text:
Why does this reaction proceed this way?
I am confused as to why this molecule changes its structure.
Also, I am confused why instead of OH, CHO is attached.

I would have thought that the double bond would break and OH would attach at each end via syn addition.The example:

Shouldn't this be the answer?

The molecule changed its structure because the oxidation snapped the whole double bond (like, for example, ozonolysis would) and not just one component of it. How that came possible is another story.

I think this is a typo. The sodium periodate (NaIO4) should be included.


text:
Why do these two IUPAC naming examples not begin numbering the chain closest to the alkyne?
I am currently reading my chemistry book which states "The chain is numbered from the end closest to the triple bond." However, then it proceeds to give two examples which I do not believe follow this rule. What is the reasoning for this?

The oversimplified statement “The chain is numbered from the end closest to the triple bond.” applies to simple alkynes with one triple bond. It is not necessarily correct for compounds that include several multiple bonds or other characteristic groups.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), low locants are assigned first to the principal characteristic group that is expressed as a suffix and then to multiple bonds.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
(i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
(ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(…)

Therefore, the correct name for the second example given in the question is ‘but-3-yn-2-ol’ (not ‘but-1-yn-3-ol’) since a low locant is given first to the suffix ‘-ol’.

If the compound contains several multiple bonds (double or triple bonds), low locants are given to multiple bonds as a set. If there is still a choice, low locants are given to the double bonds.

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
(…)

Therefore, the correct name for the first example given in the question is ‘2-methylpent-1-en-3-yne’ (not ‘4-methylpent-4-en-2-yne’) since the locant set ‘1,3’ for the multiple bonds is lower than ‘2,4’.

Shouldn't the name in second example be but-2-ol-3-yne or the alphabetical order doesn't matter in suffix?

@manshu ‘ene’ and ‘yne’ are endings, which usually replace the ending ‘ane’ of the name of a saturated parent hydride. Functional suffixes (e.g. ‘ol’) are always attached to the name of the parent hydride, modified or not by ‘ene’ and ‘yne’ endings. Only one kind of characteristic group can be cited as suffix; any other substituent groups not expressed as suffixes are cited as prefixes. Therefore, the principle of alphabetical order does not apply to suffixes.

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text:
Density of LPG in g/L
What is the density of LPG (in vapour state) in g/L?
Please note that I'm asking about the density of vapour LPG when it comes out of the cylinder. NOT the liquefied density!

LPG – propane – gas density is 1.55 times heavier than air. It is  $1.898$ g/L vs  $ 1.225$ g/L for air (both 15°C at sea level).
For more details, please visit this website.

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text:
What does the unit M stand for?
What does $\mathrm{M}$ signify in this pH calculation? To me, it just seems to be a random variable.
For example: $\ce{[H+]} = 1.0\times10^{-7}~\mathrm{M}$
I am not sure what the $\mathrm{M}$ is supposed to mean?

M = Molarity = mol per Liter of solution

It is shorthand for molarity. More precisely the amount concentration. It is used as a unit and stands for mole per litre.
The IUPAC gold book writes:

amount concentration,
  Amount of a constituent divided by the volume of the mixture. Also called amount-of-substance concentration, substance concentration (in clinical chemistry) and in older literature molarity. For entities B it is often denoted by $\ce{[B]}$. The common unit is mole per cubic decimetre ($\mathrm{mol\, dm^{-3}}$) or mole per litre ($\mathrm{mol\, L^{-1}}$) sometimes denoted by $M$.

In your example it just means the concentration of protons in solution is $\ce{[H+]} = 1.0\times 10^{-7}~\mathrm{mol/ l}$.

Martin, thanks. What do protons have to do with this considering this is regarding PH. Also, you have "mol/1" so meaning mole per 1 what exactly?

@Atticus283blink Please read the Wikipedia page on pH it answers all your questions. Please also read the answer carefully, it's states multiple times that it is mole per litre. The unit litre may be written as L or l.

"mol/l" is mole(s) per liter He used a lowercase l instead of the uppercase L.  //  On old typewriters there was no digit 1. you were expected to use the lowercase L. Most typefaces now try to have somewhat different forms for the digit 1 and the lowercase letter L. The digit 0 and the uppercase letter O are another problem.


text:
Which is better: iodine number or bromine number?
The degree of unsaturation of fats can be determined with the help of two type of scale namely iodine number and bromine number.
Generally, iodine number is more preferred over bromine number (Is this statement true? Is it because the Wikipedia article of iodine number is more elaborated than that of bromine number?)
Iodine number is typically found out by using iodine monochloride, $\ce{ICl}$ where the iodine atom gets bonded with the carbon atom having double bond. The more iodine  used up to break the $\ce{C=C}$ bonds,  higher is the iodine number.
$$\ce{-CH=CH - + ICl -> -(I)CH-(Cl)CH - }$$
But in the 4th paragraph of the 1st page of this link, it is written that:

Iodination
  is an endothermic process, resulting in vicinal diiodides that tend
  to revert to alkenes. Consequently, the most common applications of
  alkene halogenation are chlorination and bromination.

So, in this context, bromination is better and thus bromine number is better than iodine number. The statement contradicts the previous statement i.e. using iodide number.
H.P.Kaufmann method of finding iodine number uses bromine. Is this a better method than using $\ce{ICl}$ considering the disadvantage of iodination?

Bromine number (and Bromine Index) are based on theoretically good chemistry; however, that is where their virtues end. If these tests are run according to the current ASTM procedures, you are virtually guaranteed to get an incomplete reaction and low (perhaps as much as 50%) results. When properly applied, off the shelf equipment can result in exceptionally good accuracy and precision, but it does require a deviation from ASTM-accepted technology.

Which is better for measuring unsaturation, Iodine number or Bromine number?  The answer depends on how you define "better". However I can tell you which is commercially useful. If we're dealing with fatty acids or vegetable oils, the industry standard test is Iodine number. For petroleum products, the standard test is Bromine number (or bromine index, slightly different). For the two tests, Reference Bailey's Industrial Oil and Fat Products, sixth ed. vol. 1 pg. 172, and ASTM D-1159.

Could you add some information to support a conclusion and answer the question? While your post is useful, it is more appropriate as a comment or two than an answer.

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text:
Effect of passing hydrogen sulfide gas on electrode potential

Consider the cell reaction
$$\ce{2Ag+ + Pb <=> 2Ag + Pb^2+}$$
If $\ce{H2S}$ gas is passed through the solution, what will be the effect on the EMF of the cell?

My attempt:
The Nernst equation for the EMF of a cell results in
$$\mathcal{E}=\mathcal{E}^\circ-\frac{RT}{nF}\ln{\frac{[\ce{Pb^2+}_{(aq)}]}{[\ce{Ag+_{(aq)}}]^2}}$$
Now, I assumed that $\ce{Ag2S}$ was more covalent and hence less soluble in water than $\ce{PbS}$, by Fajan's rule, due to smaller size of cation and pseudo-inert gas configuration  and hence it would precipitate out more. hence the ratio
$${\frac{[\ce{Pb^2+}_{(aq)}]}{[\ce{Ag+_{(aq)}}]^2}}$$
would decrease and hence the EMF of the cell would decrease.
Is my reasoning correct? I do not have the answer with me.

It's correct. In fact $K_\mathrm{sp}(\ce{Ag2S})=10^{-50}$ and $K_\mathrm{sp}(\ce{PbS})=10^{-28}$, so the concentration of silver ion is far smaller than lead ion. Finally, the EMF of the cell would decrease.

With concentrations that low, the cell will be dead for all practical purposes. The potential of a dead cell is pretty much a thing-in-itself; you won't be able to measure it anyway, let alone make any use of it.

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text:
Is axis of symmetry considered a true symmetry?
In most of the books when i read about optical activity of a compound this is what they say:
A compound that does not posses any kind of symmetry is optically active. In other words chiral compounds without having any kind of of symmetry are optically active.
However, this holds true when we refer to plane of symmetry or centre of symmetry, because molecules having axis of symmetry or alternate axis of symmetry are still optically active.
For example this compound possesses axis of symmetry but is still optically active:
And therefore i am totally confused about relation of optical activity with symmetry.
Thanks for answering the question

It has to be non trivial. A rotation of $0$ or $2\pi$ about an axis being trivial.

No, this is not what they say. If you skip a few words from a meaningful sentence, it is quite easy to arrive at nonsense. It is not just "any kind of symmetry"; it is "any kind of inversion symmetry", or "any kind of symmetry with determinant -1". Rotational symmetry does not apply, though it is surely true and real (and non-trivial, for that matter).

A molecule is chiral (1, 2) if it does not have an improper axis of rotation. Since your drawing does not seem to show the meso form of 2,3-dichlorobutane, it is optically active.

This is to complement the answer given already and to address @Ivan Neretin's comment under OP's question. It is an example that generally (with lots of asterisks) rotations cannot be excluded. This is a copy from J. March's  6ed (p. 158).

From the text:

Such compounds possess ... an alternating axis of symmetry as in 1

Of course Ivan is correct that you only care for "determinant $-1$" motions or else orientation reversing and in this case we use a rotation by $\pi$ which has determinant 1 (or else belongs to $SO_3$). My group theory is not as good since I am an organic chemist but I think the actual symmetry element is something like $\rho_{\pi/2} \circ r = 1$ whereas the rotation element above is $\rho_\pi=(\rho_{\pi/2} \circ r)\circ (\rho'_{\pi/2} \circ r')=1 \circ 1=1$ and it does have determinant 1 but is not one of the symmetry group elements (in the sense that $x^{34}$ is in $C_4=\{1,x,x^2,x^3 \}$ but only as $x^{34}=x^2$ ).
To make things more complicated March is defining an "alternating axis of symmetry" as:

An alternating axis of symmetry 17 of order n is an axis such that
  when an object containing such an axis is rotated by 360/n about the
  axis and then reflection is effected across a plane at right angles to
  the axis, a new object is obtained that is indistinguishable from the
  original one

So it appears that in the example of (1) we talk of $Z_4$ symmetry (if we call $Z$ the alternating axis subgroup) and  $Z_4 \sim C_2$
Finally, I think that (1) is an extreme example and almost always for organic compounds chirality (or not) is determined by the lack (or not) of a plane of symmetry

isn't symmetry of axis a disqualifying property for a molecule being an isomer?


text:
Is the phenyl cation or ethynylium more stable?
I recently gave an exam and I received an interesting question to which I believe the answer is wrong.
Given two compounds, a benzene molecule with carbocation at one of its constituting carbon atoms and an ethyne carbocation. Now according to me even though ethyne carbocation is pretty unstable due to the sp hybridized carbon, but if a charge is placed on a carbon in the benzene ring, too much strain is caused on the ring which destabilizes it more. I believe that the stability of the ring is a more dominating factor when carbocations are involved,  but the answer was in favour of the ethyne carbocation.

Could someone please clear this doubt if whether any other factors were taken into consideration or my basic premise is wrong?

Ring strain is not relevant in this case - the benzene ring already has trigonal planar carbons.

@bon I believe ring strain (if you think this concept is appropriate at all) is a big issue here because the s-contents of the carbon will probably be redistributed into the sigma system hence widening the angle and introducing a deformation. A more general question, however, is, how do you judge stability in the first place?

@Martin-マーチン You're probably right. I hadn't considered that.

@bon I might have time tomorrow to look into it. I think this is something worthwhile to discuss ~ at least a nice thought experiment.

related http://pubs.acs.org/doi/pdf/10.1021/ja970808s

Preamble
As trivial as this question might seem, it is not. It should certainly never be asked in at an exam level lower than basic quantum chemistry.
Any of the tempting explanation schemes will certainly fail. A bonding picture in the Lewis formalism is almost impossible; hybridisation descriptions in these compounds is extremely complex.
As a general tip: The smaller the molecule, the more complicated is the bonding.
What is stability?
This question seems so simple, yet is incredibly complex. As a quantum chemist, I see any molecule as stable if it is an energy well (stationary point) on the potential energy surface (and this might already be a simplification). This concept does not help in any way when you want to compare stabilities, because it is just one property of a particular arrangement of nuclei in a particular state.
There is no such thing as absolute stability. We cannot measure it. It is an abstract concept, that we like to think that we understand it. An example from the real world: ice is stable below 0 °C and vapour is stable above 100 °C. Yet ice is not more stable than vapour.
Comparing stability, we always need a framework of multiple variables or constants and a reference point.
It is therefore often assumed, that the boundary conditions are the same (exactly). Even on a purely theoretical level this it is extremely difficult to find these exact boundary conditions. Sooner or later you have to introduce approximations, which eventually crash your system, especially regarding the reference point.
It is difficult, but not impossible.
In order to compare the relative stability of those two cations we can assume a hypothetical hydride transfer between them, hence building a closed reference frame. In computational chemistry we use the concept of isodesmic reactions for such purposes. It is obviously not free of any bias and a complete hypothetical, but it can give us some more insight.
In this case we can propose the following hypothetical isodesmic reaction:

Note that the number of bonds of the same type are the same on each side are equal. Close enough at least. One could argue, that we break a hydrogen sp²-carbon bond and form a hydrogen sp-carbon bond, and vice versa, therefore not having the same types of bonds on the left and right side. But, this is probably as close as we can get.
Having set up a concise set of parameters, we can now go ahead and perform a few quantum chemical calculations. I decided to use my personal standard: DF-BP86(D3BJ)/def2-SVP level of theory.
And we run into more problems
The singlet, closed shell molecules benzene (despite aromaticity) and acetylene are no problem. But when we come to the cations, things get interesting.
The phenyl-cation does not propose a big threat either, it is big enough to compensate the empty orbital in the σ- and π-system.
The high symmetry of the acetylene molecule make it impossible to treat on this level of theory. The linear form is - by my definition - not stable. At least not at this level of theory, at least not in the singlet state. What we do need is a multi-reference method like CASSCF at least. And then we still don't know which state is preferred and if we can still use our proposed hypothetical reaction. Because if the spin state is different on either side, the whole thing falls apart.
Results
Nevertheless while we're already at it, let us look at some of the results and clear up some misconceptions.
Benzene and the Phenyl-cation
Upon ionisation the system needs to compensate the change with a change in geometry. This is a purely electronic effect. In benzene all carbons are trigonal coordinated, quite nicely having three sp² hybrid orbitals; it is one of the very few examples, where symmetry dictates that. Removing a hydrid ion also means removing electrons from such an orbital, freeing up more stable s-character. As a result the bond angle widens to accommodate more s-character in the carbon-carbon bonds. Now this leads to a visibly distorted geometry - a delicate equilibrium of multiple factors.

The π-system remains largely undisturbed. The σ-bonds compensate almost completely for the change. The cationic carbon is now better described as being between linear and trigonal, hence having hybridorbitals between sp and sp².
Acetylene and the Ethynyl carbocation
Acetylene is a wonderful molecule by itself. It has extremely high symmetry, which is quite rare for four atoms. With that comes a large number of degenerate orbitals. And a highly complicated bonding situation.
You might want to regard the carbons as having sp-hybrid orbitals, and that is fine as an approximation, but the deeper you dig, the creepier it gets. (Also see Bent's rule.)
Disturbing this molecule in any way, results in breaking the symmetry. The linear version of HCC+ still has high symmetry and now we have a problem. Upon removal of the hydrid ion, the remaining electron density will try to use as much s-character as possible. As a result part of the s-character is tried to be redistributed into the π-system. These orbitals now do not match the spacial symmetry anymore and the molecule is not stable in its linear form in the singlet state. This is not the whole truth though, but it gets much more complicated and this will certainly be too much for this answer.
A stable structure on this level of theory in the singlet state is an almost three-membered ring. It is probably best to think of it as a protonated dicarbon. All bonds are elongated compared with acetylene.

Since we have a significant different bonding situation now, the previous assumptions of the same type of bonds left and right break down and we can no longer compare these two molecules in the given reference framework.
(Without respect for actual science, we could still do it. Then we would know that the left side of the equation is about 600 kJ/mol more stable than the right.)
Conclusion
The phenyl-cation might be stable, it is a local minimum at DF-BP86/def2-SVP, while the acetylene-cation does not exist. The HCC-cation is better thought of as a protonated dicarbon molecule.
If I understood you correctly, then your assumption, that the "strain" on the ring is bigger than the instability of the orbitals in the acetylene cation, is indeed wrong.

So, for the structure on the right to be the correct answer, it should be a side-on $\left(\ce{\pi_{C-C} -> H+}\right)$ bond with the hydrogen, rather than end-on? Or am I misinterpreting what you're saying?

@SendersReagent No you interpret that correctly. At least that is the easiest interpretation of the data from that calculation. The end-on might be stable as a triplet, but I did not check that.

Yeah, this question definitely doesn't belong on any lower-level chemistry exam.

In relation to the phenyl cation, see also this answer discussing the structure of the isoelectronic neutral compound borinine ($\ce{C5H5B}$).

I think this answer could be given in terms of hybridization as in benzene carbocation C is sp (%s character↑ then Electronegativity↑) and +ve charge is less stable on it and of course angle strain is also cause instability whereas in acetylene carbocation C+ has complex hybridization (I have not study yet) and HCC+ ion (hypothetically exist) then its sp mixing also involve π bonds leads to %s character↓ EN↓ +ve charge somewhat more stable.

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.


text:
If the polarity of the solvent increases, will the Rf value increase or decrease?

True or False: A sample of aspirin has an $R_f$ value of 0.20 when measured on a silica gel TLC plate with a certain solvent mixture. If the polarity of the solvent is increased, the $R_f$ value of the aspirin will become larger.

Would this be false? Because aspirin is polar (I believe would be considered polar because it dissolves in water despite some other sources saying that it is actually non-polar) and the silica in the TLC plates are also polar, aspirin would not be able to travel very far because it would interact with the silica. Thus, shorter distance over longer distance = lower $R_f$ value, with $R_f$ = distance spot moves/ distance solvent moves.

I wish we would stop teaching chromatography in terms of "polar" and "nonpolar."
The aspirin will interact fairly strongly with the silica due to hydrogen bonding/electrostatic interactions of the carboxylic acid and the ester with the silica. If you increased the "polar" component of the mobile phase, it would travel further due to the mobile phase competing more for binding spots of the silica gel.
So increasing the polarity of the mobile phase does raise $\mathrm{R_f}$ of aspirin (or anything) on regular silica gel.

Since aspirin has a carboxylic acid group on it, it would be considered polar. Silica gel, consisting of $\ce{SiO2}$, is also polar. Since polar molecules attract other polar molecules, the aspirin molecules will tend to bind to the silica and not move up the TLC plate in a nonpolar eluent, resulting in a low $R_f$ value.
When the polarity of the eluent is increased, its ability to attract polar molecules is increased. This means that it can better compete with the silica in attracting the aspirin, and the $R_f$ value will be higher.

You beat me by 7 seconds.

@SendersReagent :O


text:
Lone pair and dipole moment of ozone
(This is a question similar to this and this, but none of the answers to those two questions answer my question.)
I've seen many places explain the dipole moment of ozone from the formal charges of the O atoms. However, I don't quite feel comfortable with this explanation since I feel like formal charges are not actual charges; they're more like a bookkeeping convention.
From this question, it seems like we should take into account the lone pair on the central O when summing the dipole moments. Thus, shouldn't the central O be the negative end of the dipole (since it has the lone pair)? But why does this contradict the prediction made by formal charges?
In general, to find the overall polarity of a molecule, when is it appropriate to sum up all the dipole moments (including those from the lone pairs), and when should we use formal charge instead?

The formal charges are a part of the resonance depiction of ozone, which is a representation of the electronic structure so they are a completely acceptable way of rationalizing the dipole moment.

Ozone is not a symmetric molecule: though bent, it's not bent enough to be equilateral, there is a "middle one". The central O is in rather a crappy position for its self-respect as an electronegative atom: it is having electrons pulled from it in both directions. It has a bit of a +ve charge, at least when compared to its neighbours at either end. You can think of it in terms of resonance (easy, but not really physical), or by what they tell you (the orbitals are depleted near the middle). Either way: not many electrons there. As that middle O also sticks out, there will be a dipole moment.

Other oxygens have their lone pairs as well.


text:
Determining order of reaction when concentration of reactant is not given

Suppose reactant $\ce{A}$ decomposes in an aqueous solution to give $\ce{P}, \ce{Q}$:
  $$
\ce{A(aq) -> P(aq) + Q(aq)}
$$
Suppose further that $[\ce{A}]_0=0.42\text{M}$, and that $[\ce{Q}]$ was measured as follows (in M):
\begin{align}
t&=0 &[\ce{Q}]&=0 \\
t&=100 \quad &[\ce{Q}]&=0.101\\
t&=200 \quad &[\ce{Q}]&=0.165\\
t&=300 \quad &[\ce{Q}]&=0.206\\
t&=400 \quad &[\ce{Q}]&=0.239\\
t&=500 \quad &[\ce{Q}]&=0.263
\end{align}
Using the graph of $\ln [\ce{A}]$ and $\frac 1 {[\ce{A}]}$ vs $t$, determine if the reaction is of order $1$ or $2$.

I know how to solve these kind of problems when I'm given the concentration of $[\ce{A}]$, I just plot the graph and check which one resembles a line more, and the slope gives me the velocity constant $k$.
But here $[\ce{A}]$ is not given. I supposed it would be $[\ce{A}](t)=[\ce{A}]_0-[\ce{Q}](t)$, but I was told that was wrong. Could someone help me out?

I'd go with $A=[A]_0-Q$ anyway and see where it takes me.

It had better be $\left[A\right]!\left(t\right) = \left[A\right]_0 - \left[Q\right]!\left(t\right)$, otherwise there's a major mass balance problem in the system!

@Brian Could you elaborate? I don't understand where the mass balance problem would be.

@YoTengoUnLCD See my response in chat

@Brian Thank you for your answer,  I was able to read it just now, as for some reason my phone didn't let me get into chat. My question was more towards as why the concentration of $P$ doesn't affect $[A]$ (my understanding of this reaction is that $A$ dissolves into $P+Q$, thus $ 1 M$ of $A$ should correspond with $1 M$ of $Q$, but rather with $1M$ of $P+Q$...). Sorry, I'm really confused.

@YoTengoUnLCD Complete reaction of $1,\mathrm M~A$ would yield $1,\mathrm M~P + 1,\mathrm M~Q$. In this reaction, one could think of $A$ as being "$PQ$": $\ce{PQ->P + Q}$.

If the stoichiochemistry is as you present it:
$$\ce{A -> P + Q}$$
Then if at $t=0$, $[\ce{A}]=[\ce{A}]_0$ and $[\ce{Q}]=0$, then because one equivalent of $\ce{Q}$ is formed for each equivalent of $\ce{A}$ consumed, your method of determining $[\ce{A}]_t$ must be correct.
$$[\ce{A}]_t = [\ce{A}]_0 - [\ce{Q}]_t$$
Perhaps the error your instructor noted came in applying your relationship or interpreting your graph.


text:
Does a freezing point of a substance always increase as its boiling point increases?
I am studying for an introductory to physical sciences test and came upon a chart describing substances found in petroleum and the distillation process.
Each freezing point increases as each boiling point increases.  Each of these properties also increases with the rise in density.
Is this just true for substances in petroleum (methane, ethane, propane, butane, pentane...) or is this consistent throughout all substances?
Update:
(Pure substances)

Nope. See: methanol vs ethanol.

I don't think this is true even for all substances in petroleum. Wait, doesn't propane have lower melting point than ethane?

Since there are zillions of chemicals you should in general be skeptical when "always" is used for some trend. There always seem to be exceptions.

Yes and no.
Yes in the sense that boiling point and freezing point usually correlate. If there is a line of homologous compounds, than this properties rise in general, as both are increased with increase in number of dispersion interactions involved. However, it is likely to find several compounds with odd behavior, especially around beginning of the line.
As a counterexample. Toluene has higher boiling point than benzene, but much lower freezing point. This is because toluene has less symmetrical form and consequently has harder time packing into strictly ordered lattice.
Another counterexample. Gallium melts around 30 celsium, but boils well above 2000 celsium. However, Tallium melts around 300 celsium, but boils around 1700.
However, for similar compounds the values usually correlate in the series. Say, for alkanes both values steadily grow with increase of number of carbons... give or take occasional peaks due to details in crystal packing.
However, if you consider compounds of different families, it is not predictable without looking into structures of the compounds.

An increase in the boiling point does not always follow an increase in the freezing point - these two properties, while seemingly correlated by the fact that they are governed by intermolecular forces, are still independent of another.
Take the case of Carbon Tetrachloride, $\ce{CCl4}$ vs Ethanol, Phenol vs Aniline
$\ce{CCl4}$ boils at 76.8 degrees and freezes at –22.8 degrees Celsius.
Ethanol boils at 78.4 degrees but freezes at –114.6 degrees Celsius.
Phenol boils at 181.75 degrees and freezes at 43 degrees Celsius
Aniline boils at 184.3 degrees but freezes at - 6.0 degrees Celsius
There's really no hard-and-fast rule, but you can make an informed guess that freezing point is affected more significantly by intermolecular forces than boiling is, because in liquid state the molecules are fairly loosely bonded whereas in a solid they would be stacked and packed to optimize the intermolecular forces.

Among single substances there is water.  Increase the pressure above 4.56 torr and freezing point goes down until you get to ice (III) (>200 atmospheres), boiling point goes up.

This is true for all solutions and mixtures. Its what is called Colligative properties.
Freezing/melting point and Boiling of a homogeneously (evenly) mixed solution depends on the number of dissolved particles in the solution. How much these values change also depends specifically on the mixture.
If you think about it makes sense. Freezing point decreases because its harder for the similar species in the mixture to form nice regular crystal without other junk getting in the way. Similar boiling point increase is the result of solutes lowering the the vapor pressure of a solvent.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch15/colligative.php

I think the question is about pure compounds.

This question asks "Is this just true for substances in petroleum (methane, ethane, propane, butane, pentane...) or is this consistent throughout all substances?" Ie. Pure or not

The question is a bit confusing as to whether it is for pure substances (the methane, ethane, ... listed), or for their solutions. After re-reading it I fall firmly in the pure substance interpretation.


text:
Is this home-made pesticide from cigarette butts and battery acid safe?
An old friend once told me that a good way to kill bugs in my garden was to soak lots of cigarette butts in water to make a gallon of the brown liquid. Then add one ounce of battery acid and mix well. He said to spray it on the garden and wait two to three weeks before harvesting because it would take that long for the mixture to become Niacin. I thought it sounded cool but never tried it.
Now 40 years later I would like to, but I want to know if this is really true or would it ruin my vegetables? It's been 22 years since my chemistry classes and tried to make sense of this link.
It seems like the nicotine/acid thing would become neutral, but again, I have forgotten so much that I'd feel safer running this past a few more knowledgeable folks first.

I, uhh... "wouldn't" would be a good way to say it.

I'm voting to close this question as off-topic because what you're describing is fraught with potential harm to you or others.

@Todd This should probably not be closed. I think this question poses a bigger threat if it is not properly answered. Consider this meta posting, too. Apart from this it is about chemistry and it can be answered objectively. I personally see no reason to close it.

@Martin-マーチン - I've reconsidered based on your comment and rereading the question. It looks to me like the OP is asking for advice concerning what amounts to synthesis of niacin from nicotine and the effect on vegetables. Voting to reopen.

This seems like a bad idea to me.
The use battery acid (sulfuric acid) seems really terrible. In the past there was an awful problem with coal power plants releasing sulfur oxides into the air which formed acid rain (sulfuric acid). You're proposing making a brew that is orders of magnitude worse than acid rain to put on your garden plants. I think such a strong solution of acid would get rid of the bugs by killing your plants. ;-)
If you goggle nicotine "bug spray" you can find a number of websites that discussing seeping tobacco in water, then filtering the brew to make a spray. In such a solution the nicotine is so dilute that it would be safe to handle. However concentrated solutions of nicotine like used for various e-cigarettes could be dangerous to handle. Lookup nicotine MSDS in google (here is one such MSDS sheet).

(1)So no battery acid is needed.
(2) The various websites state that the sprayed nicotine solution decomposes fairly quickly - a couple of days, not weeks.

Frankly I don't smoke or chew tobacco. Since I'd have to buy the tobacco for such a use, I'd rather just buy a commercial pesticide and follow its use instructions. It is absolutely crazy to me that so many folks so berserk when they see the word "chemical." If you showed a strawberry to most folks they would just identify it as such. But it is nothing but a sack of hundreds of chemicals. So to me all strawberries are "organic strawberries." To someone who is allergic to strawberries one can be deadly. Did you know that there are varieties of lima beans which have a high enough concentration of cyanide that eating them can kill? (dangerous foods) So spraying a nicotine solution may seem "natural" but it is still a chemical.

The principle is simple:  The residual nicotine in the cigarette butts cause disruption to the nervous system of insects, just as caffeine would. The sulfuric acid is added to increase solubility as a salt.
My answer to efficacy is two parts:
1.) While this may kill insects if the nicotine concentration is high enough, a few butts in a gallon may not be enough.  Nicotine is potent and the reason that tobacco produces nicotine in the first place is to act as a pesticide,  recommend experimenting with number of cigarettes to find an effective number.
2.) An ounce of battery acid may be a bit much and give you a solution that is too acidic for plants. I would recommend using maybe a table spoon at most and buffering the solution with sodium bicarbonate (baking soda).
As a disclaimer I make no warranty of safety.

I am downvoting this, because safety is the prime concern of the question. It should be thoroughly addressed as such.


text:
Why are the two CH2 groups in allene perpendicular to each other?
Allene, $\ce{H2C=C=CH2}$, has two $\ce{=CH2}$ groups perpendicular to each other. This is the same as hydrogen peroxide's $\ce{H}$ atoms. Is there any reason why they adopt the perpendicular arrangement, when the VSEPR model allows both the perpendicular and planar arrangements for the groups on the ends? I don't think it's a steric effect, because the hydrogen atoms/methylene groups are facing opposite directions and far apart.

No, VSEPR theory does not allow for a coplanar arrangement. Let us consider allene and its orbitals:

Source. Green indicates $\mathrm{p}$ orbitals; blue $\mathrm{sp^2}$; and red $\mathrm{s}$.
Hybridization
The terminal carbons are $\mathrm{sp^2}$ hybridized, and form three $\sigma$-bonds each. This means that each terminal carbon has one unhybridized $\mathrm{p}$-orbital.
The central carbon atom is $\mathrm{sp}$ hybridized, and forms two $\sigma$-bonds. This means it has two unhybridized $\mathrm{p}$-orbitals. For our sake, we will call these two orbitals $\mathrm{p_x}$ and $\mathrm{p_y}$. These orbitals are perpendicular to one another.
$\mathrm{\pi}$-Bonds
A $\pi$-bond is the result of the sharing of two $\mathrm{p}$-electrons, and it is particularly important for us to consider them as belonging to a particular region of space. In order for the central carbon's $\mathrm{p_x}$-electron to form a $\pi$-bond with one of the terminal carbons, it must be with another $\mathrm{p_x}$-electron. If the two did not share the same spacial orientation no bond could form. Since the $\mathrm{x}$-dimension is occupied by the electrons of a $\pi$-bond between the central carbon and one of the terminal carbons, the only way for the central carbon to form an additional $\pi$-bond is by sharing its $\mathrm{p_y}$-electron with the other terminal carbon. Hence, the two terminal carbons are necessarily perpendicular to one another.

ChemTube3D has an animation showing allene: http://chemtube3d.com/orbitalsallene


text:
Method and Tips for Inorganic Qualitative Analysis
In a few weeks time, I will be doing an inorganic qualitative analysis test which is very pressed for time (a majority of people don't finish). The form of the test is most likely going to be that there will be around 9 unknown solutions (each contains 1 cation and 1 anion) which I have to identify. I will be given a list of all the possible cations and anions (there will be around 20 cations and  20 anions) and also we will have pH paper and a heat gun.
I wondering if anyone could suggest any tips on how to go about determining the composition of each unknown solution.
Currently, my method is to do it systematically by making a table and reacting all the solutions together. However, this can very time-consuming and probably not the most efficient way to identify the solutions.
Also, is there any distinctive reactions or properties that we will help me identify certain ions. For example, antimony and sulfide will give a unique peachy pink precipitate, or adding water to bismuth may cause a precipitate (as acid is usually required to make bismuth go into solution so by adding water, the acid is diluted and the bismuth precipitates out.)

This is so open ended it is impossible to give any guidance. There are numerous books on inorganic qual. Generally the scheme is to identify multiple anions and cations in the same solution. // Can you prep any test solutions beforehand? Can you take in notes, or reference books? What sort of equipment will you have access to? (dozens of test tubes, centrifuge, ect..) How much solution will you get for each unknown? How long will the test last?

If you can prep test solutions and take in notes, I'd be develop a scheme using a couple of drops of test solution into a plate with multiple wells. So to well 1 to you add 1.0 M HCl, well 2 1.0 M NaOH, well 3 1.0 M Na2CO3, and so on. // The immediate problem that I see is that unless you start out with several gross of test tubes you could spend the majority of your time washing glassware.

@MaxW we get around 30 test tubes so spending time washing glassware isn't really such a big problem. Like I said above, the only other equipment we get is a heat gun and pH paper. There is enough solution and if it runs out it will be refilled. We aren't allowed to prepare any test solutions and bring any notes into the exam.

This is a rough draft of the complete answer. A lot more will be added to the answer very soon. At present, it includes the common tests for sulfite and bisulfite ions. Please do not downvote just because it is not a complete answer yet.
Identification Of Sulfite Anion: ($\ce{SO3^{2-}}$)
All sulfites are insoluble except "group IA sulfites " and$\ce{(NH4)2SO4}$. Sulfite ppt. is always white in color.
Test With Acid:
$\ce{SO3^{2-} + H^{+} -> SO2 (g)}  $
Sulfur dioxide gas is evolved which gives lime water test as well as baryta water test:
$\ce{SO2 + Ca(OH)2 -> Ba(HSO3)2}[\text{soluble}]$
$\ce{SO2 + Ba(OH)2 -> Ca(HSO3)2}[\text{soluble}]$
Test with $\ce{BaCl2/CaCl2}$:
$\ce{SO3^{2-} + CaCl2 -> CaSO3 + 2Cl^{-}}$
$\ce{SO3^{2-} + BaCl2 -> BaSO3 + 2Cl^{-}}$
Note that $\ce{BaSO3}$ is slowly oxidized by atmospheric oxygen to form heavy white ppt of $\ce{BaSO4}$. Similarly:
$\ce{BaSO3 + Br2 -> BaSO4 + Br-}$
$\ce{BaSO3 + H2O2 -> BaSO4 + H2O}$
$\ce{BaSO3 + HNO3 -> BaSO4 + NO}$
Test With $\ce{Pb(NO3)2}$:
$\ce{Pb(NO3)2 + SO3^{2-} -> PbSO3 [\text{white ppt.}]}$
On boiling $\ce{PbSO3}$ in presence of $\ce{O2}$, $\ce{PbSO4}$ is produced.
(On boiling $\ce{PbS2O3}$ we get $\ce{PbS}[\text{black ppt}]$ and $\ce{H2SO4}$. Hence, this fact can be used to distinguish sulfites from thiosulfates).
Test With $\ce{AgNO3}$:

Schiff's Test/Fuschin's Test/Magenta Reagent Test:
Both $\ce{SO3^{2-}}$ and $\ce{SO2}$ are decolorized in this test.
Distinguishing b/w $\ce{SO3^{2-}}$ and $\ce{CO3^{2-}}$ ion:

Some Important Redox Reactions:
$\ce{MnO4- + SO2/H2S -> Mn^{2+} + SO4^{2-}/S}$
$\ce{Cr2O7- + SO2/H2S -> Cr^{3+} + SO4^{2-}/S}$
$\ce{Br2 + H2O + SO2/H2S -> Br- + SO4^{2-}/S}$
$\ce{Fe^{3+} + SO2/H2S -> Fe^{2+} + SO4^{2-}/S}$
$\ce{IO3- + SO2/H2S -> I2 [\text{gives deep blue color in starch}] + SO4^{2-}/S}$
$\ce{SO2}$ can either be oxidized to $\ce{SO4^{2-}}$ or reduced to $\ce{S^{2-}}$. But, $\ce{H2S}$ can only be oxidized to $\ce{S}$.
Notable Reaction (Comproportionation) : $\ce{\ce{SO4^{2-}} + \ce{H2S} -> S + H2O}$
Points to remember:

$\ce{CO2/CO/CO3^{2-}/HCO3^{-}}$ do not show any change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$
$\ce{S^{2-},SO3^{2-}/HSO3^{-}}$ do show change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$.
$\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O}$ $ [\text{no visible change is observed}]$

Dvarda Alloy ($\ce{Zn}/\ce{Al}/\ce{Cu}$):
$\ce{Zn + NaOH -> Na2ZnO2 + 2H}$
In presence of dvarda alloy the following reactions can take place:
$\ce{SO3^{2-} -> S^{2-}}$
$\ce{SO2 -> H2S}$
$\ce{NO2^{-} -> NH3}$
$\ce{NO^{3-} -> NH3}$
$\ce{Zn}$ reacts with dil. $\ce{H2SO4}$ to produce $\ce{H2}$ gas which can also reduce $\ce{SO4^{2-}}$ to $\ce{S^{2-}}$.

Test with Sodium Nitroprusside:
$\ce{Na2[Fe(CN)5NO] + ZnSO4 -> Zn[Fe(CN)5NO] [\text{salmon pink ppt }]}$
$\ce{Zn[Fe(CN)5NO] + \text{moist} SO2 -> \text{red color compound}}$
$\ce{S2O3^{2-}}$ and $\ce{S^{2-}}$ ions also give the test. So they can be removed on treating with $\ce{HgCl2}$ as follows:
$\ce{Hg^{2+} + S^{2-} -> HgS}$
$\ce{Hg^{2+} + S2O3^{2-} -> HgS2O3}$
$\ce{HgS2O3 [\text{on heating}]-> HgS [\text{black ppt}] + H2SO4}$

Identification Of Bisulfite Anion ($\ce{HSO3^{-}}$):
Test With Acid:
$\ce{HSO3^{-} + H^{+} -> SO2 (g)}$
Test With $\ce{H2O2}$:
$\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O} $
$\ce{HSO3^{-} + H2O2 -> H^{+} + SO4^{2-} + H2O} $ (acidic)

Identification Of Carbonate Anion ($\ce{CO3^{2-}}$)
All carbonates are insoluble except group $\mathrm{IA/IIA}$ carbonates ($\ce{Li2CO3}$ is sparingly soluble) & $\ce{(NH4)2CO3}$.
Test With Acid:
$\ce{CO3^{2-} + H^{+} -> CO2}$ (Carbon dioxide can be identified using lime water or baryta water test)
$\ce{CO2 + Ca(OH)2 -> CaCO3}$ (white turbidity)
$\ce{CO2 + Ba(OH)2 -> BaCO3}$ (white turbidity)
When $\ce{CO2(g)}$ is in excess, $\ce{CaCO3/BaCO3 + CO2 + H2O -> Ca(HCO3)2/Ba(HCO3)2}$. This bicarbonate is water soluble.
Test Based On ppt:
$\ce{CO3^{2-} + CaCl2/BaCl2 -> BaCO3/CaCO3 + 2Cl^{-}}$
The formed carbonates are white precipitates which are soluble in $\text{(dil)}\ce{HCl}$, $\text{(dil)}\ce{HNO3}$, $\text{(dil)}\ce{CH3COOH}$ and soda water.
Remember that $\ce{HCl}$, $\ce{HNO3}$, $\ce{H2SO4}$ and $\ce{CH3COOH}$ is stronger than $\ce{H2CO3}$. So they can decompose the carbonate salt. For example $\ce{BaCO3 + HNO3 -> Ba(NO3)2 + CO2(g) + H2O}$.
Test With $\ce{Pb(NO3)2}$:
$\ce{CO3^{2-} + Pb(NO3)_2 -> PbCO3 + 2NO3^{-}}$
$\ce{PbCO3}$ is soluble in $\text{(dil)}\ce{HCl}$, $\text{(dil)}\ce{CH3COOH}$ and excess $\ce{NaOH}$.
$\ce{PbCO3 + 2NaOH -> Pb(OH)2 + Na2CO3}$
$\ce{Pb(OH)2 + 2NaOH -> Na2[Pb(OH)4]}$
Test With AgNO3:
$\ce{CO3^{2-} + AgNO3 -> Ag2CO3} \text{[white ppt.]}$
Soluble in $\text{(dil)}\ce{HCl}$ and $\text{(dil)}\ce{NH3}$.
$\ce{Ag2CO3 + NH3 <-> [Ag(NH3)2]^{+} + CO3^{2-}}$
$\ce{[Ag(NH3)2]^{+} -> Ag3N (furminating silver) -> Ag (black) + N2}$
The ppt. becomes yellow or brown upon addition of excess reagent owing to formation of $\ce{Ag2O}$.
Note: $\ce{Ag2CO3}$ on heating at above $1000^{0} C$ produces $\ce{Ag}$, $\ce{O2}$ and $\ce{CO2}$. Below that temperature if heated it produces $\ce{Ag2O}$, $\ce{CO2}$.
Test With $\ce{Hg(NO3)2}$:
$\ce{CO3^{2-} + Hg2(NO3)2 -> Hg2CO3 [yellow ppt.] -> Hg [black] + HgO [yellow] + CO2} $ (on heating)
$\ce{Hg^{2+}_{2} -> Hg + Hg^{2+}}$
Test With $\ce{HgCl2}$:
$\ce{CO3^{2-} + 4Hg^{2+} + 3H2O -> 3HgO.HgCO3 [reddish brown ppt] + 6H^{+}}$
Note: Phenopthalein turns pink in carbonate solutions and colorless in bicarbonate solutions .


text:
How to identify a redox reaction?

Which balanced equation represents a redox reaction?
  \begin{align}
&\mathrm{A.} &\ce{AgNO3 + NaCl &-> AgCl + NaNO3} \\
&\mathrm{B.} &\ce{BaCl2 + K2CO3 &-> BaCO3 + 2KCl} \\
&\mathrm{C.} &\ce{CuO + CO &-> Cu + CO2} \\
&\mathrm{D.} &\ce{HCl + KOH &-> KCl + H2O}
\end{align}

I don't understand how to identify a redox reaction. The answer key shows the correct answer as D, but I'm confused as to why.
D shows an acid-base reaction. Are all acid-base reactions redox reactions and why aren't the other answer choices correct?

The correct answer should be C. As you can clearly see the copper ion is reduced, while carbon in carbon monoxide is oxidized into carbon dioxide. Remember redox equations involve electron exchange.

You are confused because the answer key is wrong.
The way to figure it out is to assign oxidation numbers to each species following the rules. Oxidation states are simply a convenient way to keep track of electrons without having to draw out Lewis structures, count electrons pairs and so on. The rules help make that happen.
Notice the position of the sign relative to when you indicate charge. It is not the same.
The correct answer is C though: $$\ce{CuO + CO -> Cu + CO2}$$
with oxidation states
$$\ce{\overset{(+2)}{Cu}\overset{(-2)}{O}
    + \overset{(+2)}{C}\overset{(-2)}{O}
   -> \overset{(0)}{Cu}
    + \overset{(+4)}{C}\overset{(2\times(-2))}{O2}}$$
Notice Cu went from +2 to 0. It gained 2 electrons to do that, thus it was reduced. What gave up the electrons? The other carbon, because it went from +2 to +4: its oxidation state increased by 2, thus it lost 2 electrons.
Recall OIL RIG: Oxidation is loss of electrons and reduction is gain of electrons.
Whatever is oxidized increases its oxidation state, and whatever is reduced decreases it.

Thanks, I understand now. I believe our teacher told us redox reactions are usually not double replacement and wanted us to eliminate the rest of the answers as a quicker solution than listing oxidation states for each choice.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

@Martin-マーチン awesome, thanks seems straight forward will work in getting the hang

@Evan - Most acid-base reactions can be rewritten as double displacements as well, so you should be able to rule them out:$$\ce{HCl + KOH -> KCl + HOH}$$


text:
Why does sodium sulfate initially harden instead of dissolving?
I was recently drying a product by passing it through a bed of sodium sulfate in a fritted glass Buchner funnel. Once I had finished up working my reaction, in an ill-advised attempt to clean up, I decided to wash the salt out with water before proceeding with acetone. Upon adding water, the salt starting foaming, and became hard as a rock and adhered to the fritted glass. I decided to leave it sitting in the water, and after a while it did finally turn gel-like and was able to be washed out.
My question is, why in the presence of so much water, why did the salt remain as the hydrated salt instead of dissolving, and why did it take so long to dissolve after it formed the large crystal? Also any idea what the bubbling was?

Related: Why does sodium sulfate have an unusual solubility-temperature curve?

Sodium sulfate forms various hydrates including a decahydrate. The dissolution is exothermic which saturates the solution. When the warm solution reaches the relatively cold frit the supersaturated solutions precipitates. In the glass frit the pores would clog and it would take time to dissolve the sodium sulfate out.

Isn't the dissolution endothermic? It certainly has seemed to be, in my experience.


text:
How to balance a redox equation in molecular form?
I've spent a while trying to balance this equation in molecular form with no success.
$$\ce{Cu + HNO3 -> Cu(NO3)2 + 2NO + H2O}$$
I need to find the coefficient of HNO3, which is 8. However, this was my attempt:
$$\ce{Cu^0 + H + (NO3)- -> Cu^2 + 2(NO3)- + 2NO-2 + H2O }$$
$$\ce{Cu^0 -> Cu^2 + 2e- }$$
$$\ce{2x (NO3- + 2H^1 + e--> 2NO^0 + H2O)}$$
$$\ce{Cu^0 + 4H+ +2NO3- -> Cu^2+ + 4NO^0 + 2H2O + NO3^-}$$
Image of process: https://i.stack.imgur.com/QrlST.jpg
NO3 is the spectator polyatomic ion I left out during my balancing. I have no idea whether I did it right and where to go from here.
Thanks!

Judging from the picture you attached you have successfully identified the species that gets oxidised
$$\ce{Cu -> Cu^2+ + 2e-}\tag1$$
and reduced, i.e. $\ce{NO3-}$. While balancing this reaction you have made a few mistakes. You cannot leave it out as a spectator ion, since some of it is involved in the reaction.
Consider the oxidation state in nitrate and nitrogen monoxide. How many electrons have to be transferred?

 three, nitrogen goes from $+5$ to $+2$ (you have only written one)

How many water molecules will be produced? How many protons are needed?

 Two water, four protons

Write down the full half reactions

 $$\ce{\overset{+5}{N}O3- + 4H+ + 3e- -> \overset{+2}{N}O + 2H2O}\tag2$$

Now you need to find the factors to multiply $(1)$ and $(2)$ with to balance out the electrons.

 $3\times(1)$ and $2\times(2)$

This results in the following equation

 $$\require{cancel}\ce{Cu + 2NO3- + 8H+ + \cancel{6e- } -> 3Cu^2+ + \cancel{6e- } + 2NO + 4H2O}$$

How many more nitrate ions do you need to charge balance the protons? What is the final balanced equation?

 Six more nitrate ions, giving a total of eight
 $$\ce{Cu + 8HNO3 -> \underbrace{3Cu^2+ + 6NO3- }_{3Cu(NO3)2} + 2NO + 4H2O}$$

Thanks so much! I finally understood the mistakes I made and how to solve these problems.

I'd add that in the half-reactions the charge and the atoms have to balance. The utility is that you can use electrons  which really don't just float around in aqueous solution.


text:
Standard Enthalpy of Formation of H2(g) vs H(g)
My question is somewhat related to this question. When I look at a table of standard values there are two entries for Hydrogen:

$\ce {H2(g)} $  - 0 kJ/mol
$\ce {H(g)} $  -  218 kJ/mol

I understand the $\ce{H2(g)} $ is 0 because it is used as a reference, why is monatomic $\ce{ H(g)}$ given in the table and what does it mean? When would one use it in a calculation of enthalpy change?
From what I understand Hydrogen only exists as a monatomic gas at very high temperatures, the Standard enthalpies of formation are given at 1 bar, 298.15 K.

The value for the hydrogen atom is one-half the bond dissociation energy of the hydrogen molecule.

The value may be used whenever atomic hydrogen is present as a (probably intermediate) species. E.g. take the reaction
$$\ce{2H <=> H2}$$
which is the basic reaction in atomic hydrogen welding.

This is correct - often the relevant species is a lone proton. For example, in the formation of solid metal hydrides you need H to enter, diffuse, and react with the metal. H$_{2}$ just won't do it.

@JonCuster does that mean that we use the $\ce{H(g)}$ in Hess's law when we are looking at, for instance, the combustion of hydrogen with oxygen, since at some point monatomic hydrogen is an intermediate specie in that reaction?

Well, you get to choose your standard state, but you have to be consistent. The end result is the same, how you get there is slightly different.


text:
How to calculate saturation point of a solution?
Is there a way to calculate the theoretical saturation point between two materials?
For example, let's say I want to determine the solubility of vitamin A in water (or vitamin B in olive oil or vitamin D in milk.) In these examples, if I know the molecular structure of the vitamins and the various liquids, can I figure out at what point saturation will be reached?
To be clear: If there's a way, I want to learn how to calculate these theoretical saturation values. I don't want to just get them from a table somewhere.

I am pretty sure there isn't a simple way to determine the saturation point of a solution just by knowing its structure. You need to know more information. If you know the Ksp (solubility equilibrium constant) then calculating the saturation point is pretty simply. However unfortunately you need to get the Ksp from a table somewhere.

Yes. But it would probably be faster to simply find some empirical data.
You would need to:

Approximate the hydrophobicity of your vitamins (for rather big molecules like vitamins, this would need some mathematical modeling, but there are some computer programs that can do this rather accurately.

This data could be used to approximate the solubility in water and probably also olive oil. But milk is quite complex, as there will be a solubility equilibrium in/between both the fat phase and the water phase. And all the salts and other compounds in the milk (and to a small degree in olive oil) will affect the solubility.


text:
Why are the ethyl ester methylene protons not diastereotopic in ethyl (+-)-2-methylbutyrate?
In the following compound, there is one chiral center and two methylene groups.  Why are the protons on the ethyl ester methylene not diastereotopic?

You are correct to conclude that they are diastereotopic. There appears to be a bit of splitting evident if you look closely at the NMR, but it might be expected to be very small given the distance between the ethyl group methylene protons and the chiral center. But even if the distance were greater and the resulting splitting too small to be observed, the protons would still be technically diastereotopic.


text:
How does turbidity affect dissolved oxygen?
I have to do a EEI for school about turbiduty of water and finding the most cost effective method for purifying a sample of water. My group was planning to do a standard test where we use materials such as filter paper, cloth, sand, gravel, etc. and run the turbid water through it. As our school has no turbidometers we wanted a way to have some quantitative data. To do this we were planning on conducting a dissolved oxygen test (wrinklers method) as we found some sources linking turbidity to oxygen levels. Would this be a valid test? Would the filtered water need to sit overnight, for example, before we tested the DO?
And help is appreciated!!!

I don't see why oxygen levels would be related to turbidity in any way, except maybe in some very special setting. It seems more natural to find the way of measuring turbidity. Say, aim a ray of a laser pointer through the sample, take a photo from sideways, compare the intensity in Photoshop.

According to this document, turbidity and oxygen content can be correlated because the more microbes in the water, the more turbid the water is, and the less oxygen is left (because they breathe it).
But a correlation is no cause-effect relationship. Both are linked in the particular case of microbes because they affect both by their presence.
Not a reliable method at all. More oxygen in the water does not mean it is potable. If you let sit a bucket of water contaminated with microbes, the oxygen levels will drop over time, but if you stir the water, you re-dissolve some up to saturation levels, but you still have the microbes present.
You may want to evaluate the oxygen demand, or the biomass (absorption at 600nm), centrigugate and evaluate visually, etc.


text:
basics of NMR (spin-spin splitting)
here's what I wrote from lecture,

what i'm confused about is, if two magnetic moment(B0 and the one of red H) have the same direction. wouldn't the magnetic field by red H cancel out the external magnetic field? I tried to visualized it in picture below.

2.1  if the directions of the two magnetic moment are different, is it because aromatic compound can flip upside down?
2.2 If the spin state is always same whether or not compound flip upside down, is it because of E from NMR machine?

Why do protons interact only with themselves? there are also O,C and Cl

hope someone would answer me~~

The proton magnetic moment is much smaller than the external magnetic field 2. The actual direction of the spin does not matter, it is only the direction of the spin relative to that of the other proton that matters, and no the spin state is not always the same, otherwise there is no way to record NMR spectra (the lines are transitions from one spin to another) 3. 16O and 12C nuclei have no nuclear spin. Coupling to Cl occurs but we don't see it in the spectrum https://www.chem.wisc.edu/areas/reich/nmr/07-multi-02-effects.htm

Honestly I would recommend going back again. Try using a textbook perhaps, I can recommend Spectroscopy by Pavia et al. I feel like you have a number of misconceptions. The splitting of spins is not a result of the external magnetic field - it is independent of the external magnetic field. It's wrong to say that the splitting arises because the proton spin is aligned with $B_0$ so if that's what the lecture said, stop watching it. NMR is not an easy topic so don't make it harder for yourself by not understanding the fundamentals.

what orthocresol says, and your example shows second order coupling, which is hard to understand before you've got the basics.

I can't make any sense of question 2.  I think the answer to both is the physical orientation of the molecules in solution are averaged out due to molecular tumbling and other random motion.

and about Q2, I just tried to tell if the physical orientation of molecules has to do with spin state of proton.

The physical orientation of the molecules is not likely a factor.

Regarding Question 1:
NMR measures the energy difference between the field-aligned energy of a nuclei and the field opposed energy of the nuclei.
The biggest field in the situation is the external magnet.
The next biggest field is the magnetic moments of the electrons.  Electron density leads to what we refer to as chemical shift.
Lastly, there are the magnetic moments of other nuclei.  But, the other nuclei are also undergoing their own spin flipping and such like.  So if a hydrogen has its molecular neighbor with a spin aligned with the external magnetic field while relaxing, the total field it experiences is stronger.  Otherwise it is weaker.  The difference between stronger and weaker is essentially the distance between the two peaks on the NMR spectrum.  It is important to remember the external magnetic field is the strongest here, so when the nuclear spins oppose and weaken (or align and strengthen) the magnetic field, it is only a tiny amount.
Regarding question 3:
Carbon and Oxygen have complete Nuclear quantum shells, and as a result have no net magnetic moment. The concept of 'spin up' or 'spin down' means nothing to these nuclei.
Since splitting is the result of a magnetic interaction between spins, there will be no splitting.
Regarding chlorine, The coupling constants are small, and generally not observed due to limitations in the machines making the measurements.

about the sentence you mentioned, "So if a hydrogen has its molecular neighbor with a spin aligned with the external magnetic field while relaxing, the total field it experiences is stronger. ".  ------

I don't get it. and I tried to explain why I have difficulty understanding it in a second picture. is it too messed up? T_T;

Your picture is fine.  The two vectors B0 (up) and Red (up) add together making a larger "up".  It's just vector math.

but the direction of magnetic field that came out of Red(up) becomes down at the position of blue H ;;

NO!! Coupling is a through bond phenomenon NOT through space. It arises via the Fermi contact interaction, NOT by the adjacent proton reinforcing/opposing the external magnetic field. Get that idea out of your head now!


text:
Classification of oxides,transition elements
I understand  what  basic,acidic and amphoteric oxides are.
But how do we classify oxides on the basis of metallic nature?Explain how we can deduce if the oxide is metallic or not.
For example,what does it mean when it is stated that "CrO2 is a metallic oxide."
This was an answer to one of the questions in an exam that I wrote recently.

There is no relation.

Well, there is a relation of some sort, but it is pretty vague and ambiguous. Oxides of non-metals are usually acidic, though some of them (say, $\ce{NO}$) are neither basic nor acidic nor amphoteric. Oxides of metals in lower oxidation states are basic or amphoteric. Oxides of metals in higher oxidation states are sometimes downright acidic.

@IvanNeretin Oh, oops, I had a different idea in mind... I thought he was referring to CrO2 being a metal in the sense that it displays metallic conductivity... You probably interpreted it correctly.

'Metallic oxide' has nothing to do with basicity/acidity of the oxide and is  about solid phisics.
Basically it means, that the solid is electrically conductive like a true metal. This involves a half-filled conductivity zone (that can be roughly imagined as a number of half-filled overlapping orbitals of same energy), so electrons can move freely through the crystal lattice.
There is no simple rule to guess metallicity of multi-element compounds, unfortunately; because this property is strongly tied to electronic structure of the solid and crystall lattice. As an example: of $\ce{TiO2}$, $\ce{VO2}$, $\ce{CrO2}$ and $\ce{MnO2}$ only $\ce{CrO2}$ is truly metallic, while $\ce{VO2}$ looses metallic conductivity at lower temperatures (it has two allotropes, one metallic and one insulating). For TiO2 it is, at the very least, expected, as it is a $d^0$ compound, but for other metallic oxides it is unclear, or, at least, aboive average chemist's level.
=================
As for acidic/basic classification of oxides, there are few solid rules. Generally, any oxide with oxidation state 5+ is a strong suspect for acidic nature, with 3-4 being anyware from acidic to basic and +1 usually basic except for halogens.
For an oxide to be basic, corresponding metallic cation must be reasonably stable. This implies relatively large radius, and moderate to low charge density. Consequently, +1 metal oxides are practically exclusively basic, +2 metal oxides are usually basic except for very small cations (Be and few late d-elements like copper and zinc) and +3 are usually amphoteric except for very large cations (like La) Anything of higher oxidation state is usually acidic or at most aphoteric, as corresponding cation has too high charge density to be stable in water solutions. The only true salts of +4 cation I can come with are $\ce{Zr(SO4)2}$ and $\ce{Hf(SO4)2}$ and even for them the corresponding hydroxide is classified as amphoteric.
Still, in some cases such oxides still may act as bases, but corresponding cation has lower charge. Probably, the simpliest examples are $\ce{VO2}$ and $\ce{UO3}$, with corresponding salts containing $\ce{VO^{2+}}$ and $\ce{UO2^{2+}}$ respectively. This, however, is not usually condidered as an excuse to classify these oxides as basic/amphoteric.
Some oxides are not acidic/basic/amphoteric as there is no 'simple' salt corresponding ot the oxide in question. This claim is usually debatable, as such a salt usually may be found (for example, NO, typically classified as not producing salt, may be tied to salts like $\ce{Na2N2O3}$ aka  Angeli's salt), but for simplicity are not considered. There are reasons for it, as basic/acidic oxide classification is very rough, and should not be relied upon in extreme, borderline and simply too complicated cases.

But I found a wikipedia article about half metals which mentions CrO2 as an example for half metal.I also found oyher aticles stating that CrO2 is a half metal https://en.m.wikipedia.org/wiki/Half-metal

@RamAditya Half-metal is more-or-less a subspecies of metals, even if a weird one.


text:
Metallic and magnetic nature of transition metal oxides

Which of the following oxides is both metallic and ferromagnetic:
  $\ce{VO2}$, $\ce{CrO2}$, $\ce{MnO2}$, $\ce{TiO2}$?

This is how I went about solving it:
Since ferromagnetic materials also show paramagnetism, so the condition for paramagnetism must be checked.
In $\ce{TiO2}$, $\ce{Ti}$ has no unpaired electrons, it cannot be paramagnetic and as a consequence it cannot be ferromagnetic.
$\ce{CrO2}$ is amphoteric. A metallic oxide would show basic properties in most cases. So $\ce{CrO2}$ may not be the right answer.
$\ce{VO2}$ is basic. All three except $\ce{TiO2}$ are paramagnetic.
Then I googled, I didn't get an exact answer. But I read that $\ce{CrO2}$ is termed as a "half metal" which is related to its magnetic properties, most part of which I didn't understand.
Then I also found articles about half metallic ferromagnetism of $\ce{TiO2}$. But that goes against my assumption.
So my questions are:

Explain where I'm wrong.
Why $\ce{CrO2}$ is the answer? (Found that when I finally referred to the solutions of the question paper. However the solutions given
  can have mistakes. So it is possible that $\ce{CrO2}$ may not be the
  true answer).
If you think $\ce{CrO2}$ is not the answer, then why?

Basic or amphoteric nature of a compound is not related to ferromagnetism whatsoever.

What the heck is a ‘metallic oxide’, under the assumption that not all of those oxides are metallic because the counterion is metallic?

@Jan ok, i will take care next time.

@Jan 'metallic oxide' here probably means 'oxide with metallic conductivity'

@permeakra Most metal oxides have band gaps, so I'm not sure if they can be truly metallic

@QuantumMOCHACCINO indeed, most do. Some, however, do not, and these few are said to have metallic conductivity. $\ce{CrO2}$ does not have bandgap (OK, it is a bit more complicated, but good enough).

This question should be cross-posted to Physics.SE, it is not a simple chemistry question. I'll give short rationalization of the fact, but beware: it is a rationalisation, not a deduction. It is also given in layman's terms.
We know the crystal structure of $\ce{TiO2}$, it is stereotypical and quite common. It contains octahedrons $\ce{TiO6}$ with oxygens shared between three octahedrons. Since the four compounds have similar composition and the cations have similar size, we may expect them to have similar structures. This is, however, a mere guess, but lets' go with it.
For a compound to have metallic conductivity, it must have mobile electrons, i.e. partially occupied group of orbitals. Thus, $\ce{TiO2}$ is obviously out.
It is known, that $d$-orbitals of a metal in octahedral neighbourhood split: three are lower in energy then two other.
In $\ce{MnO2}$ there are three electrons, so all three lower orbitals are occupied. This makes electrons immobile: for an electron to move to a neighboring atom, it has to pair with one of the already present electrons, which requires energy. I do not expect, however, that $\ce{MnO2}$ shall be a great insulator: the pairing energy is quite low, and furthermore, any impurity shall create charge carriers. So, $\ce{MnO2}$ is probably a semiconductor.
To choose from remaining two, we should recall that ferromagnetism is common among compounds with a lot of half-filled d-orbitals. This suggests that $\ce{CrO2}$ has higher chances, than $\ce{VO2}$ Still, the latter should be paramagnetic and conducting under right conditions.
Again, you should cross-post the question on Physics.SE and probably read a good solid state physiscs book.

$\ce{TiO2}$ is diamagnetic. $\ce{VO2}$ has only 1 unpaired electron, definitely not ferromagnetic. But I am still confused between $\ce{MnO2}$ and $\ce{CrO2}$. Both are ferromagnetic. Metallic character is decided by the strength of the Metal-Oxygen bond. In case the metal is in a very high O.S. (e.g. $\ce{MnO4-}$, $\ce{Cr2O7^2-}$) the M-O bond is more covalent and cannot be ruptured by water to give a base.
In this case however, both $\ce{MnO2}$ and $\ce{CrO2}$ seem appropriate.


text:
Why do XeO and XeF8 not exist?
Since Neil Bartlett's 1962 discovery that xenon was capable of forming chemical compounds, a large number of xenon compounds have been discovered and described. Almost all known xenon compounds contain the electronegative atoms fluorine or oxygen. The oxidation state of xenon in its compounds is generally +2, +4, +6, or +8.
Xenon is known to form three fluorides: $\ce{XeF_{$n$}} \;(n = 2, 4, 6)$ where the oxidation states of xenon are +2, +4, and +6. $\ce{XeF8}$ is not known to exist even though the oxidation state would be +8. Why is this so? Could it possibly be due to the fact that 8 fluorine atoms can't fit around a xenon atom?
Xenon is known to form three oxides: $\ce{XeO_{$n$}} \;(n = 2, 3, 4)$ where the oxidation states of xenon is +4, +6, and +8. $\ce{XeO2}$ was not known until 2011. Why did it take so long to be discovered? Moreover, the lowest oxide of xenon $\ce{XeO}$ is not known though the oxidation state of xenon would be +2. Why?

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

I think this has something to do with XeO4 having sp3 hybridization while XeF6 has sp3d2 hybridization. I've often heard the generalization that despite fluorine's higher electronegativity, oxygen can generally bring out higher oxidation states in its compounds because of its ability to form double bonds. I don't know why you couldn't just keep adding more fluorines though

@Mithoron The link does not describe why it took so long time to discovered while the other two oxides were well known.

This pdf says that:  "Our findings
imply that xenon monoxide, XeO, has no region of stability up
to a pressure of at least P = 200 GPa." And then later state that: "we do not find XeO to be stable at any pressure"

$\ce{XeF8}$ is not known to exist though O.N is +8. Why is this so?

At least 2 compounds have been reported that contain the $\ce{XeF8^{2-}}$ unit. See, for example:

$\ce{(NO^+)2[XeF8]^{2-}}$ (reference)
Metal salts of the form $\ce{(M^{+})_2[XeF8]^{2-}}$ where M is a metal salt such as $\ce{Cs, Rb}$ (see the above reference) or $\ce{Na}$ (see p. 62 in Advances in Inorganic Chemistry, Volume 46, A. G. Sykes editor; link

The $\ce{XeF8}$ portion of the molecules approximates a square-antiprism geometry.

(image source)

the lowest oxide of xenon $\ce{XeO}$ is not known though O.N is +2. Why?

Perhaps the molecule is unstable, favoring disproportionation to other xenon oxides plus oxygen. For example, Andreas Hermann and Peter Schwerdtfeger suggest the following pathway at high pressure:
$\ce{3XeO → Xe3O2 + 1/2O2}$
The authors go on to note "we do not find $\ce{XeO}$ to be stable at any pressure" (link to full paper).
Edit: See orthocresol's comment below. He argues that at ambient pressure the following decomposition pathway might be more likely:
$\ce{XeO -> Xe + 1/2 O2}$

The paper writes that XeO is "metastable toward the decomposition reaction", which implies a thermodynamic instability. 2) It's worth a mention that the calculations were ran at a whopping pressure of 75 GPa (~740,000 atm), so under more normal conditions the instability of XeO cannot be attributed to this pathway. If I have read the paper correctly, Xe3O2 is unstable wrt the elements under 75 GPa, so I would hazard a guess that the likely decomposition pathway of XeO at 1 atm is XeO -> Xe + 1/2 O2.

Thanks orthocresol, I'll edit my answer to include your thinking.

One more thing:  The higher oxides: XeO3 and XeO4 were discovered earlier but the oxide XeO2 was discovered recently! What took it so long to discover?

Wikipedia says it is prepared at 0 C and quickly decomposes to $\ce{XeO3}$ with a half-life around 2 minutes. Sounds like it may have been difficult to tell that something other than $\ce{XeO3}$ was being produced.

Half-life? That kind of thing only happens with radioactive compounds. Isn't it?

@NilayGhosh No, any chemical that decomposes or transforms can be said to have a half-life, not just radioactive compounds.

According to Heats of Formation of XeF3+, XeF3-, XeF5+, XeF7+, XeF7-, and XeF8 from High Level Electronic Structure Calculations Inorganic Chemistry 2010, vol. 49, pages 261–270:

Unlike
  the previously studied XeF2, XeF4, and XeF6,
  $\ce{XeF8}$ is
  predicted to be thermodynamically unstable with respect to
  loss of $\ce{F2}$, and the reaction is calculated to be exothermic by
  22.3 kcal/mol at 0K.

Actually, $\ce{XeO}$ is known, but as an eximer.
The first observation was the work published in 1946 as New Band System in the Green Excited in a Mixture of Xenon and Oxygen and the Energy of Dissociation of CO Phys. Rev. vol. 69, pages 36– 37.
A more-recent article about XeO is Optical and Electron Spin Resonance Studies of Xenon–Nitrogen–Helium Condensates Containing Nitrogen and Oxygen Atoms J. Phys. Chem. A, 2015, 119, pp 2438–2448

Emissions from excimer XeO* molecules have been observed and extensively studied in both gaseous(1-4) and condensed(5-11) phases. Up to now, experiments with XeO* in condensed phases were limited to solid and liquid rare gas (RG) matrices. Beams of electrons, α-particles, or protons, as well as ultraviolet (UV) irradiation (with wavelengths of 260 nm and shorter) were used to form excited xenon–oxygen complexes in neon, argon, or krypton matrices doped with xenon atoms and O2, N2O, or CO2 molecules as precursors of oxygen atoms. It is worth noting that excitation of a xenon matrix doped with O atoms causes an emission with a much simpler spectrum. It consists of two intense bands with maxima at 370 and 740 nm,(8, 12) on the edges of the visible range.


text:
1H NMR of Acetylferrocene

(source: sigmaaldrich.com)
I'm trying to analyse the proton NMR of this, and there's one singlet for the H on the non-substituted Cp ring and a singlet for the methyl group, as expected.

However, while I was expecting to see a dd signal from $\ce{H_2}$ and $\ce{H_3}$ and a doublet from $\ce{H_1}$ and $\ce{H_2}$, there is only one doublet that accounts for all the protons according to integration.
Can someone help me explain this?

Can you give the details of the spectrum, i.e. frequency, solvent, as this might be important.

Can you post a picture of it?

Roof effect, but to an extreme? I was just talking to my labmate about the fact that $\alpha$-nitrotoluene shows all aromatics as a singlet. The closer the signals of two coupling protons get to each other, the more the "lean" toward each other until they collapse into one signal that doesn't appear to show coupling.

The NMR spectrum of acetylferrocene is very well known and widely published as an undergraduate teaching resource. It would be useful for reference if you were to provide a copy or image of your spectrum. At least you should describe it including chemical shifts, integration and peak appearances. I suspect that your troubles here come from a misunderstanding of your observations of the actual spectrum.
In order to help you with your analysis, you should make sure you complete an accurate integration of all peaks. This will give you some clue as to how many protons are contributing to each of the peaks you are observing. That is, each of your peaks that you think are part of your doublet contribute two protons. Also, when you suggest you are observing a doublet - what is the splitting of this doublet. I doubt it is of a magnitude that conforms to a typical H-H coupling. It is probably closer to 100Hz. So perhaps what you are seeing is in fact two distinct singlets. Or at least two distinct resonances with poorly defined splitting.
Your analysis of the expected splitting pattern for the Me-Cp ring isn't quite correct. From your drawing, H1 and H4 are chemically equivalent, but magnetically non-equivalent. This is because coupling between H1 and H2 is not the same as the coupling between H4 and H2. The same can be said for H2 and H3; H2 and H3 are chemically equivalent, but magnetically non-equivalent.  This means you have an AA'XX' spin system, which is not a simple first order spectrum in appearance. You can read more about these types of spin systems on one of my favourite web resources, Reich. Each signal actually appears as an apparent triplet.
So, why aren't you seeing any splitting in these peaks? There will be a number of reasons why this could be the case, most of which are perfectly reasonable and are no cause for concern. Sample concentration, purity, magnetic field strength, field homogeneity (shimming), temperature and solvent will all contribute to the appearance of this spectrum, and many published spectra for this molecule have splitting which is barely resolved. Without seeing your spectrum, it is difficult to make further interpretation. However, as a suggestion, you should measure the linewidths of your peaks (the width of the peak at half height). You will probably find that the Cp peaks from Me-Cp are considerably broader than the other peaks, especially than for your residual CHCl3.


text:
Elevated boiling points in alkanes with even number of carbons
Why do alkanes with even number of carbon atoms have greater boiling points than those with odd number of carbon atoms?

Because there is always a heavier alkane with equal number of carbon atoms than one with an odd number of carbon atoms.

There is not really any difference between the boiling points of even and odd carbon alkanes.  For single-chain alkanes the boiling point just goes up as a smooth curve versus chain length.
The melting points do show a small oscillating component with the even chains being higher.  If we imagine the chains in their idealized zig-zag configuration, which would be relatively favorable for the solid state, we find that the even-length chains have a center of inversion which makes them fit better than odd-length ones.  This effect fades, however, with increasing chain length.
Reference:
https://en.wikipedia.org/wiki/Alkane

I think you asked your question poorly. Here is data from Wikipedia. Most of the BP's were given as ranges. I rounded some and some I took the average. But in all cases as the number of carbons increases, the BP increases.
 alkane        #Carbons   B.P. (°C)
 methane          1       −161.49
 ethane           2        −88.5
 propane          3        −42.2
 n-butane         4          0
 n-pentane        5         36
 n-hexane         6         69
 n-heptane        7         98
 n-octane         8        125
 nonane           9        151
 decane          10        174
 undecane        11        195
 dodecane        12        216

I also fit the data to a 3rd order polynomial.
BP=-2.262586E+002 + 7.517034E+001*C - 5.350278E+000*C^2 + 1.810153E-001*C^3
    alkane    #Carbon  BP(data)   BP(fit)  error
    methane     1        -161.5   -156.3    5.2
    ethane      2         -88.5    -95.9   -7.4
    propane     3         -42.2    -44.0   -1.8
    n-butane    4           0.0      0.4    0.4
    n-pentane   5          36.0     38.5    2.5
    n-hexane    6          69.0     71.3    2.3
    n-heptane   7          98.0     99.9    1.9
    n-octane    8         125.0    125.4    0.4
    nonane      9         151.0    148.9   -2.1
    decane     10         174.0    171.4   -2.6
    undecane   11         195.0    194.2   -0.8
    dodecane   12         216.0    218.1    2.1

Here is a plot of the residuals. As Oscar Lanzi indicated there is no evidence of an even-odd trend.


text:
How to correctly write the condensed formula for salicylic acid
I am struggling to find the right way to write the condensed formula for salicylic acid
Some people say CHCHCHCHC(OH)C(COOH)
Or (CH)4C(OH)C(COOH)
Others say C6H4(OH)COOH (which doesn't seem to make sense since it looks to me like it could represent many isomers and not just one molecule like a condensed formula is supposed to)
Which (if any) is correct?

A pretty straightforward way would be $\ce{(o-C6H4)(OH)(COOH)}$ and variations. (o-C6H4) suggests  can't be interpreted in any other way then orto-disubstituted benzene ring.

Some people say $\ce{CHCHCHCHC(OH)C(COOH)}$ Or $\ce{(CH)4C(OH)C(COOH)}$

Both of these are definitely wrong since it seems to indicate some sort of alkene.

Others say $\ce{C6H4(OH)COOH}$

This is much better since it "hints" that $\ce{C6H4}$ is a benzene ring. With only 4 hydrogens and what seems to be two substitution groups that fits. Note there is no indication of the relative position of the groups. So this would represent either 3-Hydroxybenzoic acid or 4-Hydroxybenzoic acid as well.
The basic empirical formula would be $\ce{C7H6O3}$ which would represent probably dozens of molecules.
So the best way to denote the particular molecule specifically is to name it, or to draw out the whole 2D structure.

Maybe OP is trying to reinvent SMILES?

simplified molecular-input line-entry system (SMILES)  https://en.wikipedia.org/wiki/Simplified_molecular-input_line-entry_system

Thank you very much for your answer. I agree that drawing it out or naming it would be much better, however the assignment requires the condensed formula. So I guess I will use C6H4(OH)COOH then.


text:
Specific Heat affected by Pressure
I know the specific heat depends (slightly) on the conditions present. For example, methane's specific heat is 2.5 at 400 K, but 4.5 at higher temperatures.
How might specific heat depend on pressure, though. -- if we are under super low pressure (say 6 mbar rather than earth's typical 1014 mbar), how would specific heat change. I am expecting it to differ along with $PV = nRT$, but not sure.
Also, $\Delta H = mc\Delta T$ -- would I need to plug $\Delta PV/nR$ in for $\Delta T$?

Note that ‘specific heat’ is ambiguous. Apparently, you are referring to the ‘specific heat capacity at constant pressure’ (symbol: $c_p$). Furthermore, it is not permissible to omit the unit. The given specific heat capacity of methane at constant pressure and a temperature of $T=400\ \mathrm K$ is $c_p=2.5\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$.

At low pressures like 6 mb and even 1014 mg, the heat capacity is very insensitive to pressure (and the ideal gas value can be used).  At higher pressures, where non-ideal gas effects set in, the enthalpy and the heat capacity are, in general, both functions of pressure as well as temperature.  The partial derivative of enthalpy with respect to pressure at constant temperature for an arbitrary substance is given by:
$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
This is equal to zero for an ideal gas.


text:
In Statistical Transition State Theory, why is the concentration of the transition state for the forward and reverse reaction identical?
For example on Wikipedia, and also in my lecture notes, it is assumed that for a reaction
$$\ce{A + B<=> AB^{\ddagger} <=> P}$$
at equilibrium,
$$[\ce{AB^{\ddagger}_f}] = [\ce{AB^{\ddagger}_r}],$$
but I don't see how the equation follows from the assumption, since there is no direct relation between transition state concentration and total reaction rate. And if it is, how can they simply be equal instead of depending on some constants?

In transition state theory, it is assumed that the forward and the reverse reaction occur via the same transition state. This is implied by assuming a quasi-equilibrium between the reactants and the products, which adheres to a common Boltzman statistics. Therefore the labelling $\ce{AB^\ddagger_r}$, $\ce{AB^\ddagger_f}$ is artificial, because it is the same transition state:
$$\ce{AB^\ddagger_r} = \ce{AB^\ddagger} = \ce{AB^\ddagger_f},$$
and therefore the concentration is identical, too:
$$[\ce{AB^\ddagger_r}] = [\ce{AB^\ddagger}] = [\ce{AB^\ddagger_f}].$$


text:
Why is NaHCO3 a basic salt?
When NaHCO3 is added to water should give an H$^+$ ion in water and this should make it acidic. Why is it a basic salt? Does it have something to do with H2CO3 being a weak acid?

The pKa for $\ce{NaHCO3}$ is readily available using Google

The thing is that $\ce{HCO3-}$ can act as an acid and a base. These substances are  called amphiprotic (other examples include $\ce{HSO4-}$ and $\ce{H2PO4-}$). So it can donate a proton to water to form $\ce{H3O+}$ but it can also accept a proton from water to form $\ce{OH-}$. So really, the  question is which process occurs to a greater extent?.
You can't answer that question by simply looking at the molecule. Instead we need to know the $\mathrm{K_a}$ values for $\ce{H2CO3-}$ and $\ce{HCO3-}$. By comparing these values and doing a few calculations we can determine whether $\ce{HCO3-}$ is basic or acidic. It turns out $\ce{HCO3-}$ is actually basic.
However I am assume that equilibrium constants are probably above the level that you are learning at. So this is just one of these things which you should just know. For example you should know that despite $\ce{HCO3-}$ being basic, $\ce{HSO4-}$ and $\ce{H2PO4-}$ are actually acidic.


text:
Steric hindrance of an unbonded electron pair
This question is half inspired by this question, and half inspired by the structure of molecules like hydrazine and hydrogen peroxide.

When I was looking at the aforementioned molecules, I started trying to reason as to why the lowest energy conformation of these molecules are what they are.
$\hspace{34ex}$
In hydrazine, the lowest energy conformation is not the one in which the unbonded electron pairs are anti-periplanar. This makes me think that hydrogen have a greater steric hindrance than unbonded electron pairs do.
$\hspace{34ex}$
In hydrogen peroxide, however, the lowest energy conformation is not the one in which the hydrogen are anti-periplanar. This contradicts what I reasoned form the structure of hydrazine.
Being confused by this, I tried a different approach—what about the cyclohexyl carbanion? Surely using $\Delta G = -RT \ln{K_\mathrm{eq}}$ the relative steric hindrance of an unbonded electron pair and a hydrogen atom could be determined. What I became unsure about, however, was how rapid inversion might affect this, and additionally whether or not there even is even any data regarding the major and minor confirmations of the cyclohexyl carbanion (because I am unsure if it even exists in such a way that this can even be measured).
So am I flawed in my reasonings? Does the equilibrium data for the cyclohexyl carbanion exist?

The examples you mentioned are not guarded by sterical hindrance - which is negligible in the case  - but by small and consequently rarely significant stereoelectronic effects. It may be seen as interaction of electron pair and OH(NH)-bond, strengthening slightly O-O bond and weakening slightly OH-bond.   Here is some more background https://en.wikipedia.org/wiki/Stereoelectronic_effect

Since I just covered this material in depth in my physical organic chemistry class, I figured I'd write up an answer.

Electronegative elements lower the energies of all molecular orbitals to which they contribute, of which their low-lying $\sigma ^{*}$ orbitals are of particular importance. Electronegative atom lone pair MOs, which have little bonding character and are also (relatively) high-lying in energy, are very close in energy and will donate their electrons to the low lying empty orbitals, producing a stabilizing interaction. Molecular conformations are in part shaped by these interactions, which is known as the donor-acceptor effect.
Hydrogen peroxide is a classic example of this effect. In addition to sterics predicting that the hydrogens be anti-periplanar, this conformation also minimizes the molecule's net dipole (another stabilizing effect), yet its most stable conformation adopts anti-clinal geometry. This can essentially be thought of as a compromise between between sterics and dipole minimization, and the donor-acceptor effect:
$\hspace{4.3cm}$
The $\sigma ^{*} (\ce{O-H})$ is an excellent acceptor, and oxygen's highest energy unbonded pair is one of the strongest donating. The lower energy lone pair occupies a $\sigma\text{-out}$ orbital, and the higher energy lone pair occupies a pure $\mathrm{p}$ orbital. The mixing of this orbital with the $\sigma ^{*} (\ce{O-H})$ orbital favors a $90º$ dihedral angle, but due to steric and dipole effects, the actual angle is $\approx 120º$.
$\hspace{5.3cm}$
In hydrazine, the $- 2.5 \ \mathrm{kcal \ mol^{-1}}^{[1]}$ preference for the two $\mathrm{n}(\ce{N})$ orbitals being syn-clinal over anti-periplanar is also explained by the donor-acceptor effect. When the $\mathrm{n}(\ce{N})$ are syn-clinal, they are each eclipsed by the $\sigma ^{*} (\ce{N-H})$ orbitals, and orbital mixing occurs. The effects of this are further seen in the $\ce{N-N}$ bond length difference between the two conformations: $1.448\ \mathrm{Å}$ in the syn structure and $1.489\ \mathrm{Å}$ in the anti structure. The shortening of the bond length in the syn structure is a direct result of the multiple bond character introduced by the $\mathrm{n}(\ce{N}) \rightarrow \sigma ^{*} (\ce{N-H})$ interactions$^{[1]}$.

$^{[1]}$ Wilcox, C.; Bauer, S. Journal of Molecular Structure: THEOCHEM 2003, 625 (1-3), 1–8.

And this effect is known as the gauche-effect. I’m sorry, I hadn’t seen this question. If I had, I would’ve answered earlier.


text:
Why is meta-nitrophenol less acidic than para-nitrophenol?
My professor said while explaining the acidic nature of phenol that meta-nitrophenol is less acidic than para-nitrophenol, and gave the reason that resonance doesn't play on meta, but I didn't get it.

Hi Rishabh, welcome to Chemistry Stack Exchange! Have you tried drawing the resonance structures of the meta-nitrophenol and para-nitrophenol ions? Hint: you should be able to draw 2 for the para one but none for the meta one.

The acidity of nitrophenols (or any acid for that matter) is determined by the stability of the conjugate base. In the case of m-nitrophenol and p-nitrophenol, the relative stability can be determined by looking at the resonance structures.

You can see that p-nitrophenol has an additional resonance structure where the negative charge is delocalised onto the oxygens of the nitro group, which are highly electronegative and therefore stabilise the charge effectively. This stabilisation is not possible in the case of m-nitrophenol because there is no opportunity via resonance for delocalisation onto the nitro group.

Refer to the structures the negative charge never gets over META position and hense resonance does not affect the  electron density at META.
and hense nitro on META will not be able to pull electron from ring via (-M) and hense there would be negligible efffect on polarity of O-H bond.

The oxygen in phenol definitely doesn't carry a partial positive charge. It is still very electronegative and bears a negative charge.


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Phosphates Functional Group
In my text book it states that a phosphate group is considered to be $\ce{-OPO3^{2-}}$. However, online it states that phosphate is $\ce{PO4^{3-}}$. Why do these contradict? Am i confusing phosphate the chemical with the phosphate group? Please see image for reference.

Phosphate $\ce{PO4^3-}$ is an anion of the phosphoric acid $\ce{H3PO4}$. Term phosphate comes from a salt, for instance a sodium phosphate $\ce{Na3PO4}$. $\ce{Na+}$ are bound with a $\ce{PO4^3-}$ through ionic bonds. No electron sharing.
To annotate specific moieties of a molecule skeleton, a chemical group term is used. The phosphate has donated one pair of electrons on the oxygen to have a covalent bond with other molecule, for instance sugar in the image, and becomes a phosphate group in the molecule. Since the electron pair does not sit on the oxygen any more, but is shared with carbon, the phosphate charge is lowered to −2.

Edl, please see below as I had to reply in the form of an answer to post an image in regards to your comment.

AMP is charged (-2) in physiological conditions (aqueous solution with pH~7). Both -OH groups are easily deprotonated (lost H+). The picture you have found shows a neutral molecule (which is right as well), and can be isolated in this state.

so the picture below (AMP with the OH- groups) are AMP outside of an aqueous solution where as the picture you originally gave in your answer is AMP deprotonated in water which is basically the phosphate anion?

Yes, that is right!

@ EDL, thank you. Im still confused on how the image above had a 2- charge. Im assuming the whole molecule is a formal charge of 0, correct? But if the 2 O atoms indicated in your drawing each have a 1- charge I would think the molecule would have a net charge of 2-? Also, why online does your image contradict the AMP I found. Please see below for image. Why are there H's where the phosphate group is and on your image the O's are alone with a "-" sign? Is what I have below the true AMP and what you were showing me was for the purpose of the functional group?

Please use the edit link on your question to add additional information. The Post Answer button should be used only for complete answers to the question. - From Review

Doesnt seem as efficient they need to add an option to add photos to a response hence why i added an answer

@Atticus283blink It's probably best to either integrate it into your question or host the image on imgur and link to it in the comment.


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Does hydrogen bonding contribute in solubility of a substance
If a compound can't hydrogen bond with itself but can hydrogen bond with the water, why does it still remain immiscible?
Take for example ethoxyethane. It can't hydrogen bond with itself but its oxygen atoms can participate in hydrogen bonding with the slightly charged H atoms of the water.
So why does ethoxyethane still remain immiscible with water?

Because the gain in energy with bonding with ethoxyethane isn't worth it. The molecule is big compared to water. As a result, many hydrogen bonds with water are broken just in order to solvate ethoxyethane. In addition, ethoxyethane only has one site which it can hydrogen bond.

Ethoxyethane, better known as "diethyl ether" or even just "ether", can form hydrogen bonds with water.  But fitting the ether molecules into the water solvent means you have to break up some water-water hydrogen bonds, and the ether molecules cannot replace enough of those hydrogen bonds to dissolve freely.
Of course, ether does form some hydrogen bonds, which makes is more soluble than it would be without the hydrogen bonds (compare ether with, for example, diethyl sulfide).  But most (or at least the most common) freely water-soluble organic compounds have either ions to generate ion-dipole interactions with the water, or hydroxyl groups that readily fit with the hydrogen bonding structure of the water (think of ethanol or sugars).

Okay then please enlighten me on this. Do ethanol hydrogen bond with itself?

Absolutely yes.


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Why does O2 gets less compressed as compared to CH4 when pressure is applied?
As I understand it, the compression depends on the attractive forces between the molecules. And considering that both $\ce{O2}$ and $\ce{CH4}$ do not have any dipole-dipole interaction with their respective molecules, The attractive forces should just be van der waals forces, that depend on mass and surface area of contact between the molecules only. As $\ce{O2}$ has more mass and more surface area of contact than $\ce{CH4}$ (as $\ce{CH4}$ is spherical in shape), it should be compressed more.

Do you know what shape each of those molecules are? Geometry means one type fits nicely together compared to the other, so less force is required to make it do so. CH4 is far from being spherical.

But so are diatomic molecules.