text:
Balancing REDOX reactions in a basic solution, when no H+ ions are present
I am trying to balance the following redox reaction, in basic solution. From what I have learned, to balance a redox reaction in basic solution you simply balance it as you normally would in acidic solution, then make sure the H+ ions are neutralized by adding OH- ions. In the example below, is my answer balanced in basic solution already? Or do I need to somehow cancel out the H present in H2O?

The bottom equation looks balanced to me. There are the same number of atoms of each element on each side of the equation and charge is conserved. The hydrogen in the $\ce{H2O}$ is balanced by the hydrogen in the $\ce{HAsO2}$ on the other side so it looks fine.

Thank-you. So you think this would be considered balanced in basic solution?

i don't see why not. note that i'm not familiar with the chemistry shown so i can't say whether this is a feasible reaction or not but from a balancing point of view it looks fine


text:
Balancing a REDOX reaction using the oxidation number method, in acidic solution
I am trying to balance the REDOX equation shown below, but I got stuck after balancing the major atoms. Where can I go from here? I know I need to balance the H and O atoms, and make sure the charges are balanced.

Second Attempt (Answer Only)

Is this correct?

Whenever you balance a redox reaction in acidic or basic solution, you have to account for interactions with the solution itself. The easiest way of doing this is by the half-reaction method.
This is a key detail that I find a lot of students miss - you cannot balance a reaction like this without considering the solution, because the solution itself is participating. Your clue that this is happening is the phrase "in acidic (or basic) solution" - any time you see that in a redox problem, you should be thinking "half-reaction method." Another clue is that H and O appear on only one side of the equation. If you are familiar with the algebraic method of balancing equations, another clue is that the system of equations would be unsolvable.
At any rate, once you know that you need to use the half-reaction method, the next step is to use it!
This tutorial has a good description of the method, your book probably does as well, and I'll give a brief description here for good measure.
These are the major steps:

Split the reaction into two half-reactions. Then, for each reaction
Balance everything except for H and O
Balance O by adding $\ce{H2O}$ (water is available in solution, so this is legal)
Balance H by adding $\ce{H^+}$ (it's an acidic solution - so this is also legal)
Balance charge by adding electrons ($\ce{e^-}$)
Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel - this is necessary because electrons are not created or destroyed in chemical reactions.
Add the equations together, and if the solution is basic, add $\ce{OH^-}$ to both sides to remove all protons.

I'll use an example from my copy of Brown, LeMay, Bursten and Murphy's "Chemistry: The Central Science" to demonstrate (all phases are aqueous):

Balance the following reaction in acidic solution:
$\ce{MnO4^- + C2O4^2- -> Mn^2+ + CO2}$
Break the reaction into two half-reactions:
$\ce{MnO4^- -> Mn^2+}$
$\ce{C2O4^2- ->CO2}$
It's not always easy to see how to do this. The quick way is to make sure that the elements that are not O and H are in their own equations. The proper way is to figure out what is being reduced and what is being oxidized, then split them so that the oxidation happens in one reaction and the reduction happens in the other. In this case, Mn is reduced and C is oxidized.
Balance everything except for H and O:
$\ce{MnO4^- -> Mn^2+}$
$\ce{C2O4^2- -> 2CO2}$
This step is usually pretty easy.
Balance O by adding $\ce{H2O}$:
$\ce{MnO4^- -> Mn^2+ + 4H2O }$
$\ce{C2O4^2- -> 2CO2}$
We start with O because we are using water to balance it. Each time we add a water, we also add two H, so it doesn't make sense to start with H.
Balance H by adding $\ce{H^+}$:
$\ce{8H^+ + MnO4^- -> Mn^2+ + 4H2O }$
$\ce{C2O4^2- -> 2CO2}$
Remember that we can do this because it's in acidic solution. You are probably wondering what we do for basic solution. The answer is - use protons! There is a trick we can do at the end to "fix" it for basic solutions, and this is much easier than iterating between $\ce{OH^-}$ and $\ce{H2O}$
Balance charge by adding electrons:
$\ce{5e^- + 8H^+ + MnO4^- -> Mn^2+ + 4H2O }$
$\ce{C2O4^2- -> 2CO2 + 2e^-}$
While this is easy to understand, I have found that in practice, a majority of mistakes happen here. Be careful when counting charges - remember to multiply by the coefficients, and remember that the charges don't have to add up to zero yet - they just have to have the same sum on both sides.
Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel:
$\ce{10e^- + 16H^+ + 2MnO4^- -> 2Mn^2+ + 8H2O }$
$\ce{5C2O4^2- -> 10CO2 + 10e^-}$
This part is also prone to error - be careful and remember to multiply every species by the coefficent.
Add the equations together
$\ce{16H^+ + 2MnO4^- + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2 }$
If the electrons don't cancel out here, then you made a mistake - go back and check each step.
If the solution is basic, at this point you would add $\ce{OH-}$ to both sides of the equation until the 16 protons were eliminated. We can add anything we want as long as it's to both sides (since this is, after all, just algebra). Each $\ce{OH^-}$ will react with a proton to form one water molecule, and so the waters on the right would cancel out some of the waters on the left. The result would be a similar equation, but with $\ce{OH^-}$ instead of $\ce{H^+}$.
Try out your problem using these steps, and if you get stuck, update your question.

Thanks for the very thorough answer! How does my new solution look?

You are welcome! I count 18 O on the left and 23 O on the right, so you should double-check each step (most likely the step where you added water)

1- Determining   oxidation number, Oxidation and Reduction:
$$
\ce{
\overset{\color{red}{(0)}}{Sn}+
\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{~O3^-}
->
\overset{\color{black}{(+1)}}{H2}
\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}
+\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}
}$$
The atom $\ce{N}$: went from $\color{green}{(+5)}$ to $\color{green}
{(+2)}$, it gains electrons, so it is reduced.
The atom $\ce{Sn}$: went from $\color{red}{(0)}$ to $\color{red}{(+4)}$, it lost electrons, so it is oxidized.
2- Separate the redox reaction into   two half-reactions: one for the oxidation, and one for the reduction
$$\text {Oxidation half}: \ce{
\overset{\color{red}{(0)}}{Sn}
->
\overset{\color{black}{(+1)}}{H2}
\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}
}$$
$$\text {Reduction half}:
\ce{
\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{O3^-}
->
\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}
}$$
3- Balance all elements except oxygen and hydrogen in both half equations
$$\text {Oxidation half}: \ce{
\overset{\color{red}{(0)}}{Sn}
->
\overset{\color{black}{(+1)}}{H2}
\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}
}$$
$$\text {Reduction half}:
\ce{
\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{O3^-}
->
\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}
}$$
4- Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms :
$$\text {Oxidation half}: \ce{
\overset{\color{red}{(0)}}{Sn} + 3{H2O}->
\overset{\color{black}{(+1)}}{H2}
\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}
}$$
$$\text {Reduction half}:
\ce{
\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{O3^-}
->
\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}+2{H_2O}
}$$
5- Add the appropriate number of hydrogen ions to that side of the equation required to balance the hydrogen atoms:
$$\text {Oxidation half}: \ce{
\overset{\color{red}{(0)}}{Sn} + 3{H_2O}->
\overset{\color{black}{(+1)}}{H2}
\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}+4{H^+}
}$$
$$\text {Reduction half}:
\ce{
\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{O3^-}+4{H^+}
->
\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}+2{H_2O}
}$$
6-  Use electrons ($\ce{e^-}$) to equalize the net charge on both sides of the equation:
$$\text {Oxidation half}: \ce{
\overset{\color{red}{(0)}}{Sn} + 3{H_2O}->
\overset{\color{black}{(+1)}}{H2}
\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}+4{H^+} + 4{e}^-
}$$
$$\text {Reduction half}:
\ce{
\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{O3^-}+4{H^+} +3{e}^-
->
\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}+2{H_2O}
}$$
7- Make the electrons gained equal to the electrons lost:
$$\text {Oxidation half}: \ce{
3\overset{\color{red}{(0)}}{Sn} + 9{H_2O}->
3\overset{\color{black}{(+1)}}{H2}
\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}+12{H^+} + 12{e}^-
}$$
$$\text {Reduction half}:
\ce{
4\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{O3^-}+16{H^+} +12{e}^-
->
4\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}+8{H_2O}
}$$
8- Combine the two half-reactions:
$\ce{3\overset{\color{red}{(0)}}{Sn} + 9{H_2O}  +4\overset{\color{green}{(+5)}}{N}
\overset{\color{ black }{(-2)}}{O3^-}+16{H^+} +12{e}^-->3\overset{\color{black}{(+1)}}{H2}\overset{\color{red}{(+4)}}{Sn}\overset{\color{ black }{(-2)}}{O3}+12{H^+} + 12{e}^- +4\overset{\color{green}{(+2)}}{N} \overset{\color{ black }{(-2)}}{O}+8{H_2O}  } $
9- Cancele out the electrons and anything that is the same on both sides then collect the things that is the same on the same side  to form one balanced redox equation:
$\ce{3\overset{\color{red}{(0)}}{Sn} + {H_2O}  +4\overset{\color{green}{(+5)}}{N}{O_3^-} +4{H^+} ->3{H_2}\overset{\color{red}{(+4)}}{Sn}{O_3}  +4\overset{\color{green}{(+2)}}{N} {O} } $

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text:
Why do we add copper when testing a salt for nitrate ion?
In our chemistry curriculum we are doing these lab experiments where we are supposed to test salts for certain anions,cations and other stuff.
I was told that if I dissolved the salt in concentrated $\ce{H2SO4}$ and then heated it for a while over a flame, the evolution of light brown vapours would indicate the possibility of $\ce{NO3-}$. To confirm its presence, however, I would have to repeat the test but this time also add some copper bits to the solution and check for dark brown vapours.
I figured the reason was that the copper reduced the $\ce{HNO3}$ gas to $\ce{NO2}$ which increased the amount of $\ce{NO2}$ gas produced, resulting in a darker shade of brown.
My question is that why copper, which is a very weak reducing agent, used here? Why not a real reducing agent like zinc which will give more $\ce{NO2}$ even with lesser salt?

This is a preliminary test, which is neither particularly selective nor sensitive.
With nitrate, brown vapours of $\ce{NO2}$ are released.
However, bromine or iodine could also release brownish or violet vapours.
Copper (or silver) is used to remove bromide and iodide by precipitation.

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text:
Etching a Screen Protector Film
I have to etch a pattern on a plastic film similar to the ones used to protect phone screens. I want a very thin etch, one that is only visible on close inspection. How can I get that done?

I don't know but I would guess laser "cutters" would probably be interesting to look at given the resolution you need.

If you don’t find a suitable chemical method, you may want to consider using an airbrush sandblaster. You can cover the area that you don’t want to be etched.

Assuming by thin, you mean a narrow line (not depth). It all depends on how thin you want to go. For a fine width line, a relatively cheap $\ce{CO2}$ laser cutter will get you down to 100–200 µm, but depending on the plastic and how thick it is, the heat produced may warp the plastic near the cut. I think most phone screen protectors are PET film. I've personally made ~200 µm features with one, and if the plastic is thin enough, you can keep the cut width pretty narrow. Things like pulsed UV lasers will give you much finer lines on plastics without thermal effects (as they tend to ablate rather than burn plastics), but they're much more expensive. If you don't have access to either, many companies will contract for this kind of work, even for small jobs.
Another option for coarser cuts that don't have too complex of shapes is a vinyl cutter—basically a plotter with a cutting blade instead of a pen. These are quite cheap, but can't cut all shapes and resolution is on the order of a hundred µm or so, though they don't cause distortions along the edges in most materials.
If you're really serious about it, it's certainly possible to coat photoresist on most plastics and make a mask for reactive ion etching, which will get you down to 1 or 2 µm easily, but it sounds like overkill for you.


text:
How to calculate the volume of a sodium hydroxide solution necessary to add to water, to raise the pH to a certain value?

What volume of $\pu{0.300N}\, \ce{NaOH}$ would be needed to bring $\pu{16.0 mL}$ of distilled water that has a $\mathrm{pH}$ of $7.00$ to a new $\mathrm{pH}$ of $12.42$?

This is my attempt at a solution:
$$\ce{NaOH + H2O-> Na+ + OH- + H2O}$$
\begin{align}
\mathrm{pH}(\text{initial}) &= 7.00 \\
\mathrm{pH}(\text{final}) &= 12.42 \\
\mathrm{pH} &= -\log{\ce{[H+]}} \\
\ce{[H+]}_{\text{intial}} &= 10^{-7} \\
&= \pu{10^{-7} ions} \\ \hline
V_1&= 0.016L \\
C_1 &= (0.300)(10^{-7})= 3\cdot10^{-8} N\\
\mathrm{pOH} &= 14-12.42  \\
[\ce{H+}]&= 10^{-1.58} \\
&=\pu{ 0.02630 ions} \\
C_2 &= (0.300)(0.02630) \\
&=7.89\cdot10^{-3} N \\
C_1V_1&=C_2V_2 \\
(3\cdot10^{-8})(0.016)&= (7.89\cdot 10^{-3})(V_2) \\
V_2&= 6.08\cdot10^{-8} L \\
\end{align}

$$\ce{NaOH + H2O-> Na+  +OH- + H2O}$$
\begin{align}
\mathrm{pH}_1 = 7.00 &\implies\mathrm{pOH}_1 = 7 \\
\mathrm{pH}_2 = 12.42 &\implies\mathrm{pOH}_2 = 1.58
\end{align}
so $[\ce{OH-}]_1 = 10^{-7}$ and $[\ce{OH-}]_2 = 10^{-1.58}$
So, initially, you have $16^{-3}\times 10^{-7}$ mol of $\ce{OH-}$ in your water. Your $\ce{NaOH}$ solution is $\pu{0.3 M}$ so you'll bring $\pu{0.3\times 10^{-3} mol}$ each $\mathrm{mL}$.
Then, the resulting concentration will be $$C = \frac{n_\text{initial} + (0.3\times 10^{-3})\times x}{V_\text{initial} + x\times \pu{1mL}}$$
You know $V_\text{initial}$ to be $\pu{16 mL}$
You know $n_\text{initial}$ to be $\pu{1.6\times 10^{-9} mol}$
And you know the target concentration to be $10^{-1.58}$
I let you do the solving for $x$ where $x$ is the volume of $\ce{NaOH}$ solution to be added ;) be careful with the units.

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text:
How to remove a stuck stopper from a flask?
How do I remove a stopper that has been stuck in the neck of a 100 mL round bottom flask since I forgot to wipe the neck of the flask. A salt was formed in the 100 mL flask and the stopper is just stuck like cement. Is there anyway to remove it without breaking the flask, breaking the stopper, or heating the flask (since I don't want to interfere with the product)?

Is it a hollow glas stopper?
Sometimes, it helps to hold the flask at the body and hit the corner of the lab bench with the stopper. If that doesn't help, heat the neck of the flask carefully with a heat gun.

No, it was a solid glass stopper. I tried the hitting method and it worked like a miracle.

@86BCP2432T I'm glad I could help! Btw, when silicone grease is not an option - and sometimes it isn't - teflon sleeves are a good alternative.

While @KlausWarzecha provided the working solution, I offer up my own adopted mantra for future reference.  "Most human problems can be solved by an appropriate charge of high explosives."

When I was younger I worked at a place which had a serious accident which could have been fatal. Someone tried to open a flask containing pet ether using a flame to heat the stuck ground glass joint holding the stopper. I tend to use the hitting method on a wooden surface, use many gentle taps and keep on turning the flask around. I never use a modern plastic type lab bench. This method deals with most stuck joints on flasks and bottles. Never use a flame on glassware to free a joint until you have removed all flammable solvent from the glassware.

Here is an alternative that worked like a charm for me.  I have a wide-mouth lab jar with a thick, ground glass stopper that stuck tight.  None of the other methods worked; oil, heat, tapping, etc. I had an aerosol spray used to blow dust from computer keyboards/screens (it is not compressed air but a liquid that becomes a gas as it is released - i.e. the can gets very cold as you use it). A quick shot of cold gas directly on the stopper and it immediately shrank enough to easily come out of the jar.

This is another reason not to use glass stoppers with separatory funnels. If one isn''t careful, pressure (from ether, CO2) can build up and the funnel may break.  With a neoprene stopper one can feel the pressure as the neoprene stopper moves slightly.

Another easy method is to get 2 other glass stoppers and tap them on both sides of the neck of the flask firmly, sometimes this will free it up immediately, other times you may need to have someone pulling the stopper while you hit the stoppers on the sides

I like the pulling while tapping. Requires more hands than the average lab worker has, tho.

Had great success with the upside down can of compressed air! I had 2 small (10 mL) volumetric flasks with stuck PTFE stoppers. I tried warming the neck slightly but that didn't help. With the can of air, I placed the flask on the counter, blasted the cap just above the flask neck for a couple seconds, the tried opening and both came open right away! Awesome!

+1 Nice to have a confirmation of a new technique.


text:
Is sodium hypochlorite solution an ionic solution - does it conduct current?
Does an aqueous solution of sodium hypochlorite conduct current?

Yes, it is a solution of an ionic salt. a table lists electrical conductivity at various concentrations.


text:
What is the reaction equation for a mixture of trichloroethanoic acid and dimethylpropanoic acid?

Write an equation for the reaction you might expect to take place if trichloroethanoic acid, $\ce{Cl3CCO2H}$, $\mathrm{p}K_\mathrm{a} = 0.65$, were added to dimethylpropanoic acid, $\ce{(CH3)3CCO2H}$, $\mathrm{p}K_\mathrm{a} = 5.05$. Explain your answer in terms of the Brønsted–Lowry theory.

I know that $\ce{Cl3CCO2H}$ is the stronger acid as it has a lower $\mathrm{p}K_\mathrm{a}$, however, how can dimethylpropanoic acid be a base?

Organic Chemistry generally considers that anything can be an acid or a base.

Even concentrated sulphuric acid can act as a base, e.g. $$\ce{2 H2SO4 <=> H3SO4+ + HSO4-}$$

It can be a base; but it's such a poor one that you'd better get an ultra-strong acid to protonate it. Concentrated sulfuric acid might work. Trichloroacetic acid probably won't.

TL;DR
Contrary to what the answers/comments have suggested, I would say that no reaction happens here. The acid–base reaction between the two given species is thermodynamically unfeasible, with an equilibrium constant $K \sim 10^{-9}$.

If you want an acid-base reaction to occur between A and B, it's no use comparing the $\mathrm pK_\mathrm a$ values of A and B. That tells you how acidic A and B are, but you're not interested in both of them acting as acids. You're interested in one of them being an acid, and the other being a base. So the relevant quantities are the $\mathrm pK_\mathrm a$ of A and the $\mathrm pK_\mathrm b$ of B!
In this context, let's keep trichloroacetic acid $\ce{Cl3CCOOH}$ as the acid; we already know that it has $\mathrm pK_\mathrm a = 0.65$. The question tells us that for the "base" $\ce{Me3CCO2H}$, its $\mathrm pK_\mathrm a$ is $5.05$. But we're not interested in that value; we're more interested in its $\mathrm pK_\mathrm b$.
To find that out, we need to look for data beyond the question text itself. The veritable Evans tables states that the $\mathrm pK_\mathrm a$ of $\ce{PhC(OH)+OH}$ is $-7.8$. That species is the conjugate acid of $\ce{PhCOOH}$, so the $\mathrm pK_\mathrm b$ of $\ce{PhCOOH}$ is $14 - (-7.8) = 21.8$. Recall that a larger $\mathrm pK_\mathrm b$ means that a species is a weaker base. This gigantic number should already tip you off to the fact that $\ce{PhCOOH}$ is a terrible base. The species in the question, $\ce{Me3CCOOH}$, will likely have a value that's broadly similar.
How do we use these values to find whether an acid-base reaction is plausible or not? Returning to the reaction
$$\ce{AH + B <=> A- + BH+,}$$
we have the equilibrium constant
$$\begin{align}
K &= \frac{[\ce{A-}][\ce{BH+}]}{[\ce{AH}][\ce{B}]} \\
&= \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]} \cdot \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]} \cdot \frac{1}{[\ce{H3O+}][\ce{OH-}]} \\
&= \frac{K_\mathrm a(\ce{HA}) \cdot K_\mathrm b(\ce{B})}{K_\mathrm w}
\end{align}$$
For our molecules, we know that the relevant quantities are $\mathrm pK_\mathrm a(\ce{HA}) = 0.65$ and $\mathrm pK_\mathrm b(\ce{B}) \approx 21.8$. Putting in $K_\mathrm a = 10^{-0.65}$, $K_\mathrm b = 10^{-21.8}$, and $K_\mathrm w = 10^{-14}$, we get an equilibrium constant of
$$K = \frac{10^{-0.65} \cdot 10^{-21.8}}{10^{-14}} = 3.548 \times 10^{-9}$$
which is so small that it makes very little sense to claim that this reaction "happens".

[…] I know that $\ce{Cl3CCO2H}$ is the stronger acid

With other words: This is the Brønsted donor. Show some confidence in your knowledge and draw:

I wonder if subsequent dehydration and decarbonylation might occur.


text:
Kinetics: Order of Reaction
I am confused as to how an order of a reactant (n or m) in the rate law can be negative. This means that increasing the concentration of a reactant would actually decrease its rate of disappearance. How does this make sense? Doesn't a higher concentration always allow for more collisions? Also, when the order is 0, this means that concentration has no effect on the rate of disappearance. How does this make sense chemically/at the atomic level?

related http://chemistry.stackexchange.com/questions/263/zero-order-reactions?rq=1

Increasing the concentration of an inhibitor can lower the rate of reaction.

In principle, there are no zeroth-order reaction. Experimentally, however, zeroth-order rate law occurs because of the reaction condition. For example, in catalytic reactions such as hydrogenation of an alkene using platinum as a catalyst, the hydrogenation occurs on the surface of platinum. There is a finite surface area of platinum, so at a some point, it doesn't matter at all how much more hydrogen is put in. This is because no reaction can occur if the hydrogen cannot bind to the platinum.
An example of inverse first order reaction can be found here.
Basically, you have a two-step reaction:
$$\ce{A + B <=> C}$$
$$\ce{C <=> D\ (slow)}$$
Increasing concentration of $\ce{C}$ would decrease the rate of reaction.

Can we say that before the saturation of the surface of catalyst, the rate of reaction is directly proportional to the concentration of reactant and when it reaches above the saturation level then the rates becomes independent of the concentration of reactants? So to achieve zero order kinetics, concentration of reactants should always be more than the saturation level of catalyst?


text:
Dilution terminology
For thing like 8× dilution, that would be like 1 ml of a solution and then 7 ml of something like water to dilute right? Would 50 ml solution and 50 ml water be a 2× dilution? I’m a bit unsure about the terminology. If I an OD of 0.123 from a sample diluted 5×, would multiply 0.123 by 5 to get the real OD?

It seems your right, but what's OD?

Optical density.

You are right.  The × in 8× refers to the fold-change in concentration between the concentrated source and diluted solution.
$$\mathrm{DF = \frac{C_{stock}}{C_{working}}}$$ where $\mathrm{C_{stock}}$ and $\mathrm{C_{working}}$ are the concentrations of a stock solution and the desired final concentration, respectively.
The concentration of the working solution can be expanded:
$$\mathrm{C_{working} = \frac{C_{stock} V_{stock}}{V_{working}} = \frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}}$$ where $\mathrm{V_{stock}}$ is the volume of the stock solution and $\mathrm{V_{diluent}}$ is the volume of diluent (e.g. pure water) added.
Plugging back into the DF equation:
$$\mathrm{DF = \frac{C_{stock}}{\frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}} = \frac{V_{stock}+V_{diluent}}{V_{stock}}}$$

so would 50/50 of each be a 2x dilution or..

Yep, you're right again.


text:
What are the triple point pressure and temperature for pure cyclohexane?
Here is a video that demonstrates the interesting nature of the triple point of cyclohexane, so I tried researching on the internet for a phase diagram, but only found data from a manufacture where the triple point was specified as 279.48 degK at 5.388 kPa. But in the video there doesn't seem to be a cooling plate. The flask appears to be at room temperature.
What are the triple point specs for pure cyclohexane? If they match the data above, how were they able to achieve the phase transitions shown in the video at room temperature?
Also where can I find a phase diagram for pure cyclohexane on the internet?

I suspect that with volatile solvents like cyclohexane, the vacuum pump you hear in that video can do a good job of cooling the flask via evaporative cooling.

The NIST REFPROP (Reference Fluid Thermodynamic and Transport Properties) program reports the triple-point temperature of cyclohexane as 279.47 K.
The program refers to: S. G. Penoncello, R. T. Jacobsen, and A. R. H. Goodwin: “A Thermodynamic Property Formulation for Cyclohexane”, International Journal of Thermophysics, Vol. 16., No. 2, 1995, 515–531.
This publication mentions the triple-point temperature as T = 279.47 K and the triple-point pressure as p = 5.388 kPa.
The given reference is: J. G. Aston, G. J. Szasz, and H. L. Fink, J. Am. Chem. Soc. 65  1135 (1943).

The following p-T diagram (made using REFPROP) shows the melting line (solid-liquid equilibrium, blue) and the vaporization line (liquid-gas equilibrium, red), which meet at the solid-liquid-gas triple point. Unfortunately, REFPROP does not include the sublimation line for cyclohexane.


text:
Does Uranium ore count as a mixture?
I'm doing an assignment where we must research the separation of a mixture to obtain a useful final product. I thought about the separation of uranium from its ore after it has been mined, but I did not know if the ore counted as a mixture, or a compound.

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Yes, uranium ore is a mixture.
Uranium compounds such as pitchblende (uraninite) are only present as a small part (usually $<1$%) of the ore.
Extensive purification is needed. Uranium ore purification would be a good choice for your project.
See for example:
http://www.areva.com/EN/operations-677/uranium-processing-from-ore-to-yellow-cake.html
http://www.iaea.org/sites/default/files/publications/magazines/bulletin/bull23-2/23204882428.pdf

image source: http://chemwiki.ucdavis.edu/Development_Details/Approaches/ChemCases/Nuclear_Chemistry_and_the_Community/08._Uranium_Production

The main uranium ore is pitchblende, or uraninite, which is primarily composed of $\ce{UO2}$. $\ce{UO2}$ is ionically bonded together in a crystal structure and so it is definitely a compound and not a mixture. However, uraninite does contain small amounts of other compounds, such as $\ce{U3O8}$ and trace amounts of lead oxides and so the mineral as a whole is a mixture of compounds.


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Does the catalyst alter the Potential energy Surface?
Catalyst provides alternative pathway to the reactants. Does it also mean that the potential energy surface is altered?

As you said, catalysts do not alter the PES, but extend it to other dimensions, which usually provide new reaction pathways with lower reaction barriers. If you consider a surface catalyst and an impinged molecule. Considering the full systems PES, the pure molecule PES is a part of it (for large distances of the surface to the molecule). If we bring the molecule closer to the surface, we get new partial PES which have lower reaction barriers.


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Acid-base related chemistry
What is the function of silica, $\ce{SiO2}$ in the equation
$$\ce{CaO + SiO2 -> CaSiO3}$$
  a) A basic oxide
  b) A acidic oxide
  c) A reducing agent
  d) An oxidising agent
Please explain.

Please note that the homework policy on this site requires you
to show some personal effort.

Consider to answer one... Way to go @KlausWarzecha! And, I wanna second his request Tanzeel, so I say: 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.

Since this sounds like a homework question, I will not provide a full solution!
Is $\ce{SiO2}$ an oxidizing or reducing agent in the reaction?
Figure out the oxidation state for the $\ce{Ca}$ and $\ce{Si}$ in the starting materials and the product. As a rule of thumb, consider the oxidation state of (-2) for $\ce{O}$. Unless peroxides are concerned, this works well.

 $$\ce{Ca^{(+2)}O + Si^{(+4)}O2 -> Ca^{(+2)}Si^{(+4)}O3}$$
 Apparently, there is no change - this is not a redox reaction!

Is $\ce{SiO2}$ a basic or an acidic oxide?

How are basic and acidic oxides defined?
Are there any rules (of thumb) for the oxides of metals and non-metals?

 Metal oxides are usually basic oxides. These can be considered anhydrides of the corresponding metal hydroxides, $\ce{CaO}$ is a typical example: $\ce{CaO + H2O -> Ca(OH)2}$
 What about $\ce{SiO2}$?


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Real gas expansion
Will a real gas occupy all the space available for it? An ideal gas will expand to fill its container but will a real gas? And what are the changes in a real gas when it expands freely against a vacuum?

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That's my answer to the question, which may not be correct.
Real gas molecules are bound by (attractive) Van der Waal's forces (VdW forces). When molecules move apart, part of their kinetic energy is converted to the potential energy (because Van der Waal's force is attractive), if no external source of energy (i.e. heat reservoir, external pressure sources/work done) is present. (i.e. there is no exchange of energy across the boundaries of the container).
That said, when a real gas expands into a vacuum, though there's no work done on the gas, the temperature of the gas should drop because part of the kinetic energy (which is proportional to the temperature) is converted to the potential energy.
Will the gas fill the whole container? Now first ignore surface effects (i.e. boundary walls of the container). We have to understand that although the molecules are "bound" by VdW forces, which tend to draw the molecules together, the latter tendency is balanced by the thermal motion of the gas molecules, which tends to pull the molecules apart. When the temperature is so low such that the thermal motion "dies out", the molecules may settle into a lower energy state (e.g. lattice/solid)
In most cases, real gases should fill the whole container.

If it doesn't fill the whole container, it's not a gas.


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Trying to figure out what these elements have in common
I have an old periodic table and I marked five metals from groups 1A and 2A with dots to indicate something about them. Only all this time later I can't figure out what exactly it was. I've been trying to figure it out for ages.
The elements are: Li, Na, K, Ca, Ba.
Doesn't anyone know any properties that just those elements, or maybe even just primarily those five, share that I could have been marking them for?

Their salts burn nice and colourful... red, yellow, purple, red, green (but there are many more that do that). See at wikipedia

Maybe your assumption was the exam questions are about these five? Or it could be these elements' cations are commonly seen in ionic compounds? There are countless possibilities, that would render answering (better put it as guessing) this almost impossible.

They are all member of the s block in the periodic table. It means they all have valence electron in the s orbital.

other elements are also s-block and are not marked so I suspect that this is not the reason


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At what scenario will a LCD monitor that used mercury inside their lamp be harmful
Small amount of mercury and/or mercury vapors is harmful to human as it can cause damage to vital organs such as brain, heart and so on.
However, there are many electronic devices such as LCD monitor that used lamp that contain mercury for operation.
So, in what scenario will a LCD monitor that used lamp that contain mercury harmful?

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

Most modern fluorescent lamps use little mercury and have amalgam, rather than free mercury, in the lamp (the mercury vaporizes when the amalgam is heated). The harm is from many lamps being added to landfills, but is probably less significant than mercury freed by burning coal (about 500 tonnes in 2010).

I might, somewhat tongue in cheek, suggest that the glass shards from breaking the lamp open might be more of a health hazard than the small quantity of mercury...


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Carbonic acid-hydrogencarbonate buffer system pH calculation?
I am learning about the pH of buffer solutions, and am slightly confused by my text book's calculation. For the blood's buffer system of carbonic acid and hydrogencarbonate ions as the conjugate case, the equilibrium is $$\ce{H_2CO_3 <=> H^+ + HCO_3^-}$$
The acid dissociation constant is $4.3 \times 10^{-7}~ \mathrm{mol/dm^{3}}$ and the $pH$ of healthy blood is 7.4.
So $$\ce{\frac{[H_2CO_3]}{[HCO_3^{-}]}=\frac{[H^+]}{K_{a}}}=\frac{1}{10.8}$$
I understand this calculation, however my textbook then concludes that this shows that there are about 20 times more hydrogen carbonate ions than carbonic acid. Should it not be 10 times? Why was this value doubled?

I get the same answer as you do.

Your calculation is correct. However, there needs to be more bicarbonate in the blood because the majority of all metabolic processes tend to produce more acidic than basic products. There are also other ways to help the body maintain a pH of 7.4, not only the carbonic acid/bicarbonate buffer.
So both you and the book are right. If the calculations are done assuming there is just carbonic acid/bicarbonate acting as a way to maintian the pH in the blood, the ratio should be about 10 to one. Since there is a lot more going on though, the ratio is different than that.

I would assume that it is just a typo and that it is correct.
If you are still not sure, try this: do the same problem again, but substitute a different weak acid (and be sure to use the new $K_a$ value!). Obviously the numbers will be different, but make sure that the procedure still makes sense and that you do not second-guess yourself here.


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Why do aqueous solutions of sodium chloride and silver nitrate swap ions?
So, we know that:
$$ \ce{NaCl(aq) + AgNO3(aq) -> NaNO3(aq) + AgCl(s)}$$
(In other words, the ions swap and a precipitate forms).
However, both of those compounds seem pretty stable, since $\ce{Ag}$ and $\ce{Na}$ cations both have a 1+ charge, and $\ce{Cl}$ and $\ce{NO3}$ both have a 1- charge. Since they are stable, I would assume no reaction would take place since both ions are happy with the ions they are currently bonded to (sodium and silver, respectively).
However, $\ce{Na}$ is higher on the activity series than $\ce{Ag}$, and we know that these aqueous solutions actually contain dissociated ions, so I assume this has something to do with the double-replacement. But why is this? Why does this reaction take place, even though the original compounds appear stable at first glance?
Edit
I read this question, and it says

Le Châtelier's principle only states that a system previously at equilibrium will want to stay at equilibrium - that is, if we perturb it, it will try to go back to equilibrium.

The above compounds of sodium chloride and silver nitrate appear to be at an initial equilibrium, do they not?
Edit 2
The question isn't why silver chloride is a solid, but rather, why does this reaction take place if the initial two compounds are stable to begin with?

Are you asking why silver chloride is solid or what drives the ion swapping? Because I can explain that the silver chloride forms because it makes a solid and precipitates out of solution, driving equilibrium toward the solid product, but I can't explain why silver chloride is insoluble.

@user137, No, I understand why silver chloride is a solid, that's not the question. The question is, because NaCl and AgNO3 initially appear stable, why do they exchange ions?

Well, your reasoning isn't enough for them to be stable.

@MARamezani, why is that?

I favor @user137's answer: "...So when a silver ion and chloride ion meet, they quickly fall out of solution...." This was what I had in mind. Stability is really a broad topic, and it depends on many factors, but [for example thermodynamics] allow rules like this to happen.

While solutions of silver nitrate and sodium chloride are in stable equilibrium on their own, once mixed the equilibrium is changed. When the salts are dissolved, you no longer have silver and nitrate or sodium and chloride ions associated with each other, you have a mixture of individual ions in solution.
But silver chloride has very low solubility, about 1 mg can dissolve in 1 L of water. So when a silver ion and chloride ion meet, they quickly fall out of solution. So if we had an equilibrium between the dissolved ions and the small amount of "soluble" silver chloride, that equilibrium would be driven toward the silver chloride by the precipitation. When compounds escape the solution by precipitating or evaporating, they are removed from the equilibrium, so trying to replace them to restore equilibrium requires converting more free ions into silver chloride and pushes the reaction to completion.

So basically this reaction has everything to do with the fact that silver chloride is insoluble, right? So I can say (taking a different but similar reaction into consideration) that NaCl and KNO3 would not react, since all reactants and all products are soluble; therefore, we would just have a bunch of dissociated ions floating around in a NaCl and KNO3 reaction. Am I on track with this?

If all combinations of ions are soluble, then they shouldn't react. Unless they do a real reaction to form a new ion or compound, but I don't think a basic chemistry class would trick you like that.

So, we know that:
$$\ce{NaCl (aq) + AgNO3 (aq) -> NaNO3 (aq) + AgCl (s) v}$$
(In other words, the ions swap and a precipitate forms).

Ah yes, that is how it is always written on the blackboard. But remember that the label (aq) actually carries a meaning. It means that upon dissolution, it is wrong to talk about associated sodium and chloride ions in any way, rather they are separated and go diffusing through the solution on their own.
$$\ce{NaCl (s) + H2O -> Na+ (aq) + Cl- (aq)}$$
If you took one pair of $\ce{NaCl}$ ions and coloured them both pink, you would realise that the two pink ions move through the solution entirely individually. The same thing goes for $\ce{AgNO3}$. Sometimes a sodium ion may meet a chloride ion (or a silver ion a nitrate ion) but they’ll just greet and move on.
Now we mix the two solutions. All of a sudden in this chaos of ions diffusing everywhere, silver ions can meet chloride ions. They won’t just greet but they’ll hug each other because they like to hold tight together. And thus they shall precipitate. The reaction is much better written as:
$$\ce{Cl- (aq) + Ag+ (aq) -> AgCl (s) v + H2O}$$
Think about it a different way. Mix a solution of sodium chloride and potassium bromide and then evaporate the water. You will obtain a mixture of $\ce{NaCl, NaBr, KCl}$ and $\ce{KBr}$. Why? Because all ions are equally distributed and upon removal of the solvent, they will all fall together. All salts will form in an entirely statistical distribution.


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Hydrogenation at atmospheric pressure in amateur environment
I was planning on doing some small scale hydrogenation practicals at atmospheric pressure (probably from Vogel) and ran into a couple of questions. I will be using the "balloon method" (described in Procter's Advanced Practical Organic Chemistry book) as I consider it an easier technique for small scale reactions than constructing a Brown hydrogenation apparatus. Note that I'm working in an amateur environment (i.e. my home lab which does not have a fumehood). Now for the questions:

What rule of thumb can be used to determine the amount of catalyst (in these cases $\ce{Pd/C}$) needed per mole of hydrogen used (I suppose these two variables are depending on each other)?
There are different types of $\ce{Pd/C}$ out there: 3%, 5% or 10% $\ce{Pd}$ by weight, with or without a variable amount of added $\ce{Pb}$, different solvents (mostly I see is water) and also there are different bases available, not just palladium on carbon. I understand that $\ce{Pb}$ is added to poison the catalyst such as with Lindlar catalyst, but if total reduction is required then why would this be necessary? Also how would I decide between 3%, 5% or 10% $\ce{Pd}$ by weight?
What is an effective method for generating the hydrogen to fill the balloon with? I suppose that normally a pressurized hydrogen tank would be used for this, but I find this overkill for the reaction I'm planning. Furthermore, the use of  pressurized hydrogen tanks is not quite without danger and I'm not trained to handle them safely.
Why would it be required to flush the reaction mixture with an inert gas after the hydrogenation has completed? I understand that this is necessary because the dissolved hydrogen gas can ignite and possibly explode when, while filtering the catalyst, the catalyst spontaneously ignites (e.g. when the catalyst is not suspended enough and direct contact with the atmosphere is possible). However, why is an inert gas advised? What is wrong with ordinary air?
What is the function of Celite when filtering the catalyst?
How many times can the $\ce{Pd/C}$ catalyst be reused? How would you spot whether or not your catalyst has become too poisoned to be successfully used in hydrogenation?

The answer to some (but not all!) of your questions are probably in this video.  It's a pretty cool demo of "home lab" hydrogenation. https://www.youtube.com/watch?v=oqdDWA9-DSY

I've adddressed a part of your question here.
As far as the catalyst is concerned, 3-5% Pd/C is fine.

What is an effective method for generating the hydrogen to fill the balloon with? I suppose that normally a pressurized hydrogen tank would be used for this, but I find this overkill for the reaction I'm planning. Furthermore, the use of pressurized hydrogen tanks is not quite without danger and I'm not trained to handle them safely.

In the lab, I only used pressurized tanks. While commercial hydrogen generators are available - they are mostly used to generate hydrogen gas as a carrier gas for GC/MS - these things cost a fortune and really would be overkill.
Actually, pressurized tanks are pretty safe if you follow some rules:

Obviously, you can't attach the pressure regulator and operate the whole thing unless the steel cap over the main valve is removed. Don't start anything unless the tank is secured to the lab bench or the wall with a clamp, a chain or another sturdy contraption that prevents the tank from falling over! You want hydrogenation, not a rocket ;)
Get familiar with the pressure regulator.
When you're done, remove the regulator again and put the cap over the main valve. Keep the regulator at a dry and clean place.
No grease on the threads!

When operated with a bit of common sense, a small pressurized tank is the most convenient option.

What is the function of Celite when filtering the catalyst?

When you filter off the catalyst, you'll typically use a suction filter with a base of sintered glass (fritted glass). Palladium on charcoal is a very fine powder that will clog the fritted glass. A pad of Celite in the suction will prevent this. Celite is cheap, suction filters are not!

Thanks for your answer! Could you perhaps also address the other points? I cannot really accept it as an answer to the question if not the entire question is addressed.

Make sure you do not have any flames, anything which makes sparks or other sources of ignition near you when you do hydrogenation.
I have done plenty of hydrogenations at atmospheric preassure, I tend to use a gas burrete to follow the reaction. It is possible to see the change of rate when you consume all of the vinyl groups and then start to react a trisubsistuted alkene.
I would suggest that you consider using the nickel cataylst which is formed from nickel chloride and sodium borohydride, I have found it is a cheap and nice cataylst. It also is less pyrophoric than many other cataylsts when it has been exposed to hydrogen and is then exposed to air.
I have found that the easy way to do it is to take a 50 ml burette and turn it upside down, stuff a bung in it with a hole connected to a rubber tube which goes to a separation funnel. Put copper sulfate solution in the funnel. The blue colour makes it easy to read it more quickly and stops things growing in the water. The other end of the burette needs to be connected to the reaction flask.


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Higher energy π HOMO
Which is higher in energy? Cyclohexanone enol or cyclohex-2-en-1-ol?
Okay from what I understand a higher pi HOMO is synonymous with more nucleophilic. So I would expect cyclohex-2-en-1-ol to have the higher pi HOMO because in the case of cyclohexanol enol there would be inductive withdrawal by the oxygen connected to one of the sp2 carbons involved in the pi bond.
Or would it be the other way around? Resonance donation by oxygen into the pi bond of the enol makes it more nucleophilic? How can we tell how the HOMO/LUMO energies split?

https://www.jstage.jst.go.jp/article/bcsj1926/49/11/49_11_3089/_pdf

from what I understand a higher pi HOMO is synonymous with more
  nucleophilic.

Yes, that's generally true.  If you have a substituent on a double bond that increases the electron density in the double bond, then the double bond (usually one carbon more than the other) becomes more nucleophilic and the energy of the HOMO and LUMO are raised.  Conversely, electron withdrawing groups attached to a double bond withdraw electron density, making the double bond less nucleophilic and lowering the energies of the HOMO and LUMO.
Comparing cyclohexanone enol (cyclohex-1-en-1-ol) to cyclohex-2-en-1-ol, you can draw a resonance structure for the former that increases electron density in the double bond (especially the $\beta$ carbon).  This electron donating resonance effect overwhelms the electron withdrawing inductive effect of the hydroxyl group.  Therefore, the double bond in the cyclohex-1-en-1-ol will be more nucleophilic and its HOMO and LUMO will be higher in energy than those for cyclohex-2-en-1-ol.

"Yes, that's generally true." - any exceptions?

I'm sure there are many (we're talking about chemistry after all).  When a carbonyl is attached to a double bond the double bond becomes less nucleophilic, but sometimes the HOMO will move up - not down - in energy.  Conjugation, rings, strain, etc. can sometimes cause HOMO-LUMO movement that doesn't correspond with the generalities presented above.

interesting I might open a new question about this


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Why are enthalpies of reaction or Gibbs FE change given in kJ/mol
It makes no sense to me that these thermodynamic changes are given in kJ/mol. Reactions can have several compounds on both sides and its not clear what exactly the molar variable refers to…
I've been given a homework where this is the convention followed. My temptation is to just ignore the /mol as senseless, and proceed just considering the energy change. But I'm sure I'm wrong.
Can anyone help me?
Thanks.

$\frac{KJ}{mol}$ means amount of heat [an action, e.g.: gained] per mole. I don't know why that should confuse you. Could you bring up more examples?

Yeah I'm not speaking of formation reactions. Say for example: H2S+1.5 O2 -> H2O + SO2. I'm given the enthalpy as kJ/mol.

It makes perfect sense that the units KJ/mol are used.
If you burn 2 cords of wood you get twice the heat as from 1 cord of wood.
Enthaply (and Gibbs free energy) of the reaction is proportional to the number of moles that react.
If the reaction is written as:
3A + 2B => 6C
Then the change per mole would mean: per 3 moles of A = per 2 moles of B = per 6 moles of C.

In the case of molar entropy I have JK^-1mol^-1. This is useful because I can multiply the quantity by the number of moles I am considering. Here I can't see how I could do that. "Per mole". Per mole of what? Thanks.

So in the example in the answer, "per mole of what" is per 3 moles of A or per 2 moles of B or per 6 moles of C that react, all of which are equivalent

So if I were to work with 1 mole of B, I will multiply the heat of reaction by 0.5 ?

Do you agree with me that this type of thing is a bit different from the other uses of "per mole" ?

If 1 mole of B reacts, then the enthalpy change would be 0.5 of the molar enthalpy of the reaction, as you are saying, yes

Yes, it is somewhat confusing because each reactant and product can have a different number of moles in a given reaction


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Molar absorption coefficient - why are the molar absorptivities defined to be the same?
In spectroscopy, the isosbestic point is said to be where the molar absorptivities of two species is the same.
$$A= \epsilon(X+Y) \times c(X+Y) \times l$$
reduces to $$A= \epsilon_t \times c \times l$$
the absorbances and the length are the same at this point, but the concentrations are not;
why are the molar absorptivities defined to be the same?
Is c taken to be the relative contribution, rather than the exact concentration?

Please write descriptive titles.

I'd rather say the definintion is the other way round: the point where no change in absorbance was found during the reaction got a new name.

An isosbestic point is found by maintaining the sum of the concentrations of the two (or more, though it's very unlikely) species the same and varying the individual contributions of each, and looking for the point at which the absorbance spectra intersect. This is easy for something like a pH indicator where a pH change just converts one species into the other—the sum of the two species is always the same. Under these conditions, a point where the absorbance doesn't change with the ratio of the two species can only exist if the molar absorptivities of the two species are the same, otherwise changing their ratio would change the absorbance.
So the molar absorptivities are defined to be the same when you simplify your equation above because this is the only way an isosbestic point can exist. So in your simplified equation, c is the sum of the concentrations of both species as that is what determines the net absorbance, not the concentrations of each individual species.

Actually that's not the correct definition of the isosbestic point. From IUPAC goldbook:

Isosbestic point - Wavelength, wavenumber or frequency at which the total absorbance of a sample does not change during a chemical reaction or a physical change of the sample.
A simple example occurs when one molecular entity is converted into another that has the same molar absorption coefficient at a given wavelength. As long as the sum of the concentrations of the two molecular entities in the solution is held constant there will be no change in absorbance at this wavelength as the ratio of the concentrations of the two entities is varied.
Contrary to a widely accepted idea, the existence of an isosbestic point does not prove that the reaction is a quantitative conversion of one species into a unique other species or that an equilibrium exists between only two species. The observation of isosbestic points only indicates that the stoichiometry of the reaction remains unchanged during the chemical reaction or the physical change of the sample, and that no secondary reactions occur during the considered time range, For the reaction A + B → c C + d D + e E, with c, d, and e the percentages of the products C, D, and E, an isosbestic point will be observed at every wavelength where the condition ɛ A + ɛ B = c ɛ C + d ɛ D + e ɛ E, provided that the values of the percentages c, d, and e remain constant during the chemical reaction or the physical change. The use of the obsolete term isoabsorption point is not recommended.


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Calculate the weight percent of limestone in a rock sample
A 1.245 g sample of limestone ($\ce{CaCO3}$) is pulverized and then treated with 30 mL of 0.035 M $\ce{HCl}$. The excess acid requires 11.56 mL of 1.010 M $\ce{NaOH}$ to be neutralized. Calculate the %wt of limestone in the rock.
The first balanced equation would be:
$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$
Then:
$$\ce{HCl + NaOH -> H2O + NaCl}$$
I figured out the moles of $\ce{CaCO3}$ if that's useful for this problem 1.245 g of $\ce{CaCO3}$ / 100.089 g (molar mass)= 0.0124 M
I have this but I don't know the next steps on how to solve the problem.

Please note that the homework policy on this site requires you
to show some personal effort.

I have the two balanced equations CaCO3+ 2HCl --> CaCl2+H2O+ CO2

Great.  So post them.

Please check the original question. Are you sure about the concentration of the hydrochloric acid?

The question is a bit misleading ;)
Your calculation of the moles of $\ce{CaCO3}$ would be correct if the rock sample would be pure 100% calcium carbonate. But then, all the titration would be a waste of time: There's 100 wt-% $\ce{CaCO3}$ in pure $\ce{CaCO3}$.
So, there's apparently something else in the rock sample (that doesn't disturb the titration).
However, what you know for sure is:

the stoichiometry of the reaction between $\ce{CaCO3}$ and $\ce{HCl}$. With other words: How many moles of $\ce{HCl}$ are needed to react one mole of $\ce{CaCO3}$
concentration and volume of the $\ce{HCl}$ used in excess to neutralize all the $\ce{CaCO3}$
concentration and volume of the $\ce{NaOH}$ to neutralize all the $\ce{HCl}$ that is left after reacting all $\ce{CaCO3}$

HINT:
Did you consider to solve the back titration experiment by starting with the second neutralisation step?

 If 11.56mL of a 1.010 M $\ce{NaOH}$ was used to neutralize the remaining $\ce{HCl}$, how much was left of it? Compare that with the initial amount...

I don't know how to do back titration is there a formula?

I think the exercise is making the assumption that there is acid inert components to the rock.

@Lighthart Yes, that's exactly what I tried to explain.

Calculate the amount of excess acid:

$$\mathrm{x mol HCl = 11.56mL NaOH \times \frac{1.010 mol NaOH}{1000mL}\times \frac{1 mol HCl}{1mol NaOH} = 0.01167 mol HCl }$$

Determine the amount of acid originally added:

$$\mathrm{x mol HCl = 30 mL HCl \times \frac{0.035 mol NaOH}{1000mL}= 0.0011 mol HCl }$$

Realize more acid was neutralized than was originally added.  There must be an error in the question some where.


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Is this cotton or viscose?
I have to examine the exact material of a specific fabric (non-woven), of which I know is either viscose or cotton. I know that viscose and cotton are structurally similar, but there should be a way to tell them apart.
This is the result of the microscopic examination (x 100):

My professor tells me this is viscose, but I thought of cotton because of the long fibers. But I also know that cotton has a 'twisted' look under the microscope.
My professor doesn't give any more clues, which leaves me with a bit of a problem. How can I tell the two apart? Does the picture show enough evidence for viscose (because of the linear structure)? I would like to double check this, just to be sure.

Yes, the picture should be sufficient to decide between viscose and cotton; see photo of cotton at http://whrhs-forensic-chem-spring-2010.wikispaces.com/Fiber+Analysis?responseToken=906bf6a76a5e268a00e54c8bce6927ba.
Recall that cotton was the cell wall of a plant cell, so it retains some of the shape of the cell. (Silk, on the other hand, is excreted from spinnerets, as are man-made fibers, so it looks more like viscose).


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Why do the pennies in the pre/post 1982 penny experiment sometimes dissolve completely?
USA pennies were nearly pure copper pre-1982.  Post 1982 they are copper clad zinc.  A common experiment in US high schools is to notch pennies with a file and add them to an HCl solution.  The old pennies remain untouched and the zinc of the new pennies dissolves, leaving the copper cladding.
Sometimes the copper cladding also dissolves and sometimes a considerable amount of the old penny dissolves.
Why does the copper dissolve?
I propose that the copper in the penny is reacting with atmospheric oxygen, forming copper oxide that dissolves in the acidic solution.
Another proposal is that the formation of a complex ion between copper and chloride ion drives the equilibrium, allowing the copper to react in HCl.  This is analogous to how aqua regia dissolves gold.
Another proposal is that the copper oxide layer on the penny reacts with copper metal in a comproportionation to form copper (I) ion that then reacts with oxygen to make copper (II) and dissolves?
Which is right?  Why does the copper dissolve?

Another possibility: the covering is so thin (just copper plate) that it crumbles, broken apart by H2 evolution. However, Copper will react with concentrated HCl to form a complex anion: [CuCl4>]2-(aq).

Sometimes there is a dark solid in the vessel, sometimes not.

Fine copper powder dissolves in concentrated $\ce{HCl}$ on air. So, the idea is not that far-fetched. Consider following experiment: a mix $\ce{NaCl:CuCl2:HCl, 3:1:2}$ mol is left over scraps of copper foil/wire in a sealed vial for a week or two. A significant amount of wire will dissolve, producing colorless solution. Leave it on air for some time, and it will become blue. This is an experiment I performed personally.

Your conclusion about formation of the Copper Oxide is correct, but not about the reaction with atmospheric oxygen.
The origin of the Oxygen is from the hydrochloric acid solution. The solution does not mean is pure HCl , it means is a dissolve in water.  There is a redox reaction with water
I leave to you to find the redox reaction.


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Purity of precipitate in selective precipitation

An aqueous effluent contains $12\ \mathrm{mol\ l^{-1}}$ $\ce{Cd^2+}$ and $10\ \mathrm{mol\ l^{-1}}$ $\ce{Mg^2+}$ as solution of nitrates. Current practice is use $\ce{NaOH}$ to selectively precipitate the metals.
(a) Is it feasible to get 0 $\ce{Cd}$ impurities in $\ce{Mg}$ precipitate by using $\ce{NaF}$ instead of $\ce{NaOH}$ (back up your answer with adequate calculations)
(b) What trade-off with respect to yield and purity do you have to accept if you use $\ce{NaF}$ instead of $\ce{NaOH}$?
(c) If consider switching the precipitating agent during the process which sequence should be used

Attempt at answer:
Part (a)
Using $\ce{NaF}$:

$K_\mathrm{sp}(\ce{MgF2}) = 3.7\times10^{-8}$
$K_\mathrm{sp}(\ce{CdF2}) = 6.44\times10^{-3}$
$K_\mathrm{sp}(\ce{MgF2}) = [\ce{Mg^2+}][\ce{F-}]^2$
$K_\mathrm{sp}(\ce{CdF2}) = [\ce{Cd^2+}][\ce{F-}]^2$

As the $K_\mathrm{sp}$ of $\ce{MgF2}$ is smaller it is less soluble and will precipitate first. You need to add $\ce{F-}$ to the point where $\ce{CdF2}$ will just not start precipitating. The limiting concentration is when the ion product = $K_\mathrm{sp}$.
Limiting concentration for $\ce{MgF2}$:
$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{MgF2})}{[\ce{Mg^2+}]} = \sqrt\frac{3.7\times10^{-8}}{10} = 6.08\times10^{-5}\ \mathrm{mol\ l^{-1}}$
Limiting concentration for $\ce{CdF2}$:
$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{CdF2})}{[\ce{Cd^2+}]} = \sqrt\frac{6.44\times10^{-3}}{12} = 0.023\ \mathrm{mol\ l^{-1}}$
The limiting concentration of $\ce{Mg^2+}$ is lower so this will precipitate first. By adding $\ce{F-}$ up to the limiting concentration of $\ce{Cd^2+}$ no $\ce{Cd}$ will precipitate so $\ce{MgF2}$ precipitate will be pure.
Part (b)
I assume you need to find the yield and purity for both methods (using $\ce{NaF}$ or $\ce{NaOH}$).
To find the purity of $\ce{CdF2}$ need to know how much $\ce{Mg^2+}$ is left in solution at the point when $\ce{CdF2}$ starts precipitating.
$[\ce{Mg^2+}]_\text{sol} = \frac{K_\mathrm{sp}(\ce{MgF2})}{[\ce{F-}]^2} = \frac{3.7\times10^{-8}}{0.023^2} = 6.99\times10^{-8}$
Purity of $\ce{CdF2}$: $\frac{[\ce{Cd}]}{[\ce{Cd}]+[\ce{Mg2+}]_\text{sol}} \times 100\ \% = 99.99\ \%$
I assume the yield of $\ce{CdF2}$ would be $100$ as you can keep adding $\ce{F-}$ until it’s all precipitated out.
I’m not sure if to calculate the yield of magnesium I need to do the original concentration − impurities in $\ce{CdF2}$/original concentration. If I did this then:
$\frac{10-6.99\times10^{-8}}{10} \times 100\ \% = 99.99\ \%$
To compare with using $\ce{NaOH}$ as the precipitating agent:

$K_\mathrm{sp}(\ce{Mg(OH)2}) = 1.8\times10^{-11}$
$K_\mathrm{sp}(\ce{Cd(OH)2}) = 2.5\times10^{-14}$
$K_\mathrm{sp}(\ce{Mg(OH)2}) = [\ce{Mg^2+}][\ce{OH-}]^2$
$K_\mathrm{sp}(\ce{Cd(OH)2}) = [\ce{Cd^2+}][\ce{OH-}]^2$

As the $K_\mathrm{sp}$ of $\ce{Cd(OH)2}$ is smaller it is less soluble and will precipitate first.
Limiting concentration for $\ce{Mg(OH)2}$:
$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{Mg(OH)2})}{[\ce{Mg^2+}]} = \sqrt\frac{1.8\times10^{-11}}{10} = 1.43\times10^{-6}\ \mathrm{mol\ l^{-1}}$
To find the purity of $\ce{Mg(OH)2}$ need to know how much $\ce{Cd^2+}$ is left in solution at the point when $\ce{Mg(OH)2}$ starts precipitating.
$[\ce{Cd^2+}]_\text{sol} = \sqrt\frac{K_\mathrm{sp}(\ce{Cd(OH)2})}{[\ce{F-}]^2} = \frac{2.5\times10^{-14}}{(1.43\times10^{-6})^2} = 0.0138\ \mathrm{mol\ l^{-1}}$
Purity of $\ce{Mg}$: $\frac{10}{10+0.0138}\times100\ \% = 99.86\ \%$
Yield of $\ce{Cd}$: $\frac{12-0.0138}{12}\times100\ \% = 99.86\ \%$
For part B the way the question is phrased I assume the purity or yield using $\ce{NaF}$ shouldn’t be as great as using $\ce{NaOH}$ but I don’t know if I have done my calculations right as from my results it would seem they’re both accurate methods.


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Calculation of Solute Mole Fraction
Suppose we had an aqueous solution of density 0.984 g cm$^{-3}$, containing 100.0 g ethanol per liter. How would we compute the mole fraction of ethanol in this solution?

My calculations so far propose that we examine a fixed volume, 1 L of the solution:
CH$_3$CH$_2$OH: m = 100.0g | M = 46.068g mol$^{-1}$ | n = $\frac{m}{M}$ = 2.170 ... mol
solution (incl. H$_2$O): m$_{tot}$ = 984 g | M$_{tot}$ = M$_{solute}$ + M$_{solvent}$ = 46.068 + 18.016 = 64.084 g mol$^{-1}$
$\therefore$ n$_{tot}$ = 15.35 ... mol
Hence $x$ = $\frac{n}{n_{tot}}$ = 0.1413 ... $\approx$ 0.141 (3 sig. fig.)

This is not correct because you assuming that the molecular weight rations are one to one.
The coorect formulation is
(100 g Methano/46.068)/(100/46.068 + (984-100)/18.016)


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When does 'H' Shows +I/-I effect and why does T,D show better +I effect?
When does 'H' Shows +I/-I effect and why does T,D show better +I effect?
In my book Hydrogen is given in both +I & -I effect showing group ,i was curious to know under what conditions does it shows such opposite/dual effects? Also are there any other commonly encountered groups which show both +I/-I contition to condition?


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How to determine methane percentage at home?
Suppose I have gas bottle and gas cooker. Suppose I have propane in my bottle and I suspect it is diluted with unknown gas.
How can I measure with simple instruments, how many propane I have (percentage)?

The title and text of the question don't match, are you wanting to measure propane or methane? If all you're looking to use the gas for is cooking, why does the contaminating gas matter? Remember, Propane is a clean burning and efficient fuel that allows you to taste the meat and not the heat.

'Simple' is probably not in your future for this problem.

A couple of ideas:

If you think the diluent gas is non-flammable, you should be able to use the adiabatic flame temperature of the gas when you burn it in air to estimate the composition.  You will need to carefully adjust your stove to give the optimal fuel/air blend (that maximizes the flame temperature).  Then you will need to accurately measure the flame temperature.  If you assume a heat capacity for the unknown diluent, you can estimate its concentration in the gas by comparing the measured temperature to the theoretical adiabatic flame temperature for pure propane burning in air.
Use a vacuum pump to empty a container of known volume, and then allow the cooking gas into the container at ~1 atm pressure.  Measure the density of the gas and compare to the density of pure propane.
If you think the diluent gas does not contain carbon: hook up flow meters to the gas bottle (and an air or oxygen stream) to form a torch flame with known gas flow rate.  Collect all of the the exhaust produced by the flame, pass it through a cold trap to freeze out water vapor, and then pass the cold flame exhaust over a bed of anhydrous sodium hydroxide of known initial mass.  It will absorb the carbon dioxide from the exhaust.  Measure the mass increase of the sodium hydroxide.  By comparing the moles of gas that flow into the flame to the moles of CO2 absorbed (as calculated from the mass increase) you can estimate gas purity.
Find a spectrometer and record the IR or Raman spectrum of the gas!  This will probably be more accurate than any of the other methods.


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Carbon monoxide detector reaction
I have recently bought a carbon monoxide spot detector which is not working the way I intended for it. A carbon monoxide spot detector is an analog carbon-monoxide detecting patch that changes color when it comes in contact with carbon monoxide. The problem is that it changes back to it's original color whenever the carbon monoxide leaves the room.
I'd wish for it to stay the same color it changed to after it came into contact with carbon monoxide. What should the reaction look like to fix this problem? I have copy/pasted a scholarly paper from Google to better explain what's happening inside of the patch:

Principle of the $\ce{CO}$ – Detector: The palladium (II) chloride inside is reduced into metallic palladium when exposed to $\ce{CO}$. While the $\ce{CO}$ – level in air is rising, higher amount of palladium is formed. A color change of the solution from pale yellowish orange to orange or grey is followed.
  $$\ce{CO + PdCl2.2H2O → CO2 + Pd + 2HCl + H2}$$
  Afterwards, when exposing to $\ce{CO}$ – free air, copper (II) chloride oxidizes palladium back into palladium (II) chloride. In addition, copper (II) chloride is reduced to copper (I) chloride.
  $$\ce{Pd + 2CuCl2.2H2O → PdCl2.2H2O + Cu2Cl2}$$
  Then, the regeneration of copper (II) chloride occurs.
  $$\ce{Cu2Cl2 + 2HCl + H2O + 1/2 O2 → 2CuCl2.2H2O}$$
  Thus, the solution resumes to its initial state. (the oxygen comes from the surrounding air)

One of the main characteristics of a detector function is being reversible. In order to prevent the responsible reactions  of colour changing back, you have to keep oxygen away from the detector.

Use a webcam and program that detects the change of colour and writes a warning? LED, photodetector and some analog circuitry that closes a lid on the reaction, preventing the regeneration?

I hope this post is for safety reasons and not because you're running some home experiment that will get you killed...

@Zhe
This question was for a product idea I had that I wanted to conceptualize.  I was trying to make a sticker for hotel rooms, bathrooms, rentalcars, airplane toilets, etc. that changed color permanently when somebody smoked inside this room. In the end I went a different route. I am still very much interested in the solution to the reaction though.

OK, I just wanted to make sure you weren't about to poison yourself by accident....


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What is the major product on chlorination of 2-methylbutane?
I came across a question asking for the major product on chlorination of 2-methylbutane. The answer in the back of the book says it will be 2-chloro-3-methylbutane. However, I thought the major product would be 2-chloro-2-methylbutane. The later is formed through a tertiary free radical with eight hyper-conjugative structures, whereas the former forms through a secondary free radical with only four hyper-conjugative structures. How is this possible?

Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.
According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:
22 % 2-chloro-2-methylbutane
33 % 2-chloro-3-methylbutane
30 % 1-chloro-2-methylbutane
15 % 1-chloro-3-methylbutane
The statistical factor is responsible for this.
If each hydrogen atom of isopentane could be substituted at the same rate, the ratio would be
1 hydrogen atom (8 %) → 2-chloro-2-methylbutane
2 hydrogen atoms (17 %) → 2-chloro-3-methylbutane
6 hydrogen atoms (50 %) → 1-chloro-2-methylbutane
3 hydrogen atoms (25 %) → 1-chloro-3-methylbutane
In terms per hydrogen atom, the experimental yield is
22 % / 1 = 22 % → 2-chloro-2-methylbutane
33 % / 2 = 16.5 % → 2-chloro-3-methylbutane
30 % / 6 = 5 % → 1-chloro-2-methylbutane
15 % / 3 = 5 % → 1-chloro-3-methylbutane
Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.

Thanks a lot for the help. In my book (high school standard though) it is written that the chlorination favours in the following order Teritiary hydrogen>secondary hydrogen>primary hydrogen.

Yes but selectivity factor + probability factor together decide the major and the minor product. The question you have asked shouldn't have been a high school organic chemistry question unless they mention relative reactivity of the hydrogen atoms.


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Chalky residue from melted ice cubes
I bought a new refrigerator that makes ice cubes. The instructions said to discard the first 3 full bins of ice in order to ensure the new water line is flushed clean. I dumped the full bins in my stainless steel sink each time. After the ice melted each time a chalky residue was left in the sink. I emptied a couple of extra full bins and still was left with the chalky residue. I cleaned the sink after each dumping to be certain there wasn't any left over residue from the previous dump. What is the chalky substance? Is it something to be concerned about or not?

It is possible that the white material is calcium carbonate or magnesium carbonate from your water source precipitating.
See if vinegar dissolves the white residue.
If the white residue dissolves in vinegar, it is probable calcium carbonate and/or magnesium carbonate.

Also called 'Hard Water'

If it is calcium or magnesium carbonate are they harmless?

I can't give medical advice, but from what I've read studies show a health benefit drinking hard water: http://www.mgwater.com/epidem.shtml


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Correlation between the equilibrium constant and delta G

I understand this up until the last two steps. How does $\ln(K_\mathrm{eq})$ become $2{,}303\log(K_\mathrm{eq})$ and how does $-RT\cdot2{,}303$ become $-1{,}42$?

How does ln(Keq) become 2,303log(Keq)

Because $e^{2.303} = 10$ and
$\ln(x) = \log (10^{\ln(x)})$ therefore
$\ln(x)=\log((e^{2.303})^{\ln(x)})$
$\ln(x)=\log (x^{2.303})$
$\ln(x)=2.303\log(x)$

how does -RT*2,303 become -1,42?

That would only be true at a particular temperature in a particular set of units

I forgot to say that this calculation is done at 37 degrees celcius. But thanks, I understand it now.

The first step is simple:
$$\frac{{\ln x}}{{\log x}} = \ln 10 \approx 2.303$$
thus
$$ - RT\ln {K_{{\text{eq}}}} \approx  - RT \times 2.303 \times \log {K_{{\text{eq}}}}$$
The second step is terrible because $\text{kcal}$ is used and the units are missing:
$R = 1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$
$T = 310{\text{ K}}$ (my guess)
$$\begin{aligned}
  \Delta G =  & \; {-} RT \times 2.303 \times \log {K_{{\text{eq}}}} \\
   =  & \; {-}1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mol}^{ - 1}} \times 310{\text{ K}} \times {\text{2}}{\text{.303}} \times \log {K_{{\text{eq}}}} \\
   =  & \; {-}1.42{\text{ kcal mol}^{ - 1}} \times \log {K_{{\text{eq}}}} \\
\end{aligned} $$


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Attack of amine on amide
I am confused with the last step of clozapine synthesis:

How can the amine group attack the amide group to form the amidine in such acidic environment? Is the intermolecular reaction between an amine and amide group feasible under not too extreme conditions?

There reaction environment would have to be acidic to protonate the amide, the amide is not natively electrophilic enough to be attacked by the amine.
A possible point of concern is 'won't the nuclophilic amine be protonated?'  And the answer is absolutely.  But the protonation is a reversible reaction, and at equilbrium with a catalytic amount of acid (almost always TSA is used in catalytic quantities), the proton will be exchanged with various, some of which will lead to the condensation resulting in product.

The reaction goes well because it is intramolecular I guess. Would an intermolecular reaction give a decent yield to be considered useful?

Intramolecular reactions are more favorable than similar intermolecular reactions for entropic reasons as long as the ring is small-- in this case 7 atoms works great.  Once you start making 9 or 10-member rings, the entropy gain becomes much smaller.  In this case the resulting conjugation might also be a enthalpic driving force.

I know that but Would an intermolecular reaction give a decent yield?

I'm skeptical this would work well intermolecularly, but I no longer have access to literature searching to verify

Seems like the acetals of amides are used: http://www.organic-chemistry.org/synthesis/C2N/amidines.shtm


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Why, during aspirin synthesis in acidic conditions, anydride carbonyl oxygen is protonated rather than salicylic acid carbonyl oxygen ?
The common route for aspirin synthesis (described here) employs acetic anhydride and salicylic acid, in acidic conditions.
The first step of such a reaction mechanism is the protonation of the carbonyl oxygen of acetic anhydride.
Why, instead, protonation of carbonyl oxygen in salycilic acid should not occur ? Of course, such a protonation is not desired, since it won't lead to aspirin: however, can it be neglected and why ?
Thanks for any hints.

Mechanisms are explanative, and not predictive.
The first step of the reaction mechanism is protonation of the anhydride because that is the first step of the reaction.  That just is.  If the anhydride is not protonated, step 2 simply won't happen.
This does not mean that the salicyclic acid is not protonated.  It just means that the protonation of salicyclic acid is not necessary to explain how acetylsalicyclate forms.  Proton exchanges are at equilibrium.  It is likely that the salycilic acid is protonated during this process.  However, that protonated salicyclic acid may then protonate acetic anhydride and proceed as indicated in your link.  For the purposes of describing the mechanism, it is not necessary to distinguish where the proton that protonates the acetic anhydride comes from.
So your question really is self-answered, where you say, roughly, 'Such a protonation does not lead to product directly'.

I'm curious - what is the product of protonation of salicyclic acid and why is it insignificant? I know that reactions are always at equilibrium, but if the equilibrium constant is very big shouldn't it be almost irreversible? What is the product and why is it unstable? Thanks,

I don't have easy access to a chemical sketching program, so I'll try to explain.  Draw Salicyclic acid.  Add a bond from the C=O (double bond) oxygen to a new hydrogen.  Add a positive charge to to the same oxygen.  This compound is unstable because it is highly acidic.  Any positively charged oxygen will be highly electron withdrawing, and will pull electrons away from and bonded hydrogens. Related ...  being at equilibrium, we cannot describe it as irreversible, even if the proportion of one of the species is vanishingly small.


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Is most of the weight you lose actually released as carbon dioxide when you exhale?
I've heard that most of the weight you "lose" is actually carbon dioxide released when you exhale. In percent its much more than what is released in pure heat energy and feces. Does anyone have statistics on what's the split between where the energy leaves your body?
How many percentage of food weight is actually released through carbon dioxide when its burned in your body? For example when you eat a hamburger and fries, is all the carbon released later when you exhale? How much is heat energy? Provided it does not go into fat reserves.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. Digging a bit deeper, I realize your question has something to it. While resting, it's estimated that we lose an average of 300 milliliters of $\ce{CO2}$. Doing some calculations, I can conclude that a lot of loss (that's being backed up by food and water consumption) is due to $\ce{CO2}$.

Let's start with:

How much is heat energy?

2000 kilocalories per day, considering $E=mc^2$, is
$9.3 \times 10^{-8}$ grams

is all the carbon released later when you exhale?

No.  Brainstorming, the others ways would include: feces, urine, perspiration, flatulence, shedding skin, hair and nails, mucous from nose and mouth, for women: menstruation, vaginal discharge, lactation, and giving birth, and for men ejaculation.
According to The Value of Human Feces:
Per day the amount of carbon excreted in feces is 21 grams and in urine is 7 grams.
According to The Composition of Human Perspiration Proc R Soc Med. 1934 May; 27(7): 839–848, a typical amount of perspiration a day is 700mL, and per 100mL the main carbon containing compounds are 20 mg urea, 20 mg glucose and 5mg amino acids.  So clearly less than 1 gram of carbon loss per day from perspiration. If we were considering loss of salts instead of just carbon, that would be about 3 grams per day from perspiration.
For menstruation, blood loss is about 40mL per cycle, and blood is 92% water, so maybe 1 gram of carbon on a heavy day of a period.
Skin loss per day is 1.5 grams according to Cells for life, cells for an instant.
A highly controlled study having a person in a chamber for 24 hours and analyzing the air found 653g of CO2 or 178g carbon "A Respiration Calorimeter with Appliances for the Direct Determination of Oxygen" Carnigie Institution of Washington Publication 42.
So, for a person that is not lactating or giving birth, I would estimate about 30 grams per day of carbon loss is other than from exhaling carbon dioxide.  Therefore:
About 86% of carbon loss is through exhaling carbon dioxide.
However, this will certainly vary depending upon the amount of indigestible carbon such as fiber, in ones diet.
Also, if mass loss through nitrogen and phosphorous (but not hydrogen and oxygen) were also considered the loss through breathing would come down to about 80%.  Inclusion of salts would bring this percentage down further.
The sum of water from feces, urine and respiration is about 1.8 kg per day and the total mass emission of the body is 2.6kg, according to the Carnigie source above.

You forgot to mention, that most of the weight is water, strictly speaking

I guess I didn't say that directly, but that's why I said "not hydrogen and oxygen".  I will edit to make it more clear, thanks.

Arguably, whatever leaves the intestines came directly from the stomach so faeces only contain the bits of the food we can’t do anything with and the living things from our gut that digest bits of what we can’t use. (At least according to my current knowledge according to which there is no significant transfer of mass from the bloodstream to the intestines.)

@Jan, bile enters downstream of the stomach, and the intestines themselves also emit secretions.

@DavePhD True that; it’s something I forget.

Yes, nearly all of the weight you "lose" when your body mass goes down is lost as $\ce{CO2}$.
Infintessimally small -- essentially zero -- weight is lost from "heat".  We have Einstein's theory that $E=c^2m$, but if you plug in the numbers, the mass change from burning off a pound of fat is less than a microgram.
Your body could also lose weight as water, but in the long term, if you kept doing this, you'd dry up and die from dehydration.
The only way to lose the weight is to "burn off" fat by converting it to $\ce{CO2}$ and exhaling the $\ce{CO2}$.
A great article which goes through the math -- and which also shows that many dieticians and doctors do not understand this point! -- is this one: http://www.bmj.com/content/349/bmj.g7257

It's not appropriate to use E=mc^2 for a chemical change.

@jerepierre Why not? Combustion causes a net loss to the environment of potential energy stored in the chemical bonds. $E=mc^2$ always applies, regardless of whether the mass difference is easily measurable or not.

@jerepierre "Nevertheless, when a chemical reaction emits energy to its surroundings, its reactants lose an equivalent quantity of mass in the process." http://pubs.acs.org/doi/abs/10.1021/ed082p1636

Sorry, I should have read to the end of the paragraph. Unfortunately my vote is locked unless there's an edit to the answer.

Edited to emphasize the practical irrelevance of $E=c^2m$.

There is another downvote on my answer.  Not sure why...the link I added goes to the most exhaustive answer of the question of anything posted here.

I didn't downvote, but the link only considers CO2 and ignores carbon loss as urea, creatine and other compounds in urine as well as loss through feces.  The carbon loss through feces isn't just what passes through; billirubin and other carbon containing substances are passed into the intestines as bile.  The Carnigie Institute reference is more exhaustive.

Losing a pound of fat corresponds to about 4000 kilocalories, the mass equivalence of which is about 186 nanograms.

DavePhD, I did the calculation again and am still coming up with sub-nanongram qualities.  Where am I going wrong?  >>> fat_mass = 1/2.25 #one lb in kg

fat_kcal_per_g = 9
fat_kcal_per_mass = fat_mass1000(fat_kcal_per_g)
fat_kJ_per_mass = fat_kcal_per_mass4.184
c = 3e8
fat_kJ_per_mass1000/c/c
1.8595555555555554e-10

@CurtF. I have 1 lb corresponds to 4000kcal = 16,736,000 J = 16,736,000 kg m^2/s^2.  (16,736,000 kg m^2/s^2)/(300,000,000m/s)^2 = 1.86 e-10 kg = 1.86 e-7g = 186 nanograms

You're right Dave, I forgot about the implicit need for units to be kg when using Joules.  So I'm off by 1000x...at least we agree on the rest.  I'll edit my answer.

Sorry but E=mc2 is not applicable. Chemical reactions do not result in a change in mass of the components but a change only in the molecular combinations. Yes, the heat generated in oxidation comes from the potential energy of the chemical bonds so one would have to put a lot of energy into again separating the oxygen and carbon in CO2. A change in mass only occurs in nuclear reactions which also results in a change in the atoms of the elements involved to a new element.

Hello, welcome to Chem.SE! I'm afraid this answer is partially incorrect. Just because there are experimental difficulties in measuring such slight changes in mass, that doesn't mean relativity stops applying. Check out the article (or even just abstract) linked in one of DavePhD's comments.


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Test for chlorine and chloramine in city water supply
I am looking to install a water filter system and I wanted to test the water for free and total chlorine.
I understand from the city water report that it uses chloramine and NO chlorine (test report)and that chloramine content averages around 4 ppm in the water supply. I purchased one of these to test it out. http://www.bulkreefsupply.com/insta-test-free-total-chlorine-strips-lamotte.html
The above product can measure free and total chlorine ( 0, 0.5, 1, 3, 5, 10 ) PPMs. I was surprised to find out after using these test strips that city supply has 0 PPM for both free and total chlorine.
Questions
1. Any idea how I can test if these test strips are indeed working?
2. Anything wrong with the way I have tested?

As water goes through pipes it loses active chlorine due to reaction with biological material, dissolved substances and even the pipe, so it is possible there is little left by the time it reaches your house. Your local water department should provide testing for your water so you can verify this, or you could make up a known concentration of bleach in distilled water, using serial dilution to 0.5 ppm, to check that the strips are accurate.

Tried the method that you suggested, the test strips work.

Test strips for chlorine should work for both chlorine- and chloramine-treated water. See Here-to-learn's comments to this answer for the chemical kinetic reason why chlorine test strips will also work for chloramine-treated water.
Notice that your city supplies the water with chloramine. Depending on how much chloramine is present in your tap water, leaving water to sit for days on the countertop would be necessary to remove a significant amount of chloramine.
Please see this research paper: http://hbd.org/ajdelange/Brewing_articles/BT_Chlorine.pdf
Page 17 estimates the half-lives of chlorine vs. chloramine.

This is incorrect advice and should be removed. Dr. Pippik was correct in hypothesizing that the chloramines are likely reacting with biological material (biofilms) in the distribution system, thus it is below the strip detection limit when it arrives at the tap. The test strips are for detecting free chlorine and total chlorine. Total chlorine is the sum of free chlorine (i.e., hypochlorous acid and hypochlorite) and combined chlorine (i.e., mono-, di-, and tri- chloramine), thus the strips do measure the presence of chloramine.

@Here-to-learn, thanks for the comment. I agree that my answer should be revised, but my answer also points to a good source proving that chloramine has a much longer half-life. It would be excessive to delete the entire answer. Also, I'm pretty sure chloramine is less reactive to biological material than chlorine. Are you aware of this fact?

Yes, as a disinfectant, chloramine is well known to have slower kinetics compared to other disinfectants (e.g., its CT is two orders of magnitude worse than chlorine for viruses). But, your answer is not really an answer to the original question, as it neither answered (1) or (2). Rather, you should have put it as a comment for discussion. I would also argue that a broken link to a brewing site is not a good source. And finally, you suggested the OP use "test strips that are made for testing chloramin," which they did, so that's not really a relevant answer.

I think what you might have been suggesting is that chloramine will react much slower than free chlorine and thus it should still be present at the tap - is that correct? That would only be the case if the utility was using the same molar concentrations; however, chloramine, because of its slower kinetics, is typically dosed at smaller concentrations knowing it will stay in the distribution system longer. Even if we assume that high concentrations were used, we don't know the distance to OP's house, thus we can't make kinetic conclusions. And, we don't know strip detection limit.

Ok I understand. I will edit my response. I contacted the site admin regarding the broken link, and it's simply an issue with the website going through some kind of update. The admin should fix the link soon.

The link is dead. Just requested it be brought back here http://www.hbd.org/2017/05/09/tick-tock/#comment-21796 (awaiting moderation)

Found another source here http://wetnewf.org/pdfs/Brewing_articles/BT_Chlorine.pdf

I took a screenshot and uploaded/edited as an image.


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Sulfur Doped Titanium Dioxide
I'd like to know why my $\ce{TiO2}$ sol gel did not get coated nicely on my 200 gsm fiberglass? Here my solution, 5 g of tween 80, 40.15 ml isopropanol, 3.4 ml titanium isopropoxide, 0.62 ml sulphuric acid (0.02 mol).  Beside that, my $\ce{TiO2}$ sol gel can't calcinated at 500°C.  It will turn into powders and get burned into a black colour.  What should I do to overcome this problem?

Regarding why your ceramic material did not get coated nicely on the fiberglass, I think there's a problem of adherence between the fibers and the the ceramic. You need to prepare the surface of the fibers by etching for example before coating. They are often shielded.
Regarding why your  sol-gel can't be calcinated at $500°\mathrm{C}$. You need to figure out the proper temperature of calcination by monitoring the phase appearance of the ceramic materials by XRD diffractometry. You start with the XRD diffractogram of your as-prepared sample(normally, it should be amorphous). Then, you treat your sample at different  temperatures within different periods of times (you have two parameters to find: Temperature and Time). Unless you are working on a published article that has addressed all these issues.


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Impossible pH For Aqueous Weak Acid Solution
An aqueous solution of $\ce{HCl}$ has a $\mathrm{pH}$ of $0.00$ if $[\ce{H+}] = 1.00 \; \mathrm{M}$. I tried to see if an aqueous solution of acetic acid could have a $\mathrm{pH}$ of $0.00$, given that its $K_a$ value is $1.8 \times 10^{-5}$. I tried finding the concentration $[\ce{H+}]$ of such a solution using
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{C2H3O2^-}]}{[\ce{HC2H3O2}]}$$
and then using molar mass and density conversions to determine whether such an aqueous solution exists.
Does it make sense to say that such a solution does not exist if the amount of substance of acetic acid in 1 liter of solution actually take up more than 1 liter of volume?

See also superacids on Wikipedia for more context.

Your reasoning is fine. But a major caveat (or additional factor) is that the pH scale itself no longer applies to concentrated solutions (say, more than 1 molar in acid) where the molarity of water falls significantly below 55M (pure water).
You can extend the pH scale down below 0 by empirically measuring how effective an acid solution is at protonating a weak base, but the scale deviates from the straight line predicted by the concentration of acid present.


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Does an acetal give a positive Tollens test?
In Organic Chemistry (Wade) there is a question:

Which of the following compounds would give a positive Tollens' test? (Remember that the Tollens' test involves mild basic aqueous conditions.)

The answers manual says this compound - $\ce{CH3-(CH2)3-CH(OCH3)2}$ - gives a negative result as it is an "acetal stable in base: no reaction"
Now, I would have thought this is positive. Earlier in the chapter it's taught that since the Tollens reagent removes any aldehyde present, even if at equilibrium the amount of aldehyde is very small, the Tollens reagent will nevertheless predominate due to Le Chatelier's principle (the small amount of aldehyde will be removed slowly but surely, until eventually all of the solution is converted to aldehyde and then to the tollens product).
In this example, whilst I respect that the acetal product may be by far the major product, surely a very, very small percentage of the acetal is nevertheless converted to the aldehyde, via protonation of a methoxy group, which leaves, and then attack by $\ce{H2O}$, followed by deprotonation of $\ce{H2O -> OH}$, and then protonation of the other methoxy group, which leaves, leaving a carbocation, which the $\ce{OH}$ attacks forming a double bond and a positive charge on the $\ce{OH}$, followed by deprotonation on the $\ce{OH}$  giving an aldehyde?
This may be unlikely, but according to Le Chatelier's principle, it just needs to occur at some small quantity in order to be continuously removed by the tollens reagent, and thus ends up predominating and removing all of the acetal product given enough time? Why is this not positive?

That there is a equilibrium position says nothing about the speed of attaining it, keep that in mind.

The description of the hydrolysis of an acetal is accurate. The trouble is that the initial protonation event happens to such a small amount that the reaction does not proceed. We can estimate where the equilibrium of that protonation lies by looking at the pKa's of the relevant acids [Evans pKa table]. In basic solution, water (pKa 16) is the strongest acid available to protonate the acetal. The protonated acetal will have a pKa close to protonated methanol (pKa -2). So the equilibrium for that protonation will lie to the water side by approximately $\ce{10^18}$. As Jori is alluding to in the comment, the rate moving forward is going to be exceedingly slow because the concentration of the protonated intermediate will be exceedingly low. This is where the rule of thumb that acetals are stable to base comes from. See also: Why are acetals stable to bases and nucleophiles?
If the Tollen's test was performed under acidic conditions, then you would be correct that the acetal would test positive for an aldehyde. A hemi-acetal would also test positive.

Really informative answer thanks, and that Pka table is great. Out of curiosity though, given enough time and/or heat and the presence of excess Tollens reagent, would this reaction in theory move to the left completely due to Le Chatelier's principle, even in basic conditions?

I suppose so, but I wonder if there are other decomposition pathways that would compete if we're allowed to just heat the reaction indefinitely.

@jerepierre. could you please let me know which of the following give positive tollen's test ? Hemiacetal, hemiketal, acetal, ketal. Actually i just wanted to know about hemiketals. other three i know. hemiacetals give positive. ketals and acetals give negative. what about hemiketals ?

@omjoglekar the mechanism of opening a hemiketal to open form ketone works as in aldehyde. But after that, to give a positive tollen's test, that ketone should have a terminal alpha hydroxy group. So it will depend. Not all hemiketal will give positive test. But others like  of alpha D- fructose does.


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Enantioselective organic reaction not working
I'm performing the reaction between acetone and 2-nitrobenzaldehyde catalysed by (L)-Proline, as answered to one of my questions a while ago: Simple experiments involving enantiomer synthesis
However, while the literature specified using 1 mmol of (L)-Proline in 20ml of solution, due to the cost of the materials and the budget available, I've only been able to use a fraction of that, around 0.01mmol. I've scaled down the entire experiment to 5 ml to try and compensate for this.
The issue is, I've left this solution reacting for a long period of time (12+ hours) and I'm not seeing any optical activity at all. With the reduced concentrations, I'd have expected to see a lot less activity than the original literature specifies, but to see at least some indication that my reaction is working.
Are the concentrations of (L)-Proline simply too small for any significant reaction to take place? Or can there be some other reason why my reaction isn't working? Note: I'm a chemistry 'newbie', so don't assume anything!

How are you measuring the optical activity? How are you monitoring the progress of the reaction?

I am using a polarimeter, and I was expecting to see a progressive change in optical rotation with which I could determine the rate

Can you measure the optical rotation of proline itself? Based on the optical rotation of the product, can you calculate the expected change in optical rotation for these conditions?

That's just the problem - there isn't any optical rotation, both before the reaction (aka the pro line) or afterwards

Based on the concentration of proline and the reported optical rotation, should it be measurable using your instrument? You may be below the limit of detection for the polarimeter.

The original "Rutgers' reference" deploys $2\,\mbox{mmol}$ aldehyde in $20\,\mbox{mL}$ solvent (i.e. starting at $0.1\,\mbox{mol/L}$, or $0.1\,\mbox{M}$) and recommends a stirring of $3\,\mbox{h}$.

Subsequently, you mention $1\,\mbox{mmol}$ per $20\,\mbox{mL}$ solvent (i.e. starting at $0.05\,\mbox{M}$), but without an indication of reaction time.

But if the starting concentration is lowered further ($0.01\,\mbox{mmol}$ in $5\,\mbox{mL}$ equals to $0.002\,\mbox{M}$ or $2\,\mbox{mM}$), not only the detection of the product becomes difficult. Did you take into consideration that by diluting the reaction you over-proportionally slowed down the advancement of the intended reaction?  Compared to the initially mentioned starting concentration, your reaction starts at a dilution by a factor of 50 (fifty).  This is more than a big factor for process involving a bimolecular reaction.

By the way, $0.01\,\mbox{mmol}$ of proline, molar weight of $115.13\,\mbox{g/mol}$ as stated here would correspond to little more than $1.1\,\mbox{mg}$.