text:
What is the correct way to calculate the rate constant for the reaction of NO and O2?
The following data are given for the reaction of $\ce{NO}$ and $\ce{O2}$:
$$
\ce{2NO + O2  ->  2NO2}
$$
The the reaction is second  order  in $\ce{[NO]}$ and first order in $\ce{[O2]}$, and the rate of disappearance of $\ce{NO}$ is $2.5 \times 10^{-5}~\mathrm{mol\over L\,s}$ at the instant when $\ce{[NO] = [O2]} = 0.01~\mathrm{mol\over L}$.
The question asks me to calculate the rate constant.
I've thought of two ways of approaching the calculation—which of these solutions is correct?
1) Take the rate of the reaction as one-half the rate of disappearance of $\ce{NO}$:
$$
\begin{align}
R &= {1\over 2} * 2.5 \times 10^{-5} = k \ce{[NO]^2[O2]}=k(0.01)^3 \\
k &= 12.5~\mathrm{L^2\over mol^2\,s}
\end{align}
$$
2) Take the rate of the reaction as equal to the rate of disappearance of $\ce{NO}$:
$$
\begin{align}
R &=  2.5 \times 10^{-5} = k \ce{[NO]^2[O2]} = k(0.01)^3 \\
k &= 25 ~\mathrm{L^2\over mol^2\,s}
\end{align}
$$

Either definition is acceptable, as long as you clearly identify relationship between the calculated rate value $R$ and the rates-of-change for each of the species in the system.
In case (1), you've effectively calculated the rate constant defined as:
$$
R_1 = k_1\ce{[NO]^2[O2]}
    = -{\mathrm d \ce{[O2]}\over \mathrm dt}
    = -{1\over 2}{\mathrm d \ce{[NO]}\over \mathrm dt}
    = {1\over 2}{\mathrm d \ce{[NO2]}\over \mathrm dt}
$$
In case (2), you've instead calculated the rate constant defined as:
$$
R_2 = k_2 \ce{[NO^2][O2]}
    = -2{\mathrm d \ce{[O2]}\over \mathrm dt}
    = -{\mathrm d \ce{[NO]}\over \mathrm dt}
    = {\mathrm d \ce{[NO2]}\over \mathrm dt}
$$
Both cases accurately describe the kinetics, but you have to be careful to identify which definition of the rate you've used in calculating $k$.

Which value of the rate constant (k1 = 12.5 or k2 = 25) used in calculating  the rate at the instant when [NO] = 0.015 mol/L and [O2]  = 0.005 mol/L.with explaning?

@AdnanAL-Amleh There's really not a lot more I can say. The correct answer depends on which of the two rate relationships the person who wrote the problem had in mind. If you don't have any insight into their thinking, then there's no way to know which one they want.

the problem adapted and edited from : Kotz J.C.,TriekelTreichel P.M., and Weaver G.C., Chemistry and Chemical Reactivity .6th Edition.
The Answers :
K= 12.5 M^-2 S^-1  and THE Rate at the instant when [NO] = 0.015 mol/L and [O2] = 0.005 mol/L=-1.4 ×10^-5 M/sec  – Adnan AL-Amleh 20 hours ago

@AdnanAL-Amleh Well, that would've been useful information to include in the question! If they give 12.5, then clearly they set their rate up as in my $R_1$.

I am Ambiguous with these answer , because the Rate of  disappearance NO is experimentally determined  with no  consideration  of  coefficient factor in the balance eq.(the factors in eq. useful in calculate the ratios of rates-of-change for each of the species in the reaction. ) and rate formula obtained experimentally , so , I think with the second Definition in your R2. I  want support with feed Back

@AdnanAL-Amleh Reported rate constants are often ambiguous. You're absolutely right that there is indeed only one actual, physical $\mathrm d \ce{[NO]}\over \mathrm dt$ observed in this system. But, there are an infinite number of rate expressions you can write, all related by multiplicative constants, that describe the kinetics equally well. Maybe someone else can help give you more insight, but I don't know if I can assist much beyond what's contained in my answer above.

The IUPAC Gold Book recommends using the definition as in R1 above and this is also what most, if not all textbooks, use.

@porphyrin I also prefer the $R_1$ definition. Many journal articles do not follow this recommendation, however, and it can take quite a bit of work to back out the proper form of the rate expression.

There is no simple correlation between  the stoichiometry of the reaction and the rate law, the value of K in the  rate law must be determined experimentally ,

when we refer to what given in the question ,I think  the rate of the reaction = experimentally determined  value of the rate of disappearance of NO , so  the value  of  rate constant =K2= 25

@AdnanAL-Amleh "There is no simple correlation between the stoichiometry of the reaction and the rate law" I don't know what you mean by this; to my mind, the stoichiometry of the reaction defines the rate law.

@AdnanAL-Amleh "I think the rate of the reaction = experimentally determined value of the rate of disappearance of NO." Like I said, this is a valid choice of the rate expression. But, it is definitely not the only valid choice.


text:
How to liberate propylthiouracil
Two substances undergo a reaction that gives off propylthiouracil and another compound as products.
Is there an additional compound I could throw into that sludgy product to strain off the $\ce{C9H17O6}$ such as that used in the Miller Process, but instead would leave a clear propylthiouracil for my use? The chemical reaction is given below:
$\ce{2 C11H19NO6 + C3H6OS -> C7H10N2O6 + 2 C9H17O6}$

Does it have to be a chemical reaction/destruction like the Miller Process or would something like a separation process be acceptable?  Have you tried anything along these or other lines?

@airhuff Anything at all is OK. As long as the resulting products have different viscosities. As to whether I have tried anything along those lines, none of the compounds I can think leaves the prothiouracil unaffected i.e adding an aqueous sodium hydroxide solution but that breaks the propythiouracil into ammonia, sodium thiocyanate and another compound

According to https://pubchem.ncbi.nlm.nih.gov, propylthiouracil has the following solubility properties:

1 part dissolves in about 900 parts water at 20 deg C, in 100 parts
  boiling water, in 60 parts ethyl alc, in 60 parts acetone; practically
  insol in ether, chloroform, benzene (emphasis mine); freely sol in aq soln of ammonia
  and alkali hydroxides.

I would start with a small amount of your "sludgy" product. Stir it with ether as a first choice, chloroform as a second, for a while and see first of all whether you really get multiple phases or if just forms a less viscous sludge.  In a perfect world, after letting the solution set you would see the white propylthiouracil solid settle out at the bottom, with a sludge-discolored solvent layer on top.  If you see anything resembling this, then I would pursue this technique with more thorough stirring/shaking, maybe centrifuging to get better settling of your product, then further rinsing your product on filter paper with a vacuum flask, etc.  Try the other solvents if available if ether doesn't work.  The success/failure of this approach depends on the solubility of the sludgy stuff in the solvent.  It may fail, it may not give the purity you need, it may work wonderfully, but it is a cheap and simple way to test a method with a reasonable probability of success.

source: Stack Exchange

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sort: votes


text:
Increasing available area of a solid will increase effective concentration?
If I say that making a solid reactant into a powder will increase the yield of the product, because it increases the available area of the solid, hence increasing its effective concentration. Is this claim valid?

It will increase the rate of reaction. What affect that has on the yield depends on what you're making and how.

Doesn't it effectively increase concentration?

I don't think so. If we're talking about yield, then to predict the equilibrium constant, "concentration" should really be thermodynamic activity, and the activity of a solid is usually 1.

The concentration of solid in itself is always the same, like m3wolf said in the comments (unless the solid is not pure, which is not relevant anyway, because grinding wouldn't change that either). So no, you can't change the effective concentration this way.
If anything, you will have higher surface area and hence higher overall rate of reaction (or dissolution, if the reaction happens in the solution and not on the surface, and thus has to be preceded by dissolution). In this manner, you will get the same yield, but in less time.
Secondary effects can eventually change your yield in an unpredictable way, as likely down as up.

I think it's a solid hypothesis that will probably prove to be true.  I mainly base that on the likelihood that more solid reactant will be available to react.  Also, think about the statement in reverse:
"Compressing my finely powdered solid reactant into a single hard clump should increase the yield of my reaction."
Sounds unlikely.
However, I have made some assumptions in saying the hypothesis is probably true.  What if the other reactant is a liquid that thoroughly penetrates the unground solid and is able to react with all of the solid this way?  Then what if the reaction gives better yields when it proceeds slowly?  What if it's highly exothermic and the powdered solid reacts so fast it blows all over the lab?
Of course you can control the rate of reactant addition, cool the reaction, etc. to mitigate these things, but these are just some examples of what could go wrong with the seemingly obvious "true" answer I proposed above.  Still, I am confident your claim will be valid more often than not.

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text:
Can alkyl-substituted phenols be reduced to alkyl-substituted benzenes by zinc dust?
I know that phenol can be reduced to benzene by distilling it with zinc dust. But I want to know if that concept can be expanded. E.g., take chavicol or 4-(prop-2-en-1-yl)phenol; could you still remove the hydroxy group by distilling it with zinc dust?

This is the basic principle of functional groups. Organic chemists quickly noticed historically, that some compounds reacted in the same way when presented with the same chemicals; thus allowing them to deduce that it was only a certain functional group (‘radical’ in historic terminology) that was reacting while the remainder of the molecule remained untouched.
Unless conditions or reagents are known to also attack other functional groups, they will typically be unharmed in a reaction and whichever features your molecule has, it will still react in the desired manner. For example, if phenol gets reduced to benzene by distillation with zinc dust, we can be pretty sure that cresols (e.g. paracresol, 4-methylphenol) will react in the same way, only that the final product will be toluene rather than benzene.
As I am not aware of zinc dust reacting with a double bond — the only other functional group present in chavicol — I am willing to say that it will indeed lose the hydroxy group and be reduced to allylbenzene as you desire.


text:
Molecules having higher enthalpy of vaporization and boiling point than another but with lower entropy of vaporization?
I came across the Trouton's rule that predicts the entropy of vaporization of most molecules to be around 85~88kJ/(K mol). It is said to fail when there is hydrogen bond between molecules.
When I checked the entropy of vaporization for linear alkanes, it increases with molecular mass which looks reasonable to me because a liquid with greater volume should expand more once it transits completely into gas than those with lesser volume. Eventually, the increase does slow down as the values get close to ~90kJ/(K mol).
For linear monohydric alcohols, their entropies of vaporization were higher than that predicted by Trouton's rule which I have found to be explained by their lower entropies in liquid phase due to hydrogen bond. However, I found that the entropy of vaporization of alcohols decreases instead with molecular mass. But since the boiling point and enthalpy of vaporization both increase with molecular mass. Shouldn't their entropy of vaporization increase with stability like alkanes?
So my question is: Is there any way to explain the decreasing ratio between the enthalpy of vaporization and the boiling point with increasing molecular mass for linear monohydric alcohols?

Watch your units!

Are you talking about alcohols with one OH group where you are just extending the chain length or is the number of OH groups increasing as well.

just extending the chain length for monohydric alcohols.

The enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes.  For low-molecular weight alcohols, this effect is pronounced.  The longer the alkane chain becomes, the more the compound behaves like a pure alkane.  For example, using data obtained from the NIST Webbook:
Table 1. Enthalpy of vaporization in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  42}$
$\ce{n-propanol\ \  47}$
$\ce{n-butanol\ \ \ \  52}$
$\ce{n-pentanol\ \  57}$
$\ce{n-hexanol\ \ \ \  61}$
$\ce{n-heptanol\ \  67}$
$\ce{n-octanol\ \ \ \  71}$
$\ce{n-nonanol\ \ \  77}$
$\ce{n-decanol\ \ \ \  82}$
Keeping in mind the relative molecular weights of the compounds, you can see there is a decreasing effect of the hydrogen bonding (and other) effects on the n-alcohol series as we move to larger chains and become less alcohol-like and more alkane-like.  This is much more obvious in the next table, which is simply Table 1 normalized to a per-carbon basis:
Table 2. Enthalpy of vaporization on a per-carbon basis in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  21}$
$\ce{n-propanol\ \  16}$
$\ce{n-butanol\ \ \ \  13}$
$\ce{n-pentanol\ \  11}$
$\ce{n-hexanol\ \ \ \  10}$
$\ce{n-heptanol\ \  10}$
$\ce{n-octanol\ \ \ \  9}$
$\ce{n-nonanol\ \ \  9}$
$\ce{n-decanol\ \ \ \  8}$
Table 2 even more drastically illustrates the deviation in $\mathrm{\Delta}H$ changes as we move from alcohol-like compounds to alkane-like compounds. So, given that the enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes, this should explain your findings.

I realised the increase in boiling point for alcohols decreases as well. Both boiling point and enthalpy of vaporisation should reflect the strength of intermolecular forces, but since they refer to the vapour pressure needed and energy needed for vaporisation(which might not be used only to overcome the intermolecular forces) respectively, they probably don't change by the same percentage, what exactly is behind the difference? This is what still puzzles me.

Well, I may not have done a good job of answering your question then. The main point of my answer lies in table 2, which shows that 2 different processes control the enthalpy of vaporization, and similarly the saturation vapor concentration (also known as vapor pressure) or boiling point. At the low-molecular weight end, hydrogen bonding dominates, so we see the behavior common to polar, hydrogen-bonding compounds.  At the high-MW end, we see the pattern observed for alkanes.  I'll try to think of a better way to explain, but no guarantees ;)


text:
Why does oxygen in OH- have an oxidation state of 2-?
I am watching videos on Khan Academy in order to go over redox reactions (a really weak area for me.) In Sal's video, "Oxidation state trends in periodic table" at 7:10, he says:

If you had to pretend this was an ionic bond, then maybe this hydrogen
  would fully lose an electron, so it would get an oxidation state of
  plus 1. It would be oxidized by the oxygen.
And the oxygen actually has fully gained one electron. And if we're
  forced to think about this is an ionic bond, we'd say it fully gains
  two electrons. So we'd have an oxidation state of negative 2.

I'm having a hard time understanding how he is able to deduce that oxygen has an oxidation state of 2- given the above reasoning. He does add that the states add up to the overall charge which is helpful, but I think I need a little more clarity than that. My reasoning for oxygen being 2- is that "it usually is because my textbook says it usually is," which is probably an even worse explanation.
Could anyone clarify this for me?

As Jan has said, oxidation state is the charge on an atom when the bonds are "cleaved heterolytically in such a way that the electronegative partner gets both electrons". Oxygen is the second most electronegative (after fluorine), and it forms two bonds, so it gets a $-2$ charge.

Haha, thanks for quoting my answers =) I feel happy now!

Well the oxygen in $\ce{OH}$ doesn't have a -2 charge, but the oxygen in $\ce{OH^-}$ does.

First step: draw the Lewis structure of $\ce{OH-}$ with all lone pairs. You should be getting three lone pairs on oxygen and one $\ce{O-H}$ bond.
Second step: ‘[cleave] [all bonds] heterolytically in such a way that the electronegative partner gets both electrons.’[1] You should now have eight electrons around the oxygen and zero on hydrogen.
Third step: count the electrons on both (or all) atoms and check what the difference is to their elemental valence electron count. Oxygen formally has eight electrons on it which is two more than the six it would in its elemental state. The two extraneous electrons give an oxidation state of $\mathrm{-II}$. Likewise, hydrogen has lost its electron for an oxidation state of $\mathrm{+I}$.
Control step: add up the oxidation states of all atoms. If you did everything correctly, they will add up to the entire charge of the ion. Here, this works out:
$$(-2) + (+1) = -1$$

[1]: https://chemistry.stackexchange.com/a/66242

All the numbers refer to number of valence electrons (so we are not counting the electrons in the inner shell).

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text:
Trend of second Electron Affinity for N,O,F,P,S,Cl
I am aware of the definition of electron affinity and electron gain enthalpy (EGE). N has positive EGE1, and EA of P,S,Cl is more than N,O,F respectively. I want to arrange them in order of their EA2 in periods and groups.
My attempt : F,Cl attain noble gas and it's EA would be close to IE (ionization energy) of alkali metals, so, quite small. O and S would be halogen with quite high EA. P and N would be S and O so, somewhere in between the above two.
Hence, O>N>F and S>P>Cl.
for groups, I would still say that 3rd period would have more EA than 2nd but the only data that I have (from another chemistry.stack question) is 2nd EA of O is more than S.
Some help regarding data availability of EA2 is needed and fault in my logic.

I think several of those dianions are actually unbound (i.e. no activation energy towards ionization back to the monoanion), which makes measuring the second electron affinities very hard, if not entirely ill-defined.

It seems bonkers that I could not find any data. Surely EA 2 would help out in stability of ions and ionization energy of anions. I just hope somebody has something on this.


text:
Can CO2 be cracked by electrolysis?
What can be added to liquid $\ce{CO2}$ to lower the electricity needed to crack $CO_2$ ? Answer: Zirconium dioxide
What form and temperature of $\ce{CO2}$ is best to crack with or without Zirconium dioxide?
https://space.stackexchange.com/questions/19755/can-carbon-dioxide-be-broken-down-by-electrolysis

Unfortunately, none of those sentences make any sort of chemical or electrochemical sense to me. Baking soda doesn't break water down in any manner I'm aware of. What do you mean by "break CO2 down?" Electrolysis requires an electrolyte; there do exist solid electrolytes, but they are always ionic compounds, not covalent ones like CO2, so no, electrolysis will not work in solid CO2. Neon gas will not carry electric current at all, unless you apply such a high voltage that you ionize it into plasma.

I'm pretty sure @hBy2Py adequately answered this, so it can't possibly be unclear.

@M.A.R. Especially after looking at the linked question on SX.SE, I think there's an answerable question lurking in there somewhere -- I just for the life of me can't figure out what it is. Muze, you might want to head to the Chem.SE main chat to get some help clarifying what you're looking for, and then edit the question accordingly.

@hBy I think it's the one in the title.

@Muze Ok, so ... are you assuming you've got the CO2 pressurized and cooled such that it's a liquid?

You mean like electrolysis of water?  A major difference is water's polarity which makes it electrolysable, and also liquid and a good solvent of ionic compounds as well.

@JosephHirsch my feeble attempt at remembering chemistry has something to do with the appearance of this question. While electrolysis of gaseous CO2 at high temperature is workable - see MOXIE - Muze is (seems to be) asking (roughly) if electrolysis of liquid CO2 is possible, and if something can be added to the liquid to enhance its bulk electrical conductivity. This is not to suggest it would be a good or practical way to to it. Thanks for your help!


text:
Chirality of 1-Pentanol and 3-Methyl-2-Pentanol
Q) Which of the following is chiral?
a) 1-Pentanol
b) 3-Pentanol
c) 3-Methyl-2-butanol
d) 3-Methyl-1-butanol
Answer: c)
If structure and it's mirror image are not superimposable then it's chiral.

But aren't 1st, 2nd and 4th options also not superimposable ?
I drew the 1st one '1- Pentanol' and took a mirror image of it. If I keep the second one on the 1st image it's no way superimposable. I may be wrong.
Also, is there any rule like if the molecule is optically active then it's chiral?

You need a 3D diagram to judge if its mirror image is superimposable. You are allowed to rotate the mirror also. You are also allowed to rotate the sigma bonds about the axis connecting the two centres of the bond.

"optically active" is a synonym of "chiral".

@DHMO examples 3 and 4 both have one chiral carbon which implies both are optically active. Then aren't they both chiral?

Option (d) does not have chiral carbons, nor is it chiral as a whole.

As DHMO said, you need 3D diagrams, so here they are (all taken from PubChem):
a) 1-pentanol:
A good rule of thumb is that a chiral carbon has four distinct substituents (you can't have more than four anyway).
This structure has no such carbon: every carbon has at least two hydrogen atoms attached.
None is chiral.

b) 3-pentanol:
This one has none either.
The carbon in the middle of the chain, bonded to oxygen, has two ethyl groups attached, so not chiral.

c) 3-methyl-2-butanol:
The carbon attached to the oxygen has four distinct groups: hydroxyl, hydrogen, methyl and tert-butyl.
So this carbon is chiral.

d) 3-methyl-1-butanol:
Here every single carbon has either two hydrogen atoms or two methyl groups attached, so not chiral.

Since the only chiral carbon found is in structure c), this is the correct answer.


text:
How to determine the minimum pH at which iron(III) hydroxide dissolves completely?

Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

My idea was to calculate the amount of $\ce{Fe^3+}$ and $\ce{OH-}$ ions, and then add enough acid containing $\ce{H+}$ ions in order to neutralise the $\ce{OH-}$ ions corresponding to the amount of $\ce{Fe^3+}$. To do this I used the equilibrium constant leaving the hydroxide ions as the unknown. Then I calculated the difference between the initial hydroxide concentration (corresponding to $\pu{10 mg}$ of salt) and the needed concentration to reach $K_\mathrm{s}$. This corresponds to the concentration of hydronium ions in the acidic solution, and I can calculate pH.
Doing this I get $\mathrm{pH} = 3.03$. Is this how it should be done?

The OP first stated the problem as:

Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

I took that to mean that acid was added to a solution containing 10 mg of iron (iii) hydroxide and the pH of the solution when the iron (iii) hydroxide was totally dissolved was desired.*
Knowing that $K_\mathrm{sp} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ then
\begin{align}
  K_\mathrm{sp} &= \ce{[Fe^{3+}][OH^-]^3} = 6\cdot10^{-38},&
\text{ so }&&
\ce{[OH^-]} &=\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}}
\end{align}
You need to solve for $\ce{[Fe^{3+}]}$ which you can calculating knowing that there is $\pu{10 mg}$ of $\ce{Fe(OH)_3}$ in 0.1 liters of solution. It has a molecular weight of $\pu{107 g/mol}$, so there is
$$0.010/107 = \pu{9.35* 10^-5 mol//L}\text{ of }\ce{Fe(OH)_3}.$$
Going back to the equation for the $K_\mathrm{sp}$:
$$\ce{[OH^-]} =\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}} = \sqrt[3]{\frac{6\cdot10^{-38}}{9.35\cdot10^{-5}}} = 8.63\cdot10^{-12}$$
Knowing $\ce{[OH^-]}$ you can calculate $\ce{[H^+]}$ via
$$\ce{[H^+]} = \frac{1\cdot10^{-14}}{\ce{[OH^-]}} = \frac{1\cdot10^{-14}}{8.63\cdot10^{-12}} = 1.16\cdot10^{-3}$$
and then pH:
$$\mathrm{pH} = -\log\left(1.16\cdot10^{-3}\right) = 2.94.$$

Now the OP has modified the problem statement to be:

Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $10\ \mathrm{mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

We again end up with the final pH being just acidic enough to dissolve $\pu{10 mg}$ of iron (iii) hydroxide in $\pu{100 ml}$ of water, but obviously the solution must start out more acidic. The final solution will again have a pH of $2.94$ or a $\ce{[H^+]} = 1.16\cdot10^{-3}$.
In $\pu{100 mL}$ of a solution of pH $2.94$ there are $\pu{0.116 mmol}$ of acid.
We calculated before that there were $\pu{9.35*10^{-5} mol}$ of $\ce{Fe^{3+}}$, but there are three $\ce{OH^-}$ anions for every cation of $\ce{Fe^{3+}}$, so there are $3 \times 9.35\cdot10^{-5} = \pu{2.81*10^{-4} mol}$ of $\ce{OH^-}$.
The total amount of substance (in millimoles) of acid needed is
$$0.116 + 0.281 = 0.397.$$
For $\pu{100 ml}$ to contain $\pu{0.397 mmol}$ the solution would need to be 3.97 millimolar in $\ce{[H^+]}$, or the pH = $2.40$.
If you start with $\pu{100 ml}$ of solution with pH $2.40$ and dump in $\pu{10 mg}$ of $\ce{Fe(OH)3}$, then the final pH will be $2.94$ and all of the $\ce{Fe(OH)3}$ will be just dissolved.

Thank you for your answer. I understand what the steps are, I have done more exercises like this. But what I still don't really have 100% clear is how you can calculate the concentration of Iron (III) ions without involving the constant again, as you start off with the actual $\ce{Fe(OH)3}$.

HINT - Fe has an atomic weight of 55.85 and $\ce{Fe(OH)_3}$ has a molecular weight of 106.9, so 10 mg of $\ce{Fe(OH)_3}$ has (55.85/106.9)*10 = 5.22 mg of Fe.

Okay, I think I got it. Just to be clear: whenever I have solubility equilibrium problems like this one, should I always try to find the value of the concentrations that make Q = K? Because if I am not mistaken, initially, for this exercise, Q > K. And according to my theory notes, when this happens, precipitation, not dissolution, will take place until Q = K.

Q=K is the knife's edge on these types of problems. So you'd want pH to be "less than" whatever is calculated.

Just something I noticed yesterday: in the equilibrium expression, why don't you also multiply by 27, as the concentration of hydroxide ions is three times that of iron ions? And also, why do you calculate the hydroxide ion concentrations via the water constant expression? As you are trying to neutralise, shouldn't it be the same concentration as hydroxide ions?

Just something I noticed yesterday: in the equilibrium expression, why don't you also multiply by 27, as the concentration of hydroxide ions is three times that of iron ions? --- I have no idea where you're getting 27. I assure you that the $\rm{K}_{sp}$ equation is right.

And also, why do you calculate the hydroxide ion concentrations via the water constant expression? As you are trying to neutralise, shouldn't it be the same concentration as hydroxide ions? -- I first calculated $\ce{[OH^-]}$ then calculated the corresponding $\ce{[H^+]}$ from the equation for the water equilibrium. I need $\ce{[H^+]}$ to calculate pH. // I could have calculated p(OH) and subtracted that from 14 to get the same result.

I understand this, but the way I think of the exercise is: what is the pH of the acidic solution that contains the concentration of hydronium ions  you have to add in order to neutralise the right amount of hydroxide ions so that Q = K. Then, the initial concentration of hydroxide ions would be the same as iron, and the final would be the hydronium ions concentration. Why is my interpretation wrong?

If you add an "acidic solution" then you're going to change the volume of the solution. You could bubble HCl gas through the solution and keep volume essentially the same, but why add another chemical into this?

I am really sorry: I must have written the wording in an ambiguous way. The problem is meant to say that initially, you have an acidic solution of pH unknown, and then you add 10 mg of iron hydroxide. So, the hydronium ions are already there. Would this change the solving of the exercise?

Yes that is a different problem...

So would it be solved as I proposed two comments ago? Or how would you solve it?

I've been working on all of it. I posted the new answer...


text:
Catalyst for Formic acid decomposition
I wanted to decompose formic acid into $\ce{CO2}$ and $\ce{H2}$ for a school experiment. I was wondering what would be a suitable catalyst to use?

Please have a look at decomposition of formic acid.

In A Novel Example of Molecular Hydrogen Generation from Formic Acid at
Visible-Light-Responsive Photocatalyst, the authors describe the formation of carbon dioxide and hydrogen from formic acid using visible light from a halogen lamp and freshly prepared $\ce{Cu2O}$ (or platinum-covered $\ce{Cu2O}$) as a catalyst.
The preparation of the $\ce{Cu2O}$ is described as follows:

Ten cubic centimeters of $\mathrm{1.0~mol~dm^{-3}}$
  $\ce{CuSO4*9H2O}$ aqueous solution was dropped
  into $\mathrm{100~cm^3}$ of $\mathrm{1.0~mol~dm^{-3}}$ $\ce{Na2S2O3}$ aqueous solution, and
  subsequently, the mixture solution was dropped into $\mathrm{200~cm^3}$
  of $\mathrm{1.0~mol~dm^{-3}}$ $\ce{NaOH}$ solution at 80 °C.
The resulting reddish-brown precipitate ($\ce{Cu2O}$) was separated by filtration and washed with deionized water, after which the precipitation was dried at 40 °C in an $\ce{Ar}$ atmosphere.

The conversion of formic acid is described as follows:

The photocatalytic experiment was conducted in a glass vessel (volume: $\mathrm{20~cm^3}$) containing $\mathrm{10~cm^3}$ of $\mathrm{2.5~mol~dm^{-3}}$ $\ce{HCOOH}$ and a known amount of photocatalyst (10 mg (ca. 70 μmol)) [...] in
  $\ce{Ar}$ atmosphere with stirring. The pH of the $\ce{HCOOH}$ solution was adjusted to 5.0 by the addition of $\ce{NaOH}$ solution.
A halogen lamp was used as the light source (light intensity: $\mathrm{280~mW~cm^{-2}}$ ) along with a sharp-cut filter (S76-L42, Suruga Seiki Co. Ltd.), using which the samples were irradiated with light of wavelength $\lambda > 420~\mathrm{nm}$. [...]

In summary, it seems possible to perform the experiment at school. UV filters are available from different manufacturers, a Schott N-WG320 might work.

The wikipedia article of formic acid mention that decomposition of formic acid is carried out in the presence of acid, platinum or ruthenium catalyst.

Heat and especially acids cause formic acid to decompose to carbon
  monoxide (CO) and water (dehydration). Treatment of formic acid
  with sulfuric acid is a convenient laboratory source of CO.
In the presence of platinum, it decomposes with a release of hydrogen and carbon dioxide.
$$\ce{CH2O2 → H2 + CO2}$$
Soluble ruthenium catalysts are also effective(...)

For laboratory purposes, treating it with sulfuric acid is the best way.


text:
How can the formal charge of the B in the Lewis structure of BF4(-) ion be minus one?

How can the formal charge of the $\ce{B}$ in the lewis structure of
  $\ce{BF4-}$ ion be minus one ?

I draw the lewis structure of $\ce{BF4-}$ ion such like the following:
I put $\ce{B}$ in the central position and make single bonds between $\ce{B}$ and $\ce{F}$s, then completed the octet of $\ce{F}$s.
But in such a structure, the formal charge of $\ce{B}$ is $-1$, and the FC of $\ce{F}$s is 0.However, we know that if someone is going to get a minus FC, it should be the one whose electronegativity is the highest, and in this example this rule is not followed, so what is the reason for that ?

Formal charges are assigned assuming pure covalent bonds, so it often goes against electronegativity.

Couple of comments. 1) Please don't use MathJax in titles as far as possible. 2) Please use $\ce{...}$ for chemical formulae and expressions. See here for more info.

@DHMO I didn't quite understand what you have said, could you explain what you mean with more details  ?

Formal charge is the charge an atom would have if all the bonds were cleaved homolytically, ignoring electronegativity.

There's a reason it's a called a "formal" charge and not just charge.

@Zhe So what is that reason ?

Because it's a formality and not real.

@Zhe so how does this knowledge, which I have already known,  help me with solving the problem ?

I would say you don't understand because if you're trying to use formal charge together with electronegativities, you're doing it wrong. The formal charge is a formality for keeping track of total charges in the species. It doesn't reflect the charge distribution on any atom because it assumes a equal sharing of electrons in any bond. And if you understand this is the case, why are you trying to use it in a way that it shouldn't be used?


text:
Viability of proposed carbon-carbon bond-forming reaction
Here is my proposed synthetic pathway to make propanoic acid from bromoethane:

$\ce{CH3-CH2Br ->[\ce{KCN(aq)}][\text{heat}] CH3-CH2-C#N}$
$\ce{CH3-CH2-C#N ->[\ce{KOH(aq)}][\text{heat under reflux}] CH3-CH2-COOH + NH3\uparrow}$

I have some questions about my proposed pathway:

Is it viable at all?
If so, what might be the disadvantages compared to other carbon-carbon bond-forming reactions such as the Grignard reaction?
If it is viable, what is the estimated yield?
If it is viable, how can I improve the yield? Can I use silver ions to increase the yield of the first step?
If it is viable, how might I isolate the product?
If it is not viable, can I fix it? If so, how?

Honestly, I'd just buy it ;) If I had to make it in the lab, I'd oxidize the sh*t out of n-propanol :D. If I had to make it on an industrial scale, I'd do a hydroformylation. I'd avoid bromoethane with its rather low boiling point (39 °C) and the health risks (H351).

Bromoethane is just an example. It works with any alkyl bromide.

The Grignard reaction with $\ce{CO2}$ is just one step. On the down side, you need a dry solvent.

well it's another step to prepare the Grignard reagent. Also, less steps does not mean more yield.

@KlausWarzecha I'm surprised you mentioned the toxicity of the ethyl bromide instead of the cyanide...

That's in situ. Suspend magnesium in diethyl ether or THF, add a the alkyl bromide (and maybe some iodine as a starter), stir. The let a stream of dry $\ce{CO2}$ through the solution. A flask with dry ice, a tube with a drying agent and a stream of nitrogen will do. All that is pretty much one reaction without workup in between.

The cyanide isn't that much of a problem unless the solution is acidic. The alkyl halides with high volatility may not have a high acute toxicity, but they may cause cancer.

I thought bigger alkyl bromides would have a lower yield on the first step

Concerning the proposed increased yield by adding Silver: I am certain that AgCN is not very soluble, either.

With some some minor modifications, your two stage reaction

Kolbe nitrile synthesis
alkaline hydrolysis of the nitrile to corresponding carboxylic acid

will work in principle.
In the Organikum, 21. edition, 2001, p. 256ff, a general procedure for the reaction of alkyl bromides with sodium cyanide is provided.
The procedure distinguishes between reactive and less reactive halides.
Reactive halides are benzyl halides (unsubstituted, methoxy- or methyl-substituted.
Less reactive halides are halogen-substituted benzyl halides and plain alkyl halides.
For the latter, the following procedure is described:

In einem 1-l-Dreihalskolben mit Rührer, Rückflußkühler und Innenthermometer werden 250 ml Triethylenglycol, 1,25 mol gut gepulvertes, trockenes Natriumcyanid und l mol Alkylbromid oder Alkylchlorid unter gutem Rühren vorsichtig erhitzt. Der Beginn der stark exothermen Reaktion ist bei den niederen Alkylhalogeniden daran zu erkennen, daß die Lösung aufsiedet. Man steigert die Temperatur langsam auf 14O °C, bei der Umsetzung von Benzylhalogeniden nur auf 100 °C, und rührt bei dieser Temperatur noch 30 Minuten.

(Translation: In a 1-liter three-necked flask equipped with stirrer, reflux condenser and internal thermometer, 250 ml of triethylene glycol, 1.25 mol of well-powdered and dry sodium cyanide and 1 mol of alkyl bromide or alkyl chloride are carefully heated. The beginning of the strongly exothermic reaction in the lower alkyl halides is evident from the fact that the solution settles. The temperature is slowly raised to 140 °C, while the reaction of benzyl halides is only to 100 °C, and stirring is continued at this temperature for 30 minutes.)
The solvent is triethylene glycol with a boiling point of 291 °C, and the reaction is performed under exclusion of water using dry sodium cyanide.
As I have mentioned in my comment, the low boiling point of bromoethane is a disadvantage!
In fact, the Organikum recommends the use of diethyl sulfate (bp 209 °C) for the synthesis of propionitrile (bp 118 °C) and suggests to isolate the product by distilling it off the reaction mixture.
For further purification of the more volatile nitriles with an alkyl chain shorter than $\ce{C5}$, it is mentioned to

wash then with brine
dry over calcium chloride
re-distill over a 30 cm Vigreux column

For this reaction (to propionitrile), the yield is given as 90%. (For valeronitrile, hexanenitrile and heptanenitrile, which all can be prepared from the corresponding bromides, the given yields are around 80%).
According to the Organikum (p. 500) the hydrolysis of primary nitriles is achieved by heating 1 mol of the nitrile with 2 mol of 25% aqueous sodium hydroxide solution under reflux until no more ammonia evaporates (2 to 1o hours).
Whether you really want to use this two stage procedure of perform a Grignard reaction of the bromides with $\ce{CO2}$ is up to you. As mentioned in the comments, the advantage of the Grignard pathway is that it's pretty much a one-pot reaction all the way down to the carboxylic acid. The downside is the requirement of a dry solvent.

Update
In a comment, you asked for a recommendation on which carboxylic acid to synthesize.
I would synthesize a compound if

I need it (for further studies)
it is not commercially available
it costs a fortune. (Make sure that the halide that you need as a starting material is not more expensive than the final carboxylic acid.)

If it is just for practise, I'd go for phenylacetic acid (or a donor-substituted derivative), which has the following advantages:

benzyl bromide is a (weak) lacrymator, but not a CMR compound
the reaction with sodium cyanide can be carried out in refluxing acetone (instead of triethylene glycol)
phenylacetic acid, the final product, has a melting point of 77 °C and a rather low solubility in water (< 2 g/L). The product can be purified by crystallization.

How would you lower the "yield" of the isonitrile in the first step?

@DHMO I'd just rely on the elevated temperature :) Isonitriles undergo rearrangement to nitriles upon heating.

@KlausWarzecha I would have suggested the Grignard pathway too

Perfectly good solution in my opinion.  I do think the OP's question number 5 has been ignored in the discussions and answers though (how to isolate the product).  At the end of the reaction described in this answer, the product and any by-products, etc., will still be in the triethylene glycol solution.  Separating the solvent and un-reacted reagents via simple distillation might be sufficient.  Acid-base extraction of the propanoic acid might help further separate it from other organics.  Thoughts?

@airhuff Yes, acid-base extraction would suffice here. (I mentioned it to OP in chat earlier today.)

You mentioned that I shouldn't be using this method to make propanoic acid from ethyl bromide. Which acid would you recommend me to synthesize using this method?


text:
Why does a hydride shift take place in the cannizzaro reaction?
Hydride is a poor leaving group so a hydride shift should not take place and instead the attacking hydroxyl should leave the carbonyl group. I don’t understand why this doesn’t happen?

Well if the hydroxyl group just falls off then you have no reaction, right?

Right but what about the poor leaving hydride group

A discrete hydride ion may not form. An intermolecular hydride transfer takes place instead. Just compare it with normal hydride reagent reduction (like LiAlH4).

You are correct that a hydride ion is a terrible leaving group and that it can under no circumstances be eliminated in e.g. an $\mathrm{S_N2}$ reaction. However, there are quite a few reactions where hydride does leave one atom and subsequently attaches to another; Wagner-Meerwein rearrangements jump to mind as one of the most prominent examples.
The feature of a Cannizzaro reaction is that it is not actually a leaving hydride. Rather, the reactivity can be explained in a cyclic transition state as shown in scheme 1 below.

Scheme 1: Transition state and products of a Cannizzaro reaction.
Instead of seeing the reaction as a displaced hydride that subsequently attacks another atom, we can see it as a type of pseudo-electrocyclic reaction. Breaking and forming the bonds will happen almost simultaneously. At no point in the mechanism do we actually have a ‘free’ hydride anywhere.
Note, by the way, that the Cannizzaro reaction is not the only one that features a hydride shift in a cyclic transition state; you may, for example, also be interested in the Evans-Tishchenko asymmetric reduction.

I guess, going by their formal definitions, it would not be an electrocyclic reaction (https://goldbook.iupac.org/E01948.html), but rather a group transfer reaction. Not that it matters.

@orthocresol Hence pseudo ;)


text:
How was the bottle crushed by air?
In this video, a water bottle is shown to be crushed by the surrounding air.
My understanding:
When the person pours cold water over the bottle which has hot water inside it, the hot water produces water vapor which increases the pressure inside in bottle. When the person places the bottle into the cold water bowl, the temperature of the bottle decreases, which causes the water vapor inside the bottle to get condensed, thus lowers the pressure inside the water bottle. When the pressure inside decreases, the pressure outside is greater which crushes the water bottle.
If my explanation of this is wrong in anyway, can someone correct it?

This isn't a chemistry question, the decrease in pressure isn't caused by the water vapor condensing, nor do you have a "science" of this phenomenon.

Ah I am studying this in a chemistry class it's called gases and atmospheric pressure. If I am not right what causes the bottle to get crushed?

This is just physics. The volume of the gas is lowered when the temperature is decreased. This is called Charles's law. The effect from the water vapour should be negligible.

https://youtu.be/CMplxEE4xdo here watch this

Your explanation is close, and I disagree with comments dismissing the water-vapor condensation effect.  It's not primarily about heating the air but displacing it with water vapor.  When air is cooled it contracts via the gas law as described in another answer, but when water vapor is cooled it condenses and is removed from the vapor phase altogether, just as if you had pulled the vapor out with a vacuum pump.
The experiment was not carried out very well in the video in the body of your question.  They didn't provide the time and heating conditions to thoroughly displace the air with water vapor.  You can tell that it happened to some degree though by the observation of steam in the bottle, condensing on the sides before sealing the bottle.  Because of this the effect wasn't very dramatic.
The video in your comment shows the concept much more clearly.  They are actively boiling the water such that steam is shooting out of the top and the vapor composition inside their container is almost entirely water vapor.  This is an excellent chemical phase-change demonstration (the second video).

So what causes the crushing?

As you correctly stated in your question, it is the surrounding air.  Once the vapor pressure inside the container is too small to support the container material against atmospheric pressure, it collapses.

In my first chem class in college, the prof did this with a 1-gallon F-style metal can. It collapsed to a crumpled mass. Then he said that because of severe budget restrictions, he had to reverse the process and save the can by pumping it up with air pressure from the compressed air line. Fantastic demonstration. He was a showman.

You do not need the water vapour for this effect, the air that is contained in the bottle is sufficient.
You may use the ideal gas law to estimate the amount of air in the volume of the bottle of about $V=1\ \mathrm l=0.001\ \mathrm{m^3}$ when the air is heated to about $T=60\ \mathrm{^\circ C}=333.15\ \mathrm K$ (using hot water) at normal pressure $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ (since the bottle is still open):
$$\begin{align}
pV&=nRT\\[3pt]
n&=\frac{pV}{RT}\\[3pt]
&=\frac{100\,000\ \mathrm{Pa}\times0.001\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times333.15\ \mathrm K}\\[3pt]
&=0.036\ \mathrm{mol}
\end{align}$$
If you now close the bottle, the amount of air $n$ is trapped.
When this air is cooled to about $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$ (using ice water) at an approximately constant volume of $V=1\ \mathrm l=0.001\ \mathrm{m^3}$, the pressure in the bottle decreases according to the ideal gas law:
$$\begin{align}
pV&=nRT\\[3pt]
p&=\frac{nRT}{V}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{0.001\ \mathrm{m^3}}\\[3pt]
&=8.2\times10^4\ \mathrm{Pa}=0.82\ \mathrm{bar}
\end{align}$$
Therefore, the ambient air pressure of $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ compresses the bottle until the pressures are equalized (or the bottle fails). The volume in the bottle decreases according to the ideal gas law:
$$\begin{align}
pV&=nRT\\[3pt]
V&=\frac{nRT}{p}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{100\,000\ \mathrm{Pa}}\\[3pt]
&=8.2\times10^{-4}\ \mathrm{m^3}=0.82\ \mathrm l
\end{align}$$
Nevertheless, any condensation of water vapour would help to achieve this effect. The resulting changes in pressure and volume due to condensation of water can be even more extreme than the contraction of air. For example, if a volume of $V=1\ \mathrm l$ is filled with steam at $T=120\ \mathrm{^\circ C}$ and $p=1\ \mathrm{bar}$ (without any air) and then cooled until the steam is almost completely condensed to liquid water (at about $T=99.6\ \mathrm{^\circ C}$), the resulting liquid volume is only $V=0.00058\ \mathrm l=0.58\ \mathrm{ml}$.

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


text:
Percentage of yield reaction question

By bubbling $\ce{Cl2}$ in a solution containing $176.6~\mathrm{g}$ of $\ce{MgBr2}$ are obtained [sic!] $135~\mathrm{g}$ of $\ce{Br2}$. However, the yield of this reaction is not $100~\%$. How many grams of $\ce{Br2}$ t [sic!] are formed? What is the yield of the reaction? (MW [sic!] of $\ce{MgBr{2}}\ \text{[sic!]} = 184{,}13~\mathrm{amu}$; [sic!] MW [sic!] of $\ce{Br{2}}\ \text{[sic!]} = 159{,}82~\mathrm{amu}$ [sic!]). The reaction is:

$$\ce{MgBr2 + Cl2 -> MgCl2 + Br2}$$

I need help with this question please.
My thought was like this:
$$176.6~\mathrm{g}/184.13=0.9591\ \mathrm{mol}\ \ce{MgBr2}$$
and this also equals the amount of $\ce{Br2}$ because it's the same ratio so
$$0.9591\times159.82=153.28~\mathrm{g}\ \ce{Br2}$$
now the yield:
$$135/153.28=0.88~\%$$
In my logic, the answer to how many grams are formed is $135~\mathrm{g}$ but it doesn't seem right.

Welcome to Chemistry! 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. Please [edit] in your full reasoning or thoughts on this.

ok, just did :) sorry i didnt know

You should always and consistently use correct units in your equations.

ok... but my answers are right?

I would expect a lower yield for Br2 because you are not just bubbling Cl2 in aqueous MgBr2! Once you start, there is also some now free Br2 available to react with the chlorine!
The neglected competing reaction is:
$\ce{Br2 + Cl2  ⇌ 2 BrCl}$
See this paper to get an idea of the affinity for the formation of the interhalogen compound.
But that is still only part of the story, because of a possible aqueous equilibrium reaction following the formation of any BrCl:
$\ce{BrCl(aq) + H2O ⇌ HOBr(aq) + H+ + Cl-}$
and even more active species can apparently be formed (see this paper).

As I already noted in the comments, you should always and consistently use correct units. And also, the typography of the question itself was terrible but that’s not something you can influence. Anyway to get to answering the question …

I’m assuming that ‘grams of $\ce{MgBr2}$ t’ is supposed to mean something along the lines of theoretical yield. Also, I’m going to use $\mathrm{g/mol}$ as a unit for molar mass because that is what it should be. $\mathrm{amu}$ is a deprecated unit for atomic mass and cannot be used for molar mass without a conversion factor.
We thankfully are already given the reaction$$\ce{MgBr2 + Cl2 -> MgCl2 + Br2}\tag{1}$$ and it is thankfully the case that one mole of each reactant reacts to one mole of each product; i.e. all amounts of substance are theoretically equal. Thus, we can write the following:
$$\begin{align}n(\ce{MgBr2}) &= n (\ce{Br2})\tag{2.1}\\[0.4em]
M = \frac mn\tag{3} &\Longrightarrow n = \frac mM\\[0.6em]
\frac{m(\ce{MgBr2})}{M(\ce{MgBr2})} &= \frac{m(\ce{Br2})}{M(\ce{Br2})}\tag{2.2}\\[0.6em]
m(\ce{Br2}) &= \frac{m (\ce{MgBr2}) \times M(\ce{Br2})}{M(\ce{MgBr2})}\tag{2.3}\\[0.6em]
m(\ce{Br2}) &= \frac{176.6~\mathrm{g} \times 159.82~\mathrm{g/mol}}{184.13~\mathrm{g/mol}}\tag{2.4}\\[0.4em]
&= 153.28~\mathrm{g}\tag{2.5}\end{align}$$
Therefore, your calculation luckily turned out to be correct in spite of not using units.
The question also asks for a yield. Yields are typically given in percent obtained of the theoretical maximum. In the lab, I typically calculate yields from substance amounts but here we can use mass because we already have the data. When calculating a percentage, you need to remember to multiply by $100$ before adding the $\%$ sign — which could be interpreted as a shorthand for $1/100$. Thus:
$$\frac{135~\mathrm{g}}{153.28~\mathrm{g}}= 0.88 = 88~\%\tag{4}$$
The units should, naturally, always cancel out in yields.
I mentioned before that the question is ill-phrased. In a strict literal sense, the answer to ‘How many grams of $\ce{Br2}$ t are formed?’ — without any information on what the random ‘t’ is supposed to be — would indeed read $135~\mathrm{g}$ but unless you are in a quiz show that is unlikely to be the ‘correct’ answer.

Thank you very much, i appreciate your help. Im study in Italy in english, this is why the questions are so ill phrased.

@sapir Rather than just saying ‘thanks’ you can also accept this answer if it helped you by clicking the green checkmark. And once you have 15 reputation, you can also vote it up.


text:
Distance between ions
I am doing a homework where I am supposed to calculate estimate od distance between two ions in cytoplasm. I have only been given the concentration of an ion. Any help how I should start looking this problem?
Only solution I have been able to create is the one where you force some number out from consentration by using ion radius and assume them all to have same distance. However, this feels too, I dunno, mechanical way to solve this problem.

If that's all the information you've been given then I'd probably personally go with something like your solution. Work out the available volume to each ion $V$, work out the corresponding radius $V = 4\pi r^3/3$, then the distance between each ion is $2r$. But I don't know, maybe there's something better.


text:
Why is potassium permanganate(VII) such a strong oxidising agent, why does its behaviour depend on pH of the solution it is present in?
Why is potassium permanganate such a strong oxidizing agent? Let is say we have a $\ce{KMnO4}$ aqueous solution which is acidified by adding sulfuric acid. What is going on in the solution which makes it so oxidizing that immediately if we add an organic or inorganic molecule it gets immediately oxidized?
Also why under different pH potassium permanganate behaves differently i.e. Mn reduces from +7 to +2 in acidic conditions whereas it goes to +4 in neutral conditions?

Related: http://chemistry.stackexchange.com/questions/9001

The answer in the question only explains the results and chemical equations not the reason behind it .

So I guess there is no explanation for this another exception


text:
Why does a hard-boiled egg have more entropy than a normal egg?
The question "Why does a hard-boiled egg have more entropy than a normal egg?" was recently asked on one of my tests.
I said that entropy will decrease as a hard boiled egg has molecules in a more organised way than an raw egg, but I was marked wrong with a remark that entropy will increase because the proteins and other molecules are decomposed into simpler substances due to heat.
I did ask why I was not correct but what I got was the same explanation along with the decomposition reactions.
I don't feel this explanation is satisfying.
For example, do not 10 objects on a table organised properly have more entropy than 5 objects thrown anywhere on the table?
The definition of entropy is the measure of disorder in the system. So clearly in my above example the 10 objects are more organised and thus less disordered than the 5 objects in the second case and same can be said for the main question.
Can somebody please explain to me why a hard-boiled egg has more entropy than an un-boiled egg (I like if the explanation has some math)?

See similar question answered in Physics.SE here.

Here is an explanation I found on the internet that works for me:
The protein in egg whites is albumen, which is tightly folded in an uncooked egg.  The balls will slide past each other, so it is a liquid.  The entropy is low because of the precise folding.
After cooking, the protein molecules are unfolded and tangled, forming a solid. The entropy of the egg whites is actually higher.

But why it is higher ?

Are you familiar with the term configurational entropy?

Yes after reading https://en.wikipedia.org/wiki/Configuration_entropy .

But then does not that mean more things = more entropy regardless of how things are organised ?

It means that the polymer chains have more quantum states readily  available to them in the final state than in the initial, well-organized, folded configuration.

The protein in egg white is a polymer (poly peptide a polymer) and this pokymer has a unique conformational structure. This polymeric structure of protein get decomposed into  many smaller peptides due to hydrolysis of amide linkage of polypeptude chain in protein. So overall we can say one bigger molecule gives many smaller molecules and this will increase the entropy.

I’ve come across several claims that a (hard) boiled egg has greater entropy than an unboiled egg. None of them cite any peer-reviewed research to back up their claim. While it is possible that entropy increases when you boil an egg, it is also possible that it does not. It is even possible that it stays the same.
I did my own literature search to see if I could find any publication that answered the question. What I found left the question very much up in the air. So I contacted Greg Weiss, a scientist who has done research on a related topic: ‘unboiling’ an egg. Dr. Weiss was kind enough to discuss the topic with me via email.

tl:dr
Whether a hard-boiled egg has greater entropy all boils down to the question of how structured are the assemblies found in aggregated ovalbumin. Furthermore, we can't just consider the assemblies found in the ovalbumin aggregate since we're talking about assemblies composed of all the types of protein in the overall albumen. (BTW, it's important to note that heated egg white doesn't just form any type of aggregate, it forms a gel, which presumably implies additional structure.)
For those who are interested in the details, here is the full email thread.


text:
When a molecule loses hydrogen atoms, does it become oxidised?
I had the following question on a bio test covering cell respiration and glycolysis, but I'm not sure if I agree with the answer.

When a molecule loses hydrogen atoms (as opposed to hydrogen ions), it becomes
B) oxidised

According to my textbook (Principles of Life, 2nd Edition), oxidation is defined as

Relative loss of electrons in a chemical reaction; either outright removal to form an ion, or the sharing of electrons with substances having a greater affinity for them, such as oxygen. Most oxidations, including biologial ones, are associated with the liberation of energy.

Hydrogen atoms have one proton and one electron in their typical state. If a molecule is losing hydrogen atoms, then its net charge wouldn't change. For example, if a (hypothetical) molecule had 7 protons and 6 electrons, it would have a charge of +1. If it then lost a hydrogen atom, it would now have 6 protons and 5 electrons, which still yields a net charge of +1. Am I missing some critical piece of information, or is this still considered oxidation?

Oxidation may be defined as the loss of electrons, and reduction as the gain of electrons.  An ionization (loss of a proton) is neither an oxidation or a reduction, as no atom loses or gains electrons. Similarly, hydration (the addition of a water molecule, such as occurs  in beta-oxidation: the reaction catalyzed by enoyl-CoA hydratase) is not in any sense an oxidation or a reduction, although this is often a source of confusion. When an atom loses  hydrogen atoms it looses both protons and associated electrons. Hence it is an oxidation.


text:
How "strong" is Pine Sol as a solvent?
How "strong" of a solvent is Pine Sol? Is it relatively benign like Isopropyl Alcohol, stronger like Acetone, or ridiculous like TCE?
Without a background in chemistry (mine is electrical engineering), I'm trying to get a "feel" for what materials I can use what solvents on.
For example, "strong" solvents shouldn't be used on wood, plastic, etc whereas "moderate" ones can't be used on thin/"weak" plastics but sturdier plastics are OK for short periods of time (i.e. I can use them for cleaning a plastic welding mask in the garage but shouldn't use them on the plastic LCD screen).
If I'm asking the wrong questions, please feel free to answer the question I should have been asking - but keep in mind I'm looking right now for the medium-fidelity answer, not the 100%-physically-correct-by-understanding-chemical-bonds answer.

According the Clorox Company: Water acts as a base. C10-12 alcohol ethoxylates and sodium C14-17 secondary alkyl sulfonate are grease-cutting surfactants, or cleaning agents, used in cleaning products such as laundry detergent. Glycolic acid is used to remove soap scum, discoloration and scale deposits in toilets and bathtubs. Dimethicone/silica/PEG distearates are used mainly as defoamers. Caramel is added to improve the product's appearance and to let the consumer know if the product is working when it is applied. Xanthan gum is used as a biodegradable thickener.

To be able to know how harmful a substance is, you would need to take a look at its ingredients.

Also, strong is a rather qualitative term. I assumed that you are interested in whether the substance is safe to handle. Maybe you meant something else?

Without a background in chemistry (but an electrical engineering background), I'm trying to get a "feel" for what materials I can use what solvents on - e.g. "strong" solvents shouldn't be used on wood, plastic, etc whereas "moderate" ones can't be used on thin/"weak" plastics but sturdier plastics are OK for short periods of time (i.e. I can use them for cleaning a plastic welding mask in the garage but shouldn't use them on the plastic LCD screen).

If I'm asking the wrong questions, please feel free to answer the question I should have been asking - but keep in mind I'm looking right now for the medium-fidelity answer, not the 100%-physically-correct-by-understanding-chemical-bonds answer.


text:
If the atomic number is # of protons, why does emission of a beta (electron) particle increase the atomic number?

Atomic number: the number of protons in the nucleus of an atom, which is characteristic of a chemical element and determines its place in the periodic table.

Beta emission:
$$\ce{^14_6C -> ^14_7N + ^0_{-1}\beta}$$
Why does emission of an electron increase the atomic number, when the atomic number is the number of protons, not electrons?

In a beta decay a proton may change into a neutron. Or a neutron may change into a proton. http://www2.lbl.gov/abc/wallchart/chapters/03/2.html

Maybe you are confusing it with alpha decay?

Thanks for the answer. It makes sense now. I forgot the part about a neutron changing into a proton. I thought the change in number of protons had something to do with a proton being a + and an electron being a -, rather than a loss of an electron converting a neutron into a +.

During beta particle emission one neutron is converted into proton and electron, to balance charge and electron can't remain inside the nucleus hence emitted from the nucleus.


text:
If capsaicin is insoluble in water, what makes chilli water taste spicy?
I have always been told that the 'heat' from chillies came from capsaicin which is oil based. Recently I began wondering how environmentally friendly chilli based insecticide/repellent sprays could work if they are water based. A chemist friend confirmed for me that capsaicin is not soluble in water.
I minced a chilli and added this to a coffee mug of boiling water, left overnight then strained through a coffee filter. The resulting chilli water is definitely spicy, but why? What has dissolved into the water (or is the mixture some sort of emulsion of tiny oil droplets?).

It just triggers the nerve that senses heat.

Hi DHMO, what trigger's the nerves? What is chemically active in the water? Thanks!

I think what you've shown is that our olfactory senses can detect capsaicin at levels of 10's of mg/L (solubility according to Wikipedia).  Most every organic liquid has some solubility in water, its just by degrees that we decide whether to call it soluble or not.  Your emulsion idea is also interesting.  Also, there are other oils in chili peppers that could increase the solubility of capsaicin.  Did you by chance see any oily film on top of the water?  If so, maybe try drinking your spicy tea through a straw!

No oily film visible even outside in good daylight, the mixture doesnt appear to separate at all so far. I am actually hoping it doesnt separate so I can spray it on my garden plants to keep caterpillars away. I thought the coffee filter would likely soak up any obvious globules of oil from the first steep.

The slight solubility in pure water probably explains a lot (see MaxW's answer). But don't forget the other ways to increases the amount of the capsaicin: it is more soluble in alcohols, it can be dispersed in an oily phase in a relatively stable colloid...

You got me curious, so I poked around a bit on this.
First there is insoluble and then there is insoluble. The Wikipedia article on capsaicin lists its solubility as $\pu{0.0013 g}/\pu{100 mL}$ which is $13$ parts per million. So capsaicin is "relatively insoluble", but not wholly so. Second the Wikipedia article also points out that capsaicin itself is just one of the various capsaicinoid chemicals which give peppers their "hot" taste.
At the end of the (old) Wikipedia article on the Scoville scale, a taste test for hotness, it states

[...] A measurement of one part capsaicin per million corresponds to about 16 Scoville units. [...]

So, 13 pmm multiplied by 16 Scoville units per ppm is about 200 Scoville units.
A table on hot sauces and peppers lists the Original TABASCO® brand Pepper Sauce from McIlhenny Company as being 2,500 to 5,000 Scoville units. So 200 Scoville units would be about one tenth strength sauce. That should still be enough to tingle your tongue.

Interestingly if this is hot enough to keep caterpillars off my plants then I'm not going to need a lot of chilli. On the other hand from your info above I take it that it would be impossible to get more 'heat' dissolved into the water.

There are multiple capsaicinoids in a pepper, each with their own solubility. Adding alcohol and vinegar to the water should increase the solubility of the capsaicinoids.

Lets make it simple with simple quantum mechanics. water is consists of Hydrogen and Oxygen. Oxygen have bigger nucleus so it pull electron in hydrogen but hydrogen won't let it go so Hydrogen trying to pull it in but unfortunately, Oxygen is heavier and have stronger nuclear force than hydrogen. Oxygen even able to pull 2 Hydrogen atom.
Okay, we know Oxygen is such a muscular macho savage guy with a mustache and every atoms are such a thug compton gangster but what exactly happen ?. Those make water polar. because the hydrogen electron is pulled towards oxygen. The unequal distribution of force inside hydrogen atom make it slightly positive due to the induced dipole during the bodning process. That's why we classify water as polar covalent compound. And of course oxygen have slightly negative polarity because it must maintain the stability of its own atom and bonding.
What is capsaicin ?. Capsaicin is huge hydrocarbon. based on the nucleus force analysis it is nonpolar. They have polarity but not much. they have negative polarity in C=O double bond. THey also have O-H in one of its end. And a little bit positive but in really-really tiny amount of polarity of N-H binding.
So, what will happen if I put capsaicin in the water. well, water is friendly Rey Mysterio and Capsaicin is unfriendly Big Show. Water attach to some of capsaicin polar ends and try to pull it. Unfortunately to pull one capsaicin molecule, Water need more energy because it is heavy and repulsive in much ends. But, C=O ends is seems attracted to water so, it would basically able break apart and bonding with water. C=O Double bond have negative polarity, Hydrogen in water have positive polarity so, it able to pull out Oxygen from capsaicin. Yes, without any energy given to the reaction it's can be (roughly) 95-99 % unsoluble in water.
But we know, of course, reaction sometimes requires energy. Say we need to shake it, Blend it, Boil it, Electrolysis, or even burn it. Just give it energy. The solubility able to increase. But, How far it would likely to increase ?. As you notice, in repulsion of homopolarity there's energy needed to destroy this capsaicin. It's the weakness, just put up an effort to have certain amount energy which is exert the repulsion energy between capsaicin and water.
So, what happen to you is natural. When you heating up water. The orbitals of water is changing due to the electron vibrations. There's some amount of kinetic energy absorbed and force electron to move. So, the shielding between electron in orbitals can be forced and the orbital shape is slightly changing.
That's why some of capsaicins are able to dissolved in water.
But, in your case you have coffee. Coffee is complex mixture of chemical compounds. Coffee is not only consists of caffeine it's also have a bit protein and a bit minerals. Such as potassium. So, the chemical reaction between capsaicin in coffee mixture is really complex.
But let's see why it's more dissolveable. Let's check caffeine. Caffeine is have double negative polarity C=O bond. It's enough to attract water and makes caffeine soluble in water. So, you try to dissolve capsaicin in strengthen water molecule. It's a handicap match between coffee + water againts the big show. So, who wins ? I predict the tag team win. The solubility will increase but still  I predict, it will not be completely dissolve in this mixture unless you put on some milk.

Not quite the normal way to write about chemistry, but interesting.

Interesting, but a bad way to explain a simple concept.

Nuclear Chemist, I'm not normal. matt_black, yeah  I'm bad chemist, I'm not a chemist at all. I'm a physicist. And I love to do the cheap trick. Atomic mass trick


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Question about oxygen and beta carbons in chemical shifts

Why do the methylene protons that are attached to the beta carbon for the ester group have a chemical shift of +0.6 from oxygen, whereas the methyl protons attached to the beta carbon for the ester group have a chemical shift of +0.5 from oxygen? In other words, why is the oxygen treated as an ester for the methylene protons, but treated as an ether for the methyl protons?
What am I missing? How do I determine when to interpret the beta carbon as an ester or an ether when the carbon is connected to an ester group?

What do you mean by +0.6/+0.5 from oxygen?

if an alpha proton is neighboring an oxygen of an ether group, the effect on the alpha proton is +2.5. if it's neighboring an oxygen of an ester, the effect is +3.0. if a beta proton is 1/5th that of an alpha proton, the ether should give +0.5 to chemical shift, and the ester, +0.6. the methyl groups and the bottom methylene group both look like beta protons for ester to me, but only the methylene group is considered so?

Are you  talking about Shoolery increments (and relatives)?

Which beta-carbons? There is only an alpha-carbon on the ester.

I think perhaps you are misunderstanding some basic principles of determining chemical shifts. I have no idea where you are getting your +0.5/+0.6 factors
Chemical shifts are a product of all contributing functional groups; those that contribute to the local electron density either through direct electronic effects or through steric effects. Looking at the 3 proton environments, we can roughly predict their chemical shift using additive tables (I use Pretsch, Buhlmann and Badertscher).
First, let's look at the types of protons. Methyl protons and methylene protons do not have the same basis chemical shift to work from when doing additive calculations; the shielding from a C-C bond is not the same as for a C-H bond, and so the local shielding for the proton of interest will be different. Generally, when doing additive calaculations for chemical shifts, we can use the starting shifts of 0.86 (-CH3), 1.37 (-CH2-) and 1.50 (>CH-).
For this molecule, there are 3 proton environments. I have shown their predicted chemical shifts below.

For methyl protons at site 1

Each -CH3 proton is alpha to a quaternary C, beta to an oxygen-bound ester and beta to a carbon-bound ester.
So that gives us 0.86+0.05+0.44+0.07= 1.42ppm

For methylene protons at site 2

alpha to a ether oxygen, alpha to a quaternary carbon, beta to an oxygen-bound ester and beta to 2 methyl groups.
So that gives: 1.37+2.04+0+0.24-0.06-0.06= 3.53ppm

For methylene protons at site 3

alpha to ether oxygen and alpha to carbonyl-bound ester. The ester oxygen is taken into consideration as part of the ester group as a whole. Also, the atom on the other side of the ether (in this case the carbon) is taken into consideration - The only important thing here is whether the ester is oxygen-bound or carbonyl-bound.
So, that gives us 1.37+2.04+0.92= 4.33ppm

the values are from my textbook. i had erroneously assumed these values were established benchmarks, but the values you've provided are so different from what is shown in my book i now see that is not the case. (the answer in my book is presented as thus, hence the confusion: imgur.com/a/0ReHP). when i did the problem originally, i got the correct answers for the methylene protons, but answered 1.5 ppm rather than 1.4 ppm for the methyl protons using the information presented in my textbook. thank you for the answer, however, as i see the reason is much more complicated than my book presents.


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In the photoelectric effect, why we use momentum to calculate kinetic energy?
If we know the wavelength of an ejected electron, why do we use the momentum equation ($E_\mathrm k=\frac{p^2}{2m}$, $\lambda = h/p$). Why can't we use $\frac{1}{2}mv^2$ instead, using $c/\lambda = v$?
Does it have something to do with particle-like or wave-like behavior?

Try using mv^2/2 and see what happens.

Also note that $c/\lambda=f$ where $f$ is frequency (not velocity).

Relativity
Since electrons are travelling at near the speed of light, we need to use relativity.
In relativity, the correct equation relating energy to mass and momentum is the energy-momentum relation:

$E^2 = (m_0 c^2)^2 + (p c)^2$
where:

$E$ denotes the total energy of the particle.
$m_0$ denotes the (rest) mass of the particle.
$c$ denotes the speed of light.
$p$ denotes the momentum of the particle.

The momentum of a particle can be found from its (rest) mass and its velocity:

$\displaystyle p = \gamma m_0 v$
where:
$\displaystyle \gamma = \frac {1} {\sqrt{1 - \left({ \frac v c }\right)^2}} = \left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5}$

Note that $p = m v$ is only true when $0 \lt v \ll c$, when $\gamma \approx 1$.
Kinetic energy
When the particle is not moving, $p = 0$, so $E = m_0 c^2$ (mass-energy equivalence) is denoted the rest  energy.
Substituting the definition of momentum gives:

$\displaystyle E^2 = m_0^2 c^4 + \gamma^2 m_0^2 v^2 c^2 = m_0^2 c^4 \left({\frac {c^2} {c^2 - v^2}}\right) = m_0^2 c^4 \gamma^2$

Now, the kinetic energy is defined as the total energy minus the rest energy:

$E_{\mathrm k} = E - m_0 c^2$
$\displaystyle E_{\mathrm k} = m_0 c^2 \left({\gamma - 1}\right)$

Classical approximation: $\displaystyle E_\mathrm k = \frac 12 m_0 v^2$
Now, the classical formula $\displaystyle E_\mathrm k = \frac 12 m_0 v^2$ is an approximation of $E_\mathrm k$ when $0 \lt v \ll c$:

$E_{\mathrm k}$
$\displaystyle = m_0 c^2 \left({\gamma - 1}\right)$
$\displaystyle = m_0 c^2 \left({\left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5} - 1}\right)$
$\displaystyle \approx m_0 c^2 \left({1 - (-0.5) \left({ \frac v c }\right)^2 - 1}\right)$
$\displaystyle = \frac 1 2 m_0 v^2$

However, as I have mentioned, it only works when $v$ is very small, which is not the case for electrons.
Another approximation: $\displaystyle E_{\mathrm k} = \frac {p^2} {2 m_0}$

$E_{\mathrm k}$
$\displaystyle = \sqrt{m_0^2 c^4 + p^2 c^2} - m_0 c^2$
$\displaystyle = m_0 c^2 \sqrt{1 + \frac {p^2} {m_0^2 c^2}} - m_0 c^2$
$\displaystyle \approx m_0 c^2 \left({1 + \frac {p^2} {2 m_0^2 c^2}}\right) - m_0 c^2$
$\displaystyle = \frac {p^2} {2 m}$

Comparison
Now, for reasons unknown, the second approximation is actually better than the first:

(For convenience reasons, $c$ is took to be $1$ in the graph. It can be seen that the second approximation (blue) goes to infinity with the real kinetic energy (black) as $v \to c$, while the first approximation (red) does not.)
Perhaps this is the reason why the second approximation is used instead of the first.
Note that it is still an approximation.

Another reason is that, $\displaystyle \nu = \frac c \lambda$, but that is $\nu$ (Greek letter nu) for frequency not the $v$ (Latin letter vee) for velocity. The former has a unit of $\mathrm{s^{-1}}$ while the latter has a unit of $\mathrm{m~s^{-1}}$.

oh so..actually it is nothing to do with velocity...and we can't even get the velocity directly from wavelength. And if I want to use energy-momentum equation, that would be very complicated. Therefore, we use Ek=p^2/2m. Did i understand correctly?

Actually it doesn't seem that complicated to me  though.

@pronoob reasons for using p^2/2m added.

It's just because photon in newtonian frame of reference us assumed to be having no mass at rest.....so the mass factor in momentum is undefined in case of photon so in order to measure momentum we use kinetic energy. For electron too due to reason that its mass increase on speed so above method more convenient.


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Why is Ka constant when volume is increased?
The following graph shows $\ce{Ka}$  vs ${V}$  (Volume) at constant temperature, as ethanoic acid is diluted:

However, I thought it would instead produce the following graph (see below).
$\ce{Ka=[H3O+]*[A-]/[HA]}$
As ${V}$  increases, I would think all of $\ce{[H3O+]}$,$\ce{[A-]}$ and $\ce{[HA]}$ would decrease similarly. As there are two concentrations decreasing in the numerator, and only one decreasing in the denominator, this leads me to believe I can write this as $\ce{Ka ∝ [X]}$, where $\ce{[X]}$ decreases as ${V}$ increases (i.e. that I could cancel out one of the concentration in both the numerator and denominator, leaving only $1$ concentration in the numerator; noting that during ionisation $1$ mole of $\ce{[HA]}$ turns into $1$ mole of $\ce{[H3O+]}$and $1$ mole of $\ce{[-A]}$)
Giving the following graph:

Why is the top and not the bottom graph produced?

$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}}$$
The above equation defines the $K_\mathrm{a}$, or acid dissociation constant, of an acid. The reason it is even a value worth measuring is because it is an intrinsic property of the acid, i.e. it's value does not change from sample to sample (temperature aside).
While it is true that as a solution of acid becomes more dilute, $\ce{[A-]}$, $\ce{[HA]}$ and $\ce{[H3O+]}$ all decrease, they do not all decrease by the same amount. Consider the effect of adding water to the following equilibrium:
$$\ce{HA + H2O <=>A- +H3O+}$$
The number of $\ce{HA}$ molecules will decrease, and the number of $\ce{A-}$ and $\ce{H3O+}$ molecules will increase—a conclusion we arrive at via Le Châtelier's principle. Separating the volume of water from the $K_\mathrm{a}$ equation, we get:
$$K_\mathrm{a}=\ce{\frac{mol(A-)\cdot mol(H3O+)}{mol(HA)}}\cdot\frac{1}{V}$$
What we are left with is two larger numbers over a smaller and a larger number. While it's not obvious that these numbers should change to give precisely the same quotient, I can assure you they do.

Right: the key to realize is that although for a theoretical instantaneous dilution all of the concentrations would be diluted by the same amount, after the dilution you're no longer (necessarily) at equilibrium. The compounds will then react to reestablish equilibrium, which means their concentrations will change until their equilibrium ratio is equal to the $\ce{K_a}$.

The reaction of an acid in water is of the form
$$\ce{HA + H2O <=> H3O^+ + A^-}$$
When two of the species in solution come close enough to interact, then there is a certain probability that they will react to form products. The probability mainly depends on the type of the involved species and much less on the surrounding solution.
However, the probability that two species meet is proportional to their concentration (not volume). Therefore we can formulate the rate law
$$\frac{d[\ce{HA}]}{dt} = -k_1 [\ce{HA}] [\ce{H2O}] \tag{1}$$
for the forward reaction
$$\ce{HA + H2O -> H3O^+ + A^-}$$
and the rate law
$$\frac{d[\ce{A-}]}{dt} = -k_2 [\ce{H3O+}] [\ce{A-}] \tag{2}$$
for the backward reaction
$$\ce{H3O+ + A- -> HA + H2O}.$$
The rate constants $k_1$ and $k_2$ hereby reflect the said reaction probabilities.
In the thermodynamic equilibrium the reaction rates for forward and backward reaction are the same. By setting equations $(1)$ and $(2)$ equal we obtain
$$k'_a = \frac{k_1}{k_2} = \frac{[\ce{H3O+}] [\ce{A-}]}{[\ce{HA}] [\ce{H2O}]} \tag{3}$$
Since the concentration of water in a dilute aqueous solution can be considered constant, we normally set
$$k'_a [\ce{H2O}] = k_a \tag{4}$$
to get the familiar equation
$$k_a = \frac{[\ce{H3O+}] [\ce{A-}]}{[\ce{HA}]} \tag{5}$$
Neither equation $(3)$ nor equation $(5)$ depends on the volume $V$ of the solution.

Ka of weak acid is an equilibrium constant and equilibrium constant depends upon temperature and its value increases with increase in temperature.  During dilution Qa of weak acid will change and its value will decrease. When Qa < Ka( as Ka will remain constant) reaction will not remain at equilibrium and it shift towards forward direction to attain equilibrium.

This is the better answer. Changing the volume, changes Q. Then the concenatrations shift to bring Q back to K.


text:
Why is chlorine and not fluorine labelled as the most reactive halogen for halogenation?
The reactivity of halogenation in sunlight is as follows:
$$\ce{F2}>\ce{Cl2}>\ce{Br2}>\ce{I2}$$
So why does we often label chlorine as most reactive halogen towards halogenation reaction?

Because $\ce{F2}$ is so reactive and the $\ce{C-F}$ bond is so stable, it's not considered useful in most synthetic pathways.

It’s not fully clear what you’re actually asking. The halogenation is, indeed, most reactive with fluorine. However, the halogen radical (which is what sunlight does) is most easily generated with iodine. I suspect you mean the former but could you clarify?

Why did you decide to roll back ron’s edits which were improving your post’s grammar and title?

The reactivity of halogens in sunlight

OK, this tells us that we are dealing with a free radical reaction.  Sunlight breaks the halogen bond ($\ce{X-X}$) to generate halogen free radicals.
$$\ce{X2 ->C[{h\nu}]\  2X.}$$
In the next step the halogen radical abstracts a hydrogen atom from any hydrocarbon present to generate $\ce{HX}$ and a hydrocarbon free radical (note that multiple hydrocarbon radicals may be generated depending on the structure of our hydrocarbon).
$$\ce{X. + R-H -> HX + R.}$$
Finally, the hydrocarbon radical can react with the halogen free radical to produce a halogenated product.
$$\ce{R. + X. -> R-X}$$
Let's now consider the energetics of these steps.  The first step, generation of the halogen free radical, uses the sunlight to break the $\ce{X-X}$ bond, so no additional energy is required for this step. Then we break a $\ce{C-H}$ bond (so we must add energy) and make an $\ce{H-X}$ bond and a $\ce{C-X}$ bond (so energy is given off).  The various bond strengths are contained in the following Table.
\begin{array}{|c|c|c|c|c|c|} \hline
\ce{C-H}~ \text{Bond Strength} & \ce{H-X} & \text{Bond Strength}& \ce{C-X} & \text{Bond Strength} & \text{Overall}  \\ \text{(kcal/mol)}\ & & \text{(kcal/mol)} & & \text{(kcal/mol)} & \text{(kcal/mol)} \\ \hline
\ 99 & \ce{H-F} & -135 & \ce{C-F} & -116 & -152\\ \hline
\ 99 & \ce{H-Cl} & -103 & \ce{C-Cl} & -81 & -85\\ \hline
\ 99 & \ce{H-Br} & -88 & \ce{C-Br} & -68 & -57\\ \hline
\ 99 & \ce{H-I} & -71 & \ce{C-I} & -51 & -23\\ \hline
\end{array}
(bond strengths taken from Bill Reusch's table)
We can clearly see that all of these free radical reactions will occur (e.g. they are exothermic) in the order you suggested, with fluorine being, by far, the most reactive.
$$\ce{F2 >> Cl2 > Br2 > I2}$$
Saying that chlorine is the most reactive is incorrect.
Note: 99 kcal/mol was used as a typical $\ce{C-H}$ bond strength.  We could refine our model by using more exact $\ce{C-H}$ bond strengths (see this Wikipedia table) and calculate the relative rates of halogenation at the various carbon atoms in a complex hydrocarbon.

Reaction with fluorine are strongly exothermic and mostly during reaction explosion takes place and reaction generally not used for synthetic purpose.


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Hybridisation of POCl3
I was doing lewis structures and i came across a question asking the structure of POCl3. Here, the answer is a tetrahedral shape with P as central atom connected to 3Cl atoms by a single bond and 1 oxygen atom by a double bond. The valency of P is 3. So, it needs 3 electrons to get the nobel gas configuration and become stable so it could have formed single bonds with 2 cl atoms and single bond with an oxygen atom which would be single bonded to cl atom. Why isn't this structure correct?
And how can P form 4 bonds while it needs only 3? I think it's something related to D-orbital but if it undergoes sp3d hybridisation, it would need 5 bonds. Also, the structure of this compound is tetrahedral and hybridisation of P is sp3. How can this be? How can it form 4 bonds although it has only 3 unpaired electrons? Nitrogen couldn't form such compounds.
I googled about it and could find nothing but the hybridisation and lewis structure. My actual question isn't answered anywhere

I unfortunately have no time to answer this, here are a few notes though. The phosphorus is hyper-coordinated in this structure. Think about which element is the most electronegative, try charge separation. It has been disproven that d-orbitals play a significant role in such compounds, better not learn about this. Hybridisation is not a concept that can be used for prediction, it is quite visual for interpretation though.

Think of it as $\ce{[POCl2]+[Cl]-}$. Still a wrong picture, though. You'd need resonance.

Have a look at "Phosphoryl chloride" on Wikipedia and the article "Hypervalent molecule" linked from it.

There should be a duplicate lying around here somewhere … Basically, don’t use d orbitals unless your at school and your teacher tells you to. And if that is the case, forget it the moment you leave school. Instead, you can think of $\ce{\overset{+}{P}-\overset{-}{O}}$, i.e. charge separation. Also, @DHMO’s description is wrong in this case; charge separation is a more simple and better description. 4e3c bonds are only required for pentacoordinated phosphorus atoms.

I'm still not able to get the correct picture.


text:
Concentration of gas phase in Nernst equation
What if one or both electrodes are in gas phase? Use PV=nRT? Also the definition of hydrogen electrode gets a little sketchy if we use PV=nRT to write it's concentration, the electrode potential doesn't come out 0 that way.

What do you mean by doesn't come out 0? The Hydrogen electrode potential is assigned an arbitrary zero in order to have a reference point. It doesn't have a zero value under any circumstances. In order to answer your question anyway, we use pressure values as proxies for the activity of gases, not concentration.

So we use pressure values instead of concentration even in cases where ionic and gas phase both are present?

That is correct. Pressure for gaseous reactants and concentrations for dissolved ones.


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What is the molecular structure of the fragment ion m/z 161 of rosmarinic acid?
Attached is an ESI-MS/MS spectrum of product ion scan of rosmarinic acid (m/z 359.10) (source). I'm trying to figure out the structure of the fragment m/z 161. It's not clear to me what the structure should be. Does a triple bond form perhaps?

The answer is on P.24, one page above the mass spectrum.

An answer is on that page. There's no evidence presented that it's right. But I'm glad to note that their proposal is similar to mine!

I don't get that question. You already answer your own question with the mass spectrum you provide. There it already says that it is the fragment at m/z 179 minus H2O.

The source you link to says that the spectrum was acquired in negative mode.

The CID experiment on $\ce{[M - H]-}$ ion at m/z 359.08 identified as rosmarinic acid gave the two main constituents of rosmarinic acid, namely caffeic acid at m/z, 179.0 and the 2-hydroxy derivative of hydrocaffeic acid at m/z
  197.0 as illustrated in Figure 3. Similar pattern of fragmentation of rosmarinic acid during CID analysis has been reported by several authors (17, 25, 26) in analyzing extracts of Lamiaceae spices.

This means that the parent ion is $\ce{(C18H16O8 - H)-}$, and thus the (hydrated) fragment in question is $\ce{C9H7O4-}$.
The first question in any CID mechanism is whether the fragment arises via charge migration or via charge retention.  I think in this case a charge retention mechanism is pretty likely, because phenolic hydroxyl groups are acidic enough to ionize readily via proton loss.
The next question is where the loss of water comes from.  It's worth noting that the mass spectrum of even just plain old catechol, $\ce{C6H6O2}$, shows a loss of water in negative-mode CID.  That is, the parent ion $\ce{(C6H6O2 - H)-}$ loses water somehow to form $\ce{(C6H3O)-}$.  I see no reason why the same thing couldn't happen to the rosmarinic acid fragment in question.
It's impossible to say for sure without doing a lot of esoteric spectroscopic characterization of gas-phase ions, or least the synthesis and fragmentation of a series of partially deuterated or $\ce{^{18}O}$-labeled rosmarinic acid isotopologues.  Neither of those is an easy task (far from it), so it's nearly impossible to say for sure what the structure of the fragment is.
But think a perfectly reasonable proposition is that the ion is the same as whatever the structure of the catechol dehydration fragment is, perhaps a deprotonated hydroxybenzyne or maybe an epoxybenzene anion?  Loss of aromaticity in CID fragmentation not unusual.  In fact, a previous Q&A here on Chemistry StackExchange provides yet another example of this.
High-energy gas-phase ion chemistry is weird.


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Mechanism of formation of peak of mass spectrum of isobutane at m/z = 27
Here is the mass spectrum of isobutane:

(source: nist.gov)
Clearly the peak at $\mathrm{m/z} = 27$ corresponds to the species $\ce{C2H3+}$, the "protonated acetylene".
I am interested in possible formations for its formation from the molecular ion $\ce{t-BuH+}$.

Electron ionization fragmentation spectra, unlike say CID spectra, can be reasonably predicted computationally for small molecules.  This capability is only a few years old and is a major testament to how computational chemistry has advanced in the last few years.  Here's a recent review. http://pubs.acs.org/doi/full/10.1021/acs.jpca.6b02907

Some of the first mass spectrometrists to see this ion were Stevenson and Hipple in 1942.1  They were puzzled by it.
Fifteen years later, Rosenstock and Melton explained the source of this ion in an exhaustive study.2  First, the molecular ion $\ce{C4H10+}$ dissociates into fragments via five possible pathways, each with its own activation energy.  See Table VIII in their paper, which I reproduce below.

None of these primary dissociation pathways gives rise to $\ce{C2H3+}$.  Instead, it arises from a variety of processes from a subset (the metastable subset) of these five primary transitions.  These processes are shown in Tables X and XI of their paper.

All of the reactions that give rise to $\ce{C2H3+}$ are probably happening to some extent, but the lowest-energy route there for isobutane (right column) appears to be via $\ce{C2H5+ -> C2H3+ + H2}$ (only 1.9 eV).  However, $\ce{C2H5+}$ itself requires more energy to form than $\ce{C3H7+}$ ($m/z=43$), so the route $\ce{C3H7+ -> C2H3+ + CH4}$ may also occur as well.
The general conclusions of their paper are worth quoting:

The basic concept of the quasi-equilibrium theory is
that mass spectra are formed by a series of competing
consecutive unimolecular decomposition of excited ions.
The results discussed above lend strong support to this
concept. It was found that of several possible reactions
of a given ion, the metastable transition was associated
with the reaction having the least activation energy.
The general rate equation appropriate for these systems
is the product of a frequency factor and a function of
the excitation energy $E$ and activation energy $\epsilon$. When
the value of $E$ is such that the rate constant is of the
order of $10^6~\mathrm{sec^{-1}}$, the rate constant is extremely
sensitive to the value of $\epsilon$. The result is that a second
reaction having an $\epsilon$ higher by more than a few tenths of
an eV will have a considerably smaller rate constant,
even if the frequency factor is larger by a factor of a
thousand. Thus, in general, the correlation can be
carried out in terms of activation energies alone.

References

D. P. Stevenson, J. A. Hipple, Jr., J. Am. Chem. Soc. 1942, 64 (7),  1588–1594.
H. M. Rosenstock, C. E. Melton, J. Chem. Phys. 1957, 26 (2), 314-322.


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Why is Sodium Chloride White?
Why is solid sodium chloride white - and for that matter, why are the majority of salts white in solid form?

Possible duplicate of Why is snow white?

http://chemistry.stackexchange.com/questions/42303/why-are-simple-organic-molecules-usually-white

I'll point out that in mineralogy a streak test (http://geology.com/minerals/streak-test.shtml) is used to help identify the mineral. The gist is that the color of the fine particles created by rubbing the mineral across a plate of alumina is observed. For any mineral of light color the streak will probably be white or whitish, but for very dark colored minerals a very different color can often be observed.

Possible duplicate of why are simple organic molecules usually white?

Considering that sodium chloride is neither snow nor a simple organic molecule, I'm not sure how much of a duplicate this is. The argument has already been had in the comments on the latter question.

Large perfect NaCl crystals are clear and colourless. When they are smashed about as in table salt they become fractured and surfaces scratched. Light is refracted through and reflected in all directions from the surface of the crystals. They now appear white due to this scattering provided the light source is 'white '. The reflections and scattering are caused by the difference in refractive index of the crystal and air. Put the crystals into an index matching liquid and they will appear transparent again. (There is a $1/\lambda^4$ dependence on scattering which can be seen in deep snow).

Sodium chloride is colorless because the electrons are tightly bond tho the sodium cation and the chloride anion. Visible light has not enough energy to excite the electrons to higher energy levels. So there's no absorption or emission of visible light of a certain wavelength and therefore no color.
I wouldn't say that the majority of salts is colorless, since the more complex the composition of a salt is the more likely it is not colorless.

"Majority" is perhaps the wrong word, but practically all oxides, (pseudo-)halides, sulfates, nitrates etc. of main group metals are white, plus great many more. The big exception are compounds containing late transition metals, being often coloured.


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Heating of potash alum
I have this reaction : $$\ce{K2SO4.Al2(SO4)3.24H2O -> K2SO4 + Al2O3 + 24H2O}$$
And I notice that while $\ce{Al2(SO4)3}$ becomes $\ce{Al2O3}$ it doesn't happen so with $\ce{K2SO4}$. Why ?
I guessed that it would have to do something with the oxygen affinity of K.
As $\ce{K2O}$ has less oxygen than $\ce{K2SO4}$
Am I right ?

Aluminium sulfate is an unstable compound probably due to size difference between cation and anion and thus decomposes to aluminum oxide. (for e.g. size difference of the component ions of nitrogen triiode leads to instability resulting in an explosion).
$$\ce{2Al2(SO4)3 ->[770-860 C] 2Al2O3 + 6SO2 + 3O2}$$
Some sulfur dioxide react with oxygen to form sulfur trioxide.
On the other hand, potassium sulfate is stable due to equal size of potassium ion and sulfate ion and there is much stronger ionic attraction between the ions and thus require much heat in presence of reducing agent(carbon monoxide) to decompose it to  potassium sulfide.
$$\ce{K2SO4 ->[high~temp.~,~CO] K2S}$$
Thanks to @narendrakumar who extended the reasoning of  instability of aluminum sulfate. He says because of high polarization power of aluminium ion, the sulphate ion get polarized and this make it partially covalent . Compound having partial covalent character decomposes on heating and that's why aluminium sulphate decomposes while potassium sulphate remain unaffected.

Because of high polarization power of aluminium ion large sulphate ion get polarized and this make it partially covalent . Compound having partial covalent character decomposes on heating. That's why aluminium sulphate decomposes while potassium  sulphate remain unaffected


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Valid resonance contributors for 2-methylthiazole
There's a problem from an old exam that I was trying to do. The original structure is given on the left and they want all the possible resonance forms that have closed-shell atoms (octet for S) and formal charges of -1, 0, or +1 on all atoms. There's then a bold note that says, "As always, you should remember that linear atoms (sp hybridization) are not expected in rings of this size." So, my professor said that the circled electron pair in red is localized to the nitrogen atom, which therefore means that the resonance contributor on the right is invalid.
So, just to double check my logic, I'm assuming that the circled pair on right is localized because by moving them as if they were a delocalized pair, it would change the hybridization of nitrogen from sp2 to sp, which would make it's electronic geometry linear. Therefore, the pair on the nitrogen is localized because of the bold note (As always, you should remember that linear atoms (sp hybridization) are not expected in rings of this size) correct?

More to the point: the lone pair on N is perpendicular to the pi bond and cannot overlap with it - hence no delocalisation.

I’m sure a permutation of this has been asked here before … The gist is that nitrogen’s lone pair is orthogonal and cannot participate in resonance. It’s not above or below the ring, it’s pointing away from it.