the previous process

Check your equipment and gas flow measurement set up. Storage under water will deviate from atmospheric pressure. If the gas deliveries are not properly regulated, then the partial pressures and concentrations will vary. This is the variable in the Nernst equation that will change your E value. It is highly likely to be this.

Given that the standard potential for a fuel cell is 1.229 V what is causing the difference in our measurement?

Multiple sources (e.g. here, here and there) estimate the voltage to be much less than the standard potential, especially when there is a load.

Individual fuel cells produce relatively small electrical potentials, about 0.7 volts, so cells are "stacked", or placed in series, to create sufficient voltage to meet an application's requirements.

So it is surprising that the voltage is higher than the standard potential. Perhaps two cells were stacked to provide about 1.4 V, similar to the voltage of a standard alkaline battery, and about twice what you would expect from a single cell under load.

For [H2] I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

For the standard hydrogen electrode, the standard state is defined as a partial pressure of 1 atm. This is much lower than a concentration of 1 M. In any case, the Nernst equation does not predict a voltage that is significantly higher than the standard potential.

Seems your cell is operating as,
Anode: $\ce{H2 = 2H + 2e'}$
Cathode: $\ce{2H(+) + 2e' = 2H}$, and then $\ce{2H + 1/2O2(g) = H2O(l)}$ -Heat.
E = 1.4 V gives, 1.4 = [2.303RT/F] log R, where R = ratio H activity between cathode and anode R = 10^-23.65    if your anode assumes as activity H = 1.0, at the cathode H activity is 10^-23.65 the removal H by the oxygen as, $\ce{2H + 1/2 O2(g) = H2O(l)}$.  Atomic H is highly reactive as $\ce{H + H = H2}$ or $\ce{H + O = OH}$, or $\ce{2H + 1/2O2(g) = H2O(i,g)}$, etc.
$\ce{H2O(l)}$ blocks Oxygen contacting cathode (electrode reaction), i.e., $\ce{2H+ + 2e' + 1/2O2(g) = H2O(l)}$. Actually if current is OK, it is better do Fuel Cell operation in concentration cell that the oxygen is removal hydrogen by the H2O(g) vapor which is high T fuel Cell. Therefore we need high T Fuel Cell that is more appropriate in science and technology with Fuel Efficiency, CHP module.
Currently, I am developing such as:
HECE = HFC + HTEC
[H-eChemEng = H-fuel Cell + H-Thermal to Electric converter]
you have good measurement data!


text:
How to get the reducible representations of the B-B and the B-Cl bond of B2Cl4
I just can't seem to get the correct answer to this symmetry question in my book. "Introduction to molecular symmetry J.S.Ogden."
I am trying to find the reducible representations of both the $\ce{B-Cl}$ bond and the $\ce{B-B}$ bonds in $\ce{B_2Cl_4}$ in the point group of $D_\mathrm{2d}$.
The correct answer for both is $\Gamma_{\ce{B-Cl}}=A_1+B_1+E$ and $\Gamma_{\ce{B-B}}=A_1$.
My reducible representations are:

$\Gamma_{\ce{B-Cl}}=4,0,0,0,2$ for the $\ce{B-Cl}$ bond in the order $E, 2S_4, C_2, 2C', 2\sigma_\mathrm{d}$.
$\Gamma_{\ce{B-B}}= 1,0,1,0,1$.

Clearly either of these are not correct but honestly I can't see why! Your help is much appreciated! (I worked them out based on the number of bonds that remain unshifted during the respective symmetry operation. I have also made a little model for myself in Gaussian but it didn't really help.

Why do you think that your reducible representations are not correct? Have you tried reducing them? If not, you should try it and add the result and how you got to it to your question.

yes I have tried and they do not give the correct answer, So the $B-Cl$ bond representation gives me $A_1+B_2+E$ while the $B-B$ bond gives me fractional answers so I get a half $A_1$ which I know must be wrong. Both of these answers are wrong. I think my error lies in the determination of $\Gamma$ but two days later I still can't see where I'm wrong.

There seem to be two problems: the first is that your reducible representation for the $\ce{B-B}$ bond is wrong but your reducible representation for the $\ce{B-Cl}$ bonds seems to be correct and the second is that your book is apparently completely wrong. If I didn't make a mistake and my second assumption is correct then maybe you should think about switching to another book as it seems to raise more questions than it answers. But I'm not working regularly with group theory, so my analysis of the problem might be faulty, but maybe someone else can support or refute my answer.
Ok, here is my analysis:
First, make yourself familiar with the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group. Have a look at this video for example. Now, I looked at the effect each symmetry operation has on the bonds I'm looking at. For that I will use the following depiction of $\ce{B2Cl4}$:

For the $\ce{B-B}$ bond I used an arrow to indicate the direction of the bond for better emphasis as I will need that soon. In the following I will assume that you are familiar with the way characters are related to your reducible representation and how you get those characters from analyzing the changes of your basis vectors (in this case the bonds) under the symmetry operations of the group since you seem to have already used it to determine the reducible representations in your question.
Furthermore, I use the following notation: The reducible representation of the $\ce{B-Cl}$ bonds will be denoted as $\Gamma_{\ce{B-Cl}}$ and the character of the symmetry operation $C_{2}$ for this reducible representation will be denoted as $\Gamma_{\ce{B-Cl}}(C_{2})$.
Ok, let's see what the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group do to this structure, so that we can get the reducible representations:
The $E$ operation does nothing to the structure, i.e. all four $\ce{B-Cl}$ bonds and the one $\ce{B-B}$ bond stay where they are. Thus you get a character of $4$ for $\Gamma_{\ce{B-Cl}}(E)$ and a character of $1$ for $\Gamma_{\ce{B-B}}(E)$.
Now, we come to the improper axis $S_{4}$. It rotates along the long the $\ce{B-B}$ bond and mirrors perpendicular to it:

As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(S_{4}) = 0$. Since the mirror plane is perpedicular to the $\ce{B-B}$ bond the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(S_{4}) = -1$.
Next, we look at the principle $C_{2}$ rotation along the $\ce{B-B}$ bond. It converts your molecule to the following:

As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}) = 0$. But as the rotation is conducted along the $\ce{B-B}$ bond axis the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}) = 1$.
Now, for the two $C_{2}^{\prime}$ axes which are perpendicular to $C_{2}$. One disects the dihedral angle between the dark red $\ce{Cl}$ and the cyan $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other disects the dihedral angle between the red $\ce{Cl}$ and the blue $\ce{Cl}$ - its effect on the structure is shown in the right picture:

$\qquad \qquad \qquad$

Again the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}^{\prime}) = 0$. But now the rotation is conducted along an axis perpedicular to the $\ce{B-B}$ bond and so the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}^{\prime}) = -1$.
Finally there are the two $\sigma_{\mathrm{d}}$ planes which lie along the $C_{2}$. The one plane contains the dark red and the red $\ce{Cl}$ while disecting the angle betweeen the  cyan and the blue $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other contains the cyan and the blue $\ce{Cl}$ while disecting the angle betweeen the dark red and the red $\ce{Cl}$ - its effect on the structure is shown in the right picture:

$\qquad \qquad \qquad$

In this case either the red and dark red $\ce{Cl}$ atoms or the blue and cyan ones remain unchanged while the other $\ce{Cl}$ atoms change their places. So, two $\ce{B-Cl}$ bonds keep their location and direction which gives $\Gamma_{\ce{B-Cl}}(\sigma_{\mathrm{d}}) = 2$.
The $\ce{B-B}$ bond axis lies within the mirror plane and so the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(\sigma_{\mathrm{d}}) = 1$.
So, in conclusion you end up with the reducible representations (I kept the order of the symmetry elements you chose, i.e. $E$, $2 \, S_4$, $C_2$, $2 \, C_{2}^{'}$, $2 \, \sigma_{d}$):
\begin{align}
  \Gamma_{\ce{B-Cl}} &= 4, 0, 0, 0, 2 \\
  \Gamma_{\ce{B-B}}  &= 1, -1, 1, -1, 1 \\
\end{align}
and they reduce to
\begin{align}
  \Gamma_{\ce{B-Cl}} &= A_1 + B_2 + E \\
  \Gamma_{\ce{B-B}}  &= A_2 \\
\end{align}
As you can see both results differ from the ones your book provides.

This is so helpful thank you very much! I completely agree with your working and I think the only difference between my own answer and yours is that instead of putting -1 I erroneously put a zero instead. It must be a typo in the answers! Thank you so much for your effort :)

@RedPen Glad I could help. May I ask for what future purpose you are learning group theory? Is it only to understand the basics so that the meaning of symmetry labels, e.g. in transition metal complexes become clear, or are you going to really work with group theory, e.g. in quantum mechanics or spectroscopy?

At the moment I am a masters student in physical chemistry so I am just becoming familiar with the matrix formulation of symmetry, but I am hoping to study for a PHd in theoretical chemistry such as the development of new computational algorithms and similar simulation theory. My main interests are in understanding processes in photochemistry/intersections of excited states and the time evolution of systems. So I think I'd better learn quick haha. Thanks again!

I must say though I don't know too much about actual computational chemistry, I don't have a lot of experience with the different programmes in use, I'm much more familiar with the equations and theory behind it! :)

@RedPen Then I can recommend this book by Inui. It takes a more formal approach to group theory (more mathematically oriented) but it gives you a very solid foundation.


text:
Designing synthetic route to dicarbonyl compound
It's the last one, where 1,4-dimethylcyclohexane breaks apart into... something.

Here's the workup I have. I want to put out there that I'm not concerned about getting the optimal yield yet. Just making the product.
Free radical addition of $\ce{Br}$ to the 2 position.
Removal of that $\ce{Br}$ and the formation of a double bond via E2 using tBuO-
Breaking apart the structure via ozonolysis.
However, no matter where I place the double bond, I can't get the correct configuration of double bonded Oxygens (ketones and aldehydes, right?). So, I must be doing something wrong. Thoughts?

Everything you said is correct except for the first step.  Free radical bromination is reasonably selective, tertiary positions are about 20 time more reactive than secondary carbons, so the 1-bromo, 1,4-dimethyl compound is formed.  Everything's a go from there!

Radical bromination will occur at the 1-position (tertiary carbon). Your strategy from there is correct. Elimination using a small base such as hydroxide will give 1,4-dimethylcyclohex-1-ene. Ozonolysis with reductive workup gives the product.

Ah, yep! Forming the more stable radical does indeed give the right product. Thanks!


text:
SN2: Neopentyl halides and methyl halides
Two part question for you. Let's preface this with I'm in organic chemistry 1, so we're just learning the basics of these reactions.

Are neopentyl halides completely inert to any reaction through any pathway, $\mathrm{S_N1, S_N2, E1, E2}$? For $\mathrm{S_N1/E1}$, even after solvolysis of the halide, is there too much steric strain there for anything to attack the carbocation? Even a small base like $\ce{MeO-}$?
Can a methyl iodide/bromide undergo $\mathrm{S_N2}$ with any basically any nucleophile since there's practically no steric hindrance? Even a bulky compound that's not very lewis basic like tButanol?

Carey-Sundberg has some answers (vol. A, chapter 4).  The text is a bit advanced for you at the present stage, but you should give it a try.
It gives as reference JACS 1975, 97, 3694.  That paper has rate constants for nucleophilic substitution of compounds of type $\ce{RCH2Br}$, with $\ce{R=H, Me, Et, i-Pr, t-Bu }$ with $\ce{LiCl}$ in acetone at room temperature.
Rate constants in 105 s-1:
$\ce{R=H}$: 600, $\ce{R=Me }$: 9.9, $\ce{R=Et}$: 6.5, $\ce{R=i-Pr}$: 1.5, $\ce{R=t-Bu }$: 2.6*10-4
Nucleophilic substitution does take place in neopentyl halides, but at a rate of five orders of magnitudes less than in less hindered compounds.


text:
Difference between white sugar, brown sugar and red sugar
I always get confusing why the appearance of normal sugar (the white one), brown sugar and red sugar are different although they are sugar. Is it because they each have different molecular formulae or different number of bond or maybe the brown and red sugar have some transition metal inside?
I really don't. So, I wish someone will help on this question.

The white sugar you refer to as "normal" sugar is mostly pure sucrose, which forms translucent whitish crystals.
White sugar doesn't start out white --- it's actually refined from a plant extract (primarily from sugarcane) that is a mixture of sucrose, glucose, fructose, and some minerals including calcium, iron, and magnesium. The unrefined extract is "raw" sugar, and these minerals give it a brown color.
Brown sugars are partially-refined sugars which still have some of the other components besides sucrose. When it's fully refined, the appearance is white.
When we refine the raw sugar, the remainder is referred to as molasses -- a dense, dark brown syrupy mixture of sucrose, glucose, fructose, and some minerals which give it the signature dark brown color.
Red sugarcane doesn't produce red sugar as @LDC3 stated. "Red sugar" is actually refined sugar with a red pigment added to it... unless you're talking about what the Chinese refer to as "red sugar" -- which is just their name for brown sugar!

you could extend this answer by mentioning how the sweetness compares between brown and white. P.s. the bag of undyed chinese red sugar I have looks a pastel orange!

Molasses is collected from sugarcane when it is spun. It is dark brown, and when dried, makes brown sugar. Some varieties of sugarcane are red, producing red sugar. Further processing of the molasses will remove the color and produce white sugar.
Processing of sugarcane:
Wikipedia
SweetScam

This is not a chemistry answer. Please include in your answer an explanation of what accounts for the differences in color. For example, what makes red sugar from red sugarcane red?

@StevenT.Snyder You're welcome to post your answer that includes everything you desire. If the OP wants more information tan what I provided, he can certainly ask for it.


text:
Is glycerol/glycerin soluble in ethanol?
As titled, I wish to know if glycerol/glycerin is soluble in ethanol (not soluble, slightly soluble, very soluble)

Since this is homework, you should give your opinion and reason. We will then give you information.

I imagine that it is indeed soluble, since ethanol is weakly polar and glycerol is also polar. However i just want to be sure of this, as ethanol is, as said, weakly polar.

I did a search with Google with the words "glycerol ethanol solubility" and came up with this: http://www.sbioinformatics.com/design_thesis/Glycerol/Glycerol_-2520Properties&uses.pdf

As LDC3 indicated above here it states that "Glycerol is completely soluble in water and alcohol." So, this is correct, as also indicated here.
Now, for the reason this happens it is most likely because of its 3 hydroxyl groups ($-OH$) and the fact that it is a polar molecule.