text:
Weak acid dilution

Given a solution with molarity $\pu{0.5 M}$ of a weak acid $\ce{HA}$ has $\mathrm{pH} = 1.670$. How much water in liters needs to be added to $\pu{0.150 L}$ of the first solution to get a $\mathrm{pH} = 2.5$?
Consider volumes additive.

My attempt was calculating the $K_\mathrm{a}$ of the acid using the first $\mathrm{pH}$ and then using it in the standard variation table used in these problems with the knowledge that the concentration of $\ce{H3O+}$ at the equilibrium are given by the last $\mathrm{pH}$.

Can i add a photo?

Absolutely (click the 6th icon on the left on the toolbox when you are in the edit mode), but the preferred method of sharing is text. Check out How can I format math/chemistry expressions here? for more info. I'd recommend to add picture first to prevent the question from closing, and then edit it correctly (I can help you with this if you add the picture in the next 15 min, I then have to leave)

Thank you for the edit! I'd say that generally the approach looks OK, what is the problem here? Did you get the wrong answer? Regarding the formatting, >  denotes a quotation, e.g. I used it to quote out the problem itself since it's been taken from a third party (a textbook, I presume). I gotta go; anyway, welcome to Chemistry.SE and good luck with the question!

I'd like to be optimistic about it so I will tell that I got a different answer from the one that is given! Yes the problem is from a chemistry exercise book for engineers. I was having problems with the image uplaod because  of the size limit. Thanks!

@MattiaMariantoni - you don't really need to calculate the $K_a$. What is the simplified formula for the pH of a solution of a weak monoprotic acid? If you know that, you just need to write it once with $pH = 1.67$ and $C_{acid} = 0.5$, then again with $pH = 2$ and $C_{acid}$ as the unknown. Divide the first equation by the second, and you should get an expression in $C_{acid}$ alone. Solve that and you get the concentration that the acid needs to have in the final solution. The rest is stoichiometry.

Are you talking about pH=-ln([H+])?

No. I am talking about the formula that gives $pH = f(K_a,C_{acid})$ for a weak monoprotic acid. How you are tackling this problem if you don't know that? And BTW, pH is not the natural logarithm of hydronium concentration, it's in base 10.

Found it, but never seen before... thanks but why is it not necessary to calculate the $K_\mathrm{a}$?

I explained above why not: $K_a$ goes away when you divide the first equation by the second. Maybe please show the equation you found, so this post will help other users in the future.

http://phcalculator.blogspot.it/2013/04/calculating-ph-of-monoprotic-weak-acid.html

It is true what user637 writes and I will edit this into my answer.

@MattiaMariantoni : I see; no, that's not the simplified one I had in mind. With that one you do need to calculate $K_a$ first. The simplified one is $pH = -log_{10}(\sqrt{K_a \cdot C_{acid}})$.

You're welcome. In fact if you use the equations in $pH = ...$ form you need to subtract the second from the first, so you can use the properties of logarithms to eliminate $K_a$, rather than divide. I talked about dividing because I was thinking of the $[H_3O^+] = ...$ form.

The initial acid concentration is $\ce{C(HA)}=0.5\ \mathrm{M}$.
The weak dissociation of the acid follows:
$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)}$$
We'll ignore the auto-dissociation of water as negligible. So only the dissociation of the acid plays on the $pH$:
$$K_a=\frac{[\ce{H3O+}]\times [\ce{A-}]}{[\ce{HA}]}$$
Stoichiometry now tells us:
$\ce{[HA]=C(HA)-10^{-pH}}$
and:
$\ce{[A-]=10^{-pH}}$
and:
$\ce{[H3O+]=10^{-pH}}$
Insert all this into the expression above to get the $K_a$.
Now with the second $pH$, insert the same expressions into the now known $K_a$, and calculate the new acid concentration, call it $\ce{C'(HA)}$.
Let the original volume be $V_1$ and the volume after dilution $V_2$, then because the number of moles of acid hasn't changed:
$$\ce{C(HA)}\times V_1=\ce{C'(HA)}\times V_2$$
Then $V_2-V_1$ is the volume that has to be added to increase the $pH$ as stated per the problem.

Note that:
$\ce{[HA]>>10^{-pH}}$
Which simplifies the calculation of $K_a$ because $\ce{[HA] \approx C(HA)}$.

It is also correct, if we use the simplified acid constant expression, that strictly speaking we don't need to calculate the $K_a$.
At $\ce{pH=1.670}$:
$$K_a=\frac{\Big(10^{-1.67}\Big)^2}{\ce{C(HA)}}$$
At $\ce{pH=2.5}$:
$$K_a=\frac{\Big(10^{-2.5}\Big)^2}{\ce{C'(HA)}}$$
Divide both expressions by each other, the $K_a$ drop out and find the ratio of the concentrations of acid. From there find the ratio and difference of volumes, as above.

Same result of the book, although for the second pH we cannot neglect the $10^{-pH}$ term.

Why not? It's even smaller than in the first case: 0.0032 mol/L.

But the acid concentration is also much smaller, correct me if I am wrong please.

No, I think you're right. I haven't run any numbers. In any case you can use the correct expression in the second case and get the right result.


text:
How do I solve this Ionic Equilibrium problem?
Problem :

Equal volumes of $\pu{0.02M}$ $\ce{AgNO3}$ and $\pu{0.02M}$ $\ce{HCN}$ were mixed. Calculate
  $\ce{[Ag+]}$ at equilibrium assuming no cyano-complex formation.
  $K_{sp}(\ce{AgCN})= \pu{2.2\times10^{-16}}$, $K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}$

My solution:
Concentration of $\ce{AgNO3}$ and $\ce{HCN}$ before reacting when mixed will become $\pu{0.01M}$. Since $\ce{AgNO3}$ is a strong electrolyte it dissociates completely. So, initial concentration $[Ag^+]$ is $\pu{0.01M}$.
$$\ce{HCN<=>H+ + CN-} $$
After reaction let
$[\ce{HCN}]= \pu{0.01-x}$, $[\ce{H+}]=\pu{x}$, $[\ce{CN-}]=\pu{x}$.
Now $\ce{CN-}$ reacts with $\ce{Ag+}$
$$\ce{Ag+ + CN^-<=> AgCN}$$
Initial $[\ce{Ag+}] = \pu{0.01}$, $[CN^-]=\pu{x}$.
After reaction let $[\ce{Ag+}] = \pu{0.01-y}$, $[\ce{CN-}]=\pu{x-y}$, $[\ce{AgCN}] = \pu{y}$.
Now, $K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \frac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}=\frac{x(x-y)}{0.01-x}\approx\frac{x(x-y)}{0.01} $.
And $K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}] = (0.01-y)(x-y)$.
Solving these two equations I get $\pu{(0.01-y)= 8.9\times10^{-11}=[\ce{Ag+}]}$.
But the solution in my book says the answer is $\pu{5.9\times10^{-5}}$.
I don't where am I going wrong.

Yeah, but the question has asked to assume no formation of complex.

The error is your simplification $0.01-x ≈ 0.01$. $x$ is close to 0.01, you can't ignore it. If you use the non-simplified version you get more or less the correct result. FYI, by writing all the mass balances and solving numerically, you would get: $[ [HCN]=5.939E-5, [CN^-]=3.704E-12, [H^+]=0.009941, [Ag^+]=5.939E-5, [NO_3^-]=0.01]$.

BTW, I was intrigued by Gert's comment, so I solved the system including the $Ag(CN)_2^-$ equilibrium. The result, as far as $[Ag^+]$ is concerned, is the same: $[[HCN]=5.939E-5, [CN^-]=3.704E^-12, [H^+]=0.00994, [Ag^+]=5.939E-5, [NO_3^-]=0.01, [Ag(CN)_2^-]=4.563E-9]$.

@user6376297 thanks for pointing out the mistake

The OP's mistake was assuming that $\ce{[H^+]} = x$ and  $\ce{[CN^-]} = x$. If just HCN was added to water that would be true. But $\ce{Ag^+}$ and  $\ce{CN^-}$ form a ppt so $\ce{[H^+] > [CN^-] } $.

@MaxW : sorry, this is not correct. The OP knows that $[H^+]$ and $[CN^-]$ are not the same. If you look carefully, you will see that in the final equations from the OP, $[H^+] = x$ and $[CN^-] = x-y$, and the value of $x$ and $y$ is determined by the solution of the system of equations (which is $x=0.009940608815422887, y=0.009940608811718633$ - and you can easily find that $[Ag^+] = 0.01-y = 5.939E-5$). As I pointed out above, the main error was to consider $x$ much smaller than $0.01$, thus wrongly simplifying the first equation.

@Gert : why not? It's maths we're talking about here, so there is no 'surely'. If you can, please prove me wrong with calculations and figures; then you can make a point. As far as I can tell, $\frac {[Ag(CN)_2^-]}{[Ag^+][CN^-]^2} = \frac {4.563E-9}{5.939E-5 \cdot (3.704E-12)^2} = 5.6001E-18$, which seems close enough to me...

@user6376297 - Sorry, but I got lost. The OP evidently used $x$ two different ways which is bad form. I stopped trying to unsort the OP's analysis when I got to  --- After reaction let
$[\ce{HCN}]= \pu{0.01-x}$, $[\ce{H+}]=\pu{x}$, $[\ce{CN-}]=\pu{x}$.

Chemistry
AgNO3 is a strong electrolyte it dissociates completely.
$\ce{AgNO3 -> Ag+ + NO3-}$
If it was just HCN in solution, then most of the $\ce{CN^-}$ species would be as HCN. But since $\ce{AgCN}$ precipitates, more HCN dissociates. So essentially the overall reaction is:
$\ce{HCN + Ag+ -> AgCN}\,\ce{v + H+}$
The $K_{sp}$ of $\ce{AgCN}$ is such that it is reasonable to assume that for all practical purposes that the precipitation is essentially complete.
Assumptions
(1) I assume that the correct answer should have two significant figures which are all that the equilibrium constants have. However I will carry 4 significant figures in the intermediate calculations to try to avoid significant rounding errors.
(2) $\ce{[H^+]} \approx 0.01000$ molar and $\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$ (In other words about 1 ppt error for both)
Mass balance
If we consider 1 liter of the mixture of solution then we also have the following two mass balance equations:

The volume doesn't matter. Thinking of it as 1 liter just makes it simpler to get to the final result.

$0.01000 = \ce{[CN-] + [HCN]} +\ \mathrm{moles\ \ce{AgCN}\ce{v}\tag{1}}$
$0.01000 = \ce{[Ag+] +\ \mathrm{moles}\ \ce{AgCN}\ce{v}\tag{2}}$
$\therefore \ce{[Ag+] = [CN-] + [HCN]}\tag{3} $
Equilibrium equations
$K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}\tag{4}$
$K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}]\tag{5}$
dividing (4) by (5) we can get:
$2.819\times10^6 = \dfrac{\ce{[H+]}}{\ce{[Ag+][HCN]}}\tag{6}$
Rearranging (6) and substituting $\ce{[H+] = 0.01000\ \mathrm{molar}}$
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}  }{\ce{[Ag+]}}\tag{7}$
Rearranging (5) we get:
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}}\tag{8}$
Substituting (7) and (8) into (3) we get:
$\ce{[Ag+] =} \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}} + \dfrac{3.547\times10^{-9} }{\ce{[Ag+]}}\tag{9} $
Solving:
$\ce{[Ag+]^2} \approx 3.547\times10^{-9}\tag{10} $
$\ce{[Ag+] =} 5.955 \times10^{-5}\tag{11}$
Rounding to two places... $\ce{[Ag+] =} 6.0 \times10^{-5}$
Solving for other species
Substituting for $\ce{[Ag+]}$ in (8) gives
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{5.955 \times10^{-5}} = 3.694\times 10^{-12}\tag{12}$
Substituting for $\ce{[Ag+]}$ in (7) gives
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}   }{5.955 \times10^{-5}} = 5.956\times10^{-5}\tag{13}$
Argentocyanide Complex
The problem has ignored the argentocyanide complex. Is that reasonable?
$K = 5.6\times10^{18} =\dfrac{\ce{[Ag(CN)2^-]}}{\ce{[Ag+][CN-]^2}}\tag{15}$
rearranging
$\ce{[Ag(CN)2^-]} = \ce{[Ag+][CN-]^2}\times(5.6\times10^{18})=4.794\times10^{-9}\tag{16}$
So it is reasonable to ignore the additional complication.
Checking

You always check your assumptions on this sort of problem. It is easy to go down the wrong rabbit hole.

with $K_\mathrm{a}\ce{(HCN)}$
$\pu{6.2\times10^{-10}} = \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]} = \dfrac{(0.01)(3.694\times 10^{-12})}{5.956\times10^{-5}} = 6.202\times10^{-10}\tag{14}$
mass balance
It makes sense chemically that in the acidic solution $\ce{[HCN]} \gg \ce{[CN-]}$ so that $\ce{[Ag+]} \approx \ce{[HCN]}$. Making that assumption earlier using equation (6) would have reduced the work for this problem.
[H+] = 0.01000 molar and 0.01000 moles AgCN ppt per liter of the mixture
Also assuming that $\ce{[H^+] = 0.01000\ \mathrm{molar}}$ is certainly correct to two significant figures. However I tried to carry 4 significant figures in the intermediate calculations to avoid rounding errors. The assumption isn't that good.
!! PROBLEM !!
The assumptions that: $\quad\quad\ce{[H^+]} \approx 0.01000$ molar $\quad\quad\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$
are not good to 4 significant figures. It would have been better to use:
$\quad\quad\ce{[Ag+]} \approx \ce{[HCN]}$
$\quad\quad\ce{[H^+] \approx 0.01000 - [HCN] \approx 0.010000 -[Ag+]}$
and get a quadratic equation when simplifying equation (6). Then $\ce{[Ag+]} = 5.938\times10^{-5}$ and $\ce{[H+]} = 9.941\times10^{-3}$.
So the value $\ce{[Ag+]} = 5.956\times10^{-5}$ is off by +0.303%, or 3 ppt, which I think is acceptable, but it is marginal.

Sorry but the solution by the OP, when removing the wrong simplification $0.01 - x ≈ 0.01$, gives more correct figures than the ones in this answer. This answer misleads the OP into thinking that a much more complicated approach is necessary, and actually introduces errors that may result in the answer not being accepted.

@user6376297 - Yes and no. I hate problems like this. It is never clear to me how many significant figures "0.2 M $\ce{AgNO3}$" is supposed to have. I assumed that two significant figures were good enough. In the "Checking" section I made the point that I went down the wrong rabbit hole and a better solution was possible. The point was that I was trying to make was that for this type of problem that checking your assumptions is necessary. I'll edit and make this clearer...

Complete Analysis
To beat this to death perhaps here is a more complete analysis:
Assumptions
(1) Assume that the correct answer should have two significant figures (which are all that the equilibrium constants have) but carry 4 significant figures in the intermediate calculations to try to avoid significant rounding errors.
(2) AgNO3 is a strong electrolyte it dissociates completely.
$\ce{AgNO3 -> Ag+ + NO3-}$
Thus the $\ce{[NO3^-]}$ can be assumed to be exactly 0.010000 molar.
(3) $\ce{Ag(CN)}$ is very insoluble. So the reaction should produce about 0.01 molar $\ce{H+}$ and about 0.01 moles of $\ce{Ag(CN)}$ per liter of solution.
(4) In a fairly acid solution the remaining cyanide species in solution will be mostly $\ce{HCN}$ with very little $\ce{CN^-}$.
Equations
Equilibrium equations
$K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}\tag{E1}$
$K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}]\tag{E2}$
dividing (E1) by (E2) gives another equation, but only two of the three equilibrium equations will be independent.
$2.818\times 10^6 = \dfrac{\ce{[H+]}}{\ce{[Ag+][HCN]}}\tag{E3}$
Mass Balance equations
In order to balance charges in the remaining solution:
$$\ce{[Ag^+] + [H^+] = [NO3^-] + [CN^-] }$$
but since $\ce{NO3^-} =0.01000$
$$\ce{[Ag^+] + [H^+] = 0.01000 + [CN^-]\tag{M1}}$$
To mass balance $\ce{CN^-}$ for some volume, V, we have:
$$0.01000V = \ce{[CN-]V + [HCN]V} +\ \mathrm{moles(\ce{AgCN}\ce{v})}$$
To mass balance $\ce{Ag^+}$ for some volume, V, we have:
$$0.01000V = \ce{[Ag+]V} +\ \mathrm{moles(\ce{AgCN}\ce{v})}$$
Subtracting B5 from B4, rearranging, and then dividing by $\ce{V}$ gives:
$$\ce{[Ag+] = [CN-] + [HCN]}\tag{M2}$$
Now adding (M1) and (M2) we get (M3), but again only two of the three equations M1, M2 and M3 are independent.
$$\ce{[H^+] + [HCN] =0.01000}\tag{M3}$$
Unknowns
There are five unknowns ( $\ce{[H+], [Ag+], [HCN], [CN-], AgCN(s)}$ ) in the original system.
Manipulation of the mass balances for $\ce{CN^-}$ and $\ce{Ag^+}$ eliminated the $\ce{AgCN(s)}$ as a variable and left mass balance equation M2.
Summary
Now with two equations from E1, E2, and E3 and then two equations from M1, M2 and M3 there are 4 equations and 4 unknowns ( $\ce{[H+], [Ag+], [HCN], [CN-]}$ ) so an exact solution is possible. Rather than to try to tediously solve some messy polynomial equation by hand the idea is to use some notion of chemistry to simplify the equations so that the result can be obtained with at most solving a quadratic equation.
Of course to obtain a numerical solution by computer is possible regardless how messy the equations and the computer doesn't care how tedious the calculation is.

Exact Solution
Starting with the analysis noted for this problem, rearrange equation E1 to:
$K_{a}\ce{[HCN]} = \ce{[H+][CN-]}\tag{E1}$
Now solve for each of the three unknowns in terms of $\ce{[Ag+]}$.
$\ce{[CN-]}$ in terms of $\ce{[Ag+]}$.
Just need to rearrange E2:
$\ce{[CN-]} = \dfrac{K_{sp}}{\ce{[Ag+]}}\tag{E2}$
$\ce{[H+]}$ in terms of $\ce{[Ag+]}$ using M1 and E2.
\begin{align}\\
\ce{[H^+] &= 0.010000 + [CN^-] - [Ag^+]}\tag{M1}\\
\\
\ce{ &= 0.010000 + \dfrac{K_{sp}}{\ce{[Ag+]}} - [Ag^+]}\\
\\
&=\dfrac{K_{sp} + 0.010\ce{[Ag+] - [Ag+]^2}}{\ce{[Ag+]}}\\
\end{align}
$\ce{[HCN]}$ in terms of $\ce{[Ag+]}$ using M2 and E2.
\begin{align}\\
\ce{[HCN] &= [Ag+] - [CN-]}\tag{M2}\\
 \\
&= \ce{[Ag+]} - \dfrac{K_{sp}}{\ce{[Ag+]}}\\
 \\
&= \dfrac{\ce{[Ag+]^2}-K_{sp}}{\ce{[Ag+]}}\\
\end{align}
Substituting gives
$$K_a \cdot\dfrac{\ce{[Ag+]^2}-K_{sp}}{\ce{[Ag+]}} = \dfrac{K_{sp} + 0.010\ce{[Ag+] - [Ag+]^2}}{\ce{[Ag+]}} \cdot \dfrac{K_{sp}}{\ce{[Ag+]}}$$
Final Equation
One form of the final equation is given below. Of course multiplying and dividing by constant would yield other forms.
Using any other combination of equations would yield an equivalent equation.
$$ 0 = \dfrac{K_a}{K_{sp}}\ce{[Ag+]^3} + \ce{[Ag+]^2} - (K_a+0.01)\ce{[Ag+]} - K_{sp} $$
Numerical Solution
Solving with WolframAlpha equation solver using:
$$\text{a==6.2`20e-10, b==2.2`20e-16, N[[(a/b)*x^3+x^2-(a+0.01000)*x-b=0],20]}$$
gives $\ce{[Ag+]} = 5.939\,118\,459\times10^{-5}$ as noted by user user6376297 in a comment under the OP's question.

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text:
Influence of vapor pressure on surface tension

How does a liquid's vapour pressure affect its surface tension?

i.e. to say, if a liquid's vapour pressure increases, will the surface tension increase or decrease?
The SERP for this question yields only research papers to which I don't have access. This theory isn't given in my textbook either.

Your cause and effect is backward. A decreasing surface tension increases vapor pressure. though you ussually can't control surface tension as an independent variable.

@A.K. Oh! Actually, in my test paper, the statement was "A liquid with low vapour pressure will have high surface tension ", that's why I had framed the question this way. But, what is the reason for "decreasing surface tension increases vapor pressure"?

High surface tension is caused by strong intermolecular forces, the same are responsible for a low vapour pressure.
It is however quite a different matter to put a molecule on a surface and to completely strip it from a condensed phase and put it into the gas phase. There sure is no (inverse) proportionality or similar between vapour pressure and surface tension, and they don't "affect" each other.
A really stupid test question, imo.

When another independent variable such as temperature is applied, vapor pressure would increase due to weakening of the intermolecular forces; surface tension would decrease thus allowing more molecules to escape and so the solubility would decrease. However, the mass transfer rate would increase at a fixed dissolved gas concentration. The temperature imparts kinetic energy to the liquid so the molecules are more agitated. These are the effects of temperature on both parameters. I agree with the premise that vapor pressure and surface tension do not affect each other.

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text:
Is it possible to freeze water by dissolving a salt?
Theoretically, by dissolving a salt in water the melting point lowers, approximately 1.86 K kg/mol, making it more difficult to freeze water. However, the process of dissolution of certain salts is endothermic, lowing the temperature of the water. Is it possible that the water freezes due to that change of temperature although the melting point also lowers, or is it impossible?
For instance, consider the next case. In 1 kg of liquid water at 0 °C, we put a mol of KI. The melting point of water should be now about −1.86 °C. However, the enthalpy of dissolution of the KI is about 20 kJ/mol, meaning that the temperature of water should lower till −4.8 °C. So, theoretically, it should  freeze, but does this happen in reality?

Should the last sentence read -- So, theoretically, it should freeze, but does this happen in reality?

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. Best of luck with your question.

Yes, you are right

related https://chemistry.stackexchange.com/questions/5748/why-does-ice-water-get-colder-when-salt-is-added?rq=1

https://chemistry.stackexchange.com/questions/48760/safe-endothermic-reaction-that-gets-below-freezing https://chemistry.stackexchange.com/questions/24088/what-is-an-example-of-a-very-low-temperature-endothermic-reaction

It is possible, in a significantly different way that you envision. Freezing point depression via the cryoscopic constant is an example of a colligative property, which holds only for relatively dilute solutions. Once the solutions get very concentrated, intermolecular interactions become more complicated and do not generalize, meaning different compounds will affect the solvent differently.
Therefore, with enough of the right solute, it is possible to actually increase the freezing point of water. Oscar Lanzi's example of clathrates is interesting, but if you want to use salts, then I present you tetrabutylammonium hydroxide. It is known to form stable hydrates with a defined composition and containing a large amount of water, such as $\ce{(C4H9)4N+OH^-.30 H2O}$. The 30-hydrate is a solid which melts at approximately 30 °C, containing 67.6% water by weight.
This specific compound, at this specific concentration, forms a particularly stable network of solvating water molecules. You can think of it as stabilizing the structure of ice, allowing it to occur above the normal melting point of pure water. Thus, in principle, if you add enough of the anhydrous salt to water, when you reach the right ratio, the solution will "freeze".
Many compounds actually display this behaviour of forming solid hydrates at certain defined compositions, but tetrabutylammonium hydroxide is unusual in that it forms solid hydrates with a very large amount of water.

Not a salt, but methane works by forming a different phase, https://en.wikipedia.org/wiki/Methane_clathrate. More than just a gee-whiz curiosity, methane clathrate was a big complicating factor in the Deepwater Horizon explosion in the Gulf of Mexico several years ago.

Although this is could be an interesting comment, I don't see how it answers the question.

*My answer is theory based and contains mainly of text.
Your thought is correct, that salts do require heat to dissolve. This is observed when you add a large amount of table salt to your typical cup of water, the temperature drops (only slightly though).
The heat of the solution affects the solubility of the salt, so the cooler your water, the less salt you are able to dissolve it in.
And in the world, there is still no salt that can have the ridiculous high molar solubility to lower the solution to freezing temperatures.

I don see it at all. For example, in the case given (potassium iodide), the solubility of KI at 0ºC is 7mols/(kg of water). If we put one mol in a kg of water its temperature should be around -5ºC. I am not very sure, but considering that the solubility of KI at 20ºC is about 8mols/kg, chances are that a mol of KI can be dissolved in a kg of water at -5ºC

I don't think it is possible. My reason being that the world has many oceans and seas even salt water lakes. All of which vary in different densities and different temperatures but some form ice packs, ice burgs, glaciers yet others don't freeze at all. Again it's down to the type of salt density & composite of that salt surely???


text:
Confusing stability order of free radical structures
I have a question regarding the solution that has been given.
Question:

In the given compound, what would be the order of ease with which hydrogen atom can be abstracted from carbon atoms I to VI?:

The solution that has been given is:

In the solution they've posted above, how is the 2nd structure more stable than the third structure? The third one clearly has resonance with the pi bond, while the 2nd one has no resonance with the oxygen atom (it can't "accept" oxygen's lone pair as it is not a carbocation, and oxygen does not have an empty $\mathrm{d}$ orbital either).
Please can anyone shed light on this issue. Thank you!

The order looks fine: C-H BDE of CH3OCH3 = 83 kcal/mol; allylic C-H BDE of propene = 89 kcal/mol. Blanksley and Ellison, Acct. Chem. Res., 2003, 36, 255.

@user55119 Thank you for digging up references to confirm the order! :) However, I hope you understand the aim of this question was to justify the order using logical reasoning on resonance. Thanks once again!

The non-bonding electrons on oxygen donate electron density to the half filled carbon radical p-orbital. Some what like having a carbocation stabilized by oxygen. Why is propene have a higher BDE? I wouldn't have been surprised if the BDE's were switched. Certainly, Diethyl ether and disopropyl ether undergo autooxidation. Dimethyl ether undoubtedly slower. I presume propene doesn't, or very slowly, autoxidzes.

@user55119 I am sorry sir but you're confusing me. "Why is propene have a higher BDE? I wouldn't have been surprised if the BDE's were switched" I don't understand what you're trying to say exactly.  It seems  you're counter-claiming your own data. And what is the role of autooxidation here? And what is the underlying logic behind the data you had presented? Thank you!

I am saying that a priori it is difficult to predict qualitatively whether the C-H bond of dimethyl ether or the allylic C-H bond in propene is weaker. Autooxidation refers to the tendency of diethyl ether to react with O2 by a radical pathway. As to the logic you request, read Ellison's paper. Case closed.

@user55119 Thank you for helping. "a priori it is difficult to predict" I understand the question cannot be answered using simple organic chemistry rules, am I correct? "As to the logic you request, read Ellison's paper" I hope you understand sir but not everyone has access to read research papers like you do. Thanks!

Since this question was not answered considering it was posted for a long time, I'd try to give enough reasoning for the given stabilization order.
I'd start with mass spectroscopy. The most widely used technique in mass spectrometry up to date is electron ionization (EI), which produces molecular ions from gas-phase analytes. These molecular ions then fragment in a reproducible way. The energy sufficient for ionizing and fragmenting gas-phase analyte molecules is acquired by interaction with $\pu{70 eV}$ electrons. The range of the ionization energy of most organic compounds are fall between $\pu{10 eV}$  and $\pu{20 eV}$. The most reference spectra are obtained at $\pu{70 eV}$ because at that level the perturbations in electron energy have negligible effects on ion production (Ref.1).
A neutral analyte molecule absorbs energy during the interaction of the ionizing electron with its electron cloud. The analyte molecule interacts with an electron having $\pu{70 eV}$ of kinetic energy, which quickly causes the ejection of one of the electrons from the analyte (it needs to absorbs approximately $\pu{14 eV}$ as internal energy to make this ejection). Because the analyte molecule has many electrons, it is reasonable to ask, “Which electron leaves during the ionization process first?” Whereas sufficient energy may have been absorbed by the analyte molecule to eject any one of its electrons, the most probable ejection will be of an electron that is least tightly bound. Therefore, it is predictable that the most likely site of ionization will be where electrons are loosely bound (e.g., in the nonbonding orbital of a heteroatom) (Ref.1):
$$
\begin{array}{c|ccc}
\hline
\bf{\text{Compound}} & \bf{\text{Ionization Energy, }} \pu{eV}^\mathcal{a} \\
\hline
\ce{H3C-NH-CH3} & 8.23 \\
\ce{H3C-CH2-NH2} & 8.86 \\
\ce{H3C-O-CH3} & 10.03 \\
\ce{CH2=CH2} & 10.51 \\
\ce{H3C-CH3} & 11.52 \\
\hline
\end{array}\\
^a: \text{Reference 1}
$$
The above data shows abstracting an electron from $\ce{O}$ atom is easier than that from a carbon. That means, lone pair is loosely bound than $\sigma$- or $\pi$-bond. Therefore, it is safe to assume that resonance contribution of a lone pair is more than that of a double bond. Even though, this is not full a lone pair but one electron contribution for resonance, it is possible to contribute one electron from a lone pair and the other electron stays in one empty anti-bonding orbital as shown in below:

Again, after all, molecular ion of ketone stays as $\ce{R2-C=O^{.+}}$ in EI-mass spectrum of a ketone, usually with the intensity of ~40% of the base peak. That means $\ce{R-CH=O^.-R'}$ is pretty stable contributor (Recall both carbocations and carbon radicals have the same order of stabilization). Thus, it is safe to say that stabilization by oxygen atom is better than that by a double bond. This conclusion would be supported by user55119's comment (vide supra), which states that (homolytic) bond dissociation energy  (BDE) of $\ce{CH3O\color{red}{CH2-H}} = \pu{83 kcal/mol}$ while that of $\ce{CH2=CH-\color{red}{CH2-H}} = \pu{89 kcal/mol}$ (Ref.2). Keep in mind that the easiness of BDE is due to the stabilization of the resultant radical (either by oxygen or the double bond).

References:

J. Throck Watson, O. David Sparkman, In Introduction to Mass Spectrometry: Instrumentation, Applications, and Strategies for Data Interpretation; Fourth Edition, John Wiley & Sons, Ltd.: Chichester, West Sussex, England, 2007, “Chapter 6: Electron Ionization,” pp. 315-448 (ISBN: 978-0-470-51634-8).
Stephen J. Blanksby, G. Barney Ellison, "Bond Dissociation Energies of Organic Molecules," Acc. Chem. Res. 2003, 36(4), 255–263 (https://doi.org/10.1021/ar020230d).

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


text:
Lewis Acid / Base reaction for SO2 and H2O
I am having trouble with this question:
Identify the Lewis acid and Lewis base in each of the following
reactions:
$\ce{SO2(g) + H2O(l) = H2SO3(aq)}$
What I tried:
A Lewis acid is an electron pair acceptor.
A Lewis base is an electron pair donor.
The other problems of this type involved cations or anions, so I was able to identify the cation as the Lewis acid and the anion as the Lewis base.
However, this question does not involve ions.  So I tried to look at the lone pairs on the molecules.  Since they both have lone pairs, I figured since $\ce{SO2}$ has more lone pairs, it would act as the Lewis base.  Also in the final molecule, the hydrogen atoms would be bonded to two separate oxygen atoms, so I thought of them as migrating to the lone pairs on the $\ce{SO2}$ oxygen atoms.
However, the textbook says that $\ce{SO2}$ is the Lewis acid and $\ce{H2O}$ is the Lewis base.
Why would this be the case?

Clearly it is $\ce{H2O}$ that donates an electron pair to from the bond? So it is the Lewis base.

Can you explain why it must be $\ce{H2O}$ that donates the electron pair(s) and not $\ce{SO2}$?

I'll try and answer that tomorrow.

Molecules that contain polar multiple bonds can function as Lewis acid because the central atom is electron deficient with a vacant orbital that can accept an electron pair.
When $\ce{SO2}$ dissolves in water , it forms the weak diprotic acid $\ce{H2SO3}$ (sulfurous acid) from Lewis acid/Lewis base reaction:
$$\ce{O\bond{=}S\bond{=}O + H2O <=> H2SO3}$$

The $\ce{O}$ atom of $\ce{H2O}$ molecule donates a lone pair to the $\ce{S}$ of  $\ce{SO2}$,forming a new $\ce{S\bond{-}O}$ $\sigma$-bond and breaking $\ce{S\bond{=}O}$ $\pi$-bond.

The oxygen in $\ce{H2O}$ already has its orbitals ($2s^2$,$2p^4$) filled and since oxygen is in the second row of the periodic table it can't form hypervalent molecules (That means that the oxygen would have more than 8 valence electrons).
Sulfur on the other hand is in the third row of the periodic table and can therefore form such molecules.
Because of this it is clear, that the $\ce{H2O}$ can't act as a Lewis acid as it can't have any more electrons. Therefore it has to be the Lewis-base which leaves $\ce{SO2}$ as the Lewis-acid.

I think what you say is misleading. Because you are saying that water cannot act as a Lewis acid. However, water is amphoteric and can act as a Lewis acid, as it has an electron deficient hydrogen atom capable of accepting an electron pair.

I think you are referring to hydrogen bonds aren't you? In that case water is indeed acting as a Lewis acid but I'm not sure if one would consider this as a real Lewis acid-vase-reaction... See https://www.quora.com/Can-water-be-classified-as-a-lewis-acid?share=1

Oxygen is more electronegative than sulfur, so it pulls the shared valence electrons closer to itself. This causes sulfur to have a partial positive charge and hence act as the electrophile. Furthermore, sulfur is able to have more than 8 valence electrons as it is able to expand its octet by using the d orbitals.

The funny thing is existence of H2SO3 in water has not been ever confirmed, only hydrated SO2 or HSO3-.

You should say which oxygen atom you are refering to; both water and the SO2 contain oxygen atoms.


text:
What is the valence molecular orbital of the ions of a fluorine molecule?
I have the following assignment:

Draw the molecular orbital diagram of $\ce{F2}$, $\ce{F2+}$ and $\ce{F2-}$.
  Calculate the bond order and say, if there is a bond, which compound has the strongest bond
  and which one has the longest one.

My attempting at answering has lead me to the following conclusions:
As my understanding goes, $\ce{F}$ has $7$ valence electrons in the $2s^2$ and $2p^5$ subshells. By combining the atomic orbitals, one will obtain:
$2$ $\sigma_{2s}$ orbitals (a bonding and an antibonding), 2 $\sigma_{2p}$ a bonding and an antibonding, and $4$ $\pi_{2p}$ (2 degenerate bonding and 2 degenerate antibonding).
By filling the MO, I obtained the following bond orders: 1 for $\ce{F2}$ and $\frac{1}{2}$ for $\ce{F2+}$ and $\ce{F2-}$.
First of all, I would say that there is definitely a bond in all the three compounds since the bond order is positive. Then I would say that $\ce{F2}$ has the strongest bond (since it has the highest bond order), then $\ce{F2+}$ and $\ce{F2-}$ have longer bonds than $\ce{F2}$ due to the lower bond order, however I do not know which one has the longest bond.
As for the question:
I never came across $\ce{F2+}$ and $\ce{F2-}$. Is my reasoning correct? Do these compounds exist? Furthermore, I draw the MO using an atom of $\ce{F}$ and an ion of $\ce{F}$ (cation and anion depending on the case). Is the procedure correct?

Your procedure sounds correct. If a compound has some nonzero bond order, it should be possible to form so you can say $\ce{F2+}$ and $\ce{F2-}$ could exist.


text:
Synthesis Golf VIII: 4,6-Diacetylhygrophorone A12
As usual we are a bit late on this, but here is this month's molecule:

which is known as (4R,5S,6R)-4,6-diacetylhygrophorone A12, and is an antifungal natural product isolated from mushrooms of the genus Hygrophorus.
As far as I know this compound has not yet been synthesised in the lab, but has been isolated and its relative configuration assigned by NMR measurements in anisotropic media.[1]
The criteria are the same as for the last challenge I posted:

the synthesis must be enantioselective;[2] and
no more than one chiral centre may be purchased in the form of a building block. (Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.)

Otherwise, any commercially available starting material (as usual, in the Sigma–Aldrich catalogue) is fair game.

(1) Schmidts, V.; Fredersdorf, M.; Lübken, T.; Porzel, A.; Arnold, N.; Wessjohann, L.; Thiele, C. M. RDC-Based Determination of the Relative Configuration of the Fungicidal Cyclopentenone 4,6-Diacetylhygrophorone A12. J. Nat. Prod. 2013, 76 (5), 839–844. DOI: 10.1021/np300728b.
(2) The absolute stereochemistry was not assigned in ref 1, so you may synthesise either enantiomer. Obviously, though, that doesn't make a difference to the design of the synthetic route, so I have arbitrarily decided to display this enantiomer (following ref 1).

I really want to find something that works with PLE...

Is it fair game using known procedure for making an intermediate resembling the target and with most of the stereochemistry already setup?

@Marko Yes! Why not, especially if we can all learn something from it.

Establishing the correct stereochemistry proved out to be an extremely difficult task, as I have found only one procedure for these systems. This first scheme is a published procedure:

1) Diels-Alder reaction (Paul D. Bartlett, Fred A. Tate
J. Am. Chem. Soc., 1953, 75 (1), pp 91–95);
2) Alpha-allylation and 3) ring closure by anionic addition to ester group to generate two regioisomers which are separated (Siwapinyoyos, Thebtaranonth, J. Org. Chem. 1982,47, 598-599);
4) Epoxidation of the non-conjugated enone and 5) opening of the epoxide by elimination (Chantarasiri, J. Chem. Soc., Chem. Commun., 1990,0, 286-288)
6) Kinetic resolution by enzymatic O-acetylation, followed by 7) hydrolysis of the acetate and vacuum flash pyrolysis at 450 0C and 0.1 mmHg. The acetate is needed in the target molecule, but I guess that it has to be hydrolyzed before the pyrolysis as to avoid side reactions. 8) is epoxidation with dimethyldioxirane and 9) opening of the epoxide with water provides a intermediate resembling the target molecule a lot (Klomkao, Tet. Asymm. 14, 24, 2003, 3885-3889).
From this molecule starts my approach:

10) Acetonide protection of the diol should be selective (syn 1,2 hydroxyl groups react in the presence of anti hydroxyl groups and 1,3 hydroxyl groups). The free hydroxyl group is then acetylated.
11) The acetonide is hydrolyzed. The primary hydroxyl group is oxidized to the aldehyde and the tertiary one is protected as MOM ether.
The following 3 steps are used for protection of ketone in presence of aldehyde:
12) Aldehyde is protected in presence of enone first.
13) Enone is protected by Noyori conditions.
14) Aldehyde is deprotected.
15) Organocopper addition to the aldehyde in presence of the ester to form a mixture of diastereomers (surely it would show diastereoselectivity but this would need to be tried out).
16) If the diastereoselectivity is bad, the mixture of diastereomers is oxidized to the ketone which is then stereoselectively reduced by chelate-control with zinc-borohydride. This could be problematic as the acetonide oxygens could interfere with chelation, but I cannot predict this with confidence. If they don't interfere much the intermediate would look something like in the image attached at the bottom. The hydryde attacks opposite to the acetonide and the alkyl chain. If acetonide oxygens interfere, then a screening of reducing agents is required.
17) Acetylation of the hydroxyl group and MOM and acetal deprotection should provide the target molecule.

I wonder if I chose a molecule that was too tough...

@orthocresol Seems so. The stereochemical relationship is difficult to establish.

I thought that the oxygens would provide a good handle to control stereochemistry. But I will admit, I haven't actually tried to come up with a route myself.

@orthocresol when's the next challenge coming

@Laksh I apologise, but I have been very busy in the previous months, too busy to do one of these. I will try to do one in June, possibly.

A diastereoselective synthesis (total asymmetry obtained by a Sharpless kinetic resolution)

You have put in a tremendous effort into 3 Synthesis golfs (I noticed you had answered 2 similar questions recently). However, it will be much easier for the community to go through and vote for your answer were it slightly clearer. It would be great if you could use ChemDraw/ChemSketch (if you have them) or ChemDoodle to illustrate your reactions. I'm afraid I haven't got enough time right now, or I'd have done it myself. If you find it too difficult, you could wait for someone like @andselisk to come along and help.

I would do it on ChemDraw and will at some point but don't have the time at the moment, sorry.

Alright, I'll probably be free in 2 weeks, and if you couldn't find time to do it yourself yet, I'd be willing to do it. Thanks for replying!


text:
Understanding entropy / spontaneity
I have issues understanding the spontaneity for these processes:
Which of the following processes are spontaneous:
(c) alignment of iron filings in a magnetic field;
(d) the reaction of sodium metal with chlorine gas to form sodium chloride;
The answer given for both (c) and (d) is that the entropy of the system decreases.
This has me puzzled because I don't think that even answers the question.
Clearly I would think both (c) and (d) are spontaneous processes.  For (c), the iron will naturally align itself in the magnetic field. For (d), I know the reverse reaction is non-spontaneous (table salt does not normally disassociate), so the forward reaction must be spontaneous.
Assuming I am correct, how do I reconcile this with the book answer?
The entropy of the universe increases for any spontaneous process.  If we assume the entropy of the system is negative (like the book does) then the entropy of the surroundings must be positive and greater.  I don't know what that would mean in this context.
I agree that it makes sense that the entropy of the system is negative.  The iron filings are moving to a more ordered state in (c).  For (d) there are more micro-states originally since there were more moles of reactants.

Based on your provided context, your book's "answers" are indeed misleading and non-answers.

Clearly I would think both (c) and (d) are spontaneous processes. For (c), the iron will naturally align itself in the magnetic field. For (d), I know the reverse reaction is non-spontaneous (table salt does not normally disassociate), so the forward reaction must be spontaneous.
The entropy of the universe increases for any spontaneous process. If we assume the entropy of the system is negative (like the book does) then the entropy of the surroundings must be positive and greater.
I agree that it makes sense that the entropy of the system is negative. The iron filings are moving to a more ordered state in (c). For (d) there are more micro-states originally since there were more moles of reactants.

These statements are correct. As you've argued, the change in entropy of the surroundings must be positive for these reactions to be spontaneous. The surroundings constitute a thermal reservoir, so the change in entropy is simply proportional to the heat evolved during the process, which implies that heat must be transferred from the system to the surroundings.
For the formation of sodium chloride, this is quite intuitive: sodium chloride is very stable and has a large, negative enthalpy of formation. The case of magnetization is less obvious, but it can be shown (see Ex 17.4 in Blundell and Blundell's Concepts in Thermal Physics) that an isothermal increase in the magnetic field leads to evolution of heat into the surroundings, and the same reasoning as above applies.


text:
Why does LiAlH4 reduce the double bond of cinnamaldehyde, whereas NaBH4 does not?
I read an answer here about why $\ce{LiAlH_4}$ does not reduce conjugated double bonds, and that $\ce{NaBH_4}$ does. But in the case of cinnamaldehyde, $\ce{LiAlH_4}$ reduces the conjugated double bond and $\ce{NaBH_4}$ does not. Why is it so?
Citation for $\ce{LiAlH_4}$: Quora
Citation for $\ce{NaBH_4}$:

Image Source

Cinnamaldehyde (1) does not undergo 1,4-reduction but rather 1,2-reduction with LiAlH4. The product of the reduction is 3-phenylpropan-1-ol (2). If conjugate addition were to occur, enolate 3 would be produced (stable to the reductant; enolates can be prepared to protect ketones from this reductant) and aldehyde 4 would be expected, as long as the aqueous workup were to destroy the reductant before any enolate 3 could be protonated.
All of this aside, the answer was provided experimentally in 1954.1 Reduction occurs 1,2 to afford the aluminate salt of cinnamyl alcohol 5. This species undergoes hydroalumination of the double bond by a presumed intramolecular process to give alkyl aluminate 6. This cyclic aluminate was deuterated (D2O) at the benzylic position to provide 9. In addition, trapping of 6 with benzophenone afforded diol 7 in >80% yield! Reduction of cinnamaldehyde with LiAlD4 and aqueous workup would be expected to give dideuterio 3-phenylpropan-1-ol (8).
Sodium borohydride reduces cinnamaldehyde via 1,2-reduction in hydroxylic solvent to cinnamyl alcohol in 97% yield.2 The intermediate alkoxyborane (Cf.; structure 5) either is incapable of reacting with the double as the aluminate does or the alkoxyborane rapidly exchanges with solvent.

1) Dornow, A., Winter, G., Vissering, W. Chemische Berichte, 1954, 87, 629.
2) Chaikin, S.W., Brown, W.G., J. Am. Chem. Soc., 1949, 71, 122.


text:
Is supercriticality dependent on partial presssure, or total pressure?
Supercritical CO2 is commonly used as a solvent in organic chemistry (notably in industrial processes like decaffeination). I presume that in such cases, the CO2 is typically pure, or very close to it.
But suppose you had a mixture of, e.g., 50% CO2 and 50% nitrogen. Pure CO2 becomes supercritical at approximately 31C and 73atm; particularly in terms of its solvent properties, would the CO2 in that mixture also demonstrate supercritical properties at 73atm total pressure, or would it be necessary to double the total pressure, producing 73atm partial pressure of CO2? Or is the interaction more complex than that, with mixture of CO2 and N2 having a joint critical point different from either single gas?

CO2 would get supercritical, N2 no, AFAIK. It's approximation of course.

There is no such thing as supercritical properties. By looking at a gas under given pressure and temperature, you'll never be able to tell is it supercritical or not.
If you try to lower temperature or pressure and see what happens, then yes, the interaction is more complex, and a mixture of two compounds does indeed have a joint critical point different from those of both pure compounds; moreover, the very meaning of a critical point becomes different and more complicated.

If there's "no such thing as supercritical properties", then why is ScCO2 used as a solvent for industrial processes, where subcritical gaseous CO2 is not? Would liquid CO2 work just as well, with the supercritical state being used just to avoid a first-order phase transition when releasing pressure for deposition? It's the effect of mixing gasses on the range of conditions where CO2 is useful as a supercritical solvent that I am primarily interested in.

Supercritical CO2 is used in industry because it works (that is, dissolves whatever needs to be dissolved). Whether liquid CO2 works just as well is not immediately clear; chances are it does. Whether CO2 will work if mixed with N2, and whether it will get supercritical if mixed with N2 are two different, barely related questions.


text:
Why doesn’t aniline undergo Friedel-Crafts alkylation?
Even though aniline is an activated benzene derivative, it still doesn’t undergo Friedel-Crafts alkylation. Why?
Can it undergo Friedel-Crafts acylation?

The answer lies in the fact that aniline is a Lewis base and $\ce{AlCl3}$ is a Lewis acid. The reaction between aniline and $\ce{AlCl3}$ hampers the catalytic activity of $\ce{AlCl3}$ required to perform the Friedel-Crafts alkylation and acylation.

Despite the activation of the $\ce{NH2}$ group, Friedel-Crafts alkylations and acylations fail because the $\ce{NH2}$ group acts as Lewis base and interacts with the Lewis acid catalyst. ... A way to overcome this problem is to convert the $\ce{NH2}$ group into an amido group prior to Friedel-Crafts reaction. ...The amide group can be hydrolysed back to aniline group after alkylation.
(Pg. 1167 An Introduction to Organic, Inorganic and Physical Chemistry. 4th edition. Catherine E. Housecroft, Edwin C. Constable. Pearson Education Limited)

The aniline forms a Lewis acid base adduct with $\ce{AlCl3}$ :
This prevents the interaction between alkyl or acyl chloride with $\ce{AlCl3}$.


text:
In the blue bottle experiment, will it eventually be "used up" after repeated use?
The blue bottle experiment is a bottle half full of a colorless solution (dextrose, KOH, and methylene blue, in the version that I saw) and air that, when shaken, turns blue through the methylene blue oxidizing. Over time, the blue color fades while the methylene blue is reduced and the solution goes back to colorless. This can be repeated.
My question is, can this be repeated indefinitely, or is there some chemical in the solution that is being eaten up? Is the mechanical energy of shaking enough to sustain the reaction indefinitely? If it can be used up, what exactly is being used up?

Nothing last forever.

@NilayGhosh How do you know?:)

@andselisk 2nd law of thermodynamics

@NilayGhosh Which is only applicable to isolated systems. What about the rest?

@NilayGhosh Well, how about lasts a few months to a few years ;) Without adding extra chemicals (except maybe air)

I guess I understand what @Nilay is trying to say, but I'd refine the term you are looking at: it's not exactly the 2nd law of thermodynamics, but rather chemical fluctuation. Which basically means that the degradation is a rather statistical event and may or may not occur in the nearest future.

You said it yourself. It reduces the methylene blue, which would then be the limiting component.

@andselisk I edited my post. Did I make it more clear what I'm actually asking? (which is whether or not the blue bottle experiment can be reused indefinitely)

@Zhe But the act of shaking the bottle oxidizes it again. From what I understand, oxidization and reduction are inverses. Doesn't this mean that after some time, the solution returns to its original, non-shaken state? Isn't it simply flipping back and forth between oxidization states? Or am I just misunderstanding terminology?

Then you'll run out of oxygen to oxidize. I've also noticed over time that the exact colors in the reaction aren't the same any more. It's quite possible that there are side pathways that make the system not fully reversible.

The blue bottle experiment, and other similar reactions like the "chemical traffic light", are not infinitely repeatable.  They are driven by the oxidization of glucose (or some other suitable reducing agent, such as ascorbic acid) by atmospheric oxygen, and will eventually stop working once the bottle runs out of either glucose or oxygen.  However, each cycle only consumes a rather small amount of glucose and oxygen, so the reaction can be repeated quite many times.
Also, the blue indicator dye (methylene blue) itself is not used up in the reaction.  Even if the reaction does "run out of fuel", it can be made to work again just by adding more glucose to the solution and/or simply by opening the bottle to let more oxygen in.  However, since the glucose is not getting fully oxidized all the way to water and carbon dioxide in this reaction, the various incomplete oxidization products will gradually accumulate in the solution.
I'd expect that, if you kept adding more and more glucose to the solution, eventually you'd end up with a sticky sugary alkaline goo full of partly oxidized glucose products that was too viscous to be swirled around the flask any more.  At that point, you'd really have no choice but to dispose of the experiment and start over with fresh ingredients.

I could understand the chemical traffic light not being infinitely repeatable, since there's no input of energy. The video I watched on the blue bottle experiment didn't specify any of the chemicals being eaten up in the reactions. He only talked about the oxidization/reduction of the methylene blue. So I figured the energy was being supplied through shaking. Thanks for the informative answer :)

It can be repeated as long as there is oxygen in the bottle.
This is due to the fact that the oxidation and reduction that is happening will always consume the respective chemicals and only the methylene blue is being transformed back and forth.

I am not sure.  I think the dextrose is being consumed by oxidation as well.  Moreover if it it being converted to carbon dioxide this will react in acid-base manner with the KOH.  Only the methylene blue component seems to last "forever".

Yes you are absolutely right... That's what I meant only that I couldn't remember what reacted how, so I only wrote about the oxygen

That is not true. I've done this reaction dozens of times, and eventually the colors are different and then it finally stops working.

And in how far does this contradict my answer?


text:
Mass of chromium in ammonium dichromate

What mass of chromium is in $\pu{100 mL}$ of $\pu{0.0400 mol/L}$ $\ce{(NH4)2Cr2O7}$?

The specific concept is how to find the mass of a certain ion/element in a substance, given the concentration. I am at a total loss to this.
At first I tried to find the molar mass of the chromium itself, which resulted in roughly $\pu{104 g/mol}$. I then tried to find the amount of moles in the concentration, which is
$$\pu{0.1 L} \cdot \pu{0.04 mol/L} = \pu{0.004 mol}$$
Then I multiplied the molar mass of chromium from the amount of moles in the concentration, and I got $\pu{0.208 g}$.
This is not the answer. The answer was $\pu{0.800 g}$
I then tried a different approach. I tried to find the molar mass of my original substance, then divided the molar mass of chromium from the molar mass of ammonium chromate, which will give me the amount of chromium per ammonium chromate. Then I thought all I needed to do was to multiply it by the concentration. This just messed up everything and I was back to square one.
How do I approach this problem?

I briefly did the math and I didn't get 0.8 g. It would be 0.8 g though, if you were asked about the mass of chromium(VI) oxide $\ce{CrO3}$, not the metal itself.

Sorry I messed that up. Indeed it is dichromate. The atomic weight of a single chromium atom is 51.996 g/mol, but I have roughly estimated it to be 52 g/mol, and since ammonium dichromate has 2 chromium atoms, I multiplied it by 2. I am deeply sorry if any of this is confusing, because I am confused beyond all comprehension as I try to solve this question.

Definitely just asking the mass of $Cr$. Also, where does $CrO3$ come from? I can't seem to think of a dissociation equation for that.

Done, I also edited your question a little. If you are interested, read about formatting on Chemistry.SE

To find the mass of chromium, one needs to know the amount $n$ ($m = nM$). Each formula unit of ammonium dichromate $\ce{(NH4)2Cr2O7}$ contains $2$ $\ce{Cr}$, therefore
$$n(\ce{Cr}) = 2 \cdot n(\ce{(NH4)2Cr2O7})$$
The amount of ammonium dichromate can be found from its concentration (as you correctly did):
$$n(\ce{(NH4)2Cr2O7}) = C(\ce{(NH4)2Cr2O7}) \cdot V$$
Putting it all together, the final expression for the mass of chromium is
\begin{align}
m(\ce{Cr}) &= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{Cr}) \\
&= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{51.996 g mol-1} \\
&= \pu{0.416 g}
\end{align}
The answer deviates with the one provided by the textbook; however, if you were asked to find the mass of chromium(VI) oxide $\ce{CrO3}$ instead, it fits perfectly:
\begin{align}
m(\ce{CrO3}) &= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{CrO3}) \\
&= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{99.99 g mol-1} \\
&= \pu{0.80 g}
\end{align}
I deduced $\ce{CrO3}$ by backtracking to the molecular weight of the compound, which should be around $\pu{100 g mol-1}$, and among common chromium substances this is the only one that comes to mind.


text:
Propane-water molar ratio

Propane gas, $\ce{C3H8}$, burns in oxygen to produce carbon dioxide and water vapour as follows:
$$\ce{C3H8 (g) + 5O2 (g) = 3 CO2(g) + 4 H2O (g) + \pu{2200 kJ}}$$
If $\pu{1.5 mol}$ of propane is consumed in this reaction, how many moles of $\ce{H2O}$ are produced?

I'm completely stuck on this question, I have no idea how to start. Any help would be greatly appreciated.

Let's start with the balanced reaction: how many moles of water 1 mol of propane produces? You have $1~\ce{C3H8}$ on the left (reactants), and $4~\ce{H2O}$ on the right (products). Abstract from the energy, it's irrelevant here.

Wouldn't that be 4 moles of water produced?

Yep, 4 it is. Now if there is 1.5 mol of propane (e.g. 1.5 times more), how many moles of water would it be? Remember that coefficients reflect molar ratios.

6 moles of water produced?

Exactly! $4 \cdot \pu{1.5 mol} = \pu{6.0 mol}$.

Wow. That was a lot simpler than I thought. Thank you!

No prob:) Remember to properly balance the reaction first, and then use the ratios between the participating chemicals, when you encounter this type of problems.

The ratio between propane and water can be taken out of your reaction equation.
$$\frac{n(\ce{H20})}{n(\ce{C3H8})} = \frac{4}{1}$$
So if you burn 1.5 mole propane it will result in
$$\frac{4}{1} \cdot 1.5 = 6$$
mole of water being produced


text:
Lewis dot structures of CO2
I know the general Lewis dot structure of carbon dioxide is the one where there are two double bonds connecting oxygens to carbon.
However the question is, does $\ce{CO2}$ have resonance structures?
We could have moved one of the bonds (pi bond in this case) to the other $\ce{O-C}$ bond, leaving us with a single bond and a triple bond forming $\ce{CO2}$.

Strongly related, if not a duplicate: Why is carbon dioxide considered a Lewis acid?

@andselisk Hmm, I think answers there include those  resonance structures without any good reason.

It is almost always possible to draw resonance structures. In case of $\ce{CO2}$ you could imagine a resonance structure in which the carbon doesn't have an electron octet and a positive charge and one of the oxygen atoms carrying a negative charge.
You could also draw one in which the carbon is only connected via a single bound to each oxygen and carrying two positive charges while the two oxygens carry a negative charge each.
You might come up with more of those structures.
So the question you really wanna ask is whether these resonance structures have any real contribution to the actual bonding state of $\ce{CO2}$ and for that you will see that all of the described strucutures are highly unfavourable due to their energetic state. Therefore the normal Lewis-notation of $\ce{CO2}$ decribes the real bonding state pretty well as all other resonance structures don't contribute much.

I'm just confused. There was an IB chemistry test that asked for which of the following compounds have resonance structures. However, it said that CO2 did not count as a compound having resonance structure.

The they most likely meant resonance structure that actually contribute to the structure and bounding state of the molecule. In that case (as mentioned above) CO2 does indeed not have a resonance structure.

Every single molecule can be described with more then one contributor to structure https://chemistry.stackexchange.com/questions/32306/can-100-covalent-bonds-exist/32313#32313


text:
Europium stability in +2 and +3 state
I would like to ask a question about europium's stability in the $+2$ and $+3$ oxidation state.
The electronic configuration of europium in its neutral state is $\ce{[Xe] (4f)^7 (6s)^2}$.
Now, when in the $+2$ oxidation state, the electronic configuration is $\ce{[Xe] (4f)^7}$ and in the $+3$ oxidation state, it is $\ce{[Xe] (4f)^6}$.
Now, I thought the $+2$ oxidation state is more stable because it's a half-filled $\ce{f}$ sub-shell so there is less mutual repulsion between electrons in the same sub-shell.
However, an article from nature.com reads the following:

Europium metal is now known to be highly reactive; the element's most stable oxidation state is +3, but the +2 state also occurs in solid-state compounds and water.

Now, I would like to ask, how is it possible for europium to be most stable in its $+3$ oxidation state, knowing what I have mentioned above?

We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details.

Somewhat related: Lanthanide property exceptions. It's worth noticing that the answers given there also use the same argumentation as you do, which I would also agree with.

This is going to be a rehash of Oscar Lanzi's answer, but this point needs to be driven home, so I make no apologies.
"Special" electron configurations - fully-filled or half-filled subshells - are only a relatively minor factor in determining the stability of oxidation states. I've written about this before in a slightly different context, but IMO it would be well worth reading this, as the exact same principles operate in this case: Cr(II) and Mn(III) - their oxidizing and reducing properties?
What determines the "most stable oxidation state"? We can consider what happens when we increase the oxidation state of a metal: firstly, we need to pay ionisation energies to remove electrons from the metal. On the other hand, though, metal ions in higher oxidation states can form stronger bonds (e.g. lattice energy in ionic compounds, solvation energy in solution, or more covalent bonds in molecular compounds). The balance between these two factors therefore leads to a "best" oxidation state.
For example, sodium could hypothetically go up to Na(II) and form $\ce{NaCl2}$. Theoretically, this compound would have a significantly larger lattice energy than $\ce{NaCl}$. However, sodium's second ionisation energy is prohibitively large, which means that it stays put in the +1 oxidation state. On the other hand, magnesium's second ionisation energy is not prohibitively large, and therefore Mg preferentially forms $\ce{MgCl2}$ over $\ce{MgCl}$.
For all lanthanides, the first three ionisation energies are all fairly comparable. There is a graph here which shows the variation in the ionisation energies of the lanthanides, which I reproduce below.

(source: Inorganic Chemistry 6ed, Weller et al., p 630)
For all lanthanides, these three ionisation energies are easily compensated for by the extra lattice energy / solvation energy derived from a more highly charged ion. The fact that Eu(II) has a $\mathrm{f^7}$ configuration only serves to make its third IE marginally larger than that of Gd. This difference is sufficient to make Eu(II) an accessible oxidation state (cf. different electronic behaviour of EuO and GdO), but not sufficient to make it the most stable oxidation state.
Note that "most stable oxidation state" depends on the conditions, too, but I assume we are talking about aqueous solution.

Electron configuration is not the be-all and end-all of stability.  It's just one term in the energy balance.  Other factors like the electron affinity of the anion former (think of oxygen or fluorine versus sulfur or iodine), lattice or solvation energies, etc may override the electron configuration term; if so, then you find your europium in the +3 oxidation state after all.

That first sentence needs to be bolded in all caps. If there ever was a time to use obnoxious formatting for emphasis, this is it...

One reason is that in the f orbital, the +2 oxidation state makes it that one pair of electron in the orbital is not filled, and will at least try to fill it.
And the +3 orbitals filled those pairs, even though it didn't fill the whole orbital. Of course filling the whole orbital (like noble gases) will be great but it is easier for the metals to fill up it's pair.

See Hund's Rules


text:
Can such material exist?
Can a solid material exist, such that it shatters if you hit it lightly (with a solid object) but it deforms and doesn't shatter if you hit it hard (say with a hammer)?

I'd say this is more a physics question than chemistry.

Not exactly an answer to your question; however, you might want to have a look at Bologna bottles.

@paracetamol I think it's quite straightforward and everyone understands what I'm mean. If you wish, we could define "hard and light" as pressure. It doesn't matter at what pressure the result changes, the point is that there exists such point which determines whether the object gets shattered or deformed. Where shattering occurs at lower pressure than deformation.

@paracetamol Imagine a material which shatters like glass when you drop it on the floor but it gets deformed like wax when you smash it with a force of a hammer.

Ah well, I suppose I was nitpicking more than I should have. I see your point, and I stand corrected. I voted to reopen.


text:
Is there a specific chemical reaction between capsaicin and casein? Is there a stoichiometric relationship between them?
I am trying to establish the reaction between capsaicin and casein in order to neutralize the effects of the capsaicin. Therefore, I must determine if there is a reaction and a stoichiometric relation.

I rather doubt that there's a significant binding between capsaicin and the protein casein or a stoichiometric reaction.

If you look at the structure of capsaicin, you'll recognize that it's a rather unpolar molecule. With other words, it is hardly soluble in water, but will dissolve in oil. This is exactly where the fats in milk come into play. The "neutralizing" effect of milk with respect to the sensory "hotness" of capsaicin is just the solution of the unpolar compound in an unpolar solvent.

I would be very surprised if there was absolutely no interaction between the two but I would not be surprised if the interaction were in any way specific.


text:
Can carbanion rearrange to attain more stability?
We all know that carbocations rearrange to attain more stability, but I have never heard anything about carbanion rearrangements for attaining more stability.
Is it possible for a carbanion to rearrange? If not,why?

related https://chemistry.stackexchange.com/questions/42560/why-dont-free-radicals-undergo-rearrangement-like-carbocation-carbonium-ions-do

Also, look up E1cb mechanism. Not clear that we'd call that a rearrangement, but it feels related.

Sooner or later anionic rearrangement name reactions should turn up in your organic chemistry classes. I will present two in this post.
The first and arguably most well-known are the Wittig rearrangements; these can proceed as [1,2] or as [2,3] rearrangements.[1] The Kürti/Csakó states that:[1]

The [1,2]-Wittig rearrangement proceeds via a radical-pair dissociation-recombination mechanism, while the [2,3]-Wittig rearrangement is a concerted, thermally allowed sigmatropic process proceeding via an envelop-like transition state in which the substituents are pseudo-equatorial.

This means that strictly speaking only the [2,3]-Wittig rearrangement could be considered an actual carbanion rearrangement as per your question but many chemists may also consider the [1,2]-Wittig rearrangement one since it effectively does generate a rearranged and thereby more stable anion. Both reactions are shown in scheme 1.

Scheme 1: the [1,2] (top) and [2,3] (bottom) Wittig rearrangements.
The second example has recently gained significant applications in synthetic organic chemistry; it is the same general reaction except it employs a silyl ether instead of an alkyl ether. Here, the stability of both sides of the reaction is similar so both the forward and backward direction are known; these are the Brook and retro-Brook rearrangements (see scheme 2).

Scheme 2: The retro-Brook rearrangement (top) and Brook rearrangement (bottom).[2,3]
West et al. who reported the reactions in scheme 2 note the interesting effect that the amount of butyllithium added influences the equilibrium. If substoichiometric or catalytic amounts are added, the normal Brook rearrangement takes place with the silyl ether being the more stable species. Under excess butyllithium conditions, the retro-Brook rearrangement takes place indicating that under those conditions the alcoholate is the more stable species.[3]
I mentioned that arguably the Brook rearrangement is more common in organic synthesis; this is due to Amos B. Smith’s anion relay chemistry or ARC.[4] Smith uses priviledged synthetic intermediates that can give alcoholates upon an initial reaction. The previously introduced silyl group migrates towards the alcoholate, liberating another carbanion which can subsequently participate in another nucleophilic attack. This allows multiple anion-dependent transformations to occur in a single reaction vessel.

References:
[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, p. 490.
[2]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, p. 64.
[3]: R. West, R. Lowe, H. F. Stewart, A. Wright, J. Am. Chem. Soc., 1971, 93, 282–283. DOI: 10.1021/ja00730a065.
[4]: A. B. Smith III, M. Xian, J. Am. Chem. Soc., 2006, 128, 66–67. DOI: 10.1021/ja057059w.

I've never heard of proton shifts in carbanions- at least not in a formal treatment. They seem to be fairly common; for example, this mechanism for imine formation has it explicitly written out as the second step. Are there any rules for doing these?


text:
How to visualize a crystal from a unit cell PDB and how to draw the unit cell's axes?
I am looking at the following PDB experimental data:
Length (Å)  Angle (°)
a = 155.72  α = 90.00
b = 155.72  β = 90.00
c = 106.16  γ = 120.00

I find it hard to imagine in my head what does it mean to have $\gamma = 120.00$ as an angle.

How can I see the unit cell's axes on the following structure?
How can I build a crystal with 16 or so structures like this?

How can I build the biological structure from asymmetric unit cell?

To view the $120^\circ$ angle, use any molecular viewer; one of the simplest and versatile desktop app is probably Jmol (available on Windows, Linux, MacOS). Download the PDB file, open it with the program. What you are going to see is the asymmetric unit:

Now, right click and select Symmetry > Reload {444 666 1} + Display 555. This will basically populate the entire unit cell (might take a while). This command is described in interactive scripting documentation:

{ijk i'j'k' 1} Loads a block of unit cells within the range ijk and i'j'k' (which should include 555), normalizing the operators to move the geometric center of the generated set of atoms into cell 555, then applying the lattice translation.

Finally, select Style > Scheme > Cartoon:

Can you elaborate on the parameters: Reload {444 666 1} + Display 555

@ox90 Basically, it's a Jmol code for displaying the entire content of a unit cell, disregarding the molecular boundaries.

@ox90 https://chemapps.stolaf.edu/jmol/docs/

I am trying to do the same thing with VMD, but I don't know how to do it. How do I take the asymmetric unit cell and rotate and shift it to create the total biological assembly
.

@ox90 Sorry, I don't really use VMD.


text:
Half-life equation for 2nd order kinetics
My friends and I were doing some problems from this year's IChO Preparatory Problems (PDF from the 49th International Chemistry Olympiad (2017)) when we stumbled upon a question which we had some confusion with.

Task 8. Decomposition of Nitrous Oxide
  Nitrous oxide decomposes exothermically into nitrogen and oxygen, at a temperature of approximately $\pu{565 ^\circ C}$.
  $$\ce{2N2O (g) -> 2N2 (g) + O2 (g)}$$
  This reaction follows the second-order kinetics when carried out entirely in the gas phase.
8.1) If the reaction is initiated with $[\ce {N2O}]$ equal to $\pu{0.108 mol dm-3}$, what will its concentration be after $\pu{1250 s}$ have elapsed at $\pu{565 ^\circ C}$? The rate constant for the second order decomposition of $\ce{N2O}$ is $\pu{1.10\times10^-3 dm3 mol-1 s-1}$ at this temperature.

Task 8 details the kinetics of the decomposition of nitrous oxide. We were confused with part 8.1 of the task which required us to find the concentration of the reactant after a particular duration of time has elapsed, given the temperature, initial concentration and rate constant of the reaction. We approached the question using two different methods:
I approached it using the method, as suggested in the solutions manual, of using the integrated rate law for 2nd order kinetics and substituting the values provided into that equation. My answer was exactly that stated in the solutions manual.
However, they approached it by first finding the half-life using the equation for the half-life for 2nd order kinetics and then, finding the concentration after knowing how many half-lives have passed. Their answer was slightly off.
After substituting random values into the equations for both methods, I realised that the "half-life method" which my friends used only gives a reasonable approximation when the time elapsed < half-life, gives the exact value when the time elapsed = half-life and goes completely off when the time elapsed > half-life. Why is that so? Since the half-life equation for 2nd order kinetics is derived from the integrated rate law (as shown on Chemistry LibreTexts), shouldn't they give the same answer?

Roughly speaking, half-life is not a thing at all for the second order (or any order other than first, for that matter).
What would they do after finding the number of half-lives? Raise 2 to that power? Too bad, because that only works for the exponential decay, which is the solution of the first-order kinetics and no other order. Really, what if the number of half-lives equals 2? After spending the first half-life, we now have a different half-life ahead of us, because it is concentration dependent, and the concentration has changed.
Following your link, we see the same statement, except they don't put enough emphasis on it, as to my taste:

For this reason, the concept of half-life for a second-order reaction is far less useful.

I understand what you mean. But how were they able to mathematically derive the half-life equation t1/2 = 1/(kCo) from the integrated rate law? Since the equation has been proven mathematically, it should work?

They got it by plugging the halved concentration into the exact integrated rate law and solving for time. As such, this thing is indeed proven mathematically and surely does work. It does tell you when the concentration reaches ${1\over2}c_0$. But it shouldn't and doesn't tell you when it reaches ${1\over4}c_0$. Nor does it tell what the concentration will be in $2t_{1/2}$.

So you are saying that it will only work for the 1st half-life, as I have also observed previously when trying to randomly substitute values, because the derivation has only been for the case of the 1st half-life?

Yes, exactly.$;$


text:
Chemistry of making glass

To boil glass, the lime and sand mixture of potash was heated and the off gas spent $\pu{125 mL}$ of barley hydroxide solution (density $\pu{1.14 g mL-1}$). The mass of the sediment was $\pu{4.925 g}$. It is known that the ratio of gas to alkalinity is $1 : 1$.

Find the volume of absorbed gas.

Find mass of sodium hydroxide in solution.

This is a homework, I read the book a few times but didn't understand how to solve this. Can you please help me?

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. Please don't use [tag:homework] tag in the future.

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. Best of luck and welcome to Chemistry.SE.

"...didn't understand how to solve this" Surely you must've been taught something in this regard to go about solving this problem, otherwise your teacher wouldn't hand it to you. What exactly is it that you don't understand?

The question is inconsistent! What is barley hydroxide? Is this supposed to mean barium hydroxide? In the context of the question and the compounds involved, this would make sense! If so, why is then asked for the remaining sodium hydroxide?

Apparently, a gas is released in the process of making glass. Can you figure out what that gas might be?
When let into an aqueous solution of a hydroxide (most likely barium), a precipitate is formed. This could give you some further intel on the nature of the gas.
If you know which gas was formed, you can come up with the nature of the precipitate.
From the molar mass of the precipitate and the given weight, you can calculate the number of moles.
You were given the volume and the density of the hydroxide solution. A part of the hydroxide has reacted to yield the precipitate. You should now be able to calculate how much is left in solution.


text:
Determine whether the tank will overflow with mass flowrate
Hydrocarbon mixture ($\pu{20 mol\%}$ $\ce{CH4}$ and the balance is $\ce{C2H6}$) enters a tank at a velocity of $\pu{1 m/s}$ and is withdrawn at a rate of $\pu{2 kg/s}$. The tank is initially one quarter full. The diameter of the pipe is given as $\pu{15 cm}$ and the height of the tank is $\pu{2 m}$. Density of $\ce{CH4}$ is $\pu{422.36 kg/m^3}$ and density of $\ce{C2H6}$ is $\pu{543.83 kg/m^3}$. The volumetric flowrate of the hydrogen mixture entering the tank is $\pu{6361.56 m^3/h}$
Determine the mass flowrate ($\pu{kg/h}$) of hydrocarbon entering the tank and determine how long ($\pu{s}$) does the tank take to overflow or to be emptied.
As for determining the mass flow rate, of the inlet, I used the densities of the component to help me get the total mass flowrate. This is what I have done:
$$\text{Density} = \frac{\text{Mass flowrate}}{\text{volumetric flowrate}}$$
For $\ce{CH4}$:
$$\text{Mass flowrate} =  (422.36)(6361) = \pu{2686868.482 kg/h}$$
For $\ce{C2H6}$:
$$\text{Mass flowrate} = (543.83)(6361) = \pu{3459302.63 kg/h}$$
Therefore, total mass flowrate into the tank is $\pu{6146171.112 kg/h}$.
Now I'm stuck on the question to determine how long does it take for the tank to overflow or be emptied.
Should I use the mass flowrate to determine or the volumetric flowrate?
I assumed is the mass flowrate, but I am unsure how to start.

You need to get your extraction rate and your filling rate to the same unit. The difference of both is then the rate in which the tank empties/fills. Convert that to a unit of time and calculate how long it will take to empty/fill up the tank.

I have updated your post with proper markup. If you want to know more, please have a look here and here. Please also note that the [homework] tag is deprecated.


text:
Why does Zinc not have magnetic properties despite it being a transition element?
I know Zinc has a full d-orbital which might contribute to the effect. But it is not ferromagnetic or paramagnetic like other transition element complexes. Why is that?

Well, paramagnetism and all the *ferr*magnetisms depend on unpaired electrons, so …

"has a d-orbital which might contribute to the effect" That's not the right conclusion.

Everything has magnetic properties. Zinc, in particular, is diamagnetic.

This Minute Physics video has a pretty accessible explanation of magnetism in different materials


text:
How do I perform a partial optimisation in GAMESS?
Sometimes it is necessary to perform a partial optimisation of a molecule, i.e. keeping certain variables constant. One example might be to pre-optimise a transition state, taking advantage of not having to compute the force constants in the initial set-up.
For example, let's consider the transition state of the Diels-Alder Reaction of ethene and butadiene, where we want to freeze the distance between the two molecules.
How can I achieve this in the software suite GAMESS?

Unfortunately GAMESS is not as user friendly as other quantum chemistry packages, so when it comes to running non-standard calculations a little bit more effort is necessary.
First of all I recommend optimising the reactants by themselves, so that you have a somewhat valid idea what the bond lengths on your level of theory are going to be. After that you should assemble your best guess for the structure. In this example I have chosen a separation between the molecules of about 220 pm.

The coordinates of this structure are:

! DIENE
C       -0.486294808     -0.431915730     -1.395523790
C       -1.166676574      0.550508040     -0.705577944
C       -1.166676574      0.550508040      0.705577944
C       -0.486294808     -0.431915730      1.395523790
H       -0.352988862     -0.347708199      2.465648783
H       -0.084717264     -1.289523730      0.864198908
H       -1.712490163      1.338624758      1.223514576
H       -1.712490163      1.338624758     -1.223514576
H       -0.352988862     -0.347708199     -2.465648783
H       -0.084717264     -1.289523730     -0.864198908
! DIENOPHILE
C        1.559305232     -0.085874351     -0.690651129
C        1.559305232     -0.085874351      0.690651129
H        1.514911897     -1.022640981      1.232100569
H        1.603704743      0.851022589      1.232175888
H        1.603704743      0.851022589     -1.232175888
H        1.514911897     -1.022640981     -1.232100569

There are approaches, where you would simply fix the Cartesian coordinates. Here we only want to fix two bond lengths, 1(C)-11(C) and 4(C)-12(C). If we were to fix the coordinates of these atoms, we would also fix 1(C)-4(C) and 11(C)-12(C). Therefore a better approach is to use internal coordinates, in GAMESS the keyword group $ZMAT is used.
Since there are a lot, it is tricky to set them up. We will therefore use the implemented automated procedure. Starting with a dry run, i.e. EXETYP=CHECK, so that we can check what the program actually does. We need to define $3\times N-6=42$ internal coordinates, thus NZVAR=42. Also note, that we do not use symmetry in this case. Because of the nature of the structure we need to add some bonds already manually, NONVDW(1)=1,11, 4,12.
This is the resulting input:

 $CONTRL
   SCFTYP=RHF
   RUNTYP=ENERGY
   EXETYP=CHECK
   COORD=UNIQUE
   ICHARG=0
   MULT=1
   NZVAR=42
   $END
 $BASIS
   GBASIS=STO
   NGAUSS=3
   $END
 $ZMAT
   AUTO=.TRUE.
   DLC=.TRUE.
   NONVDW(1)=1,11, 4,12
   $END
 $GUESS
   GUESS=HUCKEL
   $END
 $DATA
   C6H10
   C1
   CARBON      6.0     -0.486294808        -0.431915730        -1.395523790
   CARBON      6.0     -1.166676574         0.550508040        -0.705577944
   CARBON      6.0     -1.166676574         0.550508040         0.705577944
   CARBON      6.0     -0.486294808        -0.431915730         1.395523790
   HYDROGEN    1.0     -0.352988862        -0.347708199         2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730         0.864198908
   HYDROGEN    1.0     -1.712490163         1.338624758         1.223514576
   HYDROGEN    1.0     -1.712490163         1.338624758        -1.223514576
   HYDROGEN    1.0     -0.352988862        -0.347708199        -2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730        -0.864198908
   CARBON      6.0      1.559305232        -0.085874351        -0.690651129
   CARBON      6.0      1.559305232        -0.085874351         0.690651129
   HYDROGEN    1.0      1.514911897        -1.022640981         1.232100569
   HYDROGEN    1.0      1.603704743         0.851022589         1.232175888
   HYDROGEN    1.0      1.603704743         0.851022589        -1.232175888
   HYDROGEN    1.0      1.514911897        -1.022640981        -1.232100569
   $END

The resulting auxiliary outputfile .dat has now the group that we will use from here on in. And the .log file will give us the exact value of the internal coordinate which we want to freeze.

                     --------------------
                     INTERNAL COORDINATES
                     --------------------

                 - - ATOMS - -         COORDINATE      COORDINATE
 NO.   TYPE    I  J  K  L  M  N        (BOHR,RAD)       (ANG,DEG)
 ----------------------------------------------------------------
   1 STRETCH   1 11                     4.1406435       2.1911343
...
  10 STRETCH   4 12                     4.1406435       2.1911343
...

We can now modify our $ZMAT group accordingly. Unfortunately the input gets very long, if you would like to retain some readability. Note that we create redundant internal coordinates, so we need to increase the number of NZVAR, too, to read all of them in.
Technically we do not have to set the generation of coordinates to false, but it helps remembering what you have done if you come back in a month or so, therefore AUTO=.FALSE.. Now we add the coordinates we would like to freeze with IFZMAT(1)=<list> and the corresponding values FVALUE(1)=<list>.

 $CONTRL
   SCFTYP=RHF
   RUNTYP=OPTIMIZE
   COORD=UNIQUE
   ICHARG=0
   MULT=1
   NZVAR=89
   $END
 $BASIS
   GBASIS=STO
   NGAUSS=3
   $END
 $ZMAT
   AUTO=.FALSE.
   DLC=.TRUE.
   IZMAT(1)=
   1,   1,  11,
   1,   1,  10,
   1,   1,   9,
   1,   1,   2,
   1,   2,   8,
   1,   2,   3,
   1,   3,   7,
   1,   3,   4,
   1,   4,   6,
   1,   4,  12,
   1,   4,   5,
   1,  11,  16,
   1,  11,  15,
   1,  11,  12,
   1,  12,  14,
   1,  12,  13,
   2,   1,  11,  15,
   2,   1,  11,  16,
   2,   1,   2,   3,
   2,   1,   2,   8,
   2,   1,  11,  12,
   2,   2,   3,   7,
   2,   2,   1,  10,
   2,   2,   1,  11,
   2,   2,   1,   9,
   2,   2,   3,   4,
   2,   3,   2,   8,
   2,   3,   4,   5,
   2,   3,   4,  12,
   2,   3,   4,   6,
   2,   4,   3,   7,
   2,   4,  12,  13,
   2,   4,  12,  14,
   2,   4,  12,  11,
   2,   5,   4,  12,
   2,   5,   4,   6,
   2,   6,   4,  12,
   2,   9,   1,  10,
   2,   9,   1,  11,
   2,  10,   1,  11,
   2,  11,  12,  13,
   2,  11,  12,  14,
   2,  12,  11,  15,
   2,  12,  11,  16,
   2,  13,  12,  14,
   2,  15,  11,  16,
   3,   1,   2,   3,   4,
   3,   1,   2,   3,   7,
   3,   1,  11,  12,  14,
   3,   1,  11,  12,   4,
   3,   1,  11,  12,  13,
   3,   2,   1,  11,  15,
   3,   2,   1,  11,  16,
   3,   2,   3,   4,   5,
   3,   2,   3,   4,  12,
   3,   2,   3,   4,   6,
   3,   2,   1,  11,  12,
   3,   3,   2,   1,  11,
   3,   3,   4,  12,  13,
   3,   3,   4,  12,  14,
   3,   3,   4,  12,  11,
   3,   3,   2,   1,   9,
   3,   3,   2,   1,  10,
   3,   4,  12,  11,  15,
   3,   4,  12,  11,  16,
   3,   4,   3,   2,   8,
   3,   5,   4,  12,  11,
   3,   5,   4,  12,  14,
   3,   5,   4,  12,  13,
   3,   5,   4,   3,   7,
   3,   6,   4,  12,  11,
   3,   6,   4,  12,  14,
   3,   6,   4,  12,  13,
   3,   6,   4,   3,   7,
   3,   7,   3,   2,   8,
   3,   7,   3,   4,  12,
   3,   8,   2,   1,   9,
   3,   8,   2,   1,  10,
   3,   8,   2,   1,  11,
   3,   9,   1,  11,  16,
   3,   9,   1,  11,  15,
   3,   9,   1,  11,  12,
   3,  10,   1,  11,  12,
   3,  10,   1,  11,  15,
   3,  10,   1,  11,  16,
   3,  13,  12,  11,  15,
   3,  13,  12,  11,  16,
   3,  14,  12,  11,  15,
   3,  14,  12,  11,  16,
   IFZMAT(1)=
   1,   1,  11,
   1,   4,  12,
   FVALUE(1)=
   2.1911343,
   2.1911343,
   $END
 $GUESS
   GUESS=HUCKEL
   $END
 $DATA
   C6H10
   C1
   CARBON      6.0     -0.486294808        -0.431915730        -1.395523790
   CARBON      6.0     -1.166676574         0.550508040        -0.705577944
   CARBON      6.0     -1.166676574         0.550508040         0.705577944
   CARBON      6.0     -0.486294808        -0.431915730         1.395523790
   HYDROGEN    1.0     -0.352988862        -0.347708199         2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730         0.864198908
   HYDROGEN    1.0     -1.712490163         1.338624758         1.223514576
   HYDROGEN    1.0     -1.712490163         1.338624758        -1.223514576
   HYDROGEN    1.0     -0.352988862        -0.347708199        -2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730        -0.864198908
   CARBON      6.0      1.559305232        -0.085874351        -0.690651129
   CARBON      6.0      1.559305232        -0.085874351         0.690651129
   HYDROGEN    1.0      1.514911897        -1.022640981         1.232100569
   HYDROGEN    1.0      1.603704743         0.851022589         1.232175888
   HYDROGEN    1.0      1.603704743         0.851022589        -1.232175888
   HYDROGEN    1.0      1.514911897        -1.022640981        -1.232100569
   $END

The optimisation should look something like the following animation; and took 19 steps in my example.

Now you could use the results of this calculation to run a full fledged transition state search. Remember, that you have all necessary information to restart a run in the auxiliary output, i.e. the .dat file. You should always keep this one until you are completely done with your calculations.
If you want to calculate along and wonder what the optimised geometry is, here are the results of the final step:

----- RESULTS FROM SUCCESSFUL RHF      GEOMETRY SEARCH -----
----- COORDS, ORBS, GRADIENT, AND APPROX. HESSIAN -----
 COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS)
   ATOM   CHARGE       X              Y              Z
 ------------------------------------------------------------
 CARBON      6.0  -0.4847330011  -0.4122786646  -1.4086312155
 CARBON      6.0  -1.1902020846   0.5288839653  -0.7019679480
 CARBON      6.0  -1.1902020983   0.5288839622   0.7019679173
 CARBON      6.0  -0.4847329728  -0.4122786820   1.4086312270
 HYDROGEN    1.0  -0.4076405265  -0.3274824211   2.4860530248
 HYDROGEN    1.0  -0.4019303863  -1.4244097276   1.0396872383
 HYDROGEN    1.0  -1.5467259917   1.4139143889   1.2140730563
 HYDROGEN    1.0  -1.5467259527   1.4139143906  -1.2140730901
 HYDROGEN    1.0  -0.4076405381  -0.3274824126  -2.4860529903
 HYDROGEN    1.0  -0.4019304494  -1.4244096950  -1.0396872193
 CARBON      6.0   1.5587691569  -0.0858507824  -0.6884831136
 CARBON      6.0   1.5587691834  -0.0858507803   0.6884831288
 HYDROGEN    1.0   1.9167366148  -0.9561957837   1.2206602417
 HYDROGEN    1.0   1.6758814772   0.8473264236   1.2212666939
 HYDROGEN    1.0   1.6758814263   0.8473264124  -1.2212666800
 HYDROGEN    1.0   1.9167365785  -0.9561957779  -1.2206602275
--- OPTIMIZED RHF      MO-S --- GENERATED AT Thu Dec 14 23:42:52 2017
E=     -230.0376739583, E(NUC)=  227.2020852580
 $VEC
...

Disclaimer: There is another way of freezing coordinates. I might add that at another point as another answer. It is a weeny bit more complicated, however offers also some more flexibility.


text:
Blue litmus turning pink
I just read that carbonic acid turns blue litmus pink and not red.
Is it just because it is a weak acid. What about organic acids then? How do they respond to the litmus test?
If strength and color are related, can we determine the strength of an acid by determining if the litmus paper is more pink or more red?

Litmus paper will indeed be less or more strongly coloured depending on the pH of the solution you are measuring. Unfortunately though, there is no easy way to quantify it so you cannot use it to semi-accurately determine pH.


text:
Why does an electron-donating group reduce the polarity of the O-H bond?
Why does the polarity of $\ce{O-H}$ bond decrease if an electron-donating group is attached in the compound?
For example, I have seen that when alkyl groups are joined in phenols, the polarity of $\ce{O-H}$ bond in them decreases. Why does that happen?

How do you know that it decreases?

It is written in my book.

Because when the akyl group comes and attaches to the ring,since alkyl groups are electron releasing they give their electrons in ring, then the ring does not wants electron that much and so it does not attracts the electron of (O) so the polarisability of (O-H) bond in phenol decreases.


text:
Phosphate concentration
I have chemical analyses of groundwater samples (I got the analyses from a previous study) and I want to work with it in my master thesis.
In these analyses, phosphate concentration is mentioned in micrograms/liter  and it seems phosphate was analysed by means of IC-ICP-MS.
My question is: When measuring phosphate concentration (orthophosphate) either using IC-ICP-MS or any other method as HACH Method 8114, do we measure the three species $\ce{PO4^{-3},HPO4^{-2}}$, and $\ce{H2PO4^{-1}}$ together and report the result as phosphate or we measure just $\ce{PO4^{-3}}$?

Not sure exactly how or what analysis was done. In general phosphate in ug/l would be for all species $\ce{H2PO4^-, HPO4^{2-}, \text{ and } PO4^{3-}}$. Again I'd expect that the ground water had a pH from which you could calculate the fraction of each of the species.


text:
Can the friction of air disintegrate objects if they are accelerated enough?
Can an object be thrown so fast that the individual atoms or molecules separate in air? I mean if you could accelerate an object fast enough, would it disintegrate from the friction of air?
If so, would they separate faster in denser air? I think there would be more particles the object collides with, therefore the range after which it disintegrates is much shorter.

It is unclear what you are asking. Most of the atmosphere is nitrogen and oxygen, both of which are diatomic molecules. Are you asking if the molecules will split and be left with only atoms of nitrogen and oxygen? No projectile is going to transmute atoms.

Looks like you are looking for something like reentry. Temperature due to shock wave associated to a projectile can be high enough to ionized and dissociate air. I have no model but density should play a role as T should rise faster . https://en.m.wikipedia.org/wiki/Atmospheric_entry

I don't quit understand what you mean Alchimista i edited my post so see if that clarifies. also see if that edit helps you understand my question MaxW.

I think I understand what you are asking now (since your edit). A person may not be able to throw an object that fast, but think about a meteoroid breaking and burning up as it enters the atmosphere at something like 20 km/s.

Read comment by @airhuff than go back to my comment. A bullet cannot but the underlying idea in your Q is ok and can be answered with a Yes. A meteoroid not only breaks. The air in the shocked region around it can breaks apart ( ions and monoatomic species form from O2 and N2 depending on T. Which depends in turn on speed of the projectile whatever it is)

After reading your edit I see you are thinking of breaking the bullet not the "air". The above comments and link are still valid anyway. It is more for Physics SE without breaking molecules :)

https://what-if.xkcd.com/1/

Ah,  thank  you, im trying to see if something could bee spun or thrown in a short distance and extreme speeds then pherhaps the friction alond with the speed of the bullet would tear apart, or disenigrate.