text:
How closely do GCMS spectra typically agree to conclude a match?
I have two unknown peaks on my GC (both with the same MW or rather highest M/Z ratio peak = 98) and I am trying to confirm their identity.
Unknown-1

Unknown-2

Based on the expected chemistry I have identified two potential candidate molecules from a library of GCMS spectra (see images below; just click on the images please if you cannot read the smaller font numbers).
Compound-A

Compound-B

Is it safe to think these are matches or not really? The MS was run in EI+ mode.

Unknown-1 is Compund-B?
Unknown-2 is Compound-A?

Both likely? Neither likely? Any one likely?
Is having the same set of M/z fragments good enough (e.g. 98-83-55 similar between Unknown-1 & Compound-B) or ought the abundance ratios also match mostly?

Please, do not use external image hostings, since they are not reliable. Prefer the standard Stack Exchange way of uploading images.

I would never recommend making an assignment of identity based on Mass Spectrum alone - usually you'd want to get things like IR or NMR Spectra to further corroborate your identification but supposing that the underlying chemistry of the reaction is simple and that all you can do is mass spectroscopy you can use this to identify compounds.

Unknown-1 is Compound B

Plausible, but the base peak of Compound B is 83, but the base peak of Unkonwn-1 is 55. Need more information, but reasonable guess.

Unknown-2 is Compound A

Unlikely as the base peak of Compound A is 69, which is distinctly missing from the spectrum of Unknown-2. Unknown-2 also has significant peaks at 55-57 which are insignificant in Compound A's Spectrum.

Is having the same set of M/Z fragments good enough or ought the abundance ratios also match mostly?

You would have a much stronger case if the abundance ratios also matched. Remember that the peaks are formed by the breaking up of the compound and the relative stability of each fragment. If the compound is the same and run in similar conditions, it should, in principle, break up in the same/similar way.

Thanks for your comments IT Tsoi! For NMR as well as IR one needs to isolate pure compounds, correct? Basically this is a GCMS of a byproduct mixture where these two unknowns are most abundant but still at approx. 10% levels. In the absence of a pure isolation of these peaks is there any way to employ NMR / IR? Right now these components are only resolved on the GC column.

To clarify, I don't need 100% certainty of identification. Being reasonably sure will do. The underlying goal isn't a novel synthesis merely identifying by-products to enable process improvement.

@curious_cat since you just need to be reasonably sure, it may not be necessary to isolate the pure compound - you just need it in reasonable proportions relative to the rest. "Crude NMR" can be used to tell if there is substantial compound to purify from the reaction mixture but since yours is a byproduct this may not work. If you don't need 100% certainty and presumably you know the underlying chemistry quite well, I guess your ID of Unknown-1 as Cpd-B is acceptable but I am not convinced that Unknown-2 is Cpd-A based on the spectra provided. The pattern is just too different.

Thanks! Is it reasonable to expect that based on the MS whatever Unknown-2 is its molecular weight is also 98? Only C,H,O are expected in there.

Yes, it is reasonable to suspect that Unknown-2 has a molecular weight of 98

If you plan to publish the results in a pier reviewed journal reviewers would ask to provide three independent characterizations of the compound. 1-H, 13-C NMR and mass spec are typically expected. Based on the molecular weight of the compounds they are small and should be easy enough to synthesize. Or even purify from your reaction mixture, since you are doing it to scale up the process. Once you synthesize them you can TLC pure material and your reaction mixture to see if they have the same Rf value.


text:
Do bond energies differ for surface and bulk atoms in nanoparticles?
In nano-particles, surface atoms have fewer neighboring atoms in compare to bulk atoms.
Is there any difference between bond energy of surface atoms and bulk atoms?

Yes, there is, as surface atoms and bulk atoms have different bond energies in nanoparticles. This is a reason why nano particles coagulate easily and special measures have to be taken to prevent coagulation.

I'd point out that nanoparticles have a lot of weird behaviors that are not typical of a bulk solid. So for atom "neighborhoods" there are (1) bulk solid (2) surface of bulk solid (3) nanoparticles

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text:
Rate of O exchange with H2O
We have to find the order of rate of exchange of O with H2O of the following compounds

$\ce{CCl3CHO}$
$\ce{CH3CHO}$
$\ce{CH3COCH3}$
$\ce{CF3CHO}$

According to me, higher rates of exchange would be when electron density is be less on the carbonyl carbon
But how we will compare 1. & 2.? In first compound there is hydrogen bonding (chloral hydrate)

1 isn't chloral hydrate, it is just plain chloral.

@SendersReagent but there is hydrogen bond

No, that's not true. There is a hydrogen. There is no hydrogen bond. It is an aldehyde, which means there a hydrogen coming off the carbon.

@SendersReagent the product will have hydrogen bonding

When you talk about the rate of exchange of oxygen with water it might be instructive to realise that actually any aldehyde or ketone in the presence of water is in equilibrium with the hydrate, as shown below.

Source: UC Davis

There are a few ways we know that this is indeed an equilibrium process, but the main method that was used to fully elucidate the mechanism was by using isotopically labelled water containing 18O (the most abundant isotope of oxygen is 16O).
By doing this the scientists could observe the 16O in the starting aldehyde being slowly replaced with 18O from the enriched water.
Given that an equilibrium is involved, there are actually two pertinent questions...

What is the position of the equilibrium?
How fast do we reach the equilibrium?

The table below might be of some use to you. On the left, you can see a variety of carbonyl compounds, and on the right, the equilibrium constants for the hydration.
Two of the compounds you listed (acetaldehyde and trifluoroacetaldehyde) are listed in the table, and you can see that the acetaldehyde is significantly less hydrated at equilibrium than the trifluoroacetaldehyde.

Source: J. Am. Chem. Soc. 1983, 105, 868 (American Chemical Society)

From the table, some trends can be picked out (excluding formaldehyde which is almost exclusively hydrated). There is here a relationship between kinetics and thermodynamics (the more favourable a compound is to form the hydrate, the faster it will form it)

Aldehydes are far more likely to be hydrated than ketones. This is largely though not exclusively a steric effect.

Electron withdawing/electronegative substituents favour hydration. On a very simple level this is due to the fact that they make the carbonyl more electron poor, thus promoting the addition of water via its lone pair.

The order should therefore be 4>1>>2>3   (fluorine is more electron withdrawing than chlorine)

it means for reactivity the stability of product does not play a role

You are correct. The more electron withdrawing the moieties are the faster will be the exchange. $\ce{CCl3}$ is more electronwithdrawing than $\ce{CH3}$. So 1 will exchange oxygen faster than 2. Presence of chloral hydrate indicates fast rate of oxygen exchange.
http://chemistry2.csudh.edu/rpendarvis/imineformF99.html
By the way, note that chloral hydrate and chloral are two different compounds. Chloral hydrate is literally a chloral to which one molecule of water was added. Chloral hydrate is an intermediate in the oxygen exchange process.

I want to compare between compound 1 & 4 .

@koolman CF3 is more electronwithdrawing group than CCl3. It means that CF3COH exchanges water faster than CCl3COH

but the product with CCl3 is more stable due to hydrogen bonding

(1) CF3COH also forms floural hydrate ( you can buy it here; http://www.sigmaaldrich.com/catalog/product/aldrich/68351?lang=en&region=US). (2) Stability of the product is important in thermodynamics, but here we discuss kinetics.

it means for reactivity the stability of product does not play a role

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text:
Why don't everyday things burn?
Ok, so I learned about the equilibrium constant. Now, I've seen that the equilibrium constant of burning is extremely small $(K \ll 1)$. here, I have a question. you see, $K$ is still NOT 0, which means that the forward reactions happen at least a tiny bit. Then, shouldn't we see some parts of anything burning at least a little bit?

Where did you see that burning was an equilibrium process? One cant 'un-burn' something once its burned, there is no reversibility to the process.

Actually, if you look at the Wikipedia page for 'combustion' you can see that the combustion process is an equilibrium process (but the K is too small that we can 'consider' it as an irreversible process )

my question here is, the K of burning(at room temperature)is too tiny that it can be considered irreversible, but if at least a tiny tiny bit of something did the burning, shouldn't we be able to see it?

@NotNicolaou Well, every reaction is technically an equilibrium. The thing he's missing is activation energy.

Why do you think that things don't burn? Just because it doesn't start of a chain reaction ("a fire") doesn't mean it's not burning. One of the things that's very slowly killing you is oxidative stress, which can be considered "burning". And you've probably seen rusted iron before - that's also the result of the slow "burning" of iron in our oxidative atmosphere, though it's not quite the same as burning coal in a fire (a hunk of iron doesn't burn that way very well at room temperature - iron or aluminiumdust, on the other hand, is very flammable indeed).

to point it out simply, oxygen is killing/burning every carbon based thing by oxidizing it , from tree leaves to iron bars, fire is what happens when a catalyst speeds the process up through combustion.

https://en.wikipedia.org/wiki/Autoxidation

The equilibrium constant for combustion of organic matter in air with oxygen is not small, but extremely large ($K_\mathrm{eq} \gg 1$), as is expected from a reaction that is simultaneously very exothermic and (usually) increases entropy due to the formation of more gaseous molecules than the input oxygen.
The major reason carbon-based life can exist at ambient temperature in an oxygen atmosphere is purely kinetic, not thermodynamic. You, the tree outside and everything else made of carbon is right now undergoing continuous combustion. However, in the absence of catalyst, this process is too slow below a couple hundred degrees Celsius for it to be self-sustaining. More technically, combustion of organic matter is a highly exergonic process, but the activation energy is high. The meagre amount of heat generated by the handful of molecules reacting is too quickly diluted into the surroundings, and the reaction does not accelerate and spiral out of control (a fire, as described by the eternal Feynman).
Very luckily for us, Life figured out this vast untapped source of chemical energy held back by kinetics approximately three billion years ago and developed a whole metabolic process to extract this energy in a stepwise fashion using catalysis, which we call aerobic respiration. Without it, multicellular organisms could well never had evolved.

Note that simple piles of coal may become warm enough by this process to start a fire! (It takes time, and the right levels of moisture, piece size, etc.)

I believe that paper turning yellow/brown over time as it ages is due to this--is this true?  If so, it might make a good example in your answer.

@msouth I think that the reason paper turns yellowish over time is mostly due to the presence of impurities left over in the paper after manufacturing (e.g. acids) and air pollution, which cause its slow decomposition. This is especially true of older paper. Of course, paper still combusts ever so slowly, but I have no idea how long it would take for this process alone to discolour paper.

Yes it the pesky activation energy that keeps us together!
It is quite illuminating to calculate the rate of a reaction with different activation energies using a simple approach, such as the Arrhenius equation at a constant temperature; $\mathrm{k=A*exp(-\Delta E/RT)}$. If we assume that the pre-exponential term A (the rate constant at zero activation energy or infinite, i.e. very high, temperature ) is $\mathrm{10^{13} s^{-1}}$ then the following values are obtained with different activation energies in kJ/mol.
$\Delta E$ =10,  $k = 1.810^{11} $ s$^{-1}$
$\Delta E$ =40,  $k = 1.110^{6} $ s$^{-1}$
$\Delta E$ =80,  $k = 0.12 $  s$^{-1}$
$\Delta E$ =100, $k = 0.000038 $ s$^{-1}$   lifetime = 7.2 days
$\Delta E$ =140, $k = 4*10^{-12} $ s$^{-1}$  lifetime 7555 years
$\Delta E$ =180, $k = 4*10^{-19} $ s$^{-1}$  lifetime $6.9*10^{10}$ years
$\Delta E$ =200, $k = 1.5*10^{-22} $ s$^{-1}$  lifetime $2.1*10^{14}$ years
The age of the universe is approx $1.4*10^9$ years so even moderate activation barriers produce exceptionally slow reactions. Very many types of bonds have dissociation energies above 200 kJ /mol, e.g. C-C 350, C-H 413, but this is not the same as the activation energy which must be smaller than this, and by how much depends on the particular reaction. In unfolding a protein, the activation energy is essentially the dissociation (unfolding) energy, so if proteins are to last any appreciable time the activation energy has to be at least 100 kJ/mol.

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Determination of initial excited state via luminescence

A one electron species initially in the some excited state
  ($n_i$)
  is irradiated with a light of wavelength 121 nm when
  the electron is promoted to a further higher orbit ($n_f$).
  In
  returning back to the ground state, it gives an emission
  spectrum containing 15 bright lines. Out of these 15 lines,
  9 lines were found to have wavelengths smaller than
  121 nm while 5 lines have wavelengths greater than
  121 nm.
  What is the value of $n_i$?

I got the value of $n_f$ using this:
$$\frac{(n_f-1+1)(n_f-1)}{2}=15$$
$$\therefore n_f=6$$
We do not know the atomic number($Z$), nor $E_{ground state}$, nor $n_i$. What is the approach we should use to solve this problem?

You haven't yet used the information that the light was 121 nm in wavelength.  That is a key consideration.

@CurtF. I tried using it by substituting in Wave number equation, but it has 2 unknowns - $n_i$ and $Z$.

Sometimes formulae are more hindrance than help, and perhaps there are some red herrings in your problem.

A one electron species initially in some excited state $n_i$ is
  irradiated with a light of wavelength 121 nm when the electron is
  promoted to a further higher orbit $n_f$. In returning back to the
  ground state, it gives an emission spectrum containing 15 bright
  lines. Out of these 15 lines, 9 lines were found to have wavelengths
  smaller than 121 nm while 5 lines have wavelengths greater than 121
  nm. What is the value of $n_i$?

There being fifteen bright lines suggest that $n_f = 6$, since $15 = 5 + 4 + 3 + 2 + 1$, there being $5$ transitions from the $n = 6$ state to $n = 5,\,4,\,3,\,2,\,1$; $4$ transitions from the $n = 5$ state, and so on. This is the basis for the formula you've employed.
Long wavelengths correspond to low energy; we see that the transition from $n_i$ to $n = 6$ is the sixth lowest in energy. $E \propto n_i^{-2}-n_f^{-2}$, so we pull out our handy-dandy calculator and work out: $$ 6 \to 5: 0.012,\,5 \to 4: 0.023,\,6 \to 4: 0.035,\, 4 \to 3: 0.049,\, 5 \to 3: 0.071,\, 6 \to 3: 0.083.$$ Thus $n_i = 3$. We did not need to use the exact wavelength of light used for irradiation and we did not need to know the identity of the species, because this problem is only concerned with the relative energy differences between energy levels.

I am sorry but could u explain the sixth lowest in energy part in more detail?

Well, five lines (in the emission spectrum) have wavelengths greater than 121 nm, and the transition from $n_i$ to $n_f$ has a wavelength of 121 nm. Since the transition from $n_i$ to $n_f$ has the sixth longest wavelength, it  is the sixth lowest in energy.

So we actually have to calculate the energy of all 15 wavelengths like 6 to 5, 6 to 4 etc and then rank it and find the answer?

I dont understand why you took only specific cases like 6 to 5, and other 5 cases in ur answer

Well, you can be smart about it and: (i) graph everything, in which case you simply need to measure distances with a ruler, or (ii) note that $E_{2\to 1} < E_{3\to 1}, E_{4\to 1}, ...$ and $E_{3\to 2} <  E_{4\to 2}, E_{5\to 2}, ...$, and simply compare $E_{2\to 1}$ to $E_{3\to 2}$ and $E_{6\to 3}$. These are actually the only three energies we need compute, since we know that $E_{6\to 3}$ is at least greater than $E_{6\to 4},,E_{6\to 5},,E_{5\to 3},,E_{5\to 4}$, and $E_{4\to 3}$.

Could you please come to http://chat.stackexchange.com/rooms/3229/the-periodic-table chat so that I can actually clarify the doubt I have?

I postulated that those were the six lowest energy transitions based on the fact that $E \propto n^{-2}$, and I felt that my having obtained an answer that worked was enough justification for my postulate. I think the easiest way to see that my postulate is acceptable is to look up an energy level diagram for hydrogen and compare distances between energy levels.

Thank you so much @a-cyclohexane-molecule. I got it :) Never thought it would be simple as just comparing energies using the given information. Thanks again.


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How to calculate ion concentrations for a precipitation from the solubility product constant?

The equilibrium constant for dissolving the insoluble substance gold (I) iodide, AuI (s), in aqueous
  solution is 1.6 × 10^-23
  at 25ºC. Write the equilibrium constant expression and calculate the concentration
  of Au + (aq) and I (aq) ions in a solution in which 0.345g of AuI(s) is in equilibrium with the aqueous ions.
  The answer is 4x10^-12.

So 1.6x10^-23=products/reactants. The products = x^2, but what does the reactant equal to? I think they got the answer by ignoring the reactants, is this possible?

The concentration (the activity, to be precise) of solids is assumed 1. [Au+][I-]/[AuI]=[Au+][I-] = Ks. Here Ks is the solubility constant. If water originally didn't have any I- or Au+, then the amounts of [Au+]=[I-]=x. Then you have x*x=Ks and x= square root of Ks. The amount of AuI doesn't seem to play any role because they didn't specify the amount of water it is dissolved in.


text:
Structure of N2O2
I've drawn a structure of $\ce{N2O2}$ like this:
$$\ce{:\overset{\large.\!\!.}{\underset{\large.\!\!.}{O}}-N#N-\overset{\large.\!\!.}{\underset{\large.\!\!.}{O}}:}$$
I don't know if it's possible. Please correct me if I'm wrong. Also, what would be the nature of it?

The structure is wrong and what do you mean by "nature"?

I mean it's properties

Can I know why I'm wrong?

see http://chemistry.stackexchange.com/questions/4204/nitrogen-monoxide

Is you question about the nitric oxide dimer or about the hyponitrite dianion.

I don't know anything about these. I'm not an expert. I'm just a student who's practicing Lewis structures.

If you're just looking at the Lewis Structure from the perspective of the octet rule, it does appear that the structure is correct. Dinitrogen always has a lone pair of electrons which could conceivably be used for dative bonding as you suggest. So from that perspective there appears to be nothing wrong at all - other than that it doesn't exist in nature in this way.
If you're interested in learning about the dimer, a number of proposals have been made in this 1994 JACS paper. If you're interested in the hyponitrite anion, you can read about it on Wikipedia (always a good place to start), and for your reference the structure of the Sodium salt is below:

The proposed structure in the original question is not really 100% correct, it is missing formal charges @VlnMurthy

It's wrong in the sense that it is inconsistent with what is found experimentally, according to Structure of the NO Dimer J. Am. Chem. Soc., 1982, 104 (17), pp 4715–4716.
The experimental structure is cis-ONNO, with ONN angle of 99.6 degrees.
The N-N bond length is 2.236 Angstroms, while the N-O bond lengths are 1.161 Angstroms.
No single Lewis structure can accurately describe the electronic state of this molecule; however, considering the bond lengths, a single N-N bond and triple N-O bonds would be better.
See The 16 valence electronic states of nitric oxide dimer form more information.

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text:
Generation of singlet oxygen by irradiation
What is the mechanism involved in the generation of singlet oxygen when ground state triplet oxygen is irradiated in the presence of rose bengal?

Dear Homework-closers: What would you have expected from OP to keep this question from being closed? Voting to reopen.

I would expect the questioner to state what they know about the mechanism so far, or at least what sources they have consulted so far.

Well now we've come full circle... In light of our experiment I would not exercise my close vote here. [Although in my opinion this is a borderline case. While the intention is clearly not so, it does read like a homework dump.]

Just because a question is not actual homework doesn't mean you can be lazy and leave out your current findings and anything else that would help anwering it, like for example showing the structure of rose bengal. It's highly impolite, you are pushing your own work off to people on the internet to do it for you.

Rose bengal (pictured below) is a member of the fluorescein family.

(image source: wikipedia)
There are 3 steps in the process where rose bengal sensitizes the formation of singlet oxygen..
\begin{gather}
\ce{RB ->[$h\nu$]\  ^1RB}\tag1\\
\ce{^1RB ->T[ISC]\  ^3RB}\tag2\\
\ce{^3RB + ^3O2 ->T[sensitization]\  ^1RB + ^1O2}\tag3\\
\end{gather}
In the first step, ground state rose bengal (a ground state singlet) absorbs light producing an excited singlet state.  Because of the heavy iodine atoms in the molecule intersystem crossing (ISC) to the excited triplet state rapidly occurs (heavy atom effect). In the presence of ground state triplet dioxygen, a process called "sensitization" occurs. In this step the excited triplet rose bengal transfers its energy to the dioxygen promoting it to the excited singlet state while the rose bengal is deactivated and returns to its single ground state.  In this sensitization step, 2 triplets are converted to 2 singlets so, importantly, spin is conserved.

Although the question is about Rose Bengal the mechanism is quite general and is the basis for the first steps in photodynamic therapy as used to attempt cure certain cancers. The answer given by @ron shows the three possible steps in producing excited singlet oxygen molecules. The first condition for reaction is that the triplet of the excited dye molecule is above the energy of the oxygen $\Delta E > 0$ in figure. This is often the case, but not for molecules such as carotenes ( xanthophylls ) which have low lying excited state and so do not undergo these reactions. They are used by  photosynthetic organisms to prevent reactive oxygen molecules from doing damage, but in photodynamic therapy we want reaction.
As two triplet states collide in solution and react the total angular momentum for the complexes formed must be conserved and can have quintet, triplet or singlet character but with a different probability for each type. The total spin on each molecule 1 is S$_1$ =1 and similarly for molecules 2, thus the total spin for two molecules is found from the (Clebsch-Gordon)series |S$_1$+S$_2$| ..|S$_1$-S$_2$| with each term separated by unity. Thus the terms give total spin 2, 1, 0 with multiplicities (2S+1) of 5, 3 and 1, or quintets, triplets and singlets, making 9 types of complexes all together. Thus quintets are formed with 5/9 probability, triplets with 3/9 and singlets with 1/9.
If $^3M$ is the excited triplet state of the molecule, and $^1M_0$ the ground state,  the schemes for these types of quenching are
$^3M$+$^3$O$_2 \rightarrow  ^1(^3M ..^3$O$_2$)$\rightarrow ^1M_0 +^1$O$_2$ (as $^1\Delta +^1\Sigma$)  (Energy transfer)
$^3M$+$^3$O$_2 \rightarrow  ^3(^3M ..^3$O$_2$)$\rightarrow ^1M_0 +^3$O$_2$ (as $^3\Sigma ^+_g$) (intersystem crossing)
$^3M$+$^3$O$_2 \rightarrow  ^5(^3M ..^3$O$_2$)$\rightarrow $ no reaction
The last equation does not occur as it is not possible to form products from a quintet complex as each orbital in M or O$_2$ can have a maximum of only 2 electrons.
Thus both $^1\Delta$ and $^1\Sigma$ excited state can be produced, with total probability of 3/9 of collisions and the process is by energy transfer of the Dexter or exchange type rather than by the Forster of dipole-dipole type.
In solution the $^1\Sigma$ excited state has a shortish lifetime (many nanoseconds) relative to the $^1\Delta$ which is longer lived. Both lifetimes depend heavily on the solvent used. This triplet to singlet intersystem crossing (in O$_2$) occurs under the influence of solvent collisions causing spin-orbit coupling. The scheme below shows the energy levels.

"The last equation does not occur as it is not possible to form products from a quintet complex as each orbital in M or O2 can have a maximum of only 2 electrons." Theoretically, why can't one product be in a singlet state and the other in a quintet? The total number of electrons in the system is conserved.

To form a quintet state then four unpaired electrons are needed, so four molecular orbitals in the same molecule would have to have unpaired electrons, (instead of two for triplet), in effect a doubly excited state, theoretical yes, in practice no, the energy required would just be too large.


text:
What does it mean by 'photons of higher intensity'?
I found an Assertion and Reason question related to transition energy in Bohr's orbits.
Assertion In the H-atom, photons of higher intensity are emitted when electron falls from the 2nd to 1st orbit than when it falls from the 4th to 2nd orbit.
Reason Transition energy from the 2nd to 1st orbit is greater than for the 4th to 2nd orbit.
The answer says the Assertion is false and Reason is true.
What does it mean by photons of higher intensity? Is it talking about the frequency? But if it was talking about the frequency, then the Assertion would be true.

I don't think this is a well-formulated question.  The Assertion is indeed false and the Reason is indeed true, but the Reason does not imply the falsehood of the Assertion; it is an unconnected observation.  (Atomic orbital transitions can absorb or emit more than one photon, but only under special conditions.)

The assertion is incorrect. Intensity is related to the number of photons emitted and has nothing to do with Bohr orbit transitions.
The reason is correct. Rather than higher intensity, photons of higher energy are emitted when an electron falls from the 2nd to the 1st Bohr orbit as compared to when it falls from the 4th to the 2nd orbit.  This is because the energy emitted or absorbed in a Bohr transition is given by the following equation where Z is the atomic number of the atom, R is a constant and n represents the quantum number of the final and initial Bohr orbits.
$\mathrm{\Delta E=-Z^2R \left ( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right )}$
since
$\mathrm{\frac{1}{1^2}-\frac{1}{2^2}=3/4}$
is greater than
$\mathrm{\frac{1}{2^2}-\frac{1}{4^2}=3/16}$
more energy is emitted in the first case than the second.

Ron's answer is completely correct, but I'm gonna give some more detail about what I think is the point of the assertion part:
The purpose of the assertion is to make you distinguish between the energy of classical waves and quantum mechanical/particle/electromagnetic waves. The assertion is wrong because the energy of a photon (or the energy of the wavy part of any particle) is proportional to the frequency of the photon as demonstrated in the photoelectric effect and for massive particles, using the de Broglie wavelength. On the other hand, classically, the energy in a mechanical wave, like a sound wave or an earthquake, is proportional to the square amplitude of the wave. The intensity of this classical wave is defined as, $$I_{classical}=\frac{Power}{Area}= \frac{Energy}{time*Area}\propto \frac{A^2}{time*Area}$$where $A$ is the amplitude of the wave, and we simply used the fact that power is defined to be energy per unit time. This basically means the intensity of a mechanical wave is the amount energy being transported by that wave to a specific area per unit time.
This way of thinking about intensity no longer makes sense though when you begin to look at electronic transitions. If the assertion were true, this would mean that a larger number of photons are being emitted in the $2\rightarrow1$ transitions than in the $4\rightarrow2$ transition. This would then mean that when we shoot photons at the hydrogen atom, we should see some larger number of photons being absorbed in $1\rightarrow2$ transition than in the $2\rightarrow4$ transition. This is not what we observe. Rather we see that only specific wavelengths (or frequencies) of light cause the transition to occur, and the number of photons being absorbed in the transition is one.
Hopefully that helps explain why they would put intensity in the assertion. Classically it makes some sense to think that would be correct, but this isn't a mechanical wave, and a distinction must be made there.

The answer is correct because intensity refers to the number of photons emitted (i.e. intensity of light) and not the frequency (inverse of wavelength) of the photon being emitted.

Thank you so much for your answer... May I know how intensity of light is related to an atom (as per Bohr's model)

If it spontaneously falls, 1 fall = 1 photon

Oh I understand now. It is related with how the electron falls - straight to the said orbit or any permutation of the kind.
But if that is the case, the light intensity of n=2 to n=1 maybe equal to n=4 to n=2 depending on if the electron falls straight to n=2. Is that right? Btw, you answered the original question, thanks :)

Yes, your understanding is correct


text:
What is the correct structure of sodium thiosulfate?
Which one of the two structures of sodium thiosulfate is correct? One book says the second one (the one with co-ordinate bonds) while the other one says the first one?

According to me, the structure 2 is correct, as in that structure the octet of the central S atom is complete whereas in the first one there's an excess of it.

Take a look at the Wikipedia page. Na2S2O3 is the sodium salt of the thiosulfate dianion.  The negative charge on the anion is highly delocalised... the sodium isn't directly bonded to any oxygens or silicons, rather associated with the anion as a whole

http://chemistry.stackexchange.com/questions/24928/oxidation-state-of-the-sulfur-atoms-in-the-thiosulfate-ion

@user456 I would still like to remind you about our Be nice policy. First and foremost, if anything changes the meaning of your post, please roll it back. Second, check the edit history thoroughly before making any accusations. Thirdly, what's wrong with ... is hardly ever an appropriate response.

Both of your Lewis structures are wrong.
In both of the shown structures, the solid line from sodium to oxygen/ sulfur implies that there is a discrete bond between these atoms. This is not the case.
The sodium ions and the thiosulfate ions form ion pairs (or ion triples) and are attracted to each other via diffuse electrostatic interactions. Usually that is what we consider ionic bonding.
In aqueous solution the ions will be very mobile. Sodium ions may coordinate to one or more thiosulfate ions, and therein to more than one site of the thiosulfate ion. At the same time thiosulfate ions may coordinate to more than just two sodium ion, or fewer. In such solutions, the solutes also coordinate to the solvent.
Ignoring any external field, the oxygen within the thiosulfate ion are equal.
The Lewis structure best describing the thiosulfate molecular ion is having a +2 formal charge at the central sulfur, and a -1 charge on each of the ligand atoms oxygen/ sulfur. (See below for details.)
One possible (and more accurate) Lewis structure is therefore:

(Note that the dashed lines do not represent bonds, but weakly coordinating interactions, but not necessarily all of them.)
More common are probably the following Lewis structure representations:

In solid state coordination of the ions depend on various factors, especially how much water is incorporated into the crystal structure. In any of these cases, it would be wrong to draw any bonds between the (molecular) ions. I'll include a few links to publications at the end of the post.

Additional thoughts of your input structures

Which structure of sodium thiosulfate is correct? [...]

You noted correctly that in the first structure the central sulfur needs to "expand the octet". This would only be possible if d orbitals of said sulfur are involved. It has been proven many times that the contribution of d-orbitals in these hyper-coordinated species is minimal (usually below 1%).
The second structure is technically, or strictly speaking, not a Lewis structure, because dative bonds are not part of this description. However, if we allowed them as an extension, which has been proposed multiple times, then the arrow would need to point from the oxygen towards the central sulfur, as in those occurrences the more electronegative element would assume the higher electron density.

Disclaimer: The following section is a bit more advanced, and might not be appropriate to the casual reader, as another user pointed out that it might obfuscate the most important points about the structure of the anion.
Gas phase approach to determining the most likely Lewis structure
If we assume this molecule to be an ion-pair and calculate it in the gas phase, then we can draw a couple of conclusions, how a more accurate representation in the Lewis framework would look like.
I have performed a quick calculation on the DF-BP86/def2-SVP level of theory with Gaussian 09 and performed a natural bond order analysis (NBO 6.0) on it. The following picture shows the optimised geometry, alongside some inter-atomic distances.

As you can see, the sodium ions are coordinated to two oxygen, or one oxygen and a sulfur, respectively. The natural atomic charges reproduce well what I have initially stated, and reproduce well the Lewis structure I have suggested.
\begin{array}{rrr}
\text{Atom}& \text{No}& \text{Natural Charge}\\\hline
  \ce{S}  & 1 &   1.98 \\
  \ce{S}  & 2 &  -0.73 \\
  \ce{O}  & 3 &  -1.01 \\
  \ce{O}  & 4 &  -1.01 \\
  \ce{O}  & 5 &  -1.01 \\
  \ce{Na} & 6 &   0.91 \\
  \ce{Na} & 7 &   0.86 \\\hline
\end{array}
In terms of the NBO analysis, we can back this up, as it indicates no lone pair at the central sulfur (S1), three lone pairs at the oxygen (O3, O4, O5) and the ligand sulfur (S2), and four single bonds. There are vacant acceptor orbitals at the sodium ions (Na7, Na8). Here is an abridged version of the analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(skipping core orbitals)
  24. (1.98855) LP ( 1) S  2            s( 90.65%)p 0.10(  9.34%)d 0.00(  0.01%)
  25. (1.88728) LP ( 2) S  2            s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
  26. (1.84883) LP ( 3) S  2            s(  1.80%)p54.53( 98.09%)d 0.06(  0.11%)
  27. (1.97170) LP ( 1) O  3            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  28. (1.83558) LP ( 2) O  3            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  29. (1.82232) LP ( 3) O  3            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  30. (1.97170) LP ( 1) O  4            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  31. (1.83558) LP ( 2) O  4            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  32. (1.82232) LP ( 3) O  4            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  33. (1.96648) LP ( 1) O  5            s( 77.67%)p 0.29( 22.33%)d 0.00(  0.01%)
  34. (1.83753) LP ( 2) O  5            s(  0.02%)p99.99( 99.91%)d 4.78(  0.07%)
  35. (1.82147) LP ( 3) O  5            s(  0.00%)p 1.00( 99.93%)d 0.00(  0.07%)
  36. (1.95903) BD ( 1) S  1- S  2
               ( 56.47%)   0.7515* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 43.53%)   0.6598* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  37. (1.97694) BD ( 1) S  1- O  3
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  38. (1.97694) BD ( 1) S  1- O  4
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  39. (1.97758) BD ( 1) S  1- O  5
               ( 34.64%)   0.5886* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 65.36%)   0.8085* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)
 ---------------- non-Lewis ----------------------------------------------------
  40. (0.05958) LV ( 1)Na  6            s( 95.91%)p 0.04(  3.91%)d 0.00(  0.18%)
  41. (0.10019) LV ( 1)Na  7            s( 96.97%)p 0.03(  2.93%)d 0.00(  0.10%)
  42. (0.25853) BD*( 1) S  1- S  2
               ( 43.53%)   0.6598* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 56.47%)  -0.7515* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  43. (0.24584) BD*( 1) S  1- O  3
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  44. (0.24584) BD*( 1) S  1- O  4
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  45. (0.24610) BD*( 1) S  1- O  5
               ( 65.36%)   0.8085* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 34.64%)  -0.5886* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)

Key: LP = lone pair, BD = bond, LV = lone valency, BD* = "antibond"

Alternative coordinations
There are at least two other structures of the stoichiometry $\ce{Na2S2O3}$,  essentially with the same bonding motif, but different coordinations of the sodium atoms. The first is about $\pu{2.4 kJ mol-1}$ more stable than the previously shown.

The second is about $\pu{3.2 kJ mol-1}$ higher in energy.

These are electronic energies only, but they show that the sodium atoms are not discretely bound and may interchange positions quite freely.
Given that disodium thiosulfate probably only exists in a crystal lattice which cannot be described by Lewis structures, or in solution where we have to worry about other effects and equilibria, it is obvious, that there cannot be a single Lewis structure for this compound. It simply does not exist as a discrete molecule.

Literature (selection) of crystal structures

The crystal structure of anhydrous sodium thiosulphate. Sandor, E.; Csordas, L.; Acta Cryst. 1961, 14, 237-243. DOI: 10.1107/S0365110X61000802 Crystallography online: Crystal Structure of Na2 (S2 O3)
On Polymorphism of $\ce{Na2S2O3}$. von Benda, H.; von Benda, K.; Z. Naturforsch. B 1979, 34, 957-968. DOI:
10.1515/znb-1979-0716 Alternative direct link (pdf) Crystallography online: Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3)
Rerefinement of sodium thiosulfate pentahydrate. Prasad, S. M.; Rani, A.; Acta Cryst. E, 2001, 57, i67-i69. DOI: 10.1107/S1600536801011175 Crystallography online: Crystal Structure of Sodium thiosulfate pentahydrate
Structure of a new sodium thiosulfate hydrate. Hesse, W.; Leutner, B.; Böhn, K.-H.; Walker, N. P. C.; Acta Cryst. C 1993, 49, 363-365. DOI: 10.1107/S0108270192009399 Crystallography online: Crystal Structure of sodium thiosulphate 2/3-hydrate
The crystal structure of sodium thiosulphate pentahydrate. Taylor, P. G.; Beevers, C. A.; Acta Cryst. 1952, 5, 341-344. DOI: 10.1107/S0365110X52001027

Three cases:
In an aqueous environment, the anion $\ce{(S2O3)^2-}$ is surrounded by water as well as $\ce{Na+}$. In this case, none of the above.
In crystalline form, the $\ce{S-S}$ bond length is close to single $\ce{S-S}$ bond which implies a strong negative charge on the terminal sulfur. $\ce{S=O}$ bonds are shorter implying a primarily double bond character. Several pentahydrate and anhydrous isoforms are known. The sodium atom is interacting with several neighboring anions. This is a common situation in salt crystals. For example, in NaCl each $\ce{Na+}$ interacts with $\ce{6 Cl-}$ ions. In the case of the crystalline form, none of the above.
$\ce{NaS2O3-}$ was probably studied in a gas phase, but I haven't seen papers supporting this. All the forms are probably present in the gas phase, but I don't know the exact distribution. Given that the $\ce{S-S}$ bond in the crystal structure has a single bond character, structure 2 should be more commonly found in the gas phase.

'NaS2O3- was probably studied in a gas phase, but I haven't seen papers about it. In gas phase - probably all forms are present but I don't know the exact distribution. '     Could we try and stick to facts, hypothesising about a gas phase structure that doesn't exist really isn't helping the original poster who asked the question.

Apart from what @NotEvans. wrote, structure 2 is utterly wrong. If there were donor-acceptor bonds, they would most likely be reversed to what is shown.


text:
How to set up a home laboratory?
I have spent the last few years self-studying chemistry and I want to try doing practical experiments with chemistry in a safe manner. I don't plan on doing anything dangerous but would still like to keep precautionary measures. I allotted myself a medium-sized (8ft by 6ft) space in the garage to carry out any experiments. What equipment would you recommend me to keep for precautionary measures as well as carrying out chemistry experiments?

I want to study chemicals is incredibly vague. This sounds like you have no real idea what you are doing and is therefore dangerous.

Given the 'I study chemicals' I think its fair to say you probably dont have much knowledge yet. I'd build up some theory before going straight in and setting up a lab. Not only are you likely to seriously injure yourself (not to mention others around you), but your actions are also quite possibly illegal in many countries/states.

You should edit in the details of what exactly you want to do.

Your question instantly made me think of this book:
http://chemistry.about.com/b/2008/08/05/banned-book-the-golden-book-of-chemistry-experiments.htm

I have read that book. I just wanted a "second opinion".

You will be better off if you join a real lab. Researchers can help you get started in a safe manner. Chemistry experiments killed many professionals. Here is a list of things you would need.
A chemistry lab needs at least this:
Hood with 650 cubic feet per minute air flow. You CANNOT make it with a small pump.
Water.
Chemistry waste disposal. Flushing it down the drain is not a good approach.
Freezer and fridge for chemicals only.
Fire extinguisher.
Rotovap. Basic one costs 5000 dollars.
Source of vacuum.
Glassware.
Emergency shower.
Analytical instruments. NMR costs 100 000 dollars to buy, plus you will need to refill liquid helium once in a while. You can also ask chemistry department to run you samples for 10-20 dollars per piece.
Lab rules strictly prohibit working alone. In case of emergency someone should be around to save you.
Bottom line: you will poison yourself without a hood and set garage on fire. Find a university professor and ask to join their lab. If you show decent knowledge of chemistry they might agree to help you.

Feel free to comment your downvotes.

I'm going to be working with basic chemicals like copper sulfate.

I really doubt anyone would NMR in an amateur lab...

You will need a hood. You will need a place to wash your glassware. Copper sulfate (2 g per human, 30 mg/kg) is roughly as poisonous as arsenic acid (15 mg/kg, 1 g per human). Organic solvents create a fire hazard. Fumes of HCl, chlorine, bromine, are deadly when inhaled. I would strongly recommend to find a summer camp to learn chemistry.

If you google rotovap you will find that they start a lot cheaper than $5000.


text:
Why does my filtered water turn thick and almost slimy when refrigerated?
I just started using a water filter.
I have 2 containers to store the filtered water; one goes in the fridge and the other stays out on the filter.
The water that stays out is fine and stays good for days.  The water that is stored in the fridge gets thick and almost tastes gluey.
Why is this happening?

It would help to give the brand and model of the filter, since there's a lot of different filters out there with varying chemistry.

Can you please enlighten us with the brand and model of the filter, or at least how it works?

http://chemistry.stackexchange.com/questions/9395/does-water-really-go-bad-after-a-couple-of-days?s=1|0.3605

Sorry!  It's an aquasana claryum countertop AQ-5100 filter.  The water that goes in the fridge is filtered all at once.  The water that sits on the counter is filtered by the glass.  It uses activated carbon, catalytic carbon, ion exchange, sub micron filtration.  Hope this helps!

Your setup does not effectively filter nutrients. Nutrients can be problematic in climates like Charleston, SC which are humid.  Humid climates plus a nutrient content in the water create the potential for fungal growth.
Part of living in humid climates is dealing with fungal growth. Fungal growth starts from spores (capable of growing in water) which are probably living in the refrigerator.  Disinfecting the refrigerator should be done by the instructions in the manual which came with the refrigerator.
This should go without saying, the water with the starchy taste should probably be thrown out. Any drinking water left to sit has the potential to grow micro-organisms, so storing water in a refrigerator is a bad idea. Try making ice cubes in the freezer and adding it to water if you want cold water.
In humid climates, anything stored in the refrigerator should be sealed so spores cannot grow.  The refrigerator must be cleaned regularly.
Filtering water on an as-need basis may be best with your setup because a carbon filter is effective at removing anything biological.  Personally, I use a pitcher because I can refill it as needed, let the whole setup dry periodically, and easily throw out saturated cartridges.


text:
What are contrast agents in MRI and how do they work?
By reading some Wikipedia pages about MRI and MRI Contrast agent, I understand that contrast agents are mediums that reduce the relaxation times of the body tissues thereby increasing the contrast of the image obtained by MRI. My questions:

Are contrast agents just used for targeting a special tissue, such as a tumor, or they can be used the increase the contrast of the image of the whole body?
I know they decrease the relaxation time of the hydrogen atoms in body, but how do they do that?

Thanks

1) Contrast agent is used to create a sharp contrast on an otherwise fuzzy image. Consequently, its distribution must have sharp gradient, i.e it must have strong affinity to some type of tissue or physiological liquid. The target may be widely represented in the body, but it still should be a clear target. Or, alternatively, it should be injected into a physiological liquid and have no time to distribute evenly.
2) Wikipedia provides simplistic description. The idea is that to relax efficiently, a magnetic nucleus (including a proton from a hydrogen atom) needs to interact efficiently with chaotic microscopically inhomogeneous magnetic field. Such field may be produces by another microscopic magnetic objects. $\ce{Gd^{+3}}$ has seven unpaired electrons, giving the atom the strongest magnetic field of all single atoms available in reasonable chemical conditions.

(1) Contrast agents are typically injected in the blood stream and stay there during the MRI scan. They accelerate relaxation of hydrogen nuclei making blood easier to see. It can be blood in vessel or in the tissue if the vessel is broken.
(2) Gadolinium causes relaxation of 1H by relaxation mechanism. It was chosen because it has paramagnetic salts with low toxicity.

What is the relaxation mechanism?

Gadolinium is toxic, which necessitates the use of chelate complexes.


text:
Equilibrium Bond Dissociation
The equilibrium constant for the dissociation of dinitrogen tetraoxide is $K_c=5.9 \times 10^{-3}$.
If $\mathrm{1.00\ mol}$ of $\ce{N2O4}$ and $\mathrm{0.500\ mol}$ of $\ce{NO2}$ are initially placed in a container with a volume of $\mathrm{4.00\ L}$, calculate the concentrations of $\ce{N2O4(g)}$ and $\ce{NO2(g)}$ present when equilibrium is achieved at $\mathrm{25\ ^\circ C}$.
I created an ICE Table for $\ce{N2O4 <=> 2NO2}$:
$$\begin{array}{|c|c|c|} \hline
\ & \ce{N2O4} & \ce{NO2} \\ \hline
I & \dfrac{1.00}{4.00} & \dfrac{0.500}{4}\\ \hline
C & \dfrac{-x}{4.00} & \dfrac{+2x}{4.00} \\ \hline
E & \dfrac{1.00-x}{4.00} & \dfrac{0.500+2x}{4}\\ \hline
\end{array}$$
So then I let
$$5.9\times 10^{-3}=\dfrac{\left(\dfrac{0.5.00+2x}{4.00}\right)^2}{\dfrac{1.00-x}{4.00}}$$
So products over reactants, and I divided by 4 since to get the concentration, the number of moles needs to be divided by the volume which is 4dm^3.
After doing that I got $x=-0.167$ and $-0.339$, which is clearly wrong.
Correct answers: $\ce{[N2O4]}=0.292$ and $\ce{[NO2]}=0.0418$.

Your values for $x$ are perfectly correct and lead to the correct concentrations, though I can understand why you are perturbed by negative numbers. Let's see what's going on.
First, remember that $x$ is not the final concentration. It has to do with the change in concentration. If $x$ is negative, then that means the concentration is changing in the opposite direction than what you thought it was. We'll come back to that.
Let's take one of those values of $x$ and plug it into your equilibrium concentration expressions. Let's start with $x=-0.167$.
For $\ce{N2O4}$:
$$[\ce{N2O4}]=\dfrac{1.00-x}{4.00}=\dfrac{1.00--0.167}{4.00}=\dfrac{1.167}{4.00}=0.292$$
For $\ce{NO2}$
$$[\ce{NO2}]=\dfrac{0.500+2x}{4.00}=\dfrac{0.500+2(-0.167)}{4.00}=\dfrac{0.164}{4.00}=0.041$$
Notice anything? The first concentartion is spot on with one of the answers, and the second is probably just a rounding error somewhere (yours or mine).

Now about that negative value for $x$.

Notice that you have the concentration of $\ce{N2O4}$ decreasing in your ICE table, but the final concentration is higher than the initial (0.292 vs 0.250). Your assumption about the direction of the shift in the reaction toward was not correct. Not all reactions shift toward the right. How could we have known that the reaction would have shifted toward the left? We could have calculated the reaction quotient $Q_c$ at the initial conditions. If $Q_c > K_c $, then the products are in excess and the reaction shifts to the left. If $Q_c < K_c$, then the reactants are in excess and the reaction shifts to the right. We calculate $Q_c$ by putting the initial concentrations into the law of mass action.
$$Q_c =\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}=\dfrac{\left(\dfrac{0.500}{4.00}\right)}{\dfrac{1.00}{4.00}}=\dfrac{0.125^2}{0.250}=6.25\times 10^{-2}>5.9\times 10^{-3}$$
Thus, your ICE table should have the signs of $x$ reversed: the concentration of $\ce{N2O4}$ is increasing and the concentration of $\ce{NO2}$ is decreasing.
$$\begin{array}{|c|c|c|} \hline
\ & \ce{N2O4} & \ce{NO2} \\ \hline
I & \dfrac{1.00}{4.00} & \dfrac{0.500}{4}\\ \hline
C & \dfrac{+x}{4.00} & \dfrac{-2x}{4.00} \\ \hline
E & \dfrac{1.00+x}{4.00} & \dfrac{0.500-2x}{4}\\ \hline
\end{array}$$
$$5.9\times 10^{-3}=\dfrac{\left(\dfrac{0.5.00-2x}{4.00}\right)^2}{\dfrac{1.00+x}{4.00}}$$
Now $x=+0.167$ and $x=+0.338$. $x=+0.167$ leads to the same answer since the signs are switched everywhere.
The other value $x=\pm 0.338$, whether negative or positive, is not valid since $0.500-2(.338)<0$.

You're missing a zero in your $\ce{NO2}$ concentration. It should be $0.0410$, not $0.410$.


text:
Copper metal with silver ion
My book states that $\ce{2AgNO3(aq) +Cu ->Cu(NO3^-)2(aq) +2Ag(s)}$. In addition to the formula, it also shows the following image:

As in the image it shows the salmon colored copper atoms are donating an electron to the $\ce{Ag+}$ ions. In the 2nd picture on the page it shows the $\ce{Ag}$ atoms on top of the salmon colored copper atoms. I would assume that the $\ce{Cu}$ and $\ce{Ag}$ form somewhat of an ionic bond, is this correct to say? If that is the case, why does the formula not indicate a bond between the copper and silver instead the overall product is just copper nitrate plus silver. I would think it would be something like $\ce{CuAg}$?

The hint to answering your question yourself is that the book (correctly) does not speak of any ‘$\ce{CuAg}$ phase’ but of $\ce{Cu}$ and $\ce{Ag}$. There is no bond between the two.
The silver atoms merely aggregate on the copper metal surface to form metallic silver — it is the fluff you can see in the image. It depends on the exact conditions but it can probably easily be shaken off, would then fall to the ground of the tube and a fresh copper surface would appear.
Intermetallic phases can be made, but the phase diagram of copper/silver phases (see below) shows that no intermetallic copper/silver phases $\ce{Ag_xCu_y}$ exist.

The eutectic point E ($71.9~\%\ \ce{Ag}$) is merely the lowest possible co-melting point of copper/silver mixtures.

Actually the copper and silver aren't bonded. The silver that precipitates is pure. Also, a bond between two metals could not be ionic. Ionic bonds are always been metals and non-metals. Looking at the picture again, I think what may have thrown you off is that it looks like the silver at the end is Ag+. However, the silver is actually Ag+ in the beginning of the reaction when it is in the silver nitrate. That's how it is bonded with the nitrate. When the copper donates an electron to the silver, the silver goes from having a +1 charge to having no charge. It becomes neutral, and hence it doesn't bond with the copper.

Ionic bonds can also be between metal ions (e.g. $\ce{Cs+Au-}$), or between non-metal ions, such as pyridinium chloride.

There are exotic exceptions like( K-18-crown-6)${^+}$Na$^-$, but they can be ignored for most practical cases. Ag is just depositing on the surface of Cu, no ionic bond there.

@Noel how does Cu end up as cu2+? Also, in the third picture, (where it is blue) it depicts that in the free floating solution there are now cu2+ ions and NO3-, how does the copper become detached from the other other copper atoms that are aggregated together? From what I see they were originally fitted together or bonded in the coil?

The Cu ends up as Cu2+ because it donates electrons to the silver. If it 'gives away' two electrons, its charge will increase by 2. As for your second question, I'm afraid I don't know the answer. My guess would be that since it is suspended in the solution, the copper ions come in contact with the nitrate ions, and they only 'leave' the rest of the copper when they react with the nitrate. They don't freely float around as shown in the diagram.@Atticus283blink

@Noel Thanks so much! Yes, I knew that the loss of electrons makes it more positive but now i see that it donates 2 electrons to two Ag's. Yes, i assumed that was the case but wasn't sure, the pictures are very deceiving.

This may be new to consider.
I decanted ionic silver solution between 5 to 10 ppm to remove the silver oxide produced from the making of the ionic silver solution into a cubic shape container that had previously contained grade 1 copper metal chips. No remaining copper particles was visible at that time and the container was rinse under running tap water before use.
The solution had froze at approximately -10C temperature in an outdoor work shed and in the process the copper/silver ions have come together through some affinity process producing a spherical light brownish ball encased in clear distilled water in the center of the container. Copper contamination was detected early from color change of nearly colorless 5-10 ppm ionic silver solution to a bluish green solution 12 hours after filtering at 1ma current during the making of colloidal silver solution and then left to freeze with silver rounds left in the solution.
Totally amazing, so now on can actually filter out both copper ions and silver ions from aqueous solution.


text:
What is the Gibbs Free Energy in Zinc Electroplating?
In zinc electroplating process, where zinc metal (anode) is deposited on an iron plate (cathode), there is no net equation. Then why do we need to apply voltage to the system to produce zinc metal on iron? Does Gibbs Free Energy change, regardless of how much voltage is applied, after electrolysis?


text:
Disease predicting software for small molecules
Can anyone please help me in finding softwares for predicting diseases from small molecules?

Please clarify: do you mean "list all known associated diseases for molecule X" or "predict all possible diseases for molecule X, including known ones"?

Do you mean predicting health effects/toxicity of a given chemical? I haven't seen such software and I doubt it exists. Much easier problem (predict optimal synthetic route to compound X remains unsolved.
By hand your best way is to check toxicity of related compounds. A program that can predict health effects of compounds will revolutionize drug design (and cause job loss for 99% of RnD chemists in pharma).


text:
Calculation of an equilibrium reaction quotient

For the equilibrium
$$\ce{2 SO2(g) + O2(g) <=>  2 SO3(g)}$$
$$K_c = 245\ (\text{at}\ 1000\ \mathrm K)$$
The equilibrium concentrations are $[\ce{SO2}] = 0.102$, $[\ce{O2}] = 0.0132$, and $[\ce{SO3}] = 0.148$. Suppose that the concentration of $\ce{SO2}$ suddenly doubled. Calculate the $Q$ and use it to show that the forward reaction would take place to reach a new equilibrium.

I created an ice table so then I was able to find $x$:
$245=(0.148+2x)^2$ divided by $(0.0132-x)(0.204-2x)^2$
$x=0.01$
So the equilibrium concentrations are: $0.184$ for $\ce{SO2}$, $3.2\times10^{-3}$ for $\ce{O2}$ and $0.168$ for $\ce{SO3}$.
To check if the forward reaction is favored, $q<k$. The answer key states that $q=61.6$ but I don't know how they got it.

The reaction quotient, Q, is calculated the same way you would calculate the equilibrium constant, K. That is, Q is the product of the concentrations of the products of the reaction divided by the product of the concentration of the reactants,
$$
Q = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}
$$
Plugging the concentrations in $([SO_2] = 2 \times 0.102, [O_2] = 0.0132, [SO_3] = 0.148)$ should give you the answer for $Q$. I think you may have the concentrations written down incorrectly $([SO_3] = 0.184$ gives $Q = 61.6$, for instance$)$.

Why is the concentration of So2 multiplied by 2?

Q=[SO3]^2 / [SO2]^2 [O2]                                                                                0.148^2 / 0.102^2 x 0.0132 = 8.134 not 61.6

Since Q is a measure of how far things are from equilibrium, you want to calculate Q based off of the non-equilibrium concentrations (i.e. immediately after the concentration of SO_2 is doubled.)

Oh ok but it still doesnt work.          0.2148^2 / 0.204^2 x 0.0132 = 39.87 not 61.6


text:
High Water pressure
Say I had a laser drill that could cut a 2 km circle. I start it up in the deepest part of the ocean. I would then be cutting a 2 km circle straight for the earth core but just before I hit the liquid core. I shut off said drill. What would happen to the liquid ocean water at that pressure as it fall back in.

What is the pressure at the deepest part of the ocean? What is the temperature of the "liquid core"? Now you have a temp and pressure, look up the water phase diagram and see where you are. You are either gas, liquid, solid, supercritical fluid or somewhere on the boundaries of each of these regions.

This is not really on topic here. It might be acceptable on Earth Science SE but you will have to make it more specific and science based.

I'm voting to close this question as off-topic because it's not about chemistry.

If your interested only in pressure then read http://chemistry.stackexchange.com/questions/20009/hypothetical-what-happens-to-water-as-pressure-increases-to-infinity?rq=1

Sorry if put question in the wrong place

Do you mean "you can remove a cylinder of material going all the way through the Earth, letting water go down?
If so, well, the Earth should survive. There will be a mighty earthquake. Several things to consider. Earth is liquid inside. So water will not be able to get all the way in. Magma will get there first. Next you will have water overheating at the water/magma surface. Magma will be cooling down. Water is an efficient coolant, so magma will solidify and the heat exchange will slow down.
The exact yield of the explosion depends a lot on how you are going to remove this cylinder. We should wait for a sound explanation how you can do it before we start calculation.
Laser will not work. You probably think of laser as something that evaporates stuff. You don't have any way for steam to escape from the bottom of the ocean, so first, you r laser will have to evaporate all water from oceans. But then, what will be flowing in that hole?
What effects are you looking for? High static pressure? Check experiments in a diamond anvil. Loud boom?  Using a laser that can drill a hole in Earth for just that purpose is a waste. Meteorites is your best way to go. You can even use your laser to direct them at/away from Earth. Evaporating material should create enough propulsion.

The mantle is not liquid. This is one of the most common misconceptions about the Earth's structure. The mantle is entirely solid (apart from very local melting near the crust-mantle boundary). Also, I fail to see why this will cause an earthquake. Earthquakes are caused by brittle fracture along faults.

This is actually quite good description of what would happen if such fragment of Earth suddenly disappeared. Of course it can't be done via laser or even meteor, but is nice thought experiment. @bon It isn't liquid but the hole would fill quickly due to incredible pressure down there.

It's just I seen this in Independence Day 2 and thinking about the science and thought it might be cool to talk about it a bit


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Grapes in a Microwave - Experiment
I am considering attempting an experiment in which one puts two, almost severed, halves of a grape in a microwave with a glass over it. To my knowledge so far, the grape halves act as focal point for the radiation and the microwaves will burn the small part of grape skin that isn't severed, providing a spark which the microwaves will stimulate into plasma, contained by the glass jar. Three questions: is the experiment reasonably safe, am I correct in my understanding of the experiment, and what compounds will the plasma form once it cools?

Sadly, it's not a plasma just a flame. It's good fun and quite safe. I hear claims that it can damage your microwave oven, but mine has suffered no damage from repeated experiments with grapes. You don't need the jar.


text:
How to distinguish between Na2CO3 and NaHCO3 by a chemical test?
$\ce{Na2CO3}$ and $\ce{NaHCO3}$ have similar reactions. For example, they can react with acid to produce $\ce{CO2}$.
So my question is:

Can we use a chemical test to distinguish between $\ce{Na2CO3}$ and $\ce{NaHCO3}$?

I know that $\ce{NaHCO3}$ can be thermal decomposed into $\ce{Na2CO3}$ via this reaction: $2\ce{NaHCO3->} \ce{Na2CO3} + \ce{CO2} + \ce{H2O}$
However, I am not sure if thermal decomposition is a chemical test.

You should give more info about your sample. Is solid? Is a disolved in water?  In such case, do you know the concentration?
I can not imagine something more direct than just check the pH of a solution.

The pH of Na2CO3 at 1mM = 10.52; Molar mass of Na2CO3: 105.9888 g/mol. The pH of NaHCO3 at 1mM = 8.27; Molar mass of NaHCO3: 84.007 g/mol.

You could also test for evaporating water (for instance, by having the the gaseous products go through some waterfree copper-(II)-sulfate, which would turn blue in presence of water). This requires, of course, assurance that it can only be, in fact, sodium carbonate or sodium hydrocarbonate.

As you mentioned in your post, thermal decomposition and measurement of the resulting gas would be one way to distinguish what you have. I would call this a physical, rather than chemical method, however.
This kind of thermal decomposition isn't something most labs would consider for two reasons:

Both chemicals are cheap. They're generally bought by the kilo and looking at what we get charged for them the company is basically only charging us for the cost of the containers and shipping.
Thermal decomposition and gas measurement is, nowadays at least,  a fairly specialised technique. Most labs would struggle to piece together the required equipment and even if they could find it, would likely be reluctant to do so.

That said, your question seems purely academic rather than practical, so lets consider a possible chemical method.
Sodium carbonate is well known for its use in inorganic analysis, as it reacts with certain metals to give carbonate salts of well defined colour.

The test can distinguish between copper (Cu), iron (Fe), and calcium (Ca), zinc (Zn) or lead (Pb). Sodium carbonate solution is added to the salt of the metal. A blue precipitate indicates Cu2+ ion. A dirty green precipitate indicates Fe2+ ion. A yellow-brown precipitate indicates Fe3+ ion. A white precipitate indicates Ca2+, Zn2+, or Pb2+ ion.  (Source: Wikipedia)

IIRC the test doesn't work with sodium bicarbonate, and as such you could use the reactions with a cheap metal such as iron to determine whether you had sodium bicarbonate or sodium carbonate.

Heating salt is an important test in determining the anion in the salt by checking the gas evolved. This test is called dry heating test and is a important part of qualitative analysis of salt. But this test goes under physical examination of salt (checking color, appearence etc.).
One important chemical test to differentiate carbonate and bicarbonate is $\ce{MgSO4}$  test.
$$\ce{Na2CO3 + Mg^2+ -> MgCO3↓(white) + Na+}$$
$$\ce{NaHCO3 + Mg^2+ -> MgCO3↓(white) + H2O + CO2 ^}$$
When carbonate salt is reacted with $\ce{MgSO4}$, a white salt precipited while bicarbonate salt on reacting with $\ce{MgSO4}$, not only a white salt precipited but also a brisk effervescence occur. The gas  is detected  by passing it through lime water.

I'm a little unsure about the magnesium test. AFAIK magnesium bicarbonate is stable in aqueous solution. I would prefer calcium chloride, which selectively precipitates calcium carbonate from  a carbonate solution.

depends on your equipment. You can estimate the amount of $\ce{NaHCO3}$ by taking a known amount of sample, heating it to $\sim\pu{300 °C}$, cooling it to room temperature and measuring the weight again. Zero loss means 100% $\ce{Na2CO3}$. Loss of $37\%$ of weight means $100\%$ $\ce{NaHCO3}$. The dependence is linear within $0-37\%$ weight loss.
Better way is to react a gram of material with $\ce{HCl}$, send forming $\ce{CO2}$ through saturated $\ce{Ba(OH)2}$ ($\ce{Ca(OH)2}$ can be used, but wouldn't work as well), collect, wash dry and weight the precipitate $\ce{BaCO3}$. Both experiments tell you how much carbonate anion you have in a sample. You have to assume that $\ce{NaHCO3}$ and $\ce{Na2CO3}$ were the only components in the mix.
You can also evaluate the ratio based on pH of the resulting solution if you know how to calculate pH for buffers.

You should try formatting your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance.

Dissolve a sample in water to a known concentration and measure the pH.  Carbonate ion is a stronger base than bicarbonate ion and the pH will behave accordingly.


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Over potential -what and why ?
Value of standard electrode potential for the oxidation of Cl– ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl– oxidised at anode instead of water?
Standard electrode potentials describe the tendency of getting reduced.Positive value indicates that H2 can reduce that element or reduced form of the element is more stable than proton.
The answer to the question above is over potential
Please explain how it came into the story.My book explained it under "products of electrolysis",but the explanation is not lucid.
They explained it as :
Some electrochemical processes although feasible, are so slow that at  lower voltage these do not seem to take place and extra potential called over potential has to be applied which makes such process more difficult to occur.

When I took P-Chem, decades ago, one experiment we had was using a dropping mercury electrode, DME. I don't recall the details, but the set-up was fairly elaborate, especially compared to a Ag/AgCl electrode. The DME is fairly common (or was, before we decided that Hg was probably better avoided by the undergrad). The reason WHY we used it was it provided a CLEAN surface. Think about it. We wouldn't go to such great lengths if the problem of contamination wasn't significant.
All materials absorb contaminants, may chemically react with near-by chemicals (solvent, ions, gasses), and will form a charge polarization layer, especially in polar solvents like H2O. In fact some electrochemical methods rely on it. But for getting the "pure" ideal thermodynamic (reversible) numbers the large tendency of electrode surfaces to change is a real problem.
Overpotential is the term we use to describe the additional energy (voltage) we have to put in to these non-ideal systems (either at the anode or the cathode) to overcome these "side reactions". This energy is ultimately converted (mostly) to heat. Although maybe not an overpotential issue, surface area can be important, so smoothness does matter.
I can think of no area in chemistry (short of biochem) where you have to get into the weeds (details) so quickly as you do with electrochemistry. I wonder if its because chemistry is actually about electric interactions between atoms, molecules, ions, etc.? I should also mention that on other nice thing about the DME was that the surface was smooth - although I can't recall whether the surface was smooth at the atomic level, most solids are far from smooth, another problem in reproducibility.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help].

This is a question of kinetics vs thermodynamics. From the perspective of thermodynamics equilibrium of water, HCl, O2 and Cl2 is shifted towards HCl+O2. However kinetics dictates Cl2 to form faster than O2.
Kinetic vs thermodynamic control is a commonplace in chemistry. Thermodynamics suggests that we all should be dead, but thanks to kinetics we are still alive. Why Cl- is kinetically favorable? Because 4 electrons should be taken from 2 H2O molecules to form O2. At the same time intermediates in this oxidation (atomic O, H2O2) can easily oxidize two chloride ions and yield chlorine gas.


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Why don't the electrons of opposite spins collide in an orbital?
As I was reading through my textbook, I got to know that in an orbital, there are 2 electrons of opposite spins. My question is: Shouldn't they collide whilst they are rotating?

Excellent question, and welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center.

You may also find the following (http://physics.stackexchange.com/questions/200536/why-dont-electrons-collide-among-themselves) question on Physics.SE useful.

They don't rotate. Spin is an intrinsic property that is, perhaps, unfortunately named since it causes so much confusion in folks. And, electrons don't 'orbit' - they occupy wave functions.

In reality, electrons in a molecule are colliding all the time insofar as they are experiencing a force (the Coulomb force) due to the presence of other electrons, and are correlating with each other as a result. Orbital models like the Hartree-Fock mean field approximation mostly neglect these interactions, so the answer to your question is they don't because the model doesn't consider such dynamics. But many sophisticated methods for calculating the effects of this correlation exist. E.g. "Full Configuration Interaction" and "Density Functional Theory" are two such methods.
So keep in mind that orbital models are only approximate. It is fine for many purposes, but inadequate for many too. The true ground state wavefunction of a molecule cannot be represented by a single determinant, and therefore cannot be thought of strictly as a collection of electrons occupying orbitals.

While there is no perfect classical analogy, the spin of the electron is angular momentum, about its own axis, not a direction of rotation about the nucleus.
In comparison, Venus rotates about its axis in the opposite direction of Earth, but this does not make the planets more or less likely to collide.


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When is the product a compound and when is it 2 ions in these reactions?
When finding the half equation and overall redox equation for the following:
a) $\ce{Fe^2+(aq) + Cl2(g) <=>}$
b) $\ce{N2(g) + H2(g)  <=>}$
The solution from the textbook is given below. I do not understand why the products (in the overall reaction) for a) are 2 individual ions (and not $\ce{FeCl3}$ whilst the product for b) is a compound (and not $\ce{2N3- + 6H+}$).
One difference I noted was, when the electrons are balanced for a) it gives $\ce{2Fe3+}$ and $\ce{2Cl-}$ , which cannot go together unless its $\ce{2Fe3+}$ and $\ce{6Cl-}$, whilst for b), when the electrons are balanced we have $\ce{6H+}$ and $\ce{N2}$ which combine as is to form $\ce{2NH3}$. Is this the reason?
The solution given is:
a)
Oxidation: $\ce{2Fe^2+(s) -> 2Fe^3+ + 2e- }$
Reduction: $\ce{Cl2(g) + 2e- -> 2Cl- }$
Overall: $\ce{2Fe^2+ (aq) + Cl2(g) <=> 2Fe^3+ + 2Cl- }$
b)
Oxidation: $\ce{3H2 -> 6H+ + 6e-}$
Reduction: $\ce{N2 + 6e-  -> 2N^3-}$
Overall: $\ce{N2(g) + 3H2(g)  <=> 2NH3}$

While FeCl3 is a possible product, the reaction in A is aqueous. Wikipedia shows that is readily dissolves in water (92 g/100 mL @ 20C). The products exist as ions, and are written as such.
I would suspect you're not supposed to have solubility constants memorized though, so in general if it is in water they will exist as ions.
B is a gaseous reaction and the product is a compound, not an ion set like A.


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Predict the products of the controlled oxidation of 2-methylpentan-2-ol

Predict the products of the controlled oxidation of 2-methylpentan-2-ol.

From what I know, since 2-methylpentan-2-ol is a tertiary alcohol, there should be no reaction. However, my textbook states that the product includes 2-methylpentanone.
Edit: In response to the comments, my textbook neither specifies the isomer nor mentions any given conditions. I hate Nelson

What conditions are given for the oxidation?

Nothing will happen under "controlled" conditions with any of the common oxidizing reagents.  Under forcing conditions elimination of water to form olefins can occur along with epoxide formation and general molecular fragmentation.  BTW, which isomer of 2-methylpentanone does your textbook claim is formed?

The "reactant includes 2-methylpentanone"?  Or the product?   And what molecule is meant by "2-methylpentanone"?  4-methyl-2-pentanone?  Or 3-methyl-2-pentanone?  Or 2-methyl-3-pentanone?  It's hard to see how any of those could be produced in high yield through the oxidation of this alcohol.

I suppose under dehydrating conditions the preferred product would be 2-methyl-2-pentene?  Could that alkene be oxidized to 2-methyl-3-pentanone?

Given the level that book is aimed at, this seems far more like a typo than a genuine question. As @Ron pointed out, the most likely thing to happen is elimination.... anything after that is getting into significantly non 12th chemistry

@CurtF. I meant the product. I changed it. Thanks for pointing that out.

Tertiary alcohols are resistant to oxidation in neutral or alkaline solution, but are readily oxidized by acidified oxidizing agents.
In your case two products will be acetone and propionic acid.
http://chemistry.tutorcircle.com/organic-chemistry/tertiary-alcohol.html

This simply isn't true. Even the Jones reagent (18M sulfuric acid and stochiometric chromic anhydride ) doesn't touch tertiary alcohols.  There are indeed methods but reasily oxidised is simply nonsense

@NotNicolaou If you are unfamiliar with the topic study it before commenting/downvoting http://pubs.acs.org/doi/abs/10.1021/ja01190a040 You cannot oxidize it using a school chemistry textbook, but you are allowed to oxidize them once you go to a university.

Look at the limited substrate scope of that paper. This is not a general trivial method. I've also found that textbooks don't make very good oxidants....

@NotNicolaou Look at the link in my post. Scroll to the very bottom.
http://chemistry.tutorcircle.com/organic-chemistry/tertiary-alcohol.html This is the reaction we are talking about.

Like I said.  Look at the crappy substrate scope. If you'd like to continue this please do so in chat

@NotNicolaou http://chemistry.tutorcircle.com/organic-chemistry/tertiary-alcohol.html Read this link.

Let us continue this discussion in chat.

"The carbon atom bonded to oxygen must also bear a hydrogen atom.         Tertiary alcohols (R3C–OH) cannot be oxidized in this fashion."

"Tertiary Alcohols: These are resistant to oxidation because they have no hydrogen atoms attached to the oxygen bearing carbon
(carbinol carbon)."

"Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever"

"Tertiary alcohols cannot be oxidised (no carbinol C-H)"

"What happens when we try to oxidize a tertiary alcohol? The answer is simple: Nothing happens."

"Tertiary alcohols cannot be oxidized unless harsh conditions are used that cause complete structural degradation."

"Tertiary alcohols do not undergo oxidation."

@f'' In school they don't. In university they do.

I see a lot of hostility in this comment section and I would like to ask everyone to be nice to each other.


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Can compounds (or only elements) be referred to as oxidising and reducing agents?
In the reaction
$$\ce{3CuO (s) + 2NH3 (aq) <=> N2(g) + 3H2O(l) + 3Cu(s)}$$
the oxidation number of $\ce{Cu}$ goes from $+2$ in the reactant side to $0$ in the products, and so it is reduced.
The oxidation number of $\ce{O}$ is $-2$ and stays the same on both sides.
My textbook refers to $\ce{CuO}$ as the oxidising agent. Is this because the $\ce{Cu}$ in $\ce{CuO}$ is oxidised while $\ce{O}$ is neither oxidised or reduced, so overall it is an oxidising agent?
What happens for a compound $\ce{AB}$ where $\ce{A}$ gets oxidised and $\ce{B}$ gets reduced? Do I call the compound $\ce{AB}$ a reducing agent if the net change in oxidation number (between the two elements $\ce{A}$ and $\ce{B}$) is negative?

My textbook refers to $\ce{CuO}$ as the oxidising agent. Is this because the $\ce{Cu}$ in $\ce{CuO}$ is oxidised while $\ce{O}$ is neither oxidised or reduced, so overall it is an oxidising agent?

No. $\ce{CuO}$ is the oxidizing agent (the thing that causes something else to be oxidized) because the $\ce{Cu}$ is reduced (gains electrons, going from oxidation state $+2$ to $0$) in the course of oxidizing the nitrogen of $\ce{NH3}$, which loses electrons and goes from oxidation state $-3$ to $0$.
Conversely, $\ce{NH3}$ is the reducing agent here, because it causes something else to be reduced (here, the $\ce{Cu}$) and loses electrons in the process.

What happens for a compound $\ce{AB}$ where $\ce A$ gets oxidised and $\ce B$ gets reduced?

If only the species $\ce{AB}$ is involved, or in the case where two of the same species $\ce A$ react to form both oxidized and reduced products:
$$
\ce{2A -> Ox + Red}
$$
the reaction is called disproportionation.

Do I call the compound $\ce{AB}$ a reducing agent if the net change in oxidation number (between the two elements $\ce A$ and $\ce B$) is negative?

Strictly in the case of a $1\!:\!1$ compound of $\ce{AB}$ on a mole basis, if the net change in oxidation number is negative, then you would call $\ce{AB}$ an oxidizing agent, because it pulled electrons away from something else (it oxidized that other thing) in order to achieve the net negative change in oxidation number.
For $\ce{AB}$ compounds not in a $1\!:\!1$ molar ratio, a universal principle can't be stated $-$ it depends on the particulars of the mole ratio and the individual changes in oxidation number.

Actually it is the full reactants that are the oxidizing and reducing agents.  We just specify atoms for these roles as a simplifying assumption; for example, when $\ce{KMnO4}$ acts as an oxidizing agent in acid, we call the manganese the oxidizing agent because the change in manganese oxidation state from +7 to +2 accounts for the five electrons taken up by each permanganate ion during the course of reaction.  But, technically, it is the permanganate as a whole that consumes the electrons, not just the manganese.
Such a simplifying assumption is bound not to work all the time, and we may need to treat the whole compound as the oxidizing or reducing agent.  See this question for an example of such a situation.


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Extracting herbs with grape juice
I tried searching my question on google and did not find anything. I am looking to extract herbs with pure 100%
grape juice http://www.lakewoodjuices.com/product_detail/id-32/ ( it is pasteurized ) instead of using the old school
method of extracting with a 50% alcohol solution.My theory as of now would be that it would be more beneficial and it
would ferment or extract the same plus the extra beneficial minerals of the juice. I am wonder what your opinion on
this is. Do you think it would extract the same and be just as beneficial as a tincture? Would it be more beneficial?
How long should I let it sit for? should I let it sit at room temp or in cooler temp say, 45 degree F? Should I do neutral water dilution with the grape juice? how much, 50% neutral water 50% grape juice?
Thank you so much for your time and knowledge.

This is a little broad as it stands - you might want to break this question into smaller chunks to get more attention from those who can help.

You might consider that herbs may contain microorganisms that might  turn the grape juice into an undesired product.

From what I understand the grape juice will ferment and turn into alcohol while still maintaining elemental properties.  I am curios if I should do 70% grape juice with 30% neutral water then add the herbs and let them sit in 40 degree temperature for about a month to get full extraction like you would with pure alcohol tincture.

The point of using alcohol to extract things from herbs is that alcohol solubilises things that are not soluble in water. So using grape juice as an alternative is pretty pointless since it is just water with a lot of other water-soluble things in it (like sugars).
So using grape juice is a pretty poor alternative to using alcohol.
Grape juice will ferment under some circumstances, converting some of the sugars into alcohol. But this won't work well unless the conditions are controlled carefully and won't work at all if it is cold (45 F is cold). If you don't do what winemakers do, there is no telling whether it will work or whether it will work without producing nasty results. And pasteurised grape juice won't ferment at all if you don't add yeast and keep it clean. If you don't keep it clean you might well get some nasty strain of natural bug rather than a good wine yeast which will have unpredictable results.
Moreover, the best level of alcohol you are going to get even as an experienced winemaker is 10-15%. This is nothing like as useful as pure or 50% alcohol for solubilising or extracting things that are not water soluble. And grape juice with no alcohol isn't going to be a much better solvent than water (and most likely will be worse as it is already full of other dissolved things).
The old school recipes you are trying to avoid exist because people have tried these experiments before and found what works. Trying to invent new recipes that ignore the "old school" experience is a pretty futile task: what you propose isn't in the old school recipes because it doesn't work.

There are four main concerns with this procedure.
(1) Ethanol dissolves and extracts organic chemicals (active ingredients in herbs) better than water (which is the main component of juice).
(2) It is much easier to remove ethanol, than water making concentrating step easier.
(3) Ethanol is less chemically reactive and is less likely to destroy active ingredients of the herbs. Acidity of a juice is my main concern.
(4) Juice is a complex mixture of chemicals that varies from batch to batch. It is hard to make this procedure reproducible.

Thanks so much for your response. I am thinking I will use 80 to 90% pure grape juice and mix 10 to 20% neutral water. Put herbs in glass jars and let sit in 40 degree F temp for about a month. From I have gathered grape juice is also a good solvent. Im before it ferments into alcohol so I am definitely curious how this would turn out.

@Chris1989777 Juice might fermented over that time, bacteria will eat tasty components and you will get poisoned. Instead, you should get some (80-100 proof), grind herbs, and extract everything in the liquor. Just let it seat for a week. Then you can remove liquor by blowing air at it if you really want to.

@Chris1989777  If the answer helped, please press "accept button" located to the left of my question. If you have other questions please have a separate topic.


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In terms of remediation, what is sequestration?
I am reading about use of zero valent iron in remediation of heavy metals.
Is sequestration just a broader term for sorption?

Sequestration is any way of preventing the heavy metals from being able to harmfully react with biologically important molecules. In the case of zero-valent iron nanoparticles, the heavy metals are sequestered by sorption as well as reduction. However, there are other methods of sequestration, such as chelation.


text:
What is pairing energy of electrons?
What exactly is pairing energy when electrons are paired in orbitals and what is responsible for it.

related: http://chemistry.stackexchange.com/a/15224/189

Read the question that Philipp linked and then tell us what you don't understand. At the moment this question is rather vague.

Imagine a sky scraper where everyone wants to live on as low floor as possible, but landlord prohibits then to live more than 2 people per floor. These people will occupy lowest floors in pairs and the highest occupied level might have a single person or a pair. If someone moves out of the apartment on the 1st or 2nd floor, then someone from higher floors will move down to the vacant state. Electrons do the same. Floors are energy levels and the evil landlord is the Pauli exclusion principle.
Electrons are fermions. An electron's spin can be either +1/2 or -1/2. Two options. Two fermions (and, hence, electrons) cannot be the same. Electrons are the same if their spins are the same and they occupy the same orbital (Pauli Exclusion Principle). At lower temperatures (below red heat temperatures) electrons occupy the lowest energy level possible.
As a result, each new electron occupies the lowest allowed level. Because their spins can take two values 2 electrons are occupying each level. If you populate each energy level by a single electron you will get a high energy excited state that well emit light and go to lower energy two electrons per level state. If you try to squeeze more than two electrons it will not work because of Pauli principle. So, since they want to stay in lower energy levels and have to obey the exclusion principle we have electron pairs.

Bonus: here is why radicals recombine. For instance, why $\ce{2 H^. -> H2:}$
If an electron is the last one it is occupying the energy level alone.
So, electrons would like to all occupy the lowest level, but only two per level is allowed. In radicals only one electron occupies a level. When radicals A$^.$ and b$^.$ meet each of them sit on half empty orbital attracted to a single nucleus.  It is energetically favorable to share the same orbital (form a pair, which they are OK with) and orient fragments so that they "fly" (have the highest probability) between two nucleus (A and B). Staying between two opposed charged ions is better than staying next to just one charged ion. And electrons are allowed to stay in pairs (Pauli excluded 3+ electrons staying together).


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Kc(equilibrium constant) and Kp
Its given in the book that at 12.18kelvin Kc=Kp. How is that equated from the equation Kp= Kc(RT)^∆ng. I am unable to sort it out, please guide me in to the topic....

It seems like a homework question. Can you please give some more clarification?

Hi; you should define your variables more clearly and then use the equation formatting tools to make nice equations. The easiest way to do this is to click on some other questions with equations, and copy the syntax. Also there is some information at the header of the box you use to post equations.

From the ideal gas equation, we get
$$PV=nRT \implies P=\frac{n}{V}RT=cRT$$
where $c$ represents the concentration.
Consider an arbitrary reaction
$$n_AA\rightarrow n_BB$$
$$K_c=\frac{c_B^{n_B}}{c_A^{n_A}}$$
$$K_p=\frac{P_B^{n_B}}{P_A^{n_A}}=\frac{(c_BRT)^{n_B}}{(c_ART)^{n_A}}=K_c \times (RT)^{\Delta n}$$
Most probably, the reaction you are talking about has $\Delta n=0$.

Actually you can see here that in formula $K_p=K_c×(RT)^{\partial n}$; $K_p$ will be equal to $K_c$ if and only if $\partial n=0$ or $RT = 1$
So when $RT =1$ then $T =\frac1R$ that is $\frac{1}{0.0821}=\pu{12.18K}$.

Here when RT=1,its independent of difference of stochiometric co-efficient


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Electrolysis of molten nitrates, sulphates and carbonates
We know that nitrates, sulfates and carbonates doesn't or hardly oxidized by anode in the aqueous solution.
What will be produced on anode in the electrolysis of molten salts of these ions?

A common theme is disproportionation of the anion, and potential reactivity with the electrodes.

Nitrates
Electrochemical behavior of nickel electrodes in molten potassium nitrate with increased basicity, Electrochimica Acta, Vol. 36, No. 3/4, pp. 65S-660, 1991:

$\text{Ni}$ electrodes in a $\ce{KNO_3}$ melt [...] some ionic species of nickelite ($\ce{NiO_2^{2-}}$) and nickelate ($\ce{NiO_2^{-}}$) type

Voltammetric investigations of the oxygen electrochemical systems in molten sodium nitrate at 420°C, Electrochimica Acta, Vol. 33, No. 4, pp. 545-549, 1988:

the potential range is limited anodically by the electrochemical oxidation of nitrate ions into nitrogen dioxide and oxygen: $\ce{NO3-} \to \ce{NO2} + \frac{1}{2}\ce{O2} + \ce{e-}$ and cathodically by the electrochemical reduction of nitrate ions into nitrite ions: $\ce{NO3-} + \ce{2e-} \to \ce{NO2-} + \ce{O^2-}$

Carbonates
Electrochemical formation of carbon nano-powders with various porosities in molten alkali carbonates, Electrochimica Acta, Vol. 54, pp. 4566–4573, 2009:

reduction of carbonate ions: $\ce{CO3} + \ce{4e-} \to \ce{C} + \ce{3O^2-}$ [and reduction of metal ions, along with oxidation of carbonate ions:] $\ce{2CO3^2-} \to \ce{2CO2} + \ce{O2} + \ce{4e-}$

Sulfates
Polarization measurements on solid platinum-molten sodium sulphate-sodium chloride interfaces, EIectrochimica Acta, Vol. 23, pp. 381-388, 1978:

the limiting reactions [...] are suggested to be the $\ce{SO3}$ reduction to $\ce{S^2-}$ and $\ce{O2-}$ and the $\ce{SO4^2-}$ oxidation to $\ce{SO2}$ and $\ce{O2}$

Carbonates will lose CO2 on melting and turn into oxides.
Nitrates usually decompose/blow up on melting. Electrolysis will speed up explosion. Before they blow up O2 will form. Water solutions at low temperatures can also form NO4$^-$ (pernitrates).
Sulfates will form O2. water solutions of sulfates also form persulfates at low temperature.

"Nitrate molten salts are extensively used for sensible heat storage in Concentrated Solar Power (CSP) plants and thermal energy storage (TES) systems." - D'Aguanno et al., Scientific Reports, 8, 10485 (2018)


text:
Work is not a state function
From Work-Energy Theorem we have,
Work = Change in kinetic energy
         = Final K.E - Initial K.E
This shows that work only depends on the final and initial value of K.E independent of the path.
I'm lost. Can someone clear this up for me?

Your definition of work is not quite correct.  A better definition is**
Work w, is the energy change that drives net shift in the distribution of mass in the system.
Thus energy here means total energy which is kinetic plus potential. Imagine going to the top of a building. You gain potential energy by virtue of being at the top (m.g.h, mass m, acceleration due to gravity g, and height h). If you are not sure that you have gained energy, think about what happens if you were to walk of the top of the building!
You can walk up the stairs and gain this potential energy but you could also run up, and if you could manage to run, (huff-puff-pant-groan) the amount of work you would have to do is far more than walking, but the potential energy gained is exactly the same. Therefore in general the work depends on the 'path' taken, not just the starting and ending conditions. The word 'path' used in thermodynamics generally means the sequence of all the changes made to go from start to end. Hope this helps to clarify.
** ref A. Cooksey Physical Chemistry

Great analogy (the stairs).  You can also think of it in terms of integrating under a curve.  Exaxt integration gives the maximum value whilst taking increasingly large chunks for numeeical integration the value decreases significantly

Consider moving from A to B to A in viscous liquid. How much work did it take? A stateful function should answer "zero" because A=A. However it could have take any amount of work (depending on the distance from A to B, viscosity of liquid, speed, etc. Thus, work cannot be derived from initial and final states alone and is not stateful function.


text:
Boiling Point lower than Melting Point?
Under the Wikipedia entry for Metaldehyde is listed a Melting Point of $\pu{246^\circ C}$ and a Boiling point of $\pu{110^\circ C}$ (actually, it says it sublimes).
How can a Boiling point be lower than the Melting Point? Even if $\pu{110^\circ C}$ is actually a sublimation temperature how can it then subsequently melt at a higher temperature of $\pu{246^\circ C}$?

Sublimation is not boiling but there is a reason for listing it instead. Sublimation is direct vaporisation from the solid. The most familiar is solid dry ice producing $\ce{CO2}$ gas. So if metaldehyde melts at $\pu{246^\circ C}$ then one might suppose that the boiling point is higher. However, look at the generic $P$ vs. $T$ diagram below. If we assume that at $P = 1000$ is normal atmospheric pressure then a compound with this $P$ vs $T$ sublimes. As we define boiling to occur only at normal pressure and be the liquid to vapour (gas) transition then this compound can never boil at normal pressure. To become a liquid high temperature and pressure is needed. So you can see why the table lists sublimation at normal pressure rather than boiling.

So you mean the BP value is at 1 atm but the MP value is at a different (higher) pressure? In  other words, they have mentioned the normal BP but not the normal MP since at 1 atm a melting transition just does not exist?

yes, exactly. Normal boiling temp is measure at 1 bar . In the illustration if  P=1000 this compound will never boil no matter what the temperature is. To get it to boil the pressure would have to be at least approx 5000 and temperature approx 280.

The MP value of 246 C is obtained in a sealed container.

yes, if sealed then the pressure will be large, > 1 bar. There is no getting around the phase diagram. If you can find one for metaldehyde  it should be clear, but it will have the same forma as above, just with different values along the axes.

@porphyrin There is, though its blurry. A subliming solid heated fast enough may melt and then evaporate instead of subliming, I've seen it for iodine.

That's another story. Iodine does have a legitimate melting point at normal pressure, though an urban legend says otherwise.

@porphyrin Thanks! I still think it is rather silly  to ever mention a non-ambient MP or BP without also explicitly stating the Pressure conditions at which the temperature was measured at. That's what confused me I guess. I'm used to assuming atmospheric pressure whenever the pressure is not mentioned.

@permeakra Can you post the metaldehyde phase diagram even if blurry? I'd love to see.