text:
Deriving an expression for the constant pressure heat capacity

Given the relation $G = H − T S$, show that:
  $$\left(\frac{\partial S}{\partial T}\right)_p=\frac{C_p}T$$

Now I realise that $T\mathrm dS=\mathrm dQ$ and that would I suppose get the result. However, I have been asked to use the expression $G = H - TS$ to find the result. If I just differentiate implicitly I end up getting:
$$\left(\frac{\partial G}{\partial T}\right)_p+\left(\frac{\partial S}{\partial T}\right)_p=\frac{C_p}T$$
What am I doing wrong?

Btw, how did you get to your final equation. If I just differentiate $G = H -TS$ with respect to $T$ I get something else. Or what did you mean by "differentiate implicitly"?

I differentiated with respect to T yes. Perhaps I did it wrong; what did you get? I think I assumed that dT/dT was zero but it's actually 1 isn't it? Perhaps that's where I went wrong?

Yes, that's where the differnce between your equation and mine came from. And actually this gets you to your desired result even faster than in my answer. You'd  have $\left(\frac{\partial G}{\partial T}\right)_p+\left(\frac{\partial S}{\partial T}\right)_p + S=\frac{C_p}T$. Then just use $\left(\frac{\partial G}{\partial T}\right)_p = -S$ and you're there. I should have tried that way first :)

You could start by dividing your whole equation by $T$:
\begin{align}
 \frac{G}{T} = \frac{H}{T} - S
\end{align}
and then take the derivative with respect to $T$ at constant $p$:
\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} = \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \ .
\end{align}
That way you've already introduced the desired derivative of the entropy with respect to $T$ at constant $p$.
Then you can use that
\begin{align}
 \frac{G}{T} = \frac{H}{T} - S  \qquad &\Rightarrow& \qquad \mathrm{d} \!\left(\frac{G}{T}\right) &=  - \frac{H}{T^2} \mathrm{d} T + \frac{1}{T} \underbrace{\mathrm{d} H}_{= T \mathrm{d}S + V\mathrm{d}p} - \mathrm{d} S \\
&&&= - \frac{H}{T^2} \mathrm{d} T + \frac{V}{T} \mathrm{d}p \\
&\Rightarrow& \qquad \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= - \frac{H}{T^2}
\end{align}
and simply using the product rule
\begin{align}
 \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} = \frac{1}{T} \underbrace{\left(\frac{H}{\partial T}\right)_{p}}_{= C_{p}} + H \underbrace{\left(\frac{\partial \frac{1}{T}}{\partial T} \right)_{p}}_{=-\frac{1}{T^2}} = \frac{C_{p}}{T} - \frac{H}{T^2} \ .
\end{align}
Putting it all together gives you the desired result:
\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
- \frac{H}{T^2} &= \frac{C_{p}}{T} - \frac{H}{T^2} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
\Rightarrow \left(\frac{\partial S}{\partial T} \right)_{p} &= \frac{C_{p}}{T} \ .
\end{align}

Starting with $dG=-SdT+VdP$, we have  $$\left(\frac{\partial G}{\partial T}\right)_P=-S\tag{1}$$
But, from the equation $G = H - TS$, we also have:$$\left(\frac{\partial G}{\partial T}\right)_P=\left(\frac{\partial H}{\partial T}\right)_P-S-T\left(\frac{\partial S}{\partial T}\right)_P\tag{2}$$
If we combine Eqns. 1 and 2, we obtain:$$\left(\frac{\partial H}{\partial T}\right)_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{3}$$The left hand side of Eqn. 3 is the definition of $C_P$.  So,$$C_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{4}$$
QED


text:
Why are certain reflections forbidden in body centered cubic and face centered cubic structures?
I know that for body centered cubic structures the allowed reflections only happen if h+k+l is even and reflections are forbidden is h+k+l is odd, and for basic face centered cubic structures (excluding diamond cubic) for a reflection to happen the h,k,l terms must be all add or even and if they're a mix of odd and even then the reflection is forbidden.
I've tried looking in textbooks/lecture notes as to how/why these selection rules come about but I can't find an answer which explains, it's just kind of stated that this happens. Why is it that you get these allowed and forbidden reflections?
I know it'll probably be something to do with the group space/symmetry/destructive interference within the structures but I can't come up with a definitive explanation other than just stating that it's to do with 'symmetry' or 'destructive interference within the cube'. Is there a more detailed and specific answer that explains what actually goes on within the structure?

k-space (momentum space) is the Fourier transform of real space. Because of the high symmetry of fcc and bcc, it turns out when you crank through the math that lots of destructive interference occurs. In particular, following Ashcroft and Mermin's Solid State Physics, in chapter 6 they go through getting the Bragg structure factor of a bcc lattice, treating it as simple cubic with a basis. Problem 2 at the end of that chapter is to do it for fcc in a similar fashion.

I'm fairly sure all solid state chem texts will talk about it, imo Smart & Moore is a good option.

I will try find a copy of those somewhere, problem is the recommended textbooks for general inorganic courses don't really explain this sort of thing, it's just stated as a rule.

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text:
What does electrolysis of Cu(NO3)2 produce?
I have read that $\ce{Cu}$ is deposited at the cathode when copper (II) nitrate is electrolyzed. However, when I looked at reduction potentials, the reduction of nitrate to $\ce{NO}$ and $\ce{H2O}$ $(0.96\rm~V)$ is more spontaneous than the reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ $(0.34\rm~ V)$. So why is Cu formed rather than NO?
What I read doesn't mention what material the electrodes are, would it make a difference if they were inert or made of Cu? I supposed if they were made of Cu, then the Cu will be oxidized and the $\ce{NO3-}$ reduced:
$$\ce{Cu + NO3- -> Cu^2+ + NO + H2O} {\rm \qquad \qquad(unbalanced)}$$
However this has a positive voltage of $1.30$, so why would electrolysis even be necessary?
If the electrodes were inert, then there were be no Cu to oxidize, so what would happen? Would the water be oxidized? $$\ce{2H2O -> 4H+ + 4e- + O2} \\\ \rm V=-1.23$$

Copper is a cation which goes to the cathode where reduction takes place. Nitrate is an anion which goes to the anode where oxidation takes place.

But not according to the reduction potentials..nitrate is more likely to be reduced to NO and H2O than Cu2+ to Cu is

@user28625 Reduction potentials have nothing to do with the fact that copper is a cation, and cations tend to cathode where the reduction happens, while the anions tend away from it. Maybe nitrate would reduce quite willingly; but it is simply not there.

If it would reduce quite willingly, why wouldn't it go to the cathode, a source of electrons, to get reduced?

Because anions must go to the anode (see, they aren't called "anions" for  nothing), and breaking the law would cause a hefty penalty. That being said, I guess some tiny portion of $\ce{NO3-}$ eventually gets reduced.

This confuses me-- why would NO3- so willingly get reduced if attempting to do that (by going to the cathode) breaks Coulomb's law?

What is the charge of $\ce{NO3-}$? What is the charge of an electron (and, consequently, of anode)?

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text:
How to treat pressure in computational chemistry?
I have a reaction where a gas at high pressure and a solution with all kinds of species is involved. How do I take pressure correctly into account to get reasonable values for $G$?
Do I calculate thermochemical corrections at the given temperature and pressure (frequency calculations) and then add additional corrections for concentration ($\Delta G = RT \ln(p)$, where $p$ is (partial) pressure or mole fraction)? Or should I calculate the thermochemical corrections at the given temperature, but standard pressure? For the additional term, should I use the overall pressure of the gas, and the mole fraction of the species in solution, or should I check for the solubility of the gas at given temperature and pressure in the solvent?
Usually, I would approach it this way:
Geometry optimization at lower level of theory, then computation of $G$ as the sum of:

single point energy at higher level
$G$ thermodynamic corrections (at $\pu{298.15 K}$ and $\pu{1 bar}$)
$G_\mathrm{solv}$ corrections from COSMOtherm (at infinite dilution)
$\Delta G_p = RT \ln(p)$, where $p$ is either the molar fraction of the compound in solution or the partial pressure in gas phase

This works quite well, however, I'm not sure how to adjust the different terms when dealing with non-standard conditions. I think that as only dissolved gas participates in the reaction, I should be interested in this as a reference point for $\Delta G_\mathrm{r}$, etc., but with what temperature/pressure do I then calculate 2? And do I take the pressure of gas above the solution in 4 or the molar fraction of gas in solution?

I am not sure to understand in detail your question, can you expand a little more?. Normally standard thermodynamic functions in gas phase is what we trivially calculate, but if there is a strong interaction between molecules (different from chemical reaction) it can be not useful.

There are two complexities to this.  1.  Determining the free energy of a pure component at a specified temperature and a high pressure beyond the ideal gas region and 2.  Starting with the free energy of the pure components at the same total temperature and pressure and determining the free energy of the mixture.  Are you familiar with the concept of fugacity?  Do you know how to do the first step using dG=VdP at constant T.

I tried to clarify the problem - does that help? It's true that strong interactions between the molecules have an impact, especially at higher pressures/concentrations, but I think this can be neglected in my case.

That's a very interesting question.
What about using a dynamical approach?

Old question, but still seems to attract some interest.
In general, the total pressure of the system is not taken into considerations for corrections. The pressure at which you do the corrections is the partial pressure / concentration of the molecule.
I recommend calculating the thermal correction for $G$ at the temperature of interest and the default pressure ($e.g.$ 1 atm for Gaussian). You can then add the correction for the isothermal expansion/compression to the actual concentration $c_f$ / partial pressure $p_f$:
$$\int_{p_i}^{p_f}\bigg(\frac{\partial G}{\partial p}\bigg)_T dp = RT\ln\bigg(\frac{p_f}{p_i}\bigg) = RT\ln\bigg(\frac{c_f RT}{p_i}\bigg)$$
where $p_i$ is the pressure for the thermal correction ($e.g.$ 1 atm). Note that these formulas are based on the ideal gas approximation, but that's what software like Gaussian does under the hood anyway. Hence, you could in principle directly set the correct pressure and should get the same result, but I find it more convenient to have the same route section for all calculations.
Regarding your multiphasic systems: As you noted, the reaction will most likely take place in the solution, so you have to use concentration of the gas in the solution.

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text:
How should I name this branched alkyl substituent?
The parent chain is a decane (in black), but for the alkyl substituent (in red) I'm trying to decide between:

5-(1,2-dimethylpropyl)decane
5-(2-methylbutan-3-yl)decane
5-(3-methylbutan-2-yl)decane

Which of these is correct?

According to Section P-46 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), compound acyclic substituents consisting of a principal chain and one or more acyclic substituents can be named in two ways:

by using alkyl substituents according to method (1) of Subsection P-29.2

(1) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups and for the mononuclear hydrides of silicon, germanium, tin, and lead. Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

by using alkanyl substituents according to method (2) of Subsection P-29.2

(2) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ are added to the name of the parent hydride with elision of the terminal letter ‘e’, if present, when followed immediately by the letter ‘y’. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix ‘ylidyne’, the locant ‘1’ must be cited. This method is used to generate names of ‘alkanyl-type substituent groups’ that are simple substituent groups with free valences at positions other than ‘1’.

For both methods, the principal substituent chain has the greater number of skeletal atoms, i.e. the longest chain. Low locants are given first to the free valence; for method (1), the free valence always has the locant ‘1’.
The preferred names for simple substituent prefixes derived from saturated parent hydrides with a single free valence that is not at the end of the longest chain are formed according to method (2).
According to method (1), the alkyl-type name of compound substituent group given in the question is ‘1,2-dimethylpropyl’. The preferred alkanyl-type prefix according to method (2) is ‘3-methylbutan-2-yl’.
Therefore, the preferred name for the complete compound given in the question is ‘5-(3-methylbutan-2-yl)decane’.
Note that, according to current IUPAC nomenclature, locants are placed immediately before that part of the name to which they relate. Therefore, correct prefix is ‘3-methylbutan-2-yl’ (not ‘3-methyl-2-butanyl’).

Thanks for the answer. I just didn't understand why is method 2 preferred instead of method 1?

@K.Smith The Blue Book does not explain why method (2) is preferred for such compound substituent groups. I suppose, method (2) is preferred in order to achieve consistency with the nomenclature of the corresponding parent hydrides (e.g. butan → butan-2-yl rather than 1-methylpropyl) and because, in many cases, method (2) can lead to simple substituent groups (e.g. butan-2-yl) where method (1) yields compound substituent groups (e.g. 1-methypropyl).

I see. And one final question, if the substituent is butan-2-yl are there parentheses around it or not?

@K.Smith Yes, parentheses are used around substituent prefixes in order to separate locants referring to different structural elements, for example “5-(butan-2-yl)decane” or “5-(3-methylbutan-2-yl)decane”

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text:
How does Gaussian 09 normalize the excitation coefficients?
When I did excited state calculations with Gaussian 09, I was asking me on how those values are normalized because a simple sum over the squared of the coefficients did not yield 1.
So ... how are those coefficients normalized then?

I made a few quick calculations with very high printing levels. (I don't think that the molecule is important for this, but my calculations have been done on 2,3,6,7-tetranitronaphthalene with RPM6/UPM6 and 10 excitated states.)
The default printing level is IOp(9/40=1) which yields excitations with $c \ge 0.1$
Excited State   1:      Singlet-A      2.4747 eV  501.01 nm  f=0.0040  <S**2>=0.000
      41 -> 59        -0.10248
      42 -> 60         0.15085
      47 -> 57         0.15074
      55 -> 57         0.41262
      56 -> 58         0.46353

The sum of the squared coefficients is 0.44110, which is absolutely not 1.
The unrestricted version* for the same excitation includes a lot more coefficients:
Excited State   1:  3.000-A      -0.6305 eV    -1966.38 nm  f=-0.0000  <S**2>=2.000
     37A -> 69A       -0.16535
     41A -> 59A       -0.18261
     41A -> 68A       -0.19102
     42A -> 60A        0.28385
     42A -> 64A        0.19591
     47A -> 57A        0.18349
     48A -> 58A        0.10140
     55A -> 57A        0.46982
     55A -> 65A        0.11894
     56A -> 58A        0.65048
     56A -> 61A       -0.11129
     37B -> 69B        0.16535
     41B -> 59B        0.18261
     41B -> 68B        0.19102
     42B -> 60B       -0.28385
     42B -> 64B       -0.19591
     47B -> 57B       -0.18349
     48B -> 58B       -0.10140
     55B -> 57B       -0.46982
     55B -> 65B       -0.11894
     56B -> 58B       -0.65048
     56B -> 61B        0.11129
     37A <- 69A       -0.12501
     41A <- 59A       -0.13192
     41A <- 68A       -0.14182
     42A <- 60A        0.20466
     42A <- 64A        0.14232
     47A <- 57A        0.13721
     55A <- 57A        0.33618
     56A <- 58A        0.43137
     37B <- 69B        0.12501
     41B <- 59B        0.13192
     41B <- 68B        0.14182
     42B <- 60B       -0.20466
     42B <- 64B       -0.14232
     47B <- 57B       -0.13721
     55B <- 57B       -0.33618
     56B <- 58B       -0.43137

Now taking the sum over those squared coefficients gives 2.72735, which is also obviously not 1. But the subtraction of the de-excitations (<-) from the excitations (->) seems to be appropriate, as it yields 0.99451.
Seeing this, the sum for the restricted case somehow looks near 0.5 ... why not doubling the squared coefficients? This leads to 0.88220 which is acceptable close to 1. Maybe the printing level is too low for the restricted calculation?
Rising the printing level with IOp(9/40=2) ($c \ge 0.01$) yields this output for the first excitation:
 Excited State   1:      Singlet-A      2.4747 eV  501.01 nm  f=0.0040  <S**2>=0.000
      17 -> 66         0.01120
      18 -> 92         0.01037
      23 -> 82         0.01204
      24 -> 73        -0.01804
      24 -> 80        -0.01353
      25 -> 62        -0.01457
      25 -> 74         0.01303
      26 -> 57        -0.01038
      26 -> 69        -0.01748
      27 -> 68        -0.01840
      28 -> 75        -0.01002
      33 -> 73        -0.02580
      33 -> 80        -0.01986
      34 -> 76         0.01612
      34 -> 89        -0.02096
      35 -> 67         0.01410
      35 -> 75        -0.03109
      35 -> 84         0.01033
      35 -> 88         0.01154
      35 -> 91        -0.01028
      36 -> 81        -0.01389
      36 -> 90         0.01911
      37 -> 57         0.02399
      37 -> 63         0.01245
      37 -> 65         0.03441
      37 -> 69        -0.08550
      38 -> 72         0.03176
      38 -> 75         0.01264
      38 -> 88         0.01466
      38 -> 91        -0.01049
      39 -> 63         0.01742
      39 -> 65        -0.01751
      39 -> 69         0.01804
      39 -> 71         0.03883
      39 -> 90         0.01569
      40 -> 61         0.01120
      40 -> 66        -0.01695
      40 -> 70        -0.05758
      40 -> 73        -0.03491
      40 -> 77         0.01693
      41 -> 59        -0.10248
      41 -> 62         0.01194
      41 -> 68        -0.09264
      42 -> 60         0.15085
      42 -> 64         0.09639
      45 -> 57        -0.01773
      47 -> 57         0.15074
      47 -> 63         0.01564
      47 -> 65         0.04802
      47 -> 69        -0.02597
      48 -> 58         0.05950
      48 -> 61        -0.01527
      49 -> 60        -0.05724
      49 -> 64        -0.02758
      50 -> 59         0.04110
      50 -> 68         0.02528
      51 -> 58         0.01641
      51 -> 66        -0.02468
      51 -> 70        -0.02435
      51 -> 82         0.01341
      52 -> 60         0.02192
      52 -> 64         0.01863
      52 -> 67         0.02426
      52 -> 72         0.01922
      52 -> 75        -0.01609
      53 -> 62        -0.03163
      53 -> 74         0.01216
      54 -> 57         0.02946
      54 -> 63         0.03823
      54 -> 65        -0.01008
      54 -> 71         0.02057
      55 -> 57         0.41262
      55 -> 63         0.02503
      55 -> 65         0.09254
      55 -> 69        -0.03194
      56 -> 58         0.46353
      56 -> 61        -0.06155
      33 <- 73        -0.01421
      33 <- 80        -0.01130
      34 <- 89        -0.01340
      35 <- 75        -0.01702
      36 <- 90         0.01197
      37 <- 57         0.01598
      37 <- 69        -0.01657
      38 <- 72         0.01731
      39 <- 63         0.01032
      39 <- 71         0.02069
      40 <- 70        -0.02983
      40 <- 73        -0.01759
      41 <- 59        -0.01636
      41 <- 68        -0.01551
      42 <- 60         0.03039
      42 <- 64         0.01865
      47 <- 57         0.03857
      47 <- 65         0.01219
      48 <- 58         0.01179
      51 <- 70        -0.01217
      54 <- 71         0.01054
      55 <- 57         0.08529
      55 <- 65         0.02862
      55 <- 69        -0.01486
      56 <- 58         0.06624
      56 <- 61        -0.01293

Again, calculating the square of sums yield 0.53927 (doubled: 1.07853) which is a little bit more than 0.5. De-excitation values subtracted from excitation values now give reasonable 0.49889, which is quite as good as the unrestricted case if doubled (0.99778).
Rising the printing level to IOp(9/40=5) and calculating the sum of squared of the coefficients (de-exc. minus exc.) gives 1 for unrestricted case and 0.5 for the restricted case.
Based on that, Gaussian's normalization for excitation coefficients is most probably:
$$
\begin{cases}
2 \left( \sum_i c_{i,\text{exc}}^2 - \sum_i c_{i,\text{de-exc}}^2 \right) =1 & \text{for restricted methods}\\
\sum_i c_{i,\text{exc}}^2 - \sum_i c_{i,\text{de-exc}}^2 =1 & \text{for unrestricted methods}
\end{cases}
$$

* The same geometry as for the restricted calculation was used.


text:
What would be a safe/good surfactant for ultrasonic cleaning vinyl records + distilled water and isopropyl alcohol
Title pretty much says it all besides the type of records I believe are pvc (60s - current, someone correct me if there may be other materials but I absolutely don't have any shellac records) and I plan to use around 50:1 water/ISO. I'm not sure what to use as a surfactant or if it would be of any benefit, though I assume it would. Looking for something 100% safe for records though, and I do have several chemical supply companies around. Any suggestions?

You'd definitely want distilled water. // My other thought was wondering about the paper labels in the center of the record. I suppose that you could suspend the records vertically and spin them slowly - maybe blow liquid off to keep it from flowing down over the label. // Depending on what type of crud is on the records you might need multiple liquid baths. For instance the gunk from cigarette smoke might need some sort  of detergent to remove it effectively.

@MaxW Spray the cleaner evenly onto the record (not on the label !) while its spinning on the turntable and wipe from the centre out in a slow stroke. I use an already mixed record cleaner I'm not sure what its made of but it's blue liquid.

@Joel - The OP (Tim) wanted to use an ultrasonic bath. While i do think that would clean the records, I fear that it would destroy the labels. You'd also have to worry about the fibers from the labels getting into the bath and onto the record. I think the OP's approach was overkill and creating other problems. // Your approach seems very sensible.

@MaxW overkill I agree. I recommend my method to the OP I've been doing it for years same with crew I know. I've never heard of an ultrasonic bath? I spose it depends how dirty the records are. Be careful you don't wear away the grooves in the records with too extensive cleaning or too harsher solvent (:

Yes brushes are the standard method, however ultrasonic has been proven to be far better and I can confirm that I have had great success so far with some records i had thought were just going to sound bad no matter how much I cleaned. The problem with standard cleaning methods is that they tend to spread/move contaminates around as much as removing them. The labels will just not be immersed and will eventually have a gasket/protector, for now it's immersed by hand up to the label and rotate 1/8 turn per 30 sec. Believe me the ones I've cleaned sound 1000% better, and a surfactant should help.

Sorry for getting into a tangent on this, but yes there are ultrasonic record cleaning machines on the market for $3-5000.00. And people who own them love them - one pro reviewer even bought his review unit. They don't appear to be much more than a customized us cleaner with motor, brushes, and fans added so there's a lot of people using standard us cleaners and adding a motor etc. Would be great to have for more like 200.00$. Main thing for surfactant is it should dry residue free and be safe for pvc, but I don't know what to use for that.

Sorry if the above text is messed up - posted on iPhone and it looks like most of the spaces are missing. The messed up part isn't that important anyway.


text:
Tri-iodination of 2-methylpropane
In the reaction of tri-iodination of 2-methyl propane only five isomers can be formed, from which only the branch methyl can be iodinated if both the lateral carbond are iodinated .. Is this because of the inductive effect or the steric effect is the responsible here...if it is the steric effect which hinder the iodination of the branch carbon (2-methyl) why is it possible to tri-iodinate the lateral carbons( the primary carbons other than the branch one)  ?

The reason only five isomers can be formed is the three methyl groups are equivalent by symmetry.

I believe your arbitrary distinction of branch carbon/methyl from lateral carbon/methyl is creating confusion


text:
Junction Potential at Each End of a Salt Bridge
I am having trouble with the concept of a junction potential as it relates to the salt bridge of an electrochemical cell. I understand that a junction potential occurs between two solutions of differing concentrations and that it depends on the mobilities of the ions in solution.
For example, between a solution of sodium chloride and deionized water, a junction potential will develop when the ions of the sodium chloride solution are allowed to diffuse into the deionized water. The junction potential occurs because the chloride anions will diffuse more quickly into the deionized water (due to its greater mobility) than the sodium cations. This leaves the sodium chloride side of the junction between the two solutions more positive and the deionized water side more negative, creating a junction potential.
However, what I do not understand is how a junction potential develops in an electrochemical cell. A junction potential develops at both ends of the salt bridge. Are the junction potentials at either end different? If so, why? If not, how would a junction potential occur between two solutions that don't directly interact (the two solutions in this case being the solutions of the respective half cells)?

First you do realize that the junction potential would depend somewhat on the amount of current flowing in the cell. With modern high impedance preamps there wouldn't be much current. // Since the ions and ionic strength in the two half cells doesn't have to be the same there is reason to believe that the junction potentials at opposite ends of the salt bridge could be different. The gist is that the salt bridge would develop a small charge. Over both half cells and the salt bridge then the net charge would be zero of course.

Having a charge in the salt bridge just means that the system isn't at a real equilibrium. So long as the solutions in the two half cells and the salt bridge are different, then the system really isn't at equilibrium, but rather at a steady state.


text:
Do there exist metals that are composed of molecules?
Most metals are composed of pure elements or are alloys (mixtures of pure elements.)
Are there metals that are composed of one type of molecule that is metallically bonded together?
The difference between a metal that is composed of one type of compound and an alloy is that in an alloy different elements are attached by metallic bonding but in a metal composed of one compound covalent or ionic bonding attaches multiple atoms together into compounds and then these compounds form metallic bonds with each other.
I have read that Gallium in its pure elemental state forms pairs which are then metallically bonded together which sort of is what I'm wondering about except that I'm wondering if metals can form out of more complicated compounds.
I would also expect there to be a lot of compounds which on their own do not form metals but do not degrade the metallicity of alloys they are put in very much.
I think a more specific and low level question might be: are there covalent compounds that can make metallic bonds with other things?

Metallic bond kinda contradicts the very idea of molecules (it is hard to form covalent bonds when you are immersed in a sea of electrons). Gallium with its "almost molecules" is likely the best we can get. Also, you might want to think of large metal clusters. And then there are steel (which contains carbon, which is not a metal on its own) and other alloys containing some non-metallic elements.

How do you feel about the various allotropes of selenium?

related http://chemistry.stackexchange.com/questions/35278/what-are-the-intermolecular-and-intramolecular-forces-between-hg-molecules-atom

Organic metals may be a common reality in the not-too-distant future.

@NicolauSakerNeto I think you should expand this into an answer since the implication of Heeger's discoveries are that the answer is "yes".

Please don't bring in organic metals because is not what presumably the OP want. There is not organic metal as one composed of molecules hold togheter by a see of electrons. Nor ionic conductivity is a metallic one. Nor hopping of electrons. Organic metals and semiconductors are beast for which a terminology has still to come. One would need pages to even use a term that univocally describe something without messing up. Most of you can think of organic metals is that the material is not a good insulator. @Nicolau Saker Neto

Tough the answer by Matt Black made a nice service. However is quite possible that the behaviour is metallic but the magnitude (of conductivity) is not. I up vote that anyway because @matf_black made a nice service bringing up a perspective without mixing things.

Sort of, if you define metals as substances that exhibit some metallic behaviour
Metallic elements are, well, the metals. But other substances can exhibit many of the properties of those metals.
One well-known example if what you get when you dissolve a lot of sodium in liquid ammonia. Beyond a certain concentration, a new metallic-like phase if formed where electrons are solvated in the liquid (which contains ammonia molecules). The bronze, metallic colour of this phase is seen in this Periodic Video.
Many metallic properties are dependent on the delocalisation of electrons. Some conjugated polymers can have delocalised electrons and metallic levels of conductivity when doped. Whether these should count as metals is debatable but they do exhibit some metallic properties. And they won a Nobel prize for Heeger, et al. See Wikipedia for some discussion.
So there are molecular systems with some metallic properties, though their bonding is not necessarily because of the electron delocalisation.

According to A Conducting Crystal Based on A Single-Component Paramagnetic Molecule,
[Cu(dmdt)2] (dmdt ) Dimethyltetrathiafulvalenedithiolate) J. Am. Chem. Soc., 2002, 124 (34), pp 10002–10003 :

Single-component molecular metals should greatly extend the
  development of new types of molecular conductors. For example,
  while the first metallic molecule-based magnet has been reported
  recently, [reference 8] it may be possible to create such metallic molecular
  magnets from single-component molecules containing paramagnetic
  ions such as Cu2+ and Co2+.

Reference 8 is Coexistence of ferromagnetism and metallic conductivity in a molecule-based layered compound Nature 408, 447-449

The temperature dependence of the resistance exhibits a metallic behaviour over the entire temperature range that we investigated (down to 2 K).

Yes
Adding to the other examples, there is an elemental diatomic metal.
No that's not a typo.
I'm sure you learned your diatomics in school: H2, N2, O2, F2, Cl2, Br2, and I2. They probably gave you the rather awkward acronym "HOFBrINCl" as well. In college you may have learned that at high enough temperatures S2 and possibly even P2 can exist in equilibrium with other species.
But at STP there's still one missing from that list, right between Cl2 and Br2. That missing diatomic is gallium metal. In fact, the covalent Ga2 pairs are why gallium melts at such a low temperature.

Yes mercury can form an amalgam with ammonium that is $\ce{H3N-Hg-H}$ although it decomposes at room temperature.
See https://en.wikipedia.org/wiki/Amalgam_%28chemistry%29#Ammonium_amalgam
It might be possible to do this with more chemicals and mercury.
Theoretically cations like Ammonium might have a metallic nature but I wouldn't know the details of that.

Under sufficiently high pressure ammonium, $\ce{NH4}$, could form a metal. Such a metallic species may be present in the interiors of Uranus and Neptune [1](https://doi.org/10.1093/mnras/114.2.172).
Reference
1.
M.F.M. Bernal, H.S.W Massey, "Metallic Ammonium", Monthly Notices of the Royal Astronomical Society V114(2) (April 1954), Pages 172–179; doi: 10.1093/mnras/114.2.172 (open access).


text:
Preferred IUPAC names for branched unsaturated hydrocarbons (polyenes)?

What are the preferred IUPAC names for the following three compounds?

Compound 1 is 3-ethenylhexa-1,5-diene. 2 is 3-methylidenehexa-1,5-diene.
But I am still not sure about 3. I have narrowed it down to two choices: 3-ethylidenehexa-1,5-diene or 4-ethenylhexa-1,4-diene.

How do I decide which two of the three double bonds should be in the parent chain, considering that both choices of parent chain have the same number of complex and double bonds?
The information I used is from the following Q&As:
How is a side-group that contains a double bond named?
How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

Related: How is a side-group that contains a double bond named?

According to the general methodology described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a double bond is expressed by changing the ending ‘ane’ of the name of the corresponding saturated parent structure to ‘ene’.

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.
(…)

For example:

If the compound contains multiple double bonds, a multiplicative prefix is placed before the ending ‘ene’ and the letter ‘a’ is inserted between the root of the name of the parent structure and the multiplicative prefix.

P-31.1.1.2 The multiplying prefixes ‘di’, ‘tri’, etc., are placed before endings denoting unsaturation to indicate the number of multiple bonds of each kind, as required, for example, ‘diene’ and ‘triyne’. For euphonic reasons, when the endings ‘ene’ and ‘yne’ are preceded by a multiplying prefix and a locant the letter ‘a’ is inserted. There is no elision of the final letter ‘a’ of a multiplying prefix before ‘ene’ or ‘yne’, for example, ‘tetraene’ and ‘pentayne’.

For example:

The names of substituent groups are derived from the respective parent hydrides. The presence of free valences formally derived from the loss of one, two, or three hydrogen atoms at the end of the longest chain of a parent hydride is denoted by replacing the ending ‘ane’ of the parent hydride name by ‘yl’, ‘ylidene’, and ‘ylidyne’, respectively.

The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups (…). Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

For example:

methyl $(\ce{-CH3})$
methylidene $(\ce{=CH2})$
ethyl $(\ce{-CH2-CH3})$
ethenyl $(\ce{-CH=CH2})$
ethylidene $(\ce{=CH-CH3})$

Therefore, the name for the first compound given in the question is ‘3-ethenylhexa-1,5-diene’.

The name for the second compound is ‘3-methylidenehexa-1,5-diene’.

For the third compound, there is a choice among two different principal chains. In accordaqnce with the above-mentioned rules, the name could be either ‘4-ethenylhexa-1,4-diene’ or ‘3-ethylidenehexa-1,5-diene’.
The current IUPAC Recommendations stipulate various criteria for the selection of the principal chain. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain. If this criterion does not effect a choice of a senior parent structure, the following relevant criteria are applied successively until there are no alternatives remaining. According to Subsection P-44.4.1, the principal chain

(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)

Therefore, the correct name for the third compound is ‘4-ethenylhexa-1,4-diene’ (not ‘3-ethylidenehexa-1,5-diene’) since the locant set ‘1,4’ for the ‘ene’ ending is lower than ‘1,5’.

Thanks for the answer. And what does the (4Z) mean?

@K.Smith One of the double bonds in 4-ethenylhexa-1,4-diene gives rise to cis-trans isomerism. The stereodiscriptor ‘Z’ indicates the configuration that is shown in my image; the other one would be ‘E’. The locant ‘4’ indicates the position.


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1-bromo-2-fluorobenzene reaction with magnesium and furan
Why does bromofluorobenzene react with magnesium and furan to form a benzyne intermediate?
Does it have something to do with a Grignard being highly basic and a fluorine being electron withdrawing, thereby making elimination a likely pathway?
Then does a Diels-Alder type reaction commence, with furan acting as the diene and the benzyne acting as the dienophile?
How likely or favorable is this reaction in real life?

I think it's more because fluorine makes the ring electron-poor, allowing the first step (formation of the grignard) to happen more quickly. I'm kind of using the comparison of $\mathrm{S_N1}$ and $\mathrm{E1}$ to guide me to the thought that $\mathrm{S_NAr}$ would have some parallels with aromatic $\mathrm{E1}$, so it could be wrong. To see why fluorine is used for $\mathrm{S_NAr}$, see the answer here.

Of course, thinking about it, $\mathrm{S_NAr}$ is more like the opposite of $\mathrm{S_N1}$. So... We'll see how logical that thought is.

I'm guessing that you mean that 1-bromo-2-fluorobenzene and magnesium form a benzyne intermediate, and then furan reacts with that benzyne intermediate. Can you confirm?

Yes! You are correct @jerepierre

The overall pathway described is shown below. Treatment of the aryl bromide with magnesium gives a Grignard, which is equivalent to a carbanion. The carbanion can eliminate fluoride to give benzyne. Being highly strained, benzyne is a fantastic dienophile and reacts with furan through a Diels-Alder reaction.

Most of the techniques to generate benzyne use the same strategy: generate a carbanion next to a leaving group. The classic method taught in introductory organic chemistry is by deprotonation of an aryl halide with a very strong base (sodium amide), which eliminates the halide. The strategy shown here generates the carbanion using the milder method of forming a Grignard.
The elimination is possible because carbanions are very good at kicking out leaving groups. It's commonly taught that fluoride is a poor leaving group, but the end of that sentence is that it's a poor leaving group for SN2 reaction or compared to chloride, bromide, and iodide. Later, it's taught that enolates (a reasonably stable class of carbanions) are reactive enough to eliminate alkoxides/hydroxide. So it makes sense that an unstabilized carbanion as here should be able to kick out a relatively poor (but not that terrible) leaving group.
The Stoltz group has a nice summary of benzyne's history, generation, and reactivity.

Wouldn't the electrons in the $\mathrm{sp^2}$ orbital have a pretty hard time getting decent overlap with $\mathrm{\sigma ^*_{CF}}$?

@DGS it's a syn elimination which is less optimal than anti, but it is locked in the reactive  conformation

Don't forget the magnesium. When it us incorporated into the Grignard reagentbit becomes electrophilic and can attach to the fluorine. This facilitates the nucleophilic attack on the carbon-fluorine group by making a better leaving group.


text:
How do I open volumetric flask with with 1M HCl?
I have two volumetric flasks with ground glass stopper, and I have 1M HCl inside each of them. I can't open either, stopper is "glued" to the flask somehow. Potentially by hydrochloric acid. Or maybe not (I mean maybe it's stack on its own, without participation of HCl). What is conventional way to open it? I couldn't think about anything better than to put entire flask in water, but it didn't help so far.

I read another post on this topic: http://chemistry.stackexchange.com/questions/26882/how-to-remove-a-stuck-stopper-from-a-flask and followed advice "hit the corner of the lab bench with the stopper". I had two flasks - for one of them it worked as charm, but unfortunately I have broken another one. My conclusion is, that next time I'd better try another method. I don't have heat gun, but my colleague suggested to put flask neck under (portion with a stopper) under  hot water, so I will try that.


text:
What does a fractional order of reaction mean for the mechanism?
I'm doing a high school/sixth form college investigation of the kinetics between magnesium ribbon and hydrochloric acid. I have obtained a rate order was $1.5$ with reference to $[\ce{H+}]$ and hence the rate equation is
$$\mathrm{rate} = k[\ce{H+}]^{1.5}.\tag{1}$$
How does this fractional order correlate with the mechanism?
I remember my teacher once telling me that a fractional rate order just means that there is an intermediate stage and each step has its own rate order which contributes to the rate equation:
$$\ce{A ->[$x$] B ->[$y$] C}$$
$$x + y = n$$
$$\mathrm{rate} = k[\ce{A}]^n.\tag{2}$$
Is this right? Also, what would this mean for my rate-determining step? If the above is true, what would I choose as the RDS if they both contribute a different amount to the rate order?

$x + y = n$ certainly won't hold for all cases. For instance in this case all of the A could be quickly converted to B, but C forms very slowly.

A reaction mechanism is (generally) a very complicated thing which you can't hope to deduce from orders alone. One may say a fractional order indicates that the actual mechanism is "not simple" (that is, more complicated than "all molecules simply meet and turn to products at once, just like the equation says"). But then again, it is pretty much never simple.

Here's an example.
Let's consider a reaction of the generic form:
$$\ce{3A -> 2C}$$
Now, let's say we have some knowledge that this reaction proceeds by the following two-step mechanism:

Step one is fast and reversible.

$$\ce{A <=>[k_1][k_{-1}] 2B}$$

Step 2 is slow and irreversible.

$$\ce{A + B ->[k_2] C}$$
The rate of the reaction (expressed as the change in concentration of the product per unit time) can be expressed in terms of the concentrations of A and B.
$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}][\ce{B}]$$
However, we would prefer to not use intermediates in our rate law. Since the first step is fast and reversible, we can use the law of mass action to create a relationship between the concentrations of A and B:
$$K_1=\dfrac{k_{1}}{k_{-1}}=\dfrac{[\ce{B}]^2}{[\ce{A}]}$$
$$[\ce{B}]^2=K_1[\ce{A}]$$
$$[\ce{B}]=\sqrt{K_1[\ce{A}]}$$
If we substitute into the  rate law:
$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}]\sqrt{K_1[\ce{A}]} =k_2 K_1^{1/2}[\ce{A}]^{3/2}$$
The term $k_2 K_1^{1/2} =\dfrac{k_2 k_1^{1/2}}{k_{-1}^{1/2}}$ becomes $k_{\mathrm{obs}}$, the observed rate constant of the overall reaction. The fractional order comes from the second step being rate-determining, but the first step being an equilibrium with stoichiometry that generates the fractional tern.

This is well explained, but it should be commented that the only real conclusion that can be drawn from a fractional order coefficient is that there are multiple steps involved in the reaction.  The actual orders are not explicitly informative.

So Can I claim for an elementary reaction, the order can never be fractional or negative ( if it involves no intermediates ) Also what does it mean to sy if the order with respect to an reactant is negative ?


text:
What is the compound called when it is in solution?
When a compound precipitates out of solution, for example Calcium Carbonate from a water solution, it is called the precipitate.
What is the same compound called before it precipitates out - once it has dissolved? I found the word "electrolyte", but I'm unsure this applies to all solutions - only ionic, and I'm not sure a CaCO3 solution is ionic.

I believe the term you are looking for is solute; a solute is a substance dissolved in another substance, known as a solvent.

Having realised that Water is called a Solvent, I searched Google for "solvent and" and autocomplete told me that the word I'm after is "solute".

Also - this may be interesting (I hope doesn't confuse the question you asked)

A solution is a mixture featuring solutes that have been dissolved, while a suspension is a mixture of liquids also containing solid particles that may not completely dissolve inside the liquid.

(From Ask.com)


text:
How can I obtain oxygen gas from plagioclase
I would like to know if there is a possibility to decompose plagioclase to obtain oxygen gas to be used in a combustion reaction?
I know that oxygen represents $47\%$ of the weight of 1 mole of plagioclase (albite $\ce{NaAlSi3O8}$ – anorthite $\ce{CaAl2Si2O8}$), which equals to $\pu{130 g}$ out of $\pu{277 g}$, so it will be very nice if I could decompose plagioclase and obtain oxygen gas, but I don't know if that is possible or not?

Do you have any reason for thinking this is a possibility? If not, you might as well ask the same question about anything with oxygen in it. Oxygen makes up a significant portion of glass, but that doesn't mean it's going to be easy to get to elemental oxygen. Have you heard of this material being used to form molecular oxygen?

Why, it is possible all right, but would require quite a lot of other reagents, and energy too. It would be about as economically feasible as buying new iPhones to pluck out the glass and throw away the rest, instead of buying a new window pane. Even if you have these reagents and energy, it would be better to put plagioclase aside and get the oxygen from elsewhere. Unless you are stranded on some desolate remote planet, that is.

Yes, it is possible, and NASA is contemplating extracting oxygen from local resources on the moon: https://books.google.com/books?id=ho65r66FVEsC&pg=PA204&lpg=PA204&dq=nasa+extract+oxygen+plagioclase&source=bl&ots=63pDjRCx9L&sig=DJG40Wzbhsec4Ivx3gptR1ikjBM&hl=en&sa=X&ved=0ahUKEwjI7OChmP3LAhXFlIMKHTYHBgoQ6AEINjAE#v=onepage&q=nasa%20extract%20oxygen%20plagioclase&f=false,  http://science.nasa.gov/science-news/science-at-nasa/2006/05may_moonrocks/ and http://science.nasa.gov/science-news/science-at-nasa/2006/05may_moonrocks/.

I would try with a solar furnace.

Yes, I collect oxygen isotope data from plagioclase using the laser flourination method. To liberate the oxygen from the plagioclase, crystals are reacted with purified $\ce{BrF5}$. Then the plagioclase is hit with a $\ce{CO2}$ laser. Gasses are sent through a couple of cryogenic (liquid nitrogen) traps to filter, and then through a mercury diffusion pump to remove that nasty flourine.

That's cool (pun intended)! Not exactly what OP would hope, I guess, but still.


text:
Does changing the volume of a reactant affect the rate of reaction? Does changing the volume of HCl affect the rate of a Mg & HCl reaction?
I'm doing a experiment in which I'm testing the affect of the concentration of a reactant on the rate of reaction.
The reaction is between Magnesium and Hydrochloric Acid and so I'm changing the concentration of $\ce{HCl}$ ($\ce{0.5~M}$, $\ce{1~M}$ etc.)
In my lab report, I'm writing a list of controlled variables. So far I've mentioned the surface area and temperature of reactants. I would like to know more controlled variables.
I want to know whether the volume of Hydrochloric Acid affects the rate of reaction when added to the same amount of Magnesium metal?
According to some it doesn't and some say it does; I'm confused.
Also, please tell me why or why not the $\ce{HCl}$ volume affects the rate of reaction.
If the concentration of $\ce{HCl}$ changes between trials and the volume changes between the trial as well, then is the rate of reaction affected?
Does volume affect concentration in this case?

Related: Question regarding nitric acid, magnesium powder and initial reaction rate and Factors affecting mass of a gas-evolving reaction mixture over time

The way you describe this represents a poorly designed experiment. You essentially have 2 variables; concentration and volume, if I understand you correctly. This is not advisable. If you were to see any effect, you wouldn't know if it were a result of volume or concentration. You should add the SAME volume of HCl of different concentrations. That way, only concentration varies.

More volume of HCL does not increase the concentration and therefore does not increase the likelihood and so rate of reaction.
Mg and HCl forms a metal salt MgCl and produces Hydrogen gas.
There may be a very slight increase in rate of reaction but this is permissible/ very negligible due to the fact that metals and acids form adducts and these can further react/ delay further reactions between the acid and metal molecules.
The rate would definitely be affected by factors such as temperature of the (aq) HCL. This is until the reactant Mg is used up then the reaction will cease leaving you with a flat curve on a rate graph.
Now, for Higher concentrations- this implies that more reacting molecules are at higher proximity to each other therefore the intermolecular collisions between the Mg and HCl molecules in your case are more frequent therefore forming more product per unit time.
Control variable: mass of Mg to at least 1.d.p, concentration of acid moldm-3, volume of HCl cm3 - even though this doesn't affect the rate of reaction enough - they would want you to keep this constant for valid test etc.
Hope this helps!


text:
Common ingredient between two disinfectants?
I find one of the most powerful associations to be odour, placing me very clearly in a time and place. Recently we were using Dettol in the house and I was reminded very clearly of a another disinfectant my grandmother used to use to clean greyhound kennels called Jeyes Fluid.
Is there a common ingredient between the two of them? If, so what is it; if not why do they smell similar (IMO)?

Side note: There is no reason to use disinfectants around the house.

@Karl Curious as to how you're able to elicit my specific needs from the post above and pronounce thus? I said I was using it in the house, not around the house ;-)

Unless your house is a hospital or used for commercial food production,  disinfectants are superflous, and potentially detrimental to your health &  environment and definitely to your wallet.

@Karl If this were Wikipedia, I'd be tagging your comments under unfounded opinion or subjective observation and moving that they be deleted. They have no relevance to the original question at all so I'd suggest in future you refrain from making ones of this nature

@nonand I'm sorry if my opinion offended you, but we are talking chemistry here, and that is the wrong place to hide a safety concern out of politeness. Subtract the comment about your wallet, it is admittedly out of line.

Dettol's active ingredient is chloroxylenol.
Jeye's fluid's is 4-chloro-m-cresol.
It's still possible that both disinfectants simply have an added scent to make them smell "clean". However the active ingredients above are both phenols and will likely have a similar phenolic smell (think hospitals), though I haven't sampled them both myself.

Fascinating stuff, thank you. Some more on the characteristics of the "phenolic smell" here: http://perfumeshrine.blogspot.ie/2012/02/definition-terpenic-phenolic.html


text:
What are compounds with the same mass called?
It may be that the answer is just "no" and that's why I can't find it, but is there any common name for compounds with the same mass?
I'm thinking in the context of mass spectrometry where these compounds require additional steps to resolve.
Isomers are the obvious example, but I was wondering if there was a more general term.

By any stretch, you can't call $\ce{CO}$ and $\ce{N2}$ isomers, though they have about the same mass.

Isomers are not defined by their same mass but by their same chemical formula. So there is no reason at all to think about CO and N2 being isomers solely based on their "about same mass".

In mass spectrometry, ions with the same nominal mass (but possibly differing exact masses) are called isobaric.


text:
Why are hybrid orbitals more efficient at overlapping?
What is the reason for considering that hybrid orbitals provide more effective overlapping resulting in formation of stronger bonds?
Will this reasoning be correct that the hybrid orbitals have larger and smaller lobes than pure orbitals?

It is not so much about the size of the lobes but more about the direction. Hybrid orbitals usually have an orientation that allows them to overlap with the orbitals of their bonding partner much better, i.e. in a head-on manner.

Hybrid orbitals exist only in our head. It would be wrong to perceive an atom as a kind of amoeba which grows this or that kind of pseudopods (=orbitals) as it sees fit.


text:
Negative hyperconjugation stabilizes tricyclopentane derivate
As far as I know, negative hyperconjugation means, that we have electrons that are transferred from a p- or $\pi$- to a $\sigma^*$-orbital. In the lecture, we analyzed a TMS-substituted tricyclopentane derivate (tricyclo[2.1.0.0$^{2,5}$]-pentane-3-one). Apparently, the first one on the left can't be deprotected even by TBAF or boiling MeLi, but the second one easily. If I could deprotect the first one, we'd obtain a 3° carbanion that would do negative hyperconjugation with the TMS group. Would't this be favourable and should be readily feasible?

The sigma bonding C-Si electrons would feed up the p orbital in carbon, but as it is occupied, a net destabilization will result, as the antibonding orbital produced by the hyperconjugation would be more destabilized than the bonding orbital would be stabilized. There is big energy difference between the p orbital on carbon and the sigma antibonding orbital C-Si for hyperconjugation to take place to a significant extent,


text:
Is water committed to vapor state at boiling point or after?
Quite early into my studies, I learnt about phase changes with this graph.

I’m just going to focus on vaporization/condensation. I learnt that the plateau was caused because all the heat was going into breaking the bonds between the molecules, freeing them from a liquid state to a gaseous state. Thus, no energy was going into raising the kinetic energy of the substance or the temperature.
Later, I learnt about thermodynamics and Gibbs free energy. Spontaneity is dependent on temperature—some processes are spontaneous at some temperatures, and some are not. For a temperature above 100 °C, the boiling of water is spontaneous. For a temperature below 100 °C, the condensation of water is spontaneous. At 100 °C, both states coexist in equilibrium (i.e. $\Delta G = 0$). It is only when you raise/decrease the temperature ever so slightly that water is committed to either state.
I fail to reconcile this with what I learnt. At 100 °C, energy is going into breaking intermolecular bonds. That suggests to me that at 100 °C, water is being committed to vapor. According to the thermodynamic interpretation, at 100 °C, both liquid and gas coexist in equilibrium, and water is not committed to vapor until the temperature exceeds 100 °C.
Can anyone clear up my confusion?

What is that "equilibrium" at 100ºC, really? I'll tell you what: it is when both states coexist. Say, you have 1kg of vapor and 10kg of liquid at 100ºC, and these are in equilibrium. You add some heat, it goes straight into breaking those bonds, so after a while you have 10kg of vapor and 1kg of liquid - and that's still an equilibrium.

While the free energies of water and vapor are the same at 100C, to convert liquid to vapor requires adding the enthalpy of the phase transition (latent heat). So, you have on phase with one (H,S) combination, and a second phase with a different (H,S) combination.

@JonCuster Perhaps it would aid me to see these concepts put in terms of the equation $\Delta G = \Delta H - T\Delta S$. I’m not quite seeing it at the moment.

Start with 100% liquid at 100C. End with 100% vapor at 100C. The free energy of the whole system remains constant. So, $\Delta G$ for the whole system has to equal zero. Yet you put enthalpy in.

At equilibrium, not only are molecular bonds of the liquid breaking to form vapor.  There are also molecular bonds of the vapor reforming to produce liquid.  At equilibrium, the rates of the two processes are equal.  But, if you add heat to the liquid, the rate of breaking bonds exceeds to rate of reforming bonds (i.e., the temperature and vapor pressure of the liquid increases slightly), and some of the liquid changes to vapor.  But once you stop adding heat, equilibrium is re-established.


text:
Synthetic options for selective hydroxylation at terminal position
Are there any synthetic options to add a hydroxyl group at the terminal position to the bicyclo compound below? What might be reactions worth trying, any idea? Or is this a lost cause?

Or is this a lost cause?

I don't think so. This looks like a good candidate for an allylic oxidation with selenium dioxide, look here for some examples.

Thanks Klaus! That helps. I will explore that. Would you expect this to form as a side product? http://i.imgur.com/9hDnzzv.png

@curious_cat No, not really. If I remember correctly, the transformation is was regioselective on geraniol derivatives, but the yields differed a lot, depending on the kind of derivative (acetate, sulfone, thiophenyl ether) used. You might want to run a search with  the keywords Sharpless allylic oxidation, selenium dioxide, and geranyl acetate to find some more reliable references. Quite often, the reaction is performed with catalytic amounts of $\ce{SeO2}$ and tert.-butyl hydroperoxide.


text:
How does Drain opener decompose?
I used drain opener (sodium hydroxide and sodium hypochlorite) in an RV and accidentally caused it to drain out (mixed about 1:10 with water) onto concrete below.
There is a fairly strong smell.
How quickly will these chemicals decompose?
What will they decompose into?
How fast will the smell go away?

Was this liquid drainer cleaner or solid stuff? I'm guessing it's in solution, but want to make sure because Crystal Drano also includes aluminum metal which would change the answer.

The sodium hydroxide will not decompose appreciably, and leaves an area of concrete dangerously caustic. A person or pet walking across it could suffer serious burns.
You can neutralize the alkali with a mild acid such as vinegar or citric acid. Vinegar is only 5% acetic acid, so you'd need about 20 times as much as the amount of drain cleaner. Too much acid will attack concrete, so use pH indicator strips (e.g. for fish tanks or swimming pools) to adjust pH to neutral (about 7).
During the neutralization, a bit of chlorine gas will be driven off from the hypochlorite, but that smell should no last long. Any other smell would be from waste in the drain... that might last quite a while.

The sodium hydroxide will not decompose appreciably, || It will catch CO2 from air over time. While not strictly 'decomposition', this still renders it less harmful. And advising to acidify hidrocloride solution... no. Just, no.

There is little hypochlorite in drain cleaner, not "hidrocloride". A little chlorine gas may be given off, but not much.

@permeakra Could always use a little sulfite or bisulfite, right? Slightly harder to come by, though. Metabisulfite is easy if there's a wine store nearby.

True, or any other mild acid or acid salt. Vinegar was suggested because it's relatively cheap and readily available. Carbonated water (seltzer, tonic water etc.) would also work well, and have the virtue of no smell (an excess of vinegar would leave the floor smelling like pickles). It is so weak, though, that a large amount might be needed.

I was thinking some sort of protic reductant would keep the (small amount) of chlorine from forming while also neutralizing the base.

Medical stores may have some thiosulphate, which should work nicely on neutralizing active chlorine. And no, acidifying bleach (hypochlorite solution) is a cheap lab-grade way of obtaining chlorine. It may be much, and, without proper ventilation, it may cause serious chemical burns of respiratory system. Bad idea. Actually, acetic acid plus chlorine is doubly bad idea, because chloroacetic acid. Suggest oxalic acid at the very least - it is both a reducing agent and a mild acid. Citric acid is not safe in this setup, unfortunately.

@permeakra I had considered thiosulfate, but the last time I used it to quench iodine, I got a bunch of what I assume was sulfur in my workup. And I'm still bitter.


text:
When can we add multiple reactions to compute yield of products in an equilibrium?
It is widely known that we can add multiple equilibrium reactions to obtain a net reaction and a corresponding equilibrium constant for the net reaction. However, I am not sure of exactly when this method can be applied. For instance consider a diiprotic acid $\ce{H_2A}$
$$\ce{H_2A <=> HA^{-} + H^{+}} $$
$$\ce{HA^{-} <=> A^{2-} + H^{+}} $$
with hypothetical first and second ionization constants $10^{-5}$ and $10^{-10}$ respectively. Note that $K_{a_1} \gg K_{a_2}$ as is the case for real acids. If we start with $0.1~\rm M $ solution of the acid then the pH is straightforward to calculate since $\ce{H^{+}}$ from the second ionization is negligible we can write
$$\frac{[\ce{H^{+}}][\ce{HA^{-}}]}{[\ce{H_2A}]} = 10^{-5}$$
Since the concentration of the terms in the numerator are equal and the concentration of the un-dissociated acid remains approx constant we get
$$[\ce{H^{+}}] = 10^{-3} ~\mathrm M$$
If instead we take the net reaction with equilibrium constant $K_{a_1}\cdot K_{a_2} = 10^{-15}$
$$\ce{H_2A <=> A^{2-} + 2H^{+}}$$
and write the equation ($x$ is amount of acid dissociated):
$$\frac{(x)(2x)^2}{0.1-x} = 10^{-15}$$
This gives
$$2x = [\ce{H^{+}}] \approx 5.8 \times 10^{-6}$$
which is quite different from the result obtained using the first method.
Note that this is just one of the many cases I have encountered in which adding equations does not work.
Why does adding the equations not work in this case? Also, in general, when is adding equations to figure out equilibrium concentrations justified?

When you combine the reactions and the equilibrium constants like you do, you're calculating the amount of reaction which goes all the way to completion (the fraction which reacts in both steps).
The complication that you're seeing is that the protons are not just coming from the $\ce{H_2A}$ which completely dissociates, but also a (rather substantial) portion of the molecule which undergoes only partial dissociation.
To see this, think about what happens when the hydrogens in the molecule are distinguishable: for example, the dissociation of $\ce{HDA}$ (hemi-deuterated compound), where the reactions are
$$\ce{HDA <=> DA^{-} + H^{+}} $$
$$\ce{DA^{-} <=> A^{2-} + D^{+}} $$
(in reality this wouldn't be the obligate order, but just assume it is for a thought experiment).
The $10^{-15}$ equilibrium constant is only applicable for the formation of $\ce{D^{+}}$ - you still have the rest of the $\ce{H^{+}}$ from the first step, which will also contribute to the pH.
So your assumption in your last step that $\ce{[H^{+}] = 2 [A^{2-}]}$ is incorrect, thus resulting in an incorrect result.

Further explanation
Let's actually work through the $\ce{HDA}$ case, assuming (incorrectly) that the dissociation has an obligate order. You start off with a certain amount of $\ce{HDA}$ ($T = 0.1M$). Now, imagine that at equilibrium we get $x$ total turnovers for the first reaction. A certain amount of $\ce{DA^-}$ (call it $y$) is then going to go and dissociate further. So in the end at equilibrium, we have the following concentrations:
$$\ce{[HDA]} = T-x$$
$$\ce{[H^+]} = x$$
$$\ce{[DA^-]} = x-y$$
$$\ce{[D^+]} = y$$
$$\ce{[A^{2-}]} = y$$
Now we can set up our equilibrium equations:
$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} = \frac{(x)(x-y)}{T-x} = 10^{-5} M$$
$$\frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HD^{-}}]} = \frac{(y)(y)}{x-y}= 10^{-10} M$$
We have a system of two equations with two unknowns. We can thus solve for $x$ and $y$ and get $x=0.995\cdot10^{-3} M$ and $y = 3.15\cdot10^{-7} M$.
Now, let's do the combined equation. First off, we can see where it comes from:
$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} \cdot \frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{DA^{-}}]} = \frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = 10^{-5} M \cdot 10^{-10} M = 10^{-15} M^2$$
So we set up the equation:
$$\frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = \frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$
Um ... we have two unknowns, but only a single equation. We can't solve this without additional information. We can't assume that $[\ce{H^{+}}] = [\ce{D^{+}}]$, or that all the $\ce{HDA}$ which disappeared went completely to $\ce{A^{2-}}$ - some of it can (has to) form $\ce{DA^-}$ which doesn't react further. The equation is valid (if you substitute $x=0.995\cdot10^{-3} M$ into the equation, as determined above, you'll get the correct answer for $y$), but only if we have some way of knowing how much of the $\ce{HDA}$ reacted and didn't go completely to completion.
The picture is further complicated because we're not dealing with distinguishable $\ce{H^+}$ and $\ce{D^+}$ which dissociate in an obligate order, but a single species which is involved in both reactions. That is, instead of
$$\frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$
we would have
$$\frac{(x+y)(x+y)(y)}{(T-x)} = 10^{-15} M^2$$
Again, the combined equilibrium constant is perfectly correct, and accurately expresses the relationship between the concentrations at equilibrium. It's just that the additional assumptions about the relationships between the concentrations that you're making are not. Specifically, $[\ce{H^{+}}] \neq 2 \cdot [\ce{A^{2-}}]$. Note this isn't particularly specific to this sort of combined equilibrium. Any additional reaction you're not considering can change the concentrations and make your assumptions less than correct - not just formation of intermediates which don't go all the way to completion to side reactions that draw off substrates, reactants or products into other forms.
The equilibrium constant tells you the relationship between the species concentrations, but the concentrations as they actually are in solution, not what the reaction generates. The latter is only true when the reaction in question is the only one under consideration.

But the net reaction accounts for $H^{+}$ from both ionizations. Indeed, the net reaction in case of your hemi-deuterated compound will be $\ce{HDA <=> A^{2-} + D^{+} + H^{+}}$ ($K_a = 10^{-15}$) which includes both the proton and the deuteron.

@Gerard - Right, there's a contribution from both reactions, but only a small fraction of the starting material will go through both reactions. There's a much larger fraction which will only go through the first reaction and stop. That's not accounted for in the combined reaction rate constant, but it still contributes to $\ce{[H^{+}]}$

I can understand what you mean intuitively. However, I don't see where the mathematics goes wrong. In adding the equations we have nowehrer

@Gerard I expanded things greatly, I hope that addresses your concerns.


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Wetting properties of FTO coated glass
I am trying to spin coat a layer of polymer+Quantum Dots on an FTO coated glass slide. My material is forming agglomerates at the edges of my slide. My polymer is PVK which has a conc of$\approx 4.5mg/mL-4.25mg/mL$ while I am coating at only 500 RPM and I am still unable to form a film. Does anybody has any experience about forming films on FTO coated glass, or is it generally very reluctant to allow films to form on top of itself. Or is my polymer concentration too low?

There will be some agglomerates at edges of any slide using spin-coater. What does it mean that " I am still unable to form a film."? Could you specify  your results?

@JaroslavKotowski Yes, thank you very much for taking an interest, I measured an AFM image of the film and I saw non-uniformities and extreme roughnesses to the range of 60nm and was surprised and then did an AFM of the bare FTO slide and found that the roughness was almost similar to that of a bare slide, but since the AFM image is generally done only over a few microns of the surface, I couldn't/didn't do the entire sample but when seen with the naked eye by tilting the sample sideways, I could see a film along the edges of the sample.

Arising sidewalls after spin-coating is a quite common problem. I think that following articles could be beneficial for you: http://www.sciencedirect.com/science/article/pii/S1566119911003879
http://pubs.acs.org/doi/abs/10.1021/nl025750g


text:
Finding the Yield
In the experiment of Synthesizing t-Butyl Chloride, if you were to obtain 12g of t-butyl chloride what would the yield of the reaction be?
Amount of alcohol in the experiment is 17mL (density 0.775g/mL-1)
Molecular weight of t-Butyl alcohol = 74.12g/mol
Molecular weight of t-Butyl chloride = 92.57g/mol
I'm really confused on how to go about calculating this as the only thing I've learnt about calculating percentage yield is (experimental value/ theoretical value) x 100%.
Help would be greatly appreciated. Thank you very much!

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

You're having all the information that you need.

Amount of alcohol in the experiment is 17 mL (density 0.775 g/mL)

From volume and density, the mass can be can calculated
From mass and the molar mass of the alcohol, the amount of substance (in mol) is available.
This will define the maximal outcome of your reaction (100% yield).
Calculate the amount of substance for 12 g of the butyl chloride.

[…] yield is (experimental value/ theoretical value) x 100%.

Exactly :)

I ended up getting 73% for the yield :) Thank you so much for the help Klaus, I really appreciate it!


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Order of Bonding in Nonaqeuous Hydrated Salts
In the chemical formula of a hydrated salt, such as $\ce{CuSO_4.5H_2O}$, is it implied from the format of the name that the the Copper (II) ion is ionically bonded to the sulfate, with water molecules surrounding both, or that the Copper (II) ion forms a coordination complex with 6 water molecules and the complex ion bonds with the sulfate ion?
I think that in the case of copper (II) sulfate, there indeed is a complex ion formed, otherwise, how could there be the blue color? I thought that the convention for this kind of nomenclature was to use the dot for the first case, and then use $\ce{[Cu(H_2O)_5]SO_4}$ when there is a complex ion involved. Thus, isn't the better name for Copper (II) sulfate, $\ce{[Cu(H_2O)_5]SO_4}$?
*Update: It seems from the comments that in this case, Copper (II) sulfate is better described by the second naming type. Are there any transition metal salts that fit the first description: where it is a normal ionic bond except with water surrounding the cation and anion?

related http://chemistry.stackexchange.com/questions/38864/hydrates-in-aqueous-solutions

As for the name, the current one is better, for people are used to it; but if you want your formula to be as factually correct as possible, write it as $\ce{[Cu(H_2O)_4]SO_4.H2O}$ (which nobody normally writes like that, but... whatever).

The bond in $\ce{CuSO_4.5H_2O}$, if anything, is indeed a normal ionic bond.


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Solution , Solubility and Colligative Property?
Which of the following  0.1 M aqueous solution will have the lowest freezing point ?
1) Glucose
2) Urea
3) NaCl
4) K2SO4
How to relate Molarity with freezing point ?
Should I use the relation
Kf = MR(Tf)2/1000Hv

Let's say we have some ice.

Beautiful structure. Ice stays solid at some temperatures because of its neat, tight lattice. Ice stacks this over and over until there's no more ice left to stack.
If I were to put a lot of ions between this stack, the stack might bend or even break; it isn't as reinforcing as it once was. The stack isn't as durable anymore, so we would expect it to melt a lot easier - there's too much structural stress for it to be a solid.
The freezing point of solutions depends, at least in this case, on how much dissociation occurs. If we can push more ions into solution, it's going to have a harder time making a sturdy lattice - a harder time freezing.
$\ce{C6H6O6 -> C6H6O6}$ It doesn't dissociate at all.
I'm not sure how much urea dissociates, but I assume it would be like glucose.
$\ce{NaCl -> Na+ + Cl-}$ It dissociates into 2 ions.
$\ce{K2SO4 -> K+ + K+ + SO4^{2-}}$ It dissociates into 3 ions.
$\ce{K2SO4}$ dissociates the most, so it's freezing point is lowest. Hope this helps!


text:
Relative Solubilities of Salts in Acidic Solution
I stumbled across this problem which I don't understand the answer to:
Which solid is much more soluble in 1 M $\ce{HCl}$ than in $\ce{H2O}$?
(A) $\ce{CaHPO_4}$
(B) $\ce{CaCl_2}$
(C) $\ce{BaBr2}$
(D) $\ce{BaSO4}$

Surely that must be (A). $\ce{HCl}$ would turn it into the more soluble $\ce{Ca(H2PO4)2}$.

How is that more soluble than CaHPO4?

It's an experimental fact. You want to know how can we predict it just by looking at the formula? That can't be done reliably.

@IvanNeretin why not H3PO4 ? why would it stop at CaHPO4?

Check the answer in the back. But I would say (A), because hydrogen phosphate is the strongest base out of the other anions. If it is a strong base, that means that it is much favored to form dihydrogen phosphate, which shifts the equilibrium further to the right.

I did, but I don't understand why (A) is the answer.

@DavePhD You are right, the acid would drive it all the way to $\ce{CaCl2}$. Thus it would suffice to know just two facts: good solubility of $\ce{CaCl2}$ and not-so-good of $\ce{CaHPO4}$. Can these be explained in a rigorous way, without memorization? I don't think so.

@IvanNeretin so you disagree with my answer?

@DavePhD No, I agree quite completely, but your answer is still based upon the fact (known to any chemist, of course) that $\ce{CaHPO4}$ is poorly soluble, while $\ce{CaCl2}$ is well soluble. How did you know that?

@IvanNeretin I agree, you would need to know that

First, an example: Magnesium hydroxide, $\ce{Mg(OH)_2}$, is much more soluble in acid than pure water. This is because when it dissolves in pure water, $\ce{Mg(OH)_2 (s)<-> Mg^{2+} (aq) + 2OH^- (aq)}$, with only a little bit of the aqeuous products. ($\ce{Mg(OH)_2}$ is "insoluble".) However, if there is quite a bit of $\ce{H^+}$, then the $\ce{OH^-}$, which is a strong base, will react with $\ce{H^+}$ and form something else (in this case $\ce{H_2O}$). This is akin to taking out some $\ce{OH^-}$, and by Le-Chatelier's principle, will shift the equilibrium right, and you can dissolve more $\ce{Mg(OH)_2}$ in.
In your problem, $\ce{Cl^-}$ and $\ce{Br^-}$ are both exceedingly weak bases, so they don't react with $\ce{H^+}$ to form something else. (HCl and HBr are strong acids). Next, I know that $\ce{SO_4^{2-}}$ is a noticeable base (it will form $\ce{HSO_4^{-}}$, not a strong acid), and so is $\ce{HPO_4^{2-}}$ (will form $\ce{H_2PO_4^{-}}$, not a strong acid), so they are two potential options. Now we just need to figure out which base is stronger than the other, or which conjugate acid is weaker than the other. There is prior knowledge involved now: because $\ce{HSO_4^{-}}$ is a stronger acid than $\ce{HPO_4^{2-}}$, then the base of $\ce{H_2PO_4^{-}}$, which is $\ce{HPO_4^{-}}$, is the stronger base than $\ce{SO_4^{2-}}$, (and consequently all of the other ones). The ionic compound containing the stronger base is more soluble in acidic solution, and vice versa for basic solution. So (A)

Chloride, Bromide, and Sulfate all correspond to strong acids, while phosphate corresponds to a weak acid.
With 1M H+ available, phosphate will be essentially fully protonated, while the others will still be anions.


text:
Intuition on Gibbs Free Energy
In many places, I read that Gibbs Free Energy is called "free energy" because some of the enthalpy that comes from the chemical reaction becomes "waste heat" due to a change in the entropy of the system, and only the rest is usable as work. However, I cannot imagine how this is the case. Take an exothermic reaction with a positive change in entropy for example. Because $\Delta G = \Delta H - T \Delta S$, then we find that $\Delta G$ is more negative than $\Delta H.$ That is, more energy is released by the system than the enthalpy of reaction. How does this fit in with the "waste heat" explanation, that not all of the energy released is usable? There is more energy released now.

From Second Law of Thermodynamics viz. $$\Delta S_\textrm{sys}+ \Delta S_\textrm{res}\geq 0\,,$$ it can be concluded that $$W_\textrm{sys}\leq (U_1- U_2)- T(S_1 -S_2) \tag{1}\;.$$
Now, $$W_\textrm{sys}= \underbrace{W_T}_\textrm{pressure-volume/compression work}+ \underbrace{A_T}_\textrm{non-compression work}\;.$$
Putting the the first term on the RHS of $(1)\,,$ it yields,
\begin{align}A_T &\leq (U_1- U_2)- T(S_1 -S_2) - W_T\\ &\leq (U_1- U_2)- T(S_1 -S_2)+ P(V_1 -V_2)\;.\end{align}
Now, this is equal to the negative change of Gibbs energy $G$ provided the thermodynamic process is done at constant pressure and temperature.
So, that means,  $$A_T \leq G_1 -G_2\;.$$ Note the $\leq$ sign; this implies $(G_1-G_2)$ is the maximum value $A_T$ can attain.
Since, $\Delta G \lt 0\;,$ this implies $A_T$ is positive. Now, if the process is reversible, then the non-compression work $A_T$ is maximum viz: $$A_T= G_1 -G_2\;.$$
This definitely supports the term free energy as the decrease in $G$ gives the amount of energy "that can be 'freed' and made available for work".

That is, more energy is released by the system than the enthalpy of reaction.

Of course, the extra energy came from heat energy absorbed from the environment as the entropy of the products was greater than that of the reactants by $\Delta S\;.$

Free energy is the energy free to do work in the system.  One way for the entropy change of the system to be positive is for the products to have more moles of gas than the reactants.  The gas could be made to do work in a piston or turbine (or just push the atmosphere out of the way), allowing the system to release more energy than the heat released by the system.

The "heat tax" for an exothermic reaction comes from heating up the surroundings. Thermal energy can be converted to useful energy (e.g. heating up water to drive a steam turbine to generate electricity), but this can never be 100% efficient.
Research the Carnot cycle - a theoretical heat engine that gives the limit of efficiency in converting heat to work. Even assuming the impossible - perfect thermal isolation of components, no friction, etc. the efficiency of energy conversion is always <100%.


text:
How oil store all this power
I was software engenneering student at college, i'm so out from the chemistry domain, but in high school we studied physics and chemistry, and now i have more than 6 years i didn't study about chimestry nor physics, so i have two questions.
1) I wondred from a thing, for ex: in cars, usualy we mesure the power in hp, which is equivalent to 745,7 watt, imagining a car engine which produces 100hp so it produces around 74570 watt, this is huge power, alghout we many lose many of power produced by the gasoline as a heat, so how much hydrocarbons free power??!!
2) How does the gasoline store all this power?!!! in other words, what does happens exactly to gasoline when burning/exploding to give this huge power??
these just a general questions, in this life i really like to understand how things work, even if they're so out for me and don't mean anything, but it still funny to understand how things works, and if i'm understanding something falsy or anything, just let me know it :)

related http://chemistry.stackexchange.com/questions/3947/what-is-the-nature-of-chemical-energy

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. It appears to me you are in general asking about the combustion of hydrocarbons, i.e. fuel. This is a bit too broad for this site. Please try to narrow down the scope of your question to a specific problem you have with the process. You might want to start reading some Wikipedia article.

In chemical reactions, bonds are formed, and these involve significant amounts of energy. For example, when you burn octane in oxygen, $\ce{2C_8H_18 + 25O2 → 16CO_2 + 18H_2O}$, it releases 48 kJ per gram of octane. Energy changes accompany chemical reactions because bonds are being broken and new ones are formed. It is energy releasing to form a bond, and the stronger a bond is, the more energy is released. For hydrocarbons, the carbon to hydrogen bonds are quite weak, so when they form much stronger bonds, as in carbon dioxide and water, a lot of energy comes out.
Sometimes if you take a chemical reaction, it absorbs heat because the old bonds were stronger.


text:
Best oxidizing and reducing agents: Na, Zn^2+, Ba, Ba^2+, and Ag?
Which is the best oxidizing agent and the best reducing agent from the species listed below?
$\ce{Na,~Zn^2+,~Ba,~Ba^2+,~Ag}$
I determined that oxidizing agent as $\ce{Zn^2+}$ because it appears on the table with the most positive reduction potential and determined that $\ce{Ba^2+}$ as the best reducing agent because it has the most negative reduction potential. I am not sure of my answer.

Welcome to Chemistry.SE! This appears to be a homework question. Please share your thoughts and attempts towards the solution, otherwise your question may be closed.

I added my attempt to the question.

You have done good job.

Not that good, how you could get second wrong, while having this chart?

The chart says that reducing fluorine $\ce{F2}$ requires very little work, while reducing lithium cation $\ce{Li+}$ requires a lot of work.

Also keep in mind that a volt is really a unit of energy per mole of electrons. (OK not actually per mole but per 1/96500th of a mole...) The annoying habit of electrochemists is to write stoichiometries that involve multiple electrons but to use energy units that are normalized per-electron.  This is in stark contrast to all other areas of chemistry, where a unit like "kJ/mol" is understood to refer to a mole of reaction.  Thus to answer your question, it is irrelevant to consider the valency or charge state of any reactants/products.

It is a good idea to recap the the definitions of oxidation and reduction when approaching this kind of question, and also what it means to be an oxidising or a reducing agent:

Oxidising is defined as the loss of electrons, reduction as the gain.
An oxidising agent is good at oxidising other compounds while being reduced itself; and vice-versa.

Take another quick look at your list; all you have are neutral elements or the most common cations of said elements. If you want to oxidise or reduce, then we would always be going back and forth between either. Therefore, only sodium, barium and silver are potential reducing agents on your list (being oxidiseable themselves) while only zinc(II) and barium(II) can be oxidising agents. Here are the relevant equations:
$$\begin{align}\textbf{Reducing agents:}\\
&&\ce{Na &-> Na+ + e-}\\
&&\ce{Ag &-> Ag+ + e-}\\
&&\ce{Ba &-> Ba^2+ + 2 e-}\\
\textbf{Oxidising agents:}\\
&&\ce{Ba^2+ + 2 e- &-> Ba}\\
&&\ce{Zn^2+ + 2 e- &-> Zn}\end{align}$$
Using your table or our chemical intuition, we can then decide which reduction or oxidation is the easiest. The best oxidising agent without doubt is zinc(II) as it is much more noble than barium (although still not a noble metal). The best reducing agent is a slightly more tricky call without data, but barium(0) being further down in the periodic table allows us to assume it to be stronger. Luckily, this is supported by the data.

You have determined the correct oxidizing agent. It does however appear that you have determined the wrong reducing agent (if it is not a typo). $\ce{Ba^2+}$ is a poor reducing agent because it requires a lot of energy to remove a third electron from Barium (as it breaks the noble gas electron structure of $\ce{Ba^2+}$). The correct answer for reducing agent is barium (Ba), as can be read from the table you provided. Barium releases a lot of energy when oxidized to $\ce{Ba^2+}$ ($2.9~\mathrm{eV}$ per elevtron). This means it will easily undergo oxidation => strong reducing agent.


text:
How does metal ions assist oxidative rancidity?
Oxidative rancidity is defined as the oxidation of unsaturated lipids and the carbon-carbon double bond with the presence of typically oxygen gas. Sometimes aldehydes and ketones are released as products.
How does the presence of metal ions catalyze this reaction? What are, if any exist, the reaction intermediates?
EDIT:
Here are some of the things I thought about:
The reason why light favors the oxidative rancidity of lipids while temperature does not, (at least not as far as I know) may be an indication that the reaction can not occur with out ionized oxygen atoms, which photolysis of oxygen may produce. This will also explain why simply increasing the temperature will not work, since increase in temperature does not lead to an increase in oxygen ions. The presence of metal ions would than be hypothesized to assist the oxidative rancidity of lipids by helping with the formation of oxygen ions. However, that calls in question whether or not should the metal be ionized to start with, because if the metal ions were cations to begin with, say it is $Na^+$ from table salt, how would that increase the concentration of oxygen ions?

Whether or not this is actually homework, I think it fits the definition as far as Chem.SE is concerned, so you'll want to show your own thoughts and effort toward answering the question yourself.


text:
Cold fusion via chemical reactions?
How could a nuclear reaction possibly happen at room temperature? We know how other blasts work and what chemicals are used to cause their effect, but what chemicals could cause a nuclear reaction at room temperature or containable? It sounds like it would be the opposite of what causes them to go boom!

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I don't understand your question. There is currently no acceptable model that would allow cold fusion to happen. See also wikipedia. Also this question seems not to be about chemistry and is likely to be closed.

Cold fusion has been debunked.

A nuclear reaction (alpha decay, for instance) can perfectly well happen at room temperature, as well as at any other. But if you are specifically interested in fusion, it can't, and no chemicals can help.

There is a rigorously proven way to achieve measurable fusion of atomic nuclei at room temperature and pressure: muon-catalyzed fusion. Unfortunately the details of our Universe are such that this cannot be used to obtain virtually limitless energy. Oh well.

With an accelerator, one can certainly generate fusion or other nuclear reactions in atoms at room temperature. Of course, that is kind of cheating, since you are sending in (mono)energetic particles, but they are not a thermal distribution.

No chemical reaction can have any significant influence on the rate of nuclear reactions. The valence electrons have such a low probability of being actually in the atomic nuceus (zero for p orbitals, and virtually zero for s,.. orbitals, because the core is so small) that no interaction is observed. It a different story for inner electrons of heavy elements, but those are not much influenced by chemistry.
I'm not sure if a tiny variation of the half life of certain decays is measurable for different chemical states. But to really change something, let alone cause a fusion reaction, is totally impossible.
(Update: The website reference by Ian below gives biggest changes of halflife of less than one percent for Be-7, and less for a few other cases. As could be expected, electron capture decays are most sensitive. Removing inner electrons of heavy elements can make huge changes.)

Variations in half lives as a function of chemical state have been observed, see http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/decay_rates.html for a few examples


text:
Is micelle formation in a soap solution a chemical reaction?
I know that soap molecules have a hydrophilic end and a hydrophobic end which attract to water and oil respectively. However, does a chemical reaction take place?

Chemical 'reactions' are normally defined by changes in 'bonding'.  Related to soap micelles, for the micelle to exist and have integrity enough to be characterized from a situation where they did not previously exist, some 'bonding' must come into being.  That bonding is normally described as 'London Forces' or 'Vander Waals Bonds' or 'Hydrogen bonding' or 'electrostatic' interactions.  Since equivalent concentrations produce equivalent results, the micelle formation is also reproducible.  In the purest sense this does imply that soap micelle formation is a chemical reaction.
The question is more philosophical than anything.


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How do we find the angular, radial nodes and the quantum numbers when the radial probability distribution curve's equation is given?
I am having serious conceptual issues with the Schrödinger equations. Please note that I only have basic school level (11th grade) knowledge of these equations and we are expected to know very little about it. But we are given these type of questions so I guess I will be able to understand things within the boundary of this question.
$$\psi_r = k_1 * e^{-r/k_2} * (r^2 - 5k_3r + 6k_3^2)$$
Solving the equation we get $r = 3 k_3 , 2 k_3$
It is also given that $k_3 =1$ . So $r = 3, 2$

Now, since there are 2 nodes therefore, $n = 3$ . Does the equation $\psi_r$ represents only radial nodes or all nodes?
What is the azimuthal quantum number?
How many angular nodes are present?
What is the angular momentum of the given orbital?
Is this equation of an atom or an orbital?

This equation was given as an MCQ question and we were asked to find the quantum states of the given orbital on the basis of the equation. But I guess we must have conceptual insight too to answer the question so asked it here.

We can't really tell much about angular nodes. The angular part is simply not there. We have no idea what it might be.

For a given orbital, there are two types of modes: radial nodes and angular nodes.

The total number of nodes is $n-1$
Number of angular nodes is $l$
Number of radial nodes is $n-l-1$

In this question: If the given function represents the radial part of the wave function of an atomic orbital, then we get the number of radial nodes: $n-l-1=2$. i.e.  $n-l=3$:
We have several possibilities:

$n=3$, $l=0$: then we have the orbital $\ce{3s}$
$n=4$, $l=1$: then we have the orbital $\ce{4p}$
$n=5$, $l=2$: then we have the orbital $\ce{5d}$

and so on.

Your equation is a wavefunction. Simply put, a wavefunction is a description of the state of a particle(s). In most cases these are electrons (at least in chemistry). Wavefunctions describe particles not atoms. There are many-particle particle wavefunctions. In other words, wavefunctions can describe the state of more than one particle. An orbital, on the other hand, can only describe the state of one particle (but some people say two at most). We obtain the state by solving the Schrodinger equation. For electrons in atoms, there are these things called quantum numbers which specify the state of the electron. These are the n,l,m,s that is talked about in introductory chemistry classes: https://en.wikipedia.org/wiki/Quantum_number
The angular momentum as well as other properties can be calculated by looking at the quantum numbers (in wikipedia, you'll find that they list which quantum number is related to the magnitude of the angular momentum).
More specifically, the wavefunction gives information on the probability of finding  a particle in a certain state. Nodes, are places where particles can't be located. The probability of finding particle in a node is 0. Thus, wavefunction describing an electron with a principal quantum number 3 (the "radial" part) would be "aware" of the nodes. Otherwise, it wouldn't be a valid description for the probability of finding an electron. Look, at your equation, when will the wave function be zero? I am not sure how many angular nodes are present, since I am not familiar with wavefunction that you are providing. I think it may be helpful if you gave us the context in which you found the equation. However, I think that your wave function has no angular part.

Actually this was given as an mcq question and we were asked to find the quantum states , I don't know whether this wavefunction is real or hypothetical :/

@brainst then it only tells you information on the r part unless k1 is a function of the angular part

So in short if $\psi_r$ is given then it would provide only the information of radial part and if $\psi (R,\theta)$ is given then it would provide the information about the angular part as well ?

@brainst Also try to show some effort. Show what you attempted, what you understand.Otherwise people may not be too positive about your questions.

Actually I thought that knowing that the probability goes zero at node and solving the quadratic was the atmost that I was able to do and I did shown that in the answer , other all is conceptual which I had doubt about (I wasn't sure whether the two nodes that I found was all or there were angular nodes too) , and thats what  the question is about ... Is it not explained well?

@brainst It has no angular term. In fact, there is no other dependece other than r I think


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How much energy is converted in light when you burn a candle?
When a candle burns the reaction is C25H52 + 38 O2 → 25 CO2 + 26 H2O. So it seems equal at both sides. But isn't there any light and heat (low elektromagnetic waves) produced. So how many Joule is produced of light of each candle molecule?

It is not possible to deduce a value from this information.

Couldn't there be made an aproximation or what is needed more to know to get a value?

An ordinary candle has a luminous intenity of about 1 cd (candela). You can convert that to a power (assuming a median wavelenght), multiply it with the time your candle burns, and compare with the delta H in your reaction.

See How to Find the Molar Heat of a Combustion Candle for a detailed explanation and answer.


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Will the addition of salt further cool down ice that is already at -30 °C?
I know that ice gets colder when salt is added. But how big is the effect if it is already −30 °C?
Can it cool it down more or does it just work with −1 °C cold ice?
Another question: If I have a water cooling circuit, does adding salt help it cool down?

(1) At -30°C, common table salt ($\ce{NaCl}$) wouldn't do anything to ice; $\ce{CaCl2}$ probably would. (2) No.

If you add salt to icy water, it's the water that gets colder. Pure water under atmospheric pressures will be $0~\mathrm{^\circ C}$ - even when the ice melting in it is colder (for example if it's just out of the fridge). When you then add salt to that water, it lowers the freezing point of your water-salt solution, all the way down to $-21~\mathrm{^\circ C}$ once it contains $5.2\ \mathrm{mol}$ of NaCl per kg of water. That's the maximum amount of salt this solution can contain at that temperature. (Find the thermophysical data at NIST (page 30).
Now, if you add salt to liquid water (containing no ice), you will actually decrease its temperature slightly because of the energy required to solvate the salt in the water. This won't help you in a 'water cooling circuit' because you also lower the heat capacity.

thanks for the asnwers it helped me a lot. i wont use salt water then.